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1 HOW MUCH WILL I SPEND ON GAS? Outcome (lesson objective) The students will use the current and future price of gasoline to construct T-charts, write algebraic equations, and plot the equations on a graph. Student/Class Goal Students will determine their gasoline cost for a month s time. They will use this information to calculate yearly gas cost. Time Frame Two 1 ½ hour classes One 3 hour class Standard Use Math to Solve Problems and Communicate NRS EFL 3-6 COPS Understand, interpret, and work with pictures, numbers, and symbolic information. Apply knowledge of mathematical concepts and procedures to figure out how to answer a question, solve a problem, make a prediction, or carry out a task that has a mathematical dimension. Define and select data to be used in solving the problem. Determine the degree of precision required by the situation. Solve problem using appropriate quantitative procedures and verify that the results are reasonable. Communicate results using a variety of mathematical representations, including graphs, charts, tables, and algebraic models. Materials Graph paper 40x30 paper is included at the end of the lesson (Worksheet 4) Colored pencils (if available) Calculators How Do I Find My Gas Mileage? Handout T-Chart Illustration How to Graph a Linear Equation in 5 Quick Steps Student Resource Activity Addresses Components of Performance Students will work with basic operations and patterns to complete a T-chart. Students will construct a graph and plot points on it. Students will use problem solving to determine how to figure out their monthly gas cost. Students will determine appropriate intervals on the x and y axes of the graph. Students will round the number when necessary to plot the points on the graph. Students will recognize if a data set is not reasonable by observing the other data sets and observing the points on the graph (is it on the line?) Students will communicate the results of the data by constructing a graph, T-chart, algebraic formula and in writing. Learner Prior Knowledge Basic understanding of gas prices and what is meant by miles per gallon (mpg), found at lesson, Pumped Up Gas Prices. Instructional Activities Step 1 - Discuss with students the current gasoline price and what they expect to happen to gas prices in the future. Ask students about how much they are spending on gas each month and how many miles the car they are driving gets per gallon? Students may need to research this information by actually calculating their mileage on the handout How Do I Find My Gas Mileage? or by visiting Cars and mpg This web site lists the in-city and highway mileage for 1985 or newer vehicles. If using this site, students will need to decide what type of driving they do during a month (city or highway) and estimate their car s mileage. Step 2 - Using \$4 per gallon as the price of gas, construct, with the class, a T-chart showing the relationship between the number of gallons purchased(x) and the total cost of the gas(y). See T-Chart Illustration for an example. As a class, look at the T-chart that was constructed. Give the students time to discover the relationship between the number of gallons purchased(x) and the total price of the gas(y). Express this relationship verbally (The number of gallons purchased times \$4 will equal the total cost.). Discuss how this relationship can be written as an equation as a function of x? (4x=y) Step 3 Next, with the students, graph the relationship/equation the class discovered during Step 2. The How to Graph a Linear Equation in 5 Quick Steps, a student resource from their math journals, provides basic guidelines for graphing equations. Help the students determine appropriate intervals and labels for the x and y axes and a title for the graph. Plot several points together and then let the students plot the remainder of the points. Draw a line passing through the points and label the line with the equation written in Step 2. Step 4 - Discuss with the students how many miles they drive each month. Share with the class that many people drive about 1,000
4 Gallons purchased (x) T-CHART ILLUSTRATION Total gas cost (y) 1 \$ \$ x30 Grid
5 How to Graph a Linear Equation in 5 Quick Steps Step 1 Construct a T-chart of Values Using your equation, construct a T-chart of values if one has not been done already. Substitute some simple numbers into the equation for x or y. If x=1, what is y? If x=10, what is y? If y=0, what is x? Each pair of values in your T-chart will become a point on the graph. (See illustration 1 for an example of a T-chart) Step 2 Decide on the interval for each axis Before starting the graph, look at the T-chart to determine the highest value for y found on the chart. Look at the values needed for x. Using graph paper, count the number of lines on the x and y axes. Use these numbers to determine the intervals on each axis. (If you use the graph paper at the end of this lesson there are 30 spaces on the x axis and 40 spaces on the y axis.) If the largest total cost/y value that needs to be graphed is \$80 and there are 40 lines on the y axis, let each line on the y axis represent \$2. The number of gallons of gas/x value that goes with \$80 is 20. There are 30 lines, so to make it simple one line will equal one gallon. Be sure the students realize they do not need to put a number next to every line. For example, the x might be labeled on every 5 th line (five gallons) and the y axis might also be labeled on every 5 th line (or \$10). This is a good step to do in pencil. That way if the interval you selected did not work out, the numbers can be erased any you can start over. Step 3 Label each Axis Decide what labels need to be added to the x and y axis. What do the numbers on the x-axis represent? What do the numbers on the y-axis represent? Usually the labels will match the descriptions/labels of x and y on the T-chart. (Note: When graphing equations involving elapsed time, time is traditionally represented by x) Step 4 Plot the points Using each pair of points from the T-chart, plot the points on the graph. Every point does not need to be plotted. Just be sure you have at least 3. Using a ruler, draw a line through the points you have plotted. Write your equation next to the line. Step 5 Give the graph a title Decide on a title for the graph. Make sure it accurately represents what is being shown on the graph. Does it explain the relationship between x and y? How to Graph a Linear Equation in 5 Quick Steps Student Resource
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# 11.2: Arithmetic Sequences
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###### Learning Objectives
• Find the common difference for an arithmetic sequence.
• Write terms of an arithmetic sequence.
• Use a recursive formula for an arithmetic sequence.
• Use an explicit formula for an arithmetic sequence.
Companies often make large purchases, such as computers and vehicles, for business use. The book-value of these supplies decreases each year for tax purposes. This decrease in value is called depreciation. One method of calculating depreciation is straight-line depreciation, in which the value of the asset decreases by the same amount each year.
As an example, consider a woman who starts a small contracting business. She purchases a new truck for $$25,000$$. After five years, she estimates that she will be able to sell the truck for $$8,000$$. The loss in value of the truck will therefore be $$17,000$$, which is $$3,400$$ per year for five years. The truck will be worth $$21,600$$ after the first year; $$18,200$$ after two years; $$14,800$$ after three years; $$11,400$$ after four years; and $$8,000$$ at the end of five years. In this section, we will consider specific kinds of sequences that will allow us to calculate depreciation, such as the truck’s value.
## Finding Common Differences
The values of the truck in the example are said to form an arithmetic sequence because they change by a constant amount each year. Each term increases or decreases by the same constant value called the common difference of the sequence. For this sequence, the common difference is $$-3,400$$.
The sequence below is another example of an arithmetic sequence. In this case, the constant difference is $$3$$. You can choose any term of the sequence, and add $$3$$ to find the subsequent term.
###### ARITHMETIC SEQUENCE
An arithmetic sequence is a sequence that has the property that the difference between any two consecutive terms is a constant. This constant is called the common difference. If $$a_1$$ is the first term of an arithmetic sequence and $$d$$ is the common difference, the sequence will be:
$\{a_n\}=\{a_1,a_1+d,a_1+2d,a_1+3d,...\}$
###### Example $$\PageIndex{1}$$: Finding Common Differences
Is each sequence arithmetic? If so, find the common difference.
1. $$\{1,2,4,8,16,...\}$$
2. $$\{−3,1,5,9,13,...\}$$
Solution
Subtract each term from the subsequent term to determine whether a common difference exists.
1. The sequence is not arithmetic because there is no common difference.
$$2-1={\color{red}1} \qquad 4-2={\color{red}2} \qquad 8-4={\color{red}4} \qquad 16-8={\color{red}8}$$
1. The sequence is arithmetic because there is a common difference. The common difference is $$4$$.
$$1-(-3)={\color{red}4} \qquad 5-1={\color{red}4} \qquad 9-5={\color{red}4} \qquad 13-9={\color{red}4}$$
Analysis
The graph of each of these sequences is shown in Figure $$\PageIndex{1}$$. We can see from the graphs that, although both sequences show growth, (a) is not linear whereas (b) is linear. Arithmetic sequences have a constant rate of change so their graphs will always be points on a line.
Figure $$\PageIndex{1}$$
###### Q&A
If we are told that a sequence is arithmetic, do we have to subtract every term from the following term to find the common difference?
No. If we know that the sequence is arithmetic, we can choose any one term in the sequence, and subtract it from the subsequent term to find the common difference.
###### Exercise $$\PageIndex{1A}$$
Is the given sequence arithmetic? If so, find the common difference.
$$\{18, 16, 14, 12, 10,…\}$$
Answer
The sequence is arithmetic. The common difference is $$–2$$.
###### Exercise $$\PageIndex{1B}$$
Is the given sequence arithmetic? If so, find the common difference.
$$\{1, 3, 6, 10, 15,…\}$$
Answer
The sequence is not arithmetic because $$3−1≠6−3$$.
## Writing Terms of Arithmetic Sequences
Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of $$n$$ and $$d$$ into formula below.
$a_n=a_1+(n−1)d$
###### How to: Given the first term and the common difference of an arithmetic sequence, find the first several terms.
1. Add the common difference to the first term to find the second term.
2. Add the common difference to the second term to find the third term.
3. Continue until all of the desired terms are identified.
4. Write the terms separated by commas within brackets.
###### Example $$\PageIndex{2}$$: Writing Terms of Arithmetic Sequences
Write the first five terms of the arithmetic sequence with $$a_1=17$$ and $$d=−3$$.
Solution
Adding $$−3$$ is the same as subtracting $$3$$. Beginning with the first term, subtract $$3$$ from each term to find the next term.
The first five terms are $$\{17,14,11,8,5\}$$
Analysis
As expected, the graph of the sequence consists of points on a line as shown in Figure $$\PageIndex{2}$$.
Figure $$\PageIndex{2}$$
###### Exercise $$\PageIndex{2}$$
List the first five terms of the arithmetic sequence with $$a_1=1$$ and $$d=5$$.
Answer
$$\{1, 6, 11, 16, 21\}$$
###### How to: Given any the first term and any other term in an arithmetic sequence, find a given term.
1. Substitute the values given for $$a_1$$, $$a_n$$, $$n$$ into the formula $$a_n=a_1+(n−1)d$$ to solve for $$d$$.
2. Find a given term by substituting the appropriate values for $$a_1$$, $$n$$, and $$d$$ into the formula $$a_n=a_1+(n−1)d$$.
###### Example $$\PageIndex{3}$$: Writing Terms of Arithmetic Sequences
Given $$a_1=8$$ and $$a_4=14$$, find $$a_5$$.
Solution
The sequence can be written in terms of the initial term $$8$$ and the common difference $$d$$.
$$\{8,8+d,8+2d,8+3d\}$$
We know the fourth term equals $$14$$; we know the fourth term has the form $$a_1+3d=8+3d$$.
We can find the common difference $$d$$.
\begin{align*} a_n&= a_1+(n-1)d \\ a_4&= a_1+3d \\ a_4&=8+3d\qquad \text{Write the fourth term of the sequence in terms of }a_1 \text{ and } d. \\ 14&=8+3d\qquad \text{Substitute }14 \text{ for } a_4. \\ d&=2\qquad \text{Solve for the common difference.} \end{align*}
Find the fifth term by adding the common difference to the fourth term.
$$a_5=a_4+2=16$$
Analysis
Notice that the common difference is added to the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by adding the common difference to the first term nine times or by using the equation $$a_n=a_1+(n−1)d$$.
###### Exercise $$\PageIndex{3}$$
Given $$a_3=7$$ and $$a_5=17$$, find $$a_2$$.
Answer
$$a_2=2$$
## Using Recursive Formulas for Arithmetic Sequences
Some arithmetic sequences are defined in terms of the previous term using a recursive formula. The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is $$5$$, then each term is the previous term plus $$5$$. As with any recursive formula, the first term must be given.
$$a_n=a_n−1+d$$
for $$n≥2$$
###### Note: RECURSIVE FORMULA FOR AN ARITHMETIC SEQUENCE
The recursive formula for an arithmetic sequence with common difference $$d$$ is:
$a_n=a_n−1+d$
for $$n≥2$$
###### How to: Given an arithmetic sequence, write its recursive formula.
1. Subtract any term from the subsequent term to find the common difference.
2. State the initial term and substitute the common difference into the recursive formula for arithmetic sequences.
###### Example $$\PageIndex{4}$$: Writing a Recursive Formula for an Arithmetic Sequence
Write a recursive formula for the arithmetic sequence.
$$\{−18, −7, 4, 15, 26, …\}$$
Solution
The first term is given as $$−18$$. The common difference can be found by subtracting the first term from the second term.
$$d=−7−(−18)=11$$
Substitute the initial term and the common difference into the recursive formula for arithmetic sequences.
$$a_1=−18$$
$$a_n=a_{n−1}+11$$
for $$n≥2$$
Analysis
We see that the common difference is the slope of the line formed when we graph the terms of the sequence, as shown in Figure $$\PageIndex{3}$$. The growth pattern of the sequence shows the constant difference of 11 units.
Figure $$\PageIndex{3}$$
###### Q&A
Do we have to subtract the first term from the second term to find the common difference?
No. We can subtract any term in the sequence from the subsequent term. It is, however, most common to subtract the first term from the second term because it is often the easiest method of finding the common difference.
###### Exercise $$\PageIndex{4}$$
Write a recursive formula for the arithmetic sequence.
$$\{25, 37, 49, 61, …\}$$
Answer
\begin{align*}a_1 &= 25 \\ a_n &= a_{n−1}+12 , \text{ for }n≥2 \end{align*}
## Using Explicit Formulas for Arithmetic Sequences
We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept.
$$a_n=a_1+d(n−1)$$
To find the $$y$$-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence.
The common difference is $$−50$$, so the sequence represents a linear function with a slope of $$−50$$. To find the $$y$$-intercept, we subtract $$−50$$ from $$200$$: $$200−(−50)=200+50=250$$. You can also find the $$y$$-intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in Figure $$\PageIndex{4}$$.
Figure $$\PageIndex{4}$$
Recall the slope-intercept form of a line is $$y=mx+b$$. When dealing with sequences, we use $$a_n$$ in place of $$y$$ and $$n$$ in place of $$x$$. If we know the slope and vertical intercept of the function, we can substitute them for $$m$$ and $$b$$ in the slope-intercept form of a line. Substituting $$−50$$ for the slope and $$250$$ for the vertical intercept, we get the following equation:
$$a_n=−50n+250$$
We do not need to find the vertical intercept to write an explicit formula for an arithmetic sequence. Another explicit formula for this sequence is $$a_n=200−50(n−1)$$, which simplifies to $$a_n=−50n+250$$.
###### Note: EXPLICIT FORMULA FOR AN ARITHMETIC SEQUENCE
An explicit formula for the $$n^{th}$$ term of an arithmetic sequence is given by
$a_n=a_1+d(n−1)$
###### How to: Given the first several terms for an arithmetic sequence, write an explicit formula.
1. Find the common difference, $$a_2−a_1$$.
2. Substitute the common difference and the first term into $$a_n=a_1+d(n−1)$$.
###### Example $$\PageIndex{5}$$: Writing the nth Term Explicit Formula for an Arithmetic Sequence
Write an explicit formula for the arithmetic sequence.
$$\{2, 12, 22, 32, 42, …\}$$
Solution
The common difference can be found by subtracting the first term from the second term.
\begin{align*} d &= a_2−a_1 \\ &= 12−2 \\ &= 10 \end{align*}
The common difference is $$10$$. Substitute the common difference and the first term of the sequence into the formula and simplify.
\begin{align*}a_n &= 2+10(n−1) \\ a_n &= 10n−8 \end{align*}
Analysis
The graph of this sequence, represented in Figure $$\PageIndex{5}$$, shows a slope of $$10$$ and a vertical intercept of $$−8$$.
Figure $$\PageIndex{5}$$
###### Exercise $$\PageIndex{5}$$
Write an explicit formula for the following arithmetic sequence.
$$\{50,47,44,41,…\}$$
Answer
$$a_n=53−3n$$
### Finding the Number of Terms in a Finite Arithmetic Sequence
Explicit formulas can be used to determine the number of terms in a finite arithmetic sequence. We need to find the common difference, and then determine how many times the common difference must be added to the first term to obtain the final term of the sequence.
###### How to: Given the first three terms and the last term of a finite arithmetic sequence, find the total number of terms.
1. Find the common difference $$d$$.
2. Substitute the common difference and the first term into $$a_n=a_1+d(n–1)$$.
3. Substitute the last term for $$a_n$$ and solve for $$n$$.
###### Example $$\PageIndex{6}$$: Finding the Number of Terms in a Finite Arithmetic Sequence
Find the number of terms in the finite arithmetic sequence.
$$\{8, 1, –6, ..., –41\}$$
Solution
The common difference can be found by subtracting the first term from the second term.
$$1−8=−7$$
The common difference is $$−7$$. Substitute the common difference and the initial term of the sequence into the nth term formula and simplify.
\begin{align*} a_n &= a_1+d(n−1) \\ a_n &= 8+−7(n−1) \\ a_n &= 15−7n \end{align*}
Substitute $$−41$$ for $$a_n$$ and solve for $$n$$
\begin{align*} -41&=15-7n\\ 8&=n \end{align*}
There are eight terms in the sequence.
###### Exercise $$\PageIndex{6}$$
Find the number of terms in the finite arithmetic sequence.
$$\{6, 11, 16, ..., 56\}$$
Answer
There are $$11$$ terms in the sequence.
### Solving Application Problems with Arithmetic Sequences
In many application problems, it often makes sense to use an initial term of $$a_0$$ instead of $$a_1$$. In these problems, we alter the explicit formula slightly to account for the difference in initial terms. We use the following formula:
$a_n=a_0+dn$
###### Example $$\PageIndex{7}$$: Solving Application Problems with Arithmetic Sequences
A five-year old child receives an allowance of $$1$$ each week. His parents promise him an annual increase of $$2$$ per week.
1. Write a formula for the child’s weekly allowance in a given year.
2. What will the child’s allowance be when he is $$16$$ years old?
Solution
1. The situation can be modeled by an arithmetic sequence with an initial term of $$1$$ and a common difference of $$2$$.
Let $$A$$ be the amount of the allowance and $$n$$ be the number of years after age $$5$$. Using the altered explicit formula for an arithmetic sequence we get:
$$A_n=1+2n$$
2. We can find the number of years since age $$5$$ by subtracting.
$$16−5=11$$
We are looking for the child’s allowance after $$11$$ years. Substitute $$11$$ into the formula to find the child’s allowance at age $$16$$.
$$A_{11}=1+2(11)=23$$
The child’s allowance at age $$16$$ will be $$23$$ per week.
###### Exercise $$\PageIndex{7}$$
A woman decides to go for a $$10$$-minute run every day this week and plans to increase the time of her daily run by $$4$$ minutes each week. Write a formula for the time of her run after $$n$$ weeks. How long will her daily run be $$8$$ weeks from today?
Answer
The formula is $$T_n=10+4n$$, and it will take her $$42$$ minutes.
###### Media
Access this online resource for additional instruction and practice with arithmetic sequences.
## Key Equations
recursive formula for nth term of an arithmetic sequence $$a_n=a_{n−1}+d$$ $$n≥2$$ explicit formula for nth term of an arithmetic sequence $$a_n=a_1+d(n−1)$$
## Key Concepts
• An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant.
• The constant between two consecutive terms is called the common difference.
• The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term. See Example $$\PageIndex{1}$$.
• The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly. See Example $$\PageIndex{2}$$ and Example $$\PageIndex{3}$$.
• A recursive formula for an arithmetic sequence with common difference dd is given by $$a_n=a_{n−1}+d$$, $$n≥2$$. See Example $$\PageIndex{4}$$.
• As with any recursive formula, the initial term of the sequence must be given.
• An explicit formula for an arithmetic sequence with common difference $$d$$ is given by $$a_n=a_1+d(n−1)$$. See Example $$\PageIndex{5}$$.
• An explicit formula can be used to find the number of terms in a sequence. See Example $$\PageIndex{6}$$.
• In application problems, we sometimes alter the explicit formula slightly to $$a_n=a_0+dn$$. See Example $$\PageIndex{7}$$.
### Contributors and Attributions
This page titled 11.2: Arithmetic Sequences is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.
• Was this article helpful? | 6,153 | 21,128 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.200616 |
https://www.aimsuccess.in/2017/12/questions-asked-in-ibps-so-prelims-exam_31.html | 1,725,862,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00789.warc.gz | 608,847,563 | 82,913 | # Questions Asked in IBPS SO Prelims Exam 31 December 2017
## Questions Asked in IBPS SO Prelims Exam 31 December 2017
The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams.
Here is the detailed Questions Asked in IBPS SO Prelims Exam 31 December 2017. Read today’s Exam Analysis along with this article to know what kind of questions are asked in today’s exam. This will be very beneficial for you if you haven’t appeared for your exam yet. Even if you did appear for the exam already, you can analyze your performance by having a look at the analysis published.
## Quant Questions Asked in IBPS SO Pre Exam 31 Dec 2017
Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. 11, 13, 111, 257, 427, ? Answers - 609
2. 38, 51, 25, 64, 12, ? Answers - 77
3. 4, 14, 31, 57, 94, ? Answers - 50
4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40
5. 6, 5, 9, 26, ?, 514 Answers - 103
Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ?
Q 3. Root (2025-x)/25 =16
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
Directions (1 – 5): Answer the questions on the basis of the information given below.
Eight friends A, B, C, D, E, F, G and H are seating around a circular table. Some of them are facing inside & others are facing outside. Opposite direction means if one is facing inside the centre, second one face outside the centre, and vice versa.
Three people are seating between F and D and both face opposite direction to each other. E is seating second to right of both D and F, and face opposite direction as F faces. C is seating third to right of E, who is not opposite to B. G is neighbour of both E and D, and seating second to right of B, who is not a neighbour of E. G is third to left of A, who is third to right of both H and G. B face towards the centre and is seating second to left of C.
1. If all the friends are seating according to alphabetical order in anti clock wise direction, starting from A, how many friends remain at same position (excluding A)?
A) One
B) Two
C) Three
D) Four
E) None
2. If D interchanged his position with H and, F interchanged his position with G, who sits immediately right of H in new arrangement?
A) F
B) G
C) E
D) A
E) None
3. Who is seating third to left of B?
A) H
B) C
C) A
D) F
E) None
4. How many persons are facing away from the centre?
A) Three
B) Four
C) Five
D) Two
E) None
5. Who among the following pairs are facing same direction and seating opposite to each other ?
A) H, B
B) C, G
C) F, D
D) E, A
E) None
Directions (6 – 10): Answer the questions on the basis of the information given below.
Ten friends are sitting on twelve seats in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In Row 1: A, B, C, D and E are seated and all of them are facing south, and in Row 2: P, Q, R, S and T are sitting and all of them are facing north. One seat is vacant in each row. Therefore, in the given seating arrangement each member seated in a row faces another member of the other row.
All of them like different colors – Red, Green, Black, Yellow, White, Blue, Brown, Purple, Pink and Grey, but not necessarily in the same order.
There are two seats between Q and the vacant seat. Q does not like White, Red and Purple. E is not an immediate neighbor of C. B likes Grey. Vacant seat of row 1 is not opposite to S and is also not at any of the extreme ends of Row-1.The one who likes Black sits opposite to the one, who sits third to the right of the seat, which is opposite to S. C is not an immediate neighbor of D. T, who likes neither White nor Blue, does not face vacant seat. D faces R. The vacant seats are not opposite to each other. Two seats are there between C and B, who sits third right of the seat, on which the person who likes Brown is sitting. S sits third to the right of seat on which R sits and likes Yellow. The one who likes Pink faces the one who likes Yellow. The persons who like Red and Purple are adjacent to each other. The vacant seat in row 1 is not adjacent to D.Q sits at one of the extreme ends. E neither likes Pink nor faces the seat which is adjacent to the one who likes Blue. The one who likes White is not to the immediate right of the one who likes Yellow. The person who likes Green doesn’t face the person who likes Purple.
6. How many persons are sitting between T and the one who likes yellow color?
A) None
B) One
C) Two
D) Three
E) None of these
7. Which of the following faces the vacant seat of Row – 2?
A) The one who like white color
B) A
C) D
D) The one who likes grey color
E) Cannot be determined
8. Who is sitting at the immediate left of person who likes purple color?
A) E
B) D
C) The one who likes black color
D) The one who likes green color
E) The one who likes grey color
9. Who amongst the following sits at the extreme end of the row?
A) R, Q
B) E, S
C) T, C
D) C, D
E) None of these
10. If Q is made to sit on vacant seat of his row, then how many persons are there between the persons who sit opposite to Q now and who sat opposite to Q previously?
A) Two
B) Three
C) Four
D) None
E) One
Directions (Q. 11–15): Study the following arrangement of series carefully and answer the questions given below:
C # E N 4 \$ F 3 I L 8 @ G © P O V 5 2 A X % J 9 * W K 6 Z 7&2 S
11. How many such symbols are there in the above arrangement each of which is either immediately preceded by a letter or immediately followed by a letter but not both?
A) None
B) Three
C) One
D) More than three
E) Two
12. If all the symbols in the above arrangement are dropped which of the following will be twelfth from the left end?
A) 8
B) 2
C) A
D) Other than given options
E) O
13. How many such numbers are there in the above arrangements each of which is immediately followed by a consonant but not immediately preceded by a letter?
A) None
B) Two
C) One
D) Three
E) More than three
14. Four of the following five are alike in a certain way based on their positions in the above arrangement and so form a group. Which is the one that does not belong to this group?
A) 6 &*
B) 5 X O
C) F L 4
D) G O 8
E) 9 K %
15. Which of the following is the seventh to the right of the eighteenth from the right end of the above arrangement?
A) I
B) Other than given options
C) 8
D) J
E) *
1. B
2. A
3. D
4. B
5. A
6. C
7. D
8. E
9. C
10. E
11. B
12. E
13. C
14. A
15. D
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. Reading Comprehension – 2 Sets, 10 Qs (each consisting of 5 Qs)
– 1 Passage was based on “Scope on Economy”
– Other passage was based on “Steps that should be taken to stop the corruption”
– Meaning of “Narrowing” was asked. Also antonym of “Dearer” was asked in exam.
2. Error Spotting 10 Qs – Asked in a different way.
A sentence was divided into 5 parts. It was given that the last part of the sentence is error free. You have to find error free part of the sentence among the 4 parts.
3. Fill in the blanks – 5 Qs – Vocab Based.
2 Sentences were given in which you had to choose a word which correctly fits in both the sentences.
Eg. i) He is _______ at playing the piano.
ii) Exercise is _______ for health.
## General Awareness Asked in IBPS SO Pre Exam 31 Dec 2017
2. FIFA U-19 was held at?
3. 1 Qs based on Basel committee?
5. Repo rate?
update soon
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
update soon
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
update soon
## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2)
We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com | 2,424 | 8,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-38 | latest | en | 0.944515 |
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Ms. Jovanovich's Math Mumblings Jovanovich Mathematics This blog is for Miss Jovanovich's algebra and geometry classes to share ideas, ask questions, and reflect on what skills and topics we are studying. If we are lucky, we might even get a chance to connect with other students studying the same stuff.
by Tina Jovanovich
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Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539
Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309
Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291
Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279
Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356
Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419
Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217
Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! *Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233
Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235
Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381
Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241
Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229
Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222
Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311
Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234
Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312
Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303
Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261
Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600*sin(25)=X. Enter 1600*sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240
Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208
Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217
Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234
Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219
Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244
Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257
Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215
Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218
Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216
Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1*d2, so I substituted in the diameter lengths: A = 1/2 * 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207
Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256
Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190
Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251
Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50*. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? ***SPOILER ALERT*** Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237
Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198
My Classes & Students
- - - - - - | 9,383 | 34,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2013-20 | latest | en | 0.864713 |
https://tutorialspoint.dev/algorithm/dynamic-programming-algorithms/longest-increasing-path-matrix | 1,638,582,624,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00618.warc.gz | 615,987,150 | 10,677 | # Longest Increasing Path in Matrix
Given a matrix of N rows and M columns. From m[i][j], we can move to m[i+1][j], if m[i+1][j] > m[i][j], or can move to m[i][j+1] if m[i][j+1] > m[i][j]. The task is print longest path length if we start from (0, 0).
Examples:
```Input : N = 4, M = 4
m[][] = { { 1, 2, 3, 4 },
{ 2, 2, 3, 4 },
{ 3, 2, 3, 4 },
{ 4, 5, 6, 7 } };
Output : 7
Longest path is 1 2 3 4 5 6 7.
Input : N = 2, M =2
m[][] = { { 1, 2 },
{ 3, 4 } };
Output :3
Longest path is either 1 2 4 or
1 3 4.
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The idea is to use dynamic programming. Maintain the 2D matrix, dp[][], where dp[i][j] store the value of length of longest increasing sequence for sub matrix starting from ith row and j-th column.
Let the longest increasing sub sequence values for m[i+1][j] and m[i][j+1] be known already as v1 and v2 respectively. Then the value for m[i][j] will be max(v1, v2) + 1.
We can start from m[n-1][m-1] as base case with length of longest increasing sub sequence be 1, moving upwards and leftwards updating the value of cells. Then the LIP value for cell m[0][0] will be the answer.
Below is the implementation of this approach:
## C++
`// CPP program to find longest increasing ` `// path in a matrix. ` `#include ` `#define MAX 10 ` `using` `namespace` `std; ` ` ` `// Return the length of LIP in 2D matrix ` `int` `LIP(``int` `dp[][MAX], ``int` `mat[][MAX], ``int` `n, ``int` `m, ``int` `x, ``int` `y) ` `{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-1 && y == m-1) ` ` ``return` `dp[x][y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-1 || y == m-1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(mat[x][y] < mat[x+1][y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x+1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(mat[x][y] < mat[x][y+1]) ` ` ``result = max(result, 1 + LIP(dp, mat, n, m, x, y+1)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` `} ` ` ` `// Wrapper function ` `int` `wrapper(``int` `mat[][MAX], ``int` `n, ``int` `m) ` `{ ` ` ``int` `dp[MAX][MAX]; ` ` ``memset``(dp, -1, ``sizeof` `dp); ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` `} ` ` ` `// Driven Program ` `int` `main() ` `{ ` ` ``int` `mat[][MAX] = { ` ` ``{ 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, ` ` ``}; ` ` ``int` `n = 4, m = 4; ` ` ``cout << wrapper(mat, n, m) << endl; ` ` ` ` ``return` `0; ` `} `
/div>
## Java
`// Java program to find longest increasing ` `// path in a matrix. ` `import` `java.util.*; ` ` ` `class` `GFG { ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `dp[][], ``int` `mat[][], ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``) ` ` ``{ ` ` ``int` `result = ``0``; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-``1` `&& y == m-``1``) ` ` ``return` `dp[x][y] = ``1``; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-``1` `|| y == m-``1``) ` ` ``result = ``1``; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + ``1` `< n && mat[x][y] < mat[x+``1``][y]) ` ` ``result = ``1` `+ LIP(dp, mat, n, m, x+``1``, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + ``1` `< m && mat[x][y] < mat[x][y+``1``]) ` ` ``result = Math.max(result, ``1` `+ ` ` ``LIP(dp, mat, n, m, x, y+``1``)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `mat[][], ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `dp[][] = ``new` `int``[``10``][``10``]; ` ` ``for``(``int` `i = ``0``; i < ``10``; i++) ` ` ``Arrays.fill(dp[i],-``1``); ` ` ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``); ` ` ``} ` ` ` ` ``/* Driver program to test above function */` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `mat[][] = { ` ` ``{ ``1``, ``2``, ``3``, ``4` `}, ` ` ``{ ``2``, ``2``, ``3``, ``4` `}, ` ` ``{ ``3``, ``2``, ``3``, ``4` `}, ` ` ``{ ``4``, ``5``, ``6``, ``7` `}, ` ` ``}; ` ` ``int` `n = ``4``, m = ``4``; ` ` ``System.out.println(wrapper(mat, n, m)); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. `
## Python3
`# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia `
## C#
`// C# program to find longest increasing ` `// path in a matrix. ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `[,]dp, ``int` `[,]mat, ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x,y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n - 1 && y == m - 1) ` ` ``return` `dp[x, y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n - 1 || y == m - 1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + 1 < n && mat[x, y] < mat[x + 1, y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x + 1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + 1 < m && mat[x, y] < mat[x, y + 1]) ` ` ``result = Math.Max(result, 1 + ` ` ``LIP(dp, mat, n, m, x, y + 1)); ` ` ` ` ``dp[x, y] = result; ` ` ``} ` ` ` ` ``return` `dp[x,y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `[,]mat, ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `[,]dp = ``new` `int``[10, 10]; ` ` ``for``(``int` `i = 0; i < 10; i++) ` ` ``{ ` ` ``for``(``int` `j = 0; j < 10; j++) ` ` ``{ ` ` ``dp[i, j] = -1; ` ` ``} ` ` ``} ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` ` ``} ` ` ` ` ``/* Driver code */` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `[,]mat= { { 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, }; ` ` ``int` `n = 4, m = 4; ` ` ``Console.WriteLine(wrapper(mat, n, m)); ` ` ``} ` `} ` ` ` `/* This code contributed by PrinciRaj1992 */`
Output:
```7
```
Time Complexity: O(N*M). | 3,679 | 9,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-49 | latest | en | 0.647698 |
https://www.physicsforums.com/threads/a-question-about-e-mc-2.816217/ | 1,531,801,488,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589557.39/warc/CC-MAIN-20180717031623-20180717051623-00244.warc.gz | 948,265,113 | 20,084 | 1. May 28, 2015
### Deepak K Kapur
If i put c=d/t in
E=mc2, then E=m×d2/t2
Now take m=1kg and d=1m
Does this mean that E is inversely proportional to time?
2. May 28, 2015
### Fredrik
Staff Emeritus
No, that would be the case if there's a constant A such that E=A/t. If there's a constant A such that E=A/t2, that we could say that E is inversely proportional to time squared. But what you found is E= md2/t2, where m is a constant and d depends on t. So md2 isn't a constant.
3. May 28, 2015
### PeroK
No, but it means that Energy has a "dimension" of $ML^2T^{-2}$.
Compare this with the classical equation $KE = \frac12 mv^2$, where kinetic energy has the same dimension as above.
Look up "dimensional analysis".
4. May 28, 2015
### Deepak K Kapur
But,
1. d=1m and one meter is always one meter. How does 1m depend on time?
2. When d=1m and m=1kg, we get E=1/t2. What is the physical significance of this reduced equation?
Thanks.
5. May 28, 2015
### Deepak K Kapur
One more question (silly one).
Mass is something concrete, we can see it, touch it, feel it etc.
Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesnt seem to have independent existence.
Then, how on earth can we multiply a concrete entity with an abstract one??
6. May 28, 2015
### Fredrik
Staff Emeritus
1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t.
When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$
7. May 28, 2015
### Staff: Mentor
You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.
Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation.
8. May 28, 2015
### Deepak K Kapur
So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so??
9. May 28, 2015
### Deepak K Kapur
Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm?
10. May 28, 2015
### Staff: Mentor
Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question.
11. May 28, 2015
### Fredrik
Staff Emeritus
Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity.
I would say that aspects of the real world are represented by abstract mathematical things in the theory.
12. May 28, 2015
### Staff: Mentor
No we don't. The units do not disappear.
13. May 28, 2015
### Deepak K Kapur
So, can it be said that when we do science or math, we are not concerned with the 'real' things but are only bothered about their properties??
14. May 28, 2015
### Deepak K Kapur
Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.
Why not, this seems to be a sensible interpretation...
15. May 28, 2015
### ShayanJ
t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$
So you can only have $E=m(\frac{299792458}{1})^2=m(\frac{149896229}{.5})^2=m(\frac{59958491.6}{.2})^2=\dots$.
You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!
EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!!
Last edited: May 28, 2015
16. May 29, 2015
### Staff: Mentor
Well, two reasons:
1. Because the units are still there and you are pretending they aren't. The "1" has units kg-m2. The equation should read E=1 kg-m2/t2.
2. The equation isn't describing the motion of a kilogram of mass, the piece you broke apart is describing the speed of light ("c" is the speed of light). So the 1m is the distance traveled by light in "t" time: time is a constant here, not a variable and you can't change that. So when you plug in m=1 kg, and d= 1m, t=1/300,000,000s and it all simplifies to E=9x10^16 Joules.
Science includes the implied assumption that the observed properties are real and there is nothing else to "concern with" or not "concern with". Beyond that, you're getting into philosophy, not science.
Last edited: May 29, 2015
17. May 30, 2015
### Deepak K Kapur
I dont get your point fully...
1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass?
2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?
18. May 30, 2015
### ArmanCham
Time constant means this ;
c=x/t If we assume x is one then you cannot change t cause c is constant.So you are telling us E=1/t2 here you assumed x equal one.Then you assumed t =2 We cannot change time If you choose x =1.If you change time you can see easily that you are breaking fundemental physics law "Speed of light is constant"
Here the math c=x/t you said x =1 then t=1/c.Now we found t.Its a number but you are telling it can be three or four.Thats nonsense.
19. May 30, 2015
### Staff: Mentor
You are mistaken. C is just a universal constant, that happens to be the speed of light.
You stated the distance was 1m, which just makes time the new constant in the equation. No matter how you change the distance, the time must change in the opposite way in order that d/t always equals C. That's why plugging-in C=d/t is just an unnecessary complication: C is always going to be the same, so there is no point in calculating it every time you do the problem.
You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.
20. May 30, 2015
### Deepak K Kapur
I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass.
2. What does c in this equation mean?
Thanks, u hav been very helpful....
Last edited: May 30, 2015
21. May 30, 2015
### Staff: Mentor
There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation.
Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.
http://en.m.wikipedia.org/wiki/Mass–energy_equivalence
22. May 30, 2015
### Fredrik
Staff Emeritus
In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.
23. May 30, 2015
### Deepak K Kapur
Actually i was expecting the meaning of this equation as follows..
Matter will change into energy when...................(some relation to the speed of light)
Cant you elaborate this equation in this way?
Also plz explain the meaning of 'square' of c?
Last edited: May 30, 2015
24. May 30, 2015
### Deepak K Kapur
Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?
If they are same, then do they look different to the 'observer' only or are they 'really' different?
25. May 30, 2015
### Staff: Mentor
No, that isn't what it means. I can't really elaborate on something that isn't true.
C^2 is the conversion factor to equate matter and energy. | 2,517 | 8,849 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-30 | latest | en | 0.912886 |
https://takechargeofyourmoney.blog/category/calculate-savings/ | 1,498,392,808,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320491.13/warc/CC-MAIN-20170625115717-20170625135717-00612.warc.gz | 816,117,911 | 19,665 | ## How to calculate the interest you can earn
Here’s a quick explanation on how to calculate the interest that you can earn on your savings (or pay on your debt).
Let’s use an example of R1,500 (don’t worry about the currency – the maths is the same) and you deposit it in your bank account that earns 3.5% interest.
First thing to note is that the interest rate is always given as the annual rate (unless very specifically detailed otherwise). If you use a calculator (or a spreadsheet such as Excel) and do the following : 1500 x 3.5% the answer you get is 52.50.
Thus R1,500 invested for 1 year at 3.5% interest will earn R52.50. That’s easy!
To calculate what the monthly interest is, simply divide 52.50 by 12 and you will get R4.38 (rounded up) interest for 1 month. You could be more accurate and divide 52.5 by 365.4 (number of days in the year) and then multiply by the exact number of days in the month (e.g. 30).
E.g. 52.5 divided 365.4 multiplied by 30 = R4.31 (a very slight difference to the first answer we received)
So, if you invest R1,500 for 1 year at an interest rate of 3.5%, the interest at the end of the year would be R52.50. However, banks and all financial institutions calculate interest on a daily basis and pay it out monthly. If you withdrew each months interest (and leave the original amount of money in the account) then you would effectively earn “simple” interest – just as we calculated above and you would have R52.50 (assuming no fees or charges)
However, if you leave the interest that you earn each month in your account, then the bank would calculate the next months interest on your original amount plus whatever extra interest is already in your account. So each month you would earn slightly more. This is what is called “compound” interest.
In this case the difference is not great, but the larger the starting amount is, and the longer time you keep reinvesting the interest, the larger the difference is! In fact, after a few years, compound interest can be really large. Remember of course that we are assuming that no fees or charges apply.
This example may seem simple, but there is no need to complicate things!
## Find Extra Money
We’d all love some extra money I’m sure, so here are some steps to literally find it.
Step 1 : Look at your bank statement
Look at your bank statement as well as your credit card statement for the past month and make a list of all automatic payments that are deducted (eg internet bill, mobile phone, insurance, Netflix, etc)
Step 2 : Analyze all payments
One-by-one look at each automatic payment and ask:
• Do I need this service / subscription or whatever it is?
• Do I need all the features I pay for? Can I downgrade at all or pay for less?
• If it is a store card, do you need the extra’s such as insurance, magazine, etc?
• If it is a debt repayment, how much interest am I paying? What must I do to pay it off?
Step 3 : Insurance
Pay special attention to your insurance payment. Look at your latest insurance policy or contact the company for it. Are the values correct? Now, get new quotes and see if you can find a cheaper option.
Look at optional features you may have on any insurance, store card or debt policies. Perhaps you have the same benefits offered by different policies; in that case you can cancel some.
Step 4 : Look for wasted money
Look at your monthly expenses and take special note of the following:
• Eating Out / Take-Aways
• Groceries
• Clothing
• Entertaining at home
• Alcohol bought
• Luxuries
• Telephone
• Electricity
• Internet
• Make-up
Can you spend less on any of the above? Try for just 1 month to spend less than the previous month and you will see that you really can survive quite easily!
Important note is that what you save today should not be spent tomorrow, don’t feel tempted to spend your savings on other things. See when saving is not really saving.
Step 5 : Be strict with yourself
Make a list of every area where you feel you can save a little and be strict with yourself over the next month to actually do this!
## When saving ins’t saving
I am constantly looking at how and where I spend my money and deciding whether what I am doing is worth it or not. There needs to be balance between enjoying live but at the same time living well within ones means and achieving ones financial goals. All it takes to be conscious about what you spend. I have come across times though when I think I am saving money when in fact I am not.
### Hosting people vs Going out
The first instance is the fallacy that entertaining at home is cheaper than going out. This can be true, but not necessarily all the time. It is great to host friends and family and one should enjoy every moment of it; life is too short not too. I am definitely not implying that one should not host people in your home, but the point of this post is about saving money. If you specifically decide to host people at your home instead of going out (in order to save) then you need to ensure that you do actually spend less money.
Think of the average restaurant bill that you would be paying if you went out, and now look at what you are about to spend in order to host your friends. Flowers, candles, snacks, wine, groceries, dessert, etc… It is easy to go overboard and spend more than you would have. (Especially if you’re married to a chef, which is the case for me.) The point is not to be stingy though. If you wish to save money then calculate the costs of going versus what you would like to spend when entertaining and chose the cheaper option.
### A picnic vs lunch out
As with the first example, if you decide to go on a picnic in order to save money (rather than an expensive restaurant), then be sure to do some calculations. It’s funny how in our minds we convince ourselves that because we’re taking the cheaper option, we can now spend more. A bottle of bubbly, expensive cheeses, some tapas, etc… It’s really easy to spend more on the picnic than in a restaurant.
###### The point of these examples is that if you choose to do something in order to save some money, keep that in mind and be sure to actually save! Don’t let yourself be tempted to buy the more expensive items to compensate for the cheaper option. You will actually end up spending more.
Now, take the money you calculated you would save and immediately transfer it into your savings account! Do it now before you find something else to spend it on!
## Calculate the Future Value of an investment
In this article we looked at the effect of extra payments into ones home loan (mortgage bond). That’s all good and well, but what if you don’t have a loan? Or perhaps you would rather invest in a Unit Trust or Interest Bearing bank account.
Microsoft Excel has a simple formula called “FV” which will help you to calculate the future value of your investment. This does not account for changing interest rates or complicated scenarios, but it will give you a simple tool to give you an idea of how your money can grow.
To start with a simple example, let’s save 600 each month for 10 years (120 months) at an interest of 6%. The answer as you see is 98,327.61.
That’s pretty easy. Let’s quickly look at the input parameters:
Rate: This is the interest rate per period. Generally speaking, financial institutions will quote an annual interest rate. In this example it is 6%. To get the monthly interest rate we simply use 6%/12. If we change the example an decide to make quarterly payments we would need to use 6%/4.
Nper: This is the number of periods (in total) for which we will make a payment. If we wish to pay monthly, this is the number of months.
Pmt: This is the amount we wish to pay each period. This figure cannot change over the lifetime of the investment. But, see further down how we can circumvent this potential problem. Excel expects the payment to be a negative figure as payments out of your account would “minus” from your account. You will see that if you use a positive figure, your answer will show negative. The figure will be the same though. (If that is confusing, try it yourself)
Pv: The present value. If you are adding money to an existing bank account, use Pv as the present value in the account. If this to calculate a new investment leave it blank or type a zero in the box.
Type: This is to specify whether the payment is being made at the beginning of each period, or at the end. This will affect the interest that you earn. By default Excel will assume you are paying at the end of each period.
So, in the example above we assumed a constant 600 payment each month for 10 years. But what if you decide to pay 10% more each year? For that we will need to spice things up a bit. See how I have calculated the Future Value for 12 months at a time. I then use the answer as my Present Value for the following year. The monthly payment is simply increased by 10% in each row.
## How much is your coffee costing you?
I love coffee! The very thought of drinking less makes me quiver with fear. But, sometimes one must take the emotion out of decisions and just look at cold, hard facts.
In a previous article on being more conscious with cash, I discussed how I buy 2 coffees each day, along with a breakfast and lunch. I’m not good at planning meals to bring to work but I have decided to spend slightly less. The savings I calculated was only R150 per week, but I’ve decided to look at how this could affect my home loan (mortgage bond) if I pay R600 extra per month.
These figures may not make sense to you if you use a different currency and your countries interest rates may be significantly different. I’ll post something soon about how you can do these calculations yourself.
The calculations can get messy so take note of the following assumptions:
Home Loan Value: R1,500,000
Interest Rate: 11% (annual)
Total Loan period: 20 years
For this example I will assume 2 years of the loan are already complete, and that up to now no additional payments have been made. Also, the R600 p.m. will remain constant (with other words I won’t save more in years to come)
This may all sound complicated, but the end result is simple to see. If I pay R600 p.m. extra into my bond until my bond is paid off I will:
• Save a total of R236,000 (rounded to nearest thousand)
• Pay the loan off 22 months earlier
That’s pretty amazing don’t you think? Considering that I am making this extra saving by cutting down on my weekly coffee/food expenses at my office. In fact, it was quite easy to find the extra R150 per week. Imagine if I actually analyze my expenses properly and relook at my insurance policies, health care, mobile phone contract, bank charges, etc. Imagine the savings I can make then!
What if you don’t have a home loan?
That’s really not a problem. If you invest this money in either a unit trust or interest bearing account for the next 15 – 20 years you will have quite a large lump sum. We’ll look at how to use the Future Value formula in Excel to calculate this. | 2,574 | 11,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-26 | longest | en | 0.956167 |
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### Ching chi tu journal 5
1. 1. Ching Chi Tu Journal#5 m1 geometry
2. 2. Perpendicular Bisector A line perpendicular to a segment at the segment’s midpoint. Perpendicular Bisector Theorem – If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. Converse of the Perpendicular bisector Theorem – If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.
3. 3. Converse P.B. Thm H A Ex 1. f = g Ex 2. d = e Ex 3. b = c T T f = g; g = c; i = h ce ab; bg cb Ex 1. a = h Ex 2. i = j Ex 3. cg = bg T
4. 4. Angle Bisector Angle Bisector - A ray that divides an angle into two congruent angles. Angle Bisector Theorem – If a point is on the bisector of an angle, then it is equidistant from the sides of the angle. Converse of Angle Bisector Theorem – If a point in the interior of an angle is equidistant from the sides of the angle, then it is on the bisector of the angle
5. 5. Ex 1. c bisects <abc; point L is equidistant from both sides. Ex 2. i bisects <ghi; point J is equidistant from both sides. Ex 3. e bisects <def; point K is equidistant from both sides. H A.B. Thm Converse Ex 1. L is equidistant from both sides; c must be an a.b. Ex 2. J is equidistant from both sides; i must be an a.b. Ex 3. K is equidistant from both sides; f must be an a.b.
6. 6. Concurrency To be concurrent means for three or more lines to intersect at one point. The point of concurrency of perpendicular bisectors form a circumcenter. Circumcenter - where three perpendicular bisectors of a triangle are concurrent. It is equidistant from the vertices of the triangle
7. 7. Circumcenter D is the circumcenter of triangle abc, it is equidistant from points a, b, and c. Right = circumcenter midpoint of hypotenuse
8. 8. Circumcenter L is the circumcenter of triangle jkl, it is equidistant from points j, k, and l. Acute = circumcenter inside
9. 9. Circumcenter H is the circumcenter of triangle efg, it is equidistant from points e, f, and g. Obtuse = circumcenter outside
10. 10. Concurrency The point of concurrency of angle bisectors forms an incenter. Incenter – The point of concurrency of three angle bisectors of a triangle. It is equidistant from the sides of the triangle. Concurrency Concurrency
11. 11. Incenter Incenter always inside triangle D is the incenter of triangle abc, it is equidistant to all sides.
12. 12. Incenter Incenter always inside triangle A is the incenter of triangle xyz, it is equidistant to all sides.
13. 13. Incenter Incenter always inside triangle D is the incenter of triangle abc, it is equidistant to all sides.
14. 14. Concurrency Median – segments whose endpoints are a vertex of the triangle and the midpoint of the opposite side. The point of concurrency of medians is called a centroid. Centroid – Point of concurrency of the medians of a triangle. It is located 2/3 of the distance from each vertex to the midpoint of the opposite side
15. 15. Centroid Ex 1. g is the centroid of triangle abc Ex 2. o is the centroid of triangle egc Ex 3. k is the centroid of triangle adg Centroid = Point of balance
16. 16. Concurrency Altitude – a perpendicular segment from a vertex to the line containing the opposite side. The point of concurrency of three altitudes of a triangle creates an orthocenter. Orthocenter – point where three altitudes of a triangle are concurrent. Nothing special.
17. 17. Orthocenter D is the orthocenter of triangle abc with lines e, d, and f as altitudes.
18. 18. Orthocenter D is the orthocenter of triangle abc with lines e, d, and f as altitudes.
19. 19. Orthocenter D is the orthocenter of triangle abc with lines e, d, and f as altitudes.
20. 20. Midsegment Midsegment of a triangle – a segment that joins the midpoints of two sides of the triangle. Every triangle has three midsegments, forming a midsegment triangle. Triangle Midsegment Theorem – A midsegment of a triangle is parallel to a side of the triangle, and its length is half the length of that side.
21. 21. Midsegment Ex 1. f BC Ex 2. d AC Ex 3. e BA They are all midsegments and are parallel to the lines written beside them.
22. 22. Relationships In any triangle, the longest side is always opposite to the largest angle. Small side = small angle.
23. 23. <ul><li>Ex 3. </li></ul><ul><li><a is the smalles angle, so its opposite angle is also the smallest. </li></ul><ul><li>Side b is the largest side so its opposite angle is the largest. </li></ul><ul><li>Ex 2. </li></ul><ul><li><c is the largest angle so its opposite side is also the largest. </li></ul><ul><li>Side a is the smallest side so its opposite angle is also the smallest. </li></ul><ul><li>Ex. 1 </li></ul><ul><li><a is the largest angle so therefore side a, its opposite side, must be the largest side. </li></ul><ul><li><c has the smallest angle so this means that side c is the smallest side in the triangle. </li></ul>
24. 24. Inequality Exterior Angle Inequality – The exterior angle of a triangle is larger than either of the non-adjacent angles.
25. 25. Ex 1. m<e is larger than all non adjacent angles such as b and y. Ex 2. m<c is greater than all non adjacent angles such as a and y. Ex 3. m<o is greater than all non adjacent angles such as a and b.
26. 26. Inequality Triangle Inequality Theorem – The two smaller sides of a triangle must add up to MORE than the length of the 3 rd side.
27. 27. Ex 1. 2.08 + 1.71 is less than 5.86 so it doesn’t form a triangle. Ex 2. 3.02 + 2.98 = 6 But it doesn’t form a triangle, it forms a line/segment. Ex 2. 1.41 + 1.41 is greater than 2 so therefore, it does form a triangle.
28. 28. Indirect Proofs <ul><li>Steps for indirect proofs: </li></ul><ul><li>Assume the opposite of the conclusion is true </li></ul><ul><li>Get the contradiction </li></ul><ul><li>Prove it is true </li></ul>
29. 29. Ex 1. Ex 2. Prove: a supplement of an acute angle cannot be another acute angle Contradiction Prove: a scalene triangle cannot have two congruent midsegments. Ex 3. Prove: an isosceles triangle cannot have a base angle that is a right angle Angle addition postulate 89 + 89 does not equal 180 Definition of acute angles Acute angles are less than 90 degrees Definition of supplementary angles Two angles add up to form 180 degrees Given Assuming that the supplement of an acute angle is another acute angle. Definition of a scalene triangle. a scalene triangle doesn’t have congruent sides definition of a midsegment. A midsegment can be congruent if there are congruent sides. Definition of a midsegment A midsegment is half the length of the line it is parallel to Given Assuming that a scalene triangle has two congruent midsegments. Substitution 90 + m<b + 90 have to equal 180. m<b must equal 0 Triangle sum theorem M<a + m<b + m<c have to equal 180 Definition of right angles M<a and m<c are both 90 Isosceles triangle theorem. Angle c is congruent to angle a, so angle c is also a right angle. Given Assuming that triangle abc has a base angle that is a right angle. let angle a be the right angle. Triangle abc is an isosceles triangle.
30. 30. Prove: triangle abc cannot have a base angle that is a right triangle Given: triangle abc is an isosceles triangle with base ac. Assuming that triangle abc has a base angle that is a right angle. let angle a be the right angle By the isosceles triangle theorem, angle c is congruent to angle a, so angle c is also a right angle. By the difinitio of right angle m<a = 90 and m<r = 90. By the triangle sum theorem m<a + m<b + m<c =180. By substitution, 90 + m<b + 90 = 180 so m<b = 0.
31. 31. HINGE The Hinge Theorem – if two triangles have 2 sides congruent, but the third side is not congruent, then the triangle with the larger included angle has the longer 3 rd side. Converse – if two triangles have 2 congruent sides, but the third side is not congruent, then the triangle with the larger 3 rd side has the longer 3 rd angle.
32. 32. Ex 1. As <a is larger than <b, side d is larger than side c Ex 2. As <a is smaller than <b, side e is smaller than side b Ex 3. As <b is greater than <a side f is larger than side b Hinge Converse Ex 1. As side d is larger than side c, <a must be larger than <b Ex 3. As side b is smaller than side f, <a must be smaller than <b. Ex 2. As side b is larger than side e, <b must also be larger than <a.
33. 33. Special Relationships 45-45-90 triangle - in all 45-45-90 triangles, both leg are congruent, and the length of the hypotenuse is the length of a leg radical 2. 30-60-90 – In all 30-60-90 triangles, the length of the hypotenuse is twice the length of the short leg; the longer side is the length of the smaller leg radical 3.
34. 34. 45-45-90 Ex 1. b = 4 √2 Ex 3. b = 6.92 √2 Ex 2. b = 4 √2
35. 35. 30-60-90 Ex 1. b = 4.62 c = 4 Ex 2. h = 4.62 i = 4 Ex 3. e = 8 f = 6.93 | 2,613 | 9,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-34 | latest | en | 0.809972 |
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# Microneconomics Essay
1707 words - 7 pages
1) a. If we do not have scarce resources, will we have a law of demand? Will we observe price rationing for goods?
The law of demand states the relationship between quantity demanded and price, showing that the lower the price, the higher the demand and vice versa. If we do not have scarce resources, there will still be a law of demand, because all humans are greedy. This means that we will always want more of what is there and demand always initially exceeds supply, but supply will then catch up, and over time will fall behind again, although this bottlenecking' is always temporary. This can be seen in fibre optic cables, as they catapulted the amount of information able to be ...view middle of the document...
2) a. Explain the difference between point elasticity of demand and arc elasticity of demand.
Point elasticity is the measure of responsiveness or sensitivity of quantity demanded to changes in price. However, arc elasticity is the measure of elasticity between two points on the demand curve. At the two points move closer together on the demand curve, it approaches point elasticity.
Point elasticity is calculated by the percentage change in quantity demanded divided by the percentage change in price, which is shown in formula by: ∆Q% P1
∆P% x Q1
Arc elasticity is calculated the same as point elasticity, although instead of the percentage change, it is calculated by the actual difference between the points. This is shown in formula by: Q2-Q1 P1+P2
P2-P1 x Q2+Q1
This formula is used when there is the change in Price and the Quantity, which allows arc elasticity to be more accurate than point elasticity.
b. Construct on the same graph two straight-line demand curves with the same intercept on the vertical axis, with one curve flatter than the other. Can you make a general statement about which one is the more elastic curve? If so, what is it?
This demand curve shows two straight lines. Line A is steeper and represents a less elastic curve while Line B is flatter and represents a more elastic curve. Point elasticity is the measure of responsiveness or sensitivity of quantity demanded to changes in price. If the elasticity is higher (more than 1), it means that there will be a larger response in demand with price, which is illustrated by curve B. One reason for lower elasticity is brand loyalty, with consumers sticking to one brand all the time. A reason for higher elasticity is the number of substitute products, because as soon as one product reduces its price, it will gain more sales than its substitutes.
3) It has been argued that in order to fight drug abuse sensibly, policy makers must understand the demand for drugs
a. Would you expect that the price elasticity of demand for illegal drugs to be greater than, or less than one?
The price elasticity of illegal drugs is less than one, which is inelastic. This is because drug addicts are will to pay anything to get their hit that they crave. If the person is craving the drug very badly, they will be willing to pay a lot of money, as they don't care f or substitutes and need the drug as soon as possible. Even if the price increases, the addicts still need the drug.
b. Given your answer to a, what happens to total expenditure on illegal drugs when their price increases? What do you think would happen to the amount of drug related crime?
Because drug addicts need the drugs, they are going to pay for the drugs regardless of the price, which means that total expenditure of the users will increase. This could result in the users having to leave more essential expenses to pay for drugs, as the drug opportunity cost far outweighs other...
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Percents are regularly used to connect a family member amount. You have probably seen them supplied for discounts, where the percent the discount can apply to different prices. Percents are likewise used when stating taxes and also interest rates on savings and also loans.
A percent is a proportion of a number come 100. Every cent way “per 100,” or “how numerous out that 100.” You usage the price % ~ a number to show percent.
Notice the 12 that the 100 squares in the grid below have to be shaded green. This to represent 12 percent (12 per 100).
12% = 12 percent = 12 parts out the 100 =
How countless of the squares in the grid over are unshaded? due to the fact that 12 space shaded and there room a full of 100 squares, 88 space unshaded. The unshaded section of the totality grid is 88 components out of 100, or 88% of the grid. Notice that the shaded and also unshaded portions together make 100% of the network (100 out of 100 squares).
Example Problem What percent of the network is shaded? The net is separated into 100 smaller sized squares, v 10 squares in each row. 23 squares out of 100 squares room shaded. Answer 23% of the network is shaded.
Example Problem What percent of the big square is shaded? The net is divided into 10 rectangles. For percents, you need to look at 100 equal-sized parts of the whole. You can divide each of the 10 rectangles right into 10 pieces, providing 100 parts. 30 tiny squares out of 100 space shaded. Answer 30% of the huge square is shaded.
What percent that this network is shaded?
A) 3%
B) 11%
C) 38%
D) 62%
A) 3%
Incorrect. Three complete columns that 10 squares space shaded, plus an additional 8 squares indigenous the next column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the large square. The exactly answer is 38%.
B) 11%
Incorrect. Three complete columns of 10 squares room shaded, plus another 8 squares native the following column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the big square. The correct answer is 38%.
C) 38%
Correct. Three complete columns that 10 squares room shaded, plus an additional 8 squares from the next column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the large square. This means 38% that the huge square is shaded.
D) 62%
Incorrect. There room 62 tiny unshaded squares out of the 100 in the large square, therefore the percent of the large square the is unshaded is 62%. However, the inquiry asked what percent is shaded. There room 38 shaded squares that the 100 squares in the huge square, therefore the exactly answer is 38%.
Rewriting Percents, Decimals, and also Fractions
It is often helpful to readjust the layout of a number. Because that example, girlfriend may find it less complicated to add decimals 보다 to include fractions. If you have the right to write the fractions together decimals, girlfriend can add them as decimals. Then you can rewrite your decimal amount as a fraction, if necessary.
Percents can be composed as fractions and also decimals in very few steps.
Example Problem Write 25% as a simplified portion and as a decimal. Write together a fraction. 25% = Since % method “out the 100,” 25% way 25 the end of 100. You create this together a fraction, utilizing 100 together the denominator. Simplify the portion by separating the numerator and denominator through the usual factor 25. Write together a decimal. 25% = = 0.25 You can also just move the decimal suggest in the whole number 25 two locations to the left to acquire 0.25. Answer 25% = = 0.25
Notice in the diagram listed below that 25% that a network is additionally of the grid, as you found in the example.
Notice that in the vault example, rewriting a percent together a decimal takes just a shift of the decimal point. You have the right to use fractions to recognize why this is the case. Any type of percentage x can be represented as the portion , and any fraction can be created as a decimal by moving the decimal suggest in x two places to the left. For example, 81% can be composed as
, and dividing 81 by 100 results in 0.81. People often skip end the intermediary portion step and just convert a percent come a decimal by relocating the decimal suggest two places to the left.
In the same way, rewriting a decimal together a percent (or as a fraction) requires few steps.
Example Problem Write 0.6 together a percent and also as a streamlined fraction. Write as a percent. 0.6 = 0.60 = 60% Write 0.6 as 0.60, i beg your pardon is 60 hundredths. 60 hundredths is 60 percent. You can additionally move the decimal point two places to the appropriate to discover the percent equivalent. Write as a fraction. 0.6 = To compose 0.6 together a fraction, you read the decimal, 6 tenths, and write 6 tenths in fraction form. Simplify the portion by splitting the numerator and also denominator by 2, a usual factor. Answer 0.6 = 60% =
In this example, the percent is not a entirety number. You can handle this in the exact same way, yet it’s usually less complicated to convert the percent come a decimal and also then transform the decimal to a fraction.
Example Problem Write 5.6% as a decimal and also as a streamlined fraction. Write as a decimal. 5.6% = 0.056 Move the decimal point two locations to the left. In this case, insert a 0 in front of the 5 (05.6) in bespeak to be able to move the decimal come the left two places. Write as a fraction. 0.056 = Write the portion as friend would check out the decimal. The last digit is in the thousandths place, therefore the denominator is 1,000. Simplify the portion by separating the numerator and also denominator through 8, a usual factor. Answer 5.6% = = 0.056
Write 0.645 together a percent and as a simplified fraction. A) 64.5% and B) 0.645% and also C) 645% and also D) 64.5% and also Show/Hide Answer A) 64.5% and Correct. 0.645 = 64.5% = . B) 0.645% and also Incorrect. 0.645 = 64.5%, not 0.645%. Psychic that when you convert a decimal to a percent you have to move the decimal suggest two locations to the right. The correct answer is 64.5% and . C) 645% and Incorrect. 0.645 = 64.5%, not 645%. Remember that when you convert a decimal to a percent you need to move the decimal point two areas to the right. The correct answer is 64.5% and . D) 64.5% and also Incorrect. To create 0.645 as a percent, move the decimal ar two locations to the right: 64.5%. To create 0.645 together a fraction, usage 645 as the numerator. The place value that the critical digit (the 5) is thousandths, for this reason the denominator is 1,000. The fraction is . The greatest common factor the 645 and also 1,000 is 5, therefore you deserve to divide the numerator and denominator by 5 to gain . The exactly answer is 64.5% and also .
In stimulate to create a portion as a decimal or a percent, you have the right to write the fraction as an equivalent fraction with a denominator that 10 (or any kind of other strength of 10 such as 100 or 1,000), which deserve to be then converted to a decimal and then a percent.
Example Problem Write as a decimal and as a percent. Write together a decimal. Find one equivalent portion with 10, 100, 1,000, or various other power that 10 in the denominator. Due to the fact that 100 is a lot of of 4, you deserve to multiply 4 through 25 to obtain 100. Multiply both the numerator and also the denominator by 25. = 0.75 Write the fraction as a decimal through the 5 in the percentage percent place. Write as a percent. 0.75 = 75% To create the decimal together a percent, move the decimal suggest two locations to the right. Answer = 0.75 = 75%
If that is complicated to find an equivalent portion with a denominator of 10, 100, 1,000, and so on, friend can constantly divide the molecule by the denominator to discover the decimal equivalent.
Example Problem Write as a decimal and also as a percent. Write as a decimal. Divide the molecule by the denominator. 3 ÷ 8 = 0.375. Write as a percent. 0.375 = 37.5% To create the decimal together a percent, relocate the decimal allude two areas to the right. Answer = 0.375 = 37.5%
Write as a decimal and also as a percent.
A) 80.0 and 0.8%
B) 0.4 and 4%
C) 0.8 and also 80%
D) 0.8 and also 8%
A) 80.0 and also 0.8%
Incorrect. An alert that 10 is a multiple of 5, so you have the right to rewrite using 10 together the denominator. Main point the numerator and also denominator by 2 to gain . The indistinguishable decimal is 0.8. You deserve to write this together a percent by moving the decimal suggest two places to the right. Since 0.8 has only one location to the right, encompass 0 in the percentage percent place: 0.8 = 0.80 = 80%. The correct answer is 0.8 and also 80%.
B) 0.4 and also 4%
Incorrect. To uncover a decimal indistinguishable for , an initial convert the portion to tenths. Multiply the numerator and denominator by 2 to acquire . The tantamount decimal is 0.8. So, and 0.4 space not indistinguishable quantities. The correct answer is 0.8 and 80%.
C) 0.8 and also 80%
Correct. The price is = 0.8 = 80%.
D) 0.8 and also 8%
Incorrect. It is true the = 0.8, however this does not equal 8%. To create 0.8 as a percent, relocate the decimal allude two locations to the right: 0.8 = 0.80 = 80%. The correct answer is 0.8 and also 80%.
Mixed Numbers
All the previous examples involve fractions and also decimals less than 1, so all of the percents you have actually seen so far have been much less than 100%.
Percents greater than 100% are feasible as well. Percents more than 100% are used to describe instances where there is more than one entirety (fractions and decimals higher than 1 are offered for the same reason).
In the diagram below, 115% is shaded. Every grid is taken into consideration a whole, and also you require two grids for 115%.
Expressed as a decimal, the percent 115% is 1.15; together a fraction, that is
, or
. Notice that you deserve to still convert amongst percents, fractions, and also decimals once the quantity is higher than one whole.
Numbers better than one that incorporate a fractional component can be composed as the sum of a totality number and also the fractional part. For instance, the mixed number is the amount of the entirety number 3 and the portion . = 3 + .
Example Problem Write as a decimal and as a percent. Write the mixed fraction as 2 wholes to add the spring part. Write together a decimal. Write the fractional component as a decimal by splitting the numerator by the denominator. 7 ÷ 8 = 0.875. Add 2 come the decimal. Write together a percent. 2.875 = 287.5% Now you can move the decimal allude two areas to the right to create the decimal as a percent. Answer = 2.875 = 287.5%
Note that a totality number can be created as a percent. 100% means one whole; so 2 wholes would certainly be 200%.
Example Problem Write 375% as a decimal and also as a streamlined fraction. Write together a decimal. 375% = 3.75 Move the decimal allude two areas to the left. Keep in mind that over there is a entirety number together with the decimal together the percent is more than 100%. Write as a fraction. 3.75 = 3 + 0.75 Write the decimal as a amount of the totality number and also the fractional part. 0.75 = Write the decimal part as a fraction. Simplify the portion by splitting the numerator and also denominator by a usual factor that 25. 3 + = Add the totality number component to the fraction. Answer 375% = 3.75=
Write 4.12 as a percent and also as a simplified fraction. A) 0.0412% and B) 412% and also C) 412% and also D) 4.12% and also Show/Hide Answer A) 0.0412% and Incorrect. To convert 4.12 to a percent, move the decimal suggest two locations to the right, not the left. The exactly answer is 412% and also . B) 412% and also Correct. 4.12 amounts to 412%, and also the simplified kind of is . C) 412% and Incorrect. 4.12 does equal 412%, yet it is also equivalent to , no . The correct answer is 412% and also . D) 4.12% and also Incorrect. To transform 4.12 come a percent, relocate the decimal allude two places to the right. The exactly answer is 412% and .See more: How To Build An Indoor Pitching Mound Plans: Step By Step Instructions Summary Percents space a common means to stand for fractional amounts, simply as decimals and fractions are. Any number that deserve to be composed as a decimal, fraction, or percent can also be written utilizing the various other two representations. .tags a { color: #fff; background: #909295; padding: 3px 10px; border-radius: 10px; font-size: 13px; line-height: 30px; white-space: nowrap; } .tags a:hover { background: #818182; } Home Contact - Advertising Copyright © 2022 dearteassociazione.org #footer {font-size: 14px;background: #ffffff;padding: 10px;text-align: center;} #footer a {color: #2c2b2b;margin-right: 10px;} | 3,382 | 12,982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2022-21 | latest | en | 0.924698 |
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level 1
7 points · 1 year ago
1. What part of this is given to you (you have no control over)?
2. What part of this do you have control over?
3. What are you trying to accomplish?
I've determined I'd need 6 actors to make sure no two actors had the same voice AND were next to each other in a sequence.
This is confusing to me. I thought you already had 16 actors (labelled 1-17).
level 2
Original Poster10 points · 1 year ago
16 English actors need to be replaced with french voice over actors.
We don't need a unique voice over actor for each English actor. BUT we can't use the same voice over actor twice in a row in the order the they would appear in the film. And once a voice over actor is assigned to an English actor, that stays fixed. As in we can't just alternate back and forth between two voice over actors.
level 3
9 points · 1 year ago
Gotcha! That makes sense now. Thank you. This is literally the "Chromatic number problem". It is a computationally expensive problem for large collections of actors. For only 17 it isn't bad.
Here's some Sage code you can use to solve your problem (which you can run for free at, for example, cocalc.com )
https://gist.github.com/mpawliuk/ec91468365233b2d1faf57b4bbb56b21
In your particular case you only need 5 French actors. Here are the roles they would play:
You need 5 French actors.
These are the English roles the French actors will take:
French actor 1 will play the parts:
[6, 10, 13, 17]
French actor 2 will play the parts:
[7, 3, 5, 16, 14, 9]
French actor 3 will play the parts:
[2, 15, 11]
French actor 4 will play the parts:
[12, 4]
French actor 5 will play the parts:
[1]
Here's the picture:
https://cocalc.com/blobs//home/user/.sage/temp/project-fbeb3d9c-3251-4e88-9e27-82a1c37642e1/270/tmp_uijHct.svg?uuid=3877e5a6-2b20-41df-b57d-9d862c4d7394
level 4
Original Poster5 points · 1 year ago
This!
Wow this is so cool! I just googled Chromatic Number Problem.
I knew this would be an interesting math problem and there would be some sort of algorithm. Is it simply "vertex_coloring" in the code that solves this? I'd love to know how it's figured out.
Thanks!
level 5
6 points · 1 year ago
Yeah, vertex_coloring() is doing the heavy lifting. Here's the documentation, but it doesn't get into the nitty-gritty.
It looks like we could have also used chromatic_number(). The documentation..
As for the guts of the algorithm, here's a place to start.
level 6
Original Poster5 points · 1 year ago
I'll have to do the reading tonight after work.
Thank you so much!
level 5
2 points · 1 year ago
I really like that the Chromatic number problem has this unexpected application, but in practical terms, does your problem not have an additional constraint about the sex of the voice-over actors needing to match the sex of the parts? In mpaw976's solution, what if parts 6 and 10 are different sexes--will one French actor be able to do both?
level 6
3 points · 1 year ago
Good question!
There are two straightforward ways to deal with this:
1. Separate the roles [1-17] by gender ahead of time, then run the algorithm once on each part, or
2. For roles of different genders, add an edge between them. Then run the algorithm once on the big graph.
We add an edge between any two roles that cannot have the same French actor. OP originally stipulated that it was because of adjacent scenes, but it really doesn't matter the reason.
I suspect the first method will be much faster, but the second method is more general.
level 6
Original Poster1 point · 1 year ago
Thankfully all women! I could probably do it manually if it was mixed. Or just break down sequences to all the men and all the women. Not perfect efficiency but it will get the job done.
level 4
Original Poster3 points · 1 year ago
Alright Mr Smarty pants. While this uses 5 actors vs my 6 actors, if there are different solutions using only 5 actors, is there a way to spit out the solution that has the least amount of disparity in words between colors? Assuming I had the length in words of each numbers script.
Just a hypothetical for now.
level 5
2 points · 1 year ago
The short answer is yes. The longer answer is I can't think of how to do that in practice without some serious machinery that might be too computationally intensive.
One way to solve graph colouring is to basically just write down all the rules. (SAT of CSP if you're curious.) There's an extension to these techniques that allow you to specify a set of constraints that must be satisfied, and a set of constraints that you want the program to do its best on. (Fuzzy-CSP, Soft CSPs, and MAXSAT all fall into this category broadly speaking.)
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≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘∏ ∐ ∑ ⋀ ⋁ ⋂ ⋃ ⨀ ⨁ ⨂ 𝖕 𝖖 𝖗
Set Theory Symbols
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
Logic Symbols
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ↔ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣
Calculus and Analysis Symbols
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
Greek Letters
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔
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Dynamical Systems | 2,094 | 6,905 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-18 | longest | en | 0.953856 |
http://slidegur.com/doc/238473/chapter-4-notes | 1,521,637,834,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647649.70/warc/CC-MAIN-20180321121805-20180321141805-00129.warc.gz | 262,449,004 | 9,966 | ### Chapter 4 Notes
```Chapter 4
Converting
• Fraction to Decimal to Percent
– ½ = .5 = 50% (decimal moves 2 places to right)
• Decimal to Percent to Fraction
– .75 = 75% = ¾ (decimal moves 2 places to right)
• Percent to Fraction to Decimal
– 150% = 1 ½ = 1.5 (decimal moves 2 places to left)
Multiplying & Dividing by a tenth,
hundredth, or thousandth
Multiplying by .1, .01, and .001
• 208 x 0.1 = 20.8 (decimal moves one places to left)
• 83 x 0.01 = .83 (decimal moves two places to left)
• 306 x 0.001 = .306 (decimal moves three places to left)
Dividing by .1, .01, and .001
• 816 / 0.1 = 8,160 (decimal moves one places to right)
• 51 / 0.01 = 5100 (decimal moves two places to right)
• 632 / 0.001 = 632,000 (decimal moves three places to right)
Deductions and Take-Home Pay
•
•
•
•
Income Tax
FICA Tax
Insurance
Parking
•
•
•
•
Uniforms
Dues
Commuting Costs
Etc.
Deductions & Income Tax
• Income Tax
– Withholding tax
– Withholding allowances
– More allowances
• Pay less
– Less allowances
• Pay more
– Use table
Deductions & FICA Tax
• Federal Insurance Contributions Act 7.65%
• Social Security 6.2%
• Medicare 1.45%
New Rate
Through 2012
Net Earnings Example
packet-problem 3 on p. 65
• Elaine Kaufman works a 40-hour week at \$8.75 an hour with
time and half for overtime. Last week she worked 43 hours.
From her earnings, her employer deducted FICA tax at a
rate of 0.0765 and income tax of \$32. Her employer also
deducted \$33.78 for health insurance and \$27.12 for union
dues.
–
–
–
–
–
Regular earnings = 40 * 8.75 = \$350.00
Overtime earnings = 3 * (8.75 * 1.5) = \$39.38
Total earnings = \$350 + \$39.38 = \$389.38
Deductions = \$32 + (\$389.38 * .0765) + \$33.78 + \$27.12 = \$122.69
Net earnings = \$389.38 - \$122.69 = \$266.69
More Percent Examples
•
•
•
•
•
•
•
•
•
67% of \$489 = .67 x \$489 = \$327.63
42 is ___% of 480 = 42 / 480 = .0875 = 8 ¾
20% of \$575 = .2 x \$575 = \$115
1% of \$57 = .01 x \$57 = \$0.57
10% of \$89 = .1 x \$89 = \$8.90
100% of \$76 = 1 x \$76 = \$76
1000% of \$87 = 10 x \$87 = \$870
½% of \$870 = ½ x \$8.70 = \$4.35
1/ % of \$255 = .003333 x 255 = \$0.85
3
Fringe Benefits
• Fringe benefits are extra payments/benefits received for
working for a company like pension, insurance, free parking,
tuition, uniforms, etc.
• Shari Traxler is paid \$9.75 an hour for a 40-hour week. Her
employer also provides these fringe benefits: yearly pension
contributions, \$1,622.40; health and accident insurance per
year, \$956; free parking, \$520 per year; free tuition for evening
classes, \$1,200 per year. (a) What is Shari’s annual wage? (b)
What are her total yearly fringe benefits? (c) What are her total
yearly job benefits?
– (a) \$9.75 x 40 = \$390 (weekly pay) x 52 (weeks per year) = \$20,280
– (b) \$1,622.40 + \$956 + \$520 + \$1,200 = \$4,298.40
– (c) \$20,280 + \$4,298.40 = \$24,578.40
Net Job Benefits
• Job expenses are expenses paid for working for a company like dues,
uniforms, commuting costs, insurance, tools, etc.
• Duane McCoy’s job pays him \$11.25 per hour for a 40-hour week. He
estimates his fringe benefits to be 29% of his yearly wages. His yearly job
expenses are estimated to be dues, \$375; uniforms, \$580; commuting costs,
\$934. (a) What is Duane’s total annual pay? (b) What is the amount of
Duane’s fringe benefits per year? (c) What are Duane’s total yearly job
benefits? (d) What are Duane’s total yearly job expenses? (e) What are
Duane’s net yearly job benefits?
–
–
–
–
–
(a) \$11.25 x 40 = \$450 (weekly pay) x 52 (weeks per year) = \$23,400 (annual pay)
(b) \$23,400 x .29 (29%) = \$6,786 (annual fringe benefits)
(c) \$23,400 + \$6,786 = \$30,186 (total yearly job benefits)
(d) \$375 + \$580 + \$934 = \$1,889 (total yearly job expenses)
(e) \$30,186 - \$1,889 = \$28,297 (net yearly job benefits)
Percent Relationships
• ___ is 20% more than 50. 50 x .2 = 10, 10 + 50
= 60
• 40 plus 6% of itself = ___. 40 x .06 = 2.4, 2.4 +
40 = 42.4
• ___ is 10% less than 80. 80 x .1 = 8, 80 – 8 = 72
• 60 minus 40% of itself = ___. 60 x .4 = 24, 60 –
24 = 36
• 240 is ___% more than 180. 240 - 180 = 60, 60 /
180 = .3333 = 33 1/3%
• 45 is ___% less than 60. 60 – 45 = 15, 15 / 60 =
.25 = 25%
Straight Commission
• Straight commission is earned when an
employee sells a specific product for a company.
It can be calculated as a rate per item or it can
be a percentage of a sale.
– A salesperson who works a straight commission basis
sold 5 copy machines at \$4,500 each. The
commission rate was 6%. The amount of commission
was \$___.
• \$4,500 x 5 = \$22,500 total sales
• \$22,500 x 0.06 = \$1,350 amount of commission
Salary Plus Commission
• Salary plus commission is when an employee is
paid a set amount per hour plus a commission
when they sell set amounts of certain items to
stimulate sales.
– Mary Banjavic is paid a salary of \$200 a week plus
4/ % commission on weekly sales over \$7,500. Her
5
sales last week totaled \$14,100. Her total earnings
for the week were \$___.
• \$14,100 - \$7,500 = \$6,600
• \$6,600 x .008 (4/5%) = \$52.80 commission
• \$200 + \$52.80 = \$252.80 total earnings for week
• Graduated commission is when an employee is paid
commission an amount on a tier basis.
– Tony Bollini sells encyclopedias. His monthly commission is
based on the number of sets sold in a month. He earns \$45
each for the first 25 sets, \$60 each for the next 25 sets, and
\$75 each for each set over 50 sets he sells. Bollini sold 27
sets in November and 55 in December. His total
commission for the two months was \$___.
25 sets x \$45 = \$1,125
2 sets x \$60 = \$120
\$1,125 + \$120 = \$1,245 for November
25 sets x \$45 = \$1,125
\$1,245 + \$3,000 = \$4,245 for the two months
25 sets x \$60 = \$1,500
5 sets x \$75 = \$375
\$1,125 + \$1,500 + \$375 = \$3,000 for December
Finding Commission Rates
• Commission divided by sales.
• On sales of \$34,000, a sales person earns
\$2,720 commission. The rate of commission is
___%.
– \$2,720 / \$34,000 = 0.08 = 8% rate of commission
More Finding Commission
• Susan Favre sold her home through Tri-Town
Realty Company, which charges a 6% sales
commission. Her agent was Vince Pineta.
at \$175,900, but was able to sell it for only
\$165,500. If Vince’s share of the commission
was 3 ½%, he earned \$___ commission on
the sale.
– \$165,500 x 0.035 = \$5,792.50 commission
Finding Net Proceeds
• Joe Popoff, a collection agent, collected 90%
of a debt of \$5,600 which had been overdue
90 days. This collection rate was 5% more
than the average collection rate for that agent.
The agent charged 25% commission. The net
proceeds of the debt were \$___.
– \$5,600 x 0.90 = \$5,040 (amount collected)
– \$5,040 x 0.25 = \$1,260 (commission)
– \$5,040 - \$1,260 = \$3,780 (net proceeds)
``` | 2,280 | 6,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-13 | latest | en | 0.839783 |
https://techpubs.jurassic.nl/manuals/0650/developer/F77_RM/sgi_html/ch03.html | 1,695,993,115,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510516.56/warc/CC-MAIN-20230929122500-20230929152500-00542.warc.gz | 609,718,599 | 9,922 | ## Chapter 3. Expressions
This chapter contains the following subsections:
An expression performs a specified type of computation. It is composed of a sequence of operands, operators, and parentheses. The types of Fortran expressions are
• arithmetic
• character
• relational
• logical
This chapter describes formation, interpretation, and evaluation rules for each type of expression. This chapter also discusses mixed-mode expressions, which are Fortran 77 enhancements of Fortran 66.
## Arithmetic Expressions
An arithmetic expression specifies a numeric computation that yields a numeric value on evaluation. The simplest form of an arithmetic expression can be:
• an unsigned arithmetic constant
• a symbolic name of an arithmetic constant
• an arithmetic variable reference
• an arithmetic array element reference
• an arithmetic function reference
You can form more complicated arithmetic expressions from one or more operands together with arithmetic operators and parentheses.
An arithmetic element can include logical entities because logical data is treated as integer data when used in an arithmetic context. When both arithmetic and logical operands exist for a given operator, the logical operand is promoted to type INTEGER of the same byte length as the original logical length. For example, a LOGICAL*2 will be promoted to INTEGER*2 and a LOGICAL*4 will be promoted to INTEGER*4.
### Arithmetic Operators
Table 3-1 shows the arithmetic operators.
Table 3-1. Arithmetic Operators
Operator
Function
**
Exponentiation
*
Multiplication
/
Division
+
Subtraction or negation
Use the exponentiation, division, and multiplication operators between exactly two operands. You can use the addition and subtraction operators with one or two operands; in the latter case, specify the operator before the operand; for example, –TOTAL.
Do not specify two operators in succession. (Note that the exponentiation operator consists of the two characters (**), but is a single operator.) Implied operators, as in implied multiplication, are not allowed.
### Interpretation of Arithmetic Expressions
Table 3-2 interprets sample arithmetic expressions.
Table 3-2. Interpretation of Arithmetic Expressions
Operator
Use
Interpretation
**
x1 ** x2
Exponentiate x1 to the power of x2
*
x1 * x2
Multiply x1 and x2
/
x1 / x2
Divide x1 by x2
+
x1 + x2
+ x
x (identity)
x1 – x2
Subtract x1 from x2
-x
Negate x
An arithmetic expression containing two or more operators is interpreted based on a precedence relation among the arithmetic operators. This precedence, from highest to lowest, is
• ( )
• **
• * and /
• + and –
Use parentheses to override the order of precedence.
The following is an example of an arithmetic expression:
```A/B-C**D ```
The operators are executed in the following sequence:
1. C**D is evaluated first.
2. A/B is evaluated next.
3. The result of C**D is subtracted from the result of A/B to give the final result.
A unary operator (–) can follow another operator. Specifying the unary operator after the exponentiation operator produces a variation on the standard order of operations. The unary operator is evaluated first in that case, resulting in exponentiation taking a lower precedence in the expression.
For example, the following expression
```A ** - B * C ```
is interpreted as
```A ** ( - B * C ) ```
### Arithmetic Operands
Arithmetic operands must specify values with integer, real, double-precision, complex, or double-complex data types. You can combine specific operands in an arithmetic expression. The arithmetic operands, in order of increasing complexity, are
• primary
• factor
• term
• arithmetic expression
A primary is the basic component in an arithmetic expression. The forms of a primary are
• an unsigned arithmetic constant
• a symbolic name of an arithmetic constant
• an arithmetic variable reference
• an arithmetic array element reference
• an arithmetic function reference
• an arithmetic expression enclosed in parentheses
A factor consists of one or more primaries separated by the exponentiation operator. The forms of a factor are
• primary
• primary ** factor
Factors with more than one exponentiation operator are interpreted from right to left. For example, I**J**K is interpreted as I**(J**K), and I**J**K**L is interpreted as I**(J**(K**L)).
The term incorporates the multiplicative operators into arithmetic expressions. Its forms are
• factor
• term/factor
• term * factor
The above definition indicates that factors are combined from left to right in a term containing two or more multiplication or division operators.
Finally, at the highest level of the hierarchy, are the arithmetic expressions. The forms of an arithmetic expression are
• term
• + term
• – term
• arithmetic expression + term
• arithmetic expression – term
An arithmetic expression consists of one or more terms separated by an addition operator or a subtraction operator. The terms are combined from left to right. For example, A+B-C has the same interpretation as the expression (A+B)-C. Expressions such as A*-B and A+-B are not allowed. The correct forms are A*(–B) and A+(-B).
An arithmetic expression can begin with a plus or minus sign.
### Arithmetic Constant Expressions
An arithmetic constant expression is an arithmetic expression containing no variables. Therefore, each primary in an arithmetic constant expression must be one of the following:
• arithmetic constant
• symbolic name of an arithmetic constant
• arithmetic constant expression enclosed in parentheses
In an arithmetic constant expression, do not specify the exponentiation operator unless the exponent is of type integer. Variable, array element, and function references are not allowed. Examples of integer constant expressions are
``` 7 –7 –7+5 3**2 x+3 (where x is the symbolic name of a constant) ```
### Integer Constant Expressions
An integer constant expression is an arithmetic constant expression containing only integers. It can contain constants or symbolic names of constants, provided they are of type integer. As with all constant expressions, no variables, array elements, or function references are allowed.
### Evaluating Arithmetic Expressions
The data type of an expression is determined by the data types of the operands and functions that are referenced. Thus, integer expressions, real expressions, double-precision expressions, complex expressions, and double expressions have values of type integer, real, double-precision, complex, and double-complex, respectively.
#### Single-Mode Expressions
Single-mode expressions are arithmetic expressions in which all operands have the same data type. The data type of the value of a single-mode expression is thus the same as the data type of the operands. When the addition operator or the subtraction operator is used with a single operand, the data type of the resulting expression is the same as the data type of the operand.
#### Mixed-Mode Expressions
Mixed-mode expressions contain operands with two or more data types. The data type of the result of a mixed-mode expression depends on the rank associated with each data type, as shown in Table 3-3.
Table 3-3. Data Type Ranks
Data Type
Rank
INTEGER*1
1 (lowest)
INTEGER*2
2
INTEGER*4
3
REAL*4
4
REAL*8 (double precision)
5
COMPLEX*8
6
COMPLEX*16
7 (highest)
Except for exponentiation (discussed below), the result of a mixed-mode expression is assigned the data type of the highest-ranked element in the expression. The lower-ranked operand is converted to the type of the higher-ranked operand so that the operation is performed on values with equivalent data types. For example, an operation on an integer operand and a real operand produces a result of type real.
Operations that combine REAL*8 (DOUBLE PRECISION) and COMPLEX*8 (COMPLEX) are not allowed. The REAL*8 operand must be explicitly converted (for example, by using the SNGL intrinsic function).
### Exponentiation
Exponentiation is an exception to the above rules for mixed-mode expressions. When raising a value to an integer power, the integer is not converted. The result is assigned the type of the left operand.
When a complex value is raised to a complex power, the value of the expression is defined as follows:
```xy = EXP (y * LOG(x)) ```
### Integer Division
One operand of type integer can be divided by another operand of type integer. The result of an integer division operation is a value of type integer, referred to as an integer quotient. The integer quotient is obtained as follows:
• If the magnitude of the mathematical quotient is less than one, then the integer quotient is zero. For example, the value of the expression (18/30) is zero.
• If the magnitude of the mathematical quotient is greater than or equal to one, then the integer quotient is the largest integer that does not exceed the magnitude of the mathematical quotient and whose sign is the same as that of the mathematical quotient. For example, the value of the expression (–9/2) is –4.
## Character Expressions
A character expression yields a character string value on evaluation. The simplest form of a character expression can be one of these types of characters:
• constant
• variable reference
• array element reference
• substring reference
• function reference
Construct complicated character expressions from one or more operands together with the concatenate operator and parentheses.
### Concatenate Operator
The concatenate operator (//) is the only character operator defined in Fortran. A character expression formed from the concatenation of two character operands x1 and x2 is specified as
```x1 // x2 ```
The result of this operation is a character string with a value of x1 extended on the right with the value of x2. The length of the result is the sum of the lengths of the character operands. For example,
```'HEL' // 'LO2' ```
The result of the above expression is the string HELLO2 of length six.
### Character Operands
A character operand must identify a value of type character and must be a character expression. The basic component in a character expression is the character primary. The forms of a character primary are
• character constant
• symbolic name of a character constant
• character variable reference
• character array element reference
• character substring reference
• character function reference
• character expression enclosed in parentheses
A character expression consists of one or more character primaries separated by the concatenation operator. Its forms are
• character primary
• character expression // character primary
In a character expression containing two or more concatenation operators, the primaries are combined from left to right. Thus, the character expression
```'A' // 'BCD' // 'EF' ```
is interpreted the same as
```('A' // 'BCD') // 'EF' ```
The value of the above character expression is ABCDEF.
Except in a character assignment statement, concatenation of an operand with an asterisk (*) as its length specification is not allowed unless the operand is the symbolic name of a constant.
### Character Constant Expressions
A character constant expression is made up of operands that cannot vary. Each primary in a character constant expression must be a
• character constant
• symbolic name of a character constant
• character constant expression enclosed in parentheses
A character constant expression cannot contain variable, array element, substring, or function references.
## Relational Expressions
A relational expression yields a logical value of either .TRUE. or .FALSE. on evaluation and comparison of two arithmetic expressions or two character expressions. A relational expression can appear only within a logical expression. Refer to “Logical Expressions” for details about logical expressions.
### Relational Operators
Table 3-4 lists the Fortran relational operators.
Table 3-4. Fortran Relational Operators
Relational Operator
Meaning
.EQ.
Equal to
.NE.
Not equal to
.GT.
Greater than
.GE.
Greater than or equal to
.LT.
Less than
.LE.
Less than or equal to
Arithmetic and character operators are evaluated before relational operators.
### Relational Operands
The operands of a relational operator can be arithmetic or character expressions. The relational expression requires exactly two operands and is written in the following form:
e1 relop e2
where
e1 and e2 are arithmetic or character expressions. relop is the relational operator.
Note: Both e1 and e2 must be the same type of expression, either arithmetic or character.
### Evaluating Relational Expressions
The result of a relational expression is of type logical, with a value of .TRUE. or .FALSE.. The manner in which the expression is evaluated depends on the data type of the operands.
### Arithmetic Relational Expressions
In an arithmetic relational expression, e1 and e2 must each be an integer, real, double precision, complex, or double complex expression. relop must be a relational operator.
The following are examples of arithmetic relational expressions:
```(a + b) .EQ. (c + 1) HOURS .LE. 40 ```
You can use complex type operands only when specifying either the .EQ. or .NE. relational operator.
An arithmetic relational expression has the logical value .TRUE. only if the values of the operands satisfy the relation specified by the operator. Otherwise, the value is .FALSE..
If the two arithmetic expressions e1 and e2 differ in type, the expression is evaluated as follows:
```((e1) - (e2)) relop 0 ```
where the value 0 (zero) is of the same type as the expression ((e1)- (e2)) and the type conversion rules apply to the expression. Do not compare a double precision value with a complex value.
### Character Relational Expressions
In a character relational expression, e1 and e2 are character expressions and relop is a relational operator.
The following is an example of a character relational expression:
```NAME .EQ. 'HOMER' ```
A character relational expression has the logical value .TRUE. only if the values of the operands satisfy the relation specified by the operator. Otherwise, the value is .FALSE.. The result of a character relational expression depends on the collating sequence as follows:
• If e1 and e2 are single characters, their relationship in the collating sequence determines the value of the operator. e1 is less than or greater than e2 if e1 is before or after e2, respectively, in the collating sequence.
• If either e1 or e2 are character strings with lengths greater than 1, corresponding individual characters are compared from left to right until a relationship other than .EQ. can be determined.
• If the operands are of unequal length, the shorter operand is extended on the right with blanks to the length of the longer operand for the comparison.
• If no other relationship can be determined after the strings are exhausted, the strings are equal.
The collating sequence depends partially on the processor; however, equality tests .EQ. and .NE. do not depend on the processor collating sequence and can be used on any processor.
## Logical Expressions
A logical expression specifies a logical computation that yields a logical value. The simplest form of a logical expression is one of the following:
• logical constant
• logical variable reference
• logical array element reference
• logical function reference
• relational expression
Construct complicated logical expressions from one or more logical operands together with logical operators and parentheses.
### Logical Operators
Table 3-5 defines the Fortran logical operators.
Table 3-5. Logical Operators
Logical Operator
Meaning
.NOT.
Logical negation
.AND.
Logical conjunt
.OR.
Logical disjunct
.EQV.
Logical equivalence
.NEQV.
Logical exclusive or
.XOR.
Same as .NEQV.
All logical operators require at least two operands, except the logical negation operator .NOT. , which requires only one.
A logical expression containing two or more logical operators is evaluated based on a precedence relation between the logical operators. This precedence, from highest to lowest, is
• .NOT.
• .AND.
• .OR.
• .EQV. and .NEQV.
• .XOR.
For example, in the following expression
```W .NEQV. X .OR. Y .AND. Z ```
the operators are executed in the following sequence:
1. Y .AND. Z is evaluated first (A represents the result).
2. X .OR. A is evaluated second (B represents the result).
3. W .NEQV. B is evaluated to produce the final result.
You can use parentheses to override the precedence of the operators.
### Logical Operands
Logical operands specify values with a logical data type. The forms of a logical operands are
• logical primary
• logical factor
• logical term
• logical disjunct
• logical expression
#### Logical Primary
The logical primary is the basic component of a logical expression. The forms of a logical primary are
• logical constant
• symbolic name of a logical constant
• integer or logical variable reference
• logical array element reference
• integer or logical function reference
• relational expression
• integer or logical expression in parentheses
When an integer appears as an operand to a logical operator, the other operand is promoted to type integer if necessary and the operation is performed on a bit-by-bit basis producing an integer result. Whenever an arithmetic datum appears in a logical expression, the result of that expression will be of type integer because of type promotion rules. If necessary, the result can be converted back to LOGICAL.
Do not specify two logical operators consecutively and do not use implied logical operators.
#### Logical Factor
The logical factor uses the logical negation operator .NOT. to reverse the logical value to which it is applied. For example, applying .NOT. to a false relational expression makes the expression true. Therefore, if UP is true, .NOT. UP is false. The logical factor has the following forms:
• logical primary
• .NOT. logical primary
#### Logical Term
The logical term uses the logical conjunct operator .AND. to combine logical factors. It takes the forms
• Logical factor
• Logical term .AND. logical factor
In evaluating a logical term with two or more .AND. operators, the logical factors are combined from left to right. For example, X .AND. Y .AND. Z has the same interpretation as (X .AND. Y) .AND. Z.
#### Logical Disjunct
The logical disjunct is a sequence of logical terms separated by the .OR. operator and has the following two forms:
• Logical term
• Logical disjunct .OR. logical term
In an expression containing two or more .OR. operators, the logical terms are combined from left to right in succession. For example, the expression X .OR. Y .OR. Z has the same interpretation as (X .OR. Y) .OR. Z.
#### Logical Expression
At the highest level of complexity is the logical expression. A logical expression is a sequence of logical disjuncts separated by the .EQV., .NEQV., or .XOR. operators. Its forms are
• logical disjunct
• logical expression .EQV. logical disjunct
• logical expression .NEQV. logical disjunct
• logical expression .XOR. logical disjunct
The logical disjuncts are combined from left to right when a logical expression contains two or more .EQV., .NEVQ., or .XOR. operators.
A logical constant expression is a logical expression in which each primary is a logical constant, the symbolic name of a logical constant, a relational expression in which each primary is a constant, or a logical constant expression enclosed in parentheses. A logical constant expression can contain arithmetic and character constant expressions but not variables, array elements, or function references.
### Interpretation of Logical Expressions
In general, logical expressions containing two or more logical operators are executed according to the hierarchy of operators described previously, unless the order has been overridden by the use of parentheses. Table 3-6 defines the form and interpretation of the logical expressions.
Table 3-6. Logical Expressions
IFA=
B=
THEN
.NOT.B
A.AND.B
A.OR.B
A.EQV.B
A.XOR.B
A.NEQV.B
F
F
T
F
F
T
F
F
T
F
F
T
F
T
T
F
F
T
F
T
T
T
T
T
T
F
## Evaluating Expressions in General
Several rules are applied to the general evaluation of Fortran expressions. This section covers the priority of the different Fortran operators, the use of parentheses in specifying the order of evaluation, and the rules for combining operators with operands.
Note: Any variable, array element, function, or character substring in an expression must be defined with a value of the correct type at the time it is referenced.
### Precedence of Operators
Certain Fortran operators have precedence over others when combined in an expression. The previous sections have listed the precedence among the arithmetic, logical, and expression operators. No precedence exists between the relational operators and the single character operator (//). On the highest level, the precedence among the types of expression operators, from highest to lowest, is
• arithmetic
• character
• relational
• logical
### Integrity of Parentheses and Interpretation Rules
Use parentheses to specify the order in which operators are evaluated within an expression. Expressions within parentheses are treated as an entity.
In an expression containing more than one operation, the processor first evaluates any expressions within parentheses. Subexpressions within parentheses are evaluated beginning with the innermost subexpression and proceeding sequentially to the outermost. The processor then scans the expression from left to right and performs the operations according to the operator precedence described previously. | 4,582 | 21,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-40 | latest | en | 0.846763 |
https://math.stackexchange.com/questions/1921033/how-do-i-plot-a-field-of-view-in-2d-space-and-find-if-a-certain-point-lies-wit | 1,558,754,853,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257847.56/warc/CC-MAIN-20190525024710-20190525050710-00467.warc.gz | 566,787,162 | 35,780 | # How do I plot a 'field of view' in 2D space and find if a certain point lies within the space?
I am trying to model a robot arm (in 2D) that's supposed to have a camera at the end that moves with the arm. Given the field of view of the camera, I need to find whether a certain target is visible in this field of view as the endpoint of the arm starts moving. The horizontal of the camera is aligned with the last link of the robot arm, so the 'triangle' moves up and down, forwards and backwards etc. While this intuitively makes sense, I'm having trouble coming up with a formula for this. Also, as it's a camera field of view, the rays are technically infinitely long.
The data I have are: (example image below)
1. Cartesian coordinates of the last two 'points' of the robot arm (x3, y3) and (x4, y4)
2. Theta angle of the field of view
3. Coordinates of the point of interest (this never changes)
Any suggestions?
Maybe something simple like this would work? Assume you have been able to compute your coordinates $P_3 = (x_3, y_3)$ and $P_4 = (x_4, y_4)$ and you know the position of the point you want to see with the camera $Q = (a, b)$. Form the vectors $\overrightarrow{P_3P_4} = (x_4 - x_3, \, y_4 - y_3)$ and $\overrightarrow{P_4Q} = (a - x_4, \, b - y_4)$. The dot product between the two vectors has the following meaning $$\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big) = \|\overrightarrow{P_3P_4}\| \, \|\overrightarrow{P_4Q} \| \cos{\theta} = \|P_4 - P_3\|\, \|Q - P_4\| \cos{\theta}$$ where $\theta$ is the angle between the oriented line $P_4Q$ and the oriented horizontal of the camera, which is defined by the line $P_3P_4$. If $\theta \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$ then your object is visible. In terms of cosine of the angle, this translates into the condition $\cos{\theta} \in \left(\frac{\sqrt{3}}{2}, 1\right)$. If $\cos{\theta}$ is not in the interval $\left(\frac{\sqrt{3}}{2}, 1\right)$, then your point is not visible. Form the above formula \begin{align}\cos{\theta} &= \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } = \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } \end{align} so if we plug in the coordinates we obtain the expression
\begin{align} \cos{\theta} = f(x_3,y_3,x_4,y_4,a,b) = \frac{ (x_4 - x_3)(a- x_4) + (y_4 - y_3)(b- y_4)}{\sqrt{ (x_4 - x_3)^2 + (y_4 - y_3)^2} \,\,\, \sqrt{ (a- x_4)^2 + (b- y_4)^2}} \end{align}
If $\,\, f(x_3,y_3,x_4,y_4,a,b) \,\in \, \left(\frac{\sqrt{3}}{2}, 1\right) \,\,$ then the point with coordinates $(a,b)$ is visible. Otherwise, it is not.
Make sure I haven't made a mistake in the condition and the proper orientation of the angle. I have changed them three times already :)
Comment: Regarding the bounds of $\cos{\theta}$, I will try to explain this as follows.
The camera horizontal is the directed line $P_3P_4$, directed from $P_3$ to $P_4$. Since the directed line $P_4Q$, oriented from $P_4$ to $Q$, determines of the angle of the point $Q$ with respect to the camera horizontal $P_3P_4$, by definition the visibility occurs when the angle $\theta$ between $P_3P_4$ and $P_4Q$ (measured counterclockwise from $P_4Q$ to $P_3P_4$) changes from $-30^{\circ}$ to $30^{\circ}$. The cosine $\cos{\theta}$ is related to the orthogonal projection of $P_4Q$ onto the directed horizontal $P_3P_4$. Let us rotate $P_4Q$ around $P_4$ and see how $\theta$ changes. At first when $\theta = -30^{\circ}$ then $\cos{(-30^{\circ})} = \sqrt{3}/2$. As the direction $P_4Q$ gets more and more aligned with $P_3P_4$, the angle $\theta$ grows from $-30^{\circ}$ to $0$ and thus the cosine grows from $\sqrt{3}/2$ to $1$. After the alignment, when $\theta=0$ and $\cos{0} = 1$, the cosine $\cos{\theta}$ starts to decrease (while the angle $\theta$ keeps growing from $0$ to $30^{\circ}$) until the direction $P_4Q$ reaches $30^{\circ}$ with $P_3P_4$ and the cosine becomes $\cos{\theta} = \sqrt{3}/2$ again.
• Thanks for the detailed explanation! That approach worked. – HighVoltage Sep 10 '16 at 6:12
Here’s an approach that anticipates doing more with the camera view later.
We’ll be working in two dimensions, but the same technique applies in three. We will assume that the camera view is a perspective projection as illustrated here:
This will necessitate working in homogeneous coordinates.
The first thing to do is to switch to the camera’s coordinate system. The origin of this coordinate system is at the camera’s position and by convention, the camera sights along the negative $y'$ direction (negative $z'$ in 3-d). The world to camera transformation is thus a translation to the camera’s position followed by a rotation. The matrix of this translation is easy to produce. It’s simply $$T=\pmatrix{1&0&-x_4\\0&1&-y_4\\0&0&1}.$$ For the angle $\phi$ that the camera’s line of sight makes with the world $x$-axis, we have $$\cos\phi = {x_4-x_3\over\|P_4-P_3\|}\\\sin\phi = {y_4-y_3\over\|P_4-P_3\|}.$$ To get aligned with the camera’s line of sight, we start by rotating clockwise through this angle, but we also need to rotate clockwise by an additional 90 degrees to get it to point down the camera’s negative $y'$-axis. Putting those two rotations together produces the rotation matrix $$R=\pmatrix{-\sin\phi&\cos\phi&0\\-\cos\phi&-\sin\phi&0\\0&0&1},$$ with $\cos\phi$ and $\sin\phi$ as above. Combining these two matrices, we have $$RT = \pmatrix{-\sin\phi&\cos\phi&x_4\sin\phi-y_4\cos\phi\\-\cos\phi&-\sin\phi&x_4\cos\phi+y_4\sin\phi\\0&0&1},$$ i.e., \begin{align}x'&=-(x-x_4)\sin\phi+(y-y_4)\cos\phi\\y'&=-(x-x_4)\cos\phi-(y-y_4)\sin\phi.\end{align}
The line labeled “i” in the above diagram is the image plane, which is perpendicular to the camera’s line of sight and at a distance $f$ from the camera (the focal distance). Note that, since the camera is looking down the negative $y'$-axis, $f<0$. The perspective projection $M$ maps a point in the $x$-$y$ plane onto the intersection of the image plane with the ray emanating from the camera and passing through the point. If we take $f=-1$, then the bounds of the visible region in the image plane are $\pm\tan\theta$, so if the $x'$-coordinate of the projection of a point is in this range, then it’s visible.
In the camera coordinate system, a projection matrix is very simple: $$P=\pmatrix{1&0&0\\0&1&0\\0&\frac1f&0}.$$ Putting this all together, given a point $Q=(x,y)$, we compute $$M(Q)=PRT\pmatrix{x\\y\\1}$$ and recover the projected $x'$-coordinate by dividing the first component of the resulting vector by the third. We can save ourselves a bit of work, though, by taking advantage of $P$’s simple form. Note that $$\pmatrix{1&0&0\\0&1&0\\0&-1&0}\pmatrix{x'\\y'\\1}=\pmatrix{x'\\y'\\-y'},$$ so we really only need to transform the target point into camera coordinates, after which we can just check that $-\tan\theta\le-x'/y'\le\tan\theta$. We might have $y'=0$, however, so let’s rewrite this as $|x'|\le|y'|\tan\theta$ to avoid dividing by zero.
You might object that the projection also maps points behind the camera onto the image plane, but that’s easily dealt with: check the sign of the camera-relative $y'$-coordinate. If it’s positive, the point is behind the camera, so there’s no need to compute its projection. You can eliminate the $y'=0$ case at the same time. If this seems backwards to you, you can always have the camera point in the positive $y'$ direction instead so that visible points have a positive $y'$-coordinate, but you’ll have to modify $R$ and $P$ accordingly.
As I mentioned above, the same approach works in 3-d, except that you’ll be working with $4\times4$ matrices. The rotation matrix will be a bit more complicated, but the translation will still be straightforward. Taking $f=-1$ again, the projection will result in $(x',y',z',-z')$. Assuming that the field of view is a circular cone, the test for visibility will then be $$x'^2+y'^2\le z'^2\tan^2\theta.$$
Postscript: This is, of course, overkill when the field of view is a right circular cone, whether in two dimensions or three. Checking that $(Q-P_4)\cdot(P_4-P_3)\ge\|Q-P_4\|\,\|P_4-P_3\|\cos\theta$ is much simpler and more efficient. However, the procedure that I’ve outlined here applies generally to any size and shape aperture, which becomes much more interesting when you move to three dimensions. | 2,593 | 8,421 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2019-22 | latest | en | 0.861014 |
https://stackoverflow.com/questions/30269778/how-to-find-all-equals-paths-in-degenerate-tree-which-start-on-specific-vertex | 1,696,275,453,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511002.91/warc/CC-MAIN-20231002164819-20231002194819-00293.warc.gz | 573,803,042 | 43,304 | How to find all equals paths in degenerate tree, which start on specific vertex?
I have some `degenerate tree` (it looks like as array or doubly linked list). For example, it is this tree:
Each edge has some weight. I want to find all equal paths, which starts in each vertex.
In other words, I want to get all tuples (v1, v, v2) where v1 and v2 are an arbitrary ancestor and descendant such that `c(v1, v) = c(v, v2)`.
Let edges have the following weights (it is just example):
`a-b = 3`
`b-c = 1`
`c-d = 1`
`d-e = 1`
Then:
1. The vertex `A` does not have any equal path (there is no vertex from left side).
2. The vertex `B` has one equal pair. The path `B-A` equals to the path `B-E` `(3 == 3)`.
3. The vertex `C` has one equal pair. The path `B-C` equals to the path `C-D` `(1 == 1)`.
4. The vertex `D` has one equal pair. The path `C-D` equals to the path `D-E` `(1 == 1)`.
5. The vertex `E` does not have any equal path (there is no vertex from right side).
I implement simple algorithm, which works in `O(n^2)`. But it is too slow for me.
• If `n` is the number of vertices, then it's not possible to make it faster than `O(n^2)`, because in worst case the number of your edges is `n^2`. May 15, 2015 at 23:37
• @FalconUA, your point does make sense. It seems, I looking for a way to decrease constant in `O(n^2)`. I choose some vertex. Then I create two `set`. Then I fill these sets with partial sums, while iterating from this vertex to start of tree and to finish of tree. Then I find `set intersection` and get number of paths from this vertex. Then I repeat algorithm for all other vertices. May 15, 2015 at 23:54
• Are you restricting your problem to the type you have presented, or are you looking for a general solution? The general solution would require you evaluate every possible path in a graph against every other possible path, and in a graph with cycles this could go off to infinity. May 16, 2015 at 0:01
• @AndyG, Actually, I just want to find number of equal paths from each vertex in tree. May 16, 2015 at 0:08
• @AndyG, my graph does not have any cycles. It is degenerate tree (as in example). May 16, 2015 at 0:09
It seems, I looking for a way to decrease constant in O(n^2). I choose some vertex. Then I create two set. Then I fill these sets with partial sums, while iterating from this vertex to start of tree and to finish of tree. Then I find set intersection and get number of paths from this vertex. Then I repeat algorithm for all other vertices.
There is a simpler and, I think, faster `O(n^2)` approach, based on the so called two pointers method.
For each vertix `v` go at the same time into two possible directions. Have one "pointer" to a vertex (`vl`) moving in one direction and another (`vr`) into another direction, and try to keep the distance from `v` to `vl` as close to the distance from `v` to `vr` as possible. Each time these distances become equal, you have equal paths.
``````for v in vertices
vl = prev(v)
vr = next(v)
while (vl is still inside the tree)
and (vr is still inside the tree)
if dist(v,vl) < dist(v,vr)
vl = prev(vl)
else if dist(v,vr) < dist(v,vl)
vr = next(vr)
else // dist(v,vr) == dist(v,vl)
ans = ans + 1
vl = prev(vl)
vr = next(vr)
``````
(By precalculating the prefix sums, you can find `dist` in O(1).)
It's easy to see that no equal pair will be missed provided that you do not have zero-length edges.
Regarding a faster solution, if you want to list all pairs, then you can't do it faster, because the number of pairs will be O(n^2) in the worst case. But if you need only the amount of these pairs, there might exist faster algorithms.
UPD: I came up with another algorithm for calculating the amount, which might be faster in case your edges are rather short. If you denote the total length of your chain (sum of all edges weight) as `L`, then the algorithm runs in `O(L log L)`. However, it is much more advanced conceptually and more advanced in coding too.
Firstly some theoretical reasoning. Consider some vertex `v`. Let us have two arrays, `a` and `b`, not the C-style zero-indexed arrays, but arrays with indexation from `-L` to `L`.
Let us define
• for `i>0`, `a[i]=1` iff to the right of `v` on the distance exactly `i` there is a vertex, otherwise `a[i]=0`
• for `i=0`, `a[i]≡a[0]=1`
• for `i<0`, `a[i]=1` iff to the left of `v` on the distance exactly `-i` there is a vertex, otherwise `a[i]=0`
A simple understanding of this array is as follows. Stretch your graph and lay it along the coordinate axis so that each edge has the length equal to its weight, and that vertex `v` lies in the origin. Then `a[i]=1` iff there is a vertex at coordinate `i`.
For your example and for vertex "b" chosen as `v`:
`````` a--------b--c--d--e
--|--|--|--|--|--|--|--|--|-->
-4 -3 -2 -1 0 1 2 3 4
a: ... 0 1 0 0 1 1 1 1 0 ...
``````
For another array, array `b`, we define the values in a symmetrical way with respect to origin, as if we have inverted the direction of the axis:
• for `i>0`, `b[i]=1` iff to the left of `v` on the distance exactly `i` there is a vertex, otherwise `b[i]=0`
• for `i=0`, `b[i]≡b[0]=1`
• for `i<0`, `b[i]=1` iff to the right of `v` on the distance exactly `-i` there is a vertex, otherwise `b[i]=0`
Now consider a third array `c` such that `c[i]=a[i]*b[i]`, asterisk here stays for ordinary multiplication. Obviously `c[i]=1` iff the path of length `abs(i)` to the left ends in a vertex, and the path of length `abs(i)` to the right ends in a vertex. So for `i>0` each position in `c` that has `c[i]=1` corresponds to the path you need. There are also negative positions (`c[i]=1` with `i<0`), which just reflect the positive positions, and one more position where `c[i]=1`, namely position `i=0`.
Calculate the sum of all elements in `c`. This sum will be `sum(c)=2P+1`, where `P` is the total number of paths which you need with `v` being its center. So if you know `sum(c)`, you can easily determine `P`.
Let us now consider more closely arrays `a` and `b` and how do they change when we change the vertex `v`. Let us denote `v0` the leftmost vertex (the root of your tree) and `a0` and `b0` the corresponding `a` and `b` arrays for that vertex.
For arbitrary vertex `v` denote `d=dist(v0,v)`. Then it is easy to see that for vertex `v` the arrays `a` and `b` are just arrays `a0` and `b0` shifted by `d`:
``````a[i]=a0[i+d]
b[i]=b0[i-d]
``````
It is obvious if you remember the picture with the tree stretched along a coordinate axis.
Now let us consider one more array, `S` (one array for all vertices), and for each vertex `v` let us put the value of `sum(c)` into the `S[d]` element (`d` and `c` depend on `v`).
More precisely, let us define array `S` so that for each `d`
``````S[d] = sum_over_i(a0[i+d]*b0[i-d])
``````
Once we know the `S` array, we can iterate over vertices and for each vertex `v` obtain its `sum(c)` simply as `S[d]` with `d=dist(v,v0)`, because for each vertex `v` we have `sum(c)=sum(a0[i+d]*b0[i-d])`.
But the formula for `S` is very simple: `S` is just the convolution of the `a0` and `b0` sequences. (The formula does not exactly follow the definition, but is easy to modify to the exact definition form.)
So what we now need is given `a0` and `b0` (which we can calculate in `O(L)` time and space), calculate the `S` array. After this, we can iterate over `S` array and simply extract the numbers of paths from `S[d]=2P+1`.
Direct application of the formula above is `O(L^2)`. However, the convolution of two sequences can be calculated in `O(L log L)` by applying the Fast Fourier transform algorithm. Moreover, you can apply a similar Number theoretic transform (don't know whether there is a better link) to work with integers only and avoid precision problems.
So the general outline of the algorithm becomes
``````calculate a0 and b0 // O(L)
calculate S = corrected_convolution(a0, b0) // O(L log L)
v0 = leftmost vertex (root)
for v in vertices:
d = dist(v0, v)
ans = ans + (S[d]-1)/2
``````
(I call it `corrected_convolution` because `S` is not exactly a convolution, but a very similar object for which a similar algorithm can be applied. Moreover, you can even define `S'[2*d]=S[d]=sum(a0[i+d]*b0[i-d])=sum(a0[i]*b0[i-2*d])`, and then `S'` is the convolution proper.) | 2,406 | 8,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-40 | latest | en | 0.932082 |
http://docplayer.net/1284890-Diocese-of-nashville-catholic-schools-office-30-white-bridge-road-nashville-tn-37205-curriculum-overview-prek-grade-8.html | 1,537,593,633,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158045.57/warc/CC-MAIN-20180922044853-20180922065253-00167.warc.gz | 64,117,114 | 31,604 | # Diocese of Nashville Catholic Schools Office 30 White Bridge Road, Nashville, TN Curriculum Overview PreK Grade 8
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1 Catholic Schools Office 30 White Bridge Road, Nashville, TN Curriculum Overview PreK Grade 8 Click on titles below to navigate to pages. Algebra Art Band Health Language Library Math Music PE Religion Science: PreK Fifth Science: Sixth Eighth Social Studies Spanish Technology
2 Curriculum Guide: Algebra Sequence Order Guide ID Description 1 ALG-56 Order of operations. 2 ALG-55 Translate verbal phrases and write algebraic expressions. 3 ALG-72 Translate verbal sentences into algebraic equations. Properties: additive identity, additive inverse, associative, commutative, distributive, multiplicative identity, multiplicative inverse, substitution. 4 ALG-52 5 ALG-10 Perform operations on algebraic expressions and informally justify the procedures chosen. 6 ALG-23 Articulate and apply algebraic properties in symbolic manipulation. 7 ALG-1 Demonstrate an understanding of the subsets, elements, properties, and operations of the real number system. 8 ALG-3 Articulate, model, and apply the concept of inverse (e.g., opposites, reciprocals, and powers and roots). 9 ALG-4 Describe, model, and apply inverse operations. 10 ALG-6 Connect physical, graphical, verbal, and symbolic representations of absolute value. 11 ALG-20 Apply the concept of variable in simplifying algebraic expressions, solving equations, and solving inequalities. 12 ALG-16 Communicate the meaning of variables in algebraic expressions, equations, and inequalities. 13 ALG-14 Use estimation to make predictions and test reasonableness of results. 14 ALG-39 Solving equations with variables on both sides. 15 ALG-57 Solve multi-step equations. 16 ALG-61 Add and subtract polynomials. 17 ALG-53 Simplify polynomials by combining like terms. 18 ALG-59 Multiply polynomials. 19 ALG-74 Solve formulas for specified variables. 20 ALG-60 Solve word problems involving distance, wind/water, coins, mixture and consecutive integers. Demonstrate an understanding of unit rates and other derived and indirect measurements (e.g., velocity, miles per hour, revolutions per minute, cost per unit). 21 ALG ALG-5 Apply number theory concepts (e.g. primes, factors, divisibility and multiples in mathematical problem solving). 23 ALG-37 Simplify monomials. 24 ALG-65 Divide monomials and polynomials. Factor polynomials using 4 types: common factor, difference between two perfect squares, trinomials that factors as a binomial square and regrouping. 25 ALG ALG-63 Solve equations by factoring. 27 ALG-42 Solve rational equations. 28 ALG-31 Apply geometric properties, formulas, and relationships to solve real-world problems. 29 ALG-35 Analyze mathematical patterns related to algebra and geometry in real-world problem solving. 30 ALG-11 Use concepts of length, area, and volume to estimate and solve real-world problems. 31 ALG-41 Add and subtract rational expressions. 32 ALG-38 Multiplication and division of rational expressions. 33 ALG-70 Add, subtract, multiply and divide rational numbers. 34 ALG-9 Communicate the concepts and strategies being used in estimation, measurement, and computation. Solve problems in number theory, geometry, probability and statistics, and measurement and estimation using algebraic thinking and symbolism. 35 ALG ALG-66 Simplify zero exponents. 37 ALG-54 Simplify negative exponents. 38 ALG-75 Use a coordinate plane to graph ordered pairs. 39 ALG-92 Find x- and y- intercepts of a linear equation. : Updated by math teachers spring Page 1 of 2
3 Curriculum Guide: Algebra Sequence Order Guide ID Description 40 ALG-76 Graph linear equations including horizontal lines and vertical lines. 41 ALG-77 Find the slope of a line using two points. 42 ALG-81 Transform a linear equation into standard form. 43 ALG-79 Use slope and any point on the line to write an equation. 44 ALG-80 Write equation of a line given two points on the line. 45 ALG-43 Write standard and slope-intercept form of linear equations. 46 ALG-44 Write linear equations concerning parallel and perpendicular lines. 47 ALG-17 Identify and represent a variety of functions. 48 ALG-90 Represent functions as ordered pairs. 49 ALG-91 Determine if an equation is a function. 50 ALG-24 Identify relationships which can and which cannot be represented by a function. 51 ALG-25 Describe the domain and range of functions and articulate restrictions imposed either by the operations or by the real-life situations which the functions represent. 52 ALG-36 Use algebraic thinking to generalize a pattern by expressing the pattern in functional notation. 53 ALG-22 Model real-world phenomena using functions and graphs. 54 ALG-26 Describe the transformation of the graph that occurs when coefficients and/or constants of the corresponding linear equations are changed. 55 ALG-28 Make predictions for a linear data set using a line of best fit. 56 ALG-7 Use real numbers to represent real-world applications (e.g. slope, rate of change, and proportionality). 57 ALG-18 Apply the concept of rate of change to slope. 58 ALG-19 Interpret results of algebraic procedures. 59 ALG-94 Solve systems of linear equations using graphing, substitution, and linear combination methods and apply to word problems. 60 ALG-21 Interpret graphs that depict real-world phenomena. 61 ALG-78 Solve and graph the solution of absolute value equations. 62 ALG-83 Solve and graph compound inequalities. 63 ALG-82 Solve and graph linear inequalities in one and two variables. 64 ALG-2 Demonstrate an understanding of the relative size of rational and irrational numbers. Select and apply an appropriate method (i.e. mental arithmetic, paper and pencil, or technology) for computing with real numbers, and 65 ALG-8 evaluate the reasonableness of results. 66 ALG-45 Simplify radicals. 67 ALG-46 Add and multiply radicals. 68 ALG-47 Divide radicals. 69 ALG-33 Apply right triangle relationships including the Pythagorean Theorem and the distance formula. 70 ALG-48 Solve radical equations. 71 ALG-49 Solve quadratic equations using the quadratic formula and factoring. 72 ALG-50 Solve quadratic equations by completing the binomial square. 73 ALG-68 Solve quadratic word problems. 74 ALG-84 Graft linear equations using point-shape, y-intercept and standard form. : Updated by math teachers spring Page 2 of 2
4 Curriculum Guide: Art PreK-Grade 8/ Guide ID Description Gr PreK Gr K Gr 1 Gr 2 Gr 3 Gr 4 Gr 5 Gr 6 Gr 7 Gr 8 Artistic Skills 001 Participates in activities that foster fine motor development. 002 Make use of manipulative materials that vary in size and skill requirement. 003 Cleans up appropriate work area 004 Gains ability in using different art media in a variety of ways for creative expression and representation 005 Draws using crayons, pencils, a variety of markers, paint. 006 Holds and uses scissors and pencils correctly 007 Uses glue properly 009 Progresses in ability to create drawings, paintings, models and other art creations that have more details. 011 Identifies basic geometric shapes: circle, square, rectangle, triangle, oval 012 Identifies five basic forms: cube, cone, cylinder, pyramid and sphere 013 Identifies lines (i.e.curved, straight, wavy, zig-zag, horizontal, vertical). 014 Draws from memory, observation, and imagination 015 Draws around shapes with a pencil Recognizes drawing materials can be used in different ways: points (sharpened-blunt), 016 overlapping, pressing heavily, pressing lightly 017 Uses scissors to cut along a pre-drawn line 018 Identifies color hues: red, yellow, blue, orange, green, violet, black, white, brown 019 Identifies primary and secondary colors 020 Identifies warm colors as yellow, orange, red 021 Identifies cool colors as green, blue, violet 022 Recognizes that mixing primary colors creates secondary colors 023 Identifies art that uses more of one element than another, such as textures, lines, patterns 024 Distinguishes between visual texture and actual texture 025 Paints and draws with one color and develops rhythmic patterns 026 Demonstrates resist process in combination of crayon and thinned paint 027 Creates art from recycled or natural materials (collages, sculptures, monoprints) 030 Distinguishes paintings from sculpture 031 Recognizes that artists often mix media 032 Uses print making process 033 Locates the horizon in a picture 034 Explores clay, related themes and techniques 035 Distinguishes man-made and natural elements of design 036 Creates drawings, paintings, models and other art creations that have more details 037 Experiment with a variety of mediums and art materials for tactile experience and exploration 038 Create artistic works with intent and purpose using varying tools, texture, color, and technique Art teachers updated curriculum Italicized standards are TN Developmental Standards Page 1 of 3
5 Guide ID Curriculum Guide: Art PreK-Grade 8/ Description Identifies likenesses and differences: large-small, few-many, dark-light, bright-dull, thickthin; surface qualities (textures): rough-smooth, coarse-slick, warm-cold, soft-hard Gr PreK Gr K Gr 1 Gr 2 Gr 3 Gr 4 Gr 5 Gr 6 Gr 7 Gr Describes objects/colors/textures/lines/shapes that can be seen in art works 200 Uses negative space in designing 201 Uses shading to indicate the source of light 202 Recognizes overlapping to create depth. 203 Recognizes varying points of view in picture plans 204 Mixes secondary colors Identifies paintings and varying works of different media such as sculptures, collages, weavings Explores three dimensional media--for example, clay 207 Utilizes weaving techniques 211 Identifies symmetrical and asymmetrical shapes and designs 300 Draws straight lines with a pencil and ruler 301 Draws one point perspective 302 Identifies color qualities such as intensity, brightness, and value 303 Recognizes an arch, dome, column, etc. 304 Mounts pictures on a larger sheet of paper 400 Creates tint adding white; shade adding black; and tone using complimentary colors 401 Explores a variety of sculpture, media and techniques such as pulp, strips, and use of molds to create dimensional and relief sculptures Creates art incorporating elements and principles previously introduced (pattern, texture, line, color) Identifies positive/negative space 501 Draws two point perspective 502 Makes smooth gradations creating values 503 Distinguishes between transparent and opaque paint 504 Distinguishes between relief and free-standing sculpture 505 Recognizes art styles as realistic, non-objective, abstract, op, pop, etc Art History 208 Recognizes art created by artists introduced in the course 209 Recognizes architecture as a form of art 210 Understands weaving as an early art form to create clothing and shelter for people 305 Recognizes other art forms such as video, drama, dance, photography 306 Recognizes stained glass as a form of art 307 Recognizes kinetic sculpture 403 Recognizes calligraphy as an art form 506 Studies the development of historical art periods and representative artists of those periods Art: Personal Development 008 Expresses personal ideas, thoughts, and feelings through drawing Art teachers updated curriculum Italicized standards are TN Developmental Standards Page 2 of 3
6 Curriculum Guide: Art PreK-Grade 8/ Guide ID Description Gr PreK Gr K Gr 1 Gr 2 Gr 3 Gr 4 Gr 5 Gr 6 Gr 7 Gr Responds to artistic creations or events. 028 Develops a sense of pride, confidence, and satisfaction in creations of self and others 039 Respond and react to visual arts created by self and others 308 Learns to describe, analyze, interpret, and judge art works 402 Chooses personal classwork to put in an exhibit and explain decision 404 Chooses artist's work which is most preferred Art in Society 029 Recognizes art can tell stories 309 Recognizes different cultures have their own art forms 405 Explores forms of commercial art: product design, graphics, advertising 600 Develops an awareness of how societies express values and beliefs through visual forms Art teachers updated curriculum Italicized standards are TN Developmental Standards Page 3 of 3
7 Curriculum Guide: Band Grades 5-8/ Guide ID Description Gr 5 Gr 6 Gr 7 Gr 8 Basic Skills 500 The instrument: identify parts of the instrument 501 The instrument: assemble the instrument 502 The instrument: demonstrate proper care of the instrument 503 Reed instruments: choose and care for reeds 504 Use proper posture: total body support, arm, hand, wrist, and finger positions 505 Playing posture woodwinds, brass: properly support the instrument while sitting or standing to play Playing posture percussion: use proper grip/playing position for snare drum, bass drum, mallet percussion instruments Tone quality: produce sounds demonstrating characteristic tone quality 508 Tone quality: demonstrate an awareness of ways to improve tone quality 509 Tone quality: define and interpret through performance, variations in dynamics: forte and piano 510 Tone quality woodwinds, brass: use proper breathing techniques 511 Tone quality woodwinds, brass: play with good embouchure 512 Tone quality woodwinds, brass: produce an evenly sustained tone on a single note 513 Tone quality percussion: demonstrate an awareness of variations in sound which result from using different beaters, mallets and sticks (if available) 514 Pitch: identify and define flat, sharp, and natural signs 515 Pitch: identify and play in the key signatures of concert b flat 516 Pitch: identify and play all pitches through B flat concert scale 517 Pitch woodwinds, brass: use alternate fingerings/position introduced appropriately in exercises and repertoire 518 Tuning and intonation: identify and utilize the tuning mechanisms of the instrument 519 Tuning and intonation: demonstrate ability to tune to a given pitch 520 Tuning and intonation woodwinds, brass: improve intonation by making appropriate adjustments in embouchure, posture and breath support 801 Tone quality: demonstrate an even tone while varying the dynamic level in scale and arpeggio patterns and in repertoire 802 Tone quality: demonstrate consistent tone quality across the range while playing scales 803 Tone quality: demonstrate consistent tone quality over all dynamics from pianissimo through fortissimo 804 Tone quality: define and interpret thorough performance, variations in all dynamics (pianissimo through fortissimo) 805 Tone quality percussion: produce an even sound while playing 807 Tone quality percussion: demonstrate production of even, sustained sounds while rolling long notes 808 Tone quality percussion: recognize the produce an appropriate tone quality for each instrument studied 809 Pitch: identify and play in the key signatures of concert B flat, E flat, F as a base line 811 Pitch: perform a chromatic scale off one octave 813 Tuning and intonation: demonstrate an increasing ability to adjust the instrument to a given pitch 814 Tuning and intonation: develop the ability to tune one's instrument and correct intonation problems while playing Culminating Performance 546 Evaluate one's own performance of an exercise or solo 547 Perform as a member of the full band and demonstrate mastery of technical and musical demands previously introduced 664 Perform a solo or ensemble piece which demonstrates the technical and musical mastery learned in the second year Page 1 of 3 Updated by band teachers in 2014 Guide ID reflects grade level objective is introduced, e.g. 500 Grade 5, 800 Grade 8, etc.
8 Curriculum Guide: Band Grades 5-8/ Guide ID Description Gr 5 Gr 6 Gr 7 Gr Evaluate one's own performance of an exercise or solo and compare to others' evaluations of the same performance Perform repertoire as a member of the full band, which demonstrates the technical and musical mastery learning in 666 Standard of Excellence, Book Two 667 Demonstrate appropriate performance etiquette 736 Perform repertoire as a member of the full band, which demonstrates technical and musical mastery learning 847 Perform repertoire as a member of the full band, which demonstrates technical and musical mastery learning Musical Concepts 525 Rhythmic perception: count and perform rhythm patterns combining various notes and rests 526 Woodwinds, brass, percussion: whole note, whole rest, half note, half rest, quarter note, quarter rest, eighth note, eighth rest, dotted half note 527 Drums: whole note, whole rest, half note, half rest, quarter note, quarter rest, eighth note, eighth rest, dotted half note, sixteenth note, sixteenth rest 528 Pitch perception: name the lines and spaces on the staff of the treble or bass clef 529 Pitch perception: define and interpret through performance, accidentals and key signatures 530 Melodic perception: define and interpret through performance, phrase and breath marks 610 Rhythm perception: count and perform rhythm patterns combining various notes and rests: woodwinds, brass, mallets: whole note, whole rest, half note, half rest, quarter note, quarter rest, eighth note, eighth rest Rhythm perception: count and perform rhythm patterns combining various notes and rests: woodwinds, brass, 611 mallets: dotted half note, dotted quarter note, and syncopation (eighth-quarter-eighth) Rhythm perception: drums: count and perform rhythm patterns combining various notes and rests: whole note, whole 612 rest, half note, half rest, quarter note, quarter rest, eighth note, eighth rest 613 Rhythm perception: drums: count and perform rhythm patterns combining various notes and rests: dotted half note, sixteenth notes, eighth and two sixteenths, and an eighth dotted quarter note, and syncopation (eighth-quarter-eighth) 614 Rhythm perception: define common time, 4/4, 3/4, and 2/4 time signatures 615 Rhythm perception: define and perform in cut time 616 Rhythm perception: perform rhythm patterns incorporating syncopation 617 Rhythm perception: perform rhythm patterns incorporating fermátas 618 Rhythm perception: perform rhythmic patterns incorporating syncopation 619 Rhythm perception: count and perform rhythm patterns incorporating ties 620 Rhythm perceptions: count and perform rhythm patterns incorporating pick-up notes 621 Rhythm perceptions: replicate the two beat, three beat, four beat conducting patterns 622 Pitch perception: identify the lines and spaces of the staff in treble or bass clef 623 Pitch perception: identify and perform key signatures: flat, sharp, natural: key signatures of concert B flat 624 Melodic perception: define and interpret through performance, phrase, and breath marks 625 Melodic perception: define and recognize melodic contour by steps, skip, and leaps 629 Melodic perception: be introduced to the theme(s) in compositions 630 Melodic perception: be introduced to a countermelody 632 Melodic perception: be introduced to compositions Symbols and terms: define and perform various symbols and terms: repeat, solo, soli, tutti, unison, accent, first and 633 second endings, one measure repeat sign, long rest Symbols and terms: be introduced to d.c. al fine, syncopation, interval, d.s. al fine, enharmonic, staccato, tenuto, 634 legato, major chord, minor chord, d.c. al coda, and coda 651 Sight reading: sight read a short piece that encompasses the technical and musical scope Updated by band teachers in 2014 Page 2 of 3 Guide ID reflects grade level objective is introduced, e.g. 500 Grade 5, 800 Grade 8, etc.
9 Curriculum Guide: Band Grades 5-8/ Guide ID Description Gr 5 Gr 6 Gr 7 Gr Rhythm perception: count and perform rhythm patterns combining various notes and rests (woodwinds, brass, mallets): whole note, whole rest, half note, half rest, quarter note, quarter rest, eighth note, eighth rest (cont.) 715 Rhythm perception: count and perform rhythm patterns combining various notes and rests (woodwinds, brass, mallets): dotted half note, dotted quarter note and syncopation (eighth-quarter-eighth), sixteenth notes (cont.) 716 Rhythm perception: count and perform rhythm patterns combining various notes and rests (woodwinds, brass, mallets): eighth and two sixteenths, two sixteenth and an eighth, dotted eighth and sixteenth note, triplet (cont.) 717 Rhythm perception: count and perform rhythm patterns combining various notes and rests (woodwinds, brass, mallets): sixteenth and dotted eighth, sixteenth-eighth-sixteenth, sixteenth rest, dotted quarter rest 718 Rhythm perception: drums: whole note, whole rest, half note, half rest, quarter note, quarter rest, eighth note, eighth rest, dotted half note, sixteenth note, eighth, and two sixteenths, and an eighth dotted quarter note and syncopation (1/8-1/4-1/8) 719 Pitch perception: define and interpret through performance accidentals and key signatures: flat, sharp, natural: key signatures of concert B flat, E Flat, F 721 Symbols and terms: define and interpret through performance, variations in dynamics: Pianissimo, piano, mezzo piano, mezzo forte, forte, fortissimo, crescendo, and decrescendo, sporzando and forte-piano 722 Symbols and terms: be introduced to variations in tempo: largo, andante, andantino, moderato, allegro, allegretto, maestoso, ritardando, accelerando 733 Sight reading: sight read a short piece that encompasses the technical and musical scope of third year 821 Rhythm perception drums: define common time, 4/4, 3/4, and 2/4 time signatures 823 Rhythm perception drums: perform rhythm patterns incorporating fermatas 824 Rhythm perception drums: perform rhythmic patterns incorporating syncopation 825 Rhythm perception drums: count and perform rhythm patterns incorporating ties 826 Rhythm perception drums: count and perform rhythm patterns incorporating pick-up notes 831 Melodic perception: define and interpret through performance, phrase and breath marks 832 Melodic perception: define and recognize melodic contour by steps, skips, and leaps 845 Sight read a short piece that encompasses the technical and musical scope of the 4th year Musical Judgments 541 Play each piece in the correct style 660 Play each piece in the correct style Special Techniques 521 Woodwind, brass technique: use the proper technique in the attack and release of sounds: tongue, accent and slur 522 Brass: demonstrate lip slurs from a concert Bb (B Flat) to a concert F 523 Trombone: demonstrate legato tonguing and slide slurs 524 Snare drum: identify and rudiments (single paradiddle, flam, double strokes and buzz strokes) 600 Woodwind, brass: use the proper technique in the attack and release of sounds: tongue, accent, and slur 601 Woodwind, brass: demonstrate the use of staccato, tenuto, and legato articulations 602 Brass: demonstrate lip slur on concert B flat 603 Trombone: demonstrate legato tonguing and slide slurs 604 Percussion: demonstrate proper technique when playing snare drum, bass drum, mallet percussion instruments 605 Percussion: be introduced to snare drum: identify and play rudiments: 5 stroke, 9 stroke, flam, flam tap 701 Trombone: continue developing legato tonguing and slide slurs 704 Percussion: demonstrate proper technique when playing snare drum, bass drum, mallet percussion instruments, and if available, timpani, suspended cymbal, triangle, wood block, sleigh bells, tambourine, maracas, claves, temple blocks, crash cymbals, bongo Updated by band teachers in 2014 Page 3 of 3 Guide ID reflects grade level objective is introduced, e.g. 500 Grade 5, 800 Grade 8, etc.
10 School Name: Curriculum Overview xls Curriculum Guide: Health Teacher Name: Standard Category Objective GR K Gr 1 Gr 2 Gr 3 Gr 4 Gr 5 Gr 6 Gr 7 Gr 8 STANDARD 1: The student will demonstrate the ability to implement decision making and goal setting skills to promote his/her personal health and wellness, thereby enhancing quality of life for himself/herself and those around him/her. 1 HR Introduce the decision-making process 1 HR Demonstrates making personal choices based on reasoned arguments 1 HR Review the decision-making process 1 HR Describes individual goals and aspirations for healthy living Identifies choices and examine alternatives and consequences of each choice when 1 HR making decisions as it relates to healthy living 1 HR Differentiates between long and short-term personal goals Analyzes the effectiveness of personal decision-making as it relates to future health 1 HR/PE and wellness outcomes STANDARD 2: The student will understand the importance of personal hygiene practices as related to healthy living. Identifies proper personal hygiene skills (e.g. hand washing, shampooing, flossing, 2 PE/HR tooth brushing, bathing) 2 PE/HR Identifies the consequences of poor oral hygiene (e.g. cavities, gum disease, loss of teeth) Describes basic personal hygiene methods including hand washing, dental/oral care, 2 PE bathing/shampooing, and dressing 2 PE Explains how personal hygiene practices can affect personal health and social relationships 2 PE Identifies and evaluates basic personal hygiene habits Analyzes personal choices related to hygiene and their influences on others as they 2 PE relate to healthy living (e.g., clean clothing, body and hair, tattoos, and body piercing) STANDARD 3: The student will understand the role of body systems as related to healthy living. 3 HR/S Identifies the basic body parts; their names and location 3 HR/S Identifies the functions of the basic body systems (e.g. heart, lungs, brain, stomach) 3 HR/S Identifies and classifies components of major body systems Explains repercussions of risky behaviors on body systems (e.g. smoking, legal and 3 HR/S/R illegal drug use, alcohol, sexual activity, and high fat diet) 3 PE/HR Explains the benefits of good posture on personal health (e.g. breathing, back health) Names basic human body organs (e.g. heart, lungs, stomach, intestines, liver, and 3 S brain) Adapted from Tennessee Health Education Standards (2008) Page 1 of 9 Guide ID reflects grade level objective is introduced, 001 PK-K, 100 Grade 1, etc.
11 School Name: Curriculum Overview xls Curriculum Guide: Health Teacher Name: Standard Category Objective GR K Gr 1 Gr 2 Gr 3 Gr 4 Gr 5 Gr 6 Gr 7 Gr 8 STANDARD 4: The student will understand the relationship of physical activity and rest to healthy living. Identifies the importance of participating in the recommended one hour of daily 4 PE physical activity 4 PE Describes how getting the recommended eight to ten hours of sleep daily contributes to healthy living 4 PE Explains the role of adequate sleep in health and performance of daily activities 4 PE Identifies how poor food choices and physical inactivity contribute to the development of chronic diseases (e.g. obesity, high blood pressure, diabetes) Determines the benefits of exercise in relation to improved health during all stages of 4 PE life 4 PE Assesses the relationship of physical activity to other areas of personal health (social, mental, and emotional well-being) 4 PE Understands the link between physical activity and positive stress management and emotional well-being STANDARD 5: The student will understand the relationship of nutrition to healthy living. 5 HR/S Identifies the USDA My Plate guidelines for healthy eating 5 HR/S Describes the importance of healthy meals and snacks Uses the My Plate as a guide for choosing a variety of foods necessary for good 5 HR health 5 PE Explains how personal health and body composition is influenced by balancing diet and physical exercise 5 PE Identifies the causes of obesity Describes that adequate water intake and a nutritious breakfast are essential 5 PE components of healthy living 5 PE Explains the food and exercise pyramids 5 PE Identifies the energy nutrients (fats, carbohydrates and protein) and non-energy nutrients (vitamins, minerals and water) Understands the relationship between energy intake and energy output ( calories in = 5 PE calories out ) 5 PE Explains how the essential nutrients relate to body growth and development STANDARD 6: The student will understand the contributions of family relationships to healthy living. 6 C/HR Identifies the signs of abuse (emotional and physical abuse) 6 HR Demonstrates respect for the responsibilities of each person within the family Adapted from Tennessee Health Education Standards (2008) Page 2 of 9 Guide ID reflects grade level objective is introduced, 001 PK-K, 100 Grade 1, etc.
12 School Name: Curriculum Overview xls Curriculum Guide: Health Teacher Name: Standard Category Objective GR K Gr 1 Gr 2 Gr 3 Gr 4 Gr 5 Gr 6 Gr 7 Gr 8 6 HR Understands cultural differences exist and influence family customs Describes describe a variety of family structures (e.g. two parent, single parent, 6 HR blended, extended, foster, and adopted) and how they change over time 6 HR Identifies being loved and cared for are basic human needs 6 R/C Describes family structures, roles, and how they may change Identifies how family values impact gender discrimination, harassment, and various 6 R/C types of abuse (domestic violence, sexual, emotional) 6 R/HR/C Discusses expectations and stereotypes about the opposite sex. STANDARD 7: The student will understand the stages of human growth and development. 7 HR/R Defines adolescence, puberty and human development 7 HR/R Identifies the components of the male and female reproductive system 7 HR/R Identifies sexual feelings common to adolescents and differentiate between having sexual feelings and acting on them 7 HR/R Discuss HIV/STI (sexually transmitted infection) symptoms, treatments, and complications 7 HR/R Identifies reasons for abstaining from sexual activity 7 PE/S Identifies the components of the male and female reproductive system 7 R Identifies abstinence from sexual activity as the responsible and preferred choice for adolescents Review how physical growth and development is accompanied by changes in 7 R/HR emotions Describes how physical growth and development is accompanied by changes in 7 R/HR emotions 7 R/HR Demonstrates how to be respectful of others as they grow and develop Describes respectful ways to be supportive of their classmates (e.g. voice changes-- 7 R/HR students laugh and body odor) STANDARD 8: The student will understand the importance of positive self-concept and interpersonal relationships for healthy living. 8 HR Identifies the characteristics of a friend 8 HR Identifies list of positive traits about themselves 8 HR Identifies the characteristics of a bully Adapted from Tennessee Health Education Standards (2008) Page 3 of 9 Guide ID reflects grade level objective is introduced, 001 PK-K, 100 Grade 1, etc.
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https://de.scribd.com/document/357888887/Tension-Members | 1,561,039,955,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999218.7/warc/CC-MAIN-20190620125520-20190620151520-00229.warc.gz | 409,097,171 | 63,585 | You are on page 1of 21
CIE 428 Module B Instructor: Andrew Whittaker
This module of CIE 428 covers the following subjects
Introduction
Design strength
Net area
Staggered fasteners
Block shear
INTRODUCTION
Tension members are structural elements that are subjected to axial
tensile forces. Examples include:
Members in trusses
Bracing in frames to resist lateral forces from blast, wind,
and earthquake
8/21/2002 9:13 AM 1
CIE 428 Module B Instructor: Andrew Whittaker
Stresses (f) in axially loaded members are calculated using the
following equation
P
f =
A
Gusset plate at support
8/21/2002 9:13 AM 2
CIE 428 Module B Instructor: Andrew Whittaker
What are the gross and net areas for the bar?
Gross area:
Net area:
The design of tension members and their connections is covered by
Section B (Design Requirements), D (Tension Members), and J
(Connections) of the AISC LRFD Specification, 3rd Edition.
DESIGN STRENGTH
Excessive deformation (yielding under gravity loads)
Fracture
limiting stresses on the gross section to less than the yield stress.
For yielding on the gross section, the nominal strength is:
Pn = Fy Ag
Fracture is avoided by limiting stresses on the net section to less
than the ultimate tensile strength. For fracture on the net section,
the nominal strength is:
Pn = Fu An
8/21/2002 9:13 AM 3
CIE 428 Module B Instructor: Andrew Whittaker
Resistance (or phi) factors are applied to these nominal strengths
to calculate design strengths. The smaller of the two calculated
strengths:
0.9 Fy Ag ; 0.75 Fu An
is taken as the design strength.
Consider the example from the text that is reproduced below. The
of 35 and 15 kips, respectively. Steel is A36. Bolts are 0.875-inch
in diameter. Assume that the effective net area is 85% of the
computed net area.
8/21/2002 9:13 AM 4
CIE 428 Module B Instructor: Andrew Whittaker
NET AREA
The performance of a tension member is often governed by the
response of its connections. The LRFD Specification introduces a
measure of connection performance known as joint efficiency,
which is a function of
Material properties (ductility)
Fastener spacing
Stress concentrations
Shear lag
Addressed specifically by the LRFD Specification
8/21/2002 9:13 AM 5
CIE 428 Module B Instructor: Andrew Whittaker
The LRFD Specification introduces the concept of effective net
area to account for shear lag effects.
Ae = UAg
Ae = UAn
where
x
U = 1
L
and x is the distance from the plane of the connection to the
centroid of the connected member and L is the length of the
connection in the direction of the load. Figures from the text are
reproduced below to illustrate these definitions.
8/21/2002 9:13 AM 6
CIE 428 Module B Instructor: Andrew Whittaker
For welded connections, the commentary to the LRFD
Specification (Section B.3) gives values for U that can be used in
lieu of detailed calculation. Different values are given depending
on whether there are more or less than two fasteners per line in the
1. For W, M, and S shapes that have a width-to-depth ratio of
0.67 or more and are connected through the flanges with at
least three fasteners per line in the direction of the applied
2. For other shapes with at least three fasteners per line, U =
0.85.
8/21/2002 9:13 AM 7
CIE 428 Module B Instructor: Andrew Whittaker
3. For all members with only two fasteners per line, U = 0.75.
1. For W, M, and S shapes that have a width-to-depth ratio of
0.67 or more and are connected through the flanges, U = 0.9.
2. For all other shapes, U = 0.85.
The figure below from the textbook illustrates the calculation for
bolted connections.
8/21/2002 9:13 AM 8
CIE 428 Module B Instructor: Andrew Whittaker
The effective net area will only be less than the net area if some
elements of the cross-section are not connected. For tension
members such as single plates and bars connected by longitudinal
fillet welds,
the LRFD Specification writes that Ae = UAg , where U = 1.0, 0.87,
and 0.75 for l 2 w, 1.5w l 2 w, w l 1.5w, respectively.
For transverse welds, the LRFD Specification writes that the
effective net area is equal to the area of the connected element of
the cross section.
Calculations for effective net area are presented in the two
examples from the textbook that are reproduced on the following
two pages.
Bolted connection
Welded connection
8/21/2002 9:13 AM 9
CIE 428 Module B Instructor: Andrew Whittaker
8/21/2002 9:13 AM 10
CIE 428 Module B Instructor: Andrew Whittaker
STAGGERED FASTENERS
The presentation to date has focused on regular, non-staggered
fastener geometries.
line.
Often, for reasons of connection geometry, fasteners must be
placed in more than one line
Staggered fasteners
See F3.13 from the textbook that is reproduced below for
examples of staggered connections
8/21/2002 9:13 AM 11
CIE 428 Module B Instructor: Andrew Whittaker
Consider part (c) of the figure above. One cross-section to be
considered for calculation of the effective net area is shown in red;
another is given by line abcd. For fracture plane abcd, the equation
f=P/A does not apply because stresses on bc are a combination of
normal and shearing stresses.
For calculation of the effective net area, the Section B2 of the
LRFD Specification makes use of the product of the plate
thickness and the net width. The net width is calculated as
s2
wn = wg d +
4g
where all terms are defined above.
8/21/2002 9:13 AM 12
CIE 428 Module B Instructor: Andrew Whittaker
All possible failure patterns should be considered. For example,
consider the plate shown below for use with 1-inch diameter bolts.
What calculations are needed for design?
8/21/2002 9:13 AM 13
CIE 428 Module B Instructor: Andrew Whittaker
How is the net area calculated if connections are made in both legs
of an angle?
The angle is first unfolded along the middle surface.
The gross width is the sum of the leg lengths minus the
thickness.
Gage line crossing the heel of the angle must be reduced
by the thickness of the angle.
Consider the angle connection below. The holes are for 0.875-inch
diameter bolts. Steel is A36.
Because both legs of the angle are connected, what is the
effective net area?
8/21/2002 9:13 AM 14
CIE 428 Module B Instructor: Andrew Whittaker
What are the design strengths for:
Yielding?
Fracture?
BLOCK SHEAR
Block shear is an important consideration in the design of steel
connections. Consider the figure below that shows the connection
of a single-angle tension member. The block is shown shaded.
In this example, the block will fail in shear along ab and tension on
bc. The LRFD procedure is based on one of the two failure
surfaces yielding and the other fracturing.
Fracture on the shear surface is accompanied by yielding
on the tension surface
Fracture on the tension surface is accompanied by yielding
on the shear surface
Both surfaces contribute to the total resistance.
8/21/2002 9:13 AM 15
CIE 428 Module B Instructor: Andrew Whittaker
The nominal strength in tension is Fu Ant for fracture and Fy Agt
for yielding where the second subscript t denotes area on the
tension surface (bc in the figure above). The yield and ultimate
stresses in shear are taken as 60% of the values in tension. The
LRFD Specification considers two failure modes:
Rn = 0.75[0.6 Fu Anv + 0.6 Fy Agt ]
Because the limit state is fracture, the equation with the larger of
the two fracture values controls. Consider the single angle in
tension shown below.
What are the shear areas?
8/21/2002 9:13 AM 16
CIE 428 Module B Instructor: Andrew Whittaker
Agv = 0.375 (7.5); Anv = 0.375[7.5 2.5 1]
The design of a tension member involves selecting a member from
Gross area
Net area
Slenderness
L
300 to prevent vibration, etc; does not apply to
r
cables.
If the member has a bolted connection, the choice of cross section
must account for the area lost to the boltholes.
Because the section size is not known in advance, the default
values of U are generally used for preliminary design.
Detailing of connections is a critical part of structural steel design.
Connections to angles are generally problematic if there are two
lines of bolts. Consider the figure below that provides some
guidance on sizing angles and bolts.
Gage distance g1 applies when there is one line of bolts
8/21/2002 9:13 AM 17
CIE 428 Module B Instructor: Andrew Whittaker
Gage distances g 2 and g3 apply when there are two lines
As an example, design an equal angle tension member, 12 feet
long to resist the loads shown. Use A36 steel.
Pick a section from the LRFD manual based on Ag , r, gi :
8/21/2002 9:13 AM 18
CIE 428 Module B Instructor: Andrew Whittaker
Assume that the angle is connected through the long leg and
estimate a value of U (0.85), calculate values of An and Ae .
The nominal tensile strength of a threaded rod can be written as
Pn = As Fu 0.75 Ab Fu
where
Ab is the nominal (unthreaded area),
8/21/2002 9:13 AM 19
CIE 428 Module B Instructor: Andrew Whittaker
and 0.75 is a lower bound (conservative) factor relating As and
Ab . See Section J3.6 of the LRFD Specification for details.
The design strength of a threaded rod is calculated as Pn = 0.75 Pn
PIN-CONNECTED MEMBERS
Pinned connections transmit no moment (ideally) and often utilize
components machined to tight tolerances (plus, minus 0.001). The
figure below from the textbook shows failure modes for pin-
connected members and each failure mode must be checked for
design.
t Pn = 0.75(2tbeff Fu ) , where beff = 2t + 0.63 b
8/21/2002 9:13 AM 20
CIE 428 Module B Instructor: Andrew Whittaker
Shear on the effective area
d
sf Pn = 0.75(0.6 Asf Fu ) = 0.75(0.6[2t{a + }]Fu )
2
1.8 Apb Fy is based on a deformation limit state under
service loads producing stresses of 90% of yield.
Tension on the gross section
Pn = 0.90( Ag Fy )
- END OF MODULE -
8/21/2002 9:13 AM 21 | 2,636 | 9,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-26 | latest | en | 0.901578 |
https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&oldid=177838 | 1,679,824,089,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00164.warc.gz | 142,200,166 | 13,497 | # 1984 AIME Problems/Problem 15
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Determine $x^2+y^2+z^2+w^2$ if
$\frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1$
$\frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1$
$\frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1$
$\frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1$
## Solution 1
Rewrite the system of equations as $\frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.$ This equation is satisfied when $t = 4,16,36,64$, as then the equation is equivalent to the given equations. After clearing fractions, for each of the values $t=4,16,36,64$, we have the equation $x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)$ $+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)$. We can move the expression $(t-1)(t-9)(t-25)(t-49)$ to the left hand side to obtain the difference of the polynomials: $x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)$ $+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)$ and $(t-1)(t-9)(t-25)(t-49)$
Since the polynomials are equal at $t=4,16,36,64$, we can express the difference of the two polynomials with a quartic polynomial that has roots at $t=4,16,36,64$, so
$x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)$ $+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = -(t-4)(t-16)(t-36)(t-64)$
Note the leading coefficient of the RHS is $-1$ because it must match the leading coefficient of the LHS, which is $-1$.
Now we can plug in $t=1$ into the polynomial equation. Most terms drop, and we end up with
$$x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)$$
so that
$$x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}$$
Similarly, we can plug in $t=9,25,49$ and get
\begin{align*} y^2&=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\ z^2&=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\ w^2&=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{align*}
Now adding them up,
\begin{align*}z^2+w^2&=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\ x^2+y^2&=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}
with a sum of
$$\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.$$
Postscript for the puzzled: This solution which is seemingly unnecessarily redundant in that it computes $x^2,y^2,z^2,$ and $w^2$ separately before adding them to obtain the final answer is appealing because it gives the individual values of $x^2,y^2,z^2,$ and $w^2$ which can be plugged into the given equations to check.
## Solution 2
As in Solution 1, we have
$(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)$ $-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)$
$=(t-4)(t-16)(t-36)(t-64)$
Now the coefficient of $t^3$ on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the $t^4$ and $t^3$ terms are, so we can eventually apply Vieta's. We can write the long equation as $$(x^2 + y^2 + z^2 + w^2)t^3 + \dots = t^4 - (1^2 + 3^2 + 5^2 + 7^2)t^3 + \dots$$ Rearranging gives us $$t^4 - (1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2)t^3 \dots = 0.$$ By Vieta's, we know that the sum of the roots of this equation is $$1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2.$$ (recall that the roots of the original and this manipulated form of it had roots $2^2, 4^2, 6^2,$ and $8^2$). Thus, $$x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2 - 1^2 - 3^2 - 5^2 - 7^2 = \boxed{36}.$$
## Solution 3 (Highly Unrecommended)
Before starting this solution, I highly recommend never following this unless you have no idea what to do with an hour of your time. Even so, learning the above solutions will be more beneficial.
\begin{align*} \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1\\ \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1\\ \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1\\ \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1\\ \end{align*} can be rewritten as \begin{align*} \frac{x^2}{3}-\frac{y^2}{5}-\frac{z^2}{21}-\frac{w^2}{45}=1\\ \frac{x^2}{15}+\frac{y^2}{7}-\frac{z^2}{9}-\frac{w^2}{33}=1\\ \frac{x^2}{35}+\frac{y^2}{27}+\frac{z^2}{11}-\frac{w^2}{13}=1\\ \frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\ \end{align*} You might be able to see where this is going. First off, find $\text{lcm}(3,5,21,45),\text{lcm}(15,7,9,33), \text{lcm}(35,27,11,13),$ and $\text{lcm}(63,55,39,15)$. Then, multiply by the respective lcm to clear all of the denominators. Once you do that, maniuplate the equations to solve for $w^2+x^2+y^2+z^2$.
Now, most of this is just a brainless bash, and reemphasizing, please try to learn the above solutions. This is only a last resort and only to be used if you have too much time left. The exact amount of time this bash takes depends on the person and how quickly they can manipulate the equations. | 2,453 | 5,298 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 48, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-14 | latest | en | 0.591988 |
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https://www.physicsforums.com/threads/convolution-shortcut.770050/ | 1,508,426,662,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823309.55/warc/CC-MAIN-20171019141046-20171019161046-00569.warc.gz | 968,212,792 | 21,854 | # Convolution shortcut?
1. Sep 10, 2014
### Color_of_Cyan
1. The problem statement, all variables and given/known data
Find (use shortcut):
x(t) = 2e-4tu(t) * e2tu(t) * t2σ(t - 2)
2. Relevant equations
Convolution properties:
# "shape of Y (output) is different from x1, x2"
# x1 * x2 = x2 * x1
# x1 * (x2 + x3) = (x1 * x2) + (x1 * x3)
# x1(t) = * x2(t) * x3(t) * .....
# step * ramp = parabolic function
# eatu(t) * ebtu(t) = (1/a-b)[eat - ebt]u(t)
3. The attempt at a solution
2e-4tu(t) * e2tu(t) * t2σ(t - 2)
just doing
2e-4tu(t) * e2tu(t) for now, and using the last property listed:
= 2(1/(2+4))[e-4t - e2t]u(t)
= (1/3)[e-4t - e2t]u(t)
Last edited: Sep 10, 2014
2. Sep 10, 2014
### rude man
What is σ(t)? I've never seen that function before.
What are you trying to find? You already have x(t). Simplifying?
3. Sep 10, 2014
### Color_of_Cyan
It's a impulse function supposedly (and sadly I don't seem to have any property listed where it's involved). And yes, trying to simplify and do convolution with all of this.
4. Sep 10, 2014
### rude man
OK, I guess σ(t) is what the rest of us call δ(t)?
I suppose you can convolve the first & 2nd terms, the convolve the result with the third.
So why not write the formal convolution integral for the first two terms for a starter.
I'm afraid I don't know any 'shortcut' here.
5. Sep 11, 2014
### Color_of_Cyan
Hah, yes, δ(t). (Of course I forgot the 'δ' symbol was here and my poor eyesight never saw it.)
Okay
(f * g)(t) = 0∫tf(t - τ)g(τ)dτ then, is this it?
So if f = 2e-4tu(t) and g = e2tu(t). So it's something like
= 0∫t[2e-4(t - τ)u(t - τ)eu(τ)]dτ
I had to change all the function variables (including inside the 'u(t)') to τ also right? Do I have to also end up integratring the 'u(τ)'? If so, how?
6. Sep 11, 2014
### rude man
Permit me to write T = tau, it's a lot easier typing ...
So, proceeding from that integral, with f(t-T) = exp{2(t - T)}U(t - T) and g(T) = 2exp{-4T}U(T), I would change t - T to -(T - t) everywhere in the integrand. Now, graph U(T) and -U(T - t) vs. T and multiply them. Then get rid of the two U functions in the integrand by suitable choice of the upper and lower limits of integration. Then, perform the integration.
The second convolution is done the same way. Remember the sampling property of the delta function: ∫f(x)δ(x - a)dx with upper limit +∞ and lower limit -∞ = f(a).
7. Sep 11, 2014
### donpacino
use laplace!!!
convolution in the time domain becomes multiplication in the frequency domain.
8. Sep 11, 2014
### rude man
Laplace is limited to t > 0. This is not specified here. But you could use Fourier. I have looked at that but found it harder to do the inversion integral back to the time domain than just plowing ahead solely in the time domain.
BTW there is also the double-sided Laplace transform which no one uses.
9. Sep 13, 2014
### Color_of_Cyan
Sorry for getting back to this late
So you have the convolution for the first two terms as:
= 0t[exp(2(t - T))u(t - T)][2exp(-4T)u(T)]dT
Does changing it to -(T - t) help simply it? It would appear as this:
= 0t[exp(-2(T - t))u(-1(T - t))][2exp(-4T)u(T)]dT
Can you explain this? Graph any unit step function?
In the class this convolution problem actually came before we formally covered Laplace / Fourier (which I will also need help on later).
10. Sep 13, 2014
### rude man
I just noticed you meant the lower limit of integration to be 0. It should be -∞. (BTW the upper limit should also be ∞. You can make the upper limit t in this case, it makes no difference here). Otherwise, good.
Yes, it clarifies the next step.
Yes I can. I goofed!
I should have said "graph U(T) and U[-(T-t)] vs. T".
Can you graph U(x) vs. x?
U(x - a) vs. x?
U(-x) vs. x?
U[-(x-a)} vs. x?
You can do them all just by remembering
U(ζ) = 0, ζ < 0
U(ζ) = 1, ζ > 0.
11. Sep 15, 2014
### Color_of_Cyan
Should it be -∞ to +∞ for the whole property or just for this problem? I don't think I quite understand that. I'll take your word for it though, so I think it would now be:
-∞[exp(2(t - T))u(t - T)][2exp(-4T)u(T)]dT
Isn't this just the unit step graph property? Or what's the name or term for this if it isn't? What would be 'ζ' (if this is what you are basing the graph on)? I still can not see how this is supposed to help unless you are trying to hint at doing graphical convolution.
12. Sep 15, 2014
### rude man
That is the general definition for the convolution integral. Often, depending on the problem, -∞ can be replaced by 0 and +∞ by t, but it's always correct to use the infinite + and - limits. Just keep in mind that if the integrand is identically zero below and/or above certain values then the infinite limits can be replaced by those respective values. This is the key idea I'm trying to point out in doing the U graphs. U(T)U{-(T-t)} has zero value below and above certain values. What are those values?
No, I don't want you to do any graphical convolution. I am suggesting that by graphing the two U functions and multiplying them graphically you will know what the upper & lower limits of integration are for evaluating the actual convolution integral.
Maybe you can figure out the limits of integration by looking at U(T)U{-(T-t)} some other way, but doing it graphically is real easy.
ζ was just a dummy variable.
13. Sep 17, 2014
### Color_of_Cyan
This will sound real bad, but I'm afraid I haven't really heard of graphical multiplication before.
Are you saying you want me to graph exp(-2(T - t)) and 2exp(-4T)?
Also my bad the integral in my last post was supposed to now be this:
-∞[exp(-2(T - t))u(-1(T - t))][2exp(-4T)u(T)]dT
14. Sep 17, 2014
### rude man
No, I want you to graph U(T) and -U{(T-t)}, then multiply them into one graph.
[/quote]
That's fine, that was correct also but this way the integration becomes more apparent, with t = constant in the integration.
15. Sep 18, 2014
### Color_of_Cyan
Okay, I'm not sure if these are correct I don't really know what T is:
https://imagizer.imageshack.us/v2/547x297q90/909/MvuSyH.jpg [Broken]
https://imagizer.imageshack.us/v2/547x297q90/673/owhtZD.jpg [Broken]
Last edited by a moderator: May 6, 2017
16. Sep 18, 2014
### rude man
You got U(T) right.
You got -U{(T-t)} wrong. You graphed -U(T) instead.
When is -U(x) = 0? = 1? Now let x = T - t. t is a constant in this. T is your variable. Try again ...
17. Sep 19, 2014
### Color_of_Cyan
So it's shifted then, is that what you are saying?:
https://imagizer.imageshack.us/v2/547x297q90/537/vHAo1X.jpg [Broken]
Last edited by a moderator: May 6, 2017
18. Sep 19, 2014
### rude man
Sorry, I keep saying -U when I mean U(-).
So you want to graph U{-(T - t)}.
Change (T-t) to t on the horizontal (T) axis and you'd be there.
Now, multiply the two graphs into one new graph.
It's real easy if you understand
0 x 0 = 0
0 x 1 = 0
1 x 0 = 0
1 x 1 = 1
19. Sep 21, 2014
### Color_of_Cyan
Sorry for getting back to this super late again (site moved and I had some comp problems).
Does it mean the first graph (u(t)) is 0 and the second graph (U(-1(T - t))) is 1? I can not really see yet as to how these help with convolution integrals?
20. Sep 23, 2014
### Color_of_Cyan
Bump, also forgot to post the graph, does multiplying the two graphs mean it will be the same as u(T) if u(T) is 0?
https://imagizer.imageshack.us/v2/547x297q90/909/MvuSyH.jpg [Broken]
Last edited by a moderator: May 7, 2017 | 2,342 | 7,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-43 | longest | en | 0.889925 |
http://piracy-studies.org/2lnwo/tensor-calculus-in-machine-learning-fb5bf7 | 1,624,539,076,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00445.warc.gz | 34,732,196 | 15,511 | Write your own functions to implement each tensor operation. Jon is renowned for his compelling lectures, which he offers in-person at Columbia University, New York University, and the NYC Data Science Academy. What are Tensors? Computing derivatives of tensor expressions, also known as tensor calculus, is a fundamental task in machine learning. Expert instructions, unmatched support and a verified certificate upon completion! https://machinelearningmastery.com/introduction-to-matrix-decompositions-for-machine-learning/. In mathematics, tensor calculus, tensor analysis, or Ricci calculus is an extension of vector calculus to tensor fields. Begin with TensorFlow’s curated curriculums to improve these four skills, or choose your own learning path by exploring our resource library below. Okay. Developed by Gregorio Ricci-Curbastro and his student Tullio Levi-Civita, it was used by Albert Einstein to … “Given a tensor A with q dimensions and tensor B with r dimensions, the product of these tensors will be a new tensor with the order of q + r or, said another way, q + r dimensions.”. After this short intro to tensors, a question still remains – why TensorFlow is called like that and why does this framework need tensors at all. Perhaps check the library API on how to convert lists and arrays to that type? Recently, an algorithm for computing higher order derivatives of ten- How to implement element-wise operations with tensors. Or, m vectors of dimensions n by 1. What’s more, in 2016 Google developed TPUs (tensor processing units). Also available on the ArXiv in pdf form. The tensor product operator is often denoted as a circle with a small x in the middle. If we want to manually create the same tensor, we would need to write this line of code. For this 3D tensor, axis 0 specifies the level, axis 1 specifies the row, and axis 2 specifies the column. A Gentle Introduction to Tensors for Machine Learning with NumPyPhoto by Daniel Lombraña González, some rights reserved. Disclaimer | Best Public Datasets for Machine Learning, Why Cloud Computing Is Critical for a Data Scientist, Data Science vs Computer Science: The Best Degree For a Data Scientist, Data Science vs Machine Learning vs Data Analytics vs Business Analytics. It can be thought of as a vector of length 1, or a 1×1 matrix. Running the example prints the result of multiplying the tensors. Pick up a machine learning paper or the documentation of a library such as PyTorch and calculus comes screeching back into your life like distant relatives around the holidays. in spacetime). It is followed by a vector, where each element of that vector is a scalar. Many of the operations that can be performed with scalars, vectors, and matrices can be reformulated to be performed with tensors. In the example below, we define two order-1 tensors (vectors) with and calculate the tensor product. The most prominent example being Google’s TensorFlow. The element-wise multiplication of one tensor from another tensor with the same dimensions results in a new tensor with the same dimensions where each scalar value is the element-wise multiplication of the scalars in the parent tensors. In this section, we will work through the four main arithmetic operations. A = array([ In NumPy, we can subtract tensors directly by subtracting arrays. E.g. A key concern is the efficiency of evaluating the expressions and their derivatives that hinges on the representation of these ex-pressions. Tensors are a type of data structure used in linear algebra, and like vectors and matrices, you can calculate arithmetic operations with tensors. Tensors are mathematical objects that generalize scalars, vectors and matrices to higher dimensions. New video series. You should already have background knowledge of how ML works or completed the learning materials in the beginner curriculum Basics of machine learning with TensorFlow before continuing with this additional content. You can see that, at least in three-dimensions, the tensor is printed as a series of matrices, one for each layer. It can be helpful to understand what is NOT a tensor. In 1904, psychologist Charles Spearman tried to understand whether human intelligence is a composite of different types of measureable intelligence. Welcome! When you browse on this site, cookies and other technologies collect data to enhance your experience and personalize the content and advertising you see. I have one question about tensor conversion.İ am using attention mechanism,and I must do my operations in for loop so that i store my results in a list.At the end,i cannot convert the list into a tensor in order to make the results connected with dense layers.Can u suggest anything to come over this problem? And, they will need to know enough tensor calculus to understand why a sophisticated deep architecture may be misbehaving during learning. can you please explain how ” -1 ” came here ? The matrix m2 will be a different one with the elements: [9, 8, 7] and [1, 3, -5]. But I have a general question. This means that the lowest unit is not an integer or a float; instead, it is a vector. The mathematical concept of a tensor could be broadly explained in this way. Search, t111, t121, t131 t112, t122, t132 t113, t123, t133, T = (t211, t221, t231), (t212, t222, t232), (t213, t223, t233), t311, t321, t331 t312, t322, t332 t313, t323, t333, A = (a211, a221, a231), (a112, a122, a132), B = (b211, b221, b231), (b112, b122, b132), a111 + b111, a121 + b121, a131 + b131 a112 + b112, a122 + b122, a132 + b132, C = (a211 + b211, a221 + b221, a231 + b231), (a112 + b112, a122 + b122, a132 + b132), a111 - b111, a121 - b121, a131 - b131 a112 - b112, a122 - b122, a132 - b132, C = (a211 - b211, a221 - b221, a231 - b231), (a112 - b112, a122 - b122, a132 - b132), a111 * b111, a121 * b121, a131 * b131 a112 * b112, a122 * b122, a132 * b132, C = (a211 * b211, a221 * b221, a231 * b231), (a112 * b112, a122 * b122, a132 * b132), a111 / b111, a121 / b121, a131 / b131 a112 / b112, a122 / b122, a132 / b132, C = (a211 / b211, a221 / b221, a231 / b231), (a112 / b112, a122 / b122, a132 / b132), a11 * b11, a11 * b12, a12 * b11, a12 * b12, a11 * b21, a11 * b22, a12 * b21, a12 * b22, C = (a21 * b11, a21 * b12, a22 * b11, a22 * b12), a21 * b21, a21 * b22, a22 * b21, a22 * b22, Making developers awesome at machine learning, Click to Take the FREE Linear Algebra Crash-Course, Tensor Algebra and Tensor Analysis for Engineers, Fundamental Tensor Operations for Large-Scale Data Analysis in Tensor Train Formats, A Gentle Introduction to Matrix Factorization for Machine Learning, https://machinelearningmastery.com/introduction-to-matrix-decompositions-for-machine-learning/, https://www.youtube.com/watch?v=8ptMTLzV4-I&t=321s, How to Index, Slice and Reshape NumPy Arrays for Machine Learning, How to Calculate Principal Component Analysis (PCA) from Scratch in Python, A Gentle Introduction to Sparse Matrices for Machine Learning, Linear Algebra for Machine Learning (7-Day Mini-Course), How to Calculate the SVD from Scratch with Python. You said that “For this 3D tensor, axis 0 specifies the level, axis 1 specifies the column, and axis 2 specifies the row.”. Jason Brownlee points this out by even quoting from the “Deep Learning” book. 0scar Chang 晴れ男 Seven Myths in Machine Learning Research 16 Feb 2019. tldr; We present seven myths commonly believed to be true in machine learning research, circa Feb 2019. [[21,22,23], [24,25,26], [27,28,29]] A tensor is just a convenient data format, which allows for some very efficient operations. [[11,12,13], [14,15,16], [17,18,19]], Terms | The element-wise subtraction of one tensor from another tensor with the same dimensions results in a new tensor with the same dimensions where each scalar value is the element-wise subtraction of the scalars in the parent tensors. A tensor is a generalization of vectors and matrices and is easily understood as a multidimensional array. That’s where tensors come in handy – no matter the number of additional attributes we want to add to describe an object, we can simply add an extra dimension in our tensor. You also have the option to opt-out of these cookies. But I think I should be: But opting out of some of these cookies may have an effect on your browsing experience. I’m still confused, as other explanations mention that tensors have extra properties that are not captured by the idea that it’s just a generalization of matrices: “But [the generalized matrix] description misses the most important property of a tensor! [[1,2,3], [4,5,6], [7,8,9]], This becomes useful for operations that work with vectors or matrices as inputs. Why not just use Numpy arrays? I am no expert in math, but isn’t vector is a special type of tensor not the other way around ? Linear Algebra for Machine Learning. Hi Jason! © 2020 365 Data Science. If one transforms the other entities in the structure in a regular way, then the tensor must obey a related transformation rule.”, https://medium.com/@quantumsteinke/whats-the-difference-between-a-matrix-and-a-tensor-4505fbdc576c. So in machine learning / data processing a tensor appears to be simply defined as a multidimensional numerical array. Physicists use the term tensor to mean a geometric object that remains invariant (i.e., it retains properties like length, direction, etc) when a coordinate system changes). Its dimensions could be signified by k,m, and n, making it a KxMxN object. A key concern is the efficiency of evaluating the expressions and their derivatives that hinges on the representation of these expressions. Perhaps talk to the author about their ideas? Recently, an algorithm for computing higher order derivatives of tensor expressions like Jacobians or Hessians This website uses cookies to improve your experience while you navigate through the website. This is a fantastic introduction to tensors. It can be thought of as a vector of length 1, or a 1×1 matrix. array ( [ 1, 1, 2, 3, 5, 8 ]) print (x) print ( 'A vector is of rank %d' % (x. ndim)) [1 1 … Tensor notation is much like matrix notation with a capital letter representing a tensor and lowercase letters with subscript integers representing scalar values within the tensor. In NumPy, we can add tensors directly by adding arrays. Here, we will use the “o” operator to indicate the Hadamard product operation between tensors. It is a term and set of techniques known in machine learning in the training and operation of deep learning models can be described in terms of tensors. Address: PO Box 206, Vermont Victoria 3133, Australia. The example of this mapping is illustrated in Fig. Useful article, but it doesn’t describe what tensors represent in the machine learning domain. I think you might mean a Tensor data type for a given library? This section lists some ideas for extending the tutorial that you may wish to explore. The dimensions of a vector are nothing but Mx1 or 1xM matrices. Ask your questions in the comments below and I will do my best to answer. Discover how in my new Ebook: tensor calculus, is a fundamental task in machine learning. The key idea is to consider the tensors which are derived from multivariate moments of the observed data. Usually, we would load, transform, and preprocess the data to get tensors. We will denote it here as “(x)”. This makes them extremely scalable, too. Thank you for your blog, which is very helpful. Thanks, it is well-written. The tensor product is not limited to tensors, but can also be performed on matrices and vectors, which can be a good place to practice in order to develop the intuition for higher dimensions. Check out the complete Data Science Program today. If you search the web for the definition of a tensor, you will likely be overwhelmed by the varying explanations and heated discussions. Sometimes it even contains strings, but that’s rare. Re-upload after I caught a small mistake soon after uploading. I am totally new in tensor and this is the first time I am learning it. This free online course on the Tensor Flow machine learning will introduce you to a brief history of TensorFlow. The result is an order-2 tensor (matrix) with the lengths 2×2. print(A[0,1,0]) –> 4: Level 0, Row 2, Column 0, In all the addition, subtraction, product, and division examples, I see this: Finally, we’ve got different frameworks and programming languages. How did tensors become important you may ask? Now, a tensor is the most general concept. In the general case, an array of numbers arranged on a regular grid with a variable number of axes is known as a tensor. In this tutorial, you will discover what tensors are and how to manipulate them in Python with NumPy. A vector is a single dimension (1D) tensor, which you will more commonly hear referred to in computer science as an array. Interestingly, the meaning of this word had little to do with what we call tensors from 1898 until today. The element-wise addition of two tensors with the same dimensions results in a new tensor with the same dimensions where each scalar value is the element-wise addition of the scalars in the parent tensors. Here, we first define rows, then a list of rows stacked as columns, then a list of columns stacked as levels in a cube. tensor calculus, is a fundamental task in machine learning. An vector is made up of a series of numbers, has 1 axis, and is of rank 1. x = np. This tutorial is divided into 3 parts; they are: 1. Sir how to do that sum using for loop.Please explain? So a first-order tensor would be a vector. Developed by Gregorio Ricci-Curbastro and his student Tullio Levi-Civita, it was used by Albert Einstein to develop his general theory of relativity. An example of such a 3D tensor would be 1000 video frames of 640 × 480 size. The tensor product can be implemented in NumPy using the tensordot() function. Tensors are simply a generalization of the concepts we have seen so far. Very nice tutorial. The tensor product is the most common form of tensor multiplication that you may encounter, but there are many other types of tensor multiplications that exist, such as the tensor dot product and the tensor contraction. Build up a step-by-step experience with SQL, Python, R, Power BI, and Tableau. Good tutorial, with a very clear definition. Lately, it has joined the machine learning community?s lexicon. LinkedIn | This category only includes cookies that ensures basic functionalities and security features of the website. In the MNIST case, the first thing we immediately realise is that the input data are not in the MPS form, so the encoding mapping has to be constructed at first. However, state-of-the-art machine learning frameworks are doubling down on tensors. For this 3D tensor, axis 0 specifies the level, axis 1 specifies the row, and axis 2 specifies the column. The below content is intended to guide learners to more theoretical and advanced machine learning content. 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. Before talking about tensors, let us first see an example of how matrix factorization can be used to learn latent variable models. All Rights Reserved. Scalars, vectors, and matrices are all tensors of ranks 0, 1, and 2, respectively. Therefore it is essential for a machine learning engineer to have a good understanding of it. "Mastering Calculus for Deep learning / Machine learning / Data Science / Data Analysis / AI using Python " With this course, You start by learning the definition of function and move your way up for fitting the data to the function which is the core for any Machine learning, Deep Learning , Artificial intelligence, Data Science Application . Why do we need tensors in deep learning. B = (b211, t221, t231). Do you mean matrix factorization? Nowadays, we can argue that the word ‘tensor’ is still a bit ‘underground’. Very straightforward, great use of codes and charts. Tensors are simply a generalisation of matrices. Facebook | Simpson’s paradox explained, or when facts aren’t really facts, Backpropagation. The example below defines a 3x3x3 tensor as a NumPy ndarray. Ltd. All Rights Reserved. Einstein developed and formulated the whole theory of ‘general relativity’ entirely in the language of tensors. Running the example prints the addition of the two parent tensors. Over time, the definition of a tensor has varied across communities from mathematics to quantum physics. First of all, Einstein has successfully proven that tensors are useful. Well... may… Start with the fundamentals with our Statistics, Maths, and Excel courses. n the example below, we define two order-1 tensors (vectors) with and calculate the tensor product. Myth 1: TensorFlow is a Tensor manipulation library Myth 2: Image datasets are representative of real images found in the wild Myth 3: Machine Learning researchers do … This not only optimizes the CPU usage, but also allows us to employ GPUs to make calculations. The tensor network machine learning is illustrated on two example problems: MNIST and boundary decision. It contains two matrices, 2×3 each. Computing derivatives of tensor expressions, also known as tensor calculus, is a fundamental task in machine learning. Still not sure you want to turn your interest in data science into a career? A vector is a one-dimensional or first order tensor and a matrix is a two-dimensional or second order tensor. Let’s take a look at the tensor product for vectors. A scalar has the lowest dimensionality and is always 1×1. print(A[0,0,1]) –> 2: Level 0, Row 0, Column 1 I'm Jason Brownlee PhD © 2020 Machine Learning Mastery Pty. Given a tensor A with q dimensions and tensor B with r dimensions, the product of these tensors will be a new tensor with the order of q + r or, said another way, q + r dimensions. Introduction to Tensor Flow Machine Learning Learn about the fundamentals of machine learning and the concept of TensorFlow in this free online course. TensorFlow is a machine learning library with tools for data scientists to design intelligent systems (interface for expressing machine learning algorithms and implementation for executing such algorithms). Tensor calculus … A tensor can be defined in-line to the constructor of array() as a list of lists. Tensor even appears in name of Google’s flagship machine learning library: “TensorFlow“. These cookies will be stored in your browser only with your consent. Well done! Let’s look at that in the context of Python. This component (a rank 0 tensor) will change when the underlying coordinate system changes. Very interesting. ]), print(A[0,0,0]) –> 1: Level 0, Row 0, Column 0 Very quick read-through for beginners like me. And in fact, tensors can be stored in ndarrays and that’s how we often deal with the issue. They aren’t, really. This tutorial helped me to understand the concepts. Then we have matrices, which are nothing more than a collection of vectors. In this tutorial, you discovered what tensors are and how to manipulate them in Python with NumPy. Unlike the infinitesimal calculus, tensor calculus allows presentation of physics equations in a form that is independent of the choice of coordinates on the manifold. So a single component cannot be a tensor, even though it satisfies the definition of a multidimensional array. Running the example first prints the shape of the tensor, then the values of the tensor itself. In fact, the first use of the word ‘tensor’ was introduced by William Hamilton. You won’t hear it in high school. A key concern is the efficiency of evaluat-ing the expressions and their derivatives that hinges on the representation of these expressions. Suppose we focus on a single component in a vector. Should the “t” be “b”? Well, not without the help of one of the biggest names in science – Albert Einstein! If you explore any of these extensions, I’d love to know. Click to sign-up and also get a free PDF Ebook version of the course. This tutorial is divided into 3 parts; they are: Take my free 7-day email crash course now (with sample code). RSS, Privacy | The element-wise division of one tensor from another tensor with the same dimensions results in a new tensor with the same dimensions where each scalar value is the element-wise division of the scalars in the parent tensors. Such an object can be thought of as a collection of matrices. | ACN: 626 223 336. 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. How to Transition to Data Science from Computer Science? 2. In fact, your Math teacher may have never heard of it. Spearman’s method consisted of making his subjects take several different kin… Tensor Calculus In mathematics, tensor calculus or tensor analysis is an extension of vector calculus to tensor fields (tensors that may vary over a manifold, e.g. This is not how tensors are defined in mathematics and physics! I think any amateur in tensor could easily take over from here. He authored the 2019 book Deep Learning Illustrated, an instant #1 bestseller that was translated into six languages. Sitemap | As with matrices, we can perform element-wise arithmetic between tensors. As with matrices, the operation is referred to as the Hadamard Product to differentiate it from tensor multiplication. This is where our course "Machine Learning & Data Science Foundations Masterclass" comes in. These cookies do not store any personal information. In NumPy, we can divide tensors directly by dividing arrays. Furthermore, they will need to understand the design patterns that underlie machine learning systems (very different in style and form from classical software design patterns). Linear Regression – Machine Learning for Mathies on Tensor Calculus; Computer Science Background – Machine Learning for Mathies on Statistical Inference; Problem Set 1 – Machine Learning for Mathies on Additional Sources; Linear Regression – Machine Learning for Mathies on Parameter Estimation; Archives. Thanks Jason! If you search the web for the definition of a tensor, you will likely be overwhelmed by the varying explanations and heated discussions. Three dimensions is easier to wrap your head around. It is mandatory to procure user consent prior to running these cookies on your website. The concept for tensors was first introduced by Gregorio Ricci-Curbastro an Italian born mathematician with … It seems computer scientists have borrowed this term from physicists / mathematicians and redefined it to mean a “multidimensional array”. For instance, a photo is described by pixels. A key concern is the efficiency of evaluating the expressions and their derivatives that hinges on the representation of these ex-pressions. The Linear Algebra for Machine Learning EBook is where you'll find the Really Good stuff. Now, let’s create an array, T, with two elements: m1 and m2. Very nice, simple and well detailed introduction to one of the key mathematical tools for deep learning. and I help developers get results with machine learning. In NumPy, we can multiply tensors directly by multiplying arrays. Let’s describe a highly simplified version of his method, where the hypothesis is that there are exactly two kinds of intelligence: quantitative and verbal. After completing this tutorial, you will know: Kick-start your project with my new book Linear Algebra for Machine Learning, including step-by-step tutorials and the Python source code files for all examples. Vectors are one of the most crucial concepts within Machine Learning because many bugs are due to having matrix /vector dimensions that don't fit properly. Tensors have been around for nearly 200 years. Newsletter | Second, in machine learning, we often explain a single object with several dimensions. In the same way, TensorFlow works with tensors. We have derived a number of exciting results on the use of tensor decomposition for unsupervised learning of latent variable models. Do you have any questions? It is a 2x2x3 object. Twitter | However, it is always good to have the theoretical background. If we are talking about a 3D movie experience, a pixel could be perceived in a different way from each of our eyes. Implement three other types of tensor multiplication not covered in this tutorial with small vector or matrix data. Thanks for this. Jon Krohn is Chief Data Scientist at the machine learning company, Untapt.. Tensor Product tensor calculus for beginners provides a comprehensive and comprehensive pathway for students to see progress after the end of each module. You can explore the curriculum or sign up 12 hours of beginner to advanced video content for free by clicking on the button below. A peak into the mathematics of optimization. Element-Wise Tensor Operations 4. We also use third-party cookies that help us analyze and understand how you use this website. It is followed by a vector, where each element of that vector is a scalar. weights in a neural net. Our first matrix m1 will be a matrix with two vectors: [5, 12, 6] and [-3, 0, 14]. Running the example prints the result of the tensor product. And it's not just any old scalar calculus that pops up---you need differential matrix calculus, the shotgun wedding of linear algebra and multivariate calculus. If you are familiar with basic linear algebra, you should have no trouble understanding what tensors are. Lately, it has joined the machine learning (ML) community’s lexicon. As you can imagine, tensors with lots of elements are very hard to manually create. Read it as “order-one”, not negative one. These are processors, which consider a ‘tensor’ a building block for a calculation and not 0s and 1s as does a CPU, making calculations exponentially faster. With a team of extremely dedicated and quality lecturers, tensor calculus for beginners will not only be a place to share knowledge but also to help students get inspired to explore and discover many creative ideas from themselves. Tensors and new machine learning tools such as TensorFlow are hot topics these days, especially among people looking for ways to dive into deep learning. Data matrix is an order-2 tensor ( matrix ) with the fundamentals of machine learning m, and 2! A fundamental task in machine learning, deep learning ” book easier to your! To develop his general theory of relativity a verified certificate upon completion tensor ( matrix ) with the fundamentals machine... Engineer to have the option to opt-out of these extensions, I ’ d love to know enough tensor,... Product can be defined in-line to the constructor of array ( ndarray ) this tutorial you. Of elements are very hard to manually create block of modern machine learning library “! Tensor would be 1000 video frames of 640 × 480 size analysis, and Excel courses what... The option to opt-out of these cookies will be stored in ndarrays and that ’ s,! By adding arrays with lots of elements are very hard to manually create and arrays to that?. Of different types of tensor expressions, also known as tensor calculus to Flow... S paradox explained, or when facts aren ’ t really facts Backpropagation. Or 1xM matrices term from physicists / mathematicians and redefined it to mean a tensor is printed a!, psychologist Charles Spearman tried to understand why a sophisticated deep architecture be! That can be represented in Python both, and matrices to higher dimensions here, we can tensors! Has joined the machine learning is described by pixels will be stored your... An example of such a 3D tensor would be 1000 video frames of ×... Product in NumPy, we will use the “ o ” operator to indicate the Hadamard product differentiate... Theoretical and advanced machine learning domain in deep learning to tensor Flow machine learning with NumPyPhoto by Lombraña! Option to opt-out of these expressions of beginner to advanced video content for free clicking. Observed data ndarray ) heard of it a mxn where the matrix has... More, in machine learning, deep learning frameworks here for students to see progress after the end of module. A lot of discussion around tensors as the cornerstone data structure has 1 axis, and n.... It seems computer scientists have borrowed this term from physicists / mathematicians and it... ( vectors ) with the fundamentals with tensor calculus in machine learning Statistics, Maths, and 2, respectively toolkit! Below, we can multiply tensors directly by subtracting arrays step-by-step experience with SQL,,. Same tensor tensor calculus in machine learning even though it satisfies the definition that physicists use lists ideas. Thought of as a multidimensional array think of a multidimensional array ” the button below theoretical background time! M vectors of dimensions m by 1 function properly be simply defined as a multidimensional array ” deep may. Your confusion is warranted because this is not how tensors are defined in mathematics, tensor analysis and... You use this website includes cookies that ensures basic functionalities and security features of the course security features of operations. If you are familiar with basic Linear algebra for machine learning defined in-line to the constructor of (... The matrix a has dimension m rows and n columns Spearman tried to understand why sophisticated. To a brief history of TensorFlow in this tutorial, you will what! 1 bestseller that was translated into six languages tensor appears to be performed with tensors mxn the... Element of that vector is a scalar Python with NumPy it seems computer scientists have borrowed this term physicists! //Www.Youtube.Com/Watch? v=8ptMTLzV4-I & t=321s Power BI, and axis tensor calculus in machine learning specifies the row, and courses! Other popular deep learning illustrated, an algorithm for computing higher order derivatives of tensor expressions, also known tensor. Matrix dimenion is definedas a mxn where the matrix a has dimension m rows and columns... Up a step-by-step experience with SQL, Python, R, Power,! Hard to manually create the same tensor, you can explore the or... The result of dividing the tensors which are derived from multivariate moments of the two parent.! To opt-out of these cookies on your website must be set to 0 ’ d love know! Boundary decision by Gregorio Ricci-Curbastro and his student Tullio Levi-Civita, it has joined the tensor calculus in machine learning! Tensor Flow machine learning is illustrated in Fig of such a 3D movie experience, a tensor, we add. Good to have a good understanding of tensors, I would suggest checking out eigenchris:! N columns by adding arrays could be framed that way a Gentle introduction to tensor fields to mean tensor! Evaluating the expressions and their derivatives that hinges tensor calculus in machine learning the tensor dot product in NumPy, we realize it..., Australia tensors for machine learning the tensors optimizes the CPU usage, but it could perceived. Computing higher order derivatives of tensor multiplication not covered in this tutorial is into! Often deal with the fundamentals of machine learning, deep learning, deep learning it followed... To that type itself, both, and is of rank 1. x = np matrix! Dimensions could be broadly explained in this section provides more resources on representation. Efficiency of evaluat-ing the expressions and their derivatives that hinges on the below. This 3D tensor, you will likely be overwhelmed by the varying explanations and heated discussions, m vectors dimensions! Evaluat-Ing the expressions and their derivatives that hinges on the representation of these extensions I. Units ) them in Python any of these expressions blog, which is very helpful all, Einstein successfully. Tensors are a generalization of vectors and matrices are all tensors of ranks 0, 1, and columns. Than a NumPy ndarray Foundations Masterclass '' comes in for extending the tutorial that you may to! Context of Python this mapping is illustrated on two example problems: MNIST and boundary decision vector a., t, with two elements: m1 and m2 our Statistics Maths... A tensor, you will likely be overwhelmed by the varying explanations and heated discussions if want. Bit ‘ underground ’ up a step-by-step experience with SQL, Python, R is famously a vector-oriented programming.... Vector-Oriented programming language of 640 × 480 size component in a structure and interacts with other entities! Can perform element-wise arithmetic between tensors that the lowest dimensionality and is always good have... Extension of vector calculus to tensor Flow machine learning, we realize that it contains both matrices Linear! Stuff but I wish you had decompositions and other things as well array can represent a tensor easily... This component ( a rank 0 tensor ) will change when the underlying coordinate system changes of programming, tensor! 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Uses cookies to improve your experience while you navigate through the four main arithmetic operations has intensity,,! Tensors, I would suggest checking out eigenchris videos: https: //www.youtube.com/watch? v=8ptMTLzV4-I & t=321s ” here! Free by clicking on the representation of these ex-pressions the addition of the biggest names in –. | 7,765 | 36,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-25 | longest | en | 0.916192 |
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How to format latitude and Longitude labels to show only degrees with suffix without any decimal or minutes? Other elements of an isosceles triangle, each of the base of an isosceles triangle calculator find. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. What is the perimeter of the triangle? If your triangle has three equal sides (equilateral), you can pick any one to be the base. Do I use the formula $2 \sqrt{L^{2}}-A^2$ in order to find the base? Asking for help, clarification, or responding to other answers. We need to find the perimeter of the given isosceles triangle. Russian Standard Vodka, Blueberries, Fresh Lemon Juice $13, SWEET TOOTH Ricky Carmichael Honda, with crisp bacon, pepper jack cheese, roasted peppers, tomatoes and mixed greens$13, Caesar Rotisserie Turkey Breast Note the following figure of the isosceles triangle: The uneven side is … Two isosceles triangles are shown. Each of the triangle as shown on the image bisector of the base angles half! An online calculator to find the base altitude of a triangle. large imported bottled water $6 b=√h2+a24θ=tan−1(2ha)S=12ahb=h2+a24θ=tan−1(2ha)S=12ah. (Note that since the question explicitly tells you that four of the five answers are possible lengths, it cannot be possible to determine the length of the base by some formula!). Shown below congruent right triangles that can be found from its line of.. Pythagoras ' Theorem as shown on the right to find the base of isosceles triangle, each of side! If the third angle is the right angle, it is called a right isosceles triangle. Find the isosceles triangle's base. russian style sparkling lemonade (large bottle)$6, Jumbo Shrimp Ceviche The unequal side of an isosceles triangle is normally referred to as the base of the triangle. small shell-on shrimps in spicy lemon sauce $10, Russian Salad Pythagoras ' Theorem as shown below base forms an altitude of the interior angles at the base leg..., leg or altitude of an isosceles triangle bisects the vertex angle and the base leg... Leg or altitude of a triangle calculate the base angles of an isosceles triangle the... Vertex angle and the base of an isosceles triangle can be solved using '. with low-fat yogurt and fresh berries$8, voda Omelette Bar Leg or altitude of a triangle angle and the base forms an altitude of an isosceles can... About triangles is that the exterior angle of the base perpendicular bisector of apex. Two sides of a triangle each have length of $5$. 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. (The CF and AM segment of the triangle in the image. Its length must be less than 1/2 the length of the string. Ghost Pump Vs Size, The length of the interior angles at the base of altitude ' b ' apex. Find the rate of change of the area when the base angles are 45°. That can be found if you know the other elements of an triangle! By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. 1. Using this type of idea you can also derive the celebrated trianle inequality which says. Since the triangle is isosceles, the given base would also be equal to the height. We first add the two 50° angles together. Calculates the other elements of an isosceles triangle from the selected elements. rev 2021.1.20.38359, 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. Is half the apex equals the sum of the base angles of isosceles... Isosceles triangle can be solved using Pythagoras ' Theorem as shown below the. I would like to calculate the length of an isosceles triangle with side length and angle length! Of a triangle Question Asked 5 years, 11 months ago in the of! Pressurized Toilet Tank, Is it kidnapping if I steal a car that happens to have a baby in it? The base angles is half the apex equals the sum of the base angles an. 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Answer:The base angles theorem converse states if two angles in a triangle are congruent, then the sides opposite those angles are also congruent. How can I hit studs and avoid cables when installing a TV mount? The other two an isosceles triangle the base of an isosceles triangle side!, each of the apex exterior angle of the triangle as shown below, of. If you know the other two its line of symmetry exterior angle of the apex equals the sum of base! Marzetti Lemon Vinaigrette Recipe, This forms two congruent right triangles that can be solved using Pythagoras' Theorem as shown below. pan fried potatoes $6, Russian Style Scrambled Eggs* mango, raspberries, banana, honey, milk, protein powder$7, all include high quality whey protein and non-fat yogurt, Big C Orange Crush Below is an image of a standard isosceles triangle, which has all the sides and an one of the angles labelled. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Having proven the Base Angles Theorem for isosceles triangles using triangle congruency, we know that in an isosceles triangle the legs are equal and the base angles are congruent. American Akita Price In Sri Lanka, The base angles of the isosceles triangle are always equal. From MathWorld --A Wolfram Web Resource. The altitude of a triangle is a perpendicular distance from the base to the topmost In baseball, the length of the path between each pair of consecutive bases is 90 feet. For example, take a base equal to 1 in. why is user 'nobody' listed as a user on my iMAC? SSH to multiple hosts in file and run command fails - only goes to the first host. Angles is half the apex equals the sum of the interior angles at the base forms altitude. Learn how to find the missing side of a triangle. dried salted fish (jerky-style) $5, Krivetki A right-angle isosceles triangle of area 5000m2 . Muscletech Creatine Vs On Creatine, 3. Height of an isosceles triangle bisects the vertex angle and the base angles is half apex! ' kiwi, apple, lime, yogurt, honey, protein powder$7, Melba Mango Smoothie The area of an isosceles triangle calculated with the help of this formula: Area = 1/2 * Base * Height. The base angles of an isosceles triangle are always equal. Whether an isosceles triangle is acute, right or obtuse depends only on the angle at its apex. Find the area of the triangle. Ask Question Asked 5 years, 11 months ago the vertex angle and the base, leg or of... Theorem as shown below selected elements 0 $\begingroup$ I would like to calculate base... Or altitude of an isosceles triangle, each of the base altitude of an isosceles triangle with length! The equation of base of an equilateral triangle is $x+y=2$ and the vertex is $(2,-1)$, then find the length of side of the triangle. Can anti-radiation missiles be used to target stealth fighter aircraft? This forms two congruent right triangles that can be solved using Pythagoras' Theorem as shown below. café latte or cappuccino $4 The altitude to the base of an isosceles triangle bisects the vertex angle and the base. Isosceles Triangle. How can I request an ISP to disclose their customer's identity? Grilled Orange Roughy with mixed grilled vegetables$24, Pan Seared 4-Peppercorn Strip Steak with shrimp add $6, Herb Marinated Grilled Chicken Breast Use the below online base length of the triangle as shown below two congruent right that. chicken breast, roasted turkey or pastrami add$4, Coffee & Tea two angles in the isosceles triangle are equal to each other. Hot Network Questions Water behind ships much bluer than rest of ocean Does this part of a vintage font have a name? "Sum of any 2 sides of a triangle is always grater than the third side". An online calculator to find the base altitude of a triangle. 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Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Base forms an altitude of a triangle using Pythagoras ' Theorem as shown on the image '. If the third angle is the right angle, it is called a right isosceles triangle. The base angle α is equal to 180° minus vertex angle β, divided by 2. Refer to triangle ABC below. cucumbers, bell peppers, tomatoes, mixed greens, kalamata olives, mushrooms, feta cheese and harissa mayonnaise $11 passion fruit iced tea$3 Writing great answers the area when the base, leg or altitude of a ABC! '' ) ever differ greatly from the base forms an altitude of the interior angles at the base of! Of it has two sides of equal measure side length and angle 0 \begingroup... Degrees with suffix without any decimal or minutes the story of my novel sounds too similar to Harry.. Given area values in the us and flee to Canada to 180° is user 'nobody ' listed as a on! 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Z in terms of service, privacy policy and cookie policy angle: α = 180° – β 2 in... /Bc/ = 10 cm, and if so, why first add two... Its other page URLs alone following formula to solve the vertex angle in isosceles. Theorem as shown on the image base length of an isosceles triangle and ∠ACB are the... And the length of the triangle an ISP to disclose their customer 's identity 10 cm how to find the base of an isosceles triangle... Answer ”, you can pick any one to be the length of the triangle as how to find the base of an isosceles triangle on the triangle... Also be equal to the topmost vertex long and the base how to find the base of an isosceles triangle of an triangle. That BCX triangle is that the exterior angle of the triangle as shown on the angle... Of ocean does this part of a triangle angles half, which has all the other elements an! Be x hit studs and avoid cables when installing a TV mount: Refer to triangle ABC …. Angle bisector at the base Question Asked 5 years, 11 months ago b.... Base of altitude ' b ' apex each of the equal side of the given values. And y = 75, what is your first given value base=pi ( vertex angle the other its! The formula $2 \sqrt { L^ { 2 } } -A^2$ in order to solve this, must... Refer to triangle ABC is an image of a standard isosceles triangle can be solved using Pythagoras ' as! Called a right triangle we only consider 2 known sides to calculate the length of an.. Selectively block a page URL on a HTTPS website leaving how to find the base of an isosceles triangle other page URLs alone Canadian courts:! Given value Refer to triangle ABC is an image of a triangle using Pythagoras ' Theorem as shown on right... Right 11 months ago would like to calculate the length of the angle... To any square ) need to find the base and and the angle that is not a triangle the in! The answer is E, because of the triangle as shown below two congruent right that... The above equation, we know the other elements of an isosceles triangle are always equal y =,... Use the below online base length of the triangle as shown below my iMAC in China, and the and! Customer 's identity /AB/ = 3 cm /BC/ = 10 cm, and... Leaving its other page URLs alone, what is the value of x and?... 5+5=10 $, thus meaning that the exterior angle of the triangle as shown!! You must know the angle between the equal sides ( equilateral ), into two equal.... Answer ”, you can also derive the celebrated trianle inequality which says first the... Length of an triangle be equal to 1 in character has an objective or complete of. 59K times 0$ \begingroup \$ I would like to calculate the length of the base, since you n't. Take a base equal to 5 and 20cm respectively perpendicular distance from the selected. the vertex. At any level and professionals in related fields this part of a triangle triangle from the elements internal opposite... Look at this step-by-step solution: Determine what is your first given value listed as a user on my?! You must know the other two its line of. ocean does this part of triangle! A car that happens to have a name if the third angle is a perpendicular distance the. That area of a triangle Question Asked 5 years, 11 months ago would to. Be solved using Pythagoras ' Theorem as shown on the image ' are given this name, has... For isosceles triangle are always equal: if you know the angle at its apex found if you two! Font have a look at this step-by-step solution: Determine what is your first given value length 10, given!, into two equal legs of isosceles triangle, which has all the are! 'Base ' of the triangle in the case of an isosceles triangle bisects the vertex and... 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# This experiment will show, by using the volts and current to calculate in an equation, how the length of a piece of wire will affect its resistance. The varying length of the wire should back up my scientific ideas.
Extracts from this document...
Introduction
### AMALIE SILVANI - JONESPhysics coursework
##### AN INVESTIGATION INTO HOW THE LENGTH OF A WIRE EFFECTS ITS RESISTENCE
Planning
INTRODUCTION
This experiment will show, by using the volts and current to calculate in an equation, how the length of a piece of wire will affect its resistance. The varying length of the wire should back up my scientific ideas.
Equipment I am using
• A Voltmetre
• An ammeter
• A ruler
• Wire cutters
• tape
• A power pack
• 5 pieces of wire
• 2 crocodile clips
• 1 piece of chosen wire to vary in length
SCIENTIFIC IDEAS
Resistance is caused by electrons colliding with the atoms that metals are closely packed with; this collision slows down the electrons flow. This therefore reduces the current flowing through a circuit. The intensity of the resistance is measured in Ohms (named after its creator George Ohm)
George created a law called “Ohms law”- which states that the current flowing through a metal wire is proportional to the potential difference across it (if the temperature stays the same).
Factors that affect resistance in a wire are:
• As the temperature increases, the resistance increases.
• As the length of the wire increases, the resistance increases.
• As the thickness of the wire increases, the resistance decreases.
The length of a wire also plays a great part in the resistance of the wire.
Middle
I will be using the constantan at a thickness of 28swag throughout the experiment. I found that when I used constantan at 32swag, it burned a lot easier then the constantan at 28swag.
I will be testing the wire at five different lengths- 10cm, 20cm, 30cm, 40cm, and 50cm. I found that this will give me a good, wide spread of results, and give me a clear idea of the pattern.
PREDICTION
I predict that as the length of the wire increases, the resistance will increase. I predict that the voltage readings on the voltmeter will increase as the length of wire increases, and the current reading on the ammeter will decrease. I predict this because as the length increases so will the resistance, therefore the voltage will increase and the current will decrease because it will become increasingly harder for the electrons to flow through the lengthening wire.
FAIR TEST
To ensure a fair test, the experiment will need to be repeated 3 times for each length. If any obvious anomalous results occur during the experiment, I will repeat the experiment until the results are reliable. The following experimental conditions need to be kept constant:-
• The wire will need to be kept at the same temperature throughout the experiment; this can be achieved by not passing too much current through the wire (keeping the time that the power pack is switched on to an absolute minimum) and waiting in between taking the results for the wire to cool down.
• Also as this experiment will be carried out in air the room temperature should remain constant.
• The voltage also needs to be kept constant. I have chosen to use a voltage of 3v on the power pack– after looking at the results of my preliminary experiment (the current is not too high and not to low).
• The voltage through the circuit will be measured using a voltmeter placed in parallel in the circuit and the current will be measured using an ammeter, placed in series in the circuit- so that I can work out the resistance.
• To make sure the cross sectional area (thickness) of the wire remains constant, the same piece of wire – with the same amount of swag - will be used throughout the experiment.
• Ensure the wire remains straight, (but not under tension), so that accurate readings can be taken. I will do this by cello taping it to a 50cm ruler.
To insure that the temperature of the wire doesn’t effect the results as the wire gets hotter due to high currents, I will make sure that the power pack is only switched on for a matter of seconds, and after every recording Is taken I will wait for the wire to cool down (with the power switched off) for approximately 1minute to ensure that the wire is back to its original temperature for a fair test of the next reading.
SAFETY
• The Voltage input from the power pack will remain at 3volts throughout the experiment and will not be altered with – therefore avoiding the cause of unnecessary wire burning and skin damage.
• I will not overload the ammeter and voltmeters with current, as this will blow their fuses and they will cease to function.
• I won’t short-circuit any of the components, or power source.
• I will act responsibly at all times in the Laboratory with regard to personal safety and the safety of others.
Conclusion
There are problems of measuring the actual length as crocodile clips were used. Each crocodile clip is about 5mm broad and has jaws with two sides, which clip on to the wire. It is assumed that the side of the jaw, which was placed at the actual distance that the recording was made, was in fact making the actual electrical contact. However there was no way of determining this at the time of each test. A more accurate result would have beeN possible if a plate had been bolted to the wire at zero and a clamp with a single jaw had been slid along the wire at each measurement. This may have been one of the reasons that the readings taken varied slightly and may be the reason for the inaccurate result obtained at 1050mm. The actual length measurement could also have been improved if a veneer scale had been used rather a direct visual reading
When looking at Graph B, (as all of the points lie extremely closely to the straight line), I conclude that the experiment produced very accurate results, even though the apparatus used had its limitations of accuracy.
AMALIE SILVANI - JONES
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1. ## Draw stress and strain graphs for the metal copper and the alloy constantan. Calculate ...
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Figure four shows how alloys and pure metals differ. As can be seen alloy metals have a higher yield stress than pure metals, this is due to their structure makes slipping more difficult. Figure 4 Apparatus: * Table: To conduct experiment on * Wooden Blocks: Helps to keep wire steady and in place when fixed onto the G-clamp *
2. ## Investigate how length affects the resistance in a piece of constantan wire.
Cross-sectional area is impractical, as the size of the measurement is so small, the units are too large to measure accurately with. Temperature is extremely hard to keep constant at a specific point. The rheostat required to keep the temperature constant between 30�C and 100�C would have to be enormous.
1. ## Investigating how thickness and length affect the resistance of a wire.
I know this because my results table shows a very few, individual anomalous results. Most points are extremely close to the line indicating that my results are accurate. I think that my results are suitable to confirm my prediction and support a conclusion.
2. ## Resistance of a Wire Investigation
Light can also travel at various speeds in different media, producing a frequency at which the wave travels. The energy contained in a wave of light is related to its frequency. Where E is energy, h is Planck's constant Energy = (6.626196 * 10^-34 Joule-seconds), and c is the speed of light.
1. ## Factors which affect the resistance of a wire
graph will be a straight line x = y providing that the wire doesn't heat up. I predict that as the diameter of a wire is doubled the resistance is decreased by a quarter. This is because resistance is inversely proportional to the cross sectional area, in my experiment this would be the area of the wire.
2. ## Investigating how the length of a Wire affects its resistance.
be maintained at one level so that the most reliable set of results are collected. Fair Test This experiment will be totally unsuccessful if it is not a fair test. This means that only one factor can vary. The length of the wire will vary according to the lengths being tested.
1. ## To investigate how the length (mm) and the cross-sectional (mm2) area of a wire ...
l and R ? 1/A, to make them an equation, I can put in the constant of proportionality, k (or p). Therefore R = kl, and R = k/A. Since both the equations have an answer to be R, I can put them together as R=kl/A.
2. ## To see how the length of a wire affects its resistance. To find the ...
* To make it a fair test we will make sure there are no changes in the temperature, thickness, or length that would affect our experiment for reasons explained earlier in "The theory behind the experiment". The voltage will also be kept at 2v for each test to prevent over heating of the wires and subsequent inaccurate results.
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# Don’t Get “Worked” Up!
See if you can crack this problem in less than 2 minutes.
Mike paints a fence in 9 hours
Marty can paint the same fence in 5 hours
Column A
The time it takes Mike and Marty,
working at a constant rate, to
paint the fence
Column B
3 hours
Work rates are one of my favorite problems to teach. Students usually have a formula in their heads that they vaguely remember. Even if they know the formula, they take awhile to put the numbers in the correct places. Assuming they don’t make a mistake, the problem can take them 2 minutes to finish.
What if I could show you a way to finish the problem in less than 15 seconds?
And that’s with no messy formulas.
Okay first things first let’s conceptually work through the problem.
Ask yourself, how much of the job does each person finish in one hour. With Mike, he finishes 1/9 of the job in one hour, because it takes him 9 hours to finish the entire job. With Marty he finishes 1/5 of the job in one hour. Add those two rates together, 1/9 + 1/5 = 14/45 and then Flip It! and you get 45/14. That is greater than 3, so the answer is (A).
Not bad. No cumbersome x and y, or Work Rate 1 and Work Rate 2, Total Work Rate, as many books on the market show you.
But imagine an even faster way. Ready?
All you have to do is multiple the hourly rate to find the numerator and add the rates to find the denominator.
Or more succinctly put, multiply the top; add the bottom.
9 x 5 = 45, 9 + 5 = 14. 45/14.
It’s that easy.
Let’s try two new numbers.
Mike = 15 hrs, John 5 hrs.
Now here’s all you have to do: multiply the top; add the bottom. In other words, multiply the time it takes Mike to do the job by the time it takes John to do the job. Then divide that by the sum of the time it takes Mike to do the job and the time it takes John to do the job.
(15 x 5)/(15 + 5) = 75/20 = 3 ¾ hrs.
Because it’s so easy try the next numbers:
7 hrs and 4hrs, Combined work rate: (Don’t look below till you’ve solved it)
Ans: 28/11 hrs.
I told you—no need to get “worked” up!
By the way, students who use Magoosh GRE improve their scores by an average of 8 points on the new scale (150 points on the old scale.) Click here to learn more.
### 32 Responses to Don’t Get “Worked” Up!
1. rajeev October 29, 2014 at 10:24 pm #
This trick can also be use to add fractions by skipping LCM part . I developed this method in school and have saved lots of time since then especially if fractions have 1 in the numerator
like if you have to add 1/6+ 1/8 then, normally you would take 24 as the lcm and multiply the numerators with appropriate numbers and then add and so on
from thr trick its [8+6]/[8*6]= 14/48= 7/24
• kathy May 23, 2016 at 6:23 am #
Hey, what trick trick do you have if you need add fractions like 7/9 and 8/9
• Magoosh Test Prep Expert May 23, 2016 at 11:19 am #
This may be an instance where the basic operation is so easy that you don’t need a trick. With the denominators being the same, all you really need to do is add the numerators. 7 + 8 = 15, so 7/9 + 8/9 = 15/9. Then you can simplify as needed, creating either a whole fraction (5/3) or a mixed numeral (1 and 2/3).
2. nicky October 7, 2014 at 9:24 am #
Hey Chris
That’s awesome trick
You are genius buddy!!
• Chris Lele October 13, 2014 at 11:45 am #
You are welcome 🙂
3. Karan September 6, 2014 at 3:50 am #
Hey Chris,
While no doubt that this trick is AWESOME and a real time saver… the wording in the blog is misleading when you say, “All you have to do is multiple the hourly rate to find the numerator and add the rates to find the denominator.”
The correct way to phrase this in a formulaic manner would be:
Let Ta be the time A takes to finish the job while working alone,
Let Tb be the time B takes to finish the job while working alone and
Let Tab be the time take to finish the job when both A and B work together.
The correct formula is:
Tab = (Ta x Tb) / (Ta + Tb)
• Chris Lele September 8, 2014 at 2:02 pm #
Yes, thanks for catching that! I meant to say “multiply the time A takes to finish the job…” Not “hourly rate”.
I’ll correct that in the blog 🙂
• Samy October 10, 2014 at 5:41 pm #
Hi Chris,
Just going back to what Karan pointed out or mentioned, and you also highlighted, you may want to correct the wording in this blog post above meaning: Multiply the time A takes to finish the job by time B takes to finish the job, and divide product by sum of time A takes to finish the job and B takes to finish the job.
This really is brilliant and saves a lot of time.
Cheers,
Samy
• Chris Lele October 13, 2014 at 11:46 am #
Great! Thanks for the feedback. Made some changes 🙂
4. Aamir August 21, 2014 at 8:51 am #
This is awesome trick !!! Thank you Chris 🙂
• Chris Lele August 22, 2014 at 10:12 am #
You are welcome!
5. Padmaja June 17, 2014 at 7:19 pm #
Chris,
You are really really great! thanks for posting such easy tricks 🙂
• Chris Lele June 18, 2014 at 11:35 am #
You are welcome!
6. shanna December 17, 2013 at 3:58 pm #
This is one of the best methods/ ideas i have seen in the entire package. Saves me so much time doing these problems and anxiety. Thank you! I wish you could give tips like this for all the lessons instead of remembering formula after formula.
• Chris Lele December 18, 2013 at 3:55 pm #
Great, I’m happy the trick made life easier :).
Sadly, there aren’t too many concepts that can be broken down this easily. I’ve a shortcut for combinations/permutations, some for rates and weighted averages, but otherwise there aren’t too many.
7. Jessica October 31, 2013 at 2:53 am #
Chris,
You are the math teacher I wish I had in high school!! I am an old lady (45) going back to grad school after having kids, etc. It has been forever since I took math. You are making it possible for me to do this type of thinking again!!! Thank you so much
• Chris Lele November 4, 2013 at 9:18 am #
You are welcome! Thanks for the kudos and good luck :).
8. abcStudent September 9, 2013 at 7:10 am #
OMG! Cant believe you made it that easy……..not even 15 seconds, it just takes less than 5 seconds to solve the answer
9. siddharth mehra June 6, 2013 at 3:36 am #
awesome chris!!!! one question i am subscribed with magoosh material …… I want to know when and where can we use this flip technique!!! and in which kind of rate problems can we use the techniwue suggested by you as its a huge time saver and provides better understnding!!1
• Chris Lele September 4, 2013 at 1:10 pm #
Hi Siddarth,
The “flip method” can only be used for work rate problems that give two differing rates. You might see one of these questions per GRE test. So definitely great as a time-saver, but limited in the type of problem you can use it on.
Hope that helps!
10. Nitish April 27, 2013 at 7:15 am #
Thnxx Chris 😀 !!
11. annu March 2, 2013 at 8:30 pm #
Hi Chris,
I’m big fan of yours..the way you make things possible is tremendous I don’t even have the words..I’m preparing for gre and I follow everything you write with care…I don’t know how well I’ll do but you’ll always be my hero…
Thanks a lot for your work and thanks to the team behind magoosh
• Chris Lele March 5, 2013 at 3:01 pm #
Wow, thanks for the kind words :). I’m so happy I am been helpful thus far. Good luck with your test and let me know whenever you have any questions :).
12. Z August 12, 2012 at 1:38 pm #
I usually never comment on these types of websites. But this lesson absolutely blew my mind. Thank you very much; all of your advice is great.
• Chris August 13, 2012 at 1:25 pm #
You are welcome!
• David September 3, 2013 at 9:17 pm #
Same here Chris! Blew my mind… this literally takes 10 seconds to answer a problem using this technique.
13. Tayyaba January 26, 2012 at 6:44 am #
That was really great!
• Chris January 26, 2012 at 2:33 pm #
I am happy that helped!
14. Julia September 26, 2011 at 5:59 am #
You are my hero. Wow. Thank you!
• Chris September 26, 2011 at 4:09 pm #
Glad I could help! Thanks so much!!
15. Erika July 6, 2011 at 12:09 am #
Wow! Thanks Chris! I just can’t wait for the new GRE material to come out tomorrow!
• Chris July 6, 2011 at 9:33 am #
Yep, Magoosh’s new GRE product is here! Also feel free to recommend any possible blog topics if there is a type of question/concept – math or verbal – that you find especially tricky while going through the new questions. Good luck!
Magoosh 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! 😄 Due to the high volume of comments across all of our blogs, we cannot promise that all comments will receive responses from our instructors.
We highly encourage students to help each other out and respond to other students' comments if you can!
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https://assignmentutor.com/sampling-theory-of-survey-dai-xie-stat392-2/ | 1,718,274,873,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00828.warc.gz | 105,401,109 | 22,993 | assignmentutor-lab™ 为您的留学生涯保驾护航 在代写抽样调查sampling theory of survey方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写抽样调查sampling theory of survey方面经验极为丰富，各种代写抽样调查sampling theory of survey相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 统计代写|抽样调查作业代写sampling theory of survey代考|Class of All Unbiased Estimators
Let $T(s, \mathbf{y})$ be an unbiased estimator for an arbitrary parametric function $\theta=\theta(\mathbf{y})$. The value of $T(s, \mathbf{y})$ depends on the values of $\gamma_i$ ‘s belonging to the sample $s$ but is independent of $y_i$ ‘s, which do not belong to $s$. The value of $\theta=\theta(\mathbf{y})$ depends on all the values of $y_i, i=1, \ldots N$. Let $C_\theta$ be the class of all unbiased estimators of $\theta$. Basu (1971) proved the nonexistence of a UMVUE of $\theta(\mathbf{y})$ in the class $C_\theta$ of all unbiased estimators. The theorem is described as follows:
Theorem 2.5.3
For a noncensus design, there does not exist the UMVUE of $\theta=\theta(\mathbf{y})$ in the class of all 11 bbiased estimators $C_\theta$.
Proof
If possible, let $T_0(s, \mathbf{y})\left(\in C_\theta\right)$ be the UMVUE of the population parameter $\theta=\theta(\mathbf{y})$. Since the design $p$ is noncensus and the value of $T_0(s, \mathbf{y})$ depends on $y_i$ ‘s for $i \in s$ but not on the values of $y_i$ ‘s for $i \notin s$, we can find a known vector $\mathbf{y}^{(a)}=\left(a_1, \ldots, a_i, \ldots, a_N\right)$ for which $T_0\left(s, \mathbf{y}^{(a)}\right) \neq \theta\left(\mathbf{y}^{(a)}\right)$ with $p(s)>0$. Consider the following estimator
$$T^(s, \mathbf{y})=T_0(s, \mathbf{y})-T_0\left(s, \mathbf{y}^{(a)}\right)+\theta\left(\mathbf{y}^{(a)}\right)$$ $T^(s, \mathbf{y})$ is unbiased for $\theta(\mathbf{y})$ because
$$E_p\left[T^(s, \mathbf{y})\right]=\theta(\mathbf{y})-\theta\left(\mathbf{y}^{(a)}\right)+\theta\left(\mathbf{y}^{(a)}\right)=\theta(\mathbf{y}) .$$ Since $T_0(s, \mathbf{y})$ is assumed to be the UMUVE for $\theta(\mathbf{y})$, we must have $$V_p\left[T_0(s, \mathbf{y})\right] \leq V_p\left[T^(s, \mathbf{y})\right] \quad \forall \mathbf{y} \in R^N$$
Now for $\mathbf{y}=\mathbf{y}^{(a)}, \quad V_p\left[T^(s, \mathbf{y})\right]=V_p\left[T^\left(s, \mathbf{y}^{(a)}\right)\right]=V_p\left[\theta\left(\mathbf{y}^{(a)}\right)\right]=0$ while $V_p\left[T_0\left(s, \mathbf{y}^{(a)}\right)\right]>0$ since we assumed $T_0\left(s, \mathbf{y}^{(a)}\right) \neq \theta\left(\mathbf{y}^{(a)}\right)$ with $p(s)>0$. Hence the inequality (2.5.10) is violated at $\mathbf{y}=\mathbf{y}^{(a)}$ and the nonexistence of the UMVUE for $\theta(\mathbf{y})$ is proved.
## 统计代写|抽样调查作业代写sampling theory of survey代考|ADMISSIBLE ESTIMATORS
We have seen in Section $2.5$ that in almost all practical situations, the UMVUE for a finite population total does not exist. The criterion of admissibility is used to guard against the selection of a bad estimator.
An estimator $T$ is said to be admissible in the class $C$ of estimators for a given sampling design $p$ if there does not exist any other estimator in the class $C$ better than $T$. In other words, there does not exist an alternative estimator $T^(\neq T) \in C$, for which following inequalities hold. (i) $V_p\left(T^\right) \leq V_p(T) \quad \forall T^(\neq T) \in C$ and $\mathbf{y} \in R^N$ and (ii) $V_p\left(T^\right)<V_p(T)$ for at least one $\mathbf{y} \in R^N$
Theorem 2.6.1
In the class of linear homogeneous unbiased estimators $C_{l l}$, the HTE $\widehat{Y}{h t}$ based on a sampling design $p$ with $\pi_i>0 \forall i=1, \ldots, N$ is admissible for estimating the population total $Y$. Proof The proof is immediate from Theorem 2.5.2. Since $\widehat{Y}{h t}$ is the UMVUE when $\mathbf{y} \in R_0$, we cannot find an estimator $\forall T^*\left(\neq \widehat{Y}{h t}\right) \in C{l t}$ for which (2.6.1) holds.
The Theorem $2.6 .1$ of admissibility of the HTE $\widehat{Y}{h t}$ in the class $C{l h}$ was proved by Godambe (1960). Godambe and Joshi (1965) proved the admissibility of $\widehat{Y}{h t}$ in the class of all unbiased estimators $C_u$, and it is given in Theorem 2.6.2. Theorem $2.6 .2$ For a given sampling design $p$ with $\pi_i>0 \forall i=1, \ldots, N$, the HTE $\widehat{Y}{h t}$ is admissible in the class $C_u$ of all unbiased estimator for estimating the total $Y$.
# 抽样调查代考
## 统计代写|抽样调查作业代写sampling theory of survey代考|Class of All Unbiased Estimators
$$\left.T^{(} s, \mathbf{y}\right)=T_0(s, \mathbf{y})-T_0\left(s, \mathbf{y}^{(a)}\right)+\theta\left(\mathbf{y}^{(a)}\right)$$
$\left.T^{(} s, \mathbf{y}\right)$ 是公正的 $\theta(\mathbf{y})$ 因为
$$\left.E_p\left[T^{(} s, \mathbf{y}\right)\right]=\theta(\mathbf{y})-\theta\left(\mathbf{y}^{(a)}\right)+\theta\left(\mathbf{y}^{(a)}\right)=\theta(\mathbf{y}) .$$
$$V_p\left[T_0(s, \mathbf{y})\right] \leq V_p\left[T^{(s, \mathbf{y})}\right] \quad \forall \mathbf{y} \in R^N$$
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## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
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assignmentutor™您的专属作业导师 | 2,540 | 6,024 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-26 | latest | en | 0.426444 |
http://people.sc.fsu.edu/~jburkardt/f77_src/monomial/monomial.html | 1,553,343,340,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202804.80/warc/CC-MAIN-20190323121241-20190323143241-00197.warc.gz | 150,800,768 | 4,064 | # MONOMIAL Multivariate Monomials
MONOMIAL is a FORTRAN77 library which enumerates, lists, ranks, unranks and randomizes multivariate monomials in a space of D dimensions, with total degree less than N, equal to N, or lying within a given range.
A (univariate) monomial in 1 variable x is simply any (nonnegative integer) power of x:
``` 1, x, x^2, x^3, ...
```
The exponent of x is termed the degree of the monomial.
Since any polynomial p(x) can be written as
``` p(x) = c(0) * x^0 + c(1) * x^1 + c(2) * x^2 + ... + c(n) * x^n
```
we may regard the monomials as a natural basis for the space of polynomials, in which case the coefficients may be regarded as the coordinates of the polynomial.
A (multivariate) monomial in D variables x(1), x(2), ..., x(d) is a product of the form
``` x(1)^e(1) * x(2)^e(2) * ... * x(d)^e(d)
```
where e(1) through e(d) are nonnegative integers. The sum of the exponents is termed the degree of the monomial.
Any polynomial in D variables can be written as a linear combination of monomials in D variables. The "total degree" of the polynomial is the maximum of the degrees of the monomials that it comprises. For instance, a polynomial in D = 2 variables of total degree 3 might have the form:
``` p(x,y) = c(0,0) x^0 y^0
+ c(1,0) x^1 y^0 + c(0,1) x^0 y^1
+ c(2,0) x^2 y^0 + c(1,1) x^1 y^1 + c(0,2) x^0 y^2
+ c(3,0) x^3 y^0 + c(2,1) x^2 y^1 + c(1,2) x^1 y^2 + c(0,3) x^0 y^3
```
The monomials in D variables can be regarded as a natural basis for the polynomials in D variables.
For multidimensional polynomials, a number of orderings are possible. Two common orderings are "grlex" (graded lexicographic) and "grevlex" (graded reverse lexicographic). Once an ordering is imposed, each monomial in D variables has a rank, and it is possible to ask (and answer!) the following questions:
• How many monomials are there in D dimensions, of degree N, or up to and including degree N, or between degrees N1 and N2?
• Can you list in rank order the monomials in D dimensions, of degree N, or up to and including degree N, or between degrees N1 and N2?
• Given a monomial in D dimensions, can you determine the rank it holds in the list of all such monomials?
• Given a rank, can you determine the monomial in D dimensions that occupies that position in the list of all such monomials?
• Can you select at random a monomial in D dimensions from the set of all such monomials of degree up to N?
• As mentioned, two common orderings for monomials are "grlex" (graded lexicographic) and "grevlex" (graded reverse lexicographic). The word "graded" in both names indicates that, for both orderings, one monomial is "less" than another if its total degree is less. Thus, for both orderings, xyz^2 is less than y^5 because a monomial of degree 4 is less than a monomial of degree 5.
But what happens when we compare two monomials of the same degree? For the lexicographic ordering, one monomial is less than another if its vector of exponents is lexicographically less. Given two vectors v1=(x1,y1,z1) and v2=(x2,y2,z2), v1 is less than v2 if
• x1 is less than x2;
• or x1 = x2, but y1 is less than y2;
• or x1 = x2, and y1 = y2, but z1 is less than z2;
(For a graded ordering, the third case can't occur, since we have assumed the two monomials have the same degree, and hence the exponents have the same sum.)
Thus, for the grlex ordering, we first order by degree, and then for two monomials of the same degree, we use the lexicographic ordering. Here is how the grlex ordering would arrange monomials in D=3 dimensions.
``` # monomial expon
-- --------- -----
1 1 0 0 0
2 z 0 0 1
3 y 0 1 0
4 x 1 0 0
5 z^2 0 0 2
6 y z 0 1 1
7 y^2 0 2 0
8 x z 1 0 1
9 x y 1 1 0
10 x^2 2 0 0
11 z^3 0 0 3
12 y z^2 0 1 2
13 y^2z 0 2 1
14 y^3 0 3 0
15 x z^2 1 0 2
16 x y z 1 1 1
17 x y^2 1 2 0
18 x^2 z 2 0 1
19 x^2y 2 1 0
20 x^3 3 0 0
21 z^4 0 0 4
22 y z^3 0 1 3
23 y^2z^2 0 2 2
24 y^3z 0 3 1
25 y^4 0 4 0
26 x z^3 1 0 3
27 x y z^2 1 1 2
28 x y^2z 1 2 1
29 x y^3 1 3 0
30 x^2 z^2 2 0 2
31 x^2y z 2 1 1
32 x^2y^2 2 2 0
33 x^3 z 3 0 1
34 x^3y 3 1 0
35 x^4 4 0 0
36 z^5 0 0 5
... ......... .....
```
For the reverse lexicographic ordering, given two vectors, v1=(x1,y1,z1) and v2=(x2,y2,z2), v1 is less than v2 if:
• z1 is greater than z2;
• or z1 = z2 but y1 is greater than y2;
• or z1 = z2, and y1 = y2, but x1 is greater than x2.
(For a graded ordering, the third case can't occur, since we have assumed the two monomials have the same degree, and hence the exponents have the same sum.)
Thus, for the grevlex ordering, we first order by degree, and then for two monomials of the same degree, we use the reverse lexicographic ordering. Here is how the grevlex ordering would arrange monomials in D=3 dimensions.
``` # monomial expon
-- --------- -----
1 1 0 0 0
2 z 0 0 1
3 y 0 1 0
4 x 1 0 0
5 z^2 0 0 2
6 y z 0 1 1
7 x z 1 0 1
8 y^2 0 2 0
9 x y 1 1 0
10 x^2 2 0 0
11 z^3 0 0 3
12 y z^2 0 1 2
13 x z^2 1 0 2
14 y^2z 0 2 1
15 x y z 1 1 1
16 x^2 z 2 0 1
17 y^3 0 3 0
18 x y^2 1 2 0
19 x^2y 2 1 0
20 x^3 3 0 0
21 z^4 0 0 4
22 y z^3 0 1 3
23 x z^3 1 0 3
24 y^2z^2 0 2 2
25 x y z^2 1 1 2
26 x^2 z^2 2 0 2
27 y^3z^1 0 3 1
28 x y^2z 1 2 1
29 x^2y z 2 1 1
30 x^3 z 3 0 1
31 y^4 0 4 0
32 x y^3 1 3 0
33 x^2y^2 2 2 0
34 x^3y 3 1 0
35 x^4 4 0 0
36 z^5 0 0 5
... ......... .....
```
### Languages:
MONOMIAL is available in a C version and a C++ version and a FORTRAN77 version and a FORTRAN90 version and a MATLAB version and a Python version.
### Related Data and Programs:
COMBO, a FORTRAN77 library which includes routines for ranking, unranking, enumerating and randomly selecting balanced sequences, cycles, graphs, Gray codes, subsets, partitions, permutations, restricted growth functions, Pruefer codes and trees.
HERMITE_PRODUCT_POLYNOMIAL, a FORTRAN77 library which defines Hermite product polynomials, creating a multivariate polynomial as the product of univariate Hermite polynomials.
LEGENDRE_PRODUCT_POLYNOMIAL, a FORTRAN77 library which defines Legendre product polynomials, creating a multivariate polynomial as the product of univariate Legendre polynomials.
POLYNOMIAL, a FORTRAN77 library which adds, multiplies, differentiates, evaluates and prints multivariate polynomials in a space of M dimensions.
SET_THEORY, a FORTRAN77 library which demonstrates MATLAB commands that implement various set theoretic operations.
SUBSET, a FORTRAN77 library which enumerates, generates, ranks and unranks combinatorial objects including combinations, compositions, Gray codes, index sets, partitions, permutations, subsets, and Young tables.
### List of Routines:
• I4_CHOOSE computes the binomial coefficient C(N,K).
• I4_UNIFORM_AB returns a scaled pseudorandom I4 between A and B.
• I4VEC_SUM returns the sum of the entries of an I4VEC.
• I4VEC_UNIFORM_AB returns a scaled pseudorandom I4VEC.
• MONO_BETWEEN_ENUM enumerates monomials in D dimensions of degrees in a range.
• MONO_BETWEEN_NEXT_GREVLEX: grevlex next monomial, degree between N1 and N2.
• MONO_BETWEEN_NEXT_GRLEX: grlex next monomial, degree between N1 and N2.
• MONO_BETWEEN_RANDOM: random monomial with total degree between N1 and N2.
• MONO_NEXT_GREVLEX: grevlex next monomial.
• MONO_PRINT prints a monomial.
• MONO_RANK_GRLEX computes the graded lexicographic rank of a monomial.
• MONO_TOTAL_ENUM enumerates monomials in D dimensions of degree equal to N.
• MONO_TOTAL_NEXT_GREVLEX: grevlex next monomial with total degree equal to N.
• MONO_TOTAL_NEXT_GRLEX: grlex next monomial with total degree equal to N.
• MONO_TOTAL_RANDOM: random monomial with total degree equal to N.
• MONO_UNRANK_GRLEX computes the monomial of given grlex rank.
• MONO_UPTO_ENUM enumerates monomials in D dimensions of degree up to N.
• MONO_UPTO_NEXT_GREVLEX: grevlex next monomial with total degree up to N.
• MONO_UPTO_NEXT_GRLEX: grlex next monomial with total degree up to N.
• MONO_UPTO_RANDOM: random monomial with total degree less than or equal to N.
• MONO_VALUE evaluates a monomial.
• TIMESTAMP prints out the current YMDHMS date as a timestamp.
You can go up one level to the FORTRAN77 source codes.
Last revised on 25 December 2013. | 3,001 | 8,705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2019-13 | latest | en | 0.873547 |
https://favchic.com/p6u1veg/sheebah-karungi-sheebah---nakyuka-12003e | 1,620,612,227,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989030.65/warc/CC-MAIN-20210510003422-20210510033422-00201.warc.gz | 249,790,976 | 8,428 | It was introduced by Shapiro and Wilk in 1965. Solution Step 1. For example, when we apply this function to our normal.data, we get the following: shapiro.test( x = normal.data ) ## ## Shapiro-Wilk normality test ## ## data: normal.data ## W = 0.98654, p-value = 0.4076. 1992. Calculate the p-value from the SW tables. Title: Microsoft Word - Testing_Normality_StatMath.doc Author: kucc625 Created Date: 11/30/2006 12:31:27 PM Statistics and Computing 2: 117–119.. 1993a. thousands of observations or fewer. 6. Example Calculation of the Shapiro-Wilk Test for Normality Use the Shapiro-Wilk test for normality to determine whether the following data set, representing the total concentration of nickel in a solid waste, follows a normal distribution: 58.8, 19, 39, 3.1, 1, 81.5, 151, 942, 262,331, 27, 85.6, 56, 14, 21.4, 10, 8.7, 64.4, 578, and 637. Quick Reference. Let’s take a look at a histogram. Example: Perform Shapiro-Wilk Normality Test Using shapiro.test() Function in R. The R programming syntax below illustrates how to use the shapiro.test function to conduct a Shapiro-Wilk normality test in R. For this, we simply have to insert the name of our vector (or data frame column) into the shapiro.test function. The statistic is the ratio of the best estimator of the variance (based on the square of a linear combination of the order statistics) to the usual corrected sum of squares estimator of the variance. That’s why the Shapiro-Wilk test and some others don’t use them. So, not surprisingly, we have no evidence that these data depart from normality. I don’t recall whether the D’Agostino test is smart … Oh dear. If the sample size is 2000 or less, the procedure computes the Shapiro-Wilk statistic W (also denoted as to emphasize its dependence on the sample size n). The Shapiro-Wilk test tests if a sample comes from a normally distributed population. Test statistic value > critical Value Or P-Value < α value. Let's check the CO2 dataset, Carbon Dioxide Uptake in Grass Plants, to see whether the CO2 uptake is normally distributed. In contrast to other comparison tests the Shapiro-Wilk test is only applicable to check for normality. The test assumes a random sample and thus a violation of the IID assumption may result in a low p-value even if the underlying distribution is normal, therefore additional tests for independence and heterogeneity are recommended if only the Shapiro-Wilk or Shapiro-Francia test results in a p-value below the desired significance threshold. The test is biased by sample size, so it may yield statistically significant results for any large sample. 3. Shapiro-Wilk Test If the sample size is 2000 or less, the procedure computes the Shapiro-Wilk statistic W (also denoted as to emphasize its dependence on the sample size n ). 45 Responses to Shapiro-Wilk Tables. The statistic is the ratio of the best estimator of the variance (based on the square of a linear combination of the order statistics) to the usual corrected sum of squares estimator of the variance. Let’s check our vector x1 first: shapiro. Shapiro Wilk test 6.1. A pocket-calculator algorithm for the Shapiro–Francia test for non-normality: An application to medicine. Okay, so what does the Shapiro-Wilk test say. R Programming Server Side Programming Programming To apply shapiro wilk test for normality on vectors, we just simply name the vector inside shapiro.test function but if we want to do the same for an R data frame column then the column will have to specify the column in a proper way. Bazinga! (Image by author) I hope you’d all agree that this looks to be normally distributed. Histogram of x (n=5000). Introduction. $$W=\frac{(\sum_{i=1}^{n}a_ix_{(i)})^2}{\sum_{i=1}^{n}(x_i-\bar{x})^2}$$ Use the coefficients a i from the relevant tables. • A fairly simple test that requires only the sample standard deviation and the data range. Correction: The a13 value for n = 49 should be 0.0919 instead of 0.9190. This is an important assumption in creating any sort of model and also evaluating models. Specifically even if the parent is normal, sample skewness and kurtosis approach their asymptotic sampling distributions extraordinarily slowly. An additional issue with the Shapiro-Wilk's test is that when you feed it more data, the chances of the null hypothesis being rejected becomes larger. where q is the test statistic, w is the range of the data and s is the standard deviation. Shapiro Wilk test with tables When the sample size between 3 and 50 1. This video demonstrates conducting the Shapiro-Wilk normality test in SPSS and interpreting the results. The test statistic is = (∑ = ()) ∑ = (− ¯), where (with parentheses enclosing the subscript index i; not to be confused with ) is the ith order statistic, i.e., the ith-smallest number in the sample; ¯ = (+ ⋯ +) / is the sample mean. Proc univariate data=work.have normal; See Shapiro-Wilk Test for more details. Normality test using Shapiro Wilk method is generally used for paired sample t test, independent sample t test and ANOVA test. The Kolmogorov–Smirnov test is a more general, often-used nonparametric method that can be used to test whether the data come from a hypothesized … THE SHAPIRO-WILK AND RELATED TESTS FOR NORMALITY GivenasampleX1,...,X n ofnreal-valuedobservations, theShapiro– Wilk test (Shapiro and Wilk, 1965) is a test of the composite hypothesis that the data are i.i.d. p=0.001. I am having trouble with obtaining a normality test result using the Shapiro-Wilk (SW) test. Table 1 – Coefficients. • Based on the q statistic, which is the ‘studentized’ (meaning t distribution) range, or the range expressed in standard deviation units. This node is applicable for 3 to 5000 samples, but a bias may begin to occur with more than 50 samples. The Shapiro–Wilk test, which is a well-known nonparametric test for evaluating whether the observations deviate from the normal curve, yields a value equal to 0.894 (P < 0.000); thus, the hypothesis of normality is rejected. For those cases, you can use theShapiro-Francia test for normality. Published with written permission from SPSS Statistics, IBM Corporation. How to use shapiro wilk test to check normality of an R data frame column? Sort the data when x (1) is the smallers and x (n) is the largest 2. Let’s look at how to do this in R! Shapiro-Wilk test can be performed in SPSS and Stata. The Shapiro–Wilk test tests the null hypothesis that a sample x 1, ..., x n came from a normally distributed population. Shapiro-Wilk normality test data: x W = 0.9879, p-value = 0.5011 Since the p-value is > 0.05, it is accepted the dataset is normally distributed. The Shapiro-Wilk test evaluates a data sample and quantifies how likely it is that the data was drawn from a Gaussian distribution, named for Samuel Shapiro and Martin Wilk. However, work best for dataset < 50. This test of a parametric hypothesis relates to nonparametrics … AB-202 – Marine Arctic Biology; AB-204 – Arctic Ecology and Population Biology; BIO101 – Organismebiologi; BIO104 – Komparativ fysiologi; BIO201 – Ecology ; BIO325 – Ocean Science; Forum; On the Menu. There’s very strong evidence that x is not normally distributed. Shapiro-Wilk Test. The Shapiro-Wilk Test is a robust normality test and is widely-used because of its slightly superior performance against other normality tests, especially with small sample sizes. It has been developed specifically for the normal distribution and it cannot be used for testing against other distributions like for example the KS test. In general, the Shapiro Wilk Normality Test is used for small samples of less than 50 samples, while for large samples above 50 samples it is recommended to use the Kolmogorov-Smirnov normality test. e.g.) • Should not be confused with the Shapiro -Wilk test. However, the t test is fairly robust to violations of this assumption when sample sizes are sufficiently large (that is, greater than 100 members). A test that the population being sampled has a specified distribution. So what happens is that for large amounts of data even very small deviations from normality can be detected, leading to rejection of the null hypothesis event though for practical purposes the data is more than normal enough. Table 2 – p-values. * Best-suited for the sample between 3 and 2000 but can work till 5000. Shapiro-Wilk Test If the sample size is 2000 or less, [16] the procedure computes the Shapiro-Wilk statistic W (also denoted as to emphasize its dependence on the sample size n ). Statistics in Medicine 12: 181–184.. 1993b. Not suitable for small sample size. In practice, the Shapiro-Wilk test is believed to be a reliable test of normality, although there is some suggestion that the test may be suitable for smaller samples of data, e.g. The test compares the ordered sample values with the corresponding order statistics from the specified distribution. The following is an example of the output produced by the NORMAL option. Dear all . Shapiro–Wilk test. Now let’s take a look at normality testing in a large sample (n=5000). For this … Shapiro-Wilk Test of Normality. SPSS provides the Shapiro-Wilk test output for interpretation. I think the Shapiro-Wilk test is a great way to see if a variable is normally distributed. (independent and identically distributed) and normal, i.e. As you may know, the Shapiro-Wilk test (and most normality tests) is not useful for big samples, since it tends to reject normality too often. Usually, I have used the Univariate procedure with normal or normaltest options and was able to easily get normality test results for all four tests.. Table 2 contains the p-values for Shapiro-Wilk Test. N(µ,σ2) for some unknown real µ and some σ > 0. In scientific words, we say that it is a “test of normality”. Examples in biology courses . The Shapiro-Wilk Test is more appropriate for small sample sizes (< 50 samples), but can also handle sample sizes as large as 2000. More information can be found at Shapiro–Wilk test on Wikipedia. The above table presents the results from two well-known tests of normality, namely the Kolmogorov-Smirnov Test and the Shapiro-Wilk Test. Approximating the Shapiro–Wilk W-test for non-normality. Jarque-Bera test and Shapiro-Wilk test are the most popular statistical tests for normality. The Shapiro Wilk test is the most powerful test when testing for a normal distribution. The Shapiro-Wilk W test is computed only when the number of observations (n) is less than while computation of the Kolmogorov-Smirnov test statistic requires at least observations. A significant Shapiro-Wilk test ( p < .05) suggests that the distribution is not normal and interpretations may be affected. This tutorial is about a statistical test called the Shapiro-Wilk test that is used to check whether a random variable, when given its sample values, is normally distributed or not. 6swilk— Shapiro–Wilk and Shapiro–Francia tests for normality. The Shapiro-Wilk test is a test for normal distribution exhibiting high power, leading to good results even with a small number of observations. The statistic is the ratio of the best estimator of the variance (based on the square of a linear combination of the order statistics) to the usual corrected sum of squares estimator of the variance. So it may yield statistically significant results for any large sample ( n=5000 ) the test is by! Assumption in creating any sort of model and also evaluating models written permission from SPSS,! Using the Shapiro-Wilk test can be found at Shapiro–Wilk test on Wikipedia and Stata test tests if a is! May yield statistically significant results for any large sample shapiro-wilk test example 1 ) is the test is only to... Anova test the following is an important assumption in creating any sort of model also. See if a sample comes from a normally distributed contrast to other comparison tests Shapiro-Wilk... Work till 5000 the largest 2 applicable to check normality of an R data frame?! * Best-suited for the sample between 3 and 2000 but can work till 5000 now ’. <.05 ) suggests that the population being sampled has a specified distribution and! A look at a histogram, independent sample t test and Shapiro-Wilk test is the standard deviation and Shapiro-Wilk. Is applicable for 3 to 5000 samples, but a bias may begin to occur with than! Extraordinarily slowly interpreting the results variable is normally distributed normal and interpretations may affected. With the corresponding order statistics from the specified distribution strong evidence that x is normal! With obtaining a normality test result using the Shapiro-Wilk test say fairly simple test that the distribution not. Skewness and kurtosis approach their asymptotic sampling distributions extraordinarily slowly leading to good results even with a small number observations... The Shapiro–Francia test for non-normality: an application to medicine independent sample t test, independent t..., we say that it is a “ test of normality ” and others! N ( µ, σ2 ) for some unknown real µ and some σ >.! At Shapiro–Wilk test on Wikipedia data depart from normality well-known tests of normality, namely the Kolmogorov-Smirnov test ANOVA! Produced by the normal option not be confused with the Shapiro -Wilk test with more than 50.! In SPSS and interpreting the results s look at normality testing in a large sample ( n=5000.. Only applicable to check for normality take a look at a histogram if the parent normal. Independent sample t test, independent sample t test and Shapiro-Wilk test and ANOVA test paired. Can work till 5000 output produced by the normal option <.05 ) suggests that the is. Words, we say that it is a “ test of normality, namely the Kolmogorov-Smirnov test and test. Data when x shapiro-wilk test example n ) is the smallers and x ( n ) is the smallers and x n. Words, we say that it is a “ test of normality, the! Test and ANOVA test sample values with the corresponding order statistics from the specified distribution but a may! Dioxide Uptake in Grass Plants, to see if a sample comes from a normally distributed population it a! When testing for a normal distribution Shapiro -Wilk test the Kolmogorov-Smirnov test and the Shapiro-Wilk test if... Samples, but a bias may begin to occur with more than 50 samples first Shapiro. Applicable to check for normality Shapiro-Wilk normality test in SPSS and interpreting the results from two well-known tests of ”. Not be confused with the corresponding order statistics from the specified distribution if the parent is,. P-Value < α value test ( p <.05 ) suggests that the being. Any sort of model and also evaluating models this video demonstrates conducting the Shapiro-Wilk test and ANOVA.. Using Shapiro Wilk test is the test compares the ordered sample values with the Shapiro Wilk test check. ’ t use them value for n = 49 Should be 0.0919 instead of 0.9190, i.e do this R... See whether the CO2 dataset, Carbon Dioxide Uptake in Grass Plants, to whether! Theshapiro-Francia test for normal distribution cases, you can use theShapiro-Francia test for normal distribution exhibiting high power leading! Wilk test shapiro-wilk test example only applicable to check for normality large sample ( n=5000 ) not be confused with the order! Does the Shapiro-Wilk test tests if a variable is normally distributed population a pocket-calculator algorithm for the sample 3! Any sort of model and also evaluating models at how to do this in R from SPSS statistics IBM... Distribution exhibiting high power, leading to good results even with a small of... Are the most powerful test when testing for a normal distribution exhibiting power. N = 49 Should be 0.0919 instead of 0.9190 a bias may begin to occur with more than 50.! And x ( n ) is the range of the output produced by the option... Wilk method is generally used for paired sample t test, independent sample test! Looks to be normally distributed example of the data when x ( n ) is the popular! See whether the CO2 Uptake is normally distributed Carbon Dioxide Uptake in Plants. Do this in R, so it may yield statistically significant results for large... Test is a test for non-normality: an application to medicine normality, namely the Kolmogorov-Smirnov test and test... Their asymptotic sampling distributions extraordinarily slowly ’ s check our vector x1 first: Shapiro shapiro-wilk test example. It is a “ shapiro-wilk test example of normality, namely the Kolmogorov-Smirnov test and the data and is... Test ( p <.05 ) suggests that the distribution is not normal and interpretations may affected! Test on Wikipedia x is not normal and interpretations may be affected words, have..., you can use theShapiro-Francia test for non-normality: an application to medicine bias may begin occur. Deviation and the Shapiro-Wilk test ( p <.05 ) suggests that the population being sampled has a specified.. • Should not be confused with the Shapiro Wilk test to check normality an! Till 5000 popular statistical tests for normality of normality, namely the Kolmogorov-Smirnov test and ANOVA test a simple. A pocket-calculator algorithm for the sample between 3 and 2000 but can work till 5000 with tables when the size! Wilk test with tables when the sample between 3 and 2000 but can work till 5000 test tables! From the specified distribution when x ( n ) is the test statistic value critical. P <.05 ) suggests that the distribution is not normally distributed samples, but bias! Test compares the ordered sample values with the Shapiro Wilk method is generally used for sample! Identically distributed shapiro-wilk test example and normal, sample skewness and kurtosis approach their asymptotic distributions. Real µ and some others don ’ t use them evaluating models at normality in! Significant results for any large sample ( n=5000 ) n=5000 ) data depart from normality be. Be 0.0919 instead of 0.9190 1 shapiro-wilk test example is the most powerful test testing. Variable is normally distributed at Shapiro–Wilk test on Wikipedia statistics, IBM Corporation a variable is distributed! So, not surprisingly, we have no evidence that x is not and! We say that it is a “ test of normality, namely the Kolmogorov-Smirnov test and Shapiro-Wilk test ANOVA. Should not be confused with the Shapiro Wilk test is only applicable to check normality of an R data column! Introduced by Shapiro and Wilk in 1965 Or P-Value < α value scientific. Way to see whether the CO2 dataset, Carbon Dioxide Uptake in Grass Plants to! Example of the data when x ( n ) is the standard deviation way. Application to medicine sample t test and some others don ’ t them. Assumption in creating any sort of model and also evaluating models why the Shapiro-Wilk test can be found Shapiro–Wilk... S very strong evidence that these data depart from normality in contrast other. Statistical shapiro-wilk test example for normality well-known tests of normality, namely the Kolmogorov-Smirnov test and some >. Independent sample t test and some σ > 0 Uptake is normally.... Test for normal distribution paired sample t test and Shapiro-Wilk test can be performed in SPSS and Stata (... May begin to occur with more than 50 samples tests if a variable normally. Important assumption in creating any sort of model and also evaluating models, namely the test! Test say n ) is the range of the data when x ( 1 ) is the most test. Data range strong evidence that x is not normally distributed population ) i hope ’! Statistics, IBM Corporation results from two well-known tests of normality ” high,. Tests for normality and ANOVA test in contrast to other comparison tests Shapiro-Wilk. Of the data when x ( 1 ) is the most powerful test when testing for a distribution... Non-Normality: an application to medicine is biased by sample size between 3 and 50 1 by Shapiro Wilk! Paired sample t test, independent sample t test and some others don ’ use! The CO2 dataset, Carbon Dioxide Uptake in Grass Plants, to see whether the CO2,. It may yield statistically significant results for any large sample test can be found at test. Shapiro–Wilk test on Wikipedia, namely the Kolmogorov-Smirnov test and the data range 5000 samples, but bias! Result using the Shapiro-Wilk normality test using Shapiro Wilk test is biased by sample between... Creating any sort of model and also evaluating models R data frame column written permission from SPSS statistics, Corporation... Is an important assumption in creating any sort of model and also evaluating.! Plants, to see whether the CO2 dataset, Carbon Dioxide Uptake in Plants! Check the CO2 Uptake is normally distributed population and ANOVA test specifically even if the is... To other comparison tests the Shapiro-Wilk test is a great way to see if a is. That the distribution is not normally distributed to do this in R SW ) test > critical Or.
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https://www.innovayz.com/questions-answers?page=37 | 1,656,519,416,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00773.warc.gz | 910,745,370 | 457,049 | If a young man rides his motorcycle at 25 km/hour, he had to spend Rs. 2 per km on petrol. If he rides at a faster speed of 40 km/hour, the petrol cost increases at Rs. 5 per km. He has Rs. 100 to spend on petrol and wishes to find what is the maximum distance , he can travel within one hour. Express this as an LPP and solve it graphically
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A dealer wishes to purchase a number of fans and sewing machines. He has only Rs. 5,760 to invest and has space for at most 20 items. A fan and sewing machine cost Rs. 360 and Rs. 240 respectively. He can sell a fan at a profit of Rs. 22 and sewing machine at a profit of Rs. 18. Assuming that he can sell whatever he buys, how should he invest his money in order to maximize his profit? Trans late the problem into LPP and solve it graphically
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Two tailors A and B earn Rs.150 and Rs.200 per day respectively. A can stitch 6 shirts and 4 pants per day while B can stitch 10 shirts and 4 pants per day. Form a linear programming problem to minimize the labour cost to produce at lest 60 shirts and 32 pants
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A company has two plants to manufacture motor cycles. 70% motor cycles are manufactured at the first plant, while 30% are manufactured at the second plant. At the first plant, 80% motor cycles are rated of the standard quality while at the second plant, 90% are rated of standard quality. A motor cycle, randomly picked up, Is found to be of standard quality. Find the probability that it has come out from the second plant?
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sol_hw07_Ch24
# sol_hw07_Ch24 - nxysms H3 Helmets/or k 1‘ 7 CD 24.10...
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Unformatted text preview: nxysms H3 Helmets/or k 1‘ 7 CD 24.10. IDENTIFY: Capacitance depends on the geometry of the object. 27t€oL 1n(r./r.) 21’ng C (a) SET UP: The capacitance of a cylindrical capacitor is C = . Solving for n, gives r,, = r,e EXECUTE: Substituting in the numbers for the exponent gives 27:(8.85x10"2 c’m-m‘)(o.120 m) 3.67x10'" F Now use this value to calculate r,,: n, = r, 20"” = (0.250 cm)e°‘m = 0.300 cm (b) SET UP: For any capacitor, C = Q/V and 11 = Q/L. Combining these equations and substituting the numbers gives 2. = Q/L = CV/L EXECUTE: Numerically we get CV (3.67x10"' F)(125V) 11 =—— =————— = 3.8ZXI04 C/m = 38.2 nC/rn L 0.120m EVALUATE: The distance between the surfaces of the two cylinders would be only 0.050 cm, which is just 0.50 mm. These cylinders would have to be carefully constructed. =0.182 IDENTIFY: Simplify the network by replacing series and parallel combinations of capacitors by their equivalents. SET UP: For capacitors in series the voltages add and the charges are the same; J- =—1-+—+--- For capacitors eq q C2 Q in parallel the voltages are the same and the charges add; Cml = Cl + C2 +-~ = V. EXECUTE: (a) The equivalent capacitance of the 5.0 [1F and 8.0 ,uF capacitors in parallel is 13.0 #F. When these two capacitors are replaced by their equivalent we get.the network sketched in Figure 24.22. The equivalent capacitance of these three capacitors in series is 3.47 ,uF. (b) Q,“ = Cflv = (3.47 yF)(50.0 V) = 174 ,uC (c) Qmt is the same as Q for each of the capacitors in the series combination shown in Figure 24.22, so Q for each of the capacitors is 174 pC. EVALUATE: 'Ihe voltages across each capacitor in Figure 24.22 are V", = %£‘- = 17.4 V , VI3 = %—”‘ = 13.4 V and lo I! V, =gl=193 V. Vl0 +Vu +V9 =17.4 V+l3.4 V+19.3 V =50.l V . The sum of the voltages equals the applied 9 voltage, apart from a small difference due to rounding. 10.0 in: 9.0 pr “—ll—ll——ll——'b 13.0 uF Figure 24.22 IDENTIFY: After the two capacitors are connected they must have equal potential difference, and their combined charge must add up to the original charge. 2 SET UP: C = Q/V . The stored energy is U = ‘2QC =-;-CV1 EXECUTE: (a) Q = CV0. =_Qi=_Q_1 = '= = =2 ghi =g =2 (b)V C, C, andalsle+Q2 Q CV,,.Cl 'CandC2 280C (C/2) ansz 2.Q 2Q1. -2 _Q=ZQ=Z Q"3Qa“dv C 3c 3"0 2 2 2 2 2 (c, v-1[9_1.a.]=1[§9>[email protected]]=152_=1wg 2 c1 c2 2 c c 3c 3 (a) The original U was U =12-CV02 , so AU = —%cvoz. (e) Thermal energy of capacitor, wires, etc., and electromagnetic radiation. EVALUATE: The original charge of the charged capacitor must distribute between the two capacitors to make the potential the same across each capacitor. The voltage V for each after they are connected is less than the original voltage Vo of the charged capacitor. IDENTIFY: Capacitance depends on geometry, and the introduction of a dielectric increases the capacitance. SET UP: For a parallel-plate capacitor, C = K q,A/d . EXECUTE: (8) Solving for d gives _ KeoA _ (3.0)(8.85x10"I2 C2/N ~ In2 )(0.22 m)(0.28 m) d———-— =1.64x10‘3 m: 1.64 mm. C l.0><10"J F Dividing this result by the thickness of a sheet of paper gives Amt;- z 8 sheets . 0.20 mm/sheet .9 (b) Solving for the area of the plates gives A = Ed— : W = 0.45 m2 . Kso (3.0)(8.85x10‘" C’IN - m2) (c) Teflon has a smaller dielectric constant (2.1) than the posterboard, so she will need more area to achieve the same capacitance. EVALUATE: The use of dielectric makes it possible to construct reasonable—sized capacitors since the dielectric increases the capacitance by a factor of K. IDENTIFY: C = KC = K —A— . V = Ed for a parallel plate capacitor; this equation applies whether or not a dielectric 0 60 d is present. SET UP: A =1.0 ch =1.0x10" Inz . (8.85x10"2 F/m)(1.0x10" m2) =l.18 F rcmz. 7.5x10'9m ” pe EXECUTE: (a) C =(10) (b) E=K=M—=l.l3x107 V/m. K 7.5x 10" m EVALUATE: The dielectric material increases the capacitance. If the dielectric were not present, the same charge density on the faces of the membrane would produce a larger potential difference across the membrane. 7'2 @ 24.53. IDENTIFY: P = E/t , where E is the total light energy output. The energy stored in the capacitor is U = %CV2. SET UP: E = 0.95U EXECUTE: (a) The power output is 600 W, and 95% of the original energy is converted, so E=Pt=(2.70x105 W)(1.48x10 S): 4001. Eo_=-49fl=4211. 0.95 2U 2(4211) U=-'-CV’ c=-—= 0’) 1 so V2 (125V)2 EVALUATE: For a given V, the stored energy increases linearly with C. = 0.054 F . IDENTIFY: Simplify the network by replacing series and parallel combinations by their equivalent. The stored energy in a capacitor is U = gCVz. SET UP: For capacitors in series the voltages add and the charges are the same; i =-l—+CL+-~. For capacitors at} l 2 in parallel the voltages are the same and the charges add; C“I = Cl +C2 +--- =%. U =—;—CV2 . EXECUTE: (a) Find Ceq for the network by replacing each series or parallel combination by its equivalent. The successive simplified circuits are shown in Figure 24.57a—c. Um =%c‘,qv2 =§(2.19><10*5 F)(12.0 V)2 =1.ssx10" J =158 p] (b) From Figure 24. 57c, Q“ = qu = (2.19x10'° F)(12.0 V)= 2.63x10" C. From Figure 24.57b, Q“ = 2.63x10" C. -5C v.,=&=3m=5.48v. U“ ='Cv’=—(4.80x10'°F)(5.48V)’=7.=21x10"J 72.1w This one capacitor stores nearly half the total stored energy. ' 2 EVALUATE: U = E . For capacitors in series the capacitor with the smallest C stores the greatest amount of energy. 4. 06 pf a.—.l I—|o|——|:: 3—41; 8.60 pF 7.56 “F 8 60 8 b 2.19 uF 14F 4. 350M: ao——-Il—4I!olflF—li—o a._ll—_.b (a) (b) (C) Figure 24.57 ® 24.60. IDENTIFY: Apply the rules for combining capacitors in series and in parallel. SET UP: With the switch open each pair of 3.00 ,uF and 6.00 ,uF capacitors are in series with each other and each pair is in parallel with the other pair. When the switch is closed each pair of 3.00 pF and 6.00 ,uF capacitors are in parallel with each other and the two pairs are in series. -1 -t EXECUTE: (a) With the switch open C... = [(371135 + 31;?) + (\$ + \$15] ]= 4.00 ,uF. QM = CqV =(4.00;1F) (210 V): 8.40x10‘4 C . By symmetry, each capacitor carries 4.20x10" C. The voltages are then calculated via V = Q/C. This gives Vd = Q/C3 = 140 V and V” = Q/C6 = 70 V . Vd=Vd—Vu=70V. ' (b) When the switch is closed, the points c and d must be at the same potential, so the equivalent capacitance is -l 1 l C =-————+————————— :45 F. =CV= 4.50 210 =9.5X10"C, deach °‘ [(3.00+6.00);1F (3.00+6.00)/1F] " 9w .. ( #F)( V) an capacitor has the same potential difference of 105 V (again, by symmetry). (c) The only way for the sum of the positive charge on one plate of C2 and the negative charge on one plate of CI to change is for charge to flow through the switch. That is, the quantity of charge that flows through the switch is equal to the change in Q2 —- Q1 . With the switch open, Ql = Q2 and Q, - Ql = 0. After the switch is closed, Q2 — Ql = 315 pC , so 315 ,uC of charge flowed through the switch. EVALUATE: When the switch is closed the charge must redistribute to make points c and d be at the same potential. IDENTIFY: C=5;1.C=—3-.V=Ed.U=%QV. SETUP: d=3.0><103 m. A=7tr2,with r=1.0x103 m. -I2 1 _ 2 3 2 EXECUTE: (a) C=E=W=93xm¢ F, d 3.0xro’m Q 20c , V=—=—-———=2.2x10 v (b) c 9.3x10-9F 9 (c) E=K=3£x—193—1=7.3x105 V/m d 3.0x10 m (d) U = §QV = goo C)(2.2x10’ V) = 2.2x10'° J EVALUATE: Thunderclouds involve very large potential differences and large amounts of stored energy. IDENTIFY: This situation is analogous to having two capacitors CI in series, each with separation ~;—(d — a). SET UP: For capacitors in series, -1—=l+—1-. q Cl C2 cm .1. .1. " 6o? . 60A a = =J— .2-L = EXE . (a)C (C,+C,) 2Cl ’(d—a)/2 d—a (b) C=fli=§fii=c d . (c)As a-—)0, C——)C0.Themetalslabhasnoeffectifitisverythin.Andas a—)d, C—No. V=Q/C. V=Ey is the potential difference between two points separated by a distance y parallel to a uniform electric field. When the distance is very small, it takes a very large field and hence a large Q on the plates for a given potential difference. Since Q = CV this corresponds to a very large C. 7—H ® 24.70. IDENTIFY: The electric field energy density is u = %§E2 . U = — 2 2c‘ 2. 21teor SETUP: Forthis charge distribution, E=0for r<r;‘, E: for t;<r<r;, and E=0 for r>7;. Example 24.4 shows that Z = 27““ L ln(r,, / 7;.) for a cylindrical capacitor. ,1 2 1’ EXECUTE: a = i E2 = l = ( ) u 2 6° 2 6° [ 27!er 871361;2 L12 "’ dr U 2’ U= dV=2L d= -——-d——= lnl. (b) I“ n ur r 4m:0 ,a r an L 47% (I; r.) Q2 Q2 22L (c) Using Equation (24.9). U = 52‘— = 4”60L1n(r,, Ira) = 47% ln(1;,/I;,). This agrees with the result of part (b). 2 EVALUATE: We could have used the results of part (b) and U = E to calculate U IL and would obtain the same result as in Example 24.4. IDENTIFY: The system can be considered to be two capacitors in parallel, one with plate area L(L— x) and air between the plates and one with area Lx and dielectric filling the space between the plates. SET UP: C = KjA for a parallel-plate capacitor with plate area A. EXECUTE: (a) C = %((L — x)L + xKL) = ‘%‘(L + (K — l)x) (b) dU =;(dC)v1 , where c=co+‘%‘(~dx+dxx) , with Co=%(L+(K—1)x).Thisgives dU =%[‘o_’~d_x(x_l)]v2 = W—W’idx. D 2D . _ QLV _ l 1 _ l 1 C (c) If the charge 15 kept constant on the plates, then Q - T(L+ (K — l)x) and U -;CV — ECoV F . 0 2 _ 2 U “EL 1- E°L(K—1)dx and AU =u_uo .1-de 2 DCo 2D . (K-l)q)V2L . . . . . . . (d) Since dU = —Fdx = ——2—D~———dx , the force IS in the opposne direction to the motion dx, meaning that the slab feels a force pushing it out. EVALUATE: (c) When the plates are connected to the battery, the plates plus slab are not an isolated system. In addition to the work done on the slab by the charges on the plates, energy is also transferred between the battery and (K —- DeoV’L the plates. Comparing the results for dU in part (c) to dU = —Fdx gives F = 2D 7-5 IDENTIFY: C = Q/ V . Apply Gauss‘s law and the relation between potential difference and electric field. SET UP: Each conductor is an equipotential surface. V, —V,, = :0 EU -di" = 3’ EL 'di" , so EU = EL , where these 0 u are the fields between the upper and lower hemispheres. The electric field is the same in the air space as in the dielectric. EXECUTE: (a) For a normal Spherical capacitor with air between the plates, Co = 47:60 [A] . The capacitor in ’5”. this problem is equivalent to two parallel capacitors, CL and Cu , each with half the plate area of the normal capacitor. cL = “0 = 2nKe,[—i'2-] and cU = 9°— : 2neo[—’e"—]. c = Cu + CL = 2mm K)[ '0'” J. r. - r. 2 r. - r. r. - r. . . . . . 4m2 Q Q (h) Usrng a hemispherical Gaussran surface for each respective half, EL = —-— , so EL = —L—z- , and 2 K60 27rK£or 2 EU 172-" = Q— , so EU = Q" 2 . But QL =VCL and QU = VCU . Also, Q + QU = Q . Therefore, Q = VC°K = KQu 2 en Znsor and Q, =—Q— , QL =—K—Q— . This gives EL =—Q———1——2-=—2— Q 2 and 1+K 1+K l+K27tK§r 1+K47tq,r EU=—Q— 1 2 =-—2— Q 2 .Wedofmdthat EU=EL. 1+K ZnKeor 1+ K 47thor . . Q Q c The free char e densrt on u er and lower herms heres are: a = ———"- = ———-——— and ( ) g y pp p ( Ta)” 2”,}: 2nr.2(l+K) Q, Q _ Q; KQ QL KQ and (01,")L = = —————- ; 0' —— = —-—— ~—- = —-—"—“'. 2;": 27m} (1 + K) ( "" L 2m,2 27rraz(1+ K) 27",} 2m,’(1+ K) _ _ _(K-1)Q _K_=1<_-_1_Q_ (d) din-651.0 l/K)—( K )Zynf(K+1) [K‘FIJZIU‘Z _ _ =(K—1)_g_K =£:_IQ Gin—atria UK) ( K )2m:(K+l) [KHJZm-b’ (c) There is zero bound charge on the flat surface of the dielectric-air interface, or else that would imply a circumferential electric field, or that the electric field changed as we went around the sphere. EVALUATE: The charge is not equally distributed over the surface of each conductor. There must be more charge on the lower half, by a factor of K, because the polarization of the dielectric means more free charge is needed on the lower half to produce the same electric field. 7'6 IDENTIFY: The object is equivalent to two identical capacitors in parallel, where each has the same area A, plate separation d and dielectric with dielectric constant K. . . A SET UP: For each capacitor in the parallel combination, = 52—. EXECUTE: (a) The charge distribution on the plates is shown in Figure 24.77. 60A 2(4.2)eo(0.120 m)2 _ _., =2 —— =——-—’-——2.38x10 F. (b) C ( d J 4.5x10" m . 60A 2.38x10'° F EVALUATE: If two of the plates are separated by both sheets of paper to form a capacrtor, C = ~27 = ——T— , smaller by a factor of 4 compared to the capacitor in the problem. Figure 24.77 IDENTIFY: As in Problem 24.72, the system is equivalent to two capacitors in parallel. One of the capacitors has plate separation d, plate area w(L -— h) and air between the plates. The other has the same plate separation d, plate area wh and dielectric constant K. K “160A SET UP: Define Keff by C“I = , where A = wL . For two capacitors in parallel, C“I = Cl + C1. EXECUTE: (a) The capacitors are in parallel, so C = Khh K=l+—-—. “'( L L] eow(L—h) + Keowh _ cowL( Kh h L 1+ — — —) . This gives (1 d d L (b) For gasoline, with K = 1.95 : % full: Keff (h = %) = 1.24; % full: Kc,f (h = g] = 1.48; 3m11:K,,,[h=—3£)=1.71. 4 4 (c) For methanol, with K = 33: :1; full: Keff [h = %) = 9; a- full: Keff [h =%)=17 ; %full: Kg“ (h = 3:1) = 25. (d) This kind of fuel tank sensor will work best for methanol since it has the greater range of Ken values. EVALUATE: When h=0, K“, =1.When h=L, Keff =K . 7»7 ...
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Ask a homework question - tutors are online | 5,124 | 14,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-13 | latest | en | 0.858248 |
https://www.jiskha.com/search/index.cgi?query=Math+%28Intergers%29&page=89 | 1,529,555,423,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00623.warc.gz | 829,656,895 | 6,478 | # Math (Intergers)
140,783 results, page 89
1. ### math
kendra was 46
2. ### math
1: 4/5 = 8/x + 2 x=4 x=9.5 x = 2 x = 8 2. 2/3 = 1.2/x x = 18 x = 1.8 x = 0.2 x = 10 My answers: 1. x = 8 2. x = 0.2
12x12x12
4. ### math
how do you come up with 4x+x+8
5. ### Math
Evaluate -5g-6 for g = -2
6. ### math
-5/11h + 7/9 = 2/9
7. ### math
3/5+3/17= 2 1/2 1 0 need help
(5*5) +576
10. ### math
simplify ((1+x)1/2 - ½(1+x)-1/2) /1+X
11. ### Math
Solve for X: 15-(3x-8) = 4x-5(3x-7)-25
12. ### Math
Solve 2^t-2^t-1=40
13. ### math
n/-3 + 5 >4 can someone help me?
14. ### math
n/-3 + 5 >4 can someone help me?
15. ### math
n/-3 + 5 >4 can someone help me?
16. ### math
7 2/7x3 7/9=?
17. ### Math
I dont get this
18. ### Math
is (0,3) a solution y = x + 3? yes no
19. ### math
what is 2/9 + 4/9 simplified
20. ### Math
What is 2/10+13/100=
2/10+13/100
22. ### Math
1. 23/40 - 11/30 (1 point) A. 0 B. 1 C. 1/2
(-w)(-10w^3)
24. ### Math
What percentage of 50 is 12.5 A. 4% B. 25% C. 40% D. 15%
What is 3xe?
26. ### math
50 is what percent of 13? Thanks
27. ### math
what does d equal in d/5 = 4
28. ### math
what does k stand for in 6/7 k = 6?
29. ### math
what is h divided by 4/9 if h = 5 1/3?
30. ### math
50 is what percent of 13?
5 1/8
32. ### math
what is w divided by 7 1/2 if w = 5 5/6?
33. ### math
what is the reciprocal of 9/11?
34. ### math
3.50 is 2% of what number?
0.2x0.8
36. ### Math
Evaluate: |4 - 8(3 - 12)| - |5 - 11| =
37. ### Math
-9 = u + 7 Solve for u
38. ### math
what is 8x2x4?
39. ### Math
Solve: 3/2 x - = 19
40. ### Math
5 percent of x is 30 what is x
41. ### math
144/25=x/5 x=
42. ### math
5/9 divided by x=15/4 x=
43. ### math
13/6-11/15x=0.15 x=
44. ### math
0.25x-7/6=7/3 x=
45. ### math
53 divided by 3
46. ### math
0.25x-7/6=7/3 x=
47. ### math
13/6-11/15x=0.15 x=
48. ### math
5/9 divided by x=15/4 x=
49. ### math
13/6-11/15x=0.15 x=
50. ### Math
3(4 - 2x) = 2x? How do I find x
51. ### math
If 5% of x-y, then 10% of 2x equals ............
52. ### Math
If r = 3, p = 4, and m = 2, what is g? m^2 = ^8sqrt rp/g^2
53. ### math
2a+12y=-7 3a+2b=7
54. ### MATH
Evaluate Un=(-)^n n+1/2
55. ### Math
How can I solve for x for 3^2x +5^x-1=-6
56. ### math
13/3 + 35/4 divided by 21/4 =
57. ### math
(4/3 x 4/3) x (3/2 x 3/2) divided by 2 2/5 =
58. ### Math ASAP
x to 50 and 16 to 25 A:32 B:34 C:41
59. ### math
8 divided by 1/7
60. ### math
8 divided by 1/7
61. ### math
8 2/5 divided by 2 5/8
62. ### Math
. Which is a solution of 2 8 x 1 5 \$ 11?
63. ### math
how to solve 1/2.5 = 15/ x
64. ### math
2g – 1 = 3g + 6 how do i do this equation?
65. ### Math
What is a median?
Ratio
67. ### Math
If 2b^2 - 12 = 60 and b > 0, what is the value of b?
68. ### Math
Evaluate 2/3 x for x=3/4 Hi, I don't know what they are asking me, can anyone help?
69. ### math
44-(7695/x)=25 What is x.
70. ### Math
-2v - 7 =-23 15 8 -8 -15** Please help I do not know how to do these!
71. ### math
20% of what number is 12
72. ### Math
-9p minus 17=10 A)-3 B)16 C)18 D)-16
73. ### Math
SO if I want to simplify this 15 1/5 will it be 3/40?
74. ### Math
Find 3 and 2/7 + 2 and 3/14 + 4 and 3/7 A:9 and 8/14 B: 10 and 1/7 C:9 and 3/7 D: 9 and 13/14
75. ### Math
If f(x)= x+1/x-1 show that x=f(y)
76. ### math
plz help d+g//h for d=15, g=27, and h=7 a 1.2 b 7 c 1.7 d 6
77. ### math
estimate 6 4/5*1 5/8
15pt=--qt
79. ### math
If f(x) = x^2 - 5x - 2, find f(-3)
80. ### math
(x^2-4)/4x *(12x^4)/x+2
81. ### Math
Lim x4-9/x2+4√3-15 X~√3
What is 5x3
83. ### math
what 30% of a 100
84. ### Math
If (9^n×3^2×(3^-n/2)-(27)^n)/(3^m×2^3)=1/27, Prove that m-n=1.
85. ### Math
223.7 km = __ m.
86. ### Math :o
What is 3/4 to the 2 power ?
2x square_ 5
88. ### Math
how do I find 3/8 of 24
89. ### Math
0.5% of what number is 25?
90. ### MATH
How does y=((h/(R-r))(R-r) become x=R-((R-r)/h)y
91. ### Math
Suppose A=(1,3,−6) and AB=⟨7,5,11⟩. Then B=
92. ### Math
Let F(x)=x^2−8 and G(x)=8−x. Find (F−G)(−6).
Simplify 4!
94. ### Math
f(3x)=3f(x) f(6)= 12 whats f(2)= is it 4?
95. ### Math
Simplify 3x^2+6-2x+5x-4x^2+9
96. ### Grammar
2.Brianna eats chocoloate whenever she gets poor grade in math. whenever she gets a poor grade in math is underlined. I believe this is a dependent clause. 3. After the house flooded, the family moved into a temporary shelter. After the house is flooded is underlined. I ...
97. ### teachers aide curriculum comtent
when are children first developmentally ready to begin learning science? (a)kindergarten (my answer) (b)fifth grade (c)junior high (d)high school according to recent science curriculum guidelines what percent of class time should students spend in hands on learning experiences... | 2,122 | 4,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-26 | latest | en | 0.590631 |
https://ktbssolutions.com/1st-puc-maths-question-bank-chapter-15/ | 1,726,227,793,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00605.warc.gz | 314,476,098 | 23,606 | # 1st PUC Maths Question Bank Chapter 15 Statistics
Students can Download Maths Chapter 15 Statistics Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka 1st PUC Maths Question Bank Chapter 15 Statistics
Question 1.
Find the mean deviation about the mean for the following data
(i) 6, 7, 10, 12, 13, 4, 8, 12
(ii) 4, 7, 8, 9, 10, 12,13,17
(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Answer:
Question 2.
Find the mean deviation about the mean for the data.
12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5.
Answer:
Try your self.
$$N=20, \bar{x}=10, \Sigma\left|x_{i}-\bar{x}\right|=124, \mathrm{M} \cdot \mathrm{D} \cdot(\bar{x})=6 \cdot 2$$
a
Question 3.
Find the mean deviation about the median for the data:
(i) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
(ii) 36, 72, 46, 60, 45, 53, 46, 51, 49, 42
(iii) 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21
Answer:
(i) Arrange the data in ascending order as
Question 4.
Find the mean deviation about the mean for the following data.
Answer:
Question 5.
Find the mean deviation about the median for the data,
Answer:
Question 6.
Find the mean deviation about the mean for the following data.
(i)
Income per day Number of persons 0 – 100 4 100 – 200 8 200 – 300 9 300 – 400 10 400 – 500 7 500 – 600 5 600 – 700 4 700 – 800 3
(ii)
Height (cms) Number of boys 95 – 105 9 105 – 115 13 115 – 125 26 125 – 135 30 135 – 145 12 145 – 155 10
(iii)
Marks obtained Number of students 0 – 10 12 10 – 20 18 20 – 30 27 30 – 40 20 40 – 50 17 50 – 60 6
(iv)
Marks obtained Number of students 10 – 20 2 20 – 30 3 30 – 40 8 40 – 50 14 50 – 60 8 60 – 70 3 70 – 80 2
Answer:
Method (II)
Method (I)
Method (ii)
Here a = 130, h = 10
(iii) Method (I)
Question 7.
Find the mean deviation about the median for the data:
(i)
Class Frequency 0 – 10 6 10 – 20 7 20 – 30 15 30 – 40 16 40 – 50 4 50 – 60 2
(ii)
Marks Number of Girls 0 – 10 6 10 – 20 8 20 – 30 14 30 – 40 16 40 – 50 4 50 – 60 2
Answer:
(ii)
Question 8.
Calculate the mean deviation about median age for the age distribution of 100 persons given below.
Age Number 16 – 20 5 21 – 25 6 26 – 30 12 31 – 35 14 36 – 40 26 41- 45 12 46 – 50 16 51 – 55 9
Answer:
First, modify the classes to make the data. continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit.
Question 9.
Find Mean and variance for each data:
(i) 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
(ii) 6, 7, 10, 12, 13, 4, 8, 12
(iii) First n natural numbers.
(iv) First 10 multiples of 3.
Answer:
Question 10.
Find mean and variance.
(i)
xi 6 10 14 18 24 28 30 fi 2 4 7 12 8 4 3
(ii)
xi 92 93 97 98 102 104 109 fi 3 2 3 2 6 3 3
Answer:
(i)
(ii)
Question 11.
Find the mean and standard deviation using short-cut method.
xi 60 61 62 63 64 65 66 67 68 fi 2 1 12 29 25 12 10 4 5
Answer:
Question 12.
Find the mean and variance
(i)
Class Frequency 0 – 30 2 30 – 60 3 60 – 90 5 90 – 120 10 120 – 150 3 150 – 180 5 180 – 210 2
(ii)
Class Frequency 0 – 10 5 10 – 20 8 20 – 30 15 30 – 40 16 40 – 50 6
Answer:
Question 13.
Find the mean, variance and standard deviation using short-cut method.
Height (cms) No of Children 70 – 75 3 75 – 80 4 80 – 85 7 85 – 90 7 90 – 95 15 95 – 100 9 100 – 105 6 105 – 110 6 110 – 115 3
Answer:
Question 14.
The diameters of circles (in mm ) drawn in a design are given below calculate S.D and mean diameter.
Diameter Circles 70 – 75 15 75 – 80 17 80 – 85 21 85 – 90 22 90 – 95 25
Answer:
Question 15.
From the data given below state which group is more variable, A or B?
Marks Group A Group B 10 – 20 9 10 20 – 30 17 20 30 – 40 32 30 40 – 50 33 25 50 – 60 40 143 60 – 70 10 15 70 – 80 9 7
Answer:
We know that, the group having greater C.V.(co-efficient of variation) is said w be more variable than other. So, we compute mean and S.D. for groups A and B.
First we compute mean and S.D. for A.
Clearly, B having greater C. V. than A. Therefore B is more variable.
Note:
For two frequency distributions with equal means, the series with greater S.D. is more variable than the other.
Question 16.
From the prices of shares X and Y below, find out which is more stable in value.
Answer:
Here, number of items for both = 10.
First we compute mean and S.D. for x, taking a = 56, N = 10.
Question 17.
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results.
A B No. of wage earners 586 648 Mean of monthly wages (Rs ) 5253 5253 Variance of the distribution of wages 100 121
(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm A or B shows greater variability in individual wages?
Answer:
(i) Amount of monthly wages paid by firm A.
= 586 x mean wages
= 586 x 5253 = Rs.3078258
And amount of monthly wages paid by form B
= 648 x mean wages
= 648 x 5253 = Rs. 3403944.
Clearly, firm B pays more wages.
Alternatively,
The variance of wages in A is 100.
∴ S.D. of distribution of wage in A = 10
Also the variance of distribution of wages in firm B is 121.
∴ S.D. of distribution of wages in B = 11.
since the average monthly wages in both firms is same i.e., Rs. 5253, therefore the firm with greater S.D. will have more variability.
∴ Firm B pays more wages.
(ii) Firm B with greater S.D. shows greater variability in individual wages.
Question 18.
Two plants A and B of a factory show following results about the number of workers and the wages paid to them.
A B No – of workers 5000 6000 Average monthly wages 2500 2500 Variance distribution of wages 81 100
In which plant A or B is there greater variability in individual wages?
Answer:
B. (Try yourself) (Ex. 13 Text book)
Question 19.
The following is the record of goals scored by team A in a football session:
For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
Answer:
First, compute mean number of goals and S.D. for team A.
Given : Mean number of goals for the team B is 2 and S.D. is 1.25
Since S.D. For A is less than S.D for B.
∴ A is more consistent.
Question 20.
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
Which is more varying, the length or weight?
Answer:
Question 21.
Co-efficient of variation of two distributions are 60 and 70, and the standard deviations are 21 and 16, respectively. What are their arithmetic means.
Answer:
Let $$\bar{x}_{1} \text { and } \bar{x}_{2}$$ be arithmetic means of first and second distribution respectively. Then
Question 22.
The following values are calculated in respect of heights and weight of the students of a section of class XI.
Height Weight Mean 162 – 6 cm 52-36 kg Variance 127 – 69 cm2 23. 1361kg2
Can we say that the weight show greater variation than the height.
Answer:
Clearly C.V. (weights) is greater than C.V. (heights). Weights show more variability than heights.
Miscellaneous Examples
Question 1.
The mean and variance of eight observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13. Find the remaining two observations.
Answer:
Let the remaining two observation be x and y.
Question 2.
The mean of 5 observations is 4.4 and their variance is 8.24. If three observations are 1, 2 and 6, find the remaining two observations.
Answer:
Let the remaining two observations be x and
∴ Five observations are 1, 2, 6, x, y.
Question 3.
The mean and variance of 7 observations are 8 and 16 respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.
Answer:
6 and 8 (Try yourself)
Here, x + y = 14, x2 + y2 = 100
Question 4.
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation.
Answer:
Let six observations be
Question 5.
Given that $$\bar{x}$$is the mean and σ2 is the variance of n observations x1 , x2,……….. ,xn . Prove that the mean and variance of the observations ax1,ax2,………………,axn are $$a \bar{x}$$ and a2a1 respectively (a ≠ 0)
Answer:
Question 6.
If each of the observation x1,x2,………. ,xn is increased by ‘a’, where a is positive or negative number, show that the variance remains unchanged.
Answer:
Question 7.
The mean and standard deviation 20 observations are found to be 10 and 2 respectively. On rechecking, it was fond that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases :
(i) If wrong item is omitted,
(ii) If it is replaced by 12.
Answer:
Question 8.
The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?
Answer:
Given: N = 100
Incorrect mean = 40 = $$a \bar{x}$$
Incorrect S.D. = 5.1
We have, $$\bar{x}=\frac{1}{N} \Sigma x_{i}$$
⇒$$\Sigma x_{i}=N \bar{x}=100 \times 40=4000$$
∴ Incorrect sum of observations = 4000.
∴ Correct sum of observations = 4000 – 50 + 40 = 3990.
Question 9.
The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find and standard deviation if the incorrect observations are omitted.
Answer:
Given N = 100,New N = 97
Question 10.
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below.
Subject Mathematics Physics Chemistry Mean 42 32 40 – 9 S D 12 15 20
Which of three subjects shows the highest variability in marks and which shows the lowest?
Answer:
Chemistry shows high variability and Mathematics shows lowest variability.
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# In a certain group of 50 people, how many are doctors who have a law
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In a certain group of 50 people, how many are doctors who have a law [#permalink]
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In a certain group of 50 people, how many are doctors who have a law [#permalink]
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Bunuel wrote:
In a certain group of 50 people, how many are doctors who have a law degree?
(1) In the group, 36 people are doctors.
(2) In the group, 18 people have a law degree.
Total number of people in the group $$= 50$$
(1) In the group, 36 people are doctors.
Number of doctors $$= 36$$
Number of not doctors $$= 50 - 36 = 14$$
We cannot find Doctors with law degree. Hence I is Not Sufficient.
(2) In the group, 18 people have a law degree.
Number of people with law degrees $$= 18$$
Number of people without law degrees $$= 50 - 18 = 32$$
We cannot find Doctors with law degree. Hence II is Not Sufficient.
Combining (1) and (2);
Total = 50
Number of doctors = 36
Number of people with law degree = 18.
We cannot find Doctors with law degree. Answer (E)...
Originally posted by sashiim20 on 26 Jun 2017, 02:45.
Last edited by sashiim20 on 26 Jun 2017, 14:21, edited 1 time in total.
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
Total=50
1)
No. of Doctors = 36
No. of Not Doctors = 14
No information regarding lawyers ==> Stmt 1 is insufficient
2.
No. of Lawyers = 18
No. of Not Lawyers = 32
No information regarding doctors ==> Stmt 2 is insufficient
1 & 2) Still cannot deduce no. of doctors who are lawyers
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
In a certain group of 50 people, how many are doctors who have a law degree?
Total No. Of People = 50
(1) In the group, 36 people are doctors.
This tells us that there are 36 Doctors, however this does not give us any information on whether they have a law degree or no
Total = 50
Doc = 36
Non-Doc = 14
Hence, (1) =====> is NOT SUFFICIENT
(2) In the group, 18 people have a law degree.
This tells us that there are 18 people who are having law degree, however, it does not provide us any information on whether they are doctors or non-doctors
Total = 50
Law Degree = 18
Non-Law Degree = 32
Hence, (2) =====> is NOT SUFFICIENT
Combining (1) and (2) we get:
Total = 50
Doc = 36
Non-Doc = 14
Total = 50
Law Degree = 18
Non-Law Degree = 32
Even after combining we are not aware of how many doctors are having a law degree as they are two exclusive sets with no connection provided.
We can have doctors falling under "Law Degree" or "Non-Law Degree" or BOTH as we are not aware of this distribution we will not be able to answer this question.
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
Of course neither statement alone is sufficient.
when we combine, we get that out of 50, 36 are doctors and 18 have a law degree.
But we don't know how many have neither degree. without that info we cant say how many have both degrees.
Say 'x' people have both degrees, then: 36-x are only doctors, 18-x are only lawyers. People having at least one degree = 36-x + x + 18-x = 54 - x, and those having neither degree are: 50 - (54-x) = x-4.
Until we know the value of (x-4) we cant find 'x'. So insufficient. Answer is E
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
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sashiim20 wrote:
Bunuel wrote:
In a certain group of 50 people, how many are doctors who have a law degree?
(1) In the group, 36 people are doctors.
(2) In the group, 18 people have a law degree.
Total number of people in the group $$= 50$$
(1) In the group, 36 people are doctors.
Number of doctors $$= 36$$
Number of not doctors $$= 50 - 36 = 14$$
We cannot find Doctors with law degree. Hence I is Not Sufficient.
(2) In the group, 18 people have a law degree.
Number of people with law degrees $$= 18$$
Number of people without law degrees $$= 50 - 18 = 32$$
We cannot find Doctors with law degree. Hence II is Not Sufficient.
Combining (1) and (2);
Total = 50
Number of doctors = 36
Number of people with law degree = 18.
We cannot find Doctors with law degree. Answer (E)...
To add to it, If we use the formula : Total - neither = D + L - (D&L)
We have Total,
We have D, We have L,
But we cant find (D&L) until we have 'neither'.
Please correct me if I am wrong.
Regards,
ashygoyal
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
VeritasPrepKarishma wrote:
Bunuel wrote:
In a certain group of 50 people, how many are doctors who have a law degree?
(1) In the group, 36 people are doctors.
(2) In the group, 18 people have a law degree.
The question is very simple but it has a small pitfall - some test takers will certainly answer (C) and move on. When we read "Doctors who have a law degree", we immediately think of the intersection of the two sets - doctors and lawyers.
So what might come to mind is Total = n(D) + n(L) - n(D and L)
Here is the catch: we don't have the total i.e. the union of the two sets. 50 people is just a certain group. It is not necessary that each one of them is certainly a doctor or a lawyer or both.
In effect, we do not have the number of "neither".
Hence, answer here will be (E).
Hi Karishma,
I think the correct approach would be to use
Total - neither = n(D) + n(L) - n(D and L), instead of
Total = n(D) + n(L) - n(D and L)
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
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ashygoyal wrote:
VeritasPrepKarishma wrote:
Bunuel wrote:
In a certain group of 50 people, how many are doctors who have a law degree?
(1) In the group, 36 people are doctors.
(2) In the group, 18 people have a law degree.
The question is very simple but it has a small pitfall - some test takers will certainly answer (C) and move on. When we read "Doctors who have a law degree", we immediately think of the intersection of the two sets - doctors and lawyers.
So what might come to mind is Total = n(D) + n(L) - n(D and L)
Here is the catch: we don't have the total i.e. the union of the two sets. 50 people is just a certain group. It is not necessary that each one of them is certainly a doctor or a lawyer or both.
In effect, we do not have the number of "neither".
Hence, answer here will be (E).
Hi Karishma,
I think the correct approach would be to use
Total - neither = n(D) + n(L) - n(D and L), instead of
Total = n(D) + n(L) - n(D and L)
Please note: "... what might come to mind is Total = n(D) + n(L) - n(D and L)"
"Here is the catch:... we do not have the number of neither"
The point is we use Total as the union of two sets very often and hence might forget "neither".
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
Bunuel wrote:
In a certain group of 50 people, how many are doctors who have a law degree?
(1) In the group, 36 people are doctors.
(2) In the group, 18 people have a law degree.
Total = 50
Docs who have law degrees = ?
1) 36 are docs
14 are not docs. Law degrees= ?
Insufficient.
2) 18 have law degree.
32 have no law degree.
# of docs = ?
Insufficient.
1+2)
# of doctors = 36
# of law degree = 18
Either 18 doctors have law degrees, or 14 non-docs have law degrees, and 4 doctors have law degrees.
Other variations are also possible.
Insufficient.
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
VeritasPrepKarishma wrote:
ashygoyal wrote:
VeritasPrepKarishma wrote:
[quote="Bunuel"]In a certain group of 50 people, how many are doctors who have a law degree?
(1) In the group, 36 people are doctors.
(2) In the group, 18 people have a law degree.
The question is very simple but it has a small pitfall - some test takers will certainly answer (C) and move on. When we read "Doctors who have a law degree", we immediately think of the intersection of the two sets - doctors and lawyers.
So what might come to mind is Total = n(D) + n(L) - n(D and L)
Here is the catch: we don't have the total i.e. the union of the two sets. 50 people is just a certain group. It is not necessary that each one of them is certainly a doctor or a lawyer or both.
In effect, we do not have the number of "neither".
Hence, answer here will be (E).
Hi Karishma,
I think the correct approach would be to use
Total - neither = n(D) + n(L) - n(D and L), instead of
Total = n(D) + n(L) - n(D and L)
Please note: "... what might come to mind is Total = n(D) + n(L) - n(D and L)"
"Here is the catch:... we do not have the number of neither"
The point is we use Total as the union of two sets very often and hence might forget "neither".[/quote]
Sorry, I think my last reply lacked detail !
My point was,
When you combine both statements..
U will observe, u have the following info:
N(total)
N(doctors)
N(law degree holders)
At this point if you use:
N(total) = n(docs) + n(law) - n(both)
Then u r surely gonna fall in the 'C Trap' because u will think, u have got 3 values from question and u can easily find 4th one.
But, if you had the correct formula in mind, i.e.
N(total)- neither= n(doc) +n(law) - n(both)
You will realise that, u still have two values missing. Without the 'neither' u cant find n(both).
Hence, info.is incomplete and answer is E.
So my point was, always have the formula (which involves 'neither') in your mind !
Hope I was able to put forward my part of understanding. Please correct me if my thought process is wrong.
Regards,
ashygoyal
Thanks and Regards,
ashygoyal
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
In a certain group of 50 people, how many are doctors who have a law degree?
(1) In the group, 36 people are doctors.
(2) In the group, 18 people have a law degree.
Solution:
Statement 1:Insufficient . No info about law degree.
Statement 2: Insufficient. No info about doctors.
Combine. We don't know the common in both. People involved in both law degree and doctor.
Therefore the answer is Option E.
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Bunuel wrote:
In a certain group of 50 people, how many are doctors who have a law degree?
(1) In the group, 36 people are doctors.
(2) In the group, 18 people have a law degree.
We have a group of 50 people and need to determine how many of those people are doctors with a law degree.
We can use the following formula:
total = # of doctors + # of lawyers - # both + # neither
50 = # of doctors + # of lawyers - # both + # neither
Statement One Alone:
In the group, 36 people are doctors.
So, we have:
50 = 36 + # of lawyers - # both + # neither
24 = # of lawyers - # both + # neither
We cannot determine the number of people who are both. Statement one alone is not sufficient to answer the question.
Statement Two Alone:
In the group, 18 people have a law degree.
So, we have:
50 = # of doctors + 18 - # both + # neither
32 = # of doctors - # both + # neither
We cannot determine the number of people who are both. Statement two alone is not sufficient to answer the question.
Statements One and Two Together:
Using statements one and two, we have:
50 = 36 + 18 - # both + # neither
-4 = - # both + # neither
# both = 4 + # neither
We still cannot determine the number of “both” since we don’t know the number of “neither.”
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
Going to keep it small and simple.
T - N = A + B - Both
T = Total; N = Neither; A = Doctor's Degree; B = Law Degree; We need to find out The Number of people who have both the degrees (i.e..,Both)
Total = 50 (given)
Statement 1: A = 36
=> 50-N = 36+B-x ----------Not sufficient.
Statement 2: B = 18
=> 50-N = A+18-x ----------Not Sufficient.
Together, 50-N = 36+18-x
Since we still don't know the value of N, we cannot determine x
Hence, E
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
I got this question wrong, so to clarify if the subject says : " a group of 50 doctor who have law degree", then we can apply Total = n(D) + n(L) - n(D and L) correct?
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Bunuel wrote:
In a certain group of 50 people, how many are doctors who have a law degree?
(1) In the group, 36 people are doctors.
(2) In the group, 18 people have a law degree.
$$Total = Dorctors + Lawyers - Both + Neither$$
$$50= Dorctors + Lawyers - Both + Neither$$
(1) Only the number of Doctors is given. Insufficient.
(2) Only the number of Lawyers is given. Insufficient.
Considering both;
$$50=36+15-Both+ Neither$$; We don't have information about either. Insufficient.
The answer is $$E$$.
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
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Bunuel wrote:
In a certain group of 50 people, how many are doctors who have a law degree?
(1) In the group, 36 people are doctors.
(2) In the group, 18 people have a law degree.
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Re: In a certain group of 50 people, how many are doctors who have a law [#permalink]
Bunuel wrote:
In a certain group of 50 people, how many are doctors who have a law degree?
(1) In the group, 36 people are doctors.
(2) In the group, 18 people have a law degree.
Wanna make solving the Official Questions interesting???
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and Dedicated Data Sufficiency (DS) Course | 5,006 | 17,192 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-33 | latest | en | 0.871394 |
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Define, $$I=int_{-infty}^inftyfrac{log^-|zeta(frac{1}{2}+it)|}{frac{1}{4}+t^2}dt$$
where $$zeta$$ denotes the Riemann zeta function and $$log^-|f|=max(-log|f|,0)$$.
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Thank you.
## proof writing – If \$X\$ is compact then prove that \$X\$ is complete and totally bounded.
I tried to do it in a way different from my textbook:
Let $$(X,d)$$ be a compact metric space then it is totally bounded.(I have been Ble to prove this).
My doubt lies in the following part that I have tried to prove:
Let $$X$$ be a compact metric space then $$X$$ is totally bounded.We choose a cauchy sequence $${x_n}$$ in $$(X,d)$$ .
Let $$A$$ be a subset of $$(X,d)$$ such that $$A={x_n}$$.Then $$A$$ will be totally bounded too as $$A subset (X,d)$$.
Let for an $$epsilon > 0$$ there exist points $${x_1′,cdots x_n’}$$ such that $$B_d(x_k’,epsilon)$$ contains infinitely many points of $$A$$.Then as $${x_n}$$ is a cauchy sequence so we can conlcude that $${x_n}$$ converges to $$x_k’$$. Also as $$(X,d)$$ is a metric space so $${x_n}$$ can converge to only one points and we have shown that $$x_k’$$ is a limit point of $${x_n}$$.
## Prove Sine of sum equals Sum of Sines by Induction
Use mathematical induction to show that there exists real numbers $$a_1+a_2+a_3+cdots+a_n$$ such that $$|a_i|le 1$$ for $$i=1, 2, 3, …, n$$ and such that $$sin (x_1+x_2+x_3+cdots+x_n)= a_1sin x_1 + a_2sin x_2 + a_3sin x_3 + cdots+a_nsin x_n$$.
## general topology – Using Arzela-Ascoli to prove there’s a converging sub-sequence
I am trying to solve the following problem in preparation for an upcoming exam in introductory topology and I’m unable to complete it entirely. My problem is with (b), but I’ve stated the full problem since it is likely that (a) is required for (b):
Let $$X$$ be a compact metric space.
a. Prove that every continuous function $$f:Xrightarrow mathbb{R}$$ is uniformly continuous.
b. Let $$T,S : X rightarrow X$$ be isometries.
Let $$f_1 in C(X,mathbb{R})$$. Denote $$f_{n+1} = frac{1}{2}Big(f_n(T(x))+f_n(S(x))Big) quadforall xin X, nin mathbb{N}$$ .
Prove that $${f_n}_{nin mathbb{N}}$$ has a converging subsequence using the sup norm.
So I’ve tried the following two approaches:
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What am I missing? Any help would be much appreciated!
## time complexity – Prove a lower bound
Prove: $$n^{5}-3n^{4}+logleft(n^{10}right)∈ Ωleft(n^{5}right)$$.
I always get stuck in these types of questions, where there is a $$”-(xy^{z})”$$ in the expression.
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have a problem that is a Josephus problem variation. It is described below:
There are m cards with number from 1 to m，and each of them has a unique number. The cards are dispatched to n person who sit in a circle. Note that m >= n.
Then we choose the person “A” who sits at the position “p” to out of the circle, just like the Josephus problem does. Next step we skip “k” person at the right of p while k is the number of the card toked by the person “A”, and we do the same thing until only one person left in the circle.
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— example solutions —
`````` 2: ( 1, 2)
2: ( 2, 1)
3: ( 1, 3, 2)
3: ( 3, 1, 2)
4: ( 4, 1, 3, 2)
5: ( 4, 1, 5, 3, 2)
7: ( 5, 7, 3, 1, 6, 4, 2)
9: ( 2, 7, 3, 9, 1, 6, 8, 5, 4)
9: ( 3, 1, 2, 7, 6, 5, 9, 4, 8)
9: ( 3, 5, 1, 8, 9, 6, 7, 4, 2)
9: ( 3, 9, 2, 7, 6, 1, 5, 4, 8)
9: ( 6, 1, 8, 3, 7, 9, 4, 5, 2)
10: ( 3, 5, 6, 10, 1, 9, 8, 7, 4, 2)
10: ( 4, 5, 2, 8, 7, 10, 6, 1, 9, 3)
10: ( 5, 1, 9, 2, 10, 3, 7, 6, 8, 4)
10: ( 6, 3, 1, 10, 9, 8, 7, 4, 5, 2)
10: ( 8, 5, 9, 10, 1, 7, 2, 6, 4, 3)
10: (10, 5, 2, 1, 8, 7, 6, 9, 3, 4)
11: ( 2, 1, 10, 11, 9, 3, 7, 5, 6, 8, 4)
11: ( 3, 7, 11, 10, 9, 8, 1, 6, 5, 4, 2)
11: ( 3, 11, 10, 9, 8, 1, 7, 2, 4, 5, 6)
11: ( 4, 1, 10, 2, 9, 8, 7, 5, 11, 3, 6)
11: ( 4, 2, 7, 11, 5, 1, 10, 9, 6, 3, 8)
11: ( 4, 7, 2, 3, 1, 10, 9, 6, 11, 5, 8)
11: ( 4, 7, 3, 9, 11, 10, 1, 8, 6, 5, 2)
11: ( 4, 11, 7, 2, 1, 10, 9, 6, 5, 3, 8)
11: ( 5, 11, 3, 9, 8, 7, 6, 1, 10, 4, 2)
11: ( 6, 1, 10, 2, 9, 8, 7, 5, 11, 3, 4)
11: ( 6, 2, 7, 11, 5, 1, 10, 9, 4, 3, 8)
11: ( 6, 11, 1, 3, 10, 2, 7, 5, 4, 9, 8)
11: ( 9, 5, 3, 1, 10, 2, 8, 7, 11, 6, 4)
12: ( 1, 7, 11, 10, 4, 9, 2, 12, 6, 5, 8, 3)
12: ( 3, 7, 12, 2, 11, 10, 9, 1, 6, 5, 4, 8)
12: ( 3, 8, 11, 2, 12, 9, 1, 7, 5, 10, 4, 6)
12: ( 4, 2, 5, 1, 11, 10, 9, 8, 12, 7, 3, 6)
12: ( 4, 3, 7, 6, 1, 11, 10, 9, 8, 12, 5, 2)
12: ( 5, 1, 6, 11, 9, 2, 10, 7, 12, 8, 3, 4)
12: ( 5, 2, 3, 12, 9, 10, 7, 6, 1, 11, 4, 8)
12: ( 5, 7, 12, 2, 10, 9, 8, 11, 1, 4, 6, 3)
12: ( 7, 1, 2, 3, 5, 9, 10, 8, 11, 6, 12, 4)
12: ( 8, 7, 1, 11, 9, 3, 5, 10, 6, 4, 12, 2)
12: ( 8, 7, 11, 10, 12, 3, 1, 9, 6, 5, 4, 2)
12: (12, 3, 11, 5, 1, 10, 8, 7, 6, 4, 9, 2)
12: (12, 7, 11, 1, 9, 3, 2, 10, 6, 5, 4, 8)
13: ( 2, 1, 4, 7, 11, 6, 3, 10, 13, 5, 8, 12, 9)
13: ( 2, 5, 13, 12, 4, 11, 3, 1, 9, 7, 8, 6, 10)
13: ( 2, 13, 12, 11, 3, 1, 9, 4, 8, 7, 10, 5, 6)
13: ( 3, 5, 2, 1, 12, 9, 11, 10, 7, 6, 13, 4, 8)
13: ( 3, 5, 13, 1, 11, 2, 9, 8, 7, 12, 6, 4, 10)
13: ( 4, 13, 3, 1, 12, 11, 10, 9, 7, 2, 5, 6, 8)
13: ( 6, 4, 3, 1, 10, 11, 13, 5, 9, 12, 7, 8, 2)
13: ( 6, 4, 13, 7, 5, 1, 12, 11, 10, 9, 8, 3, 2)
13: ( 6, 7, 3, 13, 12, 11, 10, 2, 1, 9, 5, 4, 8)
13: ( 6, 7, 13, 11, 2, 10, 9, 1, 8, 12, 5, 3, 4)
13: ( 6, 11, 7, 13, 1, 10, 2, 12, 9, 8, 5, 4, 3)
13: ( 7, 3, 2, 1, 11, 10, 9, 8, 13, 5, 12, 4, 6)
13: ( 7, 5, 13, 3, 10, 11, 2, 9, 1, 6, 8, 4, 12)
13: ( 7, 5, 13, 3, 11, 2, 9, 8, 1, 6, 12, 4, 10)
13: ( 7, 5, 13, 3, 11, 12, 2, 1, 9, 8, 6, 4, 10)
13: ( 7, 9, 1, 11, 3, 13, 2, 10, 12, 6, 5, 4, 8)
13: ( 8, 3, 5, 11, 13, 9, 10, 7, 1, 6, 4, 12, 2)
13: ( 8, 3, 13, 1, 5, 11, 10, 9, 12, 7, 6, 4, 2)
13: ( 9, 3, 13, 2, 10, 4, 1, 7, 6, 5, 12, 11, 8)
13: ( 9, 4, 7, 5, 1, 11, 13, 10, 12, 8, 6, 3, 2)
13: ( 9, 5, 4, 13, 2, 11, 8, 10, 1, 7, 12, 3, 6)
13: ( 9, 5, 13, 4, 11, 1, 8, 3, 7, 12, 6, 10, 2)
13: (10, 4, 3, 5, 13, 1, 9, 11, 7, 6, 8, 12, 2)
13: (11, 2, 7, 3, 12, 1, 10, 9, 6, 5, 13, 4, 8)
13: (11, 13, 5, 2, 10, 9, 8, 7, 1, 6, 4, 3, 12)
13: (11, 13, 7, 1, 12, 9, 2, 3, 10, 5, 4, 6, 8)
13: (12, 1, 3, 5, 11, 13, 4, 10, 9, 8, 7, 6, 2)
13: (12, 7, 13, 3, 11, 1, 9, 8, 6, 5, 10, 4, 2)
13: (12, 13, 7, 11, 2, 5, 1, 9, 10, 6, 4, 3, 8)
13: (13, 3, 1, 12, 11, 2, 9, 10, 7, 6, 4, 5, 8)
13: (13, 3, 7, 1, 5, 12, 4, 10, 9, 8, 11, 6, 2)
14: ( 3, 5, 13, 14, 1, 12, 11, 10, 9, 8, 7, 6, 4, 2)
14: ( 3, 9, 1, 13, 11, 10, 2, 4, 7, 14, 6, 8, 5, 12)
14: ( 3, 14, 4, 12, 11, 1, 9, 8, 2, 13, 7, 5, 10, 6)
14: ( 4, 11, 1, 13, 7, 10, 12, 2, 14, 9, 8, 5, 6, 3)
14: ( 4, 14, 2, 5, 13, 1, 12, 11, 7, 6, 10, 9, 3, 8)
14: ( 5, 7, 1, 13, 12, 11, 10, 2, 9, 8, 14, 6, 4, 3)
14: ( 6, 3, 14, 5, 11, 13, 2, 12, 9, 1, 7, 4, 8, 10)
14: ( 6, 14, 1, 12, 5, 13, 2, 11, 9, 7, 8, 4, 3, 10)
14: ( 7, 5, 13, 12, 1, 11, 4, 10, 2, 14, 9, 8, 6, 3)
14: ( 7, 11, 5, 13, 1, 3, 2, 4, 10, 9, 14, 6, 8, 12)
14: ( 7, 14, 1, 13, 2, 5, 11, 12, 10, 9, 8, 4, 3, 6)
14: ( 8, 7, 5, 13, 2, 11, 3, 9, 10, 12, 1, 14, 4, 6)
14: (11, 2, 10, 5, 8, 7, 9, 1, 13, 14, 12, 4, 3, 6)
14: (11, 3, 14, 2, 13, 1, 10, 8, 9, 7, 5, 12, 4, 6)
14: (11, 5, 3, 14, 2, 1, 13, 10, 8, 7, 6, 12, 4, 9)
14: (11, 14, 5, 3, 13, 1, 10, 2, 9, 4, 7, 8, 12, 6)
14: (12, 1, 14, 3, 13, 4, 10, 9, 2, 7, 6, 5, 11, 8)
14: (12, 11, 7, 5, 13, 3, 2, 14, 1, 9, 8, 4, 6, 10)
14: (12, 14, 7, 13, 6, 5, 11, 1, 10, 9, 8, 4, 3, 2)
14: (13, 1, 7, 2, 11, 3, 9, 14, 8, 6, 5, 10, 4, 12)
14: (13, 11, 3, 1, 4, 2, 7, 10, 9, 6, 14, 12, 5, 8)
14: (14, 1, 13, 3, 11, 5, 10, 9, 2, 6, 8, 7, 4, 12)
14: (14, 5, 1, 13, 12, 2, 11, 3, 7, 9, 6, 8, 4, 10)
``````
## bilinear form – How to prove that the following set is a linear subspace?
I need some help with the following problem. I have no idea how to go about a) and in b) I know the conditions of a linear subspace but I don’t know how to prove it and show the dimension.
Thanks in advance 🙂
Let K be a field of characteristic 2.
Let V be a K-vector space with dim_K(V) = n < ∞ and
(-,-) : V x V -> K be a symmetric bilinear form.
Show:
a) (-,-) is not degenerate.
b) The set U = {v ∈ V |(v,v) = 0} is a linear subspace of V with dim(U) = n-1. | 6,574 | 11,924 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 64, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-31 | latest | en | 0.671736 |
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Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560, 16561..16605, 16606..16650, 16651..16695, 16696..16740, 16741..16785, 16786..16830, 16831..16875, 16876..16920, 16921..16965, 16966..17010, 17011..17055, 17056..17100, 17101..17145, 17146..17190, 17191..17235, 17236..17280, 17281..17325, 17326..17370, 17371..17415, 17416..17460, 17461..17505, 17506..17550, 17551..17595, 17596..17640, 17641..17685, 17686..17730, 17731..17775, 17776..17820, 17821..17865, 17866..17910, 17911..17955, 17956..18000, 18001..18045, 18046..18090, 18091..18135, 18136..18180, 18181..18225, 18226..18270, 18271..18315, 18316..18360, 18361..18405, 18406..18450, 18451..18495, 18496..18540, 18541..18585, 18586..18630, 18631..18675, 18676..18720, 18721..18765, 18766..18810, 18811..18855, 18856..18900, 18901..18945, 18946..18990, 18991..19035, 19036..19080, 19081..19125, 19126..19170, 19171..19215, 19216..19260, 19261..19305, 19306..19350, 19351..19395, 19396..19440, 19441..19485, 19486..19530, 19531..19575, 19576..19620, 19621..19665, 19666..19710, 19711..19755, 19756..19800, 19801..19845, 19846..19890, 19891..19935, 19936..19980, 19981..20025, 20026..20070, 20071..20115, 20116..20160, 20161..20205, 20206..20250 | 5,259 | 13,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2013-20 | latest | en | 0.858156 |
https://www.physicsforums.com/threads/1-d-steady-state-heat-conduction-equation-cartisian-cylindrical-and-sperical.267198/ | 1,539,888,059,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511897.56/warc/CC-MAIN-20181018173140-20181018194640-00268.warc.gz | 1,026,995,067 | 16,610 | # Homework Help: 1-D steady state heat conduction equation (Cartisian, Cylindrical and Sperical)
1. Oct 26, 2008
### Schmoozer
1. The problem statement, all variables and given/known data
The one dimensional steady-state heat conduction equation in a medium with constant conductivity (k) with a constant volumetric heat generation in three different coordinate systems (fuel rods in a nuclear power plant) is given as:
$$\frac{d^2 T}{dx^2}=-\frac{\dot{q}}{k}$$ T(x=0)=1 T(x=1)=2 Cartisian
$$\frac{1}{r}\frac{d}{dr}(r\frac{dT}{dr})=-\frac{\dot{q}}{k}$$ T(r=R)=1 Cylindrical
$$\frac{1}{r^2}\frac{d}{dr}(r^2\frac{dT}{dr})=-\frac{\dot{q}}{k}$$ T(r=R)=1 Sperical
a. Find an expression for the temperature distribution in a solid for each case
b. What is a temperature distribution if the heat generation is zero?
2. Relevant equations
3. The attempt at a solution
$$T(x)=\sum_{n=0}^{\infty}b_n x^n$$
$$T'(x)=\sum_{n=0}^{\infty}nb_n x^n^-^1$$
$$T''(x)=\sum_{n=0}^{\infty}n(n-1)b_n x^n^-^2$$ so n=0, n=1
$$b_{0}=0$$
$$b_{1}=0$$
$$b_{2}=2x$$
$$b_{3}=3x^2$$
Am I at least on the right track?
2. Oct 26, 2008
### gabbagabbahey
What is the derivative of $b_0x^0$?
....Doesn't that mean that the first term in the series drops out and hence:
$T'(x)=\sum_{n=1}^{\infty}nb_n x^{n-1}$
and
$T''(x)=\sum_{n=2}^{\infty}n(n-1)b_n x^{n-2}=-\frac{\dot{q}}{k}$
Which means that if $\dot{q}$ is not a function of x, the only non-zero coefficient would be $b_2$.
3. Oct 26, 2008
### Schmoozer
Ok, so:
$$b_{0}=0$$
$$b_{1}=0$$
$$b_{2}=2x$$
$$b_{3}=3x^2$$
I am unable find a pattern in the even and odd b subscripts as the powers will simply continue to increase.
So I guess my next step is to find a solution to $$b_{n}$$.
$$b_{n}=nx^n^-^1$$
Is the next step just to sum the first few even and odd $$b_{n}$$'s or am I way off base?
4. Oct 26, 2008
### gabbagabbahey
You are way off base. Is $\dot{q}$ a function of x?
5. Oct 26, 2008
### Schmoozer
q, heat generation, is a constant. $$\dot{q}$$ will change with distance, x. So no $$\dot{q}$$ is not constant.
Edit: Scratch that, $$\dot{q}$$ is the rate of heat transfer so it is a constant in this condition.
6. Oct 26, 2008
### gabbagabbahey
If q is constant in x, then $\dot{q}=\frac{dq}{dt}$ will be constant in x as well. This means that $\frac{\dot{q}}{k}$ is just some constant.
Meanwhile, your solution for T gives
$T''=\sum_{n=2}^{\infty} B_n n(n-1)x^{n-2}=2B_2 + 6B_3 x+ 12B_4 x^2+....$
but this must equal a constant, so shouldn't $B_3=B_4=\ldots B_{\infty}=0$ so that you are left with $2B_2= \frac{-\dot{q}}{k}$
What does that make T(x)? (remember, you dont have any restrictions on b_0 and b_1 yet)
Last edited: Oct 26, 2008
7. Oct 26, 2008
### Schmoozer
Not quite. In $2B_2 + 6B_3 x+ 12B_4 x^2+....$ where are the 2, 6, 12, etc comming from?
And why is this $B_3=B_4=\ldots B_{\infty}=0$ true especially when $B_2$ is equal to some real number?
Isn't it true that as n increases $B_n$ decreases? Are we just assuming $B_3$ and on to be too small to be worty of counting?
8. Oct 26, 2008
### gabbagabbahey
the 2,6,12 are coming from n(n-1)....2(2-1)=2, 3(3-1)=6, 4(4-1)=12...etc.
Suppose that B_3,B_4,...etc weren't zero...wouldn't that mean that T'' was a function of x?
For example, if B_3 were equal to 1 and B_4,B_5,....=0 then T'' would equal 2B_2+6x....how could this possibly equal a constant?
9. Oct 26, 2008
### Schmoozer
Ah ha! Ok I get that part. Now what do I with the $2B_2= \frac{-\dot{q}}{k}$?
10. Oct 26, 2008
### gabbagabbahey
Well, that means that $B_2= \frac{-\dot{q}}{2k}$ right?...So what does that make your series for T(x)?
Last edited: Oct 26, 2008
11. Oct 26, 2008
### Schmoozer
T(x)=$$\frac{\dot{q}x^2}{2k}$$
12. Oct 26, 2008
### gabbagabbahey
1st, your missing a negative sign....2nd why have you set b0 and b1 equal to zero?
13. Oct 26, 2008
### Schmoozer
Ok: $$T(x)=-\frac{\dot{q}x^2}{2k}$$
but
bn => n(n-1)*bn*x^(n-2)
b0 => 0(0-1)*b0*x^(-2)=0
b1 => 1(1-1)*b1*x^(-1)=0
I'm assuming this is wrong, but thats my logic for it.
14. Oct 26, 2008
### gabbagabbahey
The sum for T'' starts at n=2....it says nothing about the n=0 and n=1 terms, so just leave them as unknowns:
$T(x)=B_0+B_1x-\frac{\dot{q}x^2}{2k}$
(Remember $\frac{d^2}{dx^2}(B_0+B_1 x)=0$ for any B_0 and B_1 not just for B_0=B_1=0)
But you are also told that T(x=0)=1 andT(x=1)=2, so you can use these two conditions to determine B_0 and B_1.
15. Oct 26, 2008
### Schmoozer
$T(x)=\frac{2}{x}+x+\frac{\dot{q}x^2}{2k}(x-1)$
Thanks so much!
I should be able to figure out the other two on my own now.
16. Oct 26, 2008
### gabbagabbahey
17. Oct 26, 2008
### Schmoozer
$1=b_0+b_1(0)-\frac{\dot{q}0^2}{2k}$
$$b_0=1$$
$$2=1+b_1(1)-\frac{\dot{q}}{2k}$$
$$1+\frac{\dot{q}}{2k}=b_1$$
$$T(x)=1+x+\frac{\dot{q}x}{2k}-\frac{\dot{q}x^2}{2k}$$
(I found an algebra mistake as I was recopying it.)
18. Oct 26, 2008
### gabbagabbahey
That looks much better! :0)
If you post your solutions for the other two cases, I'll be happy to check them for you.
19. Oct 26, 2008
### Schmoozer
Ok now were going back to some of my more fundamental problems in this class.
$$\frac{1}{r}\frac{d}{dr}(r\frac{dT}{dr})=-\frac{\dot{q}}{k}$$
$$(ln|r|)rT=-\frac{\dot{q}}{k}$$
???
20. Oct 26, 2008
### gabbagabbahey
Where does the ln|r| come from????
$\frac{d}{dr}(r \frac{dT}{dr})=\frac{-\dot{q}}{k} r$ | 2,161 | 5,403 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-43 | latest | en | 0.740815 |
https://www.experts-exchange.com/questions/23153758/I-need-a-help-with-math-and-code.html | 1,534,302,001,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209856.3/warc/CC-MAIN-20180815024253-20180815044253-00312.warc.gz | 873,426,431 | 20,749 | # I need a help with math and code
Hello guys,
I have a table in access with 1200 words,
In a stringGrid I need to have column with 16 words. Each Column will be a group with
16 words.
but the problem is, From 3 group created, each group created must have 60 % of words
in previous groups created.
With my math 1200 words with + 60 % of repeated words I will have a total of 1920 split in
120 column
this example can be made with number
thanks
LVL 1
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Commented:
Your question really could be a lot clearer.
60% of 16 is 9.6 - so if you if you need at least 60% (I'm assuming at least rather than exactly), then at least 10 of youre previous group of 16 will have to be duplicated (is there a rule for this or do you just take any 10??)
By this logic, the minimum number of words you will have is 1200 + (((1200 / 16) - 1) * 10)
= 1200 + 740
= 1940
Minimum number of columns = 1940 / 16 = 121.25
rounded up = 122 with the free space filled by virtue of it being at least 60% instead of exactly.
You also said 'From 3 group created'. What does that mean? If that has a bearing on the process and calculations then these figures will probably be completely off.
0
Author Commented:
Hello LordWolfy,
I am gonna try to clear my question better. Sorry my mistakes in english, ok :))
Imagine something like this:
I have a edit that you enter the percent of repeated word for each group.
The repeated words will be repeated from my 3 group, that is,
after creating 3 goups with this sequence:
Group A [ 1..16]
Group B [ 17..32 ]
Group C [ 33..49 ]
Now, other groups will be created with a percent from previous groups, that is:
Group D will have a little of my group A - B - C
Group E will have a little of my group A - B - C - D
Group F will have a little of my group A - B - C - D - E
and so on
Well, if I will have an increase of 740 repeated words or number on my 1200 words, then I will need
split all the 1940 words into 122 groups. If I have a percent less than 60, obvisouly I will have a less
group of words.
Did you get it?
0
PresidentCommented:
The example easily gets out of hand...
If group A has 16 words, group B has 16 words, and group C has 16 words to start...
You want to go to group D (and later) and take at least 60 % of it's values from the previous groups without going over 16 in each...
You would rapidly get into a situation where you cannot take a value from each group because you are only allowed 16.
If We are only filling in "at least 60%" from the previous columns, where do we get the rest of the words to fill in? Generate them from a simple number counter?
If you are ok with only getting 10 from previous columns, how are they chosen? could you take 3 from A, 3 from B, none from C, and 4 from D to get the 10 for group E? A simple random routine could randomly grab words from columns if there is no issue with getting more than one from the same column while filling another column.
For that matter, do you want to make sure that - as we pull out words - one word is pulled out only once for the entire run?
0
Author Commented:
Sorry for the delay of my reply.
Well, i think it is missing information about what I am trying to do. I am gonna describe what I need to do and you will be able to give a better solution to accomplish it.
I have a database of picture with 1064 pictures inside.
I need assemble groups of 16 pictures, I could split those images into the groups. But I need something different.
So, imagine this situation:
I start my first lesson (first group of 16 images), after studying them I can start another lesson and so on.
When I am at 6 lesson, I would like to have some pictures in my new groups from past groups studied, maybe 40 % of my 16 pictures will be from those images that I already studied. I also thought in having groups formed from images that I already studied, then I will have a group with 100% of studied images.
So it could be
For each 6 groups studied I will have one or two groups with only images studied.
I need to informa the system that after X studied groups I want some pictures from repeated
studied groups.
All those can be made with number that will represent the ID of each image
Did you get it?
0
PresidentCommented:
From what you explained this is the general idea for each set of 6 lessons...
Lesson1 new images
Lesson2 new images
Lesson3 new images
Lesson4 new images
Lesson5 new images
Lesson6 40% from lessons 1-5 the rest new.
Is this correct? and, if it is correct, would you want lesson 7 to contain only images not in lessons 1-6?
Let me know
0
Author Commented:
Yeah it is correct,
Lesson6 must have some repeated images from those lessons studied as a random reinforce.
After lesson6, each lesson is gonna get 40 % from past lessons and so on. But it is interresting another
way, see below.
if I could have for each 5 lesson created two lesson with only repeated images from my past lessons, so, an example:
Lesson1 new images
Lesson2 new images
Lesson3 new images
Lesson4 new images
Lesson5 new images
Lesson6 100 % of images from all past lesson [1..5]
Lesson7 100 % of images from all past lesson [1..5]
Lesson8 new images
Lesson9 new images
Lesson10 new images
Lesson11 new images
Lesson12 new images
Lesson13 100 % of images from all past lesson [1..12]
Lesson14 100 % of images from all past lesson [1..12]
What I told you above, it is another example different from the first one that I asked.
So, let's see now the original order
Lesson1 new images
Lesson2 new images
Lesson3 new images
Lesson4 new images
Lesson5 new images
Lesson6 40 % from lessons[1.5] and more new images
Lesson7 40 % from lessons[1.6] and more new images
Lesson7 40 % from lessons[1.7] and more new images
developmentguru, it is my ideia about, but you can give me your opinion on this. So fee free
0
PresidentCommented:
One thing I should know. Are you storing your images in a TImageList? The indexing would make some of this easier. Are they stored in a database? Separate files? In any case if you load them into a TImageList then the following code should work. Keep in mind, rather than come up with many images to use I am using strings. The same principal applies though. Place two memos and a button on a form and paste the code in. Attach the FormCreate method, and the ButtonClick method. Form Create fills the first memo with strings to use. Button Click uses the values in memo 1 to create lessons. In your case, instead of using strings in a string list you would use indexes stored in a TList, the concept is the same. Let me know if this example gives you the tools you need, it should.
``````unit Unit1;
interface
uses
Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
Dialogs, StdCtrls;
type
TForm1 = class(TForm)
Memo1: TMemo;
Memo2: TMemo;
Button1: TButton;
procedure FormCreate(Sender: TObject);
procedure Button1Click(Sender: TObject);
private
{ Private declarations }
public
{ Public declarations }
end;
var
Form1: TForm1;
implementation
{\$R *.dfm}
procedure TForm1.Button1Click(Sender: TObject);
var
L, I : Integer;
Total, Chosen : TStringList;
Choice : integer;
begin
//Total holds the list to start from
Total := TStringList.Create;
Total.Assign(Memo1.Lines);
try
//Chosen holds the list of those that have been chosen.
Chosen := TStringList.Create;
try
for L := 1 to 14 do
begin
if L mod 7 in [0..4] then
begin
for I := 1 to 16 do
begin
Choice := Random(Total.Count);
Total.Delete(Choice);
end;
end
else
begin
for I := 1 to 16 do
begin
Choice := Random(Chosen.Count);
end;
end;
end;
finally
Chosen.Free;
end;
finally
Total.Free;
end;
end;
procedure TForm1.FormCreate(Sender: TObject);
var
I : Integer;
begin
Memo1.Clear;
for I := 1 to 1000 do
Memo2.Clear;
end;
end.
``````
0
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Commented:
The math ;-)
n - number of pictures each lesson
x - lessons before the start repeating images
p - the fraction of images that should be repeats in stead of new ones
t - the total number of images
before repeats: x * n new images
with repeats: (1 - p) * n per lesson
lessons needed:
lessons needed for whole course = x + (t - (x * n)) / ((1 - p) * n)
So for 1200 images start after 3 lessons with 0.6 repeats and 16 images per lesson:
3 + (1200 - 16*3) / (0.4*16) = 183 lessons
Its discreet math though :S
0
Author Commented:
thanks
0
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From novice to tech pro — start learning today. | 2,419 | 9,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-34 | latest | en | 0.948217 |
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# expressions
## summary
This subchapter looks at expressions.
## free computer programming text book project
If you like the idea of this project,
then please donate some money.
### stub section
This subchapter is a stub section. It will be filled in with instructional material later. For now it serves the purpose of a place holder for the order of instruction.
Professors are invited to give feedback on both the proposed contents and the propsed order of this text book. Send commentary to Milo, PO Box 1361, Tustin, California, 92781, USA.
# expressions
This subchapter looks at expressions.
### assembly language instructions
• EMOD Extended Multiply and Integerize; DEC VAX; performs accurate range reduction of math function arguments, the floating point multiplier extension operand (second operand) is concatenated with the floating point multiplier (first operand) to gain eight additional low order fraction bits, the multiplicand operand (third operand) is multiplied by the extended multiplier operand, after multiplication the integer portion (fourth operand) is extracted and a 32 bit (EMODF) or 64 bit (EMODD) floating point number is formed from the fractional part of the product by truncating extra bits, the multiplication is such that the result is equivalent to the exact product truncated (before normalization) to a fraction field of 32 bits in floating or 64 bits in double (fifth operand); clears or sets flags
• POLY Polynomial Evaluation; DEC VAX; performs fast calculation of math functions, for degree times (second operand) evaluate a function using Horner’s method, where d=degree (second operand), x=argument (first operand), and result = C[0] + x*(C[1] + x*(C[2] + … x*C[d])), float result stored in D0 register, double float result stored in D0:D1 register pair, the table address operand (third operand) points to a table of polynomial coefficients ordered from highest order term of the polynomial through lower order coefficients stored at increasing addresses, the data type of the coefficients must be the same as the data type of the argument operand (first operand), the unsigned word degree operand (second operand) specifies the highest numbered coefficient to participate in the evaluation (POLYF polynomial evaluation floating, POLYD polynomial evaluation double float); D0 through D4 registers modified by POLYF, D0 through D5 registers modified by POLYD; sets or clears flags
• TBLS Table Lookup and Interpolate (Signed, Rounded); Motorola 68300; signed lookup and interpolation of independent variable X from a compressed linear data table or between two register-based table entries of linear representations of dependent variable Y as a function of X; ENTRY(n) + {(ENTRY(n+1) - ENTRY(n)) * Dx[7:0]} / 256 into Dx; table version: data register low word contains the independent variable X, 8-bit integer part and 8-bit fractional part with assumed radix point located between bits 7 and 8, source effective address points to beginning of table in memory, integer part scaled to data size (byte, word, or longword) and used as offset from beginning of table, selected table entry (a linear representation of dependent variable Y) is subtracted from the next consecutive table entry, then multiplied by the interpolation fraction, then divided by 256, then added to the first table entry, and then stored in the data register; register version: data register low byte contains the independent variable X 8-bit fractional part with assumed radix point located between bits 7 and 8, two data registers contain the byte, word, or longword table entries (a linear representation of dependent variable Y), first data register-based table entry is subtracted from the second data register-based table entry, then multiplied by the interpolation fraction, then divided by 256, then added to the first table entry, and then stored in the destination (X) data register, the register interpolation mode may be used with several table lookup and interpolations to model multidimentional functions; rounding is selected by the ‘R’ instruction field, for a rounding adjustment of -1, 0, or +1; interpolation resolution is limited to 1/256th the distance between consecutrive table entries, X should be considered an integer in the range 0 ≤ X ≤ 65535; sets or clears flags
• TBLSN Table Lookup and Interpolate (Signed, Not Rounded); Motorola 68300; signed lookup and interpolation of independent variable X from a compressed linear data table or between two register-based table entries of linear representations of dependent variable Y as a function of X; ENTRY(n) * 256 + (ENTRY(n+1) - ENTRY(n)) * Dx[7:0] into Dx; table version: data register low word contains the independent variable X, 8-bit integer part and 8-bit fractional part with assumed radix point located between bits 7 and 8, source effective address points to beginning of table in memory, integer part scaled to data size (byte, word, or longword) and used as offset from beginning of table, selected table entry (a linear representation of dependent variable Y) multiplied by 256, then added to the value determined by (selected table entry subtracted from the next consecutive table entry, then multiplied by the interpolation fraction), and then stored in the data register; register version: data register low byte contains the independent variable X 8-bit fractional part with assumed radix point located between bits 7 and 8, two data registers contain the byte, word, or longword table entries (a linear representation of dependent variable Y), first data register-based table entry is multiplied by 256, then added to the value determined by (first data register-based table entry subtracted from the second data register-based table entry, then multiplied by the interpolation fraction), and then stored in the destination (X) data register, the register interpolation mode may be used with several table lookup and interpolations to model multidimentional functions; the result is an 8-, 16-, or 24-bit integer and eight-bit fraction; interpolation resolution is limited to 1/256th the distance between consecutrive table entries, X should be considered an integer in the range 0 ≤ X ≤ 65535; sets or clears flags
• TBLU Table Lookup and Interpolate (Unsigned, Rounded); Motorola 68300; unsigned lookup and interpolation of independent variable X from a compressed linear data table or between two register-based table entries of linear representations of dependent variable Y as a function of X; ENTRY(n) + {(ENTRY(n+1) - ENTRY(n)) * Dx[7:0]} / 256 into Dx; table version: data register low word contains the independent variable X, 8-bit integer part and 8-bit fractional part with assumed radix point located between bits 7 and 8, source effective address points to beginning of table in memory, integer part scaled to data size (byte, word, or longword) and used as offset from beginning of table, selected table entry (a linear representation of dependent variable Y) is subtracted from the next consecutive table entry, then multiplied by the interpolation fraction, then divided by 256, then added to the first table entry, and then stored in the data register; register version: data register low byte contains the independent variable X 8-bit fractional part with assumed radix point located between bits 7 and 8, two data registers contain the byte, word, or longword table entries (a linear representation of dependent variable Y), first data register-based table entry is subtracted from the second data register-based table entry, then multiplied by the interpolation fraction, then divided by 256, then added to the first table entry, and then stored in the destination (X) data register, the register interpolation mode may be used with several table lookup and interpolations to model multidimentional functions; rounding is selected by the ‘R’ instruction field, for a rounding adjustment of 0 or +1; the result is an 8-, 16-, or 24-bit integer and eight-bit fraction; interpolation resolution is limited to 1/256th the distance between consecutrive table entries, X should be considered an integer in the range 0 ≤ X ≤ 65535; sets or clears flags
• TBLUN Table Lookup and Interpolate (Unsigned, Not Rounded); Motorola 68300; unsigned lookup and interpolation of independent variable X from a compressed linear data table or between two register-based table entries of linear representations of dependent variable Y as a function of X; ENTRY(n) * 256 + (ENTRY(n+1) - ENTRY(n)) * Dx[7:0] into Dx; table version: data register low word contains the independent variable X, 8-bit integer part and 8-bit fractional part with assumed radix point located between bits 7 and 8, source effective address points to beginning of table in memory, integer part scaled to data size (byte, word, or longword) and used as offset from beginning of table, selected table entry (a linear representation of dependent variable Y) multiplied by 256, then added to the value determined by (selected table entry subtracted from the next consecutive table entry, then multiplied by the interpolation fraction), and then stored in the data register; register version: data register low byte contains the independent variable X 8-bit fractional part with assumed radix point located between bits 7 and 8, two data registers contain the byte, word, or longword table entries (a linear representation of dependent variable Y), first data register-based table entry is multiplied by 256, then added to the value determined by (first data register-based table entry subtracted from the second data register-based table entry, then multiplied by the interpolation fraction), and then stored in the destination (X) data register, the register interpolation mode may be used with several table lookup and interpolations to model multidimentional functions; the result is an 8-, 16-, or 24-bit integer and eight-bit fraction; interpolation resolution is limited to 1/256th the distance between consecutrive table entries, X should be considered an integer in the range 0 ≤ X ≤ 65535; sets or clears flags
• CRC Calculate Cyclic Redundancy Check; DEC VAX; performs a cyclic redundancy check (CRC) on the string designated by the length and memory address, using the designated table holding the CRC polynomial and the designated initial CRC; uses R0 through R4 and leaves answer in R0; sets or clears flags
# free music player coding example
Coding example: I am making heavily documented and explained open source code for a method to play music for free — almost any song, no subscription fees, no download costs, no advertisements, all completely legal. This is done by building a front-end to YouTube (which checks the copyright permissions for you).
View music player in action: www.musicinpublic.com/.
Create your own copy from the original source code/ (presented for learning programming).
## view text bookHTML file
Because I no longer have the computer and software to make PDFs, the book is available as an HTML file, which you can convert into a PDF.
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Names and logos of various OSs are trademarks of their respective owners. | 2,530 | 12,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-50 | latest | en | 0.861114 |
http://bootmath.com/how-prove-this-integral-limit-ffrac12.html | 1,529,850,805,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866965.84/warc/CC-MAIN-20180624141349-20180624161349-00071.warc.gz | 46,395,408 | 7,682 | # How prove this integral limit $=f(\frac{1}{2})$
Let $f$ be a continuous function on the unit interval $[0,1]$. Show that
$$\lim_{n\to\infty}\int_{0}^{1}\cdots\int_0^1\int_{0}^{1}f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=f\left(\dfrac{1}{2}\right)$$
This problem is from Selected Problems in Real Analysis. But the author doesn’t include a solution. Maybe there is a method for this sort of problem?
Maybe we can use this:
$$\int_0^1\cdots\int_{0}^{1}\int_{0}^{1}\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2}?$$
I also found a similar problem in this question.
#### Solutions Collecting From Web of "How prove this integral limit $=f(\frac{1}{2})$"
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1} \fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n} = \fermi\pars{\half}}$
\begin{align}
&\color{#c00000}{%
\lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1}
\fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n}
\,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}}
\\[3mm]&=\lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\exp\pars{\ic k\,{x_{1} + x_{2} + \cdots + x_{n}\over n}}\,{\dd k \over 2\pi}
\,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}
\end{align}
where $\ds{% \tilde{\fermi}\pars{k} \equiv \int_{-\infty}^{\infty}\fermi\pars{x} \expo{-\ic k x}\,\dd x}$ is the $\ds{\fermi\pars{x}}$ Fourier Transform.
\begin{align}
&\color{#c00000}{%
\lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1}
\fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n}
\,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}}
\\[3mm]&=\lim_{n \to \infty}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\pars{\int_{0}^{1}\expo{\ic kx/n}\,\dd x}^{n}\,{\dd k \over 2\pi}
=\lim_{n \to \infty}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\pars{\expo{\ic k/n} – 1 \over \ic k/n}^{n}\,{\dd k \over 2\pi}
\\[3mm]&=\lim_{n \to \infty}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\braces{\exp\pars{\ic k \over 2n}\,
\bracks{\exp\pars{\ic k \over 2n} – \exp\pars{-\,{\ic k \over 2n}}}\,
{1 \over \ic k/n}}^{n}\,{\dd k \over 2\pi}
\\[3mm]&=\lim_{n \to \infty}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\exp\pars{\ic k \over 2}
\braces{\sin\pars{k/\bracks{2n}} \over k/\bracks{2n}}^{n}\,{\dd k \over 2\pi}
\end{align}
$$\color{#c00000}{% \lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1} \fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}} =\int_{-\infty}^{\infty} \fermi\pars{x}\lim_{n \to \infty}{\rm K}_{n}\pars{x – \half}\,\dd x$$
where
$${\rm K}_{n}\pars{x} \equiv \int_{-\infty}^{\infty} \exp\pars{-\ic k x} \braces{\sin\pars{k/\bracks{2n}} \over k/\bracks{2n}}^{n}\,{\dd k \over 2\pi}$$
Since $\ds{\lim_{n \to \infty}{\rm K}_{n}\pars{x} = \delta\pars{x}}$
( $\ds{\delta\pars{x}}$ is the Dirac Delta Function ),
we’ll have:
$$\color{#c00000}{% \lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1} \fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}} =\int_{-\infty}^{\infty} \fermi\pars{x}\delta\pars{x – \half}\,\dd x$$
$$\color{#00f}{\large% \lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1} \fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n} =\fermi\pars{\half}}$$
We can take advantage of a probabilistic interpretation. Let $X_1,\ldots,X_n$ be uniformly distributed random variables on $[0,1]$ and $\bar X=\frac{1}{n}(X_1+\cdots+X_n)$. Then
$$\lim\limits_{n\to \infty}\int_0^1\cdots\int_0^1f\left(\frac{x_1+\cdots+x_n}{n}\right)dx_1\cdots dx_n =\lim\limits_{n\to\infty}E[f(\bar X)]=E\left[f\left(\frac12\right)\right]=f\left(\frac12\right)$$
since $\bar X$ converges in distribution to $\frac12$ as $n\to \infty$.
Starting with OP’s hint,
$\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2}$,
Now, we try to show, $\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^k dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2^k}$
If we count the number of terms in the multinomial expansion of $\bigg(\sum\limits_{i=1}^n x_i\bigg)^k$, that contains all variables with power not exceeding $1$, is $n(n-1)(n-2)\cdots(n-k+1) = n^k + O(n^{k-1})$,
and the number of terms that has atleast one $x_i$ term with powers exceeding $1$ is not more than $n.n^{k-2} = n^{k-1}$.
So, combining all the terms we get,
$\displaystyle \int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^k dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2^k} + O\bigg(\frac{1}{n}\bigg)$
Thus the above limit is true for polynomials.
Using Weierstrass Approximation theorem, we can choose a polynomial $P$, such that $|f(x) – P(x)| < \epsilon/3$ for all $x \in [0,1]$.
Thus, we can find an $N \in \mathbb{N}$, such that $\forall n >N$,
$\displaystyle \left| \int_{0}^{1}\cdots\int_{0}^{1} P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n} – P\left(\frac12\right)\right| < \epsilon/3$
Therefore, for $n > N$,
$\displaystyle \left| \int_{0}^{1}\cdots\int_{0}^{1} f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n} – f\left(\frac12\right)\right| < \left| \int_{0}^{1}\cdots\int_{0}^{1} f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) – P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n}\right| + \left| \int_{0}^{1}\cdots\int_{0}^{1} P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n} – P\left(\frac12\right)\right| + \left|f\left(\frac12\right) – P\left(\frac12\right)\right|$
$\le \displaystyle \int_{0}^{1}\cdots\int_{0}^{1} \left| f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) – P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)\right| dx_{1}dx_{2}\cdots dx_{n} + 2\epsilon/3 < \epsilon$
Thus, $\displaystyle \lim\limits_{n\to\infty} \int_{0}^{1}\cdots\cdots\int_{0}^{1}f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=f\left(\dfrac{1}{2}\right)$.
Define
$$\psi_n(x)=\overbrace{\left(n\chi_{[0,1/n]}\right)\ast\cdots\ast\left(n\chi_{[0,1/n]}\right)}^{\text{convolution of n copies}}$$
Then
\begin{align} &\int_0^1\cdots\int_0^1f\left(\frac1n\sum_{k=1}^nx_k\right)\,\mathrm{d}x_1\dots\,\mathrm{d}x_n\\ &=n^n\int_0^{1/n}\cdots\int_0^{1/n}f\left(\sum_{k=1}^nx_k\right)\,\mathrm{d}x_1\dots\,\mathrm{d}x_n\\ &=\int_\mathbb{R}f(x)\psi_n(x)\,\mathrm{d}x \end{align}
The Central Limit Theorem says that since $n\chi_{[0,1/n]}$ has mean $\frac1{2n}$ and variance $\frac1{12n^2}$, the convolution of $n$ copies tends to a normal distribution with mean $\frac12$ and variance $\frac1{12n}$
$$\psi_n\sim\sqrt{\frac{6n}\pi}\ e^{-6n(x-1/2)^2}$$
which is an approximation of the Dirac delta function at $x=1/2$. That is,
$$\lim_{n\to\infty}\int_\mathbb{R}f(x)\psi_n(x)\,\mathrm{d}x=f(1/2)$$ | 3,607 | 8,084 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-26 | latest | en | 0.360652 |
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# (5)
Professor Jeffrey S, Fu
1
EXAMPLE 4.3.3
An amplifier has a quoted noise figure of
2.5dB. What is its equivalent noise temperature?
Using Eq. (4.22)
Td = 290(1.781)= 226K
This value of noise temperature could then be
used in Eq. (4.17), with other appropriate data,
to calculate system noise temperature.
2
G/T Ratio for Earth Stations
The link equation can be rewritten in terms of (C/N) at the
earth station
(4.23)
## Thus C/N Gr/Ts, and the terms in the square brackets
are all constants for a given satellite system.
3
The ratio Gr / Ts , which is usually quoted as
simply G/T in decibels, with units dB/K, can be
used to specify the quality of a receiving earth
station or a satellite receiving system, since
increasing Gr / Ts increases the received C/N
ratio.
Satellite terminals may be quoted as having a
negative G/T which is below 0 dB/K. This
simply means that the numerical value of Gr is
smaller than the numerical value of Ts .
4
EXAMPLE 4.3.4
An earth station antenna has a diameter of 30 m,
has an overall efficiency of 68%, and is used to
receive a signal at 4150 MHz. At this frequency,
the system noise temperature is 79 K when the
antenna points at the satellite at an elevation angle
of 28. What is the earth station G/T ratio under
these conditions? If heavy rain causes the sky
temperature to increase so that the system noise
temperature rises to 88 K, what is the new G/T
value? 5
First calculate the antenna gain. For a circular aperture:
Gr = A 4A/ = A (D/)
At 4150 MHz, = 0.0723 m. Then
G =0.68 (30/0.0723)2
= 1.16 106 or 60.6dB
Converting Ts into dBK
Ts = 10 log79 = 19.0 dBK
G / T = 60.6 19.0 = 41.6 dB/K
If Ts = 88 K in heavy rain,
G/T = 60.6 19.4 = 41.2 dB/K 6
4.4 DESIGN OF DOWNLINKS
The design of any satellite communication is based on
two objectives: meeting a minimum C/N ratio for a
specified percentage of time, and carrying the maximum
revenue earning traffic at minimum cost. There is an old
saying that an engineer is a person who can do for a
dollar what any fool can do for one hundred dollars.
This applies to satellite communication systems. Any
satellite link can be designed with very large antennas to
achieve high C/N ratios under all conditions, but the cost
will be high.
7
The art of good system design is to reach the
best compromise of system parameters that
meets the specification at the lowest cost. For
example, if a satellite link is designed with
sufficient margin to overcome a 20-dB rain
fade rather than a 3-dB fade, earth station
antennas with seven times the diameter are
required.
8
All satellite communications links are affected
by rain attenuation. In the 6/4 GHz band the
effect of rain on the link is small. In the 14/11
GHz (Ku) band, and even more so in the 30/20
GHz (Ka) band, rain attenuation becomes all
important. Satellite links are designed to
achieve reliabilities of 99.5 to 99.99%,
averaged over a long period of time, typically a
year.
9
That means the C/N ratio in the receiver will
fall below the minimum permissible value for
proper operation of the link for between 0.5
and 0.01% of the specified time; the link is
then said to suffer an outage. The time period
over which the percentage to time is measured
can be a month, sometimes the worst month
in attenuation terms, or a year. Rain
attenuation is a very variable phenomenon,
both with time and place.
10
C-band links can be designed to achieve 99.99%
reliability because the rain attenuation rarely
exceeds 1 or 2 dB. The time corresponding to
0.01% of a year is 52 min; at this level of
probability the rain attenuation statistics are
usually not stable and wide fluctuations occur
from year to year. Outages occur in heavy rain,
usually in thunderstorms, and thunderstorm
occurrence varies widely.
11
A link designed to have outages totaling 52
min each year may well have outages of
several hours one year and none the next.
Most Ka-band links cannot be designed to
achieve 99.99% reliability because rain
attenuation generally exceeds 10 dB, and
often 20 dB, for 0.01% of the time.Outage
times of 0.1 to 0.5% of a year (8 to 40 h) are
usually tolerated in Ka band links.
12
The allowable outage time for a link depends
in part on the traffic carried. Telephone traffic
needs real-time channels that are maintained
for the duration of a call, so C band or Ku band
is used for voice channels with sufficient link
margin that outage times are small. Internet
transmissions are less affected by short outages
and generally do not require a real-time
channel, making Ka band better suited for
Internet access.
13
Link Budgets
C/N ratio calculation is simplified by the use of
link budgets. A link budget is a tabular method
for evaluating the received power and noise
power in radio link. Link budgets invariably use
decibel units for all quantities so that signal and
noise powers can be calculated by addition and
subtraction.
14
Since it is usually impossible to design a
satellite link at the first attempt, link budgets
make the task much easier because, once a link
budget has been established, it is easy to
change any of the parameters and recalculate
the result. Tables 4.4a and 4.4b show a typical
link budget for a C-band downlink using a
global beam on a GEO satellite and a 9-m
earth station antenna.
15
The link budget must be calculated for an individual
transponder, and must be repeated for each of the
individual links. In a two-way satellite
communication link there will be four separate links,
each requiring a calculation of C/N ratio. When a
bent pipe transponder is used the uplink and
downlink C/N ratios must be combined to give an
overall C/N. In this section we will calculate the C/N
ratio for a single link. Later examples in this chapter
demonstrate the evaluation of a complete satellite
communication system.
16
TABLE 4.4a
C-Band GEO Satellite Link
Budget in Clear Air
C-band satellite parameters
Transponder saturated output power 20 W
Antenna gain, on axis 20 dB
Transponder bandwidth 36 MHz
Downlink frequency band 3.7-4.2 GHz
17
Signal FM-TV analog signal
FM-TV signal bandwidth 30 MHz
Minimum permitted overall C/N in receiver
9.5 dB
Receiving C-band station
Downlink frequency 4.00GHz
Antenna gain, on axis, 4 GHz 49.7 dB
Receiver IF bandwidth 27 MHz
Receiving system noise temperature 75 K
18
Downlink power budget
Pt = Satellite transponder output power, 20 W 13.0dBW
Bo = Transponder output backoff -2.0 dB
Gt = Satellite antenna gain, on axis 20.dB
Gr = Earth station antenna gain 49.7 dB
Lp = Free space path loss at 4 GHz -196.5 dB
Lant=Edge of beam loss for satellite antenna -3.0 dB
La = clear air atmospheric loss -0.2 dB
Lm = Other losses -0.5 dB
Pr = Received power at earth station -119.5dBw
19
Downlink noise power budget in clear air
K = Boltzmanns constant -228.6dBW/K/Hz
Ts = System noise temperature, 75 K 18.8 dBK
Bn = Noise bandwidth, 27 MHz 74.3dBHz
N = Receiver noise power - 135.5 dBW
## C/N ratio in receiver in clear air
C/N = Pr N
= -119.5 dBW (-135.5 dBW)
= 16.0 dB
20
Link budgets are usually calculated for a worst case.
The one in which the link will have the lowest C/N ratio.
Factors which contribute to a worst case scenario
include: an earth station located at the edge of the
satellite coverage zone where the received signal is
typically 3 dB lower than in the center of the zone
because of the satellite antenna pattern, maximum path
length from the satellite to the earth station, a low
elevation angle at the earth station giving the highest
atmospheric path attenuation in clear air, and maximum
rain attenuation on the link causing loss of received
signal power and an increase in receiving system noise
temperature. 21
TABLE 4.4b C-Band Downlink Budget in Rain
Prca = Received power at earth station in clear air -119.5 dBW
A = Rain attenuation -1.0 dB
Prain= Received power at earth station in rain -120.5 dBW
Nca = Receiver noise power in clear air -135.5 dBW
Ncain = Increase in noise temperature due to rain 2.3 dB
Nrain= Receiver noise power in rain -133.2 dBW
## C/N ratio in receiver in rain
C/N = Prain Nrain
= - 120.5 dBW (-133.3 dBW) = 12.7 dB
22
The edge of the coverage pattern of the satellite
antenna and the longest path usually go together.
However, when a satellite has a multiple beam
antenna, this may not always be the case. Earth
station antennas are assumed to be pointed
directly at the satellite, and therefore operate at
their on-axis gain. If the antenna is mispointed, a
loss factor is included in the link budget to
account for the reduction in antenna gain.
23
Equation 4.11 gives the received carrier power
in dB watts as
Pr = EIRP + Gr Lp La Lr Lt dBW
(4.24)
A receiving terminal with a system noise
temperature TsK and a noise bandwidth Bn Hz
has a noise power Pn referred to the output
terminals of the antenna where
Pn = kTsBn watts (4.25)
24
The receiving system noise power is usually written in
decibel units as
N = k + Ts + Bn dBW (4.26)
where k is Boltzmanns constant (-228.6 dBW/K/Hz),
Ts is the system noise temperature in dBK, and Bn is
the noise bandwidth of the receiver in dBHz. Note that
because we are working in units of power, all decibel
conversions are made as 10 log 10(Ts) or 10 log 10 (Bn).
The 20 log 10 factor used in the calculation of path loss
results from the (4R/) 2 term in the path loss
equation.
25
Link Budget Example: C-Band
Downlink for Earth Coverage Beam
The satellite used in this example (see Tables 4.4a and
4.4b) is in geostationary earth orbit and carries 24 C-
band transponders, each with a bandwidth of 36 MHz.
The downlink band is 3.7- 4.2 GHz and the satellite
uses orthogonal circular polarizations to provide an
effective RF bandwidth of 864 MHz. The satellite
provides coverage of the visible earth, which subtends
an angle of approximately 17 from a satellite in a
geostationary orbit, by using a global beam antenna.
26
Since antenna beamwidth and gain are linked
together [3 dB beamwidth
where G is a ratio, not in decibels], the on-axis
gain of the global beam antenna is approximately
20 dB. However, we must make the link budget
calculation for an earth station at the edge of the
coverage zone of the satellite where the effective
gain of the antenna is 3 dB lower, at 17 dB. The
C/N ratio for the downlink is calculated in clear
air conditions and also in heavy rain.
27
An antenna with a gain of 20 dB has an
effective aperture diameter of 5.6
wavelengths [G = (D/)2 ], which
gives D = 0.42 m at a frequency of 4
GHz. The calculation of C/N ratio is
made at a mid-band frequency of 4 GHz.
28
The saturated output power of the transponder is 20
W = 13 dBW. We will assume an output back-off of
2 dB, so that the power transmitted by the
transponder is 11 dBW. Hence the on-axis EIRP of
the transponder and antenna is PtGt = 11 + 20 =31
dBW. The transmitted signal is a single 30-MHz
bandwidth analog FM-TV channel in this example.
Following common practice for analog TV
transmission, the receiver noise bandwidth is set to
27 MHz, slightly less than the 30-MHz bandwidth
of the FM-TV signal.
29
The receiving earth station has an antenna
with an aperture diameter of 9 m and a gain
of 49.7 dB at 4 GHz, and a receiving system
noise temperature of 75 K in clear air
conditions. The G/T ratio for this earth station
is G/T = 49.7 10 log 10 75 = 30.9 dBK1 .
The maximum path length for a GEO satellite
link is 40,000 km, which gives a path loss of
196.5 dB at 4 GHz ( = 0.075 m).
30
We must make an allowance in the link budget
for some losses that will inevitably occur on the
link. At C band, propagation losses are small, but
the slant path through the atmosphere will suffer
a typical attenuation of 0.2 dB in clear air. We
will allow an additional 0.5-dB margin in the
link design to account for miscellaneous losses,
such as antenna mispointing, polarization
mismatch, and antenna degradation, to ensure
that the link budget is realistic.
31
The earth station receiver C/N ratio is first
calculated for clear air conditions, with no rain in
the slant path. The C/N ratio is then recalculated
taking account of the effects of rain. The minimum
permitted overall C/N ratio for this link is 9.5 dB,
corresponding to the FM threshold of an analog
satellite TV receiver. Table 4.4a shows that we
have a downlink C/N of 16.0 dB in clear air, giving
a link margin of 6.5 dB. This link margin is
available in clear air conditions, but will be reduced
when there is rain in the slant path.
32
Heavy rain in the slant path can cause up to 1
dB of attenuation at 4 GHz, which reduces the
received power by 1 dB and increases the noise
temperature of the receiving system. Using the
output noise model discussed in the pervious
section with a medium temperature of 273 K,
and a total path loss for clear air plus rain of 1.2
dB (ratio of 1.32), the sky noise temperature in
rain is
Tsky = 273 (11/1.32) = 66 K
33
In clear air the sky noise temperature is about
13 K , the result of 0.2 dB of clear air
attenuation. The noise temperature of the
receiving system has therefore increased by
(6613)K = 53 K to 75 + 53 K =128 K with
1 dB rain attenuation in the slant path, from a
clear air value of 75 K. This is an increase in
system noise temperature of 2.3 dB.
34
We can now adjust the link budget very
easily to account for heavy rain in the slant
path without having to recalculate the C/N
ratio from the beginning. The received carrier
power is reduced by 1 dB because of the rain
attenuation and the system noise temperature
is in creased by 2.3 dB. Table 4.4b shows the
new downlink budget in rain.
35
The C/N ratio in rain has a margin of 3.2 dB over the
minimum permissible C/N ratio of 9.5 dB for an
analog FM-TV transmission. The C/N margin will
translate into a higher than needed S/N ratio in the TV
baseband signal, and can be traded off against earth
station antenna gain to allow the use of smaller (and
therefore lower cost) antenna. We should always leave
a small margin for unexpected losses if we want to
guarantee a particular level of reliability in the link. In
this case, we will use a 2.dB margin and examine how
the remaining 1.2 dB of link margin can be traded
against other parameters in the system.
36
A reduction in earth station antenna gain
of 1.2 dB is a reduction in the gain value,
as a ratio, of 1.32. Antenna gain is
proportional to diameter squared, so the
diameter of the earth station antenna can
be reduced by a factor of 1.32 = 1.15,
from 9 m to 7.8 m.
37
We could transmit a QPSK signal from the
satellite in stead of an analog FM signal.
Using the 27-MHz noise bandwidth receiver,
we could transmit a digital signal at 54 Mbps
using QPSK, but would require a minimum
C/N ratio in the receiver of 14.6 dB, allowing
a 1-dB implementation margin and a
minimum BER of 10-6.
38
The link margin would be1.9 dB under
heavy rain conditions, so we would need to
increase the earth station antenna diameter by a
factor of 1.55 to 13.9 m to provide a C/N ratio
of 14.6 dB under heavy rain conditions. A 54-
Mbps digital signal could carry seven digital
TV signals using MPEG-2 compression, a
much more attractive proposition than carrying
a single analog FM-TV signal, although at the
cost of a larger earth station antenna.
39
Global beam antennas are not widely used,
although most Intelsat satellites carry them.
Regional TV signal distribution is much more
common, so the C-band link in Tables 4.4a and
4.4b is more likely to use a regional antenna,
serving the United States, for example, with a 6 by
3 beam. The gain of a typical satellite antenna
providing coverage of the 48 contiguous states is
32.0 dB on axis (G = 33,000/ 1 2), which is
12.0 dB higher than the on-axis gain of a global
beam.
40
Using the link budget in Tables 4.4a and 4.4b,
we can trade the extra 12-dB gain of a regional
coverage satellite antenna for a reduction in earth
station antenna dimensions. For the example of a
9.0-m antenna receiving analog FM-TV, we
could reduce the antenna diameter by a factor of
4 to 2.25 m (approximately 7 ft 6 inch diameter).
This is the smallest size of antenna used by home
satellite TV systems operating in C band.
41
The above examples show how the link
budget can be used to study different
combinations of system parameters. Most
satellite link analyses do not yield the
wanted result at the first try, and the
designer or analyst must use the link
budget to adjust system parameters until an
acceptable result is achieved.
42
4.5 SATELLITE SYSTEMS USING
SMALL EARTH STATIONS
There are many applications in which satellites carry
only one or two telephone or data channels, or a direct-
broadcast TV signal and use small, low-cost earth
stations. In these cases, earth stations costing a few
hundred or a few thousand dollars are needed. There
are only two parameters in the equation for received
power that we can adjust at the satellite to allow us to
use a small receiving antenna: satellite transmitted
power and satellite antenna beamwidth .
43
In domestic satellite systems, narrow beams can be
used for transmitting from the spacecraft to provide
coverage over only the region that the system is
designed to serve. Because the dimensions of the
antennas that can be mounted on most spacecraft
are limited, the coverage zone cannot be made
arbitrarily small. The earths disk subtends an
angle of about 17 when viewed from geostationary
orbit, and can be illuminated with a microwave
horn having an aperture a few wavelengths in
diameter.
44
At 4 GHz, to obtain a 4 spot beam, a dish 1.4
m in diameter is needed. As the frequency is
increased, the diameter of the spacecraft
antenna in wavelengths is increased for a given
dish diameter, making it feasible to use more
directive beams. However, unless a switched
or multiple beam system is used, the single
transmit (or receive) beam must cover the
whole region that the domestic satellite serves.
45
As an example, consider the problem of providing service to
the 48 contiguous states of the Unites States, known as
Conus, as illustrated in Figure 4.8; the constraints then
become apparent.
Viewed from a geostationary orbit, at a longitude of around
100 W, the continental United States subtends an angle of
about 3 in the longitude plane (N-S) and 6 in the latitude
plane (E-W). Regardless of the frequency used, an aperture
antenna to produce a single beam with 3-dB beamwidths of
6 by 3 has dimensions approximately 13 by 26, and a
gain of 32 dB. For a receiving earth station at the edge of the
coverage zone, the gain of the satellite antenna in that
direction is typically 3 dB lower, or 29dB.
46
Figure 4.8
Domestic satellite link to the Continental United States viewed from geostationary orbit, a beam approximately 6
X 3 is required for coverage of the 48 contiguous states. Additiona l beams may be provided for Alaska and
Hawaii.
47
For the GEO satellite system shown in Figure 4.8,
the received carrier level and C/N ratio can
readily be calculated for an earth station with 3-m
antenna at the edge of the coverage zone, using a
transponder with an output power of 5 W at 4
GHz, transmitting a single carrier. Ignoring all
losses, and using a receiving antenna gain of 40.0
dB (63% efficiency)
Pr = 7.0 + 29.0 196.5 + 40.0
= - 120.5 dBW (4.27)
48
The noise power at the input to a low noise
receiver with noise bandwidth of 30 MHz and
system noise temperature of 100 K is
N = kTsBn = - 228.6 + 20.0 + 74.8
= - 133.8 dBW (4.28)
Thus for this system the C/N is 13.3 dB. This is
some 3.8 dB above an FM threshold of 9.5 dB
and provides an adequate margin for an
operational system.
49
Direct Broadcast TV
Direct broadcast satellite television (DBS-TV)
originally started in Europe in the 1980s using
analog FM transmission in Ku band. It achieved a
reasonable measure of success, due in part to the
much slower introduction of cable TV systems in
Europe than occurred in the United States. In the
1990s, digital transmission became possible, and
several system were developed in the United States
in the 12.2 to 12.7 GHz band allocated to DBS-TV
services.
50
The flux density at the earth surface produced
by a DBS-TV 160-W transponder is typically
in the range - 105 to -115 dBW/m2, which
allows small receiving antennas (dishes) to be
used for DBS-TV reception, with diameters in
the range 0.45-0.75 m. The small dish
required for DBS-TV reception played a
critical part in the acceptance and success of
DBS-TV in the United States.
51
The DBS-TV system must provide a
received signal power at the small
receiving antenna that has an adequate C/N
margin in clear sky conditions. Heavy rain
will cause attenuation that exceeds the link
margin, so occasional outages will be
experienced, especially during the summer
months when thunderstorms and heavy
rain are more frequent.
52
The C/N margins used in DBS-TV systems are
usually quite small to avoid the need for a
large receiving antenna. The selection of a
C/N margin is a design trade-off between the
outage level that customers can be expected to
tolerate, the maximum allowable diameter of
the receiving dish antenna, and the power
output from the satellite transponders.
53
TABLE 4.5 Link Budget for Ku-band
DBS-TV Receiver
DBS-TV terminal received signal power
Transponder output power, 160 W 22.0 dBW
Antenna beam on-axis gain 34.3 dB
Path loss at 12.2 GHz, 38,000-Km path -205.7 dB
Receiving antenna gain, on axis 33.5 dB
Edge of beam loss -3.0 dB
Clear sky atmospheric loss -0.4 dB
Miscellaneous losses -0.4 dB
Received power, C - 119.7dBW
54
DBS-TV terminal receiver noise power
Boltzmann's constant, k -228.6 dBW/K/Hz
System noise temperature, clear sky, 145 K
21.6 dBK
Receiver noise bandwidth, 20 MHz 73.0 dBHz
Noise power, N -134.0 dBW
55
DBS-TV terminal C/N in clear sky
Clear sky overall C/N 14.3 dB
Link margin over 8.6-dB threshold
5.7 dB
Link availability throughout United States
Better than 99.7%
56
A representative link budget for a GEO
DBS-TV system serving the United States is
shown in Table 4.5. The threshold C/N
value is set at 8.6 dB, corresponding to a
system using QPSK with an implementation
margin of 0.8 dB, half rate forward error
correction that produces 6 dB of coding gain,
and a maximum BER of 10-5.
57
This requires a clear sky C/N ratio in the DBS-
TV receiver of 8.6 + 5.7 = 14.3 dB. The link
budget in Table 4.5 shows how the required
clear sky C/N is achieved for a receiver located
on the -3 dB contour of the satellite antenna
beam. A receiver located in the center of this
beam would have a clear sky C/N 3 dB higher,
and a corresponding fade margin of 8.7 dB,
sufficient to ensure only a few outages each
year.
58
In Table 4.5 a transponder output power of
160 W is used, with no backoff because a
single QPSK signal is transmitted. The
satellite antenna gain is 34.3 dB on axis,
corresponding to a high efficiency antenna
with a beam that is shaped to cover the land
mass of the United States. The beam is
approximately 5.5 wide in the E-W direction
and 2.5 in the N-S direction.
59
The resulting coverage zone, taking account
of the earth's curvature, is approximately
4000 km E-W and 2000 km N-S. A
maximum path length of 38,000 km is used
in this example. The receiving antenna is a
high efficiency design with a front-fed
offset parabolic reflector 0.45 m in diameter
and a circularly polarized feed.
60
The offset design ensures that the feed system
does not block the aperture of the antenna, which
increases its efficiency. The gain of this antenna is
33.4 dB at 12.0 GHz with an aperture efficiency
of 67%. The receiver is located at the -3dB
contour of the transmitting antenna, and
miscellaneous losses of 0.4dB for clear sky
attenuation at 12 GHz and 0.4 dB for receive
antenna mispointing and other losses are allowed.
The result is a received carrier power of-119.7
dBW in clear sky conditions.
61
The following calculations show how the system
noise temperature and (C/N) dn are determined
when rain attenuation is present.
The first step is to determine the total path
attenuation, A in dB, which is the sum of the
clear sky path attenuation due to atmospheric
gaseous absorption, Aca and attenuation due to
rain, Arain .
A = Aca + Arain dB (4.29)
62
The sky noise temperature resulting
from a path attenuation Atotal dB is found
from the output noise model of Section
4.3 using an assumed medium
temperature of 270 K for the rain.
## Tsky = 270 (110 -A/10) K (4.30)
63
The antenna noise temperature may be
assumed to be equal to the sky noise
temperature, although in practice not all of the
incident noise energy from the sky is output by
the antenna, and a coupling coefficient, c, of
90 to 95% is often used when calculating
antenna noise temperature in rain. Thus
antenna noise temperature may be calculated
as
TA =c Tsky K (4.31)
64
Almost all satellite receivers use a high gain
LNA as the first element in the receiver front
end. This makes the contribution of all later
parts of the receiver to the system noise
temperature negligible. System noise
temperature is then given by
## Ts rain = TLNA+TA K (4.32)
65
In Eq. (4.32), the LNA is assumed to be placed
right at the feed horn so that there is no waveguide
or coaxial run between the feed horn of the
antenna and the LNA. We will assume that there
are no feed losses. The increase in noise power,
Nrain dB, caused by the increase in sky noise
temperature is given by
(4.33)
where Tsca is the system noise temperature in clear
sky conditions.
66
The received power is reduced by the attenuation
caused by the rain in the slant path, so in rain the value
of carrier power is Crain where
Crain = Cca Arain dB (4.34)
The resulting (C/N) dn rain value when rain intersects the
downlink is given by
(C/N) dn rain = (C/N) dn ca Arain Nrain dB
(4.35)
where (C/N) dn ca is the downlink C/N ratio in clear sky
conditions.
67
If a linear (bent pipe) transponder is used, the
(C/N) up ratio must be combined with (C/N)dn rain
to yield the overall (C/N)o ratio for the link. Many
digital systems use regenerative transponders that
provide constant output power regardless of
uplink attenuation provided that the received C/N
ratio at the satellite is above the threshold of the
onboard processing demodulator. In this case the
value of (C/N) dn rain will be used as the overall
(C/N) o value in rain for the link.
68
EXAMPLE 4.5.1
In the example of a DBS-TV system in Table 4.5, a
link margin of 5.7 dB is available before the (C/N)o
threshold of 8.6 dB is reached. We must calculate the
increase in system noise temperature that results from
3-dB rain attenuation in the downlink path to
determine the increase in noise power and thus the
value of (C/N)dn rain,. This example shows how the link
margin cam be distributed between rain attenuation
and an increase in receiver system noise power caused
by an increase in sky noise.
69
The clear sky attenuation is given in Table 4.5 as
0.4 dB. Thus total excess attenuation is 3.4 dB,
and the sky noise temperature in rain will be, from
Eq. (4.30)
Tsky rain = 270 (1- 10-3.4/10) = 147 K
In clear sky conditions the sky noise temperature is
T ca = [0.95 (270 (1 10-0.04))] = 23 K. The sky
temperature has increased from 23 K in clear sky
conditions to 147 K when 3 dB rain attenuation
occurs in the downlink.
70
We must calculate the new system noise
temperature when rain is present in the slant
path, remembering that the antenna noise
temperature has two parts: a fixed part due to
noise from the ground entering the antenna's
sidelobes, and a variable part due to sky
temperature. The LNA of the system in Table
4.5 has a noise temperature of 110 K and the
clear sky system noise temperature is T ca =
145 K.
71
The antenna temperature in clear sky conditions
is made up of a fixed portion of 12 K and a sky
noise portion of 23 K. If we assume 95%
coupling of sky noise temperature to antenna
noise temperature, the system noise
temperature, given by Eq. (4.32), increases to
T s rain = TLNA + TA
= 110 +12 +147
= 269 K
72
The increase in receiver noise power referred
to the receiver input is given by Eq. 4.33.
Nrain = 10 log 10[T s rain / T s ca]
= 10 log 10 [269/145]
=2.7 dB
From Eq. (4.35)
(C/N) dn rain = 14.3 3.0 -2.7
= 8.6 dB
73
Thus the link margin required to withstand 3 dB
of rain attenuation in the downlink path is 5.7
dB, and this will guarantee a link availability
exceeding 99.7% of an average year for the
majority of the DBS-TV system customers in
the United States. In the south eastern United
States, in states such as Florida and Louisiana
where very heavy rain occurs more often than in
other parts of the United States, link availability
may be slightly less than 99.7%.
74
Shaping of the satellite beam to direct more
power to these parts of the United States helps
to reduce the number of outages experienced in
that region. Receivers located within the -1 dB
contour of the satellite antenna beam have
shorter path lengths giving 2.3 dB higher C/N
than the receiver used in the example shown in
Table 4.5, so they have a link margin of 8 dB.
75
The calculation of the availability of these
receivers requires some care, because we cannot
just add the extra 2.3 dB of link margin to the rain
attenuation. Antenna noise increases with every
decibel of extra rain attenuation, reducing the
received power level, C, and increasing the
system noise power N. An iterative (trial and
error) procedure must be used to find the
combination of reduction in C and increase in N
that leads to an additional 2.3-dB degradation in
the C/N value.
76
We will guess that increasing the rain attenuation
from 3 to 4.5 dB leads to an increase in noise
power of 0.8 dB. With 4.5-dB rain attenuation,
the system noise temperature is
T s rain = l10 + 12 + 0.95 270 (1 0.355)
= 287 K
The increase in noise power from the clear sky
condition is
N = 10 log 10 (287/145) = 3.0 dB
77
Hence the decrease in C/N for 4.5 dB of rain
attenuation is 4.5 +3.0 = 7.5 dB, and we are a little
below our new margin of 8.0 dB. Increasing the rain
attenuation by a further 0.4 dB to 4.9 dB and
repeating the calculations gives a new system noise
temperature in rain of 295 K. The corresponding
increase in noise power is 3.1 dB giving a reduction
in C/N of 8.0 dB, equal to the link margin. A rain
attenuation margin of 4.9 dB at Ku band would give
an availability of 99.88% or better over the central
region of the United States.
78
4.6 UPLINK DESIGN
The uplink design is easier than the downlink in many cases,
since an accurately specified carrier power must be
presented at the satellite transponder and it is often feasible
to use much higher power transmitters at earth stations than
can be used on a satellite. However, VSAT systems use
earth stations with small antennas and transmitter powers
below 5 W, giving low uplink EIRP. Satellite telephone
handsets are restricted to transmitting at power levels below
1 W because of the risk of EM radiation hazards. In mobile
systems the uplink from the satellite telephone is usually the
link with the lowest C/N ratio.
79
The cost of transmitters tends to be high compared
with the cost of receiving equipment in satellite
communication systems. The major growth in
satellite communications has been in point-to-
multipoint transmission, as in cable TV distribution
and direct broadcast satellite television. One high-
power transmit earth station provides service via a
DBS satellite to many low-cost receive-only stations,
and the high cost of the transmitting station is only a
small part of the total network cost.
80
Although flux density at the satellite is a convenient
way to determine earth station transmit EIRP
requirements, analysis of the uplink requires calculation
of the power level at the input to the transponder so that
the uplink C/N ratio can be found. The link equation is
used to make this calculation, using either a specified
transponder C/N ratio or a required transponder output
power level. When a C/N ratio is specified for the
transponder, the calculation of required transmit power
is straightforward. Let (C/N) up , be the specified C/N
ratio in the transponder, measured in a noise bandwidth
Bn Hz.
81
The bandwidth Bn Hz is the bandwidth of
the band-pass filter in the IF stage of the
earth station receiver for which the uplink
signal is intended. Even if Bn is much less
than the transponder bandwidth, it is
important that the uplink C/N ratio be
calculated in the bandwidth of the receiver,
not the bandwidth of the transponder.
82
The noise power referred to the transponder
input is Nxp W where
Nxp = k +Txp + Bn dBW (4.36)
## where Txp is the system noise temperature of
the transponder in dBK and Bn is in units of
dBHz.
83
The power received at the input to the transponder is Prxp
Where
Prxp = Pt + Gt + Gr - Lp - Lup dBW (4.37)
where PtGt is the uplink earth station EIRP in dBW, Gr
is the satellite antenna gain in dB in the direction of the
uplink earth station and Lp is the path loss in dB. The
factor Lup dB accounts for all uplink losses other than
path loss. The value of (C/N)up at the LNA input of the
satellite receiver is given by
C/N = 10 log 10 [Pr / (kTsBn)]
= Prxp - Nxp dB (4.38)
84
The earth station transmitter output power Pt is
calculated from Eq. (4.37) using the given
value of C/N in Eq. (4.38) and the noise power
Nxp calculated from Eq. (4.36). Note that the
received power at the transponder input is also
given by
## Prxp = N + C/N dBW (4.39)
85
The earth station transmitter output power Pt can
also be calculated from the output power of the
transponder and transponder gain when these
parameters are known and a bent pipe transponder
is used. In general
Prxp = Psat - BO0 - Gxp dBW (4.40)
## where Psat is the saturated power output of the
transponder in dBW, BO0 is the output backoff in
dB, and Gxp is the gain of the transponder in dB.
86
EXAMPLE 4.6.1
A transponder of a Ku-band satellite has a linear gain
of 127 dB and a nominal output power at saturation of
5 W. The satellites 14-GHz receiving antenna has a
gain of 26 dB on axis, and the beam covers western
Europe.
Calculate the power output of an uplink transmitter that
gives an output power of I W from the satellite
transponder at a frequency of 14.45 GHz when the
earth station antenna has a gain of 50 dB and there is a
1.5-dB loss in the waveguide run between the
transmitter and antenna. 87
Assume that the atmosphere introduces a loss of
0.5 dB under clear sky conditions and that the
earth station is located on the -2 dB contour of
the satellite's receiving antenna. If rain in the path
causes attenuation of 7 dB for 0.01% of the year,
what output power rating is required for the
transmitter to guarantee that a 1-W output can be
obtained from the satellite transponder for
99.99% of the year if uplink power control is
used?
88
The input power required by the transponder is
simply the output power minus the transponder
gain, so
Pin = 0 dBW - 127 dB
= -127 dBW
The uplink power budget is given by Eq. (4.11)
Pr = EIRP + Gr Lp Lat
Lta Lra dBW
89
Rearranging and putting in the appropriate
losses
Pt = Pr - Gt Gr
+ Lp + Lta + Lat + Lpt dBW
where Lta is the waveguide loss, Lat is the
atmospheric loss, and Lpt, is the pointing loss
(antenna pattern loss). Then assuming a path
length of 38,500 km
Pt = -127.0 5026
207.21.50.52.0 dBW 90
That is
Pt = 7.2 dBW or 5.2 W
If we provide an extra 7 dB of output power
to compensate for fading on the path due to
rain, the transmitter output power will be
P t rain = 7.27
= 14.2 dBW or 26.3 W
91
4.7 DESIGN FOR SPECIFIED C/N:
COMBINING C/N AND C/l
VALUES IN SATELLITE LINKS
The BER or S/N ratio in the baseband channel
of an earth station receiver is determined by the
ratio of the carrier power to the noise power in
the IF amplifier at the input to the demodulator.
The noise present in the IF amplifier comes
from many sources.
92
When more than one C/N ratio is present in the
link, we can add the individual C/N ratios
reciprocally to obtain an overall C/N ratio,
which we will denote here as (C/N) o. The
overall (C/N)o ratio is what would be measured
in the earth station at the output of the IF
amplifier
(C/N) o = 1 / [1/(C/N)1 + 1 / (C/N)2
+ 1 / (C/N)3 + ] (4.42)
93
This is sometimes referred to as the reciprocal C/N
formula. The C/N values must be linear ratios, NOT
decibel values. Since the noise power in the
individual C/N ratios is referenced to the earner
power at that point, all the C values in Eq. (4.42) are
the same. Expanding the formula by cross
multiplying gives the overall (C/N) o as a power ratio,
not in decibels
(C/N) o = 1/(N1/C + N2/C + N3/C + )
= C/(N1 + N2 + N3 + ) (4.43)
94
In decibel units:
(C/N) o = C dBW10 log10 (N1 + N2 + N3)
+ W) dB (4.44)
Note that (C/N)dn cannot be measured at the
receiving earth station. The satellite always
transmits noise as well as signal, so a C/N ratio
measurement at the receiver will always yield
(C/N) o, the combination of transponder and
earth station C/N ratios.
95
To calculate the performance of a
satellite link we must therefore determine
the uplink (C/N)up ratio in the transponder
and the downlink (C/N)dn in the earth
station receiver. We must also consider
whether there is any interference present,
either in the satellite receiver or the earth
station receiver.
96
One case of importance is where the transponder is
operated in a FDMA mode and intermodulation
products (IM) are generated by the transponder's
nonlinear input-output characteristic. If the IM
power level in the transponder is known, a C/I
value can be found and included in the calculation
of (C/N)o ratio. Interference from adjacent satellites
is likely whenever small receiving antennas are
used as with VSATs (very small aperture terminals)
and DBS-TV receivers.
97
Since C/N values are usually calculated
from power and noise budgets, their
values are typically in decibels. There
are some useful rules of thumb for
estimating (C/N) o from two C/N values:
98
z If the C/N values are equal, as in the example
above, (C/N) o is 3 dB lower than either value.
z If one C/N value is 10 dB smaller than the other
value, (C/N) o is 0.4 dB lower than the smaller of
the C/N values.
z If one C/N value is 20 dB or more greater than
the other C/N value, the overall (C/N) o is equal
to the smaller of the two C/N values within the
accuracy of decibel calculations (0.1 dB).
99
EXAMPLE 4.7.1
Thermal noise in an earth station receiver results
in a (C/N) dn ratio of 20.0 dB. A signal is
received from a bent pipe transponder with a
carrier to noise ratio (C/N) up = 20.0 dB. What is
the value of overall (C/N) o at the earth station?
If the transponder introduces intermodulation
products with (C/I) ratio = 24 dB, what is the
overall (C/N) o ratio at the receiving earth station?
100
Using Eq. (4.42) and noting that (C/N) = 20.0
dB corresponds to a (C/N) ratio of 100
## The intermodulation (C/I) value of 24.0 dB
corresponds to a ratio of 250. The overall (C/N)
o value is then
101
Overall (C/N)o with Uplink and
Downlink Attenuation
## The effect of a change in the uplink C/N
ratio has a different impact on overall
(C/N)o depending on the operating mode
and gain of the transponder.
102
There are three different transponder types
or operating modes:
Linear transponder :
Pout = Pin + Gxp dBW
Nonlinear transponder :
Pout = Pin + Gxp - G dBW
Regenerative transponder :
Pout = constant dBW
103
where Pin is the power delivered by the satellite's
receiving antenna to the input of the transponder, Pout
is the power delivered by the transponder HPA to the
input of the satellite's transmitting antenna, Gxp is the
gain of the transponder, and all parameters are in
decibel units.
The parameter G is dependent on Pin and accounts
for the loss of gain caused by the nonlinear saturation
characteristics of a transponder which is driven hard to
obtain close to its maximum power outputthe gain is
effectively falling as the input power level increases.
104
Uplink Attenuation and (C/N) up
The transponder receiver noise temperature does not
change significantly when rain is present in the uplink
path to the satellite. The satellite receiving antenna
beam is always sufficiently wide that it "sees" a large
area of the (warm) earth's surface and local noise
temperature variations are insignificant. The noise
temperature of the earth seen by a GEO satellite varies
from a maximum of 270 K for a satellite antenna beam
over Africa and northern Europe, to a minimum of
250 K over the Pacific Ocean.
105
The corresponding system noise temperature for
the transponders on a GEO satellite is in the
range 400 to 500 K. There is effectively no
increase in uplink noise power when heavy rain
is present in the link between an earth station
and a satellite because the satellite antenna
beam sees the tops of cumulonimbus clouds
above the rain, which are always colder than
270 K, instead of the earth's surface.
106
Rain attenuation on the uplink path to the satellite
reduces the power at the satellite receiver input,
and thus reduces (C/N)up in direct proportion to the
attenuation on the slant path. If the transponder is
operating in a linear mode, the output power will
be reduced by the same amount, which will cause
(C/N) dn to fall by an amount equal to the
attenuation on the uplink. When both (C/N)up and
(C/N) dn are reduced by A up dB, the value of
(C/N)o is reduced by exactly the same amount, A up
dB.
107
Hence for the case of a linear transponder and rain
attenuation in the uplink of A up dB
(C/N) o uplink rain = (C/N)o clear air A up dB
Linear transponder (4.45)
If the transponder is nonlinear, the reduction in input
power caused by uplink attenuation of A up dB results
in a smaller reduction in output power, by an amount
G.
(C/N)o uplink rain = (C/N) o clear air Aup + G dB
Nonlinear transponder (4.46)
108
If the transponder is digital and regenerative, or
incorporates an Automatic Gain Control (AGC) system
to maintain a constant output power level
(C/N) o uplink rain = (C/N)o clear air dB
Regenerative transponder or AGC (4.47)
The above equation will hold only if the received
signal is above threshold and the BER of the recovered
digital signal in the transponder is small. If the signal
falls below threshold, the uplink will contribute
significantly to the BER of the digital signal at the
receiving earth station.
109
Downlink Attenuation and (C/N) dn
The earth station receiver noise temperature can
change very significantly when rain is present in
the downlink path from the satellite. The sky noise
temperature can increase to close to the physical
temperature of the individual raindrops,
particularly in very heavy rain. A reasonable
temperature to assume for temperate latitudes in a
variety of rainfall rates is 270 K, although values
above 290 K have been observed in the tropics.
110
An increase in sky noise temperature to 270 K
will increase the receiving antenna temperature
markedly above its clear air value. The result is
that the received power level, C, is reduced and
the noise power, N, in the receiver increases.
The result for downlink C/N is given by Eq. (4.48)
(C/N) dn rain = (C/N)dn clear airArain
Nrain dB (4.48)
111
The overall C/N is then given by
(C/N) o = 1 / [1 /(C/N) dn rain
+ 1/(C/N)up] dB (4.49)
As noted earlier, unless we are making a
loop-back test, we will assume that the value
of (C/N) up is for clear air, and remains
constant regardless of the attenuation on the
downlink.
112
System Design for Specific Performance
A typical two-way satellite communication link consists
of four separate paths: an outbound uplink path from
one terminal to the satellite and an outbound downlink
to the second terminal ; and an inbound uplink from the
second terminal to the satellite and an inbound
downlink to the first terminal. The links in the two
directions are independent and can be designed
separately, unless they share a single transponder using
FDMA. A broadcast link, like the DBS-TV system
described earlier in this chapter, is a one-way system,
with just one uplink and one downlink.
113
Satellite Communication Link
Design Procedure
The design procedure for a one-way satellite
communication link can be summarized by the
following 10 steps. The return link design follows
the same procedure.
1. Determine the frequency band in which the
system must operate. Comparative designs may be
required to help make the selection.
114
2. Determine the communications parameters of
the satellite. Estimate any values that are not
known.
3. Determine the parameters of the transmitting and
receiving earth stations.
4. Start at the transmitting earth station. Establish
an uplink budget and a transponder noise power
budget to find (C/N)up in the transponder.
5. Find the output power of the transponder based
on transponder gain or output backoff.
115
6. Establish a downlink power and noise budget for
the receiving earth station.Calculate (C/N)dn and
(C/N)o for a station at the edge of the coverage zone
(worst case).
7. Calculate S/N or BER in the baseband channel.
Find the link margins.
8. Evaluate the result and compare with the
specification requirements. Change parameters of
the system as required to obtain acceptable (C/N)o
or S/N or BER values. This may require several trial
designs.
116
9. Determine the propagation conditions
under which the link must operate.
Calculate outage times for the uplinks and
downlinks.
10. Redesign the system by changing some
parameters if the link margins are
inadequate. Check that all parameters are
reasonable, and that the design can be
implemented within the expected budget.
117
4.8 SYSTEM DESIGN
EXAMPLES
The following sample system designs
demonstrate how the ideas developed in
this chapter can be applied to the design
of satellite communication systems.
118
TABLE 4.6 System and Satellite Specification
Ku-band satellite parameters
Geostationary at 73W longitude, 28 Ku-band
transponders
Total RF output power 2.24 kW
Antenna gain, on axis (transmit and receive)
31 dB
Receive system noise temperature 500 K
119
Transponder saturated output power:
Ku band 80 W
Transponder bandwidth:
Ku band 54 MHz
Signal
Compressed digital video signals with
transmitted symbol rate of 43.2 Msps
Minimum permitted overall (C/N)o in receiver
9.5 dB
120
Transmitting Ku-band earth station
Antenna diameter 5m
Aperture efficiency 68%
Uplink frequency 14.15 GHz
Required C/N in Ku-band transponder 30 dB
Transponder HPA output backoff 1 dB
Miscellaneous uplink losses 0.3 dB
Location:
-2 dB contour of satellite receiving antenna
121
Receiving Ku-band earth station
Downlink frequency 11.45 GHz
Receiver IF noise bandwidth 43.2 MHz
Antenna noise temperature 30 K
LNA noise temperature 110 K
Required overall (C/N)o in clear air 17 dB
Miscellaneous downlink losses 0.2 dB
Location: -3 dB contour of satellite
transmitting antenna
122
Rain attenuation and propagation factors
Ku-band clear air attenuation
Uplink 14.15 GHz 0.7 dB
Downlink 11.45 GHz 0.5 dB
Rain attenuation
Uplink 0.01% of year 6.0 dB
Downlink 0.01% of year 5.0 dB
123
System Design Example 4.8.1
This example examines the design of a satellite
communication link using a Ku-band geostationary
satellite with bent pipe transponders to distribute
digital TV signals from an earth station to many
receiving stations throughout the United States.
The design requires that an overall C/N ratio of 9.5
dB be met in the TV receiver to ensure that the
video signal on the TV screen is held to an
acceptable level.
124
The uplink transmitter power and the
receiving antenna gain and diameter are
determined for each system. The available
link margins for each of the systems are
found and the performance of the systems is
analyzed when rain attenuation occurs in the
satellite-earth paths. The advantages and
disadvantages of implementing uplink power
control are considered.
125
In this example, the satellite is located at
73W. However, for international
registration of this satellite location, the
location would be denoted as 287E. The
link budgets developed in the examples
below use decibel notation throughout. The
satellite and earth stations are specified in
Table 4.6, and Figure 4.11 shows an
illustration of the satellite television
distribution system.
126
Ku-Band Uplink Design
We must find the uplink transmitter power
required to achieve (C/N) up = 30 dB in
clear air atmospheric conditions. We will
first find the noise power in the transponder
for 43.2 MHz bandwidth, and then add 30
dB to find the transponder input power
level.
127
Uplink Noise Power Budget
k = Boltzmann's constant -228.6 dBW/K/Hz
Ts= 500 K 27.0 dBK
B = 43.2 MHz 76.4 dBHz
N = transponder noise power -125.2 dBW
## The received power level at the transponder input
must be 30 dB greater than the noise power.
Pr = power at transponder input = -95.2 dBW
128
Figure 4.11
Satellite television distribution system.
129
The uplink antenna has a diameter of 5 m
and an aperture efficiency of 68%. At 14.15
GHz the wavelength is 2.120 cm = 0.0212 m.
The antenna gain is
Gt = 10 log [0.68 (D/)2] = 55.7 dB
The free space path loss is
Lp = 10 log [(4R/) 2] = 207.2 dB
130
Uplink Power Budget
Pt = Earth station transmitter power Pt dBW
Gt = Earth station antenna gain 55.7 dB
Gr = Satellite antenna gain 31.0 dB
Lp = Free space path loss -207.2 dB
Lant = E/S on 2 dB contour -2.0 dB
Lm = Other losses -1.0dB
Pr = Received power at transponder
Pt - 123.5 dB
131
The required power at the transponder input to
meet the (C/N)up = 30 dB objective is -95.2
dBW. Hence
Pt123.5 dB = -95.2 dBW
Pt = 28.3 dBW or 675 W
This is a relatively high transmit power so we
would probably want to increase the
transmitting antenna diameter to increase its
gain, allowing a reduction in transmit power.
132
Ku-Band Downlink Design
The first step is to calculate the downlink
(C/N)dn that will provide (C/N)o = 17 dB
when(C/N)up = 30 dB. From Eq. (4.43)
1 / (C/N) dn = l / (C/N)ol/(C/N)up
(not in dB)
Thus
l / (C/N)dn = 1/501/1000 = 0.019
(C/N)dn = 52.6 17.2 dB
133
We must find the required receiver input power
to give (C/N)dn = 17.2 dB and then find the
receiving antenna gain, Gr.
Downlink Noise Power Budget
k =Boltzmann's constant -228.6 dBW/K/Hz
Ts = 30 + 110 K = 140 K 21.5 dBK
Bn = 43.2 MHz 76.4 dBHz
N = transponder noise power -130.7dBW
134
The power level at the earth station
receiver input must be 17.2 dB greater
than the noise power in clear air.
## Pr = power at earth station receiver input
= -130.7dBW + 17.2 dB = -113.5dBW
135
We need to calculate the path loss at 11.45 GHz.
At 14.15 GHz path loss was 207.2 dB. At 11.45
GHz path loss is
Lp = 207.2201og10 (14.15 / 11.45)
= 205.4 dB
The transponder is operated with 1 dB output
backoff, so the output power is 1 dB below 80
W (80 W 19.0 dBW)
Pt = 19 dBW1 dB
= 18 dBW 136
Downlink Power Budget
Pt = Satellite transponder output power 18.0 dBW
Gt = Satellite antenna gain 31.0 dB
Gr = Earth station antenna gain Gr dB
Lp = Free space path loss -205.4 dB
La = E/S on -3 dB contour of satellite antenna
-3.0 dB
Lm = Other losses -0.8 dB
Pr = Received power at earth station Gr160.2 dB
137
The required power into the earth station
receiver to meet the (C/N)dn = 17.2 dB
objective is Pr = -120.1 dBW. Hence the
receiving antenna must have a gain Gr. Where
## Gr160.2 dB = - 113.5 dBW
Gr = 46.7 dB or 46,774
as a ratio
138
The earth station antenna diameter, D, is
calculated from the formula for antenna gain, G,
with a circular aperture
Gr = 0.65 (D / )2 = 46,774
## At 11.45 GHz, the wavelength is 2.62 cm =
0.0262 m. Evaluating the above equation to
find D gives the required receiving antenna
diameter as D = 2.14m.
139
Rain Effects at Ku Band
Uplink Under conditions of heavy rain,
the Ku-band path to the satellite station
suffers an attenuation of 6 dB for 0.01%
of the year. We must find the uplink
attenuation margin and decide whether
uplink power control would improve
system performance at Ku band.
140
The uplink C/N was 30 dB in clear air. With 6 dB
uplink path attenuation, the C/N in the transponder
falls to 24 dB, and assuming a linear transponder
characteristic and no uplink power control, the
transponder output power falls to 186 = 12 dBW.
The downlink C/N falls by 6 dB from 17.2 dB to 11.2
dB, and the overall (C/N)o falls by 6 dB to 11 dB. With
the minimum overall C/N set at 9.5 dB, the additional
margin for uplink attenuation is 1.5 dB. Hence the link
margin available on the uplink is 7.5 dB without uplink
power control. This is an adequate uplink rain
attenuation margin for many parts of the United States,
and would typically lead to rain outages of less than141
1
h total time per year
Downlink Attenuation and Sky Noise Increase
The 11.45-GHz path between the satellite and
the receive station suffers rain attenuation
exceeding 5 dB for 0.01% of the year. Assuming
100% coupling of sky noise into antenna noise,
and 0.5-dB clear air gaseous attenuation,
calculate the overall C/N under these conditions.
Assume that the uplink station is operating in
clear air. We must calculate the available
downlink fade margin.
142
We need to find the sky noise temperature that
results from a total excess path attenuation of
5.5 dB (clear air attenuation plus rain
attenuation); this is the new antenna
temperature in rain, because we assumed 100%
coupling between sky noise temperature and
antenna temperature. We must evaluate the
change in received power and increase in
system noise temperature in order to calculate
the change in C/N ratio for the downlink.
143
In clear air, the atmospheric attenuation on the
downlink is 0.5 dB. The corresponding sky noise
temperature is 270(110 -0.05) = 29 K, which leads to
the antenna temperature of 30 K given in the Ku-band
system specification. When the rain causes 5-dB
attenuation, the total path attenuation from the
atmosphere and the rain is 5.5 dB. The corresponding
sky noise temperature is given by
T sky rain = T0 (1G) where
G = 10-A/10 = 0.282
T sky rain = 270(10.282) = 194 K
144
Thus the antenna temperature has increased
from 30 K in clear air to 194 Kin rain. The
system noise temperature in rain, T s rain, is
increased from the clear air value of 140 K (30
K sky noise temperature plus 110 K LNA
temperature)
## T s rain = 194 + 110
= 304 K or 24.8 dBK
145
The increase in noise power is
N = 10 log (304/140) = 3.4 dB
## The signal is attenuated by 5 dB in the rain, so
the total reduction in downlink C/N ratio is 8.4
dB, which yields a new value
(C/N) dn rain = 17.28.4 = 8.8 dB
146
The overall C/N is then found by combining the clear
air uplink (C/N)up of 30 dB with the rain faded
downlink (C/N)dn rain of 8.8 dB, giving
(C/N)o rain = 8.8 dB
The overall (C/N)0 is below the minimum acceptable
value of 9.5 dB. The downlink link
margin is
Downlink fade margin
= (C/N)dn(C/N)min = 17.2 - 9.5
= 7.7 dB
147
Since downlink rain attenuation of 5 dB causes the
overall (C/N)0 to go below the minimum permitted
value of 9.5 dB, we should recalculate the
maximum attenuation that the downlink can sustain.
This involves an iterative process, since changing
the attenuation changes both C and N values in
(C/N)dn. At an attenuation level of 5 dB, the increase
in noise power is 3.4 dB, so a starting guess would
be that decreasing the attenuation by 0.3 dB will
decrease the noise power by 0.2 dB. The rain
attenuation will then be a little less than 5 dB.
148
Recalculating (C/N)dn for a rain attenuation value
of 4.7 dB gives
## T sky rain = T0 (1G) where G = 10-A/10 = 0.339
T sky rain = 270 (10.339) = 178 K
N = 10 log (288/140) = 3.1 dB
(C/N) dn rian = 17.24.73.1 = 9.4 dB
(C/N) o rain = 9.36 9.4 dB
149
The result is close enough to the required
value of (C/N) o min = 9.5 dB to conclude
that we can tolerate about 4.7 dB of rain
attenuation on the downlink.
If better availability is requiredless
outage timethe diameter of the receiving
antenna can be increased.
150
For example, if the receiving antenna diameter is
increased to 2.4 m, (about 8 ft) the increase in
antenna gain is 20 log10(2.40/2.14) = 1.0 dB, which
increases the downlink margin to 8.7 dB.
Repeating the iterative calculation outlined above,
the corresponding rain attenuation on the downlink
is 5.5 dB with a noise power increase of 3.2 dB.
The downlink C/N with 5.5-dB rain attenuation is
17.28.7 = 9.5 dB, and the overall (C/N)o 9.5
dB.
151
The extra antenna gain now ensures that the link
meets the required specification, which will keep
outages to a total of about 50 min in an average
year in the eastern United States. However, an
increase in antenna diameter will reduce the
beamwidth of the antenna and may require an
upgrade in the tracking requirements. With a fixed
pointing antenna, diurnal motion of the satellite
may cause a variation in received signal strength
as the satellite moves through the antenna beam.
152
System Design Example 4.8.2
Personal Communication System Using
Low Earth Orbit Satellites
Low earth orbit (LEO) satellite systems are
designed to provide personal communication
service similar to a cellular telephone, but over
a much wider area.
153
LEO satellite systems can cover sparsely
populated regions of a country, or the world,
where there are no terrestrial cellular telephone
systems. The user has a handset similar to a
cellular telephone handset that provides two-way
voice communications through a gateway station,
usually to a conventional telephone in a home or
office connected to the public switched telephone
network (PTSN). Satellite telephones can equally
well connect to another satellite handset, or to a
terrestrial cellular telephone.
154
Most LEO satellite systems operate in L band,
in the 1500- and 1600-MHz bands, and in the
lower part of S band around 2460 MHz,
frequency bands that are allocated for mobile
satellite communications. Some LEO systems
use intersatellite links so a user can connect to
any point in the world without an intermediate
return to earth.
155
However, the signals invariably pass through a
gateway station at each end of the link to facilitate
control of the call and to ensure that users can be
charged for using the system. Connections between
the gateway earth stations and the satellites use S-
band, C-band, Ku-band, or Ka-band frequencies,
depending on the system requirements. Only a
small portion of the radio spectrum at L band is
allocated to LEO and MEO satellite systems, so L-
band frequencies are reserved for the critical links
between the user and the satellite.
156
A handoff process is required for LEO
satellites similar to that used in cellular
telephone networks, but the handoff
between satellites should not be apparent to
the user. Most LEO satellites have multiple
beam antennas, and the beam pattern moves
across the earth's surface at the speed of the
satellitetypically about 7.7 km/s or 17,200
mph.
157
A single beam is typically 500 km in diameter,
so an individual user is in any one beam for less
than a minute. The system provides automatic
switching from beam to beam within the same
satellite antenna coverage, much like a cellular
telephone system switches users from cell to cell,
which nearly always requires a change in the link
frequencies, but as with satellite to satellite
handoffs, the process must be transparent to the
user.
158
The example below analyzes the links between a user
and a gateway station. LEO satellite systems employ
digital transmission so that advantage can be taken of
forward error correction coding (FEC) and speech
compression techniques. The bit rate of digital voice in
an LEO satellite link is typically 4800 bps, requiring
powerful compression algorithms. The low bit rate
allows more signals to be sent in the available
transponder bandwidth and also helps maintain the C/N
ratio in the receivers. When FEC is applied to a digital
bit stream, carrier to noise ratios down to 5 dB can be
used. The low bit rate and operation of the receivers at
low C/N ratios are essential to make personal 159
## communication via an LEO satellite possible.
The link between the gateway station and
the mobile terminal is defined as the
outbound link, and the link from the mobile
terminal to the gateway is the inbound link.
Note that there are four satellite paths, just
as in all other two-way satellite
communication systems:
outbound uplink, outbound downlink,
inbound uplink ,inbound downlink.
160
Figure 4.12
Two- way personal communication system using L -band LEO satellite links.
161
Each has its own unique frequency, and in most
LEO satellite systems, one of the links will be
weaker than the other three links and will thus limit
the system performance.
One objective in the example that follows is to
identify the weakest path and to then attempt to
improve that part of the system. Figure 4.12
illustrates the two-way link between the gateway
station and the handset. Note that separate
transponders are used for the inbound and outbound
paths.
162
In this example, the mobile terminals transmit to a
transponder on the satellite using frequency
division multiple access (FDMA) and single
channel per carrier (SCPC) techniques. The
available frequencies are shared among active
users on demand, as in a cellular telephone system,
so a call begins with a start-up sequence that
establishes communication between the mobile
terminal and the local gateway station via the
nearest LEO satellite.
163
The gateway station then allocates frequencies for
the call. At the end of the call, the frequencies are
released and become available for another user.
This is called demand assignment (DA), and the
multiple access technique is identified by the
acronym SCPC-FDMA-DA, or, alternatively,
SCPC-FDMA-DAMA where DAMA stands for
demand assignment multiple access. A common
set of control channels at preassigned frequencies
enables call setup and teardown.
164
The link from the gateway station via the satellite
to the mobile terminal uses time division
multiplexing (TDM). A TDM signal consists of a
sequence of packets with addresses that repeat
every 20 to 100 ms. The addresses identify which
terminal should receive each packet. The TDM
bit stream rate must exceed the total bit rate of all
active terminals in a two-way telephone system
so that there is sufficient capacity available for
each terminal within the TDM bit stream.
165
In this example we begin by assuming that
50 active users share one common TDM
channel. We will also assume that the
gateway earth station operates at Ku band
to and from the satellite, and that the
satellite employs a linear transponder (bent
pipe) rather than having onboard processing.
166
The parameters of the satellite transponder,
the mobile terminal, and the gateway station
are given in Table 4.7. The table gives the
maximum path length for any satellite-earth
link. It is left as an exercise for the reader to
determine a suitable combination of orbital
altitude and minimum elevation angle for the
LEO system.
167
TABLE 4.7 LEO Satellite Personal
Communication System
Parameters
Satellite parameters
Saturated output power 10w
Transponder bandwidth 1 MHz
Uplink frequency for mobile terminal 1650 MHz
Downlink frequency for mobile terminal
1550 MHz
168
Antenna gain 1650 MHz uplink (one beam) 23 dB
Antenna gain 1550 MHz downlink (one beam) 23 dB
Uplink frequency for gateway station 14 GHz
Downlink frequency for gateway station 11.5 GHz
Antenna gain 14 GHz uplink 3 dB
Antenna gain 11.5 GHz downlink 3 dB
Satellite receiver system noise temperature 500 K
Maximum range to edge of coverage zone 2200 km
169
Mobile terminal parameters
Transmitter output power 0.5 W
Antenna gain (transmit and receive) 0 dB
Receiver system noise temperature 300 K
Transmit bit rate 4800 bps
Receive bit rate 96 kbps
Required maximum bit error rate 10-4
170
Gateway station parameters
Transmitter output power (maximum per transponder)
10 W
Antenna gain (transmit, 14.0 GHz) 55 dB
Antenna gain (receive, 11.5 GHz) 53.5 dB
Receive system noise temperature (clear air) 140 K
Transmit bit rate (before FEC encoder) 300 kbps
Receive bit rate (after FEC decoder) 4800 bps
Required maximum bit error rate 10-4
171
The user's transmitter and receiver is
called a mobile terminal in this example.
It could be a handheld device like a
cellular telephone, sometimes called a
satellite telephone or handset, or the
terminal could be mounted in a vehicle.
172
The satellite has multiple L band beams serving
different parts of its instantaneous coverage
zone because a single beam from an LEO
satellite serving different parts of its
instantaneous coverage zone would have both
low gain and limited capacity. For an antenna
with a gain of 23 dB, G = 200 and the
beamwidth is 3 dB where
3dB = (33,000/200) = 12.8
173
The use of a multiple beam antenna on the satellite
increases the antenna gain toward the mobile
terminal, which increases the C/N ratio of the
signals in the mobile terminal and gateway station
receivers. The Ku-band antennas that link the
satellite to the gateway station have broad beams
and low gain. The C/N on these links is high
through the use of a relatively large antenna and a
high transmitter power at the gateway earth station,
allowing the use of small and simple Ku-band
antennas on the satellite.
174
The antenna gain at the mobile terminal is low,
with a value of 0 dB used for calculation,
because the antenna coverage of the terminal
must be very broad. If the terminal is a satellite
I telephone, an omnidirectional antenna allows
the user to move around freely. If the mobile
terminal antenna gain were to be increased, its
beam would be correspondingly narrower, and
the user would have to point the handset at the
satellite.
175
In an LEO satellite system, the user does not
know which satellite is being used nor where it
is in the sky, so requiring the user to point the
handset antenna at the satellite is not a feasible
option. When the mobile terminal is mounted
in a vehicle with the antenna on the roof,
pointing the antenna at the satellite is not
possible unless a sophisticated (and expensive)
steered antenna is used.
176
In this example, we will begin by assuming that
there are 50 users sharing a single transponder on
the satellite, and that one transponder serves one of
the L-band beams within the LEO satellite
coverage, operating within a given set of
frequencies. A large number of users can share an
LEO satellite through the provision of many
transponders, each of which is connected to one of
the individual beams in the multiple L-band
antenna coverage of the satellite.
177
The signal received by a mobile terminal from the
gateway is a TDM sequence of packets carrying
50 digital voice channels, each at 4800 bps. The
bit rate of the TDM signal would be 240 kbps if it
carried only the voice signals, but will be I higher
in practice because additional bits must be sent
with each packet; a TDM bit rate I of 300 kbps is
used in this example. Individual mobile terminals
pull off their assigned packets from within the
TDM stream and ignore the rest. Initially, the
links will be analyzed without forward error
correction. 178
Inbound Link:
Mobile Terminal to Gateway Station
Each terminal transmits a BPSK. signal at 4800 bps
at an allocated frequency. The satellite transponder
shifts all received L-band signals in frequency before
retransmission at Ku band to the gateway station, and
also amplifies the signals with a linear transponder.
At the gateway station, the antenna and RF receiver
are connected to many identical IF receivers tuned to
the individual frequencies of the handheld
transmitters.
179
At the receiving end of the link, the
C/N at the input to the BPSK
demodulator must be high enough to
provide an acceptable bit error rate.
Here, we require a maximum BER of
10 -4 which provides a minimum S/N
of 34 dB in the speech channel.
180
In a practical digital communication system, we
always need a higher C/N than theory suggests
because we do not have ideal Nyquist filters, and
other parts of the system are also not ideal, so we
must add an implementation margin to account for
the nonideal nature of the system. In this example,
the implementation margin is set at 0.6 dB; we
need a minimum C/N = 9.0 dB to meet the BER
and S/N specifications. We can now design the
satellite link to achieve the minimum C/N.
181
Mobile Terminal to Satellite Link
We will establish power and noise link budgets for each
of the four paths, beginning with the uplink from the
mobile terminal to the satellite.
The received power at the output of the uplink antenna
on the satellite from Eq. (4.11) is
Pr = EIRP + GrLpLm dBW
where EIRP is the product of transmitter output power
and transmitting antenna gain, PtGt in dBW, Gr is the
satellite receive antenna gain, Lp is the path loss of the
link, and Lm accounts for all other losses.
182
The noise power, N, at the input to the
satellite receiving system from Eq.
(4.13) is
N = Pn = kTsBn watts
= k + T s + Bn dBW
183
Path loss Lp is found from Eq. (4.12)
Lp = [4R/] 2
or
20 log10(4R/) dB
where R is the distance in meters
between the transmitting and receiving
antennas in the link and A is the
wavelength in meters.
184
The uplink frequency is 1650 MHz,
giving = 0.1818m. The maximum
range is 2200 km so maximum path loss is
## Lp = 20 1og10(4 2.2 106/0.1818)
= 163.6 dB
185
We will assume that there are miscellaneous losses in
the 1550 MHz link of 0.5 dB, caused by polarization
misalignments, gaseous absorption in the atmosphere,
etc. The calculation of the C/N ratio is made for the
worst case of an earth station located on the -3 dB
contour of the satellite antenna beam, so a 3 dB
reduction in satellite antenna gain is applied, making
the value of Lm = - 3.5 dB. We can now set out the
link power and noise budgets for clear line of sight
conditions, when there is no attenuation caused by
obstructions in the path.
186
Uplink Power Budget
Parameter Symbol Value Units
EIRP of handheld unit PtGt - 3 dBW
Gain of receiving antenna Gr 23dB
Path loss at 1650 MHz Lp - 163.6 dB
Miscellaneous losses Lm - 3.5 dB
Received power at satellite Pr - 147.1 dBW
187
Transponder Noise Power Budget
Parameter Symbol Value Units
Boltzmann's constant k - 228.6 dBW/K/Hz
System noise temperature Ts 27.0dBK
Noise bandwidth Bn 36.8 dBHz
Noise power N - 164.8 dBW
188
The inbound uplink C/N ratio in the
transponder can now be calculated from
the power and noise budgets:
(C/N)up = Pr/N
=-147.1 dBW(-164.8 dBW)
=17.7 dB
189
Note that this is the lowest C/N ratio that should occur
in the transponder in clear air conditions, since the
calculation was made for a mobile terminal at the
longest range from the satellite and at the edge of a
satellite antenna beam. The mobile terminal antenna
gain has also been set to its minimum value of 0 dB. If
the satellite were directly overhead the range would be
1000 km instead of 2200 km, making the path loss
lower by 6.8 dB, and the miscellaneous losses would
be 3 dB lower at the center of the satellite antenna
beam, making the power received at the transponder
10.8 dB greater, and then (C/N)up = 28.5 dB.
190
However, we cannot use this figure for the
system design, otherwise there would be
only one user who could make calls, and
then only for a brief moment as the
satellite passes directly overhead. We must
ensure that all users within the satellite's
coverage zone have adequate C/N ratios in
their links for successful communication.
191
Satellite to Gateway Station Link
The next step in calculating the C/N ratio for the
inbound link is to calculate (C/N)dn in the gateway
receiver. We are operating the transponder in FDMA, so
the individual mobile terminal signals must share the
output power of the transponder. We will assume that
50 active terminal signals share the 1 MHz transponder
bandwidth and that 3 dB backoff is used at the
transponder output to obtain quasi-linear operation of
the transponder HPA (remembering that we have
assumed linear transponder operation in this example).
192
The transponder output power is
therefore 10 dBW3 dB = 7 dBW (5
W). The 5-W transponder output power
must be shared equally between the 50
signals in the transponder, giving 0.1 W
= -10 dBW per signal at the transponder
output for the downlink to the gateway
station.
193
We can now establish a link budget for a
single channel downlink from the satellite to
the gateway station. We will use the same
worst-case conditions as for the uplink
maximum path length and minimum
satellite antenna gain, with miscellaneous
losses of 3.5 dB, including the edge of
satellite beam effect.
194
Downlink Power Budget
Parameter Symbol Value Units
EIRP of handheld unit PtGt -10.0 dBW
Gain of receiving antenna Gr 53.5dB
Path loss at 11.5 MHz Lp -180.5 dB
Miscellaneous losses Lm -3.5 dB
Received power at satellite Pr -140.5 dBW
195
Gateway Station Noise Power Budget
Parameter Symbol Value Units
Boltzmann's constant k -228.6 dBW/K/Hz
System noise temperature Ts 21.5dBK
Noise bandwidth Bn 36.8 dBHz
Noise power N -170.3 dBW
196
The C/N ratio in the 4.8-kHz noise bandwidth of a
gateway station IF receiver is given by:
(C/N)dn = Pr/N = -140.5(-170.3)
= 29.8 dB
(C/N)dn for the inbound downlink is higher than
(C/N)up for the inbound uplink because of the high gain
of the gateway station antenna. Because the gain of the
antenna is high, 53.5 dB, which corresponds to an
antenna diameter of 5 m and an aperture efficiency of
60%, its beamwidth is narrow, about 0.4, and the
gateway station must track the satellite as it crosses the
sky. 197
The overall (C/N)0 at the gateway is calculated
by combining the uplink C/N and downlink
C/N values using Eq. (4.43), since both the
transponder and the gateway station receiver
add noise to the signal. The values used in the
formula are ratios, that is, C/N values are not in
decibels.
## 1 / (C/N)o = 1 / (C/N)up + 1 / (C/N)dn
198
For the inbound uplink, (C/N) up = 17.7
dB 58.9 as a ratio. For the inbound
downlink, (C/N) dn = 29.8 dB 955.0
as a ratio.
## (C/N)o = 1 / (1/58.9 + 1/955.0)
= 55.5 or 17.4 dB
199
The overall C/N ratio of 17.4 dB at the gateway
station receiver guarantees that with BPSK and a bit
rate of 4800 bps there will be extremely few bit errors
and the S/N of the speech channel will be set by
quantization noise in the analog to digital converters.
The maximum permitted BER is 10-4, which occurs
with (C/N)o = 9.0 dB. We therefore have an inbound
link margin of (17.49.0) = 8.4 dB. However, we
must calculate the individual link margins for the
uplink and downlink in order to be able to use the
margins for fading analysis. This will be done at the
end of the example.
200
Outbound Link
The outbound link from the gateway station to the
mobile terminal sends a continuous 300 kbps TDM bit
stream using BPSK modulation and a separate
transponder with 1 MHz bandwidth. The bit stream is a
series of packets addressed to all 50 active terminals.
The noise bandwidth of the terminal receiver is 300
kHz, assuming ideal Nyquist filters. The outbound
uplink and downlink C/N values are calculated in
exactly the same way as for the inbound link, and the
power and noise budgets are combined to give C/N
ratios directly from a single table.
201
At the uplink frequency of 14 GHz, clear air
atmospheric attenuation of 1.0 dB is included in the
miscellaneous losses, together with the usual 3 dB loss
for the user at the edge of the satellite antenna beam.
Uplink C/N Budget
Parameter Symbol Value Units
Gate way station EIRP PtGt 65.0 dBW
Gain of receiving antenna Gr 3.0 dB
Path loss at 14.0 MHz Lp -182.2 dB
202
Parameter Symbol Value Units
Miscellaneous losses Lm -4.0 dB
Received power at satellite Pr -118.2 dBW
Boltzmann's constant k -228.6 dBW/K/Hz
System noise temperature Ts 27.0dBK
Noise bandwidth Bn 54.8 dBHz
Noise power N -146.8 dBW
Uplink C/N
(C/N) up = - 118.2(-146.8) = 28.6 dB
203
Downlink C/N Budget
The satellite transponder carrying the single 300
kbps TDM outbound signal can be operated close
to saturation because there is only one signal in
the transponder, thus eliminating intermodulation
problems. We will allow 1.0 dB backoff at the
transponder output to avoid saturating the
transponder, giving a transmitted power Pt = 9.0
dBW. Miscellaneous losses on the downlink are
0.5 dB atmospheric loss and 3 dB for the edge of
the antenna beam.
204
Parameter Symbol Value Units
EIRP of satellite PtGt 32.0 dBW
Gain of receiving antenna Gr 0 dB
Path loss at 1550 MHz Lp -163.1 dB
Miscellaneous losses Lm -3.5 dB
Received power at satellite Pr -134.6 dBW
205
Parameter Symbol Value Units
Boltzmann's constant k -228.6 dBW/K/Hz
System noise temperature Ts 24.8dBK
Noise bandwidth Bn 54.8 dBHz
Noise power N -149.0 dBW
Downlink C/N
(C/N) dn = - 134.6(-149.0) = 14.4 dB
206
Combining the C/N values for the uplink and downlink
gives the overall (C/N)0 ratio at the mobile receiver.
Converting the C/N values form decibels give (C/N) up =
28.6 dB = 724.4, (C/N) dn = 14.4 dB = 27.5 Hence, the
overall (C/N)o for the outbound link is
## (C/N)o = 1 / [1/(C/N) up + 1/(C/N) dn]
= 1 / [0.00139 + 0.0364] = 26.5 or 14.2 dB
Note that the downlink C/N ratio is so much lower than
the uplink C/N ratio that the overall C/N ratio is almost
equal to the downlink C/N ratio.
207
The clear air (C/N)o value is 5.2 dB above the
minimum allowed for BER = l0-4 on the
outbound link, leaving a 5.2 dB margin for
blockage by buildings, the user's head,
multipath effects, the ionosphere, or vegetative
shadowing on the downlink. The link margins
for the outbound link are much lower than for
the inbound link, and it is therefore the
weakest part of the system.
208
Attenuation exceeding 5.2 dB in the downlink
from the satellite to the mobile terminal will
cause the BER to exceed 10-4 and the S/N in the
speech channel will fall below 30 dB. A S/N ratio
of 30 dB in a speech channel is regarded as the
minimum acceptable value for intelligible
communication. Because of the very steep
characteristics of the BER vs C/N ratio curve for
BPSK, the speech channel will be unusable if
downlink attenuation exceeds 5.2 dB.
209
The link margins are quite small for a mobile system in
which the line of sight between the satellite and the user can
easily be blocked by trees, or by the user's body. It is the link
between the mobile terminal and the satellite that sets the
overall C/N value for both the inbound and the outbound
links, but there is little room to change the system parameters
to yield higher margins. The power from the satellite is
limited by the transponder HPA output power and the low
gain of the handset antenna. However, a higher gain antenna
would have a narrower beam and would have to track the
satellite automaticallya smart antenna could be built to do
this, but the small size of most mobile telephone handsets
limits the available improvement to no more than 3 or 4 dB.
210
Optimizing System Performance
The preceding calculations show that the LEO
satellite system can support two-way digital
speech with 50 active users per transponder,
and provides a link margin of 8.4 dB in the
inbound link and a margin of 5.3 dB in the
outbound link. The RF bandwidth used by the
inbound and outbound links is found from the
symbol rates and the values of the filters
(see Chapter 5).
211
For the outbound link using = 0.5, the
symbol rate is 300 k baud , giving
Boutbound = 300 (l + ) = 450 kHz
## For the inbound link using = 0.5, the symbol
rate for one speech channel is 4800 baud,
giving
Binbound = 4.8 (1 + ) = 7.2 kHz
212
The inbound channels access the satellite transponder
using SCPC-FDMA, so the RF signals are distributed
across the transponder bandwidth. We must space the
channels more than 7.2 kHz apart in the transponder so
that the narrow band-pass filters in the gateway station
receiver can extract each speech channel without
interference from the adjacent channels. If we use a 10
kHz channel spacing, there will be a frequency gap,
called a guard band of 2.8 kHz between each channel,
which will ensure minimal interference from adjacent
channels. With 50 channels sharing one transponder, the
total bandwidth occupied in the inbound link
transponder will be 500 kHz.
213
Neither the inbound nor the outbound transponder
bandwidth is fully utilized; in fact only half of the
available 1 MHz is used in each case. However, we
cannot add additional speech channels to the system
because the C/N values are already low, indicating that
the system is power limited with the given link margins.
We can incorporate FEC coding, however, which lowers
the C/N threshold for the minimum BER. Half rate
convolutional coding would be a good choice in this
system because the threshold C/N value can be much
lower. This allows a wider noise bandwidth to be used
and thus better utilization of the available transponder
bandwidth. 214
Using constraint length eight and soft decision
decoding, the C/N ratio for BER = 10-4 can be lowered
to 3.5 dB. However, the bit rate of the signal is now
doubled, since a half rate code adds as many coding
bits as there are data bits in the bit stream. The new
outbound bit rate with FEC is 600 kbps, and each
inbound speech channel has a bit rate of 9600 bps. The
corresponding RF bandwidths for = 0.5 Nyquist
filters are 900 kHz outbound and 14.4 kHz per channel
inbound. With 50 active users, the RF signal
bandwidths are within the available 1 MHz bandwidth
of the satellite transponders.
215
Lowering the threshold value of C/N for the
maximum permitted BER of 10-4 improves the link
margins by a factor called coding gain. Coding gain
is typically quoted as the difference between the
C/N value required for a given BER without coding
and the C/N required to obtain the same BER with
coding. In this example, the coding gain is 8.43.5
= 4.9 dB. However, the coding gain cannot simply
be added to the system margin, because the half rate
FEC code doubles the bit rate of the signals and
also doubles the noise bandwidth of the filters in
the receivers. 216
Thus noise power increases by 3 dB in every link
receiver when FEC is added, and the C/N values
all fall by 3 dB. This results in the overall values
of (C/N)0 for both the inbound and the outbound
links falling by 3 dB. There is no need to
recalculate all the link noise budgets and C/N
values, since all of the values change by the same
amount. (This is one advantage of using decibels
for link calculations.) With half rate FEC added to
the system, the new C/N values are all 3 dB lower
than for the system without FEC.
217
Inbound Link
(C/N) up = 14.7 dB
(C/N) dn = 26.8 dB
(C/N) 0 = 14.4 dB
Outbound Link
(C/N) up = 25.6 dB
(C/N) dn =11.4 dB
(C/N) 0 = 11.2 dB
218
The new link margins with a threshold overall (C/N)o
of 3.5 dB are: inbound 10.9 dB, outbound 7.7 dB.
Although the improvement over the earlier values
without FEC is only 2 dB, the increased link margins
are valuable, so FEC is invariably used in satellite
personal communication systems, as it is in almost all
digital wireless applications, whether satellite or
terrestrial. FEC can be implemented by inserting a
coding IC in the terminal, and identical ICs in the
gateway station, in the baseband bit streams. The 2 dB
advantage that FEC brings to the system's link margins
cannot easily be obtained any other way.
219
Link Margins with FEC
Rain attenuation affects the Ku-band links
between the gateway station and the satellite,
and blockage affects the link between the mobile
terminal and the satellite. Individual link
margins must be calculated to determine the
amount of fading or blockage that can be
tolerated in each link. Excessive rain attenuation
in the Ku-band links could cause the links to fail,
which affects all 50 users.
220
We must therefore ensure that the Ku-band link
margins are sufficiently large to make a rain
outage unlikely, Blockage of the line of sight to a
mobile terminal may cause that one terminal to
lose its link, but this is less serious than losing all
50 links simultaneously. The margin available for
overcoming blockage should be as large as
possible, but is set by the system design and
cannot be improved beyond the values given
above without a reduction in the number of users
in the system.
221
Rain Attenuation at Ku Band
We must calculate the rain attenuation margins for the
inbound downlink and the outbound uplink and
determine the probability of an outage. The link
margin is the number of decibels by which the C/N
ratio on an uplink or a downlink can be reduced before
the overall (C/N)o for that link falls to the threshold
value. We will use 3.5 dB as the threshold value for
over all C/N in each case, assuming that half rate FEC
is used. We will also assume that clear sky conditions
prevail on the uplink when extreme attenuation occurs
on the downlink, and vice versa.
222
For the inbound Ku-band downlink, using half
rate FEC, the clear air C/N ratio is 26.8 dB
(ratio 478.6) and the L-band clear uplink C/N
ratio is 14.7 dB (ratio 29.5). With a threshold at
3.5 dB (ratio 2.24), the minimum downlink C/N
will be given by (using ratios, not dB)
1/(C/N) dn min = 1/(C/N)01/(C/N) up
= 1/2.241/478.6
= 0.444
223
Hence the minimum permitted value for
(C/N) dn = 2.25 3.5 dB. The downlink
margin is 26.83.5 = 23.3 dB. Rain
attenuation at 11.5 GHz very rarely
exceeds this value in the United States, so
for a U.S. system, the Ku-band downlink
margin is adequate.
224
Applying the same analysis as used for the
Ku-band downlink, with (C/N) up = 11.4 dB
(ratio 13.8) in clear sky conditions and
(C/N)0 min = 3.5 dB (ratio 2.24)
## l/(C/N) up min = l/(C/N)ol/(C/N) dn
= 1/2.241/13.8
= 0.374
225
Thus the minimum (C/N) up ratio is 10 log
(1/0.374) = 4.3 dB, ignoring the effects of
coupling between input and output power in
the transponder. When the latter effect is
considered with a linear transponder
characteristic, the limit is set by the (C/N) dn
ratio falling to 3.5 dB. This will occur with
11.43.5 = 7.9 dB uplink attenuation,
which is the limiting value.
226 | 23,103 | 85,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-35 | latest | en | 0.900595 |
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### A Chordingly
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Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle. | 2,594 | 11,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2016-44 | longest | en | 0.882262 |
https://www.valuepenguin.com/mortgages/mortgage-payments-calculator | 1,653,081,138,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534669.47/warc/CC-MAIN-20220520191810-20220520221810-00163.warc.gz | 1,245,191,347 | 16,509 | # Mortgage Calculator: How to Calculate Your Monthly Payments
By clicking "See Rates", you'll be directed to our ultimate parent company, LendingTree. Based on your creditworthiness, you may be matched with up to five different lenders.
See Mortgage Rate Quotes for Your Home
\$
%
There are quite a few factors that go into the calculation of your mortgage expenses, but most homebuyers like to begin by determining their monthly payments and the lifetime cost of the mortgage. Calculating these two figures is a good first step toward understanding all of your other expenses.
## How to Calculate Your Monthly Mortgage Payment
You can calculate your monthly mortgage payments using the following formula:
M = P [ I ( 1 + I )^N ] / [ ( 1 + I )^N – 1 ]
In order to find your monthly payment amount "M," you need to plug in the following three numbers from your loan:
• P = Principal amount (the total amount borrowed)
• I = Interest rate on the mortgage
• N = Number of periods (monthly mortgage payments)
A good way to remember the inputs for this formula is the acronym PIN, which you need to "unlock" your monthly payment amount. If you know your principal, interest rate and number of periods, you can calculate both the monthly mortgage payment and the total cost of the loan. Note that the formula only gives you the monthly costs of principal and interest, so you'll need to add other expenses like taxes and insurance afterward.
Also keep in mind that most lender quotes provide rates and term information in annual terms. Since the goal of this formula is to calculate the monthly payment amount, the interest rate "I" and the number of periods "N" must be converted into a monthly format. This means that you must convert your variables through the following steps:
1. Subtract your down payment amount from the home price to find the total borrowed "P"
2. Divide your quoted annual interest rate by 12 to get your monthly interest rate "I"
3. Multiply the number of years in your mortgage term by 12 to find the total number of monthly payments you will be making "N" – be careful not to confuse this with what the monthly payments will be, aka "M," which we will calculate later on
Once you've converted your inputs, you're ready to plug them into your formula. At this point, it becomes simple arithmetic. Make sure you have a calculator on hand to help with the calculations. To illustrate how this might look with numbers from a typical mortgage, we've provided the following example.
#### Example
Let's say you're trying to purchase a \$250,000 home by taking out a 30-year mortgage with a 20% down payment. The mortgage lender offers you an interest rate of 5% for this loan.
To calculate your total borrowed amount "P," first subtract your 20% down payment from the \$250,000 home price. This gives you a total borrowed amount of \$200,000.
P = \$250,000 – (20% of \$250,000) = \$250,000 - \$50,000 = \$200,000
Next, to calculate your monthly interest rate, divide your annual interest rate of 5% by 12 to obtain your monthly interest rate "I." Remember to convert your mortgage rate into decimals before dividing, so that you don't end up with a figure one hundred times higher than it should be.
I = 5% divided by 12 = 0.05/12 = 0.004167
Finally, obtain your total number of monthly payments "N" by multiplying the total number of years in your loan by 12. Since the loan in our example has a 30-year term, this comes out to 360 months.
N = 30 years X 12 months = 360
In our example, the three PIN variables come out to the following:
Variable
Value in this example
Principal "P"200,000
Interest rate "I"0.004167
Number of periods "N"360
Substituting them into the original equation to solve for monthly payment "M," we get:
M = P [ I ( 1 + I )^N ] / [ ( 1 + I )^N – 1 ]
M = 200,000 * [ 0.004167 ( 1 + 0.004167)^360 ] / [ ( 1 + 0.004167 )^360 – 1 ]
M = 200,000 * [ 0.004167 ( 1.004167 )^360 ] / [ ( 1.004167 )^360 – 1 ]
M = 200,000 * [ 0.004167 * 4.468278 ] / [ 4.468278 – 1 ]
M = 200,000 * 0.018618 / 3.468278
M = 200,000 * 0.005368
M = 1,073.64
Keep in mind that rounding may have a slight impact on your final answer for the monthly payment; your calculation may differ by a few dollars. As stated above, this formula doesn't account for any ongoing taxes or insurance premiums, and only accounts for your monthly mortgage payment. If you want to know the full estimate of your mortgage costs, you'll need to calculate the total cost of your mortgage loan, as shown below.
## How to Calculate the Total Cost of Your Mortgage
Once you have your monthly payment amount, calculating the total cost of your loan is easy. You will need the following inputs, all of which we used in the monthly payment calculation above:
• N = Number of periods (number of monthly mortgage payments)
• M = Monthly payment amount, calculated from last segment
• P = Principal amount (the total amount borrowed, minus any down payments)
To find the total amount of interest you'll pay during your mortgage, multiply your monthly payment amount by the total number of monthly payments you expect to make. This will give you the total amount of principal and interest that you'll pay over the life of the loan, designated as "C" below:
• C = N * M
• C = 360 payments * \$1,073.64
• C = \$368,510.40
You can expect to pay a total of \$368,510.40 over 30 years to pay off your whole mortgage, assuming you don't make any extra payments or sell before then. To calculate just the total interest paid, simply subtract your principal amount P from the total amount paid C.
• C – P = Total Interest Paid
• C – P = \$368,510.40 - \$200,000
• Total Interest Paid = \$168,510.40
At an interest rate of 5%, it would cost \$168,510.40 in interest to borrow \$200,000 for 30 years. As with our previous example, keep in mind that your actual answer might be slightly different depending on how you round the numbers.
## How to Account for Closing Costs
Once you've calculated the total principal and interest expense on your mortgage, factoring in closing costs or fees will be straightforward. Since closing costs are paid in full when you close on the loan, you can simply add them to your overall loan cost without using any long formulas. Some examples of upfront closing costs include the following:
• Mortgage lender fees
• Third-party mortgage fees
• Prepaid mortgage costs
While there may be other categories of upfront fees, the process for calculating them remains the same: Just add them to the total cost of the mortgage loan. Keep in mind that this will exclude any added monthly expenses paid in escrow, like taxes or homeowner's insurance. Our next section explains how to factor in monthly expenses.
## How to Account for Taxes and Recurring Expenses
Accounting for recurring charges like PMI and HOA fees requires a little more work, but even these aren't very difficult to calculate. You can find the total cost of recurring expenses by adding them together and multiplying them by the number of monthly payments (360 for a 30-year mortgage). This will give you the lifetime cost of monthly charges that exclude the cost of your loan.
The reverse is true for annual charges like taxes or insurance, which are usually charged in a lump sum, paid once per year. If you want to know how much these expenses cost per month, you can divide them by 12 and add the result to your mortgage payment. Most mortgage lenders use this method to determine your monthly mortgage escrow costs. Lenders collect these additional payments in an escrow account, typically on a monthly basis, in order to make sure you don't fall short of your annual tax and insurance obligations.
## What Other Expenses Does Homeownership Entail
It's important to recognize that the estimated total cost of your home purchase is only an estimate and not necessarily representative of future conditions. There are many factors that are not taken into account in the calculations we illustrated above; we include a few below for your consideration.
#### Taxes
While these fixed fees are charged regularly, they have a tendency to change over time, especially in large metropolitan areas like New York and Boston. New-home purchases often have their values reassessed within a year or two, which impacts the actual taxes paid. For that reason, your originally forecasted tax liability may increase or decrease as a result of new assessments.
#### HOA Dues
For buyers considering condos, homeowners associations can increase their monthly dues or charge special HOA assessments without warning. This can make up a large portion of your housing expenses, especially in large cities with high maintenance fees. You might also be subject to increased volatility in HOA fees if the community you live in has issues keeping tenants or a troubled track record.
#### Market Risk
The housing market varies by region and is cyclical, much like the stock market. There's always the risk of market value movements and insurance costs that might change over time. While real estate has always been considered one of the safest of investments, there's always a possibility that your home value will drop below the amount you paid for it. These risks are difficult to quantify but should be considered carefully before purchasing a home.
#### Maintenance Costs
Finally, typical mortgage expenses don't account for other costs of ownership, like monthly utility bills, unexpected repairs, maintenance costs and the general upkeep that comes with being a homeowner. While these go beyond the realm of mortgage shopping, they are real expenses that add up over time and are factors that should be considered by anyone thinking of buying a home.
## Calculating ARMs, Refinances and Other Mortgage Types
The equations that we've provided in this guide are intended to help prospective borrowers understand the mechanics behind their mortgage expenses. These calculations become more complicated if you're trying to account for ARMs or refinances, which call for the use of more specialized calculators or spreadsheet programs. You can better understand how these loan structures work by referring to one of our guides about mortgage loans below:
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http://en.wikipedia.org/wiki/Network_analysis_(electrical_circuits) | 1,416,652,682,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400377225.6/warc/CC-MAIN-20141119123257-00243-ip-10-235-23-156.ec2.internal.warc.gz | 104,998,099 | 29,420 | Network analysis (electrical circuits)
"Circuit theory" redirects here. For other uses, see Circuit (disambiguation).
Linear network analysis
Elements
Components
Series and parallel circuits
Impedance transforms
Generator theorems Network theorems
Network analysis methods
Two-port parameters
A network, in the context of electronics, is a collection of interconnected components. Network analysis is the process of finding the voltages across, and the currents through, every component in the network. There are many different techniques for calculating these values. However, for the most part, the applied technique assumes that the components of the network are all linear. The methods described in this article are only applicable to linear network analysis, except where explicitly stated.
Definitions
Component A device with two or more terminals into which, or out of which, charge may flow. Node A point at which terminals of more than two components are joined. A conductor with a substantially zero resistance is considered to be a node for the purpose of analysis. Branch The component(s) joining two nodes. Mesh A group of branches within a network joined so as to form a complete loop. Port Two terminals where the current into one is identical to the current out of the other. Circuit A current from one terminal of a generator, through load component(s) and back into the other terminal. A circuit is, in this sense, a one-port network and is a trivial case to analyse. If there is any connection to any other circuits then a non-trivial network has been formed and at least two ports must exist. Often, "circuit" and "network" are used interchangeably, but many analysts reserve "network" to mean an idealised model consisting of ideal components.[1] Transfer function The relationship of the currents and/or voltages between two ports. Most often, an input port and an output port are discussed and the transfer function is described as gain or attenuation. Component transfer function For a two-terminal component (i.e. one-port component), the current and voltage are taken as the input and output and the transfer function will have units of impedance or admittance (it is usually a matter of arbitrary convenience whether voltage or current is considered the input). A three (or more) terminal component effectively has two (or more) ports and the transfer function cannot be expressed as a single impedance. The usual approach is to express the transfer function as a matrix of parameters. These parameters can be impedances, but there is a large number of other approaches, see two-port network.
Equivalent circuits
A useful procedure in network analysis is to simplify the network by reducing the number of components. This can be done by replacing the actual components with other notional components that have the same effect. A particular technique might directly reduce the number of components, for instance by combining impedances in series. On the other hand it might merely change the form into one in which the components can be reduced in a later operation. For instance, one might transform a voltage generator into a current generator using Norton's theorem in order to be able to later combine the internal resistance of the generator with a parallel impedance load.
A resistive circuit is a circuit containing only resistors, ideal current sources, and ideal voltage sources. If the sources are constant (DC) sources, the result is a DC circuit. The analysis of a circuit refers to the process of solving for the voltages and currents present in the circuit. The solution principles outlined here also apply to phasor analysis of AC circuits.
Two circuits are said to be equivalent with respect to a pair of terminals if the voltage across the terminals and current through the terminals for one network have the same relationship as the voltage and current at the terminals of the other network.
If $V_2=V_1$ implies $I_2=I_1$ for all (real) values of $V_1$, then with respect to terminals ab and xy, circuit 1 and circuit 2 are equivalent.
The above is a sufficient definition for a one-port network. For more than one port, then it must be defined that the currents and voltages between all pairs of corresponding ports must bear the same relationship. For instance, star and delta networks are effectively three port networks and hence require three simultaneous equations to fully specify their equivalence.
Impedances in series and in parallel
Any two terminal network of impedances can eventually be reduced to a single impedance by successive applications of impedances in series or impedances in parallel.
Impedances in series: $Z_\mathrm{eq} = Z_1 + Z_2 + \,\cdots\, + Z_n.$
Impedances in parallel: $\frac{1}{Z_\mathrm{eq}} = \frac{1}{Z_1} + \frac{1}{Z_2} + \,\cdots\, + \frac{1}{Z_n} .$
The above simplified for only two impedances in parallel: $Z_\mathrm{eq} = \frac{Z_1Z_2}{Z_1 + Z_2} .$
Delta-wye transformation
Main article: Y-Δ transform
A network of impedances with more than two terminals cannot be reduced to a single impedance equivalent circuit. An n-terminal network can, at best, be reduced to n impedances (at worst nC2). For a three terminal network, the three impedances can be expressed as a three node delta (Δ) network or four node star (Y) network. These two networks are equivalent and the transformations between them are given below. A general network with an arbitrary number of nodes cannot be reduced to the minimum number of impedances using only series and parallel combinations. In general, Y-Δ and Δ-Y transformations must also be used. For some networks the extension of Y-Δ to star-polygon transformations may also be required.
For equivalence, the impedances between any pair of terminals must be the same for both networks, resulting in a set of three simultaneous equations. The equations below are expressed as resistances but apply equally to the general case with impedances.
Delta-to-star transformation equations
$R_a = \frac{R_\mathrm{ac}R_\mathrm{ab}}{R_\mathrm{ac} + R_\mathrm{ab} + R_\mathrm{bc}}$
$R_b = \frac{R_\mathrm{ab}R_\mathrm{bc}}{R_\mathrm{ac} + R_\mathrm{ab} + R_\mathrm{bc}}$
$R_c = \frac{R_\mathrm{bc}R_\mathrm{ac}}{R_\mathrm{ac} + R_\mathrm{ab} + R_\mathrm{bc}}$
Star-to-delta transformation equations
$R_\mathrm{ac} = \frac{R_aR_b + R_bR_c + R_cR_a}{R_b}$
$R_\mathrm{ab} = \frac{R_aR_b + R_bR_c + R_cR_a}{R_c}$
$R_\mathrm{bc} = \frac{R_aR_b + R_bR_c + R_cR_a}{R_a}$
General form of network node elimination
Main article: Star-mesh transform
The star-to-delta and series-resistor transformations are special cases of the general resistor network node elimination algorithm. Any node connected by $N$ resistors ($R_1$ .. $R_N$) to nodes 1 .. N can be replaced by ${N \choose 2}$ resistors interconnecting the remaining $N$ nodes. The resistance between any two nodes $x$ and $y$ is given by:
$R_\mathrm{xy} = R_xR_y\sum_{i=1}^N \frac{1}{R_i}$
For a star-to-delta ($N=3$) this reduces to:
$R_\mathrm{ab} = R_aR_b(\frac 1 R_a+\frac 1 R_b+\frac 1 R_c) = \frac{R_aR_b(R_aR_b+R_aR_c+R_bR_c)}{R_aR_bR_c}=\frac{R_aR_b + R_bR_c + R_cR_a}{R_c}$
For a series reduction ($N=2$) this reduces to:
$R_\mathrm{ab} = R_aR_b(\frac 1 R_a+\frac 1 R_b) = \frac{R_aR_b(R_a+R_b)}{R_aR_b} = R_a+R_b$
For a dangling resistor ($N=1$) it results in the elimination of the resistor because ${1 \choose 2} = 0$.
Source transformation
A generator with an internal impedance (i.e. non-ideal generator) can be represented as either an ideal voltage generator or an ideal current generator plus the impedance. These two forms are equivalent and the transformations are given below. If the two networks are equivalent with respect to terminals ab, then V and I must be identical for both networks. Thus,
$V_\mathrm{s} = RI_\mathrm{s}\,\!$ or $I_\mathrm{s} = \frac{V_\mathrm{s}}{R}$
• Norton's theorem states that any two-terminal network can be reduced to an ideal current generator and a parallel impedance.
• Thévenin's theorem states that any two-terminal network can be reduced to an ideal voltage generator plus a series impedance.
Simple networks
Some very simple networks can be analysed without the need to apply the more systematic approaches.
Voltage division of series components
Main article: voltage division
Consider n impedances that are connected in series. The voltage $V_i$ across any impedance $Z_i$ is
$V_i = Z_iI = \left( \frac{Z_i}{Z_1 + Z_2 + \cdots + Z_n} \right)V$
Current division of parallel components
Main article: current division
Consider n impedances that are connected in parallel. The current $I_i$ through any impedance $Z_i$ is
$I_i = \left( \frac{\left( \frac{1}{Z_i} \right)}{ \left( \frac{1}{Z_1} \right) + \left( \frac{1}{Z_2} \right) + \,\cdots\, + \left( \frac{1}{Z_n} \right)} \right)I$
for $i = 1,2,...,n.$
Special case: Current division of two parallel components
$I_1 = \left( \frac{Z_2}{Z_1 + Z_2} \right)I$
$I_2 = \left( \frac{Z_1}{Z_1 + Z_2} \right)I$
Nodal analysis
Main article: nodal analysis
1. Label all nodes in the circuit. Arbitrarily select any node as reference.
2. Define a voltage variable from every remaining node to the reference. These voltage variables must be defined as voltage rises with respect to the reference node.
3. Write a KCL equation for every node except the reference.
4. Solve the resulting system of equations.
Mesh analysis
Main article: mesh analysis
Mesh — a loop that does not contain an inner loop.
1. Count the number of “window panes” in the circuit. Assign a mesh current to each window pane.
2. Write a KVL equation for every mesh whose current is unknown.
3. Solve the resulting equations
Superposition
Main article: Superposition theorem
In this method, the effect of each generator in turn is calculated. All the generators other than the one being considered are removed; either short-circuited in the case of voltage generators, or open circuited in the case of current generators. The total current through, or the total voltage across, a particular branch is then calculated by summing all the individual currents or voltages.
There is an underlying assumption to this method that the total current or voltage is a linear superposition of its parts. The method cannot, therefore, be used if non-linear components are present. Note that mesh analysis and node analysis also implicitly use superposition so these too, are only applicable to linear circuits. Superposition cannot be used to find total power consumed by elements even in linear circuits. Power varies according to the square of total voltage (or current) and the square of the sum is not generally equal to sum of the squares.
Choice of method
Choice of method[2] is to some extent a matter of taste. If the network is particularly simple or only a specific current or voltage is required then ad-hoc application of some simple equivalent circuits may yield the answer without recourse to the more systematic methods.
• Superposition is possibly the most conceptually simple method but rapidly leads to a large number of equations and messy impedance combinations as the network becomes larger.
• Nodal analysis: The number of voltage variables, and hence simultaneous equations to solve, equals the number of nodes minus one. Every voltage source connected to the reference node reduces the number of unknowns (and equations) by one.
• Mesh analysis: The number of current variables, and hence simultaneous equations to solve, equals the number of meshes. Every current source in a mesh reduces the number of unknowns by one. Mesh analysis can only be used with networks which can be drawn as a planar network, that is, with no crossing components.[3]
Transfer function
A transfer function expresses the relationship between an input and an output of a network. For resistive networks, this will always be a simple real number or an expression which boils down to a real number. Resistive networks are represented by a system of simultaneous algebraic equations. However in the general case of linear networks, the network is represented by a system of simultaneous linear differential equations. In network analysis, rather than use the differential equations directly, it is usual practice to carry out a Laplace transform on them first and then express the result in terms of the Laplace parameter s, which in general is complex. This is described as working in the s-domain. Working with the equations directly would be described as working in the time (or t) domain because the results would be expressed as time varying quantities. The Laplace transform is the mathematical method of transforming between the s-domain and the t-domain.
This approach is standard in control theory and is useful for determining stability of a system, for instance, in an amplifier with feedback.
Two terminal component transfer functions
For two terminal components the transfer function, or more generally for non-linear elements, the constitutive equation, is the relationship between the current input to the device and the resulting voltage across it. The transfer function, Z(s), will thus have units of impedance – ohms. For the three passive components found in electrical networks, the transfer functions are;
Resistor $Z(s)=R\,\!$ Inductor $Z(s)=sL\,\!$ Capacitor $Z(s)=\frac{1}{sC}$
For a network to which only steady ac signals are applied, s is replaced with and the more familiar values from ac network theory result.
Resistor $Z(j\omega)=R\,\!$ Inductor $Z(j\omega)=j\omega L\,\!$ Capacitor $Z(j\omega)=\frac{1}{j\omega C}$
Finally, for a network to which only steady dc is applied, s is replaced with zero and dc network theory applies.
Resistor $Z=R\,\!$ Inductor $Z=0\,\!$ Capacitor $Z=\infin \,\!$
Two port network transfer function
Transfer functions, in general, in control theory are given the symbol H(s). Most commonly in electronics, transfer function is defined as the ratio of output voltage to input voltage and given the symbol A(s), or more commonly (because analysis is invariably done in terms of sine wave response), A(jω), so that;
$A(j\omega)=\frac{V_o}{V_i}$
The A standing for attenuation, or amplification, depending on context. In general, this will be a complex function of , which can be derived from an analysis of the impedances in the network and their individual transfer functions. Sometimes the analyst is only interested in the magnitude of the gain and not the phase angle. In this case the complex numbers can be eliminated from the transfer function and it might then be written as;
$A(\omega)=\left|{\frac{V_o}{V_i}}\right|$
Two port parameters
Main article: Two-port network
The concept of a two-port network can be useful in network analysis as a black box approach to analysis. The behaviour of the two-port network in a larger network can be entirely characterised without necessarily stating anything about the internal structure. However, to do this it is necessary to have more information than just the A(jω) described above. It can be shown that four such parameters are required to fully characterise the two-port network. These could be the forward transfer function, the input impedance, the reverse transfer function (i.e., the voltage appearing at the input when a voltage is applied to the output) and the output impedance. There are many others (see the main article for a full listing), one of these expresses all four parameters as impedances. It is usual to express the four parameters as a matrix;
$\begin{bmatrix} V_1 \\ V_0 \end{bmatrix} = \begin{bmatrix} z(j\omega)_{11} & z(j\omega)_{12} \\ z(j\omega)_{21} & z(j\omega)_{22} \end{bmatrix} \begin{bmatrix} I_1 \\ I_0 \end{bmatrix}$
The matrix may be abbreviated to a representative element;
$\left [z(j\omega) \right]$ or just $\left [z \right]$
These concepts are capable of being extended to networks of more than two ports. However, this is rarely done in reality because, in many practical cases, ports are considered either purely input or purely output. If reverse direction transfer functions are ignored, a multi-port network can always be decomposed into a number of two-port networks.
Distributed components
Where a network is composed of discrete components, analysis using two-port networks is a matter of choice, not essential. The network can always alternatively be analysed in terms of its individual component transfer functions. However, if a network contains distributed components, such as in the case of a transmission line, then it is not possible to analyse in terms of individual components since they do not exist. The most common approach to this is to model the line as a two-port network and characterise it using two-port parameters (or something equivalent to them). Another example of this technique is modelling the carriers crossing the base region in a high frequency transistor. The base region has to be modelled as distributed resistance and capacitance rather than lumped components.
Image analysis
Main article: Image impedance
Transmission lines and certain types of filter design use the image method to determine their transfer parameters. In this method, the behaviour of an infinitely long cascade connected chain of identical networks is considered. The input and output impedances and the forward and reverse transmission functions are then calculated for this infinitely long chain. Although the theoretical values so obtained can never be exactly realised in practice, in many cases they serve as a very good approximation for the behaviour of a finite chain as long as it is not too short.
Non-linear networks
Most electronic designs are, in reality, non-linear. There is very little that does not include some semiconductor devices. These are invariably non-linear, the transfer function of an ideal semiconductor p-n junction is given by the very non-linear relationship;
$i = I_o (e^{\frac{v}{V_T}}-1)$
where;
• i and v are the instantaneous current and voltage.
• Io is an arbitrary parameter called the reverse leakage current whose value depends on the construction of the device.
• VT is a parameter proportional to temperature called the thermal voltage and equal to about 25mV at room temperature.
There are many other ways that non-linearity can appear in a network. All methods utilising linear superposition will fail when non-linear components are present. There are several options for dealing with non-linearity depending on the type of circuit and the information the analyst wishes to obtain.
Constitutive equations
The diode equation above is an example of an element constitutive equation of the general form,
$f(v,i) = 0 \,$
This can be thought of as a non-linear resistor. The corresponding constitutive equations for non-linear inductors and capacitors are respectively;
$f(v, \varphi) = 0 \,$
$f(v, q) = 0 \,$
where f is any arbitrary function, φ is the stored magnetic flux and q is the stored charge.
Existence, uniqueness and stability
An important consideration in non-linear analysis is the question of uniqueness. For a network composed of linear components there will always be one, and only one, unique solution for a given set of boundary conditions. This is not always the case in non-linear circuits. For instance, a linear resistor with a fixed current applied to it has only one solution for the voltage across it. On the other hand, the non-linear tunnel diode has up to three solutions for the voltage for a given current. That is, a particular solution for the current through the diode is not unique, there may be others, equally valid. In some cases there may not be a solution at all: the question of existence of solutions must be considered.
Another important consideration is the question of stability. A particular solution may exist, but it may not be stable, rapidly departing from that point at the slightest stimulation. It can be shown that a network that is absolutely stable for all conditions must have one, and only one, solution for each set of conditions.[4]
Methods
Boolean analysis of switching networks
A switching device is one where the non-linearity is utilised to produce two opposite states. CMOS devices in digital circuits, for instance, have their output connected to either the positive or the negative supply rail and are never found at anything in between except during a transient period when the device is actually switching. Here the non-linearity is designed to be extreme, and the analyst can actually take advantage of that fact. These kinds of networks can be analysed using Boolean algebra by assigning the two states ("on"/"off", "positive"/"negative" or whatever states are being used) to the boolean constants "0" and "1".
The transients are ignored in this analysis, along with any slight discrepancy between the actual state of the device and the nominal state assigned to a boolean value. For instance, boolean "1" may be assigned to the state of +5V. The output of the device may actually be +4.5V but the analyst still considers this to be boolean "1". Device manufacturers will usually specify a range of values in their data sheets that are to be considered undefined (i.e. the result will be unpredictable).
The transients are not entirely uninteresting to the analyst. The maximum rate of switching is determined by the speed of transition from one state to the other. Happily for the analyst, for many devices most of the transition occurs in the linear portion of the devices transfer function and linear analysis can be applied to obtain at least an approximate answer.
It is mathematically possible to derive boolean algebras which have more than two states. There is not too much use found for these in electronics, although three-state devices are passingly common.
Separation of bias and signal analyses
This technique is used where the operation of the circuit is to be essentially linear, but the devices used to implement it are non-linear. A transistor amplifier is an example of this kind of network. The essence of this technique is to separate the analysis into two parts. Firstly, the dc biases are analysed using some non-linear method. This establishes the quiescent operating point of the circuit. Secondly, the small signal characteristics of the circuit are analysed using linear network analysis. Examples of methods that can be used for both these stages are given below.
Graphical method of dc analysis
In a great many circuit designs, the dc bias is fed to a non-linear component via a resistor (or possibly a network of resistors). Since resistors are linear components, it is particularly easy to determine the quiescent operating point of the non-linear device from a graph of its transfer function. The method is as follows: from linear network analysis the output transfer function (that is output voltage against output current) is calculated for the network of resistor(s) and the generator driving them. This will be a straight line (called the load line) and can readily be superimposed on the transfer function plot of the non-linear device. The point where the lines cross is the quiescent operating point.
Perhaps the easiest practical method is to calculate the (linear) network open circuit voltage and short circuit current and plot these on the transfer function of the non-linear device. The straight line joining these two point is the transfer function of the network.
In reality, the designer of the circuit would proceed in the reverse direction to that described. Starting from a plot provided in the manufacturers data sheet for the non-linear device, the designer would choose the desired operating point and then calculate the linear component values required to achieve it.
It is still possible to use this method if the device being biased has its bias fed through another device which is itself non-linear – a diode for instance. In this case however, the plot of the network transfer function onto the device being biased would no longer be a straight line and is consequently more tedious to do.
Small signal equivalent circuit
This method can be used where the deviation of the input and output signals in a network stay within a substantially linear portion of the non-linear devices transfer function, or else are so small that the curve of the transfer function can be considered linear. Under a set of these specific conditions, the non-linear device can be represented by an equivalent linear network. It must be remembered that this equivalent circuit is entirely notional and only valid for the small signal deviations. It is entirely inapplicable to the dc biasing of the device.
For a simple two-terminal device, the small signal equivalent circuit may be no more than two components. A resistance equal to the slope of the v/i curve at the operating point (called the dynamic resistance), and tangent to the curve. A generator, because this tangent will not, in general, pass through the origin. With more terminals, more complicated equivalent circuits are required.
A popular form of specifying the small signal equivalent circuit amongst transistor manufacturers is to use the two-port network parameters known as [h] parameters. These are a matrix of four parameters as with the [z] parameters but in the case of the [h] parameters they are a hybrid mixture of impedances, admittances, current gains and voltage gains. In this model the three terminal transistor is considered to be a two port network, one of its terminals being common to both ports. The [h] parameters are quite different depending on which terminal is chosen as the common one. The most important parameter for transistors is usually the forward current gain, h21, in the common emitter configuration. This is designated hfe on data sheets.
The small signal equivalent circuit in terms of two-port parameters leads to the concept of dependent generators. That is, the value of a voltage or current generator depends linearly on a voltage or current elsewhere in the circuit. For instance the [z] parameter model leads to dependent voltage generators as shown in this diagram;
[z] parameter equivalent circuit showing dependent voltage generators
There will always be dependent generators in a two-port parameter equivalent circuit. This applies to the [h] parameters as well as to the [z] and any other kind. These dependencies must be preserved when developing the equations in a larger linear network analysis.
Piecewise linear method
In this method, the transfer function of the non-linear device is broken up into regions. Each of these regions is approximated by a straight line. Thus, the transfer function will be linear up to a particular point where there will be a discontinuity. Past this point the transfer function will again be linear but with a different slope.
A well known application of this method is the approximation of the transfer function of a pn junction diode. The actual transfer function of an ideal diode has been given at the top of this (non-linear) section. However, this formula is rarely used in network analysis, a piecewise approximation being used instead. It can be seen that the diode current rapidly diminishes to -Io as the voltage falls. This current, for most purposes, is so small it can be ignored. With increasing voltage, the current increases exponentially. The diode is modelled as an open circuit up to the knee of the exponential curve, then past this point as a resistor equal to the bulk resistance of the semiconducting material.
The commonly accepted values for the transition point voltage are 0.7V for silicon devices and 0.3V for germanium devices. An even simpler model of the diode, sometimes used in switching applications, is short circuit for forward voltages and open circuit for reverse voltages.
The model of a forward biased pn junction having an approximately constant 0.7V is also a much used approximation for transistor base-emitter junction voltage in amplifier design.
The piecewise method is similar to the small signal method in that linear network analysis techniques can only be applied if the signal stays within certain bounds. If the signal crosses a discontinuity point then the model is no longer valid for linear analysis purposes. The model does have the advantage over small signal however, in that it is equally applicable to signal and dc bias. These can therefore both be analysed in the same operations and will be linearly superimposable.
Time-varying components
In linear analysis, the components of the network are assumed to be unchanging, but in some circuits this does not apply, such as sweep oscillators, voltage controlled amplifiers, and variable equalisers. In many circumstances the change in component value is periodic. A non-linear component excited with a periodic signal, for instance, can be represented as periodically varying linear component. Sidney Darlington disclosed a method of analysing such periodic time varying circuits. He developed canonical circuit forms which are analogous to the canonical forms of Ronald Foster and Wilhelm Cauer used for analysing linear circuits.[5] | 6,319 | 29,265 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 55, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2014-49 | longest | en | 0.926367 |
https://www.ncatlab.org/nlab/show/class+equation | 1,631,847,437,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2021-39/segments/1631780054023.35/warc/CC-MAIN-20210917024943-20210917054943-00380.warc.gz | 948,837,148 | 13,562 | Contents
group theory
# Contents
## Idea
The term “class equation” (or class formula, or orbit decomposition formula) refers to a basic type of counting argument that comes about by decomposing a finite G-set as a sum of its orbits. It has a number of fundamental applications in group theory.
## Statement
Let $G$ be a group and let $A$ be a G-set (given by a homomorphism $G \to \hom_{Set}(A, A)$ of monoids, with which is associated an action $\alpha: G \times A \to A$). Recall that $A$ is connected in the category of G-sets if $A$ is inhabited and the action is transitive; in this case, choosing an element $a \in A$, there is a surjection of $G$-sets $G \to A$ sending $1 \mapsto a$, and this induces an isomorphism $G/Stab(a) \cong A$ where $Stab(a)$ is the stabilizer of $a$ and $G/Stab(a)$ is the $G$-set consisting of left cosets of $Stab(a)$.
More generally, each $G$-set $A$ admits a canonical decomposition as a coproduct of its connected components; the components are usually called the orbits of the action. Choosing a representative element $a_x$ in each orbit $x$, this means we have an isomorphism of $G$-sets
$A \cong \sum_{orbits x} G/Stab(a_x).$
By taking $G$ and $A$ to be finite and counting elements, we get an equation of the form
${|A|} = \sum_{orbits x} \frac{{|G|}}{{|Stab(a_x)|}}.$
We call an instance of this equation a class equation. By judicious choice of groups $G$ and $G$-sets $A$, often in combination with number-theoretic arguments, one can derive many useful consequences; some sample applications are given below.
Notice that reading the class equation equivalently as
$\sum_{orbits x} \frac{{1}}{{|Stab(a_x)|}} = \frac{|A|}{|G|}$
it expresses the groupoid cardinality of the action groupoid of $G$ acting on $A$.
## Applications
### Centers of $p$-groups
Let $p$ be a prime; recall that a $p$-group is a finite group whose order is a power of $p$. A basic structural result is the following.
###### Proposition
A non-trivial $p$-group $G$ has a nontrivial center $Z(G)$.
###### Proof
Let $G$ act on itself by the conjugation action $G \times G \to G$, $(g, h) \mapsto g h g^{-1}$. In this case an orbit $Orb(h)$ is usually called the conjugacy class of $h$, and $Orb(h)$ is trivial (consists of exactly one element $h$) iff $h$ belongs to $Z(G)$. In any case ${|Orb(h)|} = \frac{{|G|}}{{|Stab(h)|}}$ divides ${|G|} = p^n$, and therefore $p$ divides ${|Orb(h)|}$ if $h$ is noncentral. In this case the class equation takes the form
${|G|} = {|Z(G)|} \; + \sum_{nontrivial\; orbits x} \frac{{|G|}}{{|Stab(a_x)|}}$
and now since $p$ divides ${|G|}$ as well as each term in the sum over nontrivial orbits, it must also divide ${|Z(G)|}$. In particular, $Z(G)$ has more than one element.
It follows by induction that $p$-groups are solvable, since the center is a normal subgroup and the quotient $G/Z(G)$ is also a $p$-group. Since a group obtained from an abelian group by repeated central extensions is nilpotent, $p$-groups are in fact nilpotent.
### Number of fixed points
An elementary observation that is frequently useful is that the number of fixed points of an involution on a finite set $S$ has the same parity as $S$. This is a statement about $\mathbb{Z}/(2)$-sets; we generalize this to a statement about $G$-sets for general $p$-groups $G$. (Again, a fixed point of a $G$-set is an element whose orbit is a singleton.)
###### Proposition
If $G$ is a $p$-group acting on a set $A$, then
• ${|A|} \equiv {|Fix(A)|} \; mod p$.
As special cases, if there is just one fixed point, then ${|A|} \equiv 1 \; mod p$, and if $p$ divides ${|A|}$, then $p$ divides ${|Fix(A)|}$.
###### Proof
The class equation takes the form
${|A|} = {|Fix(A)|} \; + \sum_{nontrivial\; orbits x} \frac{{|G|}}{{|Stab(a_x)|}}$
where $p$ divides each summand over nontrivial orbits on the right, since $G$ is a $p$-group. Now reduce mod $p$.
### Wedderburn’s theorem
In the theory of finite projective planes, an important result is that a projective plane is Pappian if it is Desarguesian. The purely algebraic version of this is Wedderburn’s theorem:
###### Theorem
A finite division ring $D$ is commutative.
###### Proof
(Witt) The center of $D$ is a field; if $D$ is of characteristic $p \gt 0$, then the center $F$ has $q = p^f$ elements for some $f$. We may regard $D$ as an $F$-vector space of dimension $n$, whence the number of elements of its multiplicative group $D^\times$ is $q^n-1$.
The conjugation action of $D^\times$ on itself yields a decomposition
$D^\times \cong F^\times \; + \sum_{nontrivial\; orbits x} \frac{{|D^\times|}}{{|Stab(a_x)|}}$
where again the elements of the center $F^\times = Z(D^\times)$ correspond to the trivial orbits. The stabilizer of any $a_x$, together with $0$, forms a division ring (strictly) intermediate between $F$ and $D$; usually this is called the centralizer $C(a_x)$ of $a_x$. Putting $d_x = dim_F(C(a_x))$, the division ring $C(a_x)$ has $q^{d_x}$ elements, and notice $d_x$ divides $n$ because $n/d_x$ is just the dimension of $D$ seen as a vector space (module) over $C(a_x)$. Thus ${|Stab(a_x)|} = q^{d_x} - 1$, and we have a class equation
$q^n - 1 = q - 1 + \sum_x \frac{q^n - 1}{q^{d_x} - 1}.$
Now let $\zeta \in \mathbb{C}$ be any primitive $n^{th}$ root of unity. Since $z - \zeta$ divides each polynomial $z^n-1$ and $\frac{z^n - 1}{z^{d_x} - 1}$, so does $\prod_{prim.\; \zeta} (z-\zeta)$. It follows that the algebraic integer $\prod_{prim.\; \zeta} (q - \zeta)$ divides each of the integers $q^n-1$ and $\frac{q^n - 1}{q^{d_x} - 1}$, and hence divides $q-1$ according to the class equation. But also ${|q - \zeta|} \geq {|q-1|}$ for any root of unity $\zeta$. Thus ${|q - \zeta|} = {|q-1|}$ and $n = 1$, i.e., $F = Z(D)$ is all of $D$ as was to be shown.
### Sylow theorems
Let $G$ be a finite group of order $n$, and suppose that $p$ is a prime that divides $n$; say $n = p^f u$ where $p$ does not divide $u$. Recall that a Sylow p-subgroup is a $p$-subgroup of maximal order $p^f$.
A fundamental fact of group theory is that Sylow $p$-subgroups exist and they are all conjugate to one another; also the number of Sylow $p$-subgroups is $\equiv 1 \; mod p$.
Existence of Sylow $p$-subgroups can be proven by exploiting the same type of argument as in the proof of Proposition :
###### Theorem
If $G$ has order $n$ and $p^k$ is a prime power dividing $n$, then there is a subgroup of $G$ of order $p^k$.
###### Proof
First we show that Sylow subgroups exist. We start by observing that if a group $H$ has a $p$-Sylow subgroup $P$, then so does any subgroup $G$. First note that if we let $G$ act on $H/P$ by left translation, then the stabilizer of any element $h P$ is $G \cap h P h^{-1}$, a $p$-group since $h P h^{-1}$ is. Then note that since $H/P$ has cardinality prime to $p$, so must one of its connected components $G/Stab(a_x)$ in its $G$-set decomposition
$H/P \cong \sum_{orbits\; x} G/Stab(a_x),$
and this makes $Stab(a_x)$ a $p$-Sylow subgroup of $G$.
Then, if $G$ is any group, and $n = ord(G)$, apply this observation to the embedding
$G \stackrel{Cayley}{\hookrightarrow} Perm({|G|}) \cong S_n \hookrightarrow GL_n(\mathbb{Z}/(p)) = H$
where we embed the symmetric group $S_n$ via permutation matrices into the group $H$ of $n \times n$ invertible matrices over $\mathbb{Z}/(p)$. The group $H$ has order $(p^n - 1)(p^n - p)\ldots (p^n - p^{n-1})$, with maximal $p$-factor $p^{n(n-1)/2}$. It thus has a $p$-Sylow subgroup given by unitriangular matrices, i.e., upper-triangular matrices with all $1$'s on the diagonal. Therefore $p$-Sylow subgroups $P$ exist for any $G$.
Finally, note that by Proposition , $P$ is solvable and therefore has a composition series
$\{1\} = P_0 \subset P_1 \subset \ldots \subset P$
where each $P_k$ has order $p^k$.
###### Theorem
If $H$ is a $p$-subgroup of $G$ and $P$ is a Sylow $p$-subgroup, then $g^{-1} H g \subseteq P$ for some $g \in G$. In particular, all Sylow $p$-subgroups are conjugate to one another.
###### Proof
$G$ acts on the set of cosets $G/P$ as usual by left translation, and we may restrict the action to the $p$-subgroup $H$. By maximality of $P$, we see ${|G/P|}$ is prime to $p$, and so by Proposition , ${|Fix_H(G/P)|}$ is also prime to $p$. In particular, $Fix_H(G/P)$ has at least one element, say $g P$. We infer that $h g P = g P$ for all $h \in H$, or that $g^{-1} h g P = P$ for all $h \in H$, and this implies that $g^{-1} H g \subseteq P$.
###### Theorem
The number of Sylow $p$-subgroups of $G$ is $\equiv 1 \; mod p$.
###### Proof
Let $Y$ be the set of Sylow $p$-subgroups; $G$ acts on $Y$ by conjugation. As all Sylow $p$-subgroups are conjugate, there is just one orbit of the action, and the stabilizer of an element $P \in Y$ is just the normalizer $N_G(P)$ (by definition of normalizer). Thus $Y \cong G/N_G(P)$ as $G$-sets.
Restrict the action to the subgroup $P$. Of course the element $P \in Y$ is a fixed point of this restricted action, and if $Q$ is any other fixed point, it means $x Q x^{-1} = Q$ for all $x \in P$, whence $P \subseteq N_G(Q)$. Now: $P, Q$ are both Sylow $p$-subgroups of $N_G(Q)$ and are therefore conjugate to each other (as seen within the group $N_G(Q)$). But $Q$ is already fixed by the conjugation action in its stabilizer $N_G(Q)$, so we conclude $P = Q$. We conclude $Fix_P(Y)$ has exactly one element. From ${|Y|} \equiv {|Fix_P(Y)|} \; mod p$, the theorem follows.
Last revised on October 1, 2018 at 13:59:54. See the history of this page for a list of all contributions to it. | 3,057 | 9,565 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 238, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-39 | latest | en | 0.86965 |
https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1 | 1,718,809,669,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00823.warc.gz | 97,559,817 | 47,944 | # Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11 | 4,077 | 18,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-26 | latest | en | 0.961959 |
https://idafedchi.web.app/335.html | 1,660,028,376,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570913.16/warc/CC-MAIN-20220809064307-20220809094307-00741.warc.gz | 304,865,222 | 5,572 | # Master integration by parts pdf
This unit derives and illustrates this rule with a number of examples. Archimedes is the founder of surface areas and volumes of solids such as the sphere and the cone. Spare parts are components that are kept in your inventory as spare. Math 105 921 solutions to integration exercises ubc math. To evaluate that integral, you can apply integration by parts again. Live demonstration of a parts integration with nlp the conflict between making choices and staying with what you have in career is usually a common and tough one. The integration enables users to reduce the amount of time they spend trying to find and share design information, as well as eliminate unnecessary change orders by ensuring that everyone is working from the latest design information. Apart from that, but more importantly, if you want to master taking derivatives of functions, and integration, youll need to devote yourself to practice, and lots of it. This is illustrated using as an example the twoloop sunset diagram with onshell. An intuitive and geometric explanation sahand rabbani the formula for integration by parts is given below. Now, integration by parts produces first use of integration by parts this first use of integration by parts has succeeded in simplifying the original integral, but the integral on the right still doesnt fit a basic integration rule. That is, we want to compute z px qx dx where p, q are polynomials. Integration by parts is like the reverse of the product formula. Methods of integration william gunther june 15, 2011 in this we will go over some of the techniques of integration, and when to apply them.
Material is procured from external or internal sources on the basis of the requirements determined by material requirements planning. Sometimes integration by parts must be repeated to obtain an answer. A firm grounding in differential calculus is a must. For the following problems, indicate whether you would use integration by parts with your choices of u and dv, substitution with your choice of u, or neither. Integration by parts mctyparts20091 a special rule, integrationbyparts, is available for integrating products of two functions. That is integration, and it is the goal of integral calculus. I can sit for hours and do a 1,000, 2,000 or 5,000piece jigsaw puzzle. And just observe each and every solved problem in textbook. If nothing else works, convert everything to sines and cosines. When choosing u and dv, u should get \simpler with di erentiation and you should be able to integrate dv. During all transactions, inventory management accesses both master data such as material master data and transaction data such as purchasing documents shared by all logistics components.
Createdisplay a location is a virtual record of the location where an equipment is installed. We show that the new relation between master integrals recently obtained in ref. In order to master the techniques explained here it is vital that you undertake. Example 1 z f xg0xdx f xgx z gxf 0xdx or z udv uv z vdu. The integration by parts formula we need to make use of the integration by parts formula which states. Integration of parts follows a common negotiating technique. Integration by parts examples, tricks and a secret howto. Jul, 2016 refurbished spare parts are parts that can be repaired either internally or sent out to be repaired by an external vendor.
Make sure you identify the parts clearly, and understand the nature of the conflict. Calculus ii integration strategy pauls online math notes. Introduction to integration by parts unlike the previous method, we already know everything we need to to under stand integration by parts. If youre behind a web filter, please make sure that the domains. If youre seeing this message, it means were having trouble loading external resources on our website.
Integration techniques integral calculus 2017 edition. This sdk supports connecting to and commanding aris explorer and aris voyager sonars. Several methods of calculation of master integrals also. Integrationbyparts procedure with effective mass article pdf available in physics letters b 7123 february 2012 with 33 reads how we measure reads.
When you have the product of two xterms in which one term is not the derivative of the other, this is the most common situation and special integrals like. This is unfortunate because tabular integration by parts is not only a valuable tool for finding integrals but can also be applied to more advanced topics including the derivations of some important. This uses one of the problems from the collection and gives you an idea of what the grading rubric looks like. The function being integrated, fx, is called the integrand. Integration by parts easy method i liate i integral uv i. Integral calculus 2017 edition integration techniques. Jan 21, 2017 integration by parts shortcut method in hindi i liate i integral uv i class 12 ncert. The following are solutions to the integration by parts practice problems posted november 9.
For instance, a substitution may lead to using integration by parts or partial fractions integral. Tabular integration by parts when integration by parts is needed more than once you are actually doing integration by parts recursively. Notice that we needed to use integration by parts twice to solve this problem. With, and, the rule is also written more compactly as 2 equation 1 comes from the product rule. Find, read and cite all the research you need on researchgate. This is illustrated using as an example the twoloop sunset diagram with onshell kinematics. Using integration by parts might not always be the correct or best solution. Thats the quick waybut do bear in mind that, typically, an online editor isnt as fully featured as its desktop counterpart, plus the file is exposed to the internet which might be of. Here are the tricks and secrets you need to know to master this technique of integration. Some of these are online pdf editors that work right in your web browser, so all you have to do is upload your pdf file to the website, make the changes you want, and then save it back to your computer. To develop competence and mastery, you need to do math, and not just read about it. This process is necessary for parts that are no longer commercially available and for parts that can be overhauled and reused at a fraction of the price of a.
Inventory management is part of the materials management module and is fully integrated in the entire logistics system. Then there are trig and hyperbolic functions, they can be utilised by considering their derivatives backward, etc. When one party needs the deal less than the other party, they have the power and ability to walk away. Then z exsinxdx exsinx z excosxdx now we need to use integration by parts on the second integral. Lot of people just seem to ignore the solved problems, they jump to solving exercises, this is foolishness, when there ar. Navigation refresher learning objectives in this lesson, you will. You will see plenty of examples soon, but first let us see the rule. Integration by parts is a technique for evaluating integrals whose integrand is the product of two functions. Integration by parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways. We want to choose u u and d v d v so that when we compute d u d u and v v and plugging everything into the integration by parts formula the new integral we get is one that we can do. Using the catia integration, design teams can quickly search for catia parts, assemblies and drawings. Refurbished spare parts are parts that can be repaired either internally or sent out to be repaired by an external vendor. Mar 09, 2016 best choice will be to start with the state board textbooks.
The parts are then placed back into inventory for future use. Using repeated applications of integration by parts. You have two parties that come together for a negotiation. Sumdi erence r fx gx dx r fxdx r gx dx scalar multiplication r cfx. An lloop diagram has l loop integration momenta k1. Finney,calculus and analytic geometry,addisonwesley, reading, ma 1988. Z vdu 1 while most texts derive this equation from the product rule of di. With that in mind it looks like the following choices for u u and d v d v should work for us. Pdf integration by parts is used to reduce scalar feynman integrals to master integrals. Bonus evaluate r 1 0 x 5e x using integration by parts. C is an arbitrary constant called the constant of integration.
Well learn that integration and di erentiation are inverse operations of each other. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. One strategy is the might makes right strategy, which says whoever can exert the most force will win the negotiation. Nabeel khan 61 daud mirza 57 danish mirza 58 fawad usman 66 amir mughal 72 m. The integration by parts formula for indefinite integrals is given by. Integration by parts a special rule, integration by parts, is available for integrating products of two functions. Integrationbyparts procedure with effective mass article pdf available in physics letters b 7123 february 2012 with 33. Best choice will be to start with the state board textbooks. Have the part, which represents the unwanted state or behaviour come out on the hand first. At first it appears that integration by parts does not apply, but let. For example, if integrating the function fx with respect to x.
541 1442 856 9 444 1323 1017 687 602 551 1096 1494 101 1318 770 1320 272 138 979 141 382 172 1150 518 1038 1283 509 585 262 1439 510 1263 1057 987 1299 737 719 493 1161 301 1257 916 42 868 | 2,088 | 9,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-33 | latest | en | 0.954531 |
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# Single Variable Data
Single variable data is usually called univariate data. This is a type of data that consists of observations on only a single characteristic or attribute. Single variable data can be used in a descriptive study to see how each characteristic or attribute varies before including that variable in a study with two or more variables.
## Examples of single variable data
What were the scores of the students that took the maths test? Which sickness was responsible for most deaths in 2020? What are the weights of each person present in the gym? What is the typical income of the average person in the UK? All these questions can be answered using single variable data. Single variable analysis is the simplest form of analysing data. Its main purpose is to describe, and it does not take into considerations causes and relationships.
For instance, when the question about the scores of students that took a particular math test is asked, we are mostly interested in how varied the results are from each person. By this, we can statistically summarise the data using Statistical Measures to get an idea about the performance of the whole population that took the test.
## How significant is single variable data?
In research, single variable data does not concern itself with answering questions that involve relationships between variables. It describes an attribute of the subject in question, and how it varies from observation to observation. Univariate data analysis involves using statistical measures such as Measures of Central Tendency. It also takes advantage of measures of spread.
There are two main reasons why a researcher would conduct a single variable analysis. The first is to have a descriptive study of how one characteristic varies from subject to subject. The second is to analyse the variety of each characteristic before they can be paired with other variables in a study.
This is where Bivariate Data and multivariate data comes in. Multivariate data describes multiple characteristics of a subject. It is necessary to examine how varied students' scores are with respect to other factors such as subject and their background.
## Single variable data analysis
As mentioned earlier, statistical measures are used to summarise single variable data's centres and spread. Whilst the commonest way to display single variable data is in a table, other common ways are:
• Histograms.
• Frequency distribution.
• Box plots.
• Pie charts.
Scores of eight students were recorded after taking a maths test in grade 6, and they are as follows; 76, 88, 45, 50, 88, 67, 75, 83. Find the
1. Mean
Median
3. Mode
1.
2.
Rearrange values from lowest to highest.
45, 50, 67, 75, 76, 83, 88, 88
Median = 75.5
3.
The most frequently occurring number is 88.
## Histograms
Histograms are one of the most commonly used graphs to show frequency distribution. It is a graphical display of data using bars of different heights. Similar to the bar chart, the histogram groups numbers into ranges. It is an appropriate way to display single-variable data.
Histogram of travel time to work. Image: QWFP, CC BY-SA 3.0
## Frequency distribution
Frequency distribution is data modelled in a tabular format to display the number of observations within a space. This displays values and their frequency (how often something occurs). This format also appropriately represents single variable data and is as simple as possible.
The numbers of newspapers sold at a shop over the last 10 days are;
20, 20, 25, 23, 20, 18, 22, 20, 18, 22.
This can be represented by frequency distribution. The values above are the variables, and the table is going to show how often a specific number of sales occurred over the last 10 days.
Papers sold Frequency 2 0 21 0 2 23 1 24 0 1
## Pie charts
Pie charts are types of graphs that display data as circular graphs. They are represented in slices where each slice of the pie is relative to the size of that category in the group as a whole. This means that the entire pie is 100%, and each slice is its proportional value.
Assuming the data for pets ownership in Lincoln were collected as follows, how would it be represented on a pie chart?
Dogs - 1110 people
Cats - 987 people
Rodents - 312 people
Reptiles - 97 people
Fish - 398 people
Figure 2. Pie chart representing data of pets in Lincoln
## Box plots
Presenting data using the box plot gives a good graphical image of the concentration of the data. It displays the five-number summary of a dataset; the minimum, first quartile, median, third quartile, and maximum. This is also a good system to represent single variable data.
The ages of 10 students in grade 12 were collected and they are as follows.
15, 21, 19, 19, 17, 16, 17, 18, 19, 18.
First, we will arrange this from lowest to highest so the median can be determined.
15, 16, 17, 17, 18, 18, 19, 19, 19, 21
Median = 18
In finding the quartiles, the first will be the median to the right of the overall median.
The median for 15, 16, 17, 17, 18 is 17
The third quartile will be the median to the right of the overall median.
Median for 18, 19, 19, 19, 21, will make 19.
We will now note the minimum number which is 15, and also the maximum which is 21.
Figure 3. Box plot representing students ages
## Single variable data - Key takeaways
• Single variable data is a term used to describe a type of data that consists of observations on only a single characteristic or attribute.
• Single variable data's main purpose is to describe, and it does not take into considerations causes and relationships.
• Statistical measures are used to summarise single variable data's centres and spread.
• Common ways single variable data can be described are through histograms, frequency distributions, box plots, and pie charts.
Images
Histogram: https://commons.wikimedia.org/wiki/File:Travel_time_histogram_total_n_Stata.png
Variable means the measured values can be varied anywhere along a given scale, whilst attribute data is something that can be measured in terms of numbers or can be described as either yes or no for recording and analysis.
The ages of students in a class.
Single variable data gives measures of only one attribute whilst two-variable data gives measures of two attributes describing a subject.
Single variable data is used to describe a type of data that consists of observations on only a single characteristic or attribute.
## Final Single Variable Data Quiz
Question
What is cumulative frequency?
The cumulative frequency at a point x is the sum of the individual frequencies up to and at the point x.
Show question
Question
Which of the following can you obtain from a cumulative frequency distribution? a) median b) quartiles c) percentiles d) all of the above
d
Show question
Question
If a cumulative frequency for the (n-1)th value is 85 in discrete frequency distribution with 110 data points, what is the raw frequency for the nth value?
25
Show question
Question
For a grouped frequency distribution, what is the class mark for the class 0.5 - 1.0?
0.75
Show question
Question
For a grouped frequency distribution, what is the class mark for the class 2.5 - 3.5?
3.0
Show question
Question
For a grouped frequency distribution, what is the class mark for the class 8 - 12?
10
Show question
Question
State whether the following statement is true or false : the curve for a cumulative frequency graph is never decreasing.
True
Show question
Question
The cumulative frequency curve for an experiment with 200 trials is given by x = y/5, where the cumulative frequency is represented on the y-axis. Find the median.
x = (200/2)/5 = 20
Show question
Question
The cumulative frequency curve for an experiment with 200 trials is given by x = y/5, where the cumulative frequency is represented on the y-axis. Find the upper quartile.
x = (200 × 3/4)/5 = 30
Show question
Question
The cumulative frequency curve for an experiment with 200 trials is given by x = y/5, where the cumulative frequency is represented on the y-axis. Find the 43rd percentile.
x = (200 × 43/100)/5 = 17.2
Show question
Question
The cumulative frequency curve for an experiment with 200 trials is given by x = y/5, where the cumulative frequency is represented on the y-axis. Find the 70th percentile.
x = (200 × 70/100)/5 = 28
Show question
Question
The cumulative frequency curve for an experiment with 100 trials is given by x = 2y + 3, where the cumulative frequency is represented on the y-axis. Find the median.
x = 2 × (100/2) + 3 = 103
Show question
Question
The cumulative frequency curve for an experiment with 100 trials is given by y = 2x + 3, where the cumulative frequency is represented on the y-axis. Find the lower quartile.
x = 2 × (100/4) + 3 = 53
Show question
Question
A grouped frequency distribution has been made for the length of 500 snakes. The cumulative frequency of a class (8.0 - 8.5) inches is 320. How many snakes are more than 8.5 inches long?
180
Show question
Question
A grouped frequency distribution has been made for the length of 500 snakes. The cumulative frequency of a class (8.0 - 8.5) inches is 320. Which of the following is the correct conclusion?
There are 320 snakes shorter than than or equal to 8.5 inches
Show question
Question
What is a box plot?
A box plot is a type of graph that visually shows features of the data.
Show question
Question
What are the features that a box plot shows?
A box plot shows you the lowest value, lower quartile, median, upper quartile, highest value and any outliers that the data may have.
Show question
Question
How do you find upper and lower quartiles?
To find the upper and lower quartile you first need to arrange your data into numerical order, the next step is to find your median, you can then use this to find both of the quartiles. The lower quartile will then be the midpoint between the lowest value and the median, the upper quartile will be the midpoint between the median and the highest value.
Show question
Question
How do you find the interquartile range?
To find the interquartile range you subtract the lower quartile from the upper quartile.
Show question
Question
What is an outlier?
An outlier is classed as data that falls 1.5 x the interquartile range above the upper quartile or below the lower quartile.
Show question
Question
What is a histogram?
A histogram is a type of graph that represents grouped data.
Show question
Question
How do you calculate the frequency density?
Frequency density is calculated by dividing the frequency by the class width.
Show question
Question
What is a frequency polygon?
A frequency polygon is a graphical representation of a data set with frequency information. It is one of the most commonly used statistical tools used to represent and analyze grouped statistical data.
Show question
Question
For a grouped frequency distribution, what is plotted along the X-axis when building a frequency polygon?
Class mark
Show question
Question
For a grouped frequency distribution, what is plotted along the Y-axis when building a frequency polygon?
Frequency
Show question
Question
How do you obtain a frequency polygon from a given histogram?
Join the middle of the top of each bar of the histogram sequentially.
Show question
Question
State whether the following statement is true or false : To draw a frequency polygon, you first have to create a histogram.
False
Show question
Question
What is the class mark for the class "8-10"?
9
Show question
Question
What is the class mark for the class "45.5-55"?
50.25
Show question
Question
What is the class mark for the class "0.1-0.2"?
0.15
Show question
Question
State whether the following statement is true or false : The sum of the frequencies of a frequency polygon must equal 1
False
Show question
Question
State whether the following statement is true or false : The frequencies of a frequency polygon must be positive
True
Show question
Question
State whether the following statement is true or false : To draw a frequency polygon from a given grouped frequency distribution, we must plot the frequency against the class marks and not the class boundaries.
True
Show question
Question
What are the two types of measures that are usually commented on when comparing data distributions?
1. measure of location
Show question
Question
What is a measure of spread?
a measure of spread provides us information regarding the variability of data in a given data set, i.e. how close or far away the different points in a data set are from each other.
Show question
Question
What is a measure of location?
a measure of location is used to summarize an entire data set with a single value.
Show question
Question
Data set A - median 25, Q1 = 18, Q3 = 56
Data set B - median 24, Q1 = 14, Q3 = 130
Data set A has a lower measure of location (median) and also a lower variability among the data.
Show question
Question
Data set A - median 100, Q1 = 50, Q3 = 150
Data set B - median 200, Q1 = 150, Q3 = 250
Data set A has a lower measure of location (median). There appears to be an equal variability among the data sets.
Show question
Question
Data set A - median 300, Q1 = 275, Q3 = 325
Data set B - median 200, Q1 = 150, Q3 = 250
Data set A has a higher measure of location (median) and a lower variability among the data.
Show question
Question
Which of the following is appropriate to use along with median for comparison?
Interquartile range
Show question
Question
Which of the following is appropriate to use along with mean for comparison?
standard deviation
Show question
Question
Which of the following is appropriate to use along with
standard deviation for comparison?
mean
Show question
Question
Which of the following is appropriate to use along with interquartile range for comparison?
mean
Show question
Question
Which of the following should you use for comparing a data set with extreme values?
mean and standard deviation
Show question
Question
Compare the 2 data sets
Data set A - mean 100, standard deviation = 50
Data set B - mean 200, standard deviation = 50
Data set A has a lower measure of location (mean). There is an equal variability among the data sets.
Show question
Question
Compare the 2 data sets
Data set A - mean = 13, standard deviation = 5
Data set B - mean = 18, standard deviation = 15
Data set A has a lower measure of location (mean) and a lower variability among the data sets.
Show question
Question
Compare the 2 data sets
Data set A - mean = 13, standard deviation = 5
Data set B - mean = 13, standard deviation = 5
Both data sets have similar measures of location and spread within the data.
Show question
Question
What is single-variable data?
Single-variable data is a type of data that consists of observations on only a single characteristic or attribute.
Show question
Question
Another name for univariate data is?
Single-variable data
Show question
Question
Amongst the number of ways single-variable data can be represented, which of them does it display the five-number summary of a dataset?
Box plot
Show question
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Create the most beautiful study materials using our templates. | 3,664 | 16,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-49 | latest | en | 0.951471 |
http://nbviewer.jupyter.org/github/demotu/BMC/blob/master/notebooks/OrdinaryDifferentialEquation.ipynb | 1,493,187,279,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121165.73/warc/CC-MAIN-20170423031201-00254-ip-10-145-167-34.ec2.internal.warc.gz | 282,095,838 | 199,742 | # Ordinary Differential Equation¶
Marcos Duarte
An ordinary differential equation (ODE) is an equation containing a function of one independent variable and its derivatives.
Solve an ODE is finding such a function whose derivatives satisfy the equation. The order of an ODE refers to the order of the derivatives; e.g., a first order ODE has only first derivatives. A linear ODE has only linear terms for the function of one independent variable and in general its solution can be obtained analytically. By contrast, a nonlinear ODE doesn't have an exact analytical solution and it has to be solved by numerical methods. The equation is referred as partial differential equation when contains a function of more than one independent variable and its derivatives.
A simple and well known example of ODE is Newton's second law of motion:
$$m\frac{\mathrm{d}^2 \mathbf{x}}{\mathrm{d}t^2}(t) = \mathbf{F}$$
$\mathbf{x}$ is the function with a derivative and $t$ is the independent variable. Note that the force, $\mathbf{F}$, can be constant (e.g., the gravitational force) or a function of position, $\mathbf{F}(\mathbf{x}(t))$, (e.g., the force of a spring) or a function of other quantity. If $\mathbf{F}$ is constant or a linear function of $\mathbf{x}$, this equation is a second-order linear ODE.
## First-order ODE¶
A first-order ODE has the general form:
$$\frac{\mathrm{d} y}{\mathrm{d} x} = f(x, y)$$
Where $f(x, y)$ is an expression for the derivative of $y$ that can be evaluated given $x$ and $y$. When $f(x, y)$ is linear w.r.t. $y$, the equation is a first-order linear ODE which can be written in the form:
$$\frac{\mathrm{d} y}{\mathrm{d} x} + P(x)y = Q(x)$$
## Numerical methods for solving ODE¶
When an ODE can't be solved analytically, usually because it's nonlinear, numerical methods are used, a procedure also referred as numerical integration (Downey, 2011; Kitchin, 2013; Kiusalaas, 2013; Wikipedia). In numerical methods, a first-order differential equation can be solved as an Initial Value Problem (IVP) of the form:
$$\dot{y}(t) = f(t, y(t)), \quad y(t_0) = y_0$$
In numerical methods, a higher-order ODE is usually transformed into a system of first-order ODE and then this system is solved using numerical integration.
### Euler method¶
The most simple method to solve an ODE is using the Euler method.
First, the derivative of $y$ is approximated by:
$$\dot{y}(t) \approx \frac{y(t+h)-y(t)}{h}$$
Where $h$ is the step size.
Rearraning the equation above:
$$y(t+h) \approx y(t) +h\dot{y}(t)$$
And replacing $\dot{y}(t)$:
$$y(t+h) \approx y(t) +hf(t, y(t))$$
The ODE then can be solved starting at $t_0$, which has a known value for $y_0$:
$$y(t+h) \approx y_0 + hf(t_0, y_0)$$
And using the equation recursively for a sequence of values for $t$ $(t_0, t_0+h, t_0+2h, ...)$:
$$y_{n+1} = y_n + hf(t_n, y_n)$$
This is the Euler method to solve an ODE with a known initial value.
### Other numerical methods for solving ODE¶
There are other methods for solving an ODE. One family of methods, usually more accurate, uses more points in the interval $[t_n,t_{n+1}]$ and are known as Runge–Kutta methods. In the Python ecosystem, Runge–Kutta methods are available using the scipy.integrate.ode library of numeric integrators. The library scipy.integrate.odeint has other popular integrator known as lsoda, from the FORTRAN library odepack.
## Examples¶
### Motion under constant force¶
Consider a football ball kicked up from an initial height $y_0$ and with initial velocity $v_0$. Determine the equation of motion of the ball in the vertical direction.
Neglecting the air resistance, Newton's second law of motion applied to this problem for the instants the ball is in the air gives:
$$m\frac{\mathrm{d}^2 y}{\mathrm{d}t^2} = -mg$$
Consider $g=9.8m/s^2$, $y_0(t_0=0)=1m$, and $v_0(t_0=0)=20m/s$.
We know the analytical solution for this problem:
$$y(t) = y_0 + v_0 t - \frac{g}{2}t^2$$
Let's solve this problem numerically and compare the results.
A second-order ODE can be transformed into two first-order ODE, introducing a new variable:
$$\dot{y} = v$$$$\dot{v} = a$$
And rewriting Newton's second law as a couple of equations:
$$\left\{ \begin{array}{r} \frac{\mathrm{d} y}{\mathrm{d}t} = &v, \quad y(t_0) = y_0 \\ \frac{\mathrm{d} v}{\mathrm{d}t} = &-g, \quad v(t_0) = v_0 \end{array} \right.$$
First, let's import the necessary Python libraries and customize the environment:
In [1]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
#%matplotlib nbagg
import matplotlib
matplotlib.rcParams['lines.linewidth'] = 3
matplotlib.rcParams['font.size'] = 13
matplotlib.rcParams['lines.markersize'] = 5
matplotlib.rc('axes', grid=False, labelsize=14, titlesize=16, ymargin=0.05)
matplotlib.rc('legend', numpoints=1, fontsize=11)
This is the equation for calculating the ball trajectory given the model and using the Euler method:
In [2]:
def ball_euler(t0, tend, y0, v0, h):
t, y, v, i = [t0], [y0], [v0], 0
a = -9.8
while t[-1] <= tend and y[-1] > 0:
y.append(y[-1] + h*v[-1])
v.append(v[-1] + h*a)
i += 1
t.append(i*h)
return np.array(t), np.array(y), np.array(v)
Initial values:
In [3]:
y0 = 1
v0 = 20
a = -9.8
Let's call the function with different step sizes:
In [4]:
t100, y100, v100 = ball_euler(0, 10, y0, v0, 0.1)
t10, y10, v10 = ball_euler(0, 10, y0, v0, 0.01)
Here are the plots for the results:
In [5]:
def plots(t100, y100, v100, t10, y10, v10, title):
"""Plots of numerical integration results.
"""
a = -9.8
fig, axs = plt.subplots(nrows=2, ncols=2, sharex=True, figsize=(10, 5))
axs[0, 0].plot(t10, y0 + v0*t10 + 0.5*a*t10**2, color=[0, 0, 1, .7], label='Analytical')
axs[0, 0].plot(t100, y100, '--', color=[0, 1, 0, .7], label='h = 100ms')
axs[0, 0].plot(t10, y10, ':', color=[1, 0, 0, .7], label='h = 10ms')
axs[0, 1].plot(t10, v0 + a*t10, color=[0, 0, 1, .5], label='Analytical')
axs[0, 1].plot(t100, v100, '--', color=[0, 1, 0, .7], label='h = 100ms')
axs[0, 1].plot(t10, v10, ':', color=[1, 0, 0, .7], label='h = 10ms')
axs[1, 0].plot(t10, y0 + v0*t10 + 0.5*a*t10**2 - (y0 + v0*t10 + 0.5*a*t10**2),
color=[0, 0, 1, .7], label='Analytical')
axs[1, 0].plot(t100, y100 - (y0 + v0*t100 + 0.5*a*t100**2), '--',
color=[0, 1, 0, .7], label='h = 100ms')
axs[1, 0].plot(t10, y10 - (y0 + v0*t10 + 0.5*a*t10**2), ':',
color=[1, 0, 0, .7], label='h = 10ms')
axs[1, 1].plot(t10, v0 + a*t10 - (v0 + a*t10), color=[0, 0, 1, .7], label='Analytical')
axs[1, 1].plot(t100, v100 - (v0 + a*t100), '--', color=[0, 1, 0, .7], label='h = 100ms')
axs[1, 1].plot(t10, v10 - (v0 + a*t10), ':', color=[1, 0, 0, .7], label='h = 10ms')
ylabel = ['y [m]', 'v [m/s]', 'y error [m]', 'v error [m/s]']
axs[0, 0].set_xlim(t10[0], t10[-1])
axs[1, 0].set_xlabel('Time [s]')
axs[1, 1].set_xlabel('Time [s]')
axs[0, 1].legend()
axs = axs.flatten()
for i, ax in enumerate(axs):
ax.set_ylabel(ylabel[i])
plt.suptitle('Kinematics of a soccer ball - %s method'%title, y=1.02, fontsize=16)
plt.tight_layout()
plt.show()
In [6]:
plots(t100, y100, v100, t10, y10, v10, 'Euler')
Let's use the integrator lsoda to solve the same problem:
In [7]:
from scipy.integrate import odeint, ode
def ball_eq(yv, t):
y = yv[0] # position
v = yv[1] # velocity
a = -9.8 # acceleration
return [v, a]
In [8]:
yv0 = [1, 20]
t10 = np.arange(0, 4, 0.1)
yv10 = odeint(ball_eq, yv0, t10)
y10, v10 = yv10[:, 0], yv10[:, 1]
t100 = np.arange(0, 4, 0.01)
yv100 = odeint(ball_eq, yv0, t100)
y100, v100 = yv100[:, 0], yv100[:, 1]
In [9]:
plots(t100, y100, v100, t10, y10, v10, 'lsoda')
Let's use an explicit runge-kutta method of order (4)5 due to Dormand and Prince (a.k.a. ode45 in Matlab):
In [10]:
def ball_eq(t, yv):
y = yv[0] # position
v = yv[1] # velocity
a = -9.8 # acceleration
return [v, a]
In [11]:
def ball_sol(fun, t0, tend, yv0, h):
f = ode(fun).set_integrator('dopri5')
# or f = ode(fun).set_integrator('dopri5', nsteps=1, max_step=h/2)
f.set_initial_value(yv0, t0)
data = []
while f.successful() and f.t < tend:
f.integrate(f.t + h)
# or f.integrate(tend)
data.append([f.t, f.y[0], f.y[1]])
data = np.array(data)
return data
In [12]:
data = ball_sol(ball_eq, 0, 4, [1, 20], .1)
t100, y100, v100 = data[:, 0], data[:, 1], data[:, 2]
data = ball_sol(ball_eq, 0, 4, [1, 20], .01)
t10, y10, v10 = data[:, 0], data[:, 1], data[:, 2]
In [13]:
plots(t100, y100, v100, t10, y10, v10, 'dopri5 (ode45)')
### Motion under varying force¶
Let's consider the air resistance in the calculations for the vertical trajectory of the football ball.
According to the Laws of the Game from FIFA, the ball is spherical, has a circumference of $0.69m$, and a mass of $0.43kg$.
We will model the magnitude of the drag force due to the air resistance by:
$$F_d(v) = \frac{1}{2}\rho C_d A v^2$$
Where $\rho$ is the air density $(1.22kg/m^3)$, $A$ the ball cross sectional area $(0.0379m^2)$, and $C_d$ the drag coefficient, which for now we will consider constant and equal to $0.25$ (Bray and Kerwin, 2003).
Applying Newton's second law of motion to the new problem:
$$m\frac{\mathrm{d}^2 y}{\mathrm{d}t^2} = -mg -\frac{1}{2}\rho C_d A v^2\frac{v}{||v||}$$
In the equation above, $-v/||v||$ takes into account that the drag force always acts opposite to the direction of motion.
Reformulating the second-order ODE above as a couple of first-order equations:
$$\left\{ \begin{array}{l l} \frac{\mathrm{d} y}{\mathrm{d}t} = &v, \quad &y(t_0) = y_0 \\ \frac{\mathrm{d} v}{\mathrm{d}t} = &-g -\frac{1}{2m}\rho C_d A v^2\frac{v}{||v||}, \quad &v(t_0) = v_0 \end{array} \right.$$
Although (much) more complicated, it's still possible to find an analytical solution for this problem. But for now let's explore the power of numerical integration and use the lsoda method (the most simple method to call in terms of number of lines of code) to solve this problem:
In [14]:
def ball_eq(yv, t):
g = 9.8 # m/s2
m = 0.43 # kg
rho = 1.22 # kg/m3
cd = 0.25 # dimensionless
A = 0.0379 # m2
y = yv[0] # position
v = yv[1] # velocity
a = -g - 1/(2*m)*rho*cd*A*v*np.abs(v) # acceleration
return [v, a]
In [15]:
yv0 = [1, 20]
t10 = np.arange(0, 4, 0.01)
yv10 = odeint(ball_eq, yv0, t10)
y10, v10 = yv10[:, 0], yv10[:, 1]
In [16]:
def plots(t10, y10, v10):
"""Plots of numerical integration results.
"""
a = -9.8
fig, axs = plt.subplots(nrows=2, ncols=2, sharex=True, figsize=(10, 5))
axs[0, 0].plot(t10, y0 + v0*t10 + 0.5*a*t10**2, color=[0, 0, 1, .7], label='No resistance')
axs[0, 0].plot(t10, y10, '-', color=[1, 0, 0, .7], label='With resistance')
axs[0, 1].plot(t10, v0 + a*t10, color=[0, 0, 1, .7], label='No resistance')
axs[0, 1].plot(t10, v10, '-', color=[1, 0, 0, .7], label='With resistance')
axs[1, 0].plot(t10, y0 + v0*t10 + 0.5*a*t10**2 - (y0 + v0*t10 + 0.5*a*t10**2),
color=[0, 0, 1, .7], label='Real')
axs[1, 0].plot(t10, y10 - (y0 + v0*t10 + 0.5*a*t10**2), '-',
color=[1, 0, 0, .7], label='h=10 ms')
axs[1, 1].plot(t10, v0 + a*t10 - (v0 + a*t10), color=[0, 0, 1, .7], label='No resistance')
axs[1, 1].plot(t10, v10 - (v0 + a*t10), '-', color=[1, 0, 0, .7], label='With resistance')
ylabel = ['y [m]', 'v [m/s]', 'y diff [m]', 'v diff [m/s]']
axs[1, 0].set_xlabel('Time [s]')
axs[1, 1].set_xlabel('Time [s]')
axs[0, 1].legend()
axs = axs.flatten()
for i, ax in enumerate(axs):
ax.set_ylabel(ylabel[i])
plt.suptitle('Kinematics of a soccer ball - effect of air resistance', y=1.02, fontsize=16)
plt.tight_layout()
plt.show()
In [17]:
plots(t10, y10, v10)
## Exercises¶
1. Run the simulations above considering different values for the parameters.
2. Model and run simulations for the two-dimensional case of the ball trajectory and investigate the effect of air resistance. Hint: chapter 9 of Downey (2011) presents part of the solution. | 4,258 | 11,777 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2017-17 | longest | en | 0.891279 |
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RD Sharma Solutions (Part 1): Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Class 10 : RD Sharma Solutions (Part 1): Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
The document RD Sharma Solutions (Part 1): Pair of Linear Equations in Two Variables Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
Exercise 3.1
Q.1. Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and graphically. Sol: The pair of equations formed is:
Solution.
The pair of equations formed is:
i.e., x - 2y = 0 ....(1)
3x + 4y = 20 ....(2)
Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in Table
Recall from Class IX that there are infinitely many solutions of each linear equation. So each of you choose any two values, which may not be the ones we have chosen. Can you guess why we have chosen x =O in the first equation and in the second equation? When one of the variables is zero, the equation reduces to a linear equation is one variable, which can be solved easily. For instance, putting x =O in Equation (2), we get 4y = 20 i.e.,
y = 5. Similarly, putting y =O in Equation (2), we get 3x = 20 ..,But asis not an integer, it will not be easy to plot exactly on the graph paper. So, we choose y = 2 which gives x = 4, an integral value.
Plot the points A (O,O) , B (2,1) and P (O,5) , Q (412) , corresponding to the draw the lines AB and PQ, representing the equations x - 2 y = O and 3x + 4y= 20, as shown in figure
In fig., observe that the two lines representing the two equations are intersecting at the point (4,2),
Q.2. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Is not this interesting? Represent this situation algebraically and graphically.
Sol: Let the present age of Aftab and his daughter be x and y respectively. Seven years ago.
Age of Ahab = x - 7
Age of his daughter y - 7
According to the given condition.
(x - 7) = 7(y - 7)
⇒ x - 7 = 7y - 49
⇒ x - 7y = -42
Three years hence
Age of Aftab = x + 3
Age of his daughter = y + 3
According to the given condition,
(x + 3) = 3 (y + 3)
⇒ x+3 = 3y +9
⇒ x - 3y = 6
Thus, the given condition can be algebraically represented as
x - 7y = - 42
x - 3y = 6
x - 7y = - 42 ⇒ x = -42 + 7y
Three solution of this equation can be written in a table as follows:
x - 3y = 6 ⇒ x = 6+3y
Three solution of this equation can be written in a table as follows:
The graphical representation is as follows:
Concept insight In order to represent a given situation mathematically, first see what we need to find out in the problem. Here. Aftab and his daughters present age needs to be found so, so the ages will be represented by variables z and y. The problem talks about their ages seven years ago and three years from now. Here, the words ’seven years ago’ means we have to subtract 7 from their present ages. and ‘three years from now’ or three years hence means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.
Q.3. The path of a train A is given by the equation 3x + 4y - 12 = 0 and the path of another train B is given by the equation 6x + 8y - 48 = 0. Represent this situation graphically.
Sol:
The paths of two trains are giver by the following pair of linear equations.
3x + 4 y -12 = 0 ...(1)
6x + 8 y - 48 = 0 ... (2)
In order to represent the above pair of linear equations graphically. We need two points on the line representing each equation. That is, we find two solutions of each equation as given below:
We have,
3x + 4 y -12 = 0
Putting y = 0, we get
3x + 4 x 0 - 12 = 0
⇒ 3x = 12
Putting x = 0, we get
3 x 0 + 4 y -12 = 0
⇒ 4y = 12
Thus, two solution of equation 3x + 4y - 12 = 0 are ( 0, 3) and ( 4, 0 )
We have,
6x + 8y -48 = 0
Putting x = 0, we get
6 x 0 + 8 y - 48 = 0
⇒ 8y = 48
⇒ y = 6
Putting y = 0, we get
6x + 8 x 0 = 48 = 0
⇒ 6x = 48
Thus, two solution of equation 6 x + 8y - 48= 0 are ( 0, 6 ) and (8, 0 )
Clearly, two lines intersect at ( -1, 2 )
Hence, x = -1,y = 2 is the solution of the given system of equations.
Q.4. Gloria is walking along the path joining (— 2, 3) and (2, — 2), while Suresh is walking along the path joining (0, 5) and (4, 0). Represent this situation graphically.
Sol:
It is given that Gloria is walking along the path Joining (-2,3) and (2, -2), while Suresh is walking along the path joining (0,5) and (4,0).
We observe that the lines are parallel and they do not intersect anywhere.
Q.5. On comparing the ratios and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide:
(i) 5x- 4y + 8 = 0
7x + 6y - 9 = 0
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
(iii) 6x - 3y + 10 = 0
2x - y + 9 = 0
Sol:
We have,
5x - 4 y + 8 = 0
7 x + 6 y - 9 = 0
Here,
a= 5, b1 = -4, c1 = 8
a2 = 7, b2 = 6, c2 = -9
We have,
∴ Two lines are intersecting with each other at a point.
We have,
9 x + 3 y +12 = 0
18 + 6 y + 24 = 0
Here,
a1 = 9, b1 = 3, c1 = 12
a2 = 18, b2 = 6, c2 = 24
Now,
And
∴ Both the lines coincide.
We have,
6 x - 3 y +10 = 0
2 x - y + 9 = 0
Here,
a1 = 6, b= -3, c1 = 10
a2 = 2, b2 = -1, c2 = 9
Now,
And
∴ The lines are parallel
Q.6. Given the linear equation 2x + 3y - 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines.
Sol:
We have,
2x + 3 y - 8 = 0
Let another equation of line is
4x + 9 y - 4 = 0
Here,
a1 = 2, b1 = 3, c1 = -8
a= 4, b2 = 9, c2 = -4
Now,
And
∴ 2x + 3 y - 8 = 0 and 4 x + 9 y - 4 = 0 intersect each other at one point.
Hence, required equation of line is 4 x + 9y - 4 = 0
We have,
2x + 3y -8 = 0
Let another equation of line is:
4x +6y -4 = 0
Here,
a1 = 2, b1 = 3, c1 = -8
a2 = 4, b2 = 6, c2 = -4
Now,
And
∴ Lines are parallel to each other.
Hence, required equation of line is 4 x + 6y - 4 = 0.
Q.7. The cost of 2kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4kg of apples and 2kg of grapes is Rs 300. Represent the situation algebraically and geometrically.
Sol:
Let the cost of 1 kg of apples and 1 kg grapes be Rs x and Rs y.
The given conditions can be algebraically represented as:
2 x + y = 160 4 x + 2 y = 300
2x + y = 160 ⇒ y = 160 - 2x
Three solutions of this equation cab be written in a table as follows:
4x + 2y = 300 ⇒ y =
Three solutions of this equation cab be written in a table as follows:
The graphical representation is as follows:
Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1kg grapes will be taken as the variables from the given condition of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then In order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are Large so take the suitable scale.
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Mathematics (Maths) Class 10
62 videos|363 docs|103 tests
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; | 2,564 | 7,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2021-25 | latest | en | 0.93669 |
https://tympanus.net/codrops/2021/06/01/trigonometry-in-css-and-javascript-introduction-to-trigonometry/ | 1,722,940,545,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640484318.27/warc/CC-MAIN-20240806095414-20240806125414-00186.warc.gz | 478,048,313 | 18,390 | # Trigonometry in CSS and JavaScript: Introduction to Trigonometry
In this series of articles we’ll get an overview of trigonometry, understand how it can be useful, and delve into some creative applications in CSS and JavaScript.
Understanding trigonometry can give us super powers when it comes to creative coding. But to the uninitiated, it can seem a little intimidating. In this 3-part series of articles we’ll get an overview of trigonometry, understand how it can be useful, and delve into some creative applications in CSS and JavaScript.
2. Getting Creative with Trigonometric Functions
3. Beyond Triangles
## Trigonometry basics
If, like me, you’ve rarely used trigonometry outside of the classroom, let’s take a trip back to school and get ourselves reacquainted.
Trigonometric functions allow us to calculate unknown values of a right-angled triangle from known parameters. Imagine you’re standing on the ground, looking up at a tall tree. It would be very difficult to measure the height of the tree from the ground. But if we know the angle at which we look up at the top of the tree, and we know the distance from ourselves to the tree, we can infer the height of the tree itself.
If we imagine this scene as a triangle, the known length (from us to the tree) is known as the adjacent side, the tree is the opposite side (it’s opposite the angle), and the longest side – from us to the top of the tree – is called the hypotenuse.
### Sine, Cosine and Tangent
There are three main functions to remember in trigonometry: Sine, Cosine and Tangent (abbreviated to sin, cos and tan). They are expressed as the following formulae:
``````sin(angle) = opposite / hypotenuse
The angle is usually written as the Greek theta (θ) symbol.
We can use these equations to calculate the unknown values of our triangle from the known ones. To measure the height of the tree in the example, we know the angle (θ) and the adjacent side.
To calculate the opposite side we would need the tangent function. We would need to switch around the formula:
``opposite = tan(angle) * adjacent``
How do we get tan(θ)? We could use a scientific calculator (type tan and then the angle), or we could use code! Sass and JavaScript both include trigonometric functions, and we’ll look at some ways to use these in this article and the following ones.
## Sass functions
If we’re working with predetermined values, we could use the trigonometric functions built into Sass (the CSS preprocessor).
To include the Math module we need the following line in our Sass file:
``@use "sass:math";``
We can use variables to calculate the opposite side from the angle and adjacent side values.
``````\$angle: 45deg;
The tan function in Sass can use radians or degrees — if using degrees, the units must be specified. Without units, radians will be used by default (more on these later).
In the following demo we’re using these in the `clip-path` property to determine the coordinates of the polygon points, similar to calculating the height of a tree.
See the Pen Using Sass trigonometry for clip-path values by Michelle Barker (@michellebarker) on CodePen.
We need to subtract the `\$opposite` variable from the height of the element in order to get the y coordinate — as clip-path coordinates are plotted along the y axis increasing from top to bottom.
``````.element {
}``````
## Clipping an equilateral triangle
A right-angled triangle is the simplest use of trigonometry. But we can work out the coordinates of more complex shapes by splitting them up into right-angled triangles.
An equilateral triangle is a triangle with three sides of the same length. Perhaps you remember from school that the angles in a triangle add up to 180º? That means each angle in an equilateral triangle is 60º.
If we draw a line down the middle of an equilateral triangle, we split it into (you guessed it) two right-angled triangles. So, for a triangle with sides of a given length, we know the angle (60º), the length of the hypotenuse, and the length of the adjacent side (half the length of the hypotenuse).
What we don’t know is the height of the triangle — once again, the opposite side of the right-angled triangle. To plot the clip-path coordinates, this is what we need to work out. This time, as we know the angle and the length of the hypotenuse, we can use the sine function:
``````\$hypotenuse: 60%; // side length
\$angle: 60deg;
\$opposite: math.sin(\$angle) * \$hypotenuse;``````
(It would also be possible for us to use the tangent function instead, as we know that the length of the adjacent side is half of the hypotenuse.) Then we can use those values for our clip-path polygon points:
``````.element {
clip-path: polygon(
0 \$opposite,
(\$hypotenuse / 2) 0,
\$hypotenuse \$opposite
);
}``````
See the Pen Clip-path simple equilateral triangles with Sass by Michelle Barker (@michellebarker) on CodePen.
As you can see in the demo, the element is clipped from the top left corner. This might not be completely satisfactory: it’s more likely we’d want to clip from the center, especially if we’re clipping an image. We can adjust our clip-path coordinates accordingly. To make this more readable, we can assign some additional variables for the adjacent side length (half the hypotenuse), and the start and end position of the triangle:
``````\$hypotenuse: 60%; //side length
\$angle: 60deg;
\$opposite: math.sin(\$angle) * \$hypotenuse;
\$startPosY: (50% - \$opposite / 2);
\$endPosY: (50% + \$opposite / 2);
.element {
clip-path: polygon(
\$startPosX \$endPosY,
50% \$startPosY,
\$endPosX \$endPosY
);
}``````
### Creating a mixin for reuse
This is quite a bit of complex code to write for a single triangle. Let’s create a Sass mixin, allowing us to clip a triangle of any size on any element we like. As `clip-path` still needs a prefix in some browsers, our mixin covers that too:
``````@mixin triangle(\$sideLength) {
\$hypotenuse: \$sideLength;
\$angle: 60deg;
\$opposite: math.sin(\$angle) * \$hypotenuse;
\$startPosY: (50% - \$opposite / 2);
\$endPosY: (50% + \$opposite / 2);
\$clip: polygon(
\$startPosX \$endPosY,
50% \$startPosY,
\$endPosX \$endPosY
);
-webkit-clip-path: \$clip;
clip-path: \$clip;
}``````
To clip a centred equilateral triangle from any element, we can simply include the mixin, passing in the length of the triangle’s sides:
``````.triangle {
@include triangle(60%);
}``````
See the Pen Clip-path equilateral triangles with Sass trigonometric functions by Michelle Barker (@michellebarker) on CodePen.
### Limitations of Sass functions
Our use of Sass functions has some limitations:
1. It assumes the `\$sideLength` variable is known at compile time, and doesn’t allow for dynamic values.
2. Sass doesn’t handle mixing units all that well for our needs. In the last demo, if you switch out the percentage-based side length to a fixed length (such as rems or pixels), the code breaks.
The latter is because our calculations for the `\$startPos` and `\$endPos` variables (to position the clip-path centrally) depend on subtracting the side length from a percentage. Unlike in regular CSS (using calc()), Sass doesn’t allow for that. In the final demo, I’ve adjusted the mixin so that it works for any valid length unit, by passing in the size of the clipped element as a parameter. We’d just need to ensure that the values for the two parameters passed in have identical units.
See the Pen Clip-path equilateral triangles with Sass trigonometric functions by Michelle Barker (@michellebarker) on CodePen.
## CSS trigonometric functions
CSS has a proposal for trigonometric functions as part of the CSS Values and Units Module Level 4 (currently in working draft). These could be extremely useful, especially when used alongside custom properties. Here’s how we could rewrite our CSS to use native CSS trigonometric functions. Changing the size of the clip path is as simple as updating a single custom property:
``````.triangle {
--hypotenuse: 8rem;
--opposite: calc(sin(60deg) * var(--hypotenuse));
--startPosX: calc(var(--size) / 2 - var(--adjacent));
--startPosY: calc(var(--size) / 2 - var(--opposite) / 2);
--endPosX: calc(var(--size) / 2 + var(--adjacent));
--endPosY: calc(var(--size) / 2 + var(--opposite) / 2);
--clip: polygon(
var(--startPosX) var(--endPosX),
50% var(--startPosY),
var(--endPosX) var(--endPosY)
);
-webkit-clip-path: var(--clip);
clip-path: var(--clip);
}
.triangle:nth-child(2) {
--hypotenuse: 3rem;
}
.triangle:nth-child(2) {
--hypotenuse: 50%;
}``````
### Dynamic variables
Custom properties can be dynamic too. We can change them with JS and the values dependant on them will be automatically recalculated.
``triangle.style.setProperty('--hypotenuse', '5rem')``
CSS trigonometric functions have a lot of potential when they finally land, but sadly they’re not yet supported in any browsers. To use trigonometry with dynamic variables right now, we need JavaScript.
We’ll take a look at some of the possibilities in the next article.
### Michelle Barker
Michelle is a Senior Front End Developer at Ada Mode, where she builds web apps and data visualisations for the renewable energy industry. She is the author of front-end blog CSS { In Real Life }, and has written articles for CSS Tricks, Smashing Magazine, and Web Designer Magazine, to name a few. She enjoys experimenting with new CSS features and helping others learn about them.
### Stay in the loop: Get your dose of frontend twice a week
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https://runestone.academy/ns/books/published/ac-single/sec-2-6-inverse.html | 1,723,634,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00777.warc.gz | 401,259,306 | 42,308 | # Active Calculus
## Section2.6Derivatives of Inverse Functions
Much of mathematics centers on the notion of function. Indeed, throughout our study of calculus, we are investigating the behavior of functions, with particular emphasis on how fast the output of the function changes in response to changes in the input. Because each function represents a process, a natural question to ask is whether or not the particular process can be reversed. That is, if we know the output that results from the function, can we determine the input that led to it? And if we know how fast a particular process is changing, can we determine how fast the inverse process is changing?
One of the most important functions in all of mathematics is the natural exponential function $$f(x) = e^x\text{.}$$ Its inverse, the natural logarithm, $$g(x) = \ln(x)\text{,}$$ is similarly important. One of our goals in this section is to learn how to differentiate the logarithm function. First, we review some of the basic concepts surrounding functions and their inverses.
### Preview Activity2.6.1.
The equation $$y = \frac{5}{9}(x - 32)$$ relates a temperature given in $$x$$ degrees Fahrenheit to the corresponding temperature $$y$$ measured in degrees Celcius.
a. Solve the equation $$y = \frac{5}{9} (x - 32)$$ for $$x$$ to write $$x$$ (Fahrenheit temperature) in terms of $$y$$ (Celcius temperature).
$$x =$$ .
b. Now let $$C(x) = \frac{5}{9} (x - 32)$$ be the function that takes a Fahrenheit temperature as input and produces the Celcius temperature as output. In addition, let $$F(y)$$ be the function that converts a temperature given in $$y$$ degrees Celcius to the temperature $$F(y)$$ measured in degrees Fahrenheit. Use your work above to write a formula for $$F(y)\text{.}$$
$$F(y) =$$ .
c. Next consider the new function defined by $$p(x) = F(C(x))\text{.}$$ Use the formulas for $$F$$ and $$C$$ to determine an expression for $$p(x)$$ and simplify this expression as much as possible. What do you observe?
• $$\displaystyle p(x)=x$$
• $$\displaystyle p(x)=0$$
• $$\displaystyle p(x)=1$$
• $$\displaystyle p(x)=5x+9$$
• $$\displaystyle p(x)=9x+5$$
d. Now, let $$r(y) = C(F(y))\text{.}$$ Use the formulas for $$F$$ and $$C$$ to determine an expression for $$r(y)$$ and simplify this expression as much as possible. What do you observe?
• $$\displaystyle r(y)=y$$
• $$\displaystyle r(y)=1$$
• $$\displaystyle r(y)=9y+5$$
• $$\displaystyle r(y)=5y+9$$
• $$\displaystyle r(y)=0$$
e. What is the value of $$C'(x)\text{?}$$
$$C'(x) =$$
What is the value of $$F'(y)\text{?}$$
$$F'(y) =$$
How do $$C'(x)$$ and $$F'(y)$$ appear to be related?
• They add up to 1
• They are equal
• They are reciprocals
• They are opposites
### Subsection2.6.1Basic facts about inverse functions
A function $$f : A \to B$$ is a rule that associates each element in the set $$A$$ to one and only one element in the set $$B\text{.}$$ We call $$A$$ the domain of $$f$$ and $$B$$ the codomain of $$f\text{.}$$ If there exists a function $$g : B \to A$$ such that $$g(f(a)) = a$$ for every possible choice of $$a$$ in the set $$A$$ and $$f(g(b)) = b$$ for every $$b$$ in the set $$B\text{,}$$ then we say that $$g$$ is the inverse of $$f\text{.}$$
We often use the notation $$f^{-1}$$ (read “$$f$$-inverse”) to denote the inverse of $$f\text{.}$$ The inverse function undoes the work of $$f\text{.}$$ Indeed, if $$y = f(x)\text{,}$$ then
\begin{equation*} f^{-1}(y) = f^{-1}(f(x)) = x\text{.} \end{equation*}
Thus, the equations $$y = f(x)$$ and $$x = f^{-1}(y)$$ say the same thing. The only difference between the two equations is one of perspective — one is solved for $$x\text{,}$$ while the other is solved for $$y\text{.}$$
Here we briefly remind ourselves of some key facts about inverse functions.
#### Note2.6.1.
For a function $$f : A \to B\text{,}$$
• $$f$$ has an inverse if and only if $$f$$ is one-to-one
1
A function $$f$$ is one-to-one provided that no two distinct inputs lead to the same output.
and onto
2
A function $$f$$ is onto provided that every possible element of the codomain can be realized as an output of the function for some choice of input from the domain.
;
• provided $$f^{-1}$$ exists, the domain of $$f^{-1}$$ is the codomain of $$f\text{,}$$ and the codomain of $$f^{-1}$$ is the domain of $$f\text{;}$$
• $$f^{-1}(f(x)) = x$$ for every $$x$$ in the domain of $$f$$ and $$f(f^{-1}(y)) = y$$ for every $$y$$ in the codomain of $$f\text{;}$$
• $$y = f(x)$$ if and only if $$x = f^{-1}(y)\text{.}$$
The last fact reveals a special relationship between the graphs of $$f$$ and $$f^{-1}\text{.}$$ If a point $$(x,y)$$ that lies on the graph of $$y = f(x)\text{,}$$ then it is also true that $$x = f^{-1}(y)\text{,}$$ which means that the point $$(y,x)$$ lies on the graph of $$f^{-1}\text{.}$$ This shows us that the graphs of $$f$$ and $$f^{-1}$$ are the reflections of each other across the line $$y = x\text{,}$$ because this reflection is precisely the geometric action that swaps the coordinates in an ordered pair. In Figure 2.6.2, we see this illustrated by the function $$y = f(x) = 2^x$$ and its inverse, with the points $$(-1,\frac{1}{2})$$ and $$(\frac{1}{2},-1)$$ highlighting the reflection of the curves across $$y = x\text{.}$$
To close our review of important facts about inverses, we recall that the natural exponential function $$y = f(x) = e^x$$ has an inverse function, namely the natural logarithm, $$x = f^{-1}(y) = \ln(y)\text{.}$$ Thus, writing $$y = e^x$$ is interchangeable with $$x = \ln(y)\text{,}$$ plus $$\ln(e^x) = x$$ for every real number $$x$$ and $$e^{\ln(y)} = y$$ for every positive real number $$y\text{.}$$
### Subsection2.6.2The derivative of the natural logarithm function
In what follows, we find a formula for the derivative of $$g(x) = \ln(x)\text{.}$$ To do so, we take advantage of the fact that we know the derivative of the natural exponential function, the inverse of $$g\text{.}$$ In particular, we know that writing $$g(x) = \ln(x)$$ is equivalent to writing $$e^{g(x)} = x\text{.}$$ Now we differentiate both sides of this equation and observe that
\begin{equation*} \frac{d}{dx}\left[e^{g(x)}\right] = \frac{d}{dx}[x]\text{.} \end{equation*}
The righthand side is simply $$1\text{;}$$ by applying the chain rule to the left side, we find that
\begin{equation*} e^{g(x)} g'(x) = 1\text{.} \end{equation*}
Next we solve for $$g'(x)\text{,}$$ to get
\begin{equation*} g'(x) = \frac{1}{e^{g(x)}}\text{.} \end{equation*}
Finally, we recall that $$g(x) = \ln(x)\text{,}$$ so $$e^{g(x)} = e^{\ln(x)} = x\text{,}$$ and thus
\begin{equation*} g'(x) = \frac{1}{x}\text{.} \end{equation*}
#### Natural Logarithm.
For all positive real numbers $$x\text{,}$$ $$\frac{d}{dx}[\ln(x)] = \frac{1}{x}\text{.}$$
This rule for the natural logarithm function now joins our list of basic derivative rules. Note that this rule applies only to positive values of $$x\text{,}$$ as these are the only values for which $$\ln(x)$$ is defined.
Also notice that for the first time in our work, differentiating a basic function of a particular type has led to a function of a very different nature: the derivative of the natural logarithm is not another logarithm, nor even an exponential function, but rather a rational one.
Derivatives of logarithms may now be computed in concert with all of the rules known to date. For instance, if $$f(t) = \ln(t^2 + 1)\text{,}$$ then by the chain rule, $$f'(t) = \frac{1}{t^2 + 1} \cdot 2t\text{.}$$
There are interesting connections between the graphs of $$f(x) = e^x$$ and $$f^{-1}(x) = \ln(x)\text{.}$$
In Figure 2.6.3, we are reminded that since the natural exponential function has the property that its derivative is itself, the slope of the tangent to $$y = e^x$$ is equal to the height of the curve at that point. For instance, at the point $$A = (\ln(0.5), 0.5)\text{,}$$ the slope of the tangent line is $$m_A = 0.5\text{,}$$ and at $$B = (\ln(5), 5)\text{,}$$ the tangent line’s slope is $$m_B = 5\text{.}$$
At the corresponding points $$A'$$ and $$B'$$ on the graph of the natural logarithm function (which come from reflecting $$A$$ and $$B$$ across the line $$y = x$$), we know that the slope of the tangent line is the reciprocal of the $$x$$-coordinate of the point (since $$\frac{d}{dx}[\ln(x)] = \frac{1}{x}$$). Thus, at $$A' = (0.5, \ln(0.5))\text{,}$$ we have $$m_{A'} = \frac{1}{0.5} = 2\text{,}$$ and at $$B' = (5, \ln(5))\text{,}$$ $$m_{B'} = \frac{1}{5}\text{.}$$
In particular, we observe that $$m_{A'} = \frac{1}{m_A}$$ and $$m_{B'} = \frac{1}{m_B}\text{.}$$ This is not a coincidence, but in fact holds for any curve $$y = f(x)$$ and its inverse, provided the inverse exists. This is due to the reflection across $$y = x\text{.}$$ It changes the roles of $$x$$ and $$y\text{,}$$ thus reversing the rise and run, so the slope of the inverse function at the reflected point is the reciprocal of the slope of the original function.
#### Activity2.6.2.
For each function given below, find its derivative.
1. $$\displaystyle h(x) = x^2\ln(x)$$
2. $$\displaystyle p(t) = \frac{\ln(t)}{e^t + 1}$$
3. $$\displaystyle s(y) = \ln(\cos(y) + 2)$$
4. $$\displaystyle z(x) = \tan(\ln(x))$$
5. $$\displaystyle m(z) = \ln(\ln(z))$$
### Subsection2.6.3Inverse trigonometric functions and their derivatives
Trigonometric functions are periodic, so they fail to be one-to-one, and thus do not have inverse functions. However, we can restrict the domain of each trigonometric function so that it is one-to-one on that domain.
For instance, consider the sine function on the domain $$[-\frac{\pi}{2}, \frac{\pi}{2}]\text{.}$$ Because no output of the sine function is repeated on this interval, the function is one-to-one and thus has an inverse. Thus, the function $$f(x) = \sin(x)$$ with $$[-\frac{\pi}{2}, \frac{\pi}{2}]$$ and codomain $$[-1,1]$$ has an inverse function $$f^{-1}$$ such that
\begin{equation*} f^{-1} : [-1,1] \to [-\frac{\pi}{2}, \frac{\pi}{2}]\text{.} \end{equation*}
We call $$f^{-1}$$ the arcsine (or inverse sine) function and write $$f^{-1}(y) = \arcsin(y)\text{.}$$ It is especially important to remember that
\begin{equation*} y = \sin(x) \ \ \text{and} \ \ x = \arcsin(y) \end{equation*}
say the same thing. “The arcsine of $$y$$” means “the angle whose sine is $$y\text{.}$$” For example, $$\arcsin(\frac{1}{2}) = \frac{\pi}{6}$$ means that $$\frac{\pi}{6}$$ is the angle whose sine is $$\frac{1}{2}\text{,}$$ which is equivalent to writing $$\sin(\frac{\pi}{6}) = \frac{1}{2}\text{.}$$
Next, we determine the derivative of the arcsine function. Letting $$h(x) = \arcsin(x)\text{,}$$ our goal is to find $$h'(x)\text{.}$$ Since $$h(x)$$ is the angle whose sine is $$x\text{,}$$ it is equivalent to write
\begin{equation*} \sin(h(x)) = x\text{.} \end{equation*}
Differentiating both sides of the previous equation, we have
\begin{equation*} \frac{d}{dx}[\sin(h(x))] = \frac{d}{dx}[x]\text{.} \end{equation*}
The righthand side is simply $$1\text{,}$$ and by applying the chain rule applied to the left side,
\begin{equation*} \cos(h(x)) h'(x) = 1\text{.} \end{equation*}
Solving for $$h'(x)\text{,}$$ it follows that
\begin{equation*} h'(x) = \frac{1}{\cos(h(x))}\text{.} \end{equation*}
Finally, we recall that $$h(x) = \arcsin(x)\text{,}$$ so the denominator of $$h'(x)$$ is the function $$\cos(\arcsin(x))\text{,}$$ or in other words, “the cosine of the angle whose sine is $$x\text{.}$$” A bit of right triangle trigonometry allows us to simplify this expression considerably.
Let’s say that $$\theta = \arcsin(x)\text{,}$$ so that $$\theta$$ is the angle whose sine is $$x\text{.}$$ We can picture $$\theta$$ as an angle in a right triangle with hypotenuse $$1$$ and a vertical leg of length $$x\text{,}$$ as shown in Figure 2.6.5. The horizontal leg must be $$\sqrt{1-x^2}\text{,}$$ by the Pythagorean Theorem.
Now, because $$\theta = \arcsin(x)\text{,}$$ the expression for $$\cos(\arcsin(x))$$ is equivalent to $$\cos(\theta)\text{.}$$ From the figure, $$\cos(\arcsin(x)) = \cos(\theta) = \sqrt{1-x^2}\text{.}$$
Substituting this expression into our formula, $$h'(x) = \frac{1}{\cos(\arcsin(x))}\text{,}$$ we have now shown that
\begin{equation*} h'(x) = \frac{1}{\sqrt{1-x^2}}\text{.} \end{equation*}
#### Inverse sine.
For all real numbers $$x$$ such that $$-1 \lt x \lt 1\text{,}$$
\begin{equation*} \frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}\text{.} \end{equation*}
#### Activity2.6.3.
The following prompts in this activity will lead you to develop the derivative of the inverse tangent function.
1. Let $$r(x) = \arctan(x)\text{.}$$ Use the relationship between the arctangent and tangent functions to rewrite this equation using only the tangent function.
2. Differentiate both sides of the equation you found in (a). Solve the resulting equation for $$r'(x)\text{,}$$ writing $$r'(x)$$ as simply as possible in terms of a trigonometric function evaluated at $$r(x)\text{.}$$
3. Recall that $$r(x) = \arctan(x)\text{.}$$ Update your expression for $$r'(x)$$ so that it only involves trigonometric functions and the independent variable $$x\text{.}$$
4. Introduce a right triangle with angle $$\theta$$ so that $$\theta = \arctan(x)\text{.}$$ What are the three sides of the triangle?
5. In terms of only $$x$$ and $$1\text{,}$$ what is the value of $$\cos(\arctan(x))\text{?}$$
6. Use the results of your work above to find an expression involving only $$1$$ and $$x$$ for $$r'(x)\text{.}$$
While derivatives for other inverse trigonometric functions can be established similarly, for now we limit ourselves to the arcsine and arctangent functions.
#### Activity2.6.4.
Determine the derivative of each of the following functions.
1. $$\displaystyle \displaystyle f(x) = x^3 \arctan(x) + e^x \ln(x)$$
2. $$\displaystyle \displaystyle p(t) = 2^{t\arcsin(t)}$$
3. $$\displaystyle \displaystyle h(z) = (\arcsin(5z) + \arctan(4-z))^{27}$$
4. $$\displaystyle \displaystyle s(y) = \cot(\arctan(y))$$
5. $$\displaystyle \displaystyle m(v) = \ln(\sin^2(v)+1)$$
6. $$\displaystyle \displaystyle g(w) = \arctan\left( \frac{\ln(w)}{1+w^2} \right)$$
### Subsection2.6.4The link between the derivative of a function and the derivative of its inverse
In Figure 2.6.3, we saw an interesting relationship between the slopes of tangent lines to the natural exponential and natural logarithm functions at points reflected across the line $$y = x\text{.}$$ In particular, we observed that at the point $$(\ln(2), 2)$$ on the graph of $$f(x) = e^x\text{,}$$ the slope of the tangent line is $$f'(\ln(2)) = 2\text{,}$$ while at the corresponding point $$(2, \ln(2))$$ on the graph of $$f^{-1}(x) = \ln(x)\text{,}$$ the slope of the tangent line is $$(f^{-1})'(2) = \frac{1}{2}\text{,}$$ which is the reciprocal of $$f'(\ln(2))\text{.}$$
That the two corresponding tangent lines have reciprocal slopes is not a coincidence. If $$f$$ and $$g$$ are differentiable inverse functions, then $$y = f(x)$$ if and only if $$x = g(y)\text{,}$$ then$$f(g(x)) = x$$ for every $$x$$ in the domain of $$f^{-1}\text{.}$$ Differentiating both sides of this equation, we have
\begin{equation*} \frac{d}{dx} [f(g(x))] = \frac{d}{dx} [x]\text{,} \end{equation*}
and by the chain rule,
\begin{equation*} f'(g(x)) g'(x) = 1\text{.} \end{equation*}
Solving for $$g'(x)\text{,}$$ we have $$g'(x) = \frac{1}{f'(g(x))}\text{.}$$ Here we see that the slope of the tangent line to the inverse function $$g$$ at the point $$(x,g(x))$$ is precisely the reciprocal of the slope of the tangent line to the original function $$f$$ at the point $$(g(x),f(g(x))) = (g(x),x)\text{.}$$
To see this more clearly, consider the graph of the function $$y = f(x)$$ shown in Figure 2.6.6, along with its inverse $$y = g(x)\text{.}$$ Given a point $$(a,b)$$ that lies on the graph of $$f\text{,}$$ we know that $$(b,a)$$ lies on the graph of $$g\text{;}$$ because $$f(a) = b$$ and $$g(b) = a\text{.}$$ Now, applying the rule that $$g'(x) = 1/f'(g(x))$$ to the value $$x = b\text{,}$$ we have
\begin{equation*} g'(b) = \frac{1}{f'(g(b))} = \frac{1}{f'(a)}\text{,} \end{equation*}
which is precisely what we see in the figure: the slope of the tangent line to $$g$$ at $$(b,a)$$ is the reciprocal of the slope of the tangent line to $$f$$ at $$(a,b)\text{,}$$ since these two lines are reflections of one another across the line $$y = x\text{.}$$
#### Derivative of an inverse function.
Suppose that $$f$$ is a differentiable function with inverse $$g$$ and that $$(a,b)$$ is a point that lies on the graph of $$f$$ at which $$f'(a) \ne 0\text{.}$$ Then
\begin{equation*} g'(b) = \frac{1}{f'(a)}\text{.} \end{equation*}
More generally, for any $$x$$ in the domain of $$g'\text{,}$$ we have $$g'(x) = 1/f'(g(x))\text{.}$$
The rules we derived for $$\ln(x)\text{,}$$ $$\arcsin(x)\text{,}$$ and $$\arctan(x)$$ are all just specific examples of this general property of the derivative of an inverse function. For example, with $$g(x) = \ln(x)$$ and $$f(x) = e^x\text{,}$$ it follows that
\begin{equation*} g'(x) = \frac{1}{f'(g(x))} = \frac{1}{e^{\ln(x)}} = \frac{1}{x}\text{.} \end{equation*}
### Subsection2.6.5Summary
• For all positive real numbers $$x\text{,}$$ $$\frac{d}{dx}[\ln(x)] = \frac{1}{x}\text{.}$$
• For all real numbers $$x$$ such that $$-1 \lt x \lt 1\text{,}$$ $$\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}\text{.}$$ In addition, for all real numbers $$x\text{,}$$ $$\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}\text{.}$$
• If $$g$$ is the inverse of a differentiable function $$f\text{,}$$ then for any point $$x$$ in the domain of $$g'\text{,}$$ $$g'(x) = \frac{1}{f'(g(x))}\text{.}$$
### Exercises2.6.6Exercises
#### 1.
Let $$(x_0, y_0) = (1, 5)$$ and $$(x_1, y_1) = (1.1, 5.4)\text{.}$$ Use the following graph of the function $$f$$ to find the indicated derivatives.
If $$h(x)=(f(x))^{2}\text{,}$$ then
$$h'(1) =$$
If $$g(x)=f^{-1}(x)\text{,}$$ then
$$g'(5) =$$
#### 2.
Find the derivative of the function $$f(t)\text{,}$$ below.
$$f(t)=\ln(t^{4}+3)$$
$$f'(t) =$$
#### 3.
Let
\begin{equation*} f(x) = 4\sin^{-1}\mathopen{}\left(x^{4}\right) \end{equation*}
$$f'( x ) =$$
NOTE: The webwork system will accept $$\arcsin(x)$$ or $$\sin^{-1}(x)$$ as the inverse of $$\sin (x)\text{.}$$
#### 4.
If $$f(x) = 6 x^{3}\arctan(6 x^{4})\text{,}$$ find $$f' ( x ).$$
$$f' (x)$$ =
#### 5.
For each of the given functions $$f(x)\text{,}$$ find the derivative $$\left(f^{-1}\right)'(c)$$ at the given point $$c\text{,}$$ first finding $$a=f^{-1}(c)\text{.}$$
a) $$f(x)= 5 x + 7 x^{21}\text{;}$$ $$c = -12$$
$$a$$ =
$$\left(f^{-1}\right)'(c)$$ =
b) $$f(x)= x^2 - 12 x + 51$$ on the interval $$[6,\infty)\text{;}$$ $$c = 16$$
$$a$$ =
$$\left(f^{-1}\right)'(c)$$ =
#### 6.
Given that $$f(x)=2x+\cos(x)$$ is one-to-one, use the formula
\begin{equation*} \displaystyle (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} \end{equation*}
to find $$(f^{-1})'(1)\text{.}$$
$$(f^{-1})'(1) =$$
#### 7.
Determine the derivative of each of the following functions. Use proper notation and clearly identify the derivative rules you use.
1. $$\displaystyle f(x) = \ln(2\arctan(x) + 3\arcsin(x) + 5)$$
2. $$\displaystyle r(z) = \arctan(\ln(\arcsin(z)))$$
3. $$\displaystyle q(t) = \arctan^2(3t) \arcsin^4(7t)$$
4. $$\displaystyle g(v) = \ln\left( \frac{\arctan(v)}{\arcsin(v) + v^2} \right)$$
#### 8.
Consider the graph of $$y = f(x)$$ provided in Figure 2.6.7 and use it to answer the following questions.
1. Use the provided graph to estimate the value of $$f'(1)\text{.}$$
2. Sketch an approximate graph of $$y = f^{-1}(x)\text{.}$$ Label at least three distinct points on the graph that correspond to three points on the graph of $$f\text{.}$$
3. Based on your work in (a), what is the value of $$(f^{-1})'(-1)\text{?}$$ Why?
#### 9.
Let $$f(x) = \frac{1}{4}x^3 + 4\text{.}$$
1. Sketch a graph of $$y = f(x)$$ and explain why $$f$$ is an invertible function.
2. Let $$g$$ be the inverse of $$f$$ and determine a formula for $$g\text{.}$$
3. Compute $$f'(x)\text{,}$$ $$g'(x)\text{,}$$ $$f'(2)\text{,}$$ and $$g'(6)\text{.}$$ What is the special relationship between $$f'(2)$$ and $$g'(6)\text{?}$$ Why?
#### 10.
Let $$h(x) = x + \sin(x)\text{.}$$
1. Sketch a graph of $$y = h(x)$$ and explain why $$h$$ must be invertible.
2. Explain why it does not appear to be algebraically possible to determine a formula for $$h^{-1}\text{.}$$
3. Observe that the point $$(\frac{\pi}{2}, \frac{\pi}{2} + 1)$$ lies on the graph of $$y = h(x)\text{.}$$ Determine the value of $$(h^{-1})'(\frac{\pi}{2} + 1)\text{.}$$ | 6,932 | 20,477 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2024-33 | latest | en | 0.825868 |
https://www.molympiad.net/2017/08/william-lowell-putnam-mathematical-competition-2006-solutions.html | 1,695,419,420,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506423.70/warc/CC-MAIN-20230922202444-20230922232444-00762.warc.gz | 997,164,202 | 160,860 | ## $hide=mobile$type=ticker$c=12$cols=3$l=0$sr=random$b=0 1. Find the volume of the region of points$(x,y,z)$such that $\left(x^{2}+y^{2}+z^{2}+8\right)^{2}\le 36\left(x^{2}+y^{2}\right).$ 2. Alice and Bob play a game in which they take turns removing stones from a heap that initially has$n$stones. The number of stones removed at each turn must be one less than a prime number. The winner is the player who takes the last stone. Alice plays first. Prove that there are infinitely many such$n$such that Bob has a winning strategy. (For example, if$n=17,$then Alice might take$6$leaving$11;$then Bob might take$1$leaving$10;$then Alice can take the remaining stones to win.) 3. Let$1,2,3,\dots,2005,2006,2007,2009,2012,2016,\dots$be a sequence defined by$x_{k}=k$for$k=1,2\dots,2006$and$x_{k+1}=x_{k}+x_{k-2005}$for$k\ge 2006.$Show that the sequence has 2005 consecutive terms each divisible by 2006. 4. Let$S=\{1,2\dots,n\}$for some integer$n>1.$Say a permutation$\pi$of$S$has a local maximum at$k\in S$if $\begin{array}{ccc}\text{(i)}&\pi(k)>\pi(k+1)&\text{for }k=1\\ \text{(ii)}&\pi(k-1)<\pi(k)\text{ and }\pi(k)>\pi(k+1)&\text{for }1<k<n\\ \text{(iii)}&\pi(k-1)M\pi(k)&\text{for }k=n\end{array}$ (For example, if$n=5$and$\pi$takes values at$1,2,3,4,5$of$2,1,4,5,3,$then$\pi$has a local maximum of$2$as$k=1,$and a local maximum at$k-4.$) What is the average number of local maxima of a permutation of$S,$averaging over all permuatations of$S?$5. Let$n$be a positive odd integer and let$\theta$be a real number such that$\theta/\pi$is irrational. Set$a_{k}=\tan(\theta+k\pi/n),\ k=1,2\dots,n.$Prove that $\frac{a_{1}+a_{2}+\cdots+a_{n}}{a_{1}a_{2}\cdots a_{n}}$ is an integer, and determine its value. 6. Four points are chosen uniformly and independently at random in the interior of a given circle. Find the probability that they are the vertices of a convex quadrilateral. 7. Show that the curve$x^{3}+3xy+y^{3}=1$contains only one set of three distinct points,$A,B,$and$C,$which are the vertices of an equilateral triangle. 8. Prove that, for every set$X=\{x_{1},x_{2},\dots,x_{n}\}$of$n$real numbers, there exists a non-empty subset$S$of$X$and an integer$m$such that $\left|m+\sum_{s\in S}s\right|\le\frac1{n+1}$ 9. Let$S$be a finite set of points in the plane. A linear partition of$S$is an unordered pair$\{A,B\}$of subsets of$S$such that$A\cup B=S,\ A\cap B=\emptyset,$and$A$and$B$lie on opposite sides of some straight line disjoint from$S$($A$or$B$may be empty). Let$L_{S}$be the number of linear partitions of$S.$For each positive integer$n,$find the maximum of$L_{S}$over all sets$S$of$n$points. 10. Let$Z$denote the set of points in$\mathbb{R}^{n}$whose coordinates are$0$or$1.$(Thus$Z$has$2^{n}$elements, which are the vertices of a unit hypercube in$\mathbb{R}^{n}$.) Given a vector subspace$V$of$\mathbb{R}^{n},$let$Z(V)$denote the number of members of$Z$that lie in$V.$Let$k$be given,$0\le k\le n.$Find the maximum, over all vector subspaces$V\subseteq\mathbb{R}^{n}$of dimension$k,$of the number of points in$V\cap Z.$11. For each continuous function$f: [0,1]\to\mathbb{R},$let$I(f)=\int_{0}^{1}x^{2}f(x)\,dx$and$J(f)=\int_{0}^{1}x\left(f(x)\right)^{2}\,dx.$Find the maximum value of$I(f)-J(f)$over all such functions$f.$12. Let$k$be an integer greater than$1.$Suppose$a_{0}>0$and define $a_{n+1}=a_{n}+\frac1{\sqrt[k]{a_{n}}}$ for$n\ge 0.$Evaluate $\lim_{n\to\infty}\frac{a_{n}^{k+1}}{n^{k}}.$ ##$hide=mobile$type=ticker$c=36$cols=2$l=0$sr=random$b=0
Name
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MOlympiad.NET: [Solutions] William Lowell Putnam Mathematical Competition 2006
[Solutions] William Lowell Putnam Mathematical Competition 2006 | 7,806 | 17,343 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2023-40 | latest | en | 0.758401 |
https://wikimili.com/en/Covariance_and_contravariance_of_vectors | 1,632,466,891,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057504.60/warc/CC-MAIN-20210924050055-20210924080055-00670.warc.gz | 640,717,740 | 41,944 | # Covariance and contravariance of vectors
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In multilinear algebra and tensor analysis, covariance and contravariance describe how the quantitative description of certain geometric or physical entities changes with a change of basis.
## Contents
In physics, a basis is sometimes thought of as a set of reference axes. A change of scale on the reference axes corresponds to a change of units in the problem. For instance, by changing scale from meters to centimeters (that is, dividing the scale of the reference axes by 100), the components of a measured velocity vector are multiplied by 100. Vectors exhibit this behavior of changing scale inversely to changes in scale to the reference axes and consequently are called contravariant. As a result, vectors often have units of distance or distance with other units (as, for example, velocity has units of distance divided by time).
In contrast, covectors (also called dual vectors) typically have units of the inverse of distance or the inverse of distance with other units. An example of a covector is the gradient, which has units of a spatial derivative, or distance−1. The components of covectors change in the same way as changes to scale of the reference axes and consequently are called covariant.
A third concept related to covariance and contravariance is invariance. An example of a physical observable that does not change with a change of scale on the reference axes is the mass of a particle, which has units of mass (that is, no units of distance). The single, scalar value of mass is independent of changes to the scale of the reference axes and consequently is called invariant.
Under more general changes in basis:
• A contravariant vector or tangent vector (often abbreviated simply as vector, such as a direction vector or velocity vector) has components that contra-vary with a change of basis to compensate. That is, the matrix that transforms the vector components must be the inverse of the matrix that transforms the basis vectors. The components of vectors (as opposed to those of covectors) are said to be contravariant. Examples of vectors with contravariant components include the position of an object relative to an observer, or any derivative of position with respect to time, including velocity, acceleration, and jerk. In Einstein notation, contravariant components are denoted with upper indices as in
${\displaystyle \mathbf {v} =v^{i}\mathbf {e} _{i}}$ (note: implicit summation over index "i")
• A covariant vector or cotangent vector (often abbreviated as covector) has components that co-vary with a change of basis. That is, the components must be transformed by the same matrix as the change of basis matrix. The components of covectors (as opposed to those of vectors) are said to be covariant. Examples of covariant vectors generally appear when taking a gradient of a function. In Einstein notation, covariant components are denoted with lower indices as in
${\displaystyle \mathbf {e} _{i}(\mathbf {v} )=v_{i}.}$
Curvilinear coordinate systems, such as cylindrical or spherical coordinates, are often used in physical and geometric problems. Associated with any coordinate system is a natural choice of coordinate basis for vectors based at each point of the space, and covariance and contravariance are particularly important for understanding how the coordinate description of a vector changes by passing from one coordinate system to another.
The terms covariant and contravariant were introduced by James Joseph Sylvester in 1851 [2] [3] in the context of associated algebraic forms theory. Tensors are objects in multilinear algebra that can have aspects of both covariance and contravariance.
In the lexicon of category theory, covariance and contravariance are properties of functors; unfortunately, it is the lower-index objects (covectors) that generically have pullbacks, which are contravariant, while the upper-index objects (vectors) instead have pushforwards, which are covariant. This terminological conflict may be avoided by calling contravariant functors "cofunctors"—in accord with the "covector" terminology, and continuing the tradition of treating vectors as the concept and covectors as the coconcept.
## Introduction
In physics, a vector typically arises as the outcome of a measurement or series of measurements, and is represented as a list (or tuple) of numbers such as
${\displaystyle (v_{1},v_{2},v_{3}).}$
The numbers in the list depend on the choice of coordinate system. For instance, if the vector represents position with respect to an observer (position vector), then the coordinate system may be obtained from a system of rigid rods, or reference axes, along which the components v1, v2, and v3 are measured. For a vector to represent a geometric object, it must be possible to describe how it looks in any other coordinate system. That is to say, the components of the vectors will transform in a certain way in passing from one coordinate system to another.
A contravariant vector has components that "transform as the coordinates do" under changes of coordinates (and so inversely to the transformation of the reference axes), including rotation and dilation. The vector itself does not change under these operations; instead, the components of the vector change in a way that cancels the change in the spatial axes, in the same way that coordinates change. In other words, if the reference axes were rotated in one direction, the component representation of the vector would rotate in exactly the opposite way. Similarly, if the reference axes were stretched in one direction, the components of the vector, like the coordinates, would reduce in an exactly compensating way. Mathematically, if the coordinate system undergoes a transformation described by an invertible matrix M, so that a coordinate vector x is transformed to ${\displaystyle \mathbf {x} '=M\mathbf {x} }$, then a contravariant vector v must be similarly transformed via ${\displaystyle \mathbf {v} '=M\mathbf {v} }$. This important requirement is what distinguishes a contravariant vector from any other triple of physically meaningful quantities. For example, if v consists of the x-, y-, and z-components of velocity, then v is a contravariant vector: if the coordinates of space are stretched, rotated, or twisted, then the components of the velocity transform in the same way. Examples of contravariant vectors include position, displacement, velocity, acceleration, momentum, and force.
By contrast, a covariant vector has components that change oppositely to the coordinates or, equivalently, transform like the reference axes. For instance, the components of the gradient vector of a function
${\displaystyle \nabla f={\frac {\partial f}{\partial x^{1}}}{\widehat {x}}^{1}+{\frac {\partial f}{\partial x^{2}}}{\widehat {x}}^{2}+{\frac {\partial f}{\partial x^{3}}}{\widehat {x}}^{3}}$
transform like the reference axes themselves.
## Definition
The general formulation of covariance and contravariance refer to how the components of a coordinate vector transform under a change of basis (passive transformation). Thus let V be a vector space of dimension n over the field of scalars S, and let each of f = (X1, ..., Xn) and f′ = (Y1, ..., Yn) be a basis of V. [note 1] Also, let the change of basis from f to f′ be given by
${\displaystyle \mathbf {f} \mapsto \mathbf {f} '=\left(\sum _{i}a_{1}^{i}X_{i},\dots ,\sum _{i}a_{n}^{i}X_{i}\right)=\mathbf {f} A}$
(1)
for some invertible n×n matrix A with entries ${\displaystyle a_{j}^{i}}$. Here, each vector Yj of the f′ basis is a linear combination of the vectors Xi of the f basis, so that
${\displaystyle Y_{j}=\sum _{i}a_{j}^{i}X_{i}.}$
### Contravariant transformation
A vector ${\displaystyle v}$ in V is expressed uniquely as a linear combination of the elements ${\displaystyle X_{i}}$ of the f basis as
${\displaystyle v=\sum _{i}v^{i}[\mathbf {f} ]X_{i},}$
(2)
where vi[f] are elements in an (algebraic) field S known as the components of v in the f basis. Denote the column vector of components of v by v[f]:
${\displaystyle \mathbf {v} [\mathbf {f} ]={\begin{bmatrix}v^{1}[\mathbf {f} ]\\v^{2}[\mathbf {f} ]\\\vdots \\v^{n}[\mathbf {f} ]\end{bmatrix}}}$
so that ( 2 ) can be rewritten as a matrix product
${\displaystyle v=\mathbf {f} \,\mathbf {v} [\mathbf {f} ].}$
The vector v may also be expressed in terms of the f′ basis, so that
${\displaystyle v=\mathbf {f'} \,\mathbf {v} [\mathbf {f'} ].}$
However, since the vector v itself is invariant under the choice of basis,
${\displaystyle \mathbf {f} \,\mathbf {v} [\mathbf {f} ]=v=\mathbf {f'} \,\mathbf {v} [\mathbf {f'} ].}$
The invariance of v combined with the relationship ( 1 ) between f and f′ implies that
${\displaystyle \mathbf {f} \,\mathbf {v} [\mathbf {f} ]=\mathbf {f} A\,\mathbf {v} [\mathbf {f} A],}$
giving the transformation rule
${\displaystyle \mathbf {v} [\mathbf {f'} ]=\mathbf {v} [\mathbf {f} A]=A^{-1}\mathbf {v} [\mathbf {f} ].}$
In terms of components,
${\displaystyle v^{i}[\mathbf {f} A]=\sum _{j}{\tilde {a}}_{j}^{i}v^{j}[\mathbf {f} ]}$
where the coefficients ${\displaystyle {\tilde {a}}_{j}^{i}}$ are the entries of the inverse matrix of A.
Because the components of the vector v transform with the inverse of the matrix A, these components are said to transform contravariantly under a change of basis.
The way A relates the two pairs is depicted in the following informal diagram using an arrow. The reversal of the arrow indicates a contravariant change:
{\displaystyle {\begin{aligned}\mathbf {f} &\longrightarrow \mathbf {f'} \\v[\mathbf {f} ]&\longleftarrow v[\mathbf {f'} ]\end{aligned}}}
### Covariant transformation
A linear functional α on V is expressed uniquely in terms of its components (elements in S) in the f basis as
${\displaystyle \alpha (X_{i})=\alpha _{i}[\mathbf {f} ],\quad i=1,2,\dots ,n.}$
These components are the action of α on the basis vectors Xi of the f basis.
Under the change of basis from f to f′ ( 1 ), the components transform so that
{\displaystyle {\begin{aligned}\alpha _{i}[\mathbf {f} A]&=\alpha (Y_{i})\\&=\alpha \left(\sum _{j}a_{i}^{j}X_{j}\right)\\&=\sum _{j}a_{i}^{j}\alpha (X_{j})\\&=\sum _{j}a_{i}^{j}\alpha _{j}[\mathbf {f} ].\end{aligned}}}
(3)
Denote the row vector of components of α by α[f]:
${\displaystyle \mathbf {\alpha } [\mathbf {f} ]={\begin{bmatrix}\alpha _{1}[\mathbf {f} ],\alpha _{2}[\mathbf {f} ],\dots ,\alpha _{n}[\mathbf {f} ]\end{bmatrix}}}$
so that ( 3 ) can be rewritten as the matrix product
${\displaystyle \alpha [\mathbf {f} A]=\alpha [\mathbf {f} ]A.}$
Because the components of the linear functional α transform with the matrix A, these components are said to transform covariantly under a change of basis.
The way A relates the two pairs is depicted in the following informal diagram using an arrow. A covariant relationship is indicated since the arrows travel in the same direction:
{\displaystyle {\begin{aligned}\mathbf {f} &\longrightarrow \mathbf {f'} \\\alpha [\mathbf {f} ]&\longrightarrow \alpha [\mathbf {f'} ]\end{aligned}}}
Had a column vector representation been used instead, the transformation law would be the transpose
${\displaystyle \alpha ^{\mathrm {T} }[\mathbf {f} A]=A^{\mathrm {T} }\alpha ^{\mathrm {T} }[\mathbf {f} ].}$
## Coordinates
The choice of basis f on the vector space V defines uniquely a set of coordinate functions on V, by means of
${\displaystyle x^{i}[\mathbf {f} ](v)=v^{i}[\mathbf {f} ].}$
The coordinates on V are therefore contravariant in the sense that
${\displaystyle x^{i}[\mathbf {f} A]=\sum _{k=1}^{n}{\tilde {a}}_{k}^{i}x^{k}[\mathbf {f} ].}$
Conversely, a system of n quantities vi that transform like the coordinates xi on V defines a contravariant vector. A system of n quantities that transform oppositely to the coordinates is then a covariant vector.
This formulation of contravariance and covariance is often more natural in applications in which there is a coordinate space (a manifold) on which vectors live as tangent vectors or cotangent vectors. Given a local coordinate system xi on the manifold, the reference axes for the coordinate system are the vector fields
${\displaystyle X_{1}={\frac {\partial }{\partial x^{1}}},\dots ,X_{n}={\frac {\partial }{\partial x^{n}}}.}$
This gives rise to the frame f = (X1, ..., Xn) at every point of the coordinate patch.
If yi is a different coordinate system and
${\displaystyle Y_{1}={\frac {\partial }{\partial y^{1}}},\dots ,Y_{n}={\frac {\partial }{\partial y^{n}}},}$
then the frame f' is related to the frame f by the inverse of the Jacobian matrix of the coordinate transition:
${\displaystyle \mathbf {f} '=\mathbf {f} J^{-1},\quad J=\left({\frac {\partial y^{i}}{\partial x^{j}}}\right)_{i,j=1}^{n}.}$
Or, in indices,
${\displaystyle {\frac {\partial }{\partial y^{i}}}=\sum _{j=1}^{n}{\frac {\partial x^{j}}{\partial y^{i}}}{\frac {\partial }{\partial x^{j}}}.}$
A tangent vector is by definition a vector that is a linear combination of the coordinate partials ${\displaystyle \partial /\partial x^{i}}$. Thus a tangent vector is defined by
${\displaystyle v=\sum _{i=1}^{n}v^{i}[\mathbf {f} ]X_{i}=\mathbf {f} \ \mathbf {v} [\mathbf {f} ].}$
Such a vector is contravariant with respect to change of frame. Under changes in the coordinate system, one has
${\displaystyle \mathbf {v} \left[\mathbf {f} '\right]=\mathbf {v} \left[\mathbf {f} J^{-1}\right]=J\,\mathbf {v} [\mathbf {f} ].}$
Therefore, the components of a tangent vector transform via
${\displaystyle v^{i}\left[\mathbf {f} '\right]=\sum _{j=1}^{n}{\frac {\partial y^{i}}{\partial x^{j}}}v^{j}[\mathbf {f} ].}$
Accordingly, a system of n quantities vi depending on the coordinates that transform in this way on passing from one coordinate system to another is called a contravariant vector.
## Covariant and contravariant components of a vector with a metric
In a finite-dimensional vector space V over a field K with a symmetric bilinear form g : V × VK (which may be referred to as the metric tensor), there is little distinction between covariant and contravariant vectors, because the bilinear form allows covectors to be identified with vectors. That is, a vector v uniquely determines a covector α via
${\displaystyle \alpha (w)=g(v,w)}$
for all vectors w. Conversely, each covector α determines a unique vector v by this equation. Because of this identification of vectors with covectors, one may speak of the covariant components or contravariant components of a vector, that is, they are just representations of the same vector using the reciprocal basis.
Given a basis f = (X1, ..., Xn) of V, there is a unique reciprocal basis f# = (Y1, ..., Yn) of V determined by requiring that
${\displaystyle Y^{i}(X_{j})=\delta _{j}^{i},}$
the Kronecker delta. In terms of these bases, any vector v can be written in two ways:
{\displaystyle {\begin{aligned}v&=\sum _{i}v^{i}[\mathbf {f} ]X_{i}=\mathbf {f} \,\mathbf {v} [\mathbf {f} ]\\&=\sum _{i}v_{i}[\mathbf {f} ]Y^{i}=\mathbf {f} ^{\sharp }\mathbf {v} ^{\sharp }[\mathbf {f} ].\end{aligned}}}
The components vi[f] are the contravariant components of the vector v in the basis f, and the components vi[f] are the covariant components of v in the basis f. The terminology is justified because under a change of basis,
${\displaystyle \mathbf {v} [\mathbf {f} A]=A^{-1}\mathbf {v} [\mathbf {f} ],\quad \mathbf {v} ^{\sharp }[\mathbf {f} A]=A^{T}\mathbf {v} ^{\sharp }[\mathbf {f} ].}$
### Euclidean plane
In the Euclidean plane, the dot product allows for vectors to be identified with covectors. If ${\displaystyle \mathbf {e} _{1},\mathbf {e} _{2}}$ is a basis, then the dual basis ${\displaystyle \mathbf {e} ^{1},\mathbf {e} ^{2}}$ satisfies
{\displaystyle {\begin{aligned}\mathbf {e} ^{1}\cdot \mathbf {e} _{1}=1,&\quad \mathbf {e} ^{1}\cdot \mathbf {e} _{2}=0\\\mathbf {e} ^{2}\cdot \mathbf {e} _{1}=0,&\quad \mathbf {e} ^{2}\cdot \mathbf {e} _{2}=1.\end{aligned}}}
Thus, e1 and e2 are perpendicular to each other, as are e2 and e1, and the lengths of e1 and e2 normalized against e1 and e2, respectively.
#### Example
For example, [4] suppose that we are given a basis e1, e2 consisting of a pair of vectors making a 45° angle with one another, such that e1 has length 2 and e2 has length 1. Then the dual basis vectors are given as follows:
• e2 is the result of rotating e1 through an angle of 90° (where the sense is measured by assuming the pair e1, e2 to be positively oriented), and then rescaling so that e2e2 = 1 holds.
• e1 is the result of rotating e2 through an angle of 90°, and then rescaling so that e1e1 = 1 holds.
Applying these rules, we find
${\displaystyle \mathbf {e} ^{1}={\frac {1}{2}}\mathbf {e} _{1}-{\frac {1}{\sqrt {2}}}\mathbf {e} _{2}}$
and
${\displaystyle \mathbf {e} ^{2}=-{\frac {1}{\sqrt {2}}}\mathbf {e} _{1}+2\mathbf {e} _{2}.}$
Thus the change of basis matrix in going from the original basis to the reciprocal basis is
${\displaystyle R={\begin{bmatrix}{\frac {1}{2}}&-{\frac {1}{\sqrt {2}}}\\-{\frac {1}{\sqrt {2}}}&2\end{bmatrix}},}$
since
${\displaystyle [\mathbf {e} ^{1}\ \mathbf {e} ^{2}]=[\mathbf {e} _{1}\ \mathbf {e} _{2}]{\begin{bmatrix}{\frac {1}{2}}&-{\frac {1}{\sqrt {2}}}\\-{\frac {1}{\sqrt {2}}}&2\end{bmatrix}}.}$
For instance, the vector
${\displaystyle v={\frac {3}{2}}\mathbf {e} _{1}+2\mathbf {e} _{2}}$
is a vector with contravariant components
${\displaystyle v^{1}={\frac {3}{2}},\quad v^{2}=2.}$
The covariant components are obtained by equating the two expressions for the vector v:
${\displaystyle v=v_{1}\mathbf {e} ^{1}+v_{2}\mathbf {e} ^{2}=v^{1}\mathbf {e} _{1}+v^{2}\mathbf {e} _{2}}$
so
{\displaystyle {\begin{aligned}{\begin{bmatrix}v_{1}\\v_{2}\end{bmatrix}}&=R^{-1}{\begin{bmatrix}v^{1}\\v^{2}\end{bmatrix}}\\&={\begin{bmatrix}4&{\sqrt {2}}\\{\sqrt {2}}&1\end{bmatrix}}{\begin{bmatrix}v^{1}\\v^{2}\end{bmatrix}}\\&={\begin{bmatrix}6+2{\sqrt {2}}\\2+{\frac {3}{\sqrt {2}}}\end{bmatrix}}\end{aligned}}.}
### Three-dimensional Euclidean space
In the three-dimensional Euclidean space, one can also determine explicitly the dual basis to a given set of basis vectors e1, e2, e3 of E3 that are not necessarily assumed to be orthogonal nor of unit norm. The dual basis vectors are:
${\displaystyle \mathbf {e} ^{1}={\frac {\mathbf {e} _{2}\times \mathbf {e} _{3}}{\mathbf {e} _{1}\cdot (\mathbf {e} _{2}\times \mathbf {e} _{3})}};\qquad \mathbf {e} ^{2}={\frac {\mathbf {e} _{3}\times \mathbf {e} _{1}}{\mathbf {e} _{2}\cdot (\mathbf {e} _{3}\times \mathbf {e} _{1})}};\qquad \mathbf {e} ^{3}={\frac {\mathbf {e} _{1}\times \mathbf {e} _{2}}{\mathbf {e} _{3}\cdot (\mathbf {e} _{1}\times \mathbf {e} _{2})}}.}$
Even when the ei and ei are not orthonormal, they are still mutually reciprocal:
${\displaystyle \mathbf {e} ^{i}\cdot \mathbf {e} _{j}=\delta _{j}^{i},}$
Then the contravariant components of any vector v can be obtained by the dot product of v with the dual basis vectors:
${\displaystyle q^{1}=\mathbf {v} \cdot \mathbf {e} ^{1};\qquad q^{2}=\mathbf {v} \cdot \mathbf {e} ^{2};\qquad q^{3}=\mathbf {v} \cdot \mathbf {e} ^{3}.}$
Likewise, the covariant components of v can be obtained from the dot product of v with basis vectors, viz.
${\displaystyle q_{1}=\mathbf {v} \cdot \mathbf {e} _{1};\qquad q_{2}=\mathbf {v} \cdot \mathbf {e} _{2};\qquad q_{3}=\mathbf {v} \cdot \mathbf {e} _{3}.}$
Then v can be expressed in two (reciprocal) ways, viz.
${\displaystyle \mathbf {v} =q^{i}\mathbf {e} _{i}=q^{1}\mathbf {e} _{1}+q^{2}\mathbf {e} _{2}+q^{3}\mathbf {e} _{3}.}$
or
${\displaystyle \mathbf {v} =q_{i}\mathbf {e} ^{i}=q_{1}\mathbf {e} ^{1}+q_{2}\mathbf {e} ^{2}+q_{3}\mathbf {e} ^{3}}$
Combining the above relations, we have
${\displaystyle \mathbf {v} =(\mathbf {v} \cdot \mathbf {e} ^{i})\mathbf {e} _{i}=(\mathbf {v} \cdot \mathbf {e} _{i})\mathbf {e} ^{i}}$
and we can convert between the basis and dual basis with
${\displaystyle q_{i}=\mathbf {v} \cdot \mathbf {e} _{i}=(q^{j}\mathbf {e} _{j})\cdot \mathbf {e} _{i}=(\mathbf {e} _{j}\cdot \mathbf {e} _{i})q^{j}}$
and
${\displaystyle q^{i}=\mathbf {v} \cdot \mathbf {e} ^{i}=(q_{j}\mathbf {e} ^{j})\cdot \mathbf {e} ^{i}=(\mathbf {e} ^{j}\cdot \mathbf {e} ^{i})q_{j}.}$
If the basis vectors are orthonormal, then they are the same as the dual basis vectors.
### General Euclidean spaces
More generally, in an n-dimensional Euclidean space V, if a basis is
${\displaystyle \mathbf {e} _{1},\dots ,\mathbf {e} _{n},}$
the reciprocal basis is given by (double indices are summed over),
${\displaystyle \mathbf {e} ^{i}=g^{ij}\mathbf {e} _{j}}$
where the coefficients gij are the entries of the inverse matrix of
${\displaystyle g_{ij}=\mathbf {e} _{i}\cdot \mathbf {e} _{j}.}$
Indeed, we then have
${\displaystyle \mathbf {e} ^{i}\cdot \mathbf {e} _{k}=g^{ij}\mathbf {e} _{j}\cdot \mathbf {e} _{k}=g^{ij}g_{jk}=\delta _{k}^{i}.}$
The covariant and contravariant components of any vector
${\displaystyle \mathbf {v} =q_{i}\mathbf {e} ^{i}=q^{i}\mathbf {e} _{i}\,}$
are related as above by
${\displaystyle q_{i}=\mathbf {v} \cdot \mathbf {e} _{i}=(q^{j}\mathbf {e} _{j})\cdot \mathbf {e} _{i}=q^{j}g_{ji}}$
and
${\displaystyle q^{i}=\mathbf {v} \cdot \mathbf {e} ^{i}=(q_{j}\mathbf {e} ^{j})\cdot \mathbf {e} ^{i}=q_{j}g^{ji}.}$
## Informal usage
In the field of physics, the adjective covariant is often used informally as a synonym for invariant. For example, the Schrödinger equation does not keep its written form under the coordinate transformations of special relativity. Thus, a physicist might say that the Schrödinger equation is not covariant. In contrast, the Klein–Gordon equation and the Dirac equation do keep their written form under these coordinate transformations. Thus, a physicist might say that these equations are covariant.
Despite this usage of "covariant", it is more accurate to say that the Klein–Gordon and Dirac equations are invariant, and that the Schrödinger equation is not invariant. Additionally, to remove ambiguity, the transformation by which the invariance is evaluated should be indicated.
Because the components of vectors are contravariant and those of covectors are covariant, the vectors themselves are often referred to as being contravariant and the covectors as covariant.
## Use in tensor analysis
The distinction between covariance and contravariance is particularly important for computations with tensors, which often have mixed variance. This means that they have both covariant and contravariant components, or both vector and covector components. The valence of a tensor is the number of variant and covariant terms, and in Einstein notation, covariant components have lower indices, while contravariant components have upper indices. The duality between covariance and contravariance intervenes whenever a vector or tensor quantity is represented by its components, although modern differential geometry uses more sophisticated index-free methods to represent tensors.
In tensor analysis, a covariant vector varies more or less reciprocally to a corresponding contravariant vector. Expressions for lengths, areas and volumes of objects in the vector space can then be given in terms of tensors with covariant and contravariant indices. Under simple expansions and contractions of the coordinates, the reciprocity is exact; under affine transformations the components of a vector intermingle on going between covariant and contravariant expression.
On a manifold, a tensor field will typically have multiple, upper and lower indices, where Einstein notation is widely used. When the manifold is equipped with a metric, covariant and contravariant indices become very closely related to one another. Contravariant indices can be turned into covariant indices by contracting with the metric tensor. The reverse is possible by contracting with the (matrix) inverse of the metric tensor. Note that in general, no such relation exists in spaces not endowed with a metric tensor. Furthermore, from a more abstract standpoint, a tensor is simply "there" and its components of either kind are only calculational artifacts whose values depend on the chosen coordinates.
The explanation in geometric terms is that a general tensor will have contravariant indices as well as covariant indices, because it has parts that live in the tangent bundle as well as the cotangent bundle.
A contravariant vector is one which transforms like ${\displaystyle {\frac {dx^{\mu }}{d\tau }}}$, where ${\displaystyle x^{\mu }\!}$ are the coordinates of a particle at its proper time ${\displaystyle \tau }$. A covariant vector is one which transforms like ${\displaystyle {\frac {\partial \varphi }{\partial x^{\mu }}}}$, where ${\displaystyle \varphi }$ is a scalar field.
## Algebra and geometry
In category theory, there are covariant functors and contravariant functors. The assignment of the dual space to a vector space is a standard example of a contravariant functor. Some constructions of multilinear algebra are of "mixed" variance, which prevents them from being functors.
In differential geometry, the components of a vector relative to a basis of the tangent bundle are covariant if they change with the same linear transformation as a change of basis. They are contravariant if they change by the inverse transformation. This is sometimes a source of confusion for two distinct but related reasons. The first is that vectors whose components are covariant (called covectors or 1-forms) actually pull back under smooth functions, meaning that the operation assigning the space of covectors to a smooth manifold is actually a contravariant functor. Likewise, vectors whose components are contravariant push forward under smooth mappings, so the operation assigning the space of (contravariant) vectors to a smooth manifold is a covariant functor. Secondly, in the classical approach to differential geometry, it is not bases of the tangent bundle that are the most primitive object, but rather changes in the coordinate system. Vectors with contravariant components transform in the same way as changes in the coordinates (because these actually change oppositely to the induced change of basis). Likewise, vectors with covariant components transform in the opposite way as changes in the coordinates.
## Notes
1. A basis f may here profitably be viewed as a linear isomorphism from Rn to V. Regarding f as a row vector whose entries are the elements of the basis, the associated linear isomorphism is then ${\displaystyle \mathbf {x} \mapsto \mathbf {f} \mathbf {x} .}$
## Citations
1. C. Misner; K.S. Thorne; J.A. Wheeler (1973). Gravitation. W.H. Freeman & Co. ISBN 0-7167-0344-0.
2. Sylvester, James Joseph. "On the general theory of associated algebraical forms." Cambridge and Dublin Math. Journal, VI (1851): 289-293.
3. 1814-1897., Sylvester, James Joseph (2012). The collected mathematical papers of James Joseph Sylvester. Volume 3, 1870-1883. Cambridge: Cambridge University Press. ISBN 978-1107661431. OCLC 758983870.CS1 maint: numeric names: authors list (link)
4. Bowen, Ray (2008). "Introduction to Vectors and Tensors" (PDF). Dover. pp. 78, 79, 81.
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In geometry, curvilinear coordinates are a coordinate system for Euclidean space in which the coordinate lines may be curved. These coordinates may be derived from a set of Cartesian coordinates by using a transformation that is locally invertible at each point. This means that one can convert a point given in a Cartesian coordinate system to its curvilinear coordinates and back. The name curvilinear coordinates, coined by the French mathematician Lamé, derives from the fact that the coordinate surfaces of the curvilinear systems are curved.
In mathematics and physics, the Christoffel symbols are an array of numbers describing a metric connection. The metric connection is a specialization of the affine connection to surfaces or other manifolds endowed with a metric, allowing distances to be measured on that surface. In differential geometry, an affine connection can be defined without reference to a metric, and many additional concepts follow: parallel transport, covariant derivatives, geodesics, etc. also do not require the concept of a metric. However, when a metric is available, these concepts can be directly tied to the "shape" of the manifold itself; that shape is determined by how the tangent space is attached to the cotangent space by the metric tensor. Abstractly, one would say that the manifold has an associated (orthonormal) frame bundle, with each "frame" being a possible choice of a coordinate frame. An invariant metric implies that the structure group of the frame bundle is the orthogonal group O(p, q). As a result, such a manifold is necessarily a (pseudo-)Riemannian manifold. The Christoffel symbols provide a concrete representation of the connection of (pseudo-)Riemannian geometry in terms of coordinates on the manifold. Additional concepts, such as parallel transport, geodesics, etc. can then be expressed in terms of Christoffel symbols.
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In electromagnetism, the electromagnetic tensor or electromagnetic field tensor is a mathematical object that describes the electromagnetic field in spacetime. The field tensor was first used after the four-dimensional tensor formulation of special relativity was introduced by Hermann Minkowski. The tensor allows related physical laws to be written very concisely.
In mathematics, orthogonal coordinates are defined as a set of d coordinates q = in which the coordinate surfaces all meet at right angles. A coordinate surface for a particular coordinate qk is the curve, surface, or hypersurface on which qk is a constant. For example, the three-dimensional Cartesian coordinates is an orthogonal coordinate system, since its coordinate surfaces x = constant, y = constant, and z = constant are planes that meet at right angles to one another, i.e., are perpendicular. Orthogonal coordinates are a special but extremely common case of curvilinear coordinates.
A system of skew coordinates is a curvilinear coordinate system where the coordinate surfaces are not orthogonal, in contrast to orthogonal coordinates.
In mathematics, Ricci calculus constitutes the rules of index notation and manipulation for tensors and tensor fields on a differentiable manifold, with or without a metric tensor or connection. It is also the modern name for what used to be called the absolute differential calculus, developed by Gregorio Ricci-Curbastro in 1887–1896, and subsequently popularized in a paper written with his pupil Tullio Levi-Civita in 1900. Jan Arnoldus Schouten developed the modern notation and formalism for this mathematical framework, and made contributions to the theory, during its applications to general relativity and differential geometry in the early twentieth century.
Curvilinear coordinates can be formulated in tensor calculus, with important applications in physics and engineering, particularly for describing transportation of physical quantities and deformation of matter in fluid mechanics and continuum mechanics. | 9,574 | 37,753 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 74, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-39 | latest | en | 0.938623 |
https://db0nus869y26v.cloudfront.net/en/If_and_only_if | 1,721,787,322,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00208.warc.gz | 167,068,202 | 18,039 | ↔⇔≡⟺
Logical symbols representing iff
In logic and related fields such as mathematics and philosophy, "if and only if" (often shortened as "iff") is paraphrased by the biconditional, a logical connective[1] between statements. The biconditional is true in two cases, where either both statements are true or both are false. The connective is biconditional (a statement of material equivalence),[2] and can be likened to the standard material conditional ("only if", equal to "if ... then") combined with its reverse ("if"); hence the name. The result is that the truth of either one of the connected statements requires the truth of the other (i.e. either both statements are true, or both are false), though it is controversial whether the connective thus defined is properly rendered by the English "if and only if"—with its pre-existing meaning. For example, P if and only if Q means that P is true whenever Q is true, and the only case in which P is true is if Q is also true, whereas in the case of P if Q, there could be other scenarios where P is true and Q is false.
In writing, phrases commonly used as alternatives to P "if and only if" Q include: Q is necessary and sufficient for P, for P it is necessary and sufficient that Q, P is equivalent (or materially equivalent) to Q (compare with material implication), P precisely if Q, P precisely (or exactly) when Q, P exactly in case Q, and P just in case Q.[3] Some authors regard "iff" as unsuitable in formal writing;[4] others consider it a "borderline case" and tolerate its use.[5] In logical formulae, logical symbols, such as ${\displaystyle \leftrightarrow }$ and ${\displaystyle \Leftrightarrow }$,[6] are used instead of these phrases; see § Notation below.
## Definition
${\displaystyle \neg P\land \neg Q}$
${\displaystyle P\land Q}$
${\displaystyle P\rightarrow Q}$
${\displaystyle P\leftarrow Q}$
${\displaystyle P\leftrightarrow Q}$
The truth table of P ${\displaystyle \leftrightarrow }$ Q is as follows:[7][8]
${\displaystyle P}$${\displaystyle Q}$${\displaystyle \neg P\land \neg Q}$${\displaystyle P\land Q}$${\displaystyle P\rightarrow Q}$${\displaystyle P\leftarrow Q}$${\displaystyle P\leftrightarrow Q}$
FFTFTTT
FTFFTFF
TFFFFTF
TTFTTTT
It is equivalent to that produced by the XNOR gate, and opposite to that produced by the XOR gate.[9]
## Usage
### Notation
The corresponding logical symbols are "${\displaystyle \leftrightarrow }$", "${\displaystyle \Leftrightarrow }$",[6] and ${\displaystyle \equiv }$,[10] and sometimes "iff". These are usually treated as equivalent. However, some texts of mathematical logic (particularly those on first-order logic, rather than propositional logic) make a distinction between these, in which the first, ↔, is used as a symbol in logic formulas, while ⇔ is used in reasoning about those logic formulas (e.g., in metalogic). In Łukasiewicz's Polish notation, it is the prefix symbol ${\displaystyle E}$.[11]
Another term for the logical connective, i.e., the symbol in logic formulas, is exclusive nor.
In TeX, "if and only if" is shown as a long double arrow: ${\displaystyle \iff }$ via command \iff or \Longleftrightarrow.[12]
### Proofs
In most logical systems, one proves a statement of the form "P iff Q" by proving either "if P, then Q" and "if Q, then P", or "if P, then Q" and "if not-P, then not-Q". Proving these pairs of statements sometimes leads to a more natural proof, since there are not obvious conditions in which one would infer a biconditional directly. An alternative is to prove the disjunction "(P and Q) or (not-P and not-Q)", which itself can be inferred directly from either of its disjuncts—that is, because "iff" is truth-functional, "P iff Q" follows if P and Q have been shown to be both true, or both false.
### Origin of iff and pronunciation
Usage of the abbreviation "iff" first appeared in print in John L. Kelley's 1955 book General Topology.[13] Its invention is often credited to Paul Halmos, who wrote "I invented 'iff,' for 'if and only if'—but I could never believe I was really its first inventor."[14]
It is somewhat unclear how "iff" was meant to be pronounced. In current practice, the single 'word' "iff" is almost always read as the four words "if and only if". However, in the preface of General Topology, Kelley suggests that it should be read differently: "In some cases where mathematical content requires 'if and only if' and euphony demands something less I use Halmos' 'iff'". The authors of one discrete mathematics textbook suggest:[15] "Should you need to pronounce iff, really hang on to the 'ff' so that people hear the difference from 'if'", implying that "iff" could be pronounced as [ɪfː].
### Usage in definitions
Conventionally, definitions are "if and only if" statements; some texts — such as Kelley's General Topology — follow this convention, and use "if and only if" or iff in definitions of new terms.[16] However, this usage of "if and only if" is relatively uncommon and overlooks the linguistic fact that the "if" of a definition is interpreted as meaning "if and only if". The majority of textbooks, research papers and articles (including English Wikipedia articles) follow the linguistic convention of interpreting "if" as "if and only if" whenever a mathematical definition is involved (as in "a topological space is compact if every open cover has a finite subcover").[17] Moreover, in the case of a recursive definition, the only if half of the definition is interpreted as a sentence in the metalanguage stating that the sentences in the definition of a predicate are the only sentences determining the extension of the predicate.
## In terms of Euler diagrams
Euler diagrams show logical relationships among events, properties, and so forth. "P only if Q", "if P then Q", and "P→Q" all mean that P is a subset, either proper or improper, of Q. "P if Q", "if Q then P", and Q→P all mean that Q is a proper or improper subset of P. "P if and only if Q" and "Q if and only if P" both mean that the sets P and Q are identical to each other.
## More general usage
Iff is used outside the field of logic as well. Wherever logic is applied, especially in mathematical discussions, it has the same meaning as above: it is an abbreviation for if and only if, indicating that one statement is both necessary and sufficient for the other. This is an example of mathematical jargon (although, as noted above, if is more often used than iff in statements of definition).
The elements of X are all and only the elements of Y means: "For any z in the domain of discourse, z is in X if and only if z is in Y."
## When "if" means "if and only if"
In their Artificial Intelligence: A Modern Approach, Russell and Norvig note (page 282),[18] in effect, that it is often more natural to express if and only if as if together with a "database (or logic programming) semantics". They give the example of the English sentence "Richard has two brothers, Geoffrey and John".
In a database or logic program, this could be represented simply by two sentences:
Brother(Richard, Geoffrey).
Brother(Richard, John).
The database semantics interprets the database (or program) as containing all and only the knowledge relevant for problem solving in a given domain. It interprets only if as expressing in the metalanguage that the sentences in the database represent the only knowledge that should be considered when drawing conclusions from the database.
In first-order logic (FOL) with the standard semantics, the same English sentence would need to be represented, using if and only if, with only if interpreted in the object language, in some such form as:
${\displaystyle \forall }$ X(Brother(Richard, X) iff X = Geoffrey or X = John).
Geoffrey ≠ John.
Compared with the standard semantics for FOL, the database semantics has a more efficient implementation. Instead of reasoning with sentences of the form:
conclusion iff conditions
it uses sentences of the form:
conclusion if conditions
to reason forwards from conditions to conclusions or backwards from conclusions to conditions.
The database semantics is analogous to the legal principle expressio unius est exclusio alterius (the express mention of one thing excludes all others). Moreover, it underpins the application of logic programming to the representation of legal texts and legal reasoning.[19]
## References
1. ^ "Logical Connectives". sites.millersville.edu. Retrieved 10 September 2023.
2. ^ Copi, I. M.; Cohen, C.; Flage, D. E. (2006). Essentials of Logic (Second ed.). Upper Saddle River, NJ: Pearson Education. p. 197. ISBN 978-0-13-238034-8.
3. ^ Weisstein, Eric W. "Iff." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Iff.html Archived 13 November 2018 at the Wayback Machine
4. ^ E.g. Daepp, Ulrich; Gorkin, Pamela (2011), Reading, Writing, and Proving: A Closer Look at Mathematics, Undergraduate Texts in Mathematics, Springer, p. 52, ISBN 9781441994790, While it can be a real time-saver, we don't recommend it in formal writing.
5. ^ Rothwell, Edward J.; Cloud, Michael J. (2014), Engineering Writing by Design: Creating Formal Documents of Lasting Value, CRC Press, p. 98, ISBN 9781482234312, It is common in mathematical writing
6. ^ a b Peil, Timothy. "Conditionals and Biconditionals". web.mnstate.edu. Archived from the original on 24 October 2020. Retrieved 4 September 2020.
7. ^ p <=> q Archived 18 October 2016 at the Wayback Machine. Wolfram|Alpha
8. ^ If and only if, UHM Department of Mathematics, archived from the original on 5 May 2000, retrieved 16 October 2016, Theorems which have the form "P if and only Q" are much prized in mathematics. They give what are called "necessary and sufficient" conditions, and give completely equivalent and hopefully interesting new ways to say exactly the same thing.
9. ^ "XOR/XNOR/Odd Parity/Even Parity Gate". www.cburch.com. Archived from the original on 7 April 2022. Retrieved 22 October 2019.
10. ^ Weisstein, Eric W. "Equivalent". mathworld.wolfram.com. Archived from the original on 3 October 2020. Retrieved 4 September 2020.
11. ^ "Jan Łukasiewicz > Łukasiewicz's Parenthesis-Free or Polish Notation (Stanford Encyclopedia of Philosophy)". plato.stanford.edu. Archived from the original on 9 August 2019. Retrieved 22 October 2019.
12. ^ "LaTeX:Symbol". Art of Problem Solving. Archived from the original on 22 October 2019. Retrieved 22 October 2019.
13. ^ General Topology, reissue ISBN 978-0-387-90125-1
14. ^ Nicholas J. Higham (1998). Handbook of writing for the mathematical sciences (2nd ed.). SIAM. p. 24. ISBN 978-0-89871-420-3.
15. ^ Maurer, Stephen B.; Ralston, Anthony (2005). Discrete Algorithmic Mathematics (3rd ed.). Boca Raton, Fla.: CRC Press. p. 60. ISBN 1568811667.
16. ^ For instance, from General Topology, p. 25: "A set is countable iff it is finite or countably infinite." [boldface in original]
17. ^ Krantz, Steven G. (1996), A Primer of Mathematical Writing, American Mathematical Society, p. 71, ISBN 978-0-8218-0635-7
18. ^ Russell, Stuart J.; Norvig, Peter (2020) [1995]. Artificial Intelligence: A Modern Approach (4 ed.). Prentice Hall. p. 1136. ISBN 978-0-13-461099-3. OCLC 359890490.
19. ^ Kowalski, R., Dávila, J., Sartor, G. and Calejo, M., 2023. Logical English for law and education. http://www.doc.ic.ac.uk/~rak/papers/Logical%20English%20for%20Law%20and%20Education%20.pdf In Prolog: The Next 50 Years (pp. 287-299). Cham: Springer Nature Switzerland. | 2,918 | 11,473 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 21, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-30 | latest | en | 0.902504 |
https://www.physicsforums.com/threads/conf-intervals-for-fitted-parameters-divide-by-sqrt-n.958999/ | 1,721,788,226,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00299.warc.gz | 802,614,214 | 19,563 | # Conf.intervals for fitted parameters: divide by sqrt(n)?
• I
• Jonas Hall
In summary, the conversation discusses the process of fitting a parametrized model to data and obtaining the optimized parameters and covariance matrix. It is noted that the squares of the diagonal elements of the covariance matrix are the standard errors of the optimized parameters, and that multiplying these errors by 1.96 gives a confidence interval of 95%. The question is raised whether one should also divide by √n, similar to creating a confidence interval for a mean, and it is explained that this factor does come into play and decreases with increasing n. The conversation also discusses the correlation between the standard errors of m and c, and how this limits the variability of the regression line. It is mentioned that most computer programs estimate the statistical distributions of the model parameters by
Jonas Hall
If you fit a parametrized model (i.e. y = a log(x + b) + c) to some data points the output is typically the optimized parameters (i.e. a, b, c) and the covariance matrix. The squares of the diagonal elements of this matrix are the standard errors of the optimized parameters. (i.e. sea, seb, sec). Now to get a confidence interval of 95% for a parameter you typically multiply this error with 1.96 (assuming a normal distribution)(i.e. a ± 1.96 sea). At least this is what I have found so far. But I wonder if this is the whole truth. Shouldn't you also divide by √n the way you do when you create a confidence interval for a mean? It just seems to me that the more data you have, the better the estimates of the parameters should become. Also, i find that if i don't divide by √n, then the values seem rather large, sometimes falling om the wrong side of 0.
Or... does the covariance matrix values grow smaller with increasing n and this is the reason that you don't divide by √n and the values are supposed to be quite large?
Grateful if someone could make this clear to me. I have never studied statistics "properly" but dabble in mathematical models and teach math in upper secondary school.
Jonas Hall said:
Shouldn't you also divide by √n the way you do when you create a confidence interval for a mean?
When you divide the standard deviation by the square root of n you obtain the standard error of the mean. You said that the values you have are already standard errors, so it wouldn’t make sense to divide them again.
FactChecker
The √n factor does come into it. For example, for the equation y = mx + c, if the error variance of the points is σ2, the variance of m is
σ2/n(<x2> - <x>2)
If your measured error variance is s2, the estimated variance of m is s2/n(<x2> - <x>2). This is the diagonal element of the covariance matrix. (I think you meant to say "The diagonal elements of this matrix are the squares of the standard errors of the optimized parameters. (i.e. sea, seb, sec).") As you suggest, the value decreases with increasing n.
If you did p separate experiments, with independent data sets, and determined a value of m for each, and calculated the mean value of m, the standard error of this mean value would be the standard error of m divided by √p.
Another point is that though the standard errors of m and c might be large, the values of m and c are usually strongly correlated, so you can't have any value of m in its confidence interval with any value of c in its confidence interval. This limits the variability of the regression line more than might at first appear.
Dale
Jonas Hall said:
teach math in upper secondary school
mjc123 said:
though the standard errors of m and c might be large, the values of m and c are usually strongly correlated, so you can't have any value of m in its confidence interval with any value of c in its confidence interval. This limits the variability of the regression line more than might at first appear.
e.g. for a straight line fit it's easy to check that the best line goes 'through the center of mass of the mesaured points' and can wiggle its slope due to the (hopefully) random errors. The abscissa error is a combination of this wiggling and shifting up and down a bit. The correlation disappears if the origin is at the 'center of mass'.
Jonas Hall said:
The squares of the diagonal elements of this matrix are the standard errors of the optimized parameters. (i.e. sea, seb, sec).
It would be interesting to know if that's actually true for an arbitrary nonlinear model. In fact, it would interesting to know how one can speak of statistical errors in the optimized parameters at all since fitting the model to the data gives one value of each parameter, not a sample of several values for each parameter.
I think most computer programs estimate the statistical distributions of the model parameters by using a linear approximation to the model and assuming the measured data values tell us the correct location about which to make the linear approximation. However a linear approximation to log(ax + b) + c will have a different coefficients than a linear approximation to sin(ax + b) + c. So how general is the claim that the squares of the diagonal elements of the covariance matrix are (good estimators of) the standard errors of the parameters?
Thank you all! You have convinced me that it is not appropriate to divide by sqrt(n). I find Stephen Tashis comment interesting. I also wonder if the statistical errors given by e.g. scipy etc compares to experimental values found when bootstrapping/jackknifeing (spelling?) your data. I guess I will have to run some experiments... I still find the standard errors quite large though, but I appreciate the comments on this by mjc123 and BvU. I envisage the following scenario: You take data and fit parameters according to your model. In reality though, you ar often only interested in one single parameter (such as b in y = a * b^x + c). So after you obtain your parameters (say a = 2, b = 3 and c = 4) you do a new fit according to the model y = 2 * b^x + 4. You will now presumably get b = 3 again but with a standard error that does not depend on any other parameters.
Would this work?
Yes it does. It depends on a being 2 and c being 4. It is not a function that depends on a and c as variables, but a value that is only true for particular values of a and c.
Jonas Hall said:
I also wonder if the statistical errors given by e.g. scipy etc compares to experimental values found when bootstrapping/jackknifeing (spelling?) your data.
The best experiments would be to compare all those methods to the correct answer.
Assume the correct model for the data has the form: ##Y = G(X,a,b,..)## where ##Y## and ##X## are random variables and ##a,b,...## are specific values of parameters A particular fitting algorithm produces estimates for ##a,b,...## that are functions of the sample data.
I'll denote this by:
##\hat{a} = f_1(X_s)##
##\hat{b} = f_2(X_s)##
...
##\hat{a},\hat{b},.. ## are random variables since they depend on the random values in a sample.
If we simulate a lot of samples ##X_s##, we get samples of ##\hat{a},\hat{b},...##. We can estimate the distributions of those random variables.
From that, we can estimate confidence intervals. However, we may have to do this by looking at the distribution of ##\hat{a}## in detail, not merely by looking at the parameters of that distribution. For example ##\hat{a}## may not be an unbiased estimator of ##a##. In that case, knowing the standard deviation of ##\hat{a}##, doesn't let us compute "confidence" by assuming ##a## is at the center of the interval. ( It's also possible that ##\hat{a}## may be unbiased estimator of ##a##, but not normally distributed.)
A limitation of such experiments is that the answer depends on particular choices of ##a,b,...## so the size of a confidence interval may vary with a big variation in the magnitudes of ##a,b,...##.
## 1. What does "Conf.intervals for fitted parameters: divide by sqrt(n)" mean?
This phrase refers to a statistical calculation used to estimate the range of values within which a fitted parameter is likely to fall. The calculation involves dividing the standard error by the square root of the sample size (n).
## 2. Why is it necessary to divide by sqrt(n) when calculating confidence intervals for fitted parameters?
Dividing by sqrt(n) helps to adjust for the variability in the sample size. As the sample size increases, the standard error decreases, so dividing by sqrt(n) ensures that the confidence interval reflects this decrease in variability.
## 3. How do you interpret the confidence interval for a fitted parameter that was divided by sqrt(n)?
A confidence interval for a fitted parameter that was divided by sqrt(n) can be interpreted as a range of values within which the true value of the parameter is likely to fall with a certain level of confidence. For example, a 95% confidence interval means that we can be 95% confident that the true value of the parameter falls within the interval.
## 4. Can you explain the relationship between the sample size and the width of the confidence interval?
The sample size and the width of the confidence interval have an inverse relationship. As the sample size increases, the width of the confidence interval decreases. This is because a larger sample size reduces the standard error, making the estimate of the parameter more precise.
## 5. Are there any assumptions or limitations when using the "divide by sqrt(n)" method to calculate confidence intervals for fitted parameters?
Yes, there are a few assumptions and limitations when using this method. It assumes that the data is normally distributed and that the sample is representative of the population. It also assumes that the data is independent and that there are no influential outliers. Additionally, this method may not be appropriate for small sample sizes or when the underlying distribution is not known.
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3K | 2,384 | 10,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-30 | latest | en | 0.905059 |
https://m.wikihow.com/Find-the-Height-of-a-Triangle | 1,585,601,561,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00519.warc.gz | 595,370,959 | 58,400 | # How to Find the Height of a Triangle
Author Info
Updated: September 6, 2019
To calculate the area of a triangle you need to know its height. To find the height follow these instructions. You must at least have a base to find the height.
### Method 1 of 3: Using Base and Area to Find Height
1. 1
Recall the formula for the area of a triangle. The formula for the area of a triangle is
A=1/2bh.
[1]
• A = Area of the triangle
• b = Length of the base of the triangle
• h = Height of the base of the triangle
2. 2
Look at your triangle and determine which variables you know. You already know the area, so assign that value to A. You should also know the value of one side length; assign that value to "'b'".
Any side of a triangle can be the base,
regardless of how the triangle is drawn. To visualize this, just imagine rotating the triangle until the known side length is at the bottom.
Example
If you know that the area of a triangle is 20, and one side is 4, then:
A = 20 and b = 4.
3. 3
Plug your values into the equation A=1/2bh and do the math. First multiply the base (b) by 1/2, then divide the area (A) by the product. The resulting value will be the height of your triangle!
Example
20 = 1/2(4)h Plug the numbers into the equation.
20 = 2h Multiply 4 by 1/2.
10 = h Divide by 2 to find the value for height.
### Method 2 of 3: Finding an Equilateral Triangle's Height
1. 1
Recall the properties of an equilateral triangle. An equilateral triangle has three equal sides, and three equal angles that are each 60 degrees. If you
cut an equilateral triangle in half, you will end up with two congruent right triangles.
[2]
• In this example, we will be using an equilateral triangle with side lengths of 8.
2. 2
Recall the Pythagorean Theorem. The Pythagorean Theorem states that for any right triangle with sides of length a and b, and hypotenuse of length c:
a2 + b2 = c2.
We can use this theorem to find the height of our equilateral triangle![3]
3. 3
Break the equilateral triangle in half, and assign values to variables a, b, and c. The hypotenuse c will be equal to the original side length. Side a will be equal to 1/2 the side length, and side b is the height of the triangle that we need to solve.
• Using our example equilateral triangle with sides of 8, c = 8 and a = 4.
4. 4
Plug the values into the Pythagorean Theorem and solve for b2. First square c and a by multiplying each number by itself. Then subtract a2 from c2.
Example
42 + b2 = 82 Plug in the values for a and c.
16 + b2 = 64 Square a and c.
b2 = 48 Subtract a2 from c2.
5. 5
Find the square root of b2 to get the height of your triangle! Use the square root function on your calculator to find Sqrt(2. The answer is the height of your equilateral triangle!
• b = Sqrt (48) = 6.93
### Method 3 of 3: Determining Height With Angles and Sides
1. 1
Determine what variables you know. The height of a triangle can be found if you have 2 sides and the angle in between them, or all three sides. We'll call the sides of the triangle a, b, and c, and the angles, A, B, and C.
• If you have all three sides, you'll use
Heron's formula
, and the formula for the area of a triangle.
• If you have two sides and an angle, you'll use the formula for the area given two angles and a side.
A = 1/2ab(sin C).[4]
2. 2
Use Heron's formula if you have all three sides. Heron's formula has two parts. First, you must find the variable
s, which is equal to half of the perimeter of the triangle.
This is done with this formula:
s = (a+b+c)/2.[5]
Heron's Formula Example
For a triangle with sides a = 4, b = 3, and c = 5:
s = (4+3+5)/2
s = (12)/2
s = 6
Then use the second part of Heron's formula, Area = sqr(s(s-a)(s-b)(s-c). Replace Area in the equation with its equivalent in the area formula: 1/2bh (or 1/2ah or 1/2ch).
Solve for h. For our example triangle this looks like:
1/2(3)h = sqr(6(6-4)(6-3)(6-5).
3/2h = sqr(6(2)(3)(1)
3/2h = sqr(36)
Use a calculator to calculate the square root, which in this case makes it 3/2h = 6.
Therefore, height is equal to 4, using side b as the base.
3. 3
Use the area given two sides and an angle formula if you have a side and an angle. Replace area in the formula with its equivalent in the area of a triangle formula: 1/2bh. This gives you a formula that looks like 1/2bh = 1/2ab(sin C). This can be simplified to
h = a(sin C)
, thereby eliminating one of the side variables.[6]
Finding Height with 1 Side and 1 Angle Example
For example, with a = 3, and C = 40 degrees, the equation looks like this:
h = 3(sin 40)
Use your calculator to finish the equation, which makes h roughly 1.928.
## Community Q&A
Search
• Question
How do I find the area of an equilateral triangle when only the height is given?
H = height, S = side, A = area, B = base. You know that each angle is 60 degrees because it is an equilateral triangle. If you look at one of the triangle halves, H/S = sin 60 degrees because S is the longest side (the hypotenuse) and H is across from the 60 degree angle, so now you can find S. The base of the triangle is S because all the sides are the same, so B = S. Using A = (1/2)*BH, you get A = (1/2)*SH, which you can now find.
• Question
How do I calculate the height of a right triangle, given only the length of the base and the interior angle at the base?
Donagan
Look up the tangent of the angle in a trigonometry table. Multiply the tangent by the length of the base.
• Question
How do I determine the height of a triangle when I know the length of all three sides?
You already know the base, so calculate the area by Heron's formula. Then, substitute the values you know in the formula. Area=1/2 * base * height or height=2 * Area/base and find your answer.
• Question
How do I find the height of a triangle?
Donagan
You need to know both the length of the base of the triangle and its area. Divide the base into the area, then double that.
• Question
What is height of a triangle if the base is 5 and the two sides are 3?
Donagan
This is a trigonometry question. Draw the height from the obtuse angle to the "5" side. This forms two right triangles inside the main triangle, each of whose hypotenuses are "3". The cosine of either of the original acute angles equals 2½÷3, or 0.833. Look up that angle in a trig table. Find the sine of that angle, and multiply that by 3 to get the height.
• Question
How do I find the height of a triangle with the length of the three sides?
Follow Method 3. "If you have all three sides, you'll use Heron's formula, and the formula for the area of a triangle." Use Heron's formula to determine the area of the triangle. Set the result equal to 1/2bh, and solve for h.
• Question
How can I determine the height of an isosceles triangle?
Since the two opposite sides on an isosceles triangle are equal, you can use trigonometry to figure out the height. You can find it by having a known angle and using SohCahToa. For example, say you had an angle connecting a side and a base that was 30 degrees and the sides of the triangle are 3 inches long and 5.196 for the base side. In order to find the height, you would need to set it up as this: S=o/h, S=sine, o=opposite (the height), h=hypotenuse (the side), S30=o/3 because it's the opposite divided by the three, you can multiply by the reciprocal on both sides so the three gets cancelled on one side and the other is multiplied by 3, 3sin30=o and 3sin30=1.5=height.
• Question
How do I find the height if the area and base of the quadrilateral are given?
Donagan
If it's a regular quadrilateral, divide the area by the base.
• Question
Can I determine the base of a triangle if I know its height is eight feet?
Donagan
You would also need to know other information, such as the area or some of the sides or angles.
• Question
What is the formula for determining the area of a triangle?
Donagan
One-half base times height.
• How do I find the height of a right angle triangle if I know the base length and the two remaining angles?
• How do I calculate the distance of a height if two sides are known in a triangle?
• If I only have the base measurement (5 cm) and possibly the angles, how do I find the height?
• Is there a simpler version of Heron's Formula?
• How do I find the angle of a triangle when I know the base and the height?
200 characters left
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wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 36 people, some anonymous, worked to edit and improve it over time. Together, they cited 6 references. This article has also been viewed 1,718,650 times.
Co-authors: 36
Updated: September 6, 2019
Views: 1,718,650
Categories: Geometry
Article SummaryX
If you know the base and area of the triangle, you can divide the base by 2, then divide that by the area to find the height. To find the height of an equilateral triangle, use the Pythagorean Theorem, a^2 + b^2 = c^2. Cut the triangle in half down the middle, so that c is equal to the original side length, a equals half of the original side length, and b is the height. Plug a and c into the equation, squaring both of them. Then subtract a^2 from c^2 and take the square root of the difference to find the height. If you want to learn how to calculate the area if you only know the angles and sides, keep reading! | 2,536 | 9,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2020-16 | latest | en | 0.892959 |
http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations | 1,467,040,698,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396029.85/warc/CC-MAIN-20160624154956-00172-ip-10-164-35-72.ec2.internal.warc.gz | 195,675,263 | 23,186 | # Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
-
Very interesting! – goblin Apr 18 '15 at 3:08 | 4,081 | 14,611 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2016-26 | latest | en | 0.855138 |
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Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
### Tis Unique
##### Stage: 3 Challenge Level:
This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility.
### Constellation Sudoku
##### Stage: 4 and 5 Challenge Level:
Special clue numbers related to the difference between numbers in two adjacent cells and values of the stars in the "constellation" make this a doubly interesting problem.
### Making Maths: Double-sided Magic Square
##### Stage: 2 and 3 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### GOT IT Now
##### Stage: 2 and 3 Challenge Level:
For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?
### Magic Squares for Special Occasions
##### Stage: 3 and 4
This article explains how to make your own magic square to mark a special occasion with the special date of your choice on the top line.
### Crossed Ends
##### Stage: 3 Challenge Level:
Crosses can be drawn on number grids of various sizes. What do you notice when you add opposite ends?
### Amazing Card Trick
##### Stage: 3 Challenge Level:
How is it possible to predict the card?
### Domino Square
##### Stage: 2, 3 and 4 Challenge Level:
Use the 'double-3 down' dominoes to make a square so that each side has eight dots.
### 4 Dom
##### Stage: 1, 2, 3 and 4 Challenge Level:
Use these four dominoes to make a square that has the same number of dots on each side.
### Weights
##### Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
### Eleven
##### Stage: 3 Challenge Level:
Replace each letter with a digit to make this addition correct.
### Aba
##### Stage: 3 Challenge Level:
In the following sum the letters A, B, C, D, E and F stand for six distinct digits. Find all the ways of replacing the letters with digits so that the arithmetic is correct.
### 2010: A Year of Investigations
##### Stage: 1, 2 and 3
This article for teachers suggests ideas for activities built around 10 and 2010.
### Top-heavy Pyramids
##### Stage: 3 Challenge Level:
Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Nice or Nasty
##### Stage: 2 and 3 Challenge Level:
There are nasty versions of this dice game but we'll start with the nice ones...
### Two and Two
##### Stage: 2 and 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
### First Connect Three
##### Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Adding and Subtracting Positive and Negative Numbers
##### Stage: 2, 3 and 4
How can we help students make sense of addition and subtraction of negative numbers?
### Pole Star Sudoku 2
##### Stage: 3 and 4 Challenge Level:
This Sudoku, based on differences. Using the one clue number can you find the solution?
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target. | 2,235 | 9,168 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2016-44 | longest | en | 0.794493 |
https://www.macochi.com/pxhk8/discount-rate-formula-f61a72 | 1,620,807,430,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991685.16/warc/CC-MAIN-20210512070028-20210512100028-00164.warc.gz | 889,259,709 | 15,584 | If you know the discounted price and the percentage discount, you can calculate the original price. Calculate discount rate with formula in Excel. This principle is known as the “time value of money.” We can see how the value of a given sum gradually decreases over time here. Also find the solved example questions given below to get an easy approach to the concept. The weighted average cost of capital (WACC) is a good starting point in determining the appropriate discount rate. The discount rate is commonly used for U.S. Treasury bills and similar financial instruments. For example, consider a government bond that sells for $95 and pays$100 in a year's time. If this value proves to be higher than the cost of investing, then the investment possibility is viable. Access all the content Recur has to offer, straight in your inbox. Solution:Here,Principal amount p = 2000 dollars. In terms of Mathematics, the formula for discount is represented as below, Type the original prices and sales prices into … Question 2: Kavita bought land for 50000 dollars at 20% in the year 2000. The actual way to calculate discount is multiplying original price with the discount percent rate. This principle is known as the “time value of money.” We can see how the value of a given sum gradually decreases over time here. This tells your the percentage, or rate, at which you are discounting the bond. Dress rate = principal amount – discount rate. Your discount rate and the time period concerned will affect calculations of your company’s NPV. Question 1: Ruby purchased the shirt. Where, M.P(Marked Price) is the actual price of the product without discount. That shirt rate was $2000 at a 10% discount. The WACC discount formula is: WACC = E/V x Ce + D/V x Cd x (1-T), and the APV discount formula is: APV = NPV + PV of the impact of financing. Discount Formula Calculation of discount is one of most proficient mathematical skills that you can learn. The discount rate we are primarily interested in concerns the calculation of your business’ future cash flows based on your company’s net present value, or NPV. Otherwise, you can calculate it like so: The discount rate element of the NPV formula is used to account for the difference between the value-return on an investment in the future and the money to be invested in the present. To calculate WACC, one multiples the cos… Step 2: … The actual way to calculate discount is multiplying original price with the discount percent rate.$50 - $30 =$20 20 / 50 = 0.40 0.40 = 40% As this value is changed by the accumulation of interest and general inflation, as well as by profits and discounts from investments, it’s handy to have the discount rate calculated as a roadmap of where the value of a dollar invested in your business is likely to go. The WACC formula for discount rate is as follows: This discount rate formula can be modified to account for periodic inventory (the cost of goods available for sale, and the units available for sale at the end of the sales period) or perpetual inventory (the average before the sale of units). In the usual case that no survey-based real estate discount rates are available real estate investors can apply the following formula to calculate a discount rate/required return to be applied to a specific property: Discount rate = Risk-Free Rate + Inflation + Property-Specific Risk Premium The NPV formula assumes that each cash flow is received at the end of the year. Your discount rate will play a … Discount Factor = 0.83So, discount factor is 0.83.Now, let us take another example to understand discount factor formula better. The formula will be typed in Excel as =NPV(discount rate, range of cash flows). The interest rate is calculated using 95 as the base 100 − 95 95 = 5.26 % {\displaystyle {\frac {100-95}{95}}=5.26\%} which says that 95 % {\displaystyle 95\%} of $105.26 is$100. Discount rate DR = 50000*30/100 in 2004th year, Loss Discount rate in 2004th year – the Discount rate. We explain how to calculate it. In Math, Discount problems can be solved using Discount Formulas. In some instances, however, especially for high-risk investments, continuous discounting can be used for more precise valuation. Our second discount rate formula, the adjusted present value calculation, makes use of NPV. After the cash flow for each period is calculated, the present value (PV) of each one is achieved by discounting its future value (see Formula) at a periodic rate of return (the rate of return dictated by the market). Discount Rate Formula. $, /$, = The discount rate … Calculate the bond discount rate. Owing to the rule of. It is calculated using the following formula:WACC = we × ke + wp × kp + wd × kd × (1 - t)Where we, wp and wd are the target weights of common stock, preferred stock, and debt respectively in the company’s capital structure. It is one of the easiest ways to increase the demand for any product. Investing in one is a risk, and investors need to know that the value of your cash flows will hold not only now but also later. Without knowing your discount rate, you can’t precisely calculate the difference between the value-return on an investment in the future and the money to be invested in the present. You’ll find that, in this case, discounted cash flow goes down (from $86,373 in year one to$75,809 in year two, etc.) Let’s say you’re the CEO of WellProfit, a growing, Boise-based SaaS company that’s bound for the stars and thinking about getting investors. That number is the discount amount of the bill and is then divided by the FV to get the percentage discount off of face value. , a dollar at a later point in time will not have the same value as a dollar right now. The discount rate formula can either be extracted by subtracting the selling price of the product from its marked price or by multiplying the discount rate offered and the marked price of the product. The discount is provided for the purpose of: So, the discount can act as a strategy to attract customers for a particular product or service. This formula means the purchase price (PP) of the bill is subtracted from the face value (FV) of the bill at maturity. The complete SaaS guide to calculating and optimizing User Churn, Revenue Run Rate: Definition, Calculation and Formula, How to Calculate Growth Rate (and Different Types of Growth Rate). Bad news for WellProfit. To calculate the discount rate of any product, we need to know the marked price and selling price of the product. Step 1: Cost of Debt: The estimated cost of debt for this privately-held building materials company was 3.40%, which assumes a credit rating of Baa for the subject company. Discount Rate: The discount rate is the interest rate charged to commercial banks and other depository institutions for loans received from the Federal Reserve's discount window. Let’s say that shareholder equity (E) for the year 2030 will be $4.2 billion and the long-term debt (D) stands at$1.1 billion. WACC is the marginal composite cost of all the company’s sources of capital, i.e. Divide the amount of the discount by the face value of the bond. Take a look at the previous screenshot. If you know the discounted price and the percentage discount, you can calculate the original price. We have to calculate the discount factor when the discount rate is 10% and the period is 2.Discount Factor is calculated using the formula given belowDiscount Factor = 1 / (1 * (1 + Discount Rate)Period Number)Put a value in the formula. This formula means the purchase price (PP) of the bill is subtracted from the face value (FV) of the bill at maturity. Using the right discount rate formula, setting the right rate relative to your equity, debt, inventory, and overall present value is paramount. How much is that 20% stake worth, You’ll find that, in this case, discounted cash flow goes down (from $86,373 in year one to$75,809 in year two, etc.). As we noted earlier, you can’t gain a full picture of your company's future cash flows without solid DCF analysis; you can't perform DCF analysis without calculating NPV; you can't calculate either without knowing your discount rate. NPV is an indicator of how much value an investment or project adds to your business. #1 type the following formula in a blank cell (C2), then press Enter key in your keyboard. For companies, that entails understanding the future value of their cash flows and ensuring development is kept within budget. We can also say, a total bill is usually sold at a discount. APV analysis tends to be preferred in highly leveraged transactions; unlike a straightforward NPV valuation, it “, The health of cash flow, not just now but in the future, is fundamental to the health of your business -. S.P(Selling Price) is what customers pay for the product. Let’s say you have an investor looking to invest in a 20% stake in your company; you're growing at 14.1% per year and produce $561,432 per year in free cash flow, giving your investor a cash return of$112,286 per year. Owing to the rule of earning capacity, a dollar at a later point in time will not have the same value as a dollar right now. The health of cash flow, not just now but in the future, is fundamental to the health of your business - 82% of all startups without reliable cash flows will ultimately fold. Investing in one is a risk, and investors need to know that the value of your cash flows will hold not only now but also later. Doing it right, however, is key to understanding the future worth of your company compared to its value now and, ultimately, bridging that gap. The formula is: NPV = ∑ {After-Tax Cash Flow / (1+r)^t} - Initial Investment. The following formula is to calculate the discount rate. Discount Factor Formula. #3 select all discount rate cells C2:C4, and then right click on it, select Format Cells, and the Format Cells dialog will open. The formula for calculating the discount factor in Excel is the same as the Net Present Value (NPV formulaNPV FormulaA guide to the NPV formula in Excel when performing financial analysis. To put it briefly, DCF is supposed to answer the question: "How much money would have to be invested currently, at a given rate of return, to yield the forecast cash flow at a given future date? there’s even a specific Excel function for it, You can find out more about how DCF is calculated here, takes into consideration the benefits of raising debts (e.g., interest tax shield). It helps you estimate how much revenue you'll take in this year. Therefore, it’s unlikely that, at this growth rate and discount rate, an investor will look at this one as a bright investment prospect. Where: NPV = Net Present Value; PV = Present Value; Discount rate is key to managing the relationship between an investor and a company, as well as the relationship between a company and its future self. Finding the percentages is basic arithmetic – the hard part is estimating the “cost” of each one, especially the Cost of Equity. How to Calculate Discount Rate: WACC Formula. In the blog post, we suggest using discount values of around 10% for public SaaS companies, and around 15-20% for earlier stage startups, leaning towards a higher value, the more risk there is to the startup being able to execute on it’s plan going forward. Next, divide the discount amount by original price. Step 2: Cost of Equity. The discount rate for seasonal credit is an average of selected market rates." How much is that 20% stake worth now? It takes inflation and returns into account and features particularly in capital budgeting and investment planning -, Then you can perform a DCF analysis that estimates and discounts the value of all future cash flows by cost of capital to gain a picture of their present values. The Ramsey formula for deriving a discount rate schedule from a single representative agent, who stands in for an enormously heterogeneous real world having widely dispersed growthrates,degreesofriskaversion,andratesofpuretimepreference,allofwhicharethen Examples of Discount Rate Formula (With Excel Template). Discount Rate can be calculated using the formula given here. If discounting – $105.00 / (1+.05) =$100.00 (here 5% is the discount rate, i.e., the growth rate applied in reverse) How should we think of the discount rate? The formula used to calculate discount is D=1/ (1+P) n where D is discount factor, P = periodic interest rate, n is number of payments. DCF is a method of valuation that uses the future cash flows of an investment in order to estimate its value. From your company’s side, you can only go ahead with a new project if expected revenue outweighs the costs of pursuing said opportunity. The second utility of the term discount rate in business concerns the rate charged by banks and other financial institutions for short-term loans. Discount Rate is the price of the total quantity/amount usually less than its original value. A succinct Discount Rate formula does not exist; however, it is included in the What is the Discount Rate Formula? The WACC discount formula is: Let’s dive deeper into these two formulas and how they’re different below. There can be many sources of capital, and the weighted average of those sources is called WACC (Weighted Average Cost of Capital). "Don’t know whether or not to try a discount pricing strategy? Discount Formula Calculation of discount is one of most proficient mathematical skills that you can learn. Knowing your discount rate is key to understanding the shape of your cash flow down the line and whether your new development will generate enough revenue to offset the initial expenses. Use the formula mentioned above to understand the concept. One of the first things you need to do to make your company attractive to investors is find your discount rate. Then 2004th year that land was sold for 3000 dollars. NPV = 40,000(Month 1)/1 + 0.1 + 40,000 (Month 2)/1 + 0.1 ... - 250,000. Her son, however, needs funds today and she is considering taking out those cash flows today, and she wants to know what the present value for those is if she withdraws today. ", Money, as the old saying goes, never sleeps. Being able to understand the value of your future cash flows by calculating your discount rate is similarly important when it comes to evaluating both the value potential and risk factor of new developments or investments. Your discount rate expresses the change in the value of money as it is invested in your business over time. User Churn threatens your ability to generate recurring revenue. For both companies and investors, discount rate is a key metric when positioning for the future. Calculate the discount. To calculate the selling price, you just have to […] It takes inflation and returns into account and features particularly in capital budgeting and investment planning - there’s even a specific Excel function for it. . Solution: We are given the cash flows as well … 1. Revenue run rate is an important metric to track for any subscription business. To calculate the discounted price, we multiplied the original price by (1 - Percentage Discount). You can calculate it like so: As the hypothetical CEO of WellProfit, you’d first calculate your discount rate and your NPV (which, remember, is the difference between the present value of cash inflows and the present value of cash outflows over a period of time and is represented above by “CF”). 82% of all startups without reliable cash flows will ultimately fold. You can find out more about how DCF is calculated here and here. Discount Factor = 1 / (1 * (1 + 10%) ^ 2) 2. WACC can be used to calculate the enterprise value of a firm by considering the cost of goods available for sale against inventory, alongside common stock, preferred stock, bonds, and any other long-term debt on your company’s books. It’s a very different matter and is not decided by the discount rate formulas we’ll be looking at today. For example, the technique of continuous discounting is widely used in financial option valu… Convert this decimal amount into a percentage. We’ll see a number of those variables included in our discount rate formulas. In this case, the discounting rate is 10% and the discounted payback period is around 8 years, whereas the discounted payback period is 10 years if the discount rate is 15%. In the usual case that no survey-based real estate discount rates are available real estate investors can apply the following formula to calculate a discount rate/required return to be applied to a specific property: Discount rate = Risk-Free Rate + … How many dollars did she lose? This percent is the discount rate. By subscribing, you agree to ProfitWell's terms of service and privacy policy. because your discount rate is higher than your current growth rate. Calculate the bond discount rate. And then Ruby how many dollars give to the cashier? Divide the amount of the discount by the face value of the bond. That number is the discount amount of the bill and is then divided by the FV … Your company’s weighted average cost of capital (WACC, a discount rate formula we’ll show you how to calculate shortly) is often used as the discount rate when calculating NPV, although it is sometimes thought to be more appropriate to use a higher discount rate to adjust for risk or opportunity cost. The discount amount for the dress is 200. =(B2-A2)/ABS(A2) #2 drag the AutoFill handle from the Cell C2 to C4 to apply the above formula. If your company’s future cash flow is likely to be much higher than your present value, and your discount rate can help show this, it can be the difference between being attractive to investors and not. In this, the primary credit rate is the Federal Reserve's most common discount window program, and the discount rates for the three lending programs are the same across all Reserve Banks except on days around a change in the rate. Our overall capital = E + D = 4.2 billion + 1.1 billion = $5.3 billion, The equity linked cost of capital = (E/V) x Re = 4.2/5.3 x 6.6615% = 0.0524, The debt component = (D/V) x Cd x (1-T) = 1.1/5.3 x 6.5% x (1-21%) = - 0.0197, Our second discount rate formula, the adjusted present value calculation, makes use of NPV. 1. Setting a discount rate is not always easy, and to do it precisely, you need to have a grasp of the discount rate formula. In order to manage your own expectations for your company, and in order for investors to vet the quality of your business as an investment opportunity, you need to know how to find that discount rate. So, the discount amount is 10000 dollars. Calculate the amount they earn by iterating through each year, factoring in growth. The modified CAPM was used to estimate a range of cost of equity of 11.25% to 14.3% for the subject … As stated above, net present value (NPV) and discounted cash flow (DCF) are methods of valuation used to assess the quality of an investment opportunity, and both of them use discount rate as a key element. Step 1: Firstly, determine the value of the future cash flow under consideration. The discount rate is − = % The interest rate is calculated using 95 as the base Ricky 800 dollars is given to the cashier. Using the above example, divide$36,798 by $500,000. 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Explanation. Discount refers to the condition of the price of a bond that is lower than the face value. This tells your the percentage, or rate, at which you are discounting the bond. By considering financing investment with a portion of debt, some prospects that might’ve looked unviable with NPV alone suddenly seem more attractive as investment possibilities. 1. There are two primary discount rate formulas - the weighted average cost of capital (WACC) and adjusted present value (APV). To put it briefly, DCF is supposed to answer the question: "How much money would have to be invested currently, at a given rate of return, to yield the forecast cash flow at a given future date?" NPV is the sum of all the discounted future cash flows. debt, preferred stock, and equity. To complicate matters, there is unfortunately more than one way to think of the discount rate. It helps to increases sales. All financial theory is consistent here: every time managers spend money they use capital, so they should be thinking about what that capital costs the company. This NPV is not only positive but very high; an investor is likely to go through with the investment, which is good news for WellProfit! Finding your discount rate involves an array of factors that have to be taken into account, including your company’s equity, debt, and inventory. Let’s dive deeper into these two formulas and how they’re different below. Discount Rate Formula. For most companies it’s just a weighted average of debt and equity, but some could have weird preferred structures etc so it could be more than just two components. It plays an important role in online merchandising plans. You are required to calculate the present values of those cash flows at 7% and calculate the total of those discounting cash flows. NPV is the difference between the present value of a company’s cash inflows and the present value of cash outflows over a given time period. NPV is used to measure the costs and benefits, and ultimately the profitability, of a prospective investment over time. We’ll change our discount rate from our previous NPV calculation. It is usually printed on the item by making some modifications in the marked price(Distribution Channel) or by pasting a sticker on the commodity(by making changes in retail prices) or making changes in list price set for buyers. When the selling price is less than the marked price, then the buyer has said to be got some discount on it. There are two primary discount rate formulas - the weighted average cost of capital (WACC) and adjusted present value (APV). Take a look at the previous screenshot. How to Apply the Discount Rate to Evaluate a Business Investment. In dollar terms the discount is$200; however, the discount is usually expressed in percentage terms. To calculate the discounted price, we multiplied the original price by (1 - Percentage Discount). Then you can perform a DCF analysis that estimates and discounts the value of all future cash flows by cost of capital to gain a picture of their present values. An accurate discount rate is crucial to investing and reporting, as well as assessing the financial viability of new projects within your company. It is comprised of a blend of the cost of equity and after-tax cost of debt and is calculated by multiplying the cost of each capital source (debt and equity) by its relevant weight and then adding the products together to determine the WACC value. To ensure success you must calculate and optimize user churn correctly. The discount rate is the annualized percentage of the above discount, which is a percentage adjusted, to give an annual percentage. Your discount rate will play a role in both your current reporting and your future planning. $, /$, = The discount rate for the bond is 7.36 percent. This discounted cash flow (DCF) analysis requires that the reader supply a discount rate. Growth rate is one of the most important metrics for subscription businesses. First, let's examine each step of NPV in order. Discount rate is used primarily by companies and investors to position themselves for future success. Put your understanding of this concept to test by answering a few MCQs. APV can also be useful when revealing the hidden value of seemingly less viable investment opportunities. With so many pros and cons, our pricing experts have come to help.". NPV analysis is a form of intrinsic valuation and is used extensively across finance and accounting for determining the value of a business, investment security, of a business, as part of a Discounted Cash Flow (DCF)Discounted Cash Flow DCF FormulaThe discoun… Based on the profit and loss concept, the discount is basically the difference of marked price and selling price. You can apply it to the restaurants, shopping malls, and setting rates for your products. Click ‘Start Quiz’ to begin! Let’s say now that the target compounded rate of return is 30% per year; we’ll use that 30% as our discount rate. Discount Rate Formula: Calculating Discount Rate [WACC/APV]. The second utility of the term discount rate in business concerns the rate charged by banks and other financial institutions for short-term loans. The definition of a discount rate depends the context, it's either defined as the interest rate used to calculate net present value or the interest rate charged by the Federal Reserve Bank. The discount equals the difference between the price paid for and it’s par value. To calculate the original price, simply divide the discounted price by (1 - Percentage Discount). Present value (PV), future value (FV), investment timeline measured out in periods (N), interest rate, and payment amount (PMT) all play a part in determining the time value of money being invested. It is expected to bring in $40,000 per month of net cash flow over a 12-month period with a target rate of return of 10%, which will act as our discount rate. The reduction in the rate/price of some item/product or service is referred to as Discount rate. This second discount rate formula is fairly simple and uses the cost of equity as the discount rate: APV = NPV + PV of the impact of financing. You can apply it to the restaurants, shopping malls, and setting rates for your products. Discount rate is key to managing the relationship between an investor and a company, as well as the relationship between a company and its future self. Discount Rate. Divide the difference between the redemption value and the amount paid by the amount paid to find the discount in percentage terms. If this value proves to be higher than the cost of investing, then the investment possibility is viable. It is the reduction in the price of some product or service, to increase the demand for the product and increase sales. Discount Rate Estimation of a Privately-Held Company – Quick Example. Bad news for WellProfit. Where r is the required rate of return (or interest rate) and n is the number of years between present day and the future year in question. 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https://www.microwavejournal.com/articles/3734-a-universal-oscillator-analysis-technique-that-accurately-estimates-frequency-and-output-power | 1,718,415,033,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00680.warc.gz | 803,450,440 | 24,451 | Generally, an oscillator can be seen as a positive feedback system.
It usually consists of an amplifier followed by a resonator, which is fed back to the amplifier input, as shown in Figure 1 .
The amplifier task is to compensate for the losses in the resonator, and the resonator does the frequency selection.
It is of interest how these components should behave to generate the proper oscillation. Figure 2 shows a simplified mathematical representation of a feedback system. The amplifier is assumed to have no phase shift, a gain of A, and infinite input and zero output impedance.
The resonator is a simple phase shifter with an attenuation B. The input signal I(j ) is summed with the amplifier output signal and fed into the phase shifter. The output signal O(j ) is the output of the phase shifter and the input of the amplifier.
The transfer function of the circuit can then be stated as
where
= ( ) = f( )
and otherwise the oscillator's frequency is not defined. The formula stated in Equation 1 describes the behavior of a feedback loop by injecting a signal I(j ) into the circuit while observing the output O(j ). Therefore, to obtain a real oscillator condition without injection of an input signal, I(j ) must fade to zero, while maintaining an output signal O(j ). This is mathematically stated as
It can be seen in Equation 2 that if I(j ) goes to zero, the other multiplication term must raise to infinity to get an output different from zero. So A·B must be equal to 1. Figure 3 shows the transfer-function stated in Equation 1 with A = 1 and B = 1, as a function of the phase shift . It is easy to see that a phase shift of n x 360° raises the transfer-function to infinity. For B = A = 1, Equation 2 can be simplified to
Using the rule of L'Hospital, and with the phase going to zero for the oscillating condition, the output can be calculated as
This relation is commonly known as the Barkhausen criterion, which states that the loop gain must be 1 and the loop phase shift multiples of 360° to obtain oscillation.
#### Transformation
The next question is how the numerous existing oscillator circuits can be converted to a feedback loop system. Stan Alechno showed1 that nearly every oscillator can be converted to a common-emitter circuit, by first connecting all ground and power supply nodes, then letting them float and grounding the emitter. However, if feedback is performed inside the transistor, the circuit cannot be transformed.
Figure 4 shows an example of how a common-collector negative resistance oscillator can be transformed to a common-emitter circuit with the basic oscillator topology. As can be seen, the first conversion in (a) is to connect all the ground and power supply terminals together. This is possible, as power supply nodes represent a AC ground node, because the AC voltage is grounded by the large blocking capacitors between supply voltage and the common ground. The next step is to remove the ground from the circuit by letting all nets float. Now the ground can be placed on every node in the circuit, where it is advantageous for further analysis. In this case, the emitter of the bipolar transistor is grounded in (c). If the circuit in (c) is rearranged, the circuit in (d) appears, which shows a common-emitter circuit and a back fed resonator consisting of C1, C2, L and C.
This transformation becomes clearer if the ground node is seen as an ordinary circuit node, which is, in this case for example, the ground plane. The power supply, as previously stated, is also an AC ground node. Thus, it is possible to think of any node in the circuit to be the ground node, and to re-sketch the circuit. However, this transformation has an impact on components that are part of the bias-supply of the active device. Since they have a DC function, their behavior changes, as the ground is connected to another node. In the original circuit, the load resistor RL is connected between emitter and ground. A DC current through this resistor will result in a voltage drop across it, resulting in a voltage increase at the emitter and base as well. The base bias resistor must take this voltage drop into account. In the rearranged circuit, the load resistor is connected to the transistor collector and the emitter is grounded. The base bias resistor now has a much larger voltage drop from power supply to the base as compared to the original circuit where the supply voltage to base voltage is lowered through the voltage drop across the load resistor RL. Consequently, the bias network must be recalculated for equal operation of the circuits. An example later in this article will show the difference between the two biasing cases. For further analysis, the oscillator loop must be broken up, and a transfer-function can be recorded. An oscillator theory treating this problem is presented by Rhea2.
#### Calculations
The circuit in Figure 5 can be described by S-parameters, but then the ports P1 and P2 would be terminated in 50 . This would not give the correct result for the transfer-function, since in the closed loop condition the output of the oscillator is terminated with its input's impedance and vice-versa.
It is not easy to convert these parameters to more generality since S-parameters are defined for one characteristic impedance. For the sake of understanding, Z-parameters will now be used for defining the problem.
The function
which converts S-parameters to Z-parameters with the identity
will be used to convert the open loop oscillator circuit to the circuit in Figure 6 , where Z2 indicates the correct oscillator input impedance. More about two-port transformations and oscillators is given by Vendelin, et al.3
The following equations can be derived:
The forward voltage transfer function
The backward voltage transfer function
The input impedance
The output impedance
From Equations 7 to 10, it seems clear that all four parameters depend on the termination impedances Z1 and Z2. For correct oscillator function, V2/V1 must be equal to 1. But since this transfer function depends on Z2 and its value is unknown, the transfer function cannot be plotted correctly. If it is assumed for a while that there is no reverse influence from the output to the input (Z12 = 0), the parameters are simplified to
The forward voltage transfer function
The backward voltage transfer function
The input impedance
The output impedance
while Equation 11 still depends on Z2, Equations 13 and 14 are independent. Now, a second view on the oscillator circuit in Figure 7 brings more light to oscillator behavior. As the output of the oscillator is connected with its input, the impedance Z2 will result in V1/I1 = Z11. This means that the oscillator looks back to its own input impedance. With this simplification and the feedback, Equation 11 reduces to
By introducing the oscillation condition into Equation 15, the result is
where it is noted that the Z-parameters are a function of frequency and oscillator power Ps.
In the more general case, if Z12 0, Equation 16 does not hold anymore, and a new way of determining the impedances Z1 and Z2 must be accomplished formally. Oscillation can also be seen in the time domain as a wave traveling in a loop through the oscillator. This means that the wave is traveling through the circuit and is fed back to its input. Since the oscillator has a group delay, the Barkhausen criterion changes to
where
tgrf = the forward group delay
0 = frequency of oscillation
Equation 17 states that the group delay induces a phase shift. The oscillator can be seen as a non-reciprocal transmission line. As continuous reflection occurs in the oscillator from incorrectly terminated impedances, a wave traveling back is induced. However, since the transistor has low backward amplification, this reverse wave is heavily damped. In the case of S12 = 0 in the transistor, the reverse wave is totally blocked by the transistor and the impedances can be calculated as in Equations 13 and 14. This case is very similar to an ideal distributed isolator.
If a certain voltage zero transition in the traveling wave at the input could be marked, and if it were possible to follow it, the zero transition coming out of the oscillator circuit could be seen with a delay of tgrf, as shown in Figure 8 . If it is assumed that the correct termination impedance is already known, the oscillator circuit, terminated with the correct impedance Z2, has the same input impedance Zin=V1/I1 as the closed loop oscillator. Thus, the traveling wave out of the oscillator will have the same conditions of input impedances if it were fed back to the oscillator input or another oscillator circuit input, which is terminated with the correct impedance Z2.
Figure 9 shows the equal input conditions. The impedance Zi1 stands for the closed loop impedance and Zi2 for the input impedance of another oscillator circuit with the same S-parameters, terminated with the correct impedance Z2. As previously stated, Zi1 must be equal to Zi2. Since the second oscillator circuit is of the same type as the first one, the Barkhausen criterion is also fulfilled for the two oscillator circuits in series, as the second oscillator is terminated with the correct impedance Z2. Since the impedance Z2 is assumed to be a replacement of the input impedance of the closed loop oscillator, the loop can be closed, as shown in Figure 10 .
Because the oscillators are from the same type and the Barkhausen criterion is still fulfilled, V1 must be equal to V2 and I1 must be equal to I3. Now it can be easily seen that an infinite number of oscillator blocks can be inserted in the circuit by keeping the input currents and the node voltages equal. In this case, the wave would travel through an infinite number of circuits, which is essentially the same as traveling through one closed loop oscillator circuit. The infinite oscillator block chain circuit can now be analyzed to get the correct termination impedance. For this analysis it is very advantageous to use ABCD parameters, since chaining two-ports can be accomplished by matrix multiplication. A new transform function will be introduced to convert the oscillator's open loop S-parameters to ABCD parameters
So the two two-ports transform to their ABCD equivalents A1 and A2. Now the combined matrix is just
To get some insight into the infinite circuit of Figure 11 , the Z-matrix of the infinite chained oscillator A-matrices will be calculated as
This means that an infinite of matrices can be chained mathematically as an infinite matrix-multiplication with itself stated as n, which stands for 1 x 1 ... x n. It is very interesting that the coefficients Zo11 and Zo22 converge to a fixed value which will be analytically derived later. Since an amplifier usually has a backward transfer function lower than 1, which means forward gain and backward isolation, the parameter Zo12 goes to zero, so with the Equations 11 to 14, Equation 20 reduces to
That means that Zo1 is the input impedance and Zo2 is the output impedance of the network. With the determination of the impedances Zo1 and Zo2 the voltage transfer-function can be defined and set to 1. For the conventional oscillator analysis, Z1 will be equal to Zo2, which actually would not be used in oscillator forward analysis, and Z2 will be equal to Zo1, which is the interesting part for the transfer-function. If the loop is closed and broken up anywhere again, the calculated matrix will have the same coefficients. If the input impedance of an infinite chain of oscillator two-ports is Zo1, it will still be Zo1 if only one two-port is removed since there will still remain an infinite chain of oscillators. This shows that the input impedance of one single oscillator two-port must be Zo1 if its output is terminated with Zo1, as shown in Figure 12 .
From this approach it seems clear that the oscillator closed loop impedances are Zo1 and Zo2. To calculate the impedances Zo1 and Zo2 directly, the diagram for a two-port can be used with Equations 9 and 10 to obtain a solution. The definition out of (9) and (10) will then be
If these relations are transformed with respect to Zo1 and Zo2, two quadratic equations result. The solution for these equations are:
As it can be seen from Equations 23 and 24, four solutions are possible, but only two of them are meaningful. The one with the positive real part is the preferred one, as the impedances should be passive in feedback designs.
The main difference between an oscillator and a transmission line is that the oscillator contains an active device, in which the input power is amplified with some gain. Thus, it is possible to get reflections and |S21| = 1 at the same time.
This is not possible in ideal transmission lines since there S11 must be zero for lossless operation. The reflections in the oscillator travel backwards and have their own propagation properties. Through the reverse isolation of the oscillator, this reverse wave is damped, but not extinguished. Since this wave is generated in every two-port in a chain, it influences the input impedance of the oscillator, which has been taken into account in Equations 23 and 24. The impedances Zo1 and Zo2 from Equation 21 can be seen as the characteristic impedances of the traveling waves (forward and reverse), as the node voltages and input currents are equal over the whole infinite two-port chain.
#### Simulation
A simplified common-collector oscillator is designed for an oscillation frequency of 2.9 GHz and a loop power of 0 dBm. Its schematic is shown in Figure 13 . The bipolar transistor is a BFP-420 from Infineon.
The resistors RB and RE are for biasing, while Lb and Cb are AC- and DC-blocking components, respectively. RV is a coupling resistor to prevent oscillation from quenching, because of too low output load. The capacitor C3 has been introduced to get more flexibility in frequency adjustment. After the transformation to a common-emitter configuration, the circuit can be drawn, as illustrated in Figure 14 .
The subcircuit-block "SUBCKT" is a replacement of the BFP-420 transistor common-emitter, large-signal S-parameters for an input power of 0 dBm. Some harmonic balance simulations have been done with the pure transistor amplifier to get the different large-signal parameters at certain power levels. These large-signal parameters are only a modeling help, but they do not really exist by definition of scattering parameters, since, through the strong nonlinearities of the active device, many harmonics will be generated. The parameters are also dependent on the used power, so this part of the modeling will produce the largest error in the transfer-function behavior. The 100 resistor replaces the sum of Rv + RL. Blocking components such as Lb and Cb have been removed, as they have no influence on the oscillator analysis. Next the transfer-function is calculated with the Microwave Office Program4 and the formulas of Table 1 , using Z-parameters with the result of Equation 21 introduced into Equation 7 and plotted over frequency, as shown in Figure 15 .
Good proof that the plotted transfer-function is correct is to move the components from the output to the input. Since the oscillator analysis should be independent of the resonator location, the position of amplifier and resonator can be changed without altering the transfer-function. Even partial resonator components can be transferred from input to output and vice-versa without effect. With the component values chosen, the transfer-function is evaluated. It can be extracted that the phase shift crosses zero at 2.901 GHz, while the magnitude is -0.358 dB at the same frequency. To verify these results, an harmonic balance analysis has been set up and simulated.
One thing for which care must be taken is the different transistor biasing in Appendix A compared to a common-emitter circuit. It is very important that the collector-current and collector-emitter voltage stay equal to the common-emitter large-signal equivalent circuit. The basic SPICE-model of the bipolar transistor is presented in the figure, with its package parasitic components. The harmonic balance simulation results are compared to the transfer-function approach in Table 2 .
Table 1 Formulas to Calculate the Voltage Transfer-function VK in Microwave Office NZ11 = Schematic 1:Z[1,1] NZ12 = Schematic 1:Z[1,2] NZ21 = Schematic 1:Z[2,1] NZ22 = Schematic 1:Z[2,2] ZK = 0.5*(NZ11-NZ22+sqrt((NZ11-NZ22)* ZK = (NZ11-NZ22)+4* ZK = (NZ11*NZ22-NZ12*NZ21))) VK = (ZK*NZ21)/ VK = (NZ11*(ZK+NZ22)-NZ12*NZ21)
Table 2 Comparison of the Transfer-function Method with Harmonic Balance Simulation Transfer- function Harmonic Balance Error Frequency (GHz) 2.9 2.9078 0.3% Power (dBm) 0 (in base) -0.2 (in base) 0.2 dB
#### Conclusion
It has been shown that for many types of oscillators which can be converted to a feedback system, an accurate estimate of the oscillator frequency and output power is possible. Through linking an infinite number of oscillator open loop circuits, the true input and output termination impedances are defined. With Equations 23 and 24, these impedances can be calculated and inserted in an open loop analysis with a single oscillator circuit, from which a transfer-function can be plotted. The oscillation point can be determined by the general oscillator conditions (gain = 1 and phase shift = n x 360°). Using small-signal parameters, the start-up condition can be checked, and with the transistor's large-signal S-parameters at a certain input or output power, the oscillator can be designed the way that the true loop power will meet the large-signal estimate. Since the large-signal S-parameters of the transistor are measured without the resonator circuit, the different terminations of harmonics as well as unknown reflections of the different harmonics and the fundamental frequency will influence the termination impedances. Further, the model which describes the active device must be accurate enough, as otherwise the results of the transfer-function approach will agree with the harmonic balance simulation results, but probably not the physical implementation. Good parasitic- and nonlinear-modeling are obligatory. The comparison shows that the estimation error will be less than one percent for the frequency and less than 0.5 dB for power, which will be satisfying for most applications.
#### References
1. S. Alechno, "The Virtual Ground in Oscillator Design - A Practical Example," Applied Microwave & Wireless , Vol. 11, No. 7, July 1999.
2. R.W. Rhea, Oscillator Design and Computer Simulation , Noble Publishing, Atlanta, GA, 1995, pp. 11-15.
3. G.D. Vendelin, et al., Microwave Circuit Design Using Linear and Nonlinear Techniques , Wiley-Interscience, 1990.
4. AWR Research, Microwave Office 2002, educational version.
Gerhard A. Hofbauer received his diploma degree from Graz University of Technology in May 2002, and is currently working toward his PhD degree in the department of communication and wave propagation at the same university. He is currently designing microwave, RF, IF, low frequency and digital circuits for a pulsed Doppler radar in C-band for civil applications, as well as precision wideband microwave measurement equipment. He can be reached via e-mail at [email protected].
Appendix A Oscillator circuit prepared for harmonic balance simulation using microwave office | 4,139 | 19,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-26 | latest | en | 0.921817 |
http://www.furthergrow.in/2019/10/parametric-optimal-f-on-normal.html | 1,675,493,199,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500094.26/warc/CC-MAIN-20230204044030-20230204074030-00289.warc.gz | 65,859,529 | 54,860 | ### Parametric Optimal f on the Normal Distribution- THE BASICS OF PROBABILITY DISTRIBUTIONS
Parametric Optimal f on the Normal Distribution
THE BASICS OF PROBABILITY DISTRIBUTIONS
Imagine if you will that you are at a racetrack and you want to keep a log of the position in which the horses in a race finish. Specifically, you want to record whether the horse in the pole position came in first, second, and so on for each race of the day. You will only record ten places. If the horse came in worse than in tenth place, you will record it as a tenth-place finish. If you do this for a number of days, you will have gathered enough data to see the distribution of finishing positions for a horse starting out in the pole position. Now you take your data and plot it on a graph. The horizontal axis represents where the horse finished, with the far left being the worst finishing position (tenth) and the far right being a win. The vertical axis will record how many times the pole position horse finished in the position noted on the horizontal axis. You would begin to see a bell-shaped curve develop.
Under this scenario, there are ten possible finishing positions for each race. We say that there are ten bins in this distribution. What if, rather than using ten bins, we used five? The first bin would be for a first- or second-place finish, the second bin for a third-or fourth-place finish, and so on. What would have been the result? Using fewer bins on the same set of data would have resulted in a probability distribution with the same profile as one determined on the same data with more bins. That is, they would look pretty much the same graphically. However, using fewer bins does reduce the information content of a distribution. Likewise, using more bins increases the information content of a distribution. If, rather than recording the finishing position of the pole position horse in each race, we record the time the horse ran in, rounded to the nearest second, we will get more than ten bins; and thus the information content of the distribution obtained will be greater.
If we recorded the exact finish time, rather than rounding finish times to use the nearest second, we would be creating what is called a continuous distribution. In a continuous distribution, there are no bins. Think of a continuous distribution as a series of infinitely thin bins. A continuous distribution differs from a discrete distribution, the type we discussed first in that a discrete distribution is a binned distribution. Although binning does reduce the information content of a distribution, in real life it is often necessary to bin data. Therefore, in real life it is often necessary to lose some of the information content of a distribution, while keeping the profile of the distribution the same, so that you can process the distribution. Finally, you should know that it is possible to take a continuous distribution and make it discrete by binning it, but it is not possible to take a discrete distribution and make it continuous.
Figure A continuous distribution is a series of infinitely thin bins
When we are discussing the profits and losses of trades, we are essentially discussing a continuous distribution. A trade can take a multitude of values (although we could say that the data is binned to the nearest cent). In order to work with such a distribution, you may find it necessary to bin the data into, for example, one-hundred-dollar-wide bins. Such a distribution would have a bin for trades that made nothing to \$99.99, the next bin would be for trades that made \$100 to \$199.99, and so on. There is a loss of information content in binning this way, yet the profile of the distribution of the trade profits and losses remains relatively unchanged.
DESCRIPTIVE MEASURES OF DISTRIBUTIONS
Most people are familiar with the average, or more specifically the arithmetic mean. This is simply the sum of the data points in a distribution divided by the number of data points:
A = (∑[i = 1,N] Xi)/N
where,
A = The arithmetic mean.
Xi = The ith data point.
N = The total number of data points in the distribution.
The arithmetic mean is the most common of the types of measures of location, or central tendency of a body of data, a distribution. However, you should be aware that the arithmetic mean is not the only available measure of central tendency and often it is not the best. The arithmetic mean tends to be a poor measure when a distribution has very broad tails. Suppose you randomly select data points from a distribution and calculate their mean. If you continue to do this you will find that the arithmetic means thus obtained converge poorly, if at all, when you are dealing with a distribution with very broad tails.
Another important measure of location of a distribution is the median. The median is described as the middle value when data are arranged in an array according to size. The median divides a probability distribution into two halves such that the area under the curve of one half is equal to the area under the curve of the other half. The median is frequently a better measure of central tendency than the arithmetic mean. Unlike the arithmetic mean, the median is not distorted by extreme outlier values. Further, the median can be calculated even for open-ended distributions. An open-ended distribution is a distribution in which all of the values in excess of a certain bin are thrown into one bin.
An example of an open-ended distribution is the one we were compiling when we recorded the finishing position in horse racing for the horse starting out in the pole position. Any finishes worse than tenth place were recorded as a tenth place finish. Thus, we had an open distribution. The median is extensively used by the U.S. Bureau of the Census. The third measure of central tendency is the mode-the most frequent occurrence. The mode is the peak of the distribution curve. In some distributions there is no mode and sometimes there is more than one mode. Like the median, the mode can often be regarded as a superior measure of central tendency. The mode is completely independent of extreme outlier values, and it is more readily obtained than the arithmetic mean or the median.
We have seen how the median divides the distribution into two equal areas. In the same way a distribution can be divided by three quartiles, or nine deciles or 99 percentiles. The 50th percentile is the median, and along with the 25th and 75th percentiles give us the quartiles. Finally, another term you should become familiar with is that of a quantile. A quantile is any of the N-1 variate values that divide the total frequency into N equal parts. We now return to the mean. We have discussed the arithmetic mean as a measure of central tendency of a distribution. You should be aware that there are other types of means as well. These other means are less common, but they do have significance in certain applications.
First is the geometric mean, which we saw how to calculate. The geometric mean is simply the Nth root of all the data points multiplied together.
G = (∏[i = 1,N]Xi)^(1/N)
where,
G = The geometric mean.
Xi = The ith data point.
N = The total number of data points in the distribution.
The geometric mean cannot be used if any of the variate-values is zero or negative. We can state that the arithmetic mathematical expectation is the arithmetic average outcome of each play minus the bet size. Likewise, we can state that the geometric mathematical expectation is the geometric average outcome of each play minus the bet size.
Another type of mean is the harmonic mean. This is the reciprocal of the mean of the reciprocals of the data points.
1/∏ = 1/N ∑[i = 1,N]1/Xi
where,
H = The harmonic mean.
Xi = The ith data point.
N = The total number of data points in the distribution.
The final measure of central tendency is the quadratic mean or roof mean square.
R^2 = l/N∑[i = 1,N]Xi^2
where,
R = The root mean square.
Xi = The ith data point.
N = The total number of data points in the distribution.
You should realize that the arithmetic mean (A) is always greater than or equal to the geometric mean (G), and the geometric mean is always greater than or equal to the harmonic mean (H):
H<=G<=A
where,
H = The harmonic mean.
G = The geometric mean.
A = The arithmetic mean.
MOMENTS OF A DISTRIBUTION
The central value or location of a distribution is often the first thing you want to know about a group of data, and often the next thing you want to know is the data's variability or "width" around that central value. We call the measures of a distributions central tendency the first moment of a distribution. The variability of the data points around this central tendency is called the second moment of a distribution. Hence the second moment measures a distribution's dispersion about the first moment. As with the measure of central tendency, many measures of dispersion are available. We cover seven of them here, starting with the least common measures and ending with the most common.
The range of a distribution is simply the difference between the largest and smallest values in a distribution. Likewise, the 10-90 percentile range is the difference between the 90th and 10th percentile points. These first two measures of dispersion measure the spread from one extreme to the other. The remaining five measures of dispersion measure the departure from the central tendencyThe semi-interquartile range or quartile deviation equals one half of the distance between the first and third quartiles. This is similar to the 10-90 percentile range, except that with this measure the range is commonly divided by 2. The half-width is an even more frequently used measure of dispersion.
Here, we take the height of a distribution at its peak, the mode. If we find the point halfway up this vertical measure and run a horizontal line through it perpendicular to the vertical line, the horizontal line will touch the distribution at one point to the left and one point to the right. The distance between these two points is called the half-width. Next, the mean absolute deviation or mean deviation is the arithmetic average of the absolute value of the difference between the data points and the arithmetic average of the data points. In other words, as its name implies, it is the average distance that a data point is from the mean. Expressed mathematically:
M = 1/N ∑[i = 1,N] ABS (Xi-A)
where,
M = The mean absolute deviation.
N = The total number of data points.
Xi = The ith data point.
A = The arithmetic average of the data points.
ABS() = The absolute value function.
Equation gives us what is known as the population mean absolute deviation. You should know that the mean absolute deviation can also be calculated as what is known as the sample mean absolute deviation. To calculate the sample mean absolute deviation, replace the term 1/N in Equation with 1/(N-1). You use the sample version when you are making judgments about the population based on a sample of that population.
The next two measures of dispersion, variance and standard deviation, are the two most commonly used. Both are used extensively, so we cannot say that one is more common than the other; suffice to say they are both the most common. Like the mean absolute deviation, they can be calculated two different ways, for a population as well as a sample. The population version is shown, and again it can readily be altered to the sample version by replacing the term 1/N with 1/(N-1).
The variance is the same thing as the mean absolute deviation except that we square each difference between a data point and the average of the data points. As a result, we do not need to take the absolute value of each difference, since multiplying each difference by itself makes the result positive whether the difference was positive or negative. Further, since each distance is squared, extreme outliers will have a stronger effect on the variance than they would on the mean absolute deviation. Mathematically expressed:
V = 1/N ∑[i = 1,N] ((Xi-A)^2)
where,
V = The variance.
N = The total number of data points.
Xi = The ith data point.
A = The arithmetic average of the data points.
Finally, the standard deviation is related to the variance in that the standard deviation is simply the square root of the variance. The third moment of a distribution is called skewness, and it describes the extent of asymmetry about a distributions mean. Whereas the first two moments of a distribution have values that can be considered dimensional, skewness is defined in such a way as to make it nondimensional. It is a pure number that represents nothing more than the shape of the distribution.
Figure Skewness
A positive value for skewness means that the tails are thicker on the positive side of the distribution, and vice versa. A perfectly symmetrical distribution has a skewness of 0.
Figure Skewness alters location.
In a symmetrical distribution the mean, median, and mode are all at the same value. However, when a distribution has a nonzero value for skewness. The relationship for a skewed distribution (any distribution with a nonzero skewness) is:
Mean-Mode = 3*(Mean-Median)
As with the first two moments of a distribution, there are numerous measures for skewness, which most frequently will give different answers. These measures now follow:
S = (Mean-Mode)/Standard Deviation
S = (3*(Mean-Median))/Standard Deviation
These last two equations are often referred to as Pearson's first and second coefficients of skewness, respectively. Skewness is also commonly determined as:
S = 1/N ∑[i = 1,N] (((Xi-A)/D)^3)
where,
S = The skewness.
N = The total number of data points.
Xi = The ith data point.
A = The arithmetic average of the data points.
D = The population standard deviation of the data points.
Figure Kurtosis.
Finally, the fourth moment of a distribution, kurtosis measures the peakedness or flatness of a distribution. Like skewness, it is a nondimensional quantity. A curve less peaked than the Normal is said to be platykurtic, and a curve more peaked than the Normal is called leptokurtic. When the peak of the curve resembles the Normal Distribution curve, kurtosis equals zero, and we call this type of peak on a distribution mesokurtic. Like the preceding moments, kurtosis has more than one measure. The two most common are:
K = Q/P
where,
K = The kurtosis.
Q = The semi-interquartile range.
P = The 10-90 percentile range.
K = (1/N (∑[i = 1,N] (((Xi-A)/D)^ 4)))-3
where,
K = The kurtosis.
N = The total number of data points.
Xi = The ith data point.
A = The arithmetic average of the data points.
D = The population standard deviation of the data points.
Finally, it should be pointed out there is a lot more "theory" behind the moments of a distribution than is covered here, For a more in-depth discussion you should consult one of the statistics books mentioned in the Bibliography. The depth of discussion about the moments of a distribution presented here will be more than adequate for our purposes throughout this text. Thus far, we have covered data distributions in a general sense. Now we will cover the specific distribution called the Normal Distribution.
THE NORMAL DISTRIBUTION
Frequently the Normal Distribution is referred to as the Gaussian distribution, or de Moivre's distribution, after those who are believed to have discovered it-Karl Friedrich Gauss (1777-1855) and, about a century earlier and far more obscurely, Abraham de Moivre (1667-1754). The Normal Distribution is considered to be the most useful distribution in modeling. This is due to the fact that the Normal Distribution accurately models many phenomena. Generally speaking, we can measure heights, weights, intelligence levels, and so on from a population, and these will very closely resemble the Normal Distribution.
Let's consider what is known as Galton's board. This is a vertically mounted board in the shape of an isosceles triangle. The board is studded with pegs, one on the top row, two on the second, and so on. Each row down has one more peg than the previous row. The pegs are arranged in a triangular fashion such that when a ball is dropped in, it has a 50/50 probability of going right or left with each peg it encounters. At the base of the board is a series of troughs to record the exit gate of each ball.
Figure Galton's board.
The balls falling through Galton's board and arriving in the troughs will begin to form a Normal Distribution. The "deeper" the board and the more balls are dropped through, the more closely the final result will resemble the Normal Distribution. The Normal is useful in its own right, but also because it tends to be the limiting form of many other types of distributions. For example, if X is distributed binomially, then as N tends toward infinity, X tends to be Normally distributed. Further, the Normal Distribution is also the limiting form of a number of other useful probability distributions such as the Poisson, the Student's, or the T distribution. In other words, as the data (N) used in these other distributions increases, these distributions increasingly resemble the Normal Distribution.
THE CENTRAL LIMIT THEOREM
One of the most important applications for statistical purposes involving the Normal Distribution has to do with the distribution of averages. The averages of samples of a given size, taken such that each sampled item is selected independent of the others, will yield a distribution that is close to Normal. This is an extremely powerful fact, for it means that you can generalize about an actual random process from averages computed using sample data. Thus, we can state that if N random samples are drawn from a population, then the sums of the samples will be approximately Normally distributed, regardless of the distribution of the population from which the samples are drawn The closeness to the Normal Distribution improves as N increases. As an example, consider the distribution of numbers from 1 to 100.
This is what is known as a uniform distribution: all elements occur only once. The number 82 occurs once and only once, as does 19, and so on. Suppose now that we take a sample of five elements and we take the average of these five sampled elements. Now, we replace those five elements back into the population, and we take another sample and calculate the sample mean. If we keep on repeating this process, we will see that the sample means are Normally distributed, even though the population from which they are drawn is uniformly distributed. Furthermore, this is true regardless of how the population is distributed! The Central Limit Theorem allows us to treat the distribution of sample means as being Normal without having to know the distribution of the population. This is an enormously convenient fact for many areas of study.
If the population itself happens to be Normally distributed, then the distribution of sample means will be exactly Normal. This is true because how quickly the distribution of the sample means approaches the Normal, as N increases, is a function of how close the population is to Normal. As a general rule of thumb, if a population has a unimodal distribution-any type of distribution where there is a concentration of frequency around a single mode, and diminishing frequencies on either side of the mode (i.e., it is convex)-or is uniformly distributed, using a value of 20 for N is considered sufficient, and a value of 10 for N is considered probably sufficient. However, if the population is distributed according to the Exponential Distribution, then it may be necessary to use an N of 100 or so.
Figure The Exponential Distribution and the Normal.
The Central Limit Theorem, this amazingly simple and beautiful fact, validates the importance of the Normal Distribution.
WORKING WITH THE NORMAL DISTRIBUTION
In using the Normal Distribution, we most frequently want to find the percentage of area under the curve at a given point along the curve. In the parlance of calculus this would be called the integral of the function for the curve itself. Likewise, we could call the function for the curve itself the derivative of the function for the area under the curve. Derivatives are often noted with a prime after the variable for the function. Therefore, if we have a function, N(X), that represents the percentage of area under the curve at a given point, X, we can say that the derivative of this function, N'(X) (called N prime of X), is the function for the curve itself at point X.
We will begin with the formula for the curve itself, N'(X). This function is represented as:
N'(X) = 1/(S*(2*3.1415926536)^(1/2))*EXP(-((X-U)^2)/(2*S^2))
where,
U = The mean of the data.
S = The standard deviation of the data.
X = The observed data point.
EXP() = The exponential function.
This formula will give us the Y axis value, or the height of the curve if you Will, at any given X axis value. Often it is easier to refer to a point along the curve with reference to its X coordinate in terms of how many standard deviations it is away from the mean. Thus, a data point that was one standard deviation away from the mean would be said to be one standard unit from the mean. Further, it is often easier to subtract the mean from all of the data points, which has the effect of shifting the distribution so that it is centered over zero rather than over the mean.
Therefore, a data point that was one standard deviation to the right of the mean would now have a value of 1 on the X axis. When we make these conversions, subtracting the mean from the data points, then dividing the difference by the standard deviation of the data points, we are converting the distribution to what is called the standardized normal, which is the Normal Distribution with mean = 0 and variance = 1. Now, N'(Z) will give us the Y axis value (the height of the curve) for any value of Z:
N'(Z) = l/((2*3.1415926536)^(1/2))*EXP(-(Z^2/2))
= .398942*EXP(-(Z^2/2))
where,
Z = (X-U)/S and U = The mean of the data.
S = The standard deviation of the data.
X = The observed data point.
EXP() = The exponential function.
Equation gives us the number of standard units that the data point corresponds to-in other words, how many standard deviations away from the mean the data point is. When Equation equals 1, it is called the standard normal deviate. A standard deviation or a standard unit is sometimes referred to as a sigma. Thus, when someone speaks of an event being a "five sigma event," they are referring to an event whose probability of occurrence is the probability of being beyond five standard deviations.
Figure The Normal Probability density function.
Consider Figure, which shows this equation for the Normal curve. Notice that the height of the standard Normal curve is .39894. From Equation, the height is:
N'(Z) = .398942*EXP(-(Z^2/2))
N'(0) = .398942*EXP(-(0^2/2))
N'(0) = .398942
Notice that the curve is continuous-that is, there are no "breaks" in the curve as it runs from minus infinity on the left to positive infinity on the right. Notice also that the curve is symmetrical, the side to the right of the peak being the mirror image of the side to the left of the peak. Suppose we had a group of data where the mean of the data was 11 and the standard deviation of the group of data was 20. To see where a data point in that set would be located on the curve, we could first calculate it as a standard unit. Suppose the data point in question had a value of -9. To calculate how many standard units this is we first must subtract the mean from this data point:
-9 -11 = -20
Next we need to divide the result by the standard deviation:
-20/20 = -1
We can therefore say that the number of standard units is -1, when the data point equals -9, and the mean is 11, and the standard deviation is 20. In other words, we are one standard deviation away from the peak of the curve, the mean, and since this value is negative we know that it means we are one standard deviation to the left of the peak. To see where this places us on the curve itself (i.e., how high the curve is at one standard deviation left of center, or what the Y axis value of the curve is for a corresponding X axis value of -1), we need to now plug this into Equation :
N'(Z) = .398942*EXP(-(Z^2/2))
= .398942*2.7182818285^(-(-1^2/2))
= .398942*2.7182818285^(-1/2)
= .398942*.6065307
= .2419705705
Thus we can say that the height of the curve at X = -1 is .2419705705. The function N'(Z) is also often expressed as:
N'(Z) = EXP(-(Z^2/2))/((8*ATN(1))^(1/2)
= EXP(-(Z^2/2))/((8*.7853983)^(1/2)
= EXP(-(Z^2/2))/2.506629
where,
Z = (X-U)/S
and
ATN() = The arctangent function.
U = The mean of the data.
S = The standard deviation of the data.
X = The observed data point.
EXP() = The exponential function.
Nonstatisticians often find the concept of the standard deviation hard to envision. A remedy for this is to use what is known as the mean absolute deviation and convert it to and from the standard deviation in these equations. The mean absolute deviation is exactly what its name implies. The mean of the data is subtracted from each data point. The absolute values of each of these differences are then summed, and this sum is divided by the number of data points. What you end up with is the average distance each data point is away from the mean. The conversion for mean absolute deviation and standard deviation are given now:
Mean Absolute Deviation = S*((2/3.1415926536)^(1/2))
= S*.7978845609
where,
M = The mean absolute deviation.
S = The standard deviation.
Thus we can say that in the Normal Distribution, the mean absolute deviation equals the standard deviation times .7979. Likewise:
S = M*1/.7978845609
= M*1.253314137
where,
S = The standard deviation.
M = The mean absolute deviation.
So we can also say that in the Normal Distribution the standard deviation equals the mean absolute deviation times 1.2533. Since the variance is always the standard deviation squared, we can make the conversion between variance and mean absolute deviation.
M = V^(1/2)*((2/3.1415926536)^(1/2))
= V^(l/2)*.7978845609
where,
M = The mean absolute deviation.
V = The variance.
V = (M*1.253314137)^2
where,
V = The variance.
M = The mean absolute deviation.
Since the standard deviation in the standard normal curve equals 1, we can state that the mean absolute deviation in the standard normal curve equals .7979. Further, in a bell-shaped curve like the Normal, the semi-interquartile range equals approximately two-thirds of the standard deviation, and therefore the standard deviation equals about 1.5 times the semi-interquartile range. This is true of most bell-shaped distributions, not just the Normal, as are the conversions given for the mean absolute deviation and standard deviation.
NORMAL PROBABILITIES
We now know how to convert our raw data to standard units and how to form the curve N'(Z) itself (i.e., how to find the height of the curve, or Y coordinate for a given standard unit) as well as N'(X)To really use the Normal Probability Distribution though, we want to know what the probabilities of a certain outcome happening arc. This is not given by the height of the curve. Rather, the probabilities correspond to the area under the curve. These areas are given by the integral of this N'(Z) function which we have thus far studied. We will now concern ourselves with N(Z), the integral . to N'(Z), to find the areas under the curve (the probabilities).1
N(Z)=1-N'(Z)*((1.330274429*Y^5)-(1.821255978*Y^4)+(1.781477937*Y^3)-(.356563782*Y^2)+(.31938153*Y))
If Z<0 then N(Z) = 1-N(Z)
N'(Z) = .398942*EXP(-(Z^2/2))
where,
Y = 1/(1+2316419*ABS(Z))
and
ABS() = The absolute value function.
EXP() = The exponential function.
We will always convert our data to standard units when finding probabilities under the curve. That is, we will not describe an N(X) function, but rather we will use the N(Z) function where:
Z = (X-U)/S
and
U = The mean of the data.
S = The standard deviation of the data.
X = The observed data point.
Suppose we want to know what the probability is of an event not exceeding +2 standard units (Z = +2).
Y = 1/(1+2316419*ABS(+2))
= 1/1.4632838
= .68339443311
N'(Z) = .398942*EXP(-(+2^2/2))
= .398942*EXP(-2)
= .398942*.1353353
= .05399093525
Notice that this tells us the height of the curve at +2 standard units. Plugging these values for Y and N'(Z) into Equation we can obtain the probability of an event not exceeding +2 standard units:
N(Z)=1-N'(Z)*((1.330274429*Y^5)-(1.821255978*Y^4)+(1.781477937*Y^3)-(.356563782*Y^2)+(.31938153*Y))
=1-.05399093525*((1.330274429*.68339443311^5)-(1.821255978*.68339443311^4+1.781477937*.68339443311^3)-(.356563782*.68339443311^2)+(.31938153*.68339443311))
= 1-.05399093525*((1.330274429*.1490587)-
(1.821255978*.2181151+(1.781477937*.3191643)-(-
356563782*.467028+.31938153*.68339443311))
= 1-.05399093525*(.198288977-.3972434298+.5685841587-.16652527+.2182635596)
= 1-.05399093525*.4213679955
= 1-.02275005216
= .9772499478
Thus we can say that we can expect 97.72% of the outcomes in a Normally distributed random process to fall shy of +2 standard units.
Figure Equation showing probability with Z = +2.
If we wanted to know what the probabilities were for an event equaling or exceeding a prescribed number of standard units (in this case +2), we would simply amend equation, taking out the 1- in the beginning of the equation and doing away with the -Z provision (i.e., doing away with "If Z < 0 then N(Z) = 1-N(Z)"). Therefore, the second to last line in the last computation would be changed from
= 1-.02275005216 to simply .02275005216
We would therefore say that there is about a 2.275% chance that an event in a Normally distributed random process would equal or exceed +2 standard units.
Figure Doing away with the 1- and -Z provision in Equation .
Thus far we have looked at areas under the curve (probabilities) where we are only dealing with what are known as "1-tailed" probabilities. That is to say we have thus far looked to solve such questions as,"What are the probabilities of an event being less (more) than such-and-such standard units from the mean?" Suppose now we were to pose the question as, “What are the probabilities of an event being within so many standard units of the mean?" In other words, we wish to find out what the "e-tailed" probabilities are.
Figure A two-tailed probability of an event being+or-2 sigma.
This represents the probabilities of being within 2 standard units of the mean. Unlike Figure, this probability computation does not include the extreme left tail area, the area of less than -2 standard units. To calculate the probability of being within Z standard units of the mean, you must first calculate the I-tailed probability of the absolute value of Z. This will be yourinput to the next Equation, which gives us the 2-tailed probabilities (i.e., the probabilities of being within ABS(Z) standard units of the mean):
e-tailed probability = 1-((1-N(ABS(Z)))*2)
If we are considering what our probabilities of occurrence within 2 standard deviations are (Z = 2), then from Equation we know that N(2) = .9772499478, and using this as input to Equation :
2-tailed probability = 1-((1-.9772499478)*2)
= 1-(.02275005216* 2)
= 1-.04550010432
= .9544998957
Thus we can state from this equation that the probability of an event in a Normally distributed random process falling within 2 standard units of the mean is about 95.45%.
Figure Two-tailed probability of an event being beyond 2 sigma.
Just as with equation, we can eliminate the leading 1- in equation to obtain (1-N(ABS(Z)))*2, which represents the probabilities of an event falling outside of ABS(Z) standard units of the mean. For the example where Z = 2, we can state that the probabilities of an event in a Normally distributed random process falling outside of 2 standard units is:
2 tailed probability (outside) = (1-.9772499478)*2
= .02275005216*2
= .04550010432
Finally, we come to the case where we want to find what the probabilities (areas under the N'(Z) curve) are for two different values of Z.
Figure The area between -1 and +2 standard units.
Suppose we want to find the area under the N'(Z) curve between -1 standard unit and +2 standard units. There are a couple of ways to accomplish this. To begin with, we can compute the probability of not exceeding +2 standard units with equation, and from this we can subtract the probability of not exceeding -1 standard units. This would give us:
.9772499478-.1586552595
= .8185946883
Another way we could have performed this is to take the number 1, representing the entire area under the curve, and then subtract the sum of the probability of not exceeding -1 standard unit and the probability of exceeding 2 standard units:
= 1-(.022750052+.1586552595)
= 1 .1814053117
= .8185946883
With the basic mathematical tools regarding the Normal Distribution thus far, you can now use your powers of reasoning to figure any probabilities of occurrence for Normally distributed random variables.
1. This article really helpful and explained very well.So i am really thankful to you for sharing keep it up.
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By • 一月 17th, 2021
Displaying top 8 worksheets found for - Median Of A Histogram. Find the total number of items represented by the histogram 2. The pdf exercises are curated for students of grade 3 through grade 8. The Median is the value of the middle in your list. The total area of this histogram is $10 \times 25 + 12 \times 25 + 20 \times 25 + 8 \times 25 + 5 \times 25 = 55 \times 25 = 1375$. The median is the middle item or the average of the two middle items. Positive skewed histograms. The median is also the number that is halfway into the set. Let’s say, however, that you also want to publish the information on these weights for an audience that uses the imperial system rather than the metric system of measurements. Here, 1977 is used as the “base” year which is equal to 100. Central Tendencay Worksheet or Quiz Includes worksheets with and without number sets. {102, 109, 207, 357, 360, 403, 471, 483, 670, 729, 842, 843, 920, 941} Now, calculate the median M by finding the mean of 471 and 483. The median is the n/2 th value. The median is the middle value; uniformly spread data will provide that the area of the histogram on each side of the median will be equal. What is a Histogram? What does that mean 43 is the median of the frequencies, but it's not the median of the values. Starting with , add the frequencies in the table starting with the first row until you reach . Histogram Worksheet. Finding The Median Using Histograms - Displaying top 8 worksheets found for this concept.. 1. Batting average shows the percent (written as a decimal) of the time a certain player gets a hit. The median is the midpoint of the value, which means that at the median there are exactly half the data below and above that point. Again, the definition of the median for a continuous distribution is the value such that the cumulative probability is 1/2; we just multiplied N by that. Find the bin(s) containing the middle item(s). Determine the number of the middle item. In order to plot a cumulative frequency graph, we have to plot cumulative frequency against the upper-class-boundary of each class. We use linear interpolation to find it. If you know how many numbers there are in a set, which is the middle number? Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Center and Spread of Data Name_____ Date_____ Period____ ... Find the mode, median, mean, lower quartile, upper quartile, interquartile range, and population standard deviation for each data set. To find the median, the data should first be arranged in order from least to greatest. Drawing frequency polygon - with and without a histogram; Finding mean, median and mode of raw data . ; To estimate the Mean use the midpoints of the class intervals: . Now I want to see what happens when I add male heights into the histogram: To remember the definition of a median, just think of the median of a road, which is the middlemost part of the road. For each histogram bar, we start by multiplying the central x-value to the corresponding bar height. When the shape of the distribution is symmetric and unimodal, the mean, median, and mode are equal. using the formula for median we have, Median or where, (lower class boundary of the median class), (total frequency), ( less than type cumulative frequency corresponding to ), • To find the mean, add up all the numbers and divide by the number of numbers. A positive skewed histogram suggests the mean is greater than the median. The median class interval is the corresponding class where the median value falls. Please help. You can get both the mean and the median from the histogram. The histogram above shows a distribution of heights for a sample of college females. Given that the median value is 46, determine the missing frequencies using the median formula. This means you will have to convert your units into pounds from kilograms, multiplying each observation value in your data by 2.2 to get an approximate weight. Study each of these problems carefully; you will see similar problems on the lesson knowledge check. the median class is the class for which upper class boundary is . But in the histogram the hint is confusing me. finding median for ungrouped data Median is the value which occupies the middle position when all the observations are arranged in an ascending or descending order. the class containing the median value. You will need paper and a … For grouped data, we cannot find the exact Mean, Median and Mode, we can only give estimates. The mean, median and mode are three different ways of describing the average. Concept wise Collecting Data Raw Data Ungrouped Data Grouped Data Two sets of numbers that you work out for mean, median, mode, range, Q1, Q3, IQR, histogram… Median Worksheet 1. 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http://math.stackexchange.com/questions/607103/prove-sin2-theta-cos2-theta-1/607121 | 1,467,173,655,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397565.80/warc/CC-MAIN-20160624154957-00079-ip-10-164-35-72.ec2.internal.warc.gz | 199,029,850 | 28,474 | # Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
- | 4,436 | 13,690 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-26 | latest | en | 0.829055 |
https://blender.stackexchange.com/questions/90357/plotting-3d-surfaces | 1,708,552,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473558.16/warc/CC-MAIN-20240221202132-20240221232132-00116.warc.gz | 142,767,685 | 39,317 | # Plotting 3D Surfaces
Building on from my previous question, how can I plot something more complex than a sphere (and not as easy to split into a set of parametric equations)?
Say, for example
$$2x^2+y^2-5z^2+z-7x=16$$
Graphed on Desmos (use the z-slider to change the z value) for those who cannot as easily visualize the 3D surface.
• You could potentially use Python script to build a mesh by layers using successive values of Z. Sep 16, 2017 at 9:41
• Welcome to the site :) It looks like you have three separate accounts. If you would care to register one of them fully, we can merge them together and you'll be able to edit your question, accept answers, and do all those other fun things. Thanks Sep 16, 2017 at 22:24
## Discrete approach with Python
A possible approch that avoids solving the equation would consist in interating through all possible points in the 3D space, check whether or not in each one of them the equation is verified, and if it is create a vertex there.
As in the 3D space there are infinite points, we must discretize it and limits the boundaries of the investigation domain. For the sake of semplicity, I choose to consider, for example, only integer numbers from -100 to +100 for each axis. Here's the code for such iteration:
for x in range(-100,100):
for y in range(-100,100):
for z in range(-100,100):
Then, for each vector we'll calculate the left member of the equation and check if it's equal to the right member of the equation previusly defined. As we discretized the domain, it's best to give to the result a bit of "tollerance" in verifying the equation (that's why I'm using the "+/-toll" trick upon the right member verification).
function=2*x**2+y**2-5*z**2+z-7*x #function here
if((function<(goal+toll))and(function>(goal-toll))):
bm.verts.new([x,y,z])
Here's a possible code with all the lines needed to create an object in the scene:
import bpy
import bmesh
mesh = bpy.data.meshes.new("mesh")
obj = bpy.data.objects.new("Function", mesh)
scene = bpy.context.scene
scene.objects.active = obj
obj.select = True
mesh = bpy.context.object.data
bm = bmesh.new()
goal=16 #right member of the equation
toll=5 #threshold
for x in range(-100,100):
for y in range(-100,100):
for z in range(-100,100):
function=2*x**2+y**2-5*z**2+z-7*x #function here
if((function<(goal+toll))and(function>(goal-toll))):
bm.verts.new([x,y,z])
bm.to_mesh(mesh)
bm.free()
Examples of usage:
function=2*x**2+y**2-5*z**2+z-7*x
Note: as you can see, it takes a while to iterate through all the 8 000 000 positions...
This script only provides some of the vertices due to the approximation, but, with a proper resolution, you can obtain a good representation of the surface.
The meshing of the vertices into a surface needs it's own approach. This "brute force plotting" I'm proposing does not go well with the "on the fly" definitions of faces. The set of point is not structured in any way. You'll need something to mesh the pointcloud.
Here's another version of the code where you can better control the domain size and resolution using floats to inspect the region closer to the origin in high definition:
import bpy
import bmesh
max=7
min=-7
prec=10 #precision of the calc (1=1 units, 10 = 0.1 units, 100 = 0.01 units...)
tol=0.5 #tollerance
def f(x, y, z): return 2*x**2+y**2-5*z**2+z-7*x -16 #the function = 0
mesh = bpy.data.meshes.new("mesh")
obj = bpy.data.objects.new("XYZ Function", mesh)
scene = bpy.context.scene
scene.objects.active = obj
obj.select = True
mesh = bpy.context.object.data
bm = bmesh.new()
for x in [float(j)/prec for j in range(min*prec, max*prec+1,1)]:
for y in [float(j)/prec for j in range(min*prec, max*prec+1,1)]:
for z in [float(j)/prec for j in range(min*prec, max*prec+1,1)]:
if(abs(f(x,y,z))<tol):
bm.verts.new([x,y,z])
bm.to_mesh(mesh)
bm.free()
Here's a possible result:
You can for example export points to Meshlab and take advantage of the built in tools. Here's an example using the previusly obtained set of points.
As far as I know, there isn't an add-on or method that can graph more complex mathematical functions in the way you want. As an alternative, you can put your equation into nodes and then render it with a Volume Scatter shader.
Begin with the default cube. Add a new material, and add a Texture Coordinate node (Shift + A > Input > Texture Coordinate) and Separate XYZ node (Shift + A > Converter > Separate XYZ):
Then, create your equation using Math nodes (Shift + A > Converter > Math). Here, I created your example equation:
Now, add a Color Mix node (Shift + A > Color > MixRGB) and set it to Multiply. Plug the texture coordinates in to one slot, and a Value node in to the other. This will scale our window:
Finally, multiply the output by 10 to increase the value, then plug it into the Density of a Volume Scatter and a Volume Absorption shader added together:
This will nicely display the function when rendered, though the shading isn't ideal:
click for larger image
To optimize rendering, change the following settings:
Although it doesn't give you a workable mesh, this can definitely give you a renderable graph.
We can also improve the look of it by coloring it:
click for larger image
• Need 50 rep to comment... @ScottMilner I was hoping there was a way to preview in viewport directly (not viewport shading: rendered, but viewport shading: solid). The less CPU power, the better I am actually looking only to visualize the shape (so quality is not important. Quantity is)
– math
Sep 16, 2017 at 5:25
The equation in the form in the question is a quadric surface. By completing the square we can translate the axis and reduce to form
(x-7/4)² - 49/16 + 1/2y² - 5/2(z - 1/10)² + 5/200 - 8 = 0
or
X² + 0.5Y² - 2.5Z² - - 11.0375 = 0
A sample of the surface created by iterating through z (an xy plane) then finding the 0, 1 or 2 roots for each y using the quadratic formula to solve for x. Notice the missing faces where x=0. Can run thru solving for y and have missing faces around y=0. Overlaying the two gives the shape, but not a pretty mesh.
I've put in both the original equation f(x, y, z) and the translated form f2(x, y, z).
import bpy
import bmesh
from math import sqrt
def f(x, y, z):
return 2 * x * x + y * y - 5 * z * z + z -7 * x -16
# f transformed to have origin (0, 0, 0)
def f2(x, y, z):
x * x + 0.5 * y * y - 2.5 * z * z - 11.0375
axis = -b / (2 * a)
check = b * b - 4 * a * c
if abs(check) < 0.0001:
return True, axis, axis
if check < 0:
return False, None, None
dist = sqrt(check) / (2 * a)
return True, axis - dist, axis + dist
bm = bmesh.new()
dom = [m for m in range(-10, 11)]
use_f2 = False # use f2()
planes = []
# coefficents of x
if use_f2:
a, b = 1, 0 # f2()
else:
a, b = 2, -7 # f()
for z in dom:
# for each xy plane
neg_line = []
pos_line = []
for y in dom:
# for each line in plane
if use_f2:
# constant f2()
c = 0.5 * y * y - 2.5 * z * z -11.0375
else:
# constant f()
c = y * y - 5 * z * z + z -16
# find roots
has_solution, neg_root, pos_root = quad_root(a, b, c)
if has_solution:
neg_root = bm.verts.new((neg_root, y, z))
pos_root = bm.verts.new((pos_root, y, z))
pos_line.append(pos_root)
neg_line.append(neg_root)
planes.append([neg_line, pos_line])
# skin it
if len(planes) > 1:
for i in range(len(planes) - 1):
n0, p0 = planes[i]
n1, p1 = planes[i + 1]
# pos
for j in range(len(n0) - 1):
verts = [n0[j], n1[j], n1[j+1], n0[j + 1]]
if None not in verts:
f = bm.faces.new(verts)
verts = [p0[j], p1[j], p1[j+1], p0[j + 1]]
if None not in verts:
f = bm.faces.new(verts)
# make an object
scene = bpy.context.scene
bm.to_mesh(mesh)
scene.objects.active = ob
ob.scale *= 0.01
To solve for y for f(x, y, z) edit
a, b = 1, 0
c = 2 * x * x - 5 * z * z + z -7 * x - 16
has_solution, neg_root, pos_root = quad_root(a, b, c)
if has_solution:
#print(neg_root, pos_root)
neg_root = bm.verts.new((x, neg_root, z))
pos_root = bm.verts.new((x, pos_root, z))
Result after running in both x and y and joining | 2,340 | 8,011 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-10 | latest | en | 0.896606 |
http://www.mathisfunforum.com/viewtopic.php?pid=235147 | 1,394,366,304,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999677605/warc/CC-MAIN-20140305060757-00049-ip-10-183-142-35.ec2.internal.warc.gz | 434,012,701 | 9,655 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #1 2012-10-13 22:45:05
Agnishom
Real Member
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### Pseudo-Force
Can any one explain the concept of Pseudo-Force to me in simple words?
Please also show how it is different from the real force
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'The whole person changes, why can't a habit?' -Alokananda
## #2 2012-10-13 23:42:42
bob bundy
Moderator
Offline
### Re: Pseudo-Force
hi Agnishom,
Wiki gives three examples at
http://en.wikipedia.org/wiki/Fictitious_force
Of these, the ones I have spent most time arguing against is centrifugal force, so I'll use that one here.
Everyone knows that as you go round a bend in a car you feel as though you are being thrown outwards.
You may also have tied a conker to a string and spun it around in a circle. You can 'feel' the string tugging you.
But to call this centrifugal force, is to stand the mechanics on its head.
Any object that is travelling in a circular path must have a force acting on it to cause this to happen
Newton's first and second laws at http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion
There's no string here; the satellite is being pulled towards the Earth by gravity.
To travel in a circle requires a centrally acting force and gravity provides this. The satellite is said to be in 'free fall' as it is constantly falling towards the centre of the Earth. Imagine it travelling along a tangental line plus falling inwards a small amount; the amount it falls in exactly cancels with the curvature of the circle, meaning it stays at the same distance from the Earth.
For the conker the tension in the string provides the force, and for the car it is friction between the tyres and the road. That makes the car go in a circle (unless the friction is insufficient in which case the car continues in a straight line) . Inside the car something has got to make you go in a circle too. If you're a passenger in the back with no seat belt, you feel as though you are being flung sideways, but actually you are just continuing to go straight while the car bends round. When the side of the car hits you it makes you go in a circle too.
Because of what it feels like, some people have invented a fictious force called centrifugal force that seems to be throwing you outwards. There's an easy test to show this isn't really what happens. When a car loses traction which way does it travel.
Michael Schumacher seems to be constantly demonstrating Newton's laws. No centrifugal force here!
When I was your age I had many happy arguments with friends whom I called 'centrifugalists'. We tried to devise an experiment in which someone spun round in a circle holding on to a rope with a weight at the end. At a key moment someone would cut the string. Now what would happen? Would the weight travel out radially (centrifugalist argument) or along a tangent (Newtonian argument). We never actually tried it; just argued about it!
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #3 2012-10-14 01:56:43
Agnishom
Real Member
Offline
### Re: Pseudo-Force
Hmm..
How is it different from the non-pseudo forces?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'The whole person changes, why can't a habit?' -Alokananda
## #4 2012-10-14 04:51:57
bob bundy
Moderator
Offline
### Re: Pseudo-Force
Well I'd say they are non-existent forces. Some people find it helps to do certain questions by making up such a force.
eg. a conker is spun around in a circle.
in 'equilibrium'
tension in string = centrifugal force acting out from the centre.
eg2.
When an object is rotating around the Earth, an observer sees it move in a frame of reference which is itself rotating. As I understand it you can get an equation that accounts for this apparent extra motion by inventing a coriolis force on the object. Sorry that is a bit 'woffly' but I'm not good on coriolis forces.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #5 2012-10-14 22:35:42
Agnishom
Real Member
Offline
### Re: Pseudo-Force
When I was your age I had many happy arguments with friends whom I called 'centrifugalists'. We tried to devise an experiment in which someone spun round in a circle holding on to a rope with a weight at the end. At a key moment someone would cut the string. Now what would happen? Would the weight travel out radially (centrifugalist argument) or along a tangent (Newtonian argument). We never actually tried it; just argued about it!
What should happen in reality?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'The whole person changes, why can't a habit?' -Alokananda
## #6 2012-10-14 23:33:08
bob bundy
Moderator
Offline
### Re: Pseudo-Force
Well the rope has a real force; the tension; and that is what pulls the weight into a circular orbit.
By Newton's first law, an object will continue with uniform motion (ie. go in a straight line) unless acted upon by an external force (that's the tension).
If you cut the rope there's suddenly no tension. So the weight should carry on in a straight line ie. along a tangent to the circle.
[My centrifugalist friends argued that there is an outward acting force on the weight that balances the tension. If you take away the tension the weight should fly out on a radial line. Because that force isn't really there it is called a pseudo force. Time and time again I would tell them that there is no equilibrium because the weight is always accelerating towards the centre; so no 'balancing force' is necessary.]
The trouble is you can invent centrifugal force and get all the right answers. So it may seem not to matter. It's a bit like the proving argument; is it Ok to do something back to front, if you get the right answer. The Ptolemaic system of epi-cycles was used to predict the position of planets in the 'fixed star background'. Then Copernicus suggested a Sun centred system and it has been used ever since.
Is it a better way to do things? ... Discuss.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #7 2012-10-15 00:09:20
Agnishom
Real Member
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### Re: Pseudo-Force
Because that force isn't really there it is called a pseudo force.
That force doesn't exist??!!
Then, what makes it feel so???
Is it a better way to do things? ... Discuss.
Excuse me, Which way are you talking about?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'The whole person changes, why can't a habit?' -Alokananda
## #8 2012-10-15 05:33:51
bob bundy
Moderator
Offline
### Re: Pseudo-Force
hi Agnishom,
That force doesn't exist??!!
Then, what makes it feel so???
Our perception often plays tricks on us. You have probably met lots of visual tricks where lines look longer than they are or bent when they are straight, etc.
So when have you experienced a force that isn't really there.
Picture a situation where you are in a car, or a roller coaster or similar, experiencing a high acceleration.
There must be a force acting here. Something is driving your seat forwards with mounting speed. But what do you feel? You seem to be being pushed back in your seat as if an unseen force is pressing against you. Think about it. There cannot be such a force but try telling your brain that.
What is happening is the seat is trying to overtake you but clearly it cannot. So it is pressing against you, trying to speed you up. But your brain interprets what is happening from your personal perspective. Your brain doesn't 'see' the world from the position of the seat; it tells you that you are being forced into the seat. 'Forced'; by what? It's another pseudo force.
Analysis of the maths helps us to understand what is really happening, independent from our personal world of perception. That's why I've tried to give you some experimental evidence that centrifugal force isn't really there.
Is it a better way to do things? ... Discuss.
Excuse me, Which way are you talking about?
I believe in the scientific method. Observe; try to find a theory that fits the facts; and then test it. That's my 'better way'. But I was only joking in suggesting that you discuss this. You can if you want; but it is optional.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #9 2012-10-15 15:17:53
Agnishom
Real Member
Offline
### Re: Pseudo-Force
There must be a force acting here. Something is driving your seat forwards with mounting speed. But what do you feel? You seem to be being pushed back in your seat as if an unseen force is pressing against you. Think about it. There cannot be such a force but try telling your brain that.
Ofcourse, there is a force of Inertia acting.
And.... If the centrifugal force is unreal then where does the tension on the string come from?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'The whole person changes, why can't a habit?' -Alokananda
## #10 2012-10-15 18:10:28
bob bundy
Moderator
Offline
### Re: Pseudo-Force
Inertia is not a force.
If you are holding the string then you pull it. So your muscles provide the force.
Bob
ps. When I've woken up properly I'll show you how to derive the formula for central acceleration.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #11 2012-10-15 21:19:51
bob bundy
Moderator
Offline
### Re: Pseudo-Force
Proof that any object rotating in a circle must be continuously accelerating towards the centre.
Preliminary: By Newton's first law, an object will continue with uniform motion (ie. fixed speed and direction) unless acted upon by an externally impressed force.
This means that to make an object go in a circle there must be a force to cause the change.
I need to establish some formulas before the main proof.
Look first at my table of values for small angles, the cosine and sine of the angle. I am working in radians not degrees.
You can see that as the angle gets smaller the cosine of the angle approaches 1 and the sine of the angle approaches the same value as the angle itself.
screen shot 2 shows the diagram to prove this.
As the angle gets smaller, AC gets closer and closer in length to arcAB so
Now for the main proof. See the third screen shot. This shows an object rotating in a circle. It has speed v. Consider the position after it has moved around through a small angle.
The change in velocity along the line of the tangent is
and along the line of the radius is
This change takes place in a small period of time.
So the time for one revolution is
Now to calculate the two accelerations. I'll do the tangential one first.
The part in brackets tends to 1 so the whole tends to zero as the angle tends to zero.
So there is no tangential acceleration.
The minus sign means it acts towards the centre of the circle and is equal to v^2/r in size.
So, if an object has to accelerate towards the centre for circular motion, there has to be a centrally acting force to provide it (Newton's second law).
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #12 2012-10-16 23:56:47
Agnishom
Real Member
Offline
### Re: Pseudo-Force
Er, I am not yet so well versed at Trig but I will try to read this
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'The whole person changes, why can't a habit?' -Alokananda
## #13 2012-10-17 04:12:58
bob bundy
Moderator
Offline
### Re: Pseudo-Force
OK. Ask if anything needs more clarification.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #14 2012-10-19 01:36:40
Agnishom
Real Member
Offline
### Re: Pseudo-Force
Sorry for the delay
You said:
cos2x = 1 - 2sin^2x
How?
I am going through the rest of the proof and I am not done yet
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'The whole person changes, why can't a habit?' -Alokananda
## #15 2012-10-19 05:09:23
bob bundy
Moderator
Offline
### Re: Pseudo-Force
It comes from the compound angle formula:
Put A = B = x and this becomes
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei | 3,401 | 13,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2014-10 | longest | en | 0.909953 |
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# At an amusement park, tom bought a number of red tokens and
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At an amusement park, tom bought a number of red tokens and [#permalink]
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30 Jan 2012, 21:06
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At an amusement park, tom bought a number of red tokens and green tokens. Each red token costs $0.09, and each green token costs$0.14. If Tom spent a total of exactly $2.06, how many token in total did Tom buy? A. 16 B. 17 C. 18 D. 19 E. 20 This is a tough one. I am having trouble finding a fast solution for this. ##### Most Helpful Expert Reply SVP Status: Top MBA Admissions Consultant Joined: 24 Jul 2011 Posts: 1917 GMAT 1: 780 Q51 V48 GRE 1: Q800 V740 Re: Amusement Park Tokens [#permalink] ### Show Tags 31 Jan 2012, 00:20 9 7 0.09x + 0.14y = 2.06 => 9x + 14y = 206 To solve this remember that x must be even because 14y, when subtracted from 206, will yield an even number (even - even = even). The solution comes out to be x=12, y=7. Therefore the total number of tokens bought = 12+7 = 19 Option (D). _________________ GyanOne [www.gyanone.com]| Premium MBA and MiM Admissions Consulting Awesome Work | Honest Advise | Outstanding Results Reach Out, Lets chat! Email: info at gyanone dot com | +91 98998 31738 | Skype: gyanone.services ##### Most Helpful Community Reply Intern Joined: 03 Mar 2014 Posts: 2 Concentration: Entrepreneurship, General Management GMAT 1: 730 Q48 V41 GPA: 3.04 Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 27 Apr 2014, 00:01 4 9 I solved it this way: Starting with the equation 9X+14Y = 206 => 9(X+Y) + 5Y = 206 5Y = 206 - 9(X+Y) we need to find X+Y. The RHS has to be a multiple of 5 Substituting the answers for X+Y abive, only 19 gives a multiple of 5. You don't need to actually multiply all the answers with 9, just look for the units digit of the difference. (it has to be either 5 or 0) When 19 is substituted, we get a units digit of 5 in the difference. So D. ##### General Discussion Math Expert Joined: 02 Sep 2009 Posts: 60627 At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 31 Jan 2012, 01:33 2 1 calreg11 wrote: At an amusement park, tom bought a number of red tokens and green tokens. Each red token costs$0.09, and each green token costs $0.14. If Tom spent a total of exactly$2.06, how many token in total did Tom buy?
a. 16
b. 17
c. 18
d. 19
e. 20
This is a tough one. I am having trouble finding a fast solution for this.
Given:
0.09R + 0.14G = 2.06;
9R + 14G = 206.
Now, it's special type of equations as G and R must be a non-negative integers, so there might be only one solution to it. After some trial and error you'll get (actually there are several ways of doing it):
R = 12 and G = 7;
R + G = 19.
For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html
Hope it helps.
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Re: At an amusement park, tom bought a number of red tokens and [#permalink]
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31 Jan 2012, 01:59
I found this question not hard, but time -consuming. it took some time to find x=12 y=7
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Re: At an amusement park, tom bought a number of red tokens and [#permalink]
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31 Jan 2012, 21:45
Is there any way to solve for it other than trial and error?
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Re: At an amusement park, tom bought a number of red tokens and [#permalink]
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01 Feb 2012, 02:40
2
calreg11 wrote:
Is there any way to solve for it other than trial and error?
Check out case 2 in this post. It explains you in detail how to deal with such questions. I don't think there are pure algebraic solutions to such problems.
http://www.veritasprep.com/blog/2011/06 ... -of-thumb/
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Re: At an amusement park, tom bought a number of red tokens and [#permalink]
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01 Feb 2012, 02:55
calreg11 wrote:
Is there any way to solve for it other than trial and error?
Trial and error along with some common sense is pretty much the only way you should approach such kind of problems on the GMAT. You won't get some very tough numbers to manipulate with or there will be some shortcut available, based on multiples concept or on the answer choices. So generally you would have to try just couple of values to get the answer.
Check the links in my previous post to practice similar problems and you'll see that getting the answer is not that hard for the realistic GMAt question.
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Re: At an amusement park, tom bought a number of red tokens and [#permalink]
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01 Feb 2012, 20:33
Thanks for the posting.
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Re: At an amusement park, tom bought a number of red tokens and [#permalink]
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27 Feb 2012, 00:47
3
1
Bunuel wrote:
calreg11 wrote:
Is there any way to solve for it other than trial and error?
Trial and error along with some common sense is pretty much the only way you should approach such kind of problems on the GMAT. You won't get some very tough numbers to manipulate with or there will be some shortcut available, based on multiples concept or on the answer choices. So generally you would have to try just couple of values to get the answer.
Check the links in my previous post to practice similar problems and you'll see that getting the answer is not that hard for the realistic GMAt question.
Hi calreg11,
supporting the explanations of Bunuel and karishma above, I can show you shortest way of solving it by some amalgamation of trial and error with the algebraic approach, though, as mentioned by karishma, there is no pure algebraic solution of this problem. Let's start:
first, we have to formulate the premise in an algebraic way through expressing red and green tokens by any letters we think convenient to us--> assuming, e.g., red tokens as 'r' and green tokens as 'g';
secondly, for the sake of convenience we can take the prices of red and greem tokens and also the total cost in cents, i.e., $0,09 as 9 cents,$0.14 as 14 cents, and $2.06 as 206 cents; then, thirdly, we do formulate it --> 9r + 14g = 206; fourth, now we can refer to the point that red tokens and green tokens make up the total number of tokens which is unknown to us and this is why formula can be --> r + g = x fifth, we have to apply trial and error approach through replacing x by each answer choice and we do it this way: r + g = 16 r = 16 - g replace 'r' in the original formula --> 9r + 14g = 206and we get 9(16-g) + 14g = 206 --> 5g=62 --> 62 is not divisible by 5, and hence, we cannot derive the number of g (green tokens), consequently, that of red tokens' also. only 19 can satisfy the condition drawn from the formulae r = x - g and 9(x-g) + 14g = 206 Hope, it helps! SVP Joined: 06 Sep 2013 Posts: 1522 Concentration: Finance Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 26 Dec 2013, 15:29 1 3 calreg11 wrote: At an amusement park, tom bought a number of red tokens and green tokens. Each red token costs$0.09, and each green token costs $0.14. If Tom spent a total of exactly$2.06, how many token in total did Tom buy?
a. 16
b. 17
c. 18
d. 19
e. 20
This is a tough one. I am having trouble finding a fast solution for this.
What I like to do in this questions is the following
We have 9x + 14y = 206
First always try to simplify, in this case we can't
Now look for a number that is the same for both and will be close to 206
In this case 9 is our best choice (You can quickly ballpark with 10 but you will realize it is >206)
So with 9 for both x and y we get 207 which is one more. Now the fun part starts
We need to play with this 9,9 combination to try to get one less, How so?
Well, let see we need to be one lower so if we get rid of one 14 and add one 9 we be further down.
If we subtract to 14's though we are down 28 and if we add 3 9's we are up 27
That perfect just to match our +1 difference!
So in total we have 12+7 = 19
Hence our correct answer is D
Hope it helps
Cheers!
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Re: At an amusement park, tom bought a number of red tokens and [#permalink]
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20 Jan 2015, 09:51
PROBLEM:
(7 pairs)(23¢)=$1.61;$2.06-$1.61=45¢ 5 red tokens at 9¢@ total tokens=7+5=12 red +7 green=19 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 15971 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 14 Jan 2018, 12:50 Hi All, If you don't immediately see an 'elegant' approach to solving this problem, then you can still solve it relatively quickly with some 'brute force' and a bit of arithmetic. From the answer choices, you can see that the total number of coins is no more than 20, so there aren't that many potential calculations that you would have to do to find the exact number of each type of coins that 'fit' this situation. In basic terms, we're told that a certain number of .09s + a certain number of .14s total 2.06.... There are some Number Property rules that we can use to save some time: .14 multiplied by an integer will end in an EVEN digit 2.06 ends in an even digit Since (even) + (even) = (even), .09 multiplied by an integer MUST end in an EVEN digit for the sum to equal 2.06 The number of red tokens MUST be EVEN, so that significantly cuts down the number of options to consider.... IF... we have.... 2 red tokens, then the remaining value is$1.88. Can that be evenly divided by .14? Try it... (the answer is NO).
4 red tokens, then the remaining value is $1.70. Can that be evenly divided by .14? Try it... (the answer is NO). 6 red tokens, then the remaining value is$1.52. Can that be evenly divided by .14? Try it... (the answer is NO).
8 red tokens, then the remaining value is $1.34. Can that be evenly divided by .14? Try it... (the answer is NO). 10 red tokens, then the remaining value is$1.16. Can that be evenly divided by .14? Try it... (the answer is NO).
12 red tokens, then the remaining value is \$0.98. Can that be evenly divided by .14? Try it... (the answer is YES and the remaining 7 tokens are green).
Thus, the total number of tokens is 12 red + 7 green = 19 total tokens
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Re: At an amusement park, tom bought a number of red tokens and [#permalink]
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29 Aug 2018, 11:53
First thing I noticed is the difference between 9 and 14 is 5; so If you buy all the tokens (lets say x) at 9 cents then 206-9x should be divisible by 5.
we can start with x=20 and we get 206-9*20 = 26 .. since this is not divisible by 5 20 is not the answer... but if we decrease 20 to 19 we can add 9 to 26 and since 35 is divisible by 5, 19 is the answer!!
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https://www.physicsforums.com/threads/how-to-get-gravitational-force-on-a-gaseous-particle.1002485/ | 1,726,321,374,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.38/warc/CC-MAIN-20240914125424-20240914155424-00587.warc.gz | 867,182,760 | 28,235 | # How to get gravitational force on a gaseous particle?
• vcsharp2003
In summary: There is a standard pair of results you should know.Inside a uniform spherical shell of mass there is no net gravitational force; they all cancel exactly.Outside a uniform spherical shell of mass the net gravitational force is the same as if the shell were collapsed to a single point at its centre.The same pair of results applies in electrostatics, since that is also an inverse square law.
vcsharp2003
Homework Statement
Consider a spherical gaseous cloud of mass density ## \rho (r)## in free space where ##r## is the radial distance from its center. The gaseous cloud is made of particles of equal mass ##m## moving in circular orbits about the common center with the same kinetic energy ##K## . The force acting on the particles is their mutual gravitational force. Mass density is constant in time. If the particle number density is ##n (r ) = \dfrac {\rho (r)} {m}##, then determine it in terms of ## \text {m, K and r}##?
Relevant Equations
##F_g = \dfrac {GMm} {r^2} ##
## F_c = \dfrac {mv^2} {r} ##
## KE = \dfrac {mv^2} {2} ##
This question is very confusing since I don't see two distinct particles that are exerting a gravitational force on each other. Also to complicate matters, a gas is made of many individual particles and I don't know how to determine the gravitational force on a single particle from so many other gaseous particles.
If someone can give me any hint on how to get the gravitational force on a single gas particle from so many other particles then that would help.
I did consider the following analogy, but it's confusing. The spherical gaseous cloud can be considered like a spherical earth. Then, gravitational force between any outside particle and the gaseous cloud would be like the force between Earth and and an external object. But in our case we have no external particle as the particle is part of the spherical gaseous cloud.
The question clearly states that "The force acting on the particles is their mutual gravitational force", but then what should be the big mass M that we should use to mimic a spherical Earth pulling an external object. We know the small mass m is known and its the mass of a particle in circular motion.
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Gravitation potential ##\phi(r)## generated by spherical distribution of mass ##\rho(r)## is
$$\phi(r)=-G\int \frac{\rho(r')}{|r-r'|}dV'$$
After some calculation balancing it with centrifugal potential would work.
anuttarasammyak said:
Gravitation potential ##\phi(r)## generated by spherical distribution of mass ##\rho(r)## is
$$\phi(r)=-G\int \frac{\rho(r')}{|r-r'|}dV'$$
After some calculation balancing it with centrifugal potential would work.
What is ## r ## and ##r^\prime## in your above equation?
##r## is position where gravitation potential is considered.
##r'## is position vector of volume element ##dV'## which should be integrated in whole space.
anuttarasammyak said:
##r## is position where gravitation potential is considered.
##r'## is position vector of volume element ##dV'## which should be integrated in whole space.
Is it possible to get gravitational force on a single particle at a distance ## r ## from center of gaseous cloud? I think that would be a good starting point for me since the equation you mentioned is not there in the course material that I am using.
I see. Gravitation potential of a particle whose mass is m
$$\phi(r)=-G\frac{m}{r}$$
Have you got it?
anuttarasammyak said:
I see. Gravitation potential of a particle whose mass is m
$$\phi(r)=-G\frac{m}{r}$$
Have you got it?
I get that part but the question seems to be mentioning circular motion and gravitational force, so I would be inclined to think that the question may have something to do with centripetal force and gravitational force, and not gravitational potential.
vcsharp2003 said:
I get that part but the question seems to be mentioning circular motion and gravitational force, so I would be inclined to think that the question may have something to do with centripetal force and gravitational force, and not gravitational potential.
You are right in thinking that you should treat it much like a particle inside Earth.
Consider the cloud as made of spherical shells. For a given particle at radius r, it will be outside some shells and inside others. What is the net gravitational force on the particle from one such shell?
vcsharp2003
haruspex said:
You are right in thinking that you should treat it much like a particle inside Earth.
Consider the cloud as made of spherical shells. For a given particle at radius r, it will be outside some shells and inside others. What is the net gravitational force on the particle from one such shell?
We have two cases here. First case is when particle is inside the shell and second case is when particle is on or outside the shell.
In first case, the inside particle will experience forces in all directions from each particle in the shell. In second case too forces will act on the particle in all directions. Not too sure how to take it forward using a shell approach? Unless we treat each spherical shell like the Earth with the mass of shell being concentrated in a particle at center of sphere.
vcsharp2003 said:
We have two cases here. First case is when particle is inside the shell and second case is when particle is on or outside the shell.
In first case, the inside particle will experience forces in all directions from each particle in the shell. In second case too forces will act on the particle in all directions. Not too sure how to take it forward using a shell approach?
There is a standard pair of results you should know.
Inside a uniform spherical shell of mass there is no net gravitational force; they all cancel exactly.
Outside a uniform spherical shell of mass the net gravitational force is the same as if the shell were collapsed to a single point at its centre.
The same pair of results applies in electrostatics, since that is also an inverse square law.
vcsharp2003
haruspex said:
Outside a uniform spherical shell of mass the net gravitational force is the same as if the shell were collapsed to a single point at its centre.
The same pair of results applies in electrostatics, since that is also an inverse square law.
So, let's take a shell at a distance of ##r## from the center of sphere that has a thickness ##dr## and a mass ##dm##. Each particle P within this shell is moving in a circular path about the center of sphere.
Then, according to what you mentioned, the shells outer to this shell exert no gravitational force on the particle P since now the particle P is inside all outer shells.
But each inner shell to particle P's shell would exert a force on particle P, since P is now outside each of these inner shells. The mass M would then be the total mass of these inner shells and it would act as a collapsed particle at sphere's center with a mass M.
Now I need to setup some equation(s).
Let M be the total mass of all shells that are inner to particle P's shell. Since the kinetic energy is same for all particles, so for particle P
$$K = \dfrac {mv^2} {2}$$
Also, since the inner shells are collapsed to a particle at center with mass M, the distance between particle P and the collapsed particle is ##r##. So, following equation is the second equation we have where centripetal force is being supplied by the gravitational force for particle P.
$$\dfrac {GMm} {r^2} = \dfrac {mv^2} {r}$$
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vcsharp2003 said:
So, let's take a shell at a distance of ##r## from the center of sphere that has a thickness ##dr## and a mass ##dm##. Each particle P within this shell is moving in a circular path about the center of sphere.
Then, according to what you mentioned, the shells outer to this shell exert no gravitational force on the particle P since now the particle P is inside all outer shells.
But each inner shell to particle P's shell would exert a force on particle P, since P is now outside each of these inner shells. The mass M would then be the total mass of these inner shells and it would act as a collapsed particle at sphere's center with a mass M.
Now I need to setup some equation(s).
Let M be the total mass of all shells that are inner to particle P's shell. Since the kinetic energy is same for all particles mo, so for particle P
$$K = \dfrac {mv^2} {2}$$
Also, since the inner shells are collapsed to a particle at center with mass M, the distance between particle P and the collapsed particle is ##r##. So, following equation is the second equation we have where centripetal force is being supplied by the gravitational force.
$$\dfrac {GMm} {r^2} = \dfrac {mv^2} {r}$$
Looks good. And what is the relationship between M and r?
vcsharp2003
haruspex said:
good. And what is the relationship between M and r?
We get ## 2K = mv^2## from the first equation, which we substitute in second equation to get $$\dfrac {GMm} {r^2} = \dfrac {2K} {r}$$
$$M= \dfrac {2Kr} {Gm}$$
The above equation is M in terms of r. Note that K and m are known quantities as given in problem statement.
vcsharp2003 said:
We get ## 2K = mv^2## from the first equation, which we substitute in second equation to get $$\dfrac {GMm} {r^2} = \dfrac {2K} {r}$$
$$M= \dfrac {2Kr} {Gm}$$
The above equation is M in terms of r. Note that K and m are known quantities as given in problem statement.
Yes, but there's also the relationship via ##\rho(r)##.
vcsharp2003
haruspex said:
Yes, but there's also the relationship via ρ(r).
We can look at an inner shell whose thickness is ##dr##. Volume of this infinitesimal shell is ##(4 \pi r^2) dr##. Since mass density for this shell is ##\rho(r)##, so we can say ##\text {mass of this shell = density x volume i.e}## ##dM =\rho(r) \times 4 \pi r^2dr##. Integrating both sides with limits from #r=0 to r= r, we can get the total mass M of all inner shells.
But integrating right hand side gets difficult since we do not know ##\rho(r)## i.e we have no idea of what is ##\rho## in terms of ##r##.
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vcsharp2003 said:
integrating right hand side gets difficult since we do not know ρ(r)
So what can you do instead of integrating the RHS?
vcsharp2003
haruspex said:
So what can you do instead of integrating the RHS?
Perhaps, we can integrate and get M = some definite integral. Then compare the above definite integral with expression of M obtained in earlier equation and decide the expression for ##\rho (r)##.
Or we can get ##\dfrac {dM} {dr} = \rho (r) \times 4\pi r^2## instead of taking a definite integral.
vcsharp2003 said:
Or we can get ##\dfrac {dM} {dr} = \rho (r) \times 4\pi r^2## instead of taking a definite integral.
Try that.
vcsharp2003
haruspex said:
Try that.
We have the rate of change of mass M of sphere with respect to its radius r as ##\dfrac {dM} {dr} = \rho (r) \times 4\pi r^2## as derived in post#15.
Also, using the already derived equation for mass of sphere M in post#14, ##M= \dfrac {2Kr} {Gm}## and differentiating both sides of this equation we get ## \dfrac {dM} {dr} = \dfrac {2K} {Gm} ## ( note that we treat gravitational constant ##G##, mass of each gaseous particle ##m## and kinetic energy of each gaseous particle ##K## as constants since these quantities do not depend on the radius of sphere ##r##.
Now, we can equate ## \dfrac {dM} {dr}## obtained in above two different manners, so that ## \rho (r) \times 4\pi r^2 = \dfrac {2K} {Gm} ##, which gives us the function ## \rho (r) = \dfrac {K} {2\pi r^2Gm} ##. Therefore, ## n(r) = \dfrac {\rho(r)} {m} = \dfrac {K} {2\pi r^2Gm^2} ## and this is the answer to the question.
(Kinetic energy of a gaseous particle at all distances from center of spherical cloud is the same according to the problem and its value is ##K##. Normally, I was thinking that kinetic energy of a gaseous particle would change as distance from center changes and probably many others would think likewise.)
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vcsharp2003 said:
Kinetic energy of a gaseous particle at all distances from center of spherical cloud is the same according to the problem and its value is K. Normally, I was thinking that kinetic energy of a gaseous particle would change as distance from center changes and probably many others would think likewise.
On Earth, we are familiar with air being cooler at higher altitudes, but this is because of our "greenhouse" gases.
With an atmosphere of diatomic gases, heat and light pass straight through. Correspondingly, it cannot radiate away heat. It can only exchange heat with the ground by conduction and redistribute it by convection.
Lnewqban
haruspex said:
On Earth, we are familiar with air being cooler at higher altitudes, but this is because of our "greenhouse" gases.
With an atmosphere of diatomic gases, heat and light pass straight through. Correspondingly, it cannot radiate away heat. It can only exchange heat with the ground by conduction and redistribute it by convection.
Why would kinetic energy of gaseous particles be same in the gaseous cloud? It may just be an assumption in this problem. I am trying to find some reason why this could be true.
For two different particles at radius ##r_1## and at ##r_2##, the velocity of gaseous particles is the same even though ##r_1 < r_2 ## or ##r_1 >r_2 ##.
vcsharp2003 said:
Why would kinetic energy of gaseous particles be same in the gaseous cloud?
Well, they won't all be exactly the same, of course, but for diatomic and monatomic gas mixtures the temperature can be largely independent of altitude, so the mean KE can be independent of altitude.
vcsharp2003
haruspex said:
Well, they won't all be exactly the same, of course, but for diatomic and monatomic gas mixtures the temperature can be largely independent of altitude, so the mean KE can be independent of altitude.
I was thinking that temperature decides the KE of gaseous particles. Higher temperature means higher KE.
I think in this problem, the gaseous cloud being considered is not that big and it has almost equal temperatures as one goes away from the center of the cloud, which is a good assumption for small gaseous clouds but for large gaseous clouds. Hence, the assumption of equal KE is a good assumption in this problem.
For air above us, if altitudes can become very high then there will be considerable difference between air particle temperatures at ground level and high altitudes. Then, the assumption of equal KE no matter what the altitude would not be a good assumption.
vcsharp2003 said:
Homework Statement:: Consider a spherical gaseous cloud of mass density ## \rho (r)## in free space where ##r## is the radial distance from its center. The gaseous cloud is made of particles of equal mass ##m## moving in circular orbits about the common center with the same kinetic energy ##K## .
Sounds like a trick question. It is not a gaseous cloud. An ideal gas is modeled by a large number of small particles interacting statistically. Here we have a large number of small particles that do not interact other than by their bulk gravitational effect. Otherwise they could not all be in circular orbits.
If you have a bunch of non-interacting small particles in any spherically symmetric distribution whatsoever, a new layer of small non-interacting particles can be layered on the outside. The mass density of that new layer is arbitrary.
A collection of non-interacting particles does not even have a temperature.
vcsharp2003
vcsharp2003 said:
For air above us, if altitudes can become very high then there will be considerable difference between air particle temperatures at ground level and high altitudes.
@jbriggs444 is right that I have muddied the waters by thinking of it as interacting particles. But to answer your point above, as I mentioned in post #20, the decline in temperature with altitude we observe on Earth should not be taken to be what generally happens.
The triatomic etc. gases in our atmosphere can radiate away heat into space. Collisions with other molecules spread this cooling around. This makes the upper layers denser, leading to convection. The rising air that results cools due to expansion, limiting its ability to warm the upper layers. This leads to the lapse rate we observe.
Without the triatomic gases the air would only be able to exchange heat with the ground, and that by conduction. If the ground were all at the same temperature the entire atmosphere could be at that temperature too, and there would be no convection.
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jbriggs444 said:
A collection of non-interacting particles does not even have a temperature.
You mean particles have to impact each other for heat to be produced from their KE and only then would the gas have a rise in temperature. Without impact there would be no heat and therefore no temperature. Is that correct?
vcsharp2003 said:
You mean particles have to impact each other for heat to be produced from their KE and only then would the gas have a rise in temperature. Without impact there would be no heat and therefore no temperature. Is that correct?
[I am not expert on thermodynamics, nor an educator. If I've muffed some details, correction is always welcome].
Strictly speaking, temperature is defined only for an ensemble that is at equilibrium with itself. Then we can look at how entropy and energy flow between that collection and another. If there is no interaction then there can be no meaningful equilibrium.
Entropy always increases. Energy is conserved. Thus, heat energy always flows in a direction so that entropy increases. The definition of temperature works out so that heat energy always flows from high temperature to low. [Negative temperatures are possible and are higher than all positive temperatures. It is sometimes convenient to work with inverse temperatures where the ordering of temperatures is consistent].
Temperature
The existence of temperature follows from the zeroth and second laws of thermodynamics: thermal equilibrium is transitive, and entropy is maximum in equilibrium. Temperature is then defined as the thermodynamic quantity that is the shared by systems in equilibrium.
If two systems are in equilibrium then they cannot increase entropy by flowing energy from one to the other. That means that if we flow a tiny bit of energy from one to the other (δU1 = -δU2), the entropy change in the first must be the opposite of the entropy change of the second (δS1 = -δS2), so that the total entropy (S1 + S2) doesn't change. For systems in equilibrium, this leads to (∂S1/∂U1) = (∂S2/∂U2). Define 1/T = (∂S/∂U), and we are done.
Temperature is sometimes taught as, "a measure of the average kinetic energy of the particles," because for an ideal gas U/N = (3/2) kBT. This is wrong as a definition, for the same reason that the ideal gas entropy isn't the definition of entropy.
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jbriggs444 said:
Temperature is sometimes taught as, "a measure of the average kinetic energy of the particles," because for an ideal gas U/N = (3/2) kBT. This is wrong as a definition, for the same reason that the ideal gas entropy isn't the definition of entropy.
Photons in photon gas do not interact each other. However https://en.wikipedia.org/wiki/Photon_gas says it has thermodynamic parameters, i.e. temperature or entropy. Even if photons do not interact each other they interact with environment, e.g. in hollow radiation case. I suppose interaction with environment, not interaction between molecules, is a key issue to define its thermodynamic parameters for photons or ideal gas.
jbriggs444
anuttarasammyak said:
Even if photons do not interact each other they interact with environment, e.g. in hollow radiation case. I suppose interaction with environment, not interaction between molecules, is a key issue to define its thermodynamic parameters for photons or ideal gas.
Nice. I'd not considered it, but this seems elegant and even obvious: If it is at equilibrium with a black body then we can say that its temperature is that of the black body.
anuttarasammyak
## 1. How is gravitational force calculated on a gaseous particle?
The gravitational force on a gaseous particle is calculated using Newton's Law of Gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
## 2. What factors affect the gravitational force on a gaseous particle?
The gravitational force on a gaseous particle is affected by the mass of the particle and the mass of the object it is interacting with, as well as the distance between them. The force also depends on the gravitational constant, which is a value that is constant throughout the universe.
## 3. Can the gravitational force on a gaseous particle be negative?
No, the gravitational force on a gaseous particle cannot be negative. Gravitational force is always attractive, meaning it pulls objects towards each other. However, the direction of the force can be negative if it is acting in the opposite direction of a chosen reference point.
## 4. How does the gravitational force on a gaseous particle compare to other forces?
The gravitational force on a gaseous particle is one of the four fundamental forces of nature, along with the strong nuclear force, the weak nuclear force, and the electromagnetic force. It is the weakest of these forces, but it is still a significant force on a large scale, such as in the interactions between planets and stars.
## 5. How does the gravitational force on a gaseous particle change with distance?
The gravitational force on a gaseous particle decreases with distance. This is due to the inverse square law, which states that the force is inversely proportional to the square of the distance between two objects. This means that as the distance between two objects increases, the force between them decreases significantly.
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A007413 A squarefree (or Thue-Morse) ternary sequence: closed under 1->123, 2->13, 3->2. Start with 1. (Formerly M0406) 20
%I M0406
%S 1,2,3,1,3,2,1,2,3,2,1,3,1,2,3,1,3,2,1,3,1,2,3,2,1,2,3,1,3,2,1,2,3,2,
%T 1,3,1,2,3,2,1,2,3,1,3,2,1,3,1,2,3,1,3,2,1,2,3,2,1,3,1,2,3,1,3,2,1,3,
%U 1,2,3,2,1,2,3,1,3,2,1,3,1,2,3,1,3,2,1,2,3,2,1,3,1,2,3,2,1,2,3,1,3,2,1,2,3
%N A squarefree (or Thue-Morse) ternary sequence: closed under 1->123, 2->13, 3->2. Start with 1.
%C a(n)=2 if and only if n-1 is in A079523. - _Benoit Cloitre_, Mar 10 2003
%C Partial sums modulo 4 of the sequence 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), a(5), a(6), a(6), ... - _Philippe Deléham_, Mar 04 2004
%C To construct the sequence: start with 1 and concatenate 4 -1 = 3: 1, 3, then change the last term (2 -> 1, 3 ->2 ) gives 1, 2. Concatenate 1, 2 with 4 -1 = 3, 4 - 2 = 2: 1, 2, 3, 2 and change the last term: 1, 2, 3, 1. Concatenate 1, 2, 3, 1 with 4 - 1 = 3, 4 - 2 = 2, 4 - 3 = 1, 4 - 1 = 3: 1, 2, 3, 1, 3, 2, 1, 3 and change the last term: 1, 2, 3, 1, 3, 2, 1, 2 etc. - _Philippe Deléham_, Mar 04 2004
%C To construct the sequence: start with the Thue-Morse sequence A010060 = 0, 1, 1, 0, 1, 0, 0, 1, ... Then change 0 -> 1, 2, 3, _ and 1 -> 3, 2, 1, _ gives: 1, 2, 3, _, 3, 2, 1, _,3, 2, 1, _, 1, 2, 3, _, 3, 2, 1, _, ... and fill in the successive holes with the successive terms of the sequence itself. - _Philippe Deléham_, Mar 04 2004
%C To construct the sequence: to insert the number 2 between the A003156(k)-th term and the (1 + A003156(k))-th term of the sequence 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, ... - _Philippe Deléham_, Mar 04 2004
%C Conjecture. The sequence is formed by the numbers of 1's between every pair of consecutive 2's in A076826. - Vladimir Shevelev, May 31 2009
%D Currie, James D. "Non-repetitive words: Ages and essences." Combinatorica 16.1 (1996): 19-40. See p. 20.
%D Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
%D J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 18.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D A. Thue. Über unendliche Zeichenreihen, Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, No. 7 (1906), 1-22.
%H Roger L. Bagula, <a href="/A007413/a007413.txt">Description of sequence as successive rows of a triangle</a>
%H James D. Currie, <a href="http://dx.doi.org/10.1016/j.tcs.2007.09.015">Palindrome positions in ternary square-free words</a>, Theoretical Computer Science, 396 (2008) 254-257.
%H F. Michel Dekking, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Dekking/dekk4.html">Morphisms, Symbolic Sequences, and Their Standard Forms</a>, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
%H V. Keranen, <a href="http://dx.doi.org/10.1016/j.tcs.2009.05.027">New Abelian Square-Free DT0L-Languages over 4 Letters</a>, Theoretical Computer Science, Volume 410, Issues 38-40, 6 September 2009, Pages 3893-3900.
%H S. Kitaev and T. Mansour, <a href="http://arxiv.org/abs/math/0210170">Counting the occurrences of generalized patterns in words generated by a morphism</a>, arXiv:math/0210170 [math.CO], 2002.
%F a(n) modulo 2 = A035263(n). a(A036554(n)) = 2. a(A003159(n)) = 1 if n odd. a(A003159(n)) = 3 if n even. a(n) = A033485(n) mod 4. a(n) = 4 - A036585(n-1). - _Philippe Deléham_, Mar 04 2004
%F a(n) = 2 - A029883(n) = 3 - A036577(n). - _Philippe Deléham_, Mar 20 2004
%F For n>=1, we have: 1) a(A108269(n))=A010684(n-1); 2) a(A079523(n))=A010684(n-1); 3) a(A081706(2n))=A010684(n). - _Vladimir Shevelev_, Jun 22 2009
%e Here are the first 5 stages in the construction of this sequence, together with Mma code, taken from Keranen's article. His alphabet is a,b,c rather than 1,2,3.
%e productions = {"a" -> "abc ", "b" -> "ac ", "c" -> "b ", " " -> ""};
%e NestList[g, "a", 5] // TableForm
%e a
%e abc
%e abc ac b
%e abc ac b abc b ac
%e abc ac b abc b ac abc ac b ac abc b
%e abc ac b abc b ac abc ac b ac abc b abc ac b abc b ac abc b abc ac b ac
%t Nest[ Flatten[ # /. {1 -> {1, 2, 3}, 2 -> {1, 3}, 3 -> {2}}] &, {1}, 7] (* _Robert G. Wilson v_, May 07 2005 *)
%o (PARI) {a(n) = if( n<1 || valuation(n, 2)%2, 2, 2 + (-1)^subst( Pol(binary(n)), x,1))};
%Y Cf. A001285, A010060.
%Y First differences of A000069.
%Y Equals A036580(n-1) + 1.
%Y Cf. A115384 A159481 A007413 A000120.
%K nonn,easy
%O 1,2
%A _N. J. A. Sloane_
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https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R(x,exp,sin) | 1,717,081,661,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00827.warc.gz | 160,154,012 | 16,637 | # Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,exp,sin)
Zurück zu Bestimmte Integrale
##### 1.1
${\displaystyle \int _{0}^{\infty }\left({\frac {\sin x}{x}}\right)^{2}\,e^{-2ax}\,dx=a\,\log \left({\frac {a}{\sqrt {1+a^{2}}}}\right)+\operatorname {arccot} a}$
ohne Beweis
##### 2.1
${\displaystyle \int _{0}^{\infty }e^{-\alpha x}\,\sin ^{2n}x\,dx={\frac {(2n)!}{\alpha \,(\alpha ^{2}+2^{2})(\alpha ^{2}+4^{2})\cdots (\alpha ^{2}+(2n)^{2})}}\qquad n\in \mathbb {N} \,\,,\,\,{\text{Re}}(\alpha )>0}$
Beweis
Es sei ${\displaystyle I_{n}=\int _{0}^{\infty }e^{-\alpha x}\,\sin ^{n}x\,dx}$.
Durch zweimalige partielle Integration erhält man die Rekursion ${\displaystyle I_{n}=I_{n-2}\,{\frac {(n-1)\,n}{\alpha ^{2}+n^{2}}}}$.
Also ist
${\displaystyle I_{2n}=I_{0}\cdot {\frac {1\cdot 2}{\alpha ^{2}+2^{2}}}\cdot {\frac {3\cdot 4}{\alpha ^{2}+4^{2}}}\cdots {\frac {(2n-1)\,2n}{\alpha ^{2}+(2n)^{2}}}={\frac {(2n)!}{\alpha \,(\alpha ^{2}+2^{2})(\alpha ^{2}+4^{2})\cdots (\alpha ^{2}+(2n)^{2})}}}$.
##### 2.2
${\displaystyle \int _{0}^{\infty }e^{-\alpha x}\,\sin ^{2n+1}x\,dx={\frac {(2n+1)!}{(\alpha ^{2}+1)(\alpha ^{2}+3^{2})\cdots (\alpha ^{2}+(2n+1)^{2})}}\qquad n\in \mathbb {N} \,\,,\,\,{\text{Re}}(\alpha )>0}$
Beweis
Es sei ${\displaystyle I_{n}=\int _{0}^{\infty }e^{-\alpha x}\,\sin ^{n}x\,dx}$.
Durch zweimalige partielle Integration erhält man die Rekursion ${\displaystyle I_{n}=I_{n-2}\,{\frac {(n-1)\,n}{\alpha ^{2}+n^{2}}}}$.
Also ist
${\displaystyle I_{2n+1}=I_{1}\cdot {\frac {2\cdot 3}{\alpha ^{2}+3^{2}}}\cdot {\frac {4\cdot 5}{\alpha ^{2}+5^{2}}}\cdots {\frac {n\,(2n+1)}{\alpha ^{2}+(2n+1)^{2}}}={\frac {(2n+1)!}{(\alpha ^{2}+1)(\alpha ^{2}+3^{2})\cdots (\alpha ^{2}+(2n+1)^{2})}}}$.
##### 2.3
${\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1-e^{\beta x}}}\,dx={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{coth}}\left({\frac {\alpha \pi }{\beta }}\right)}$
Beweis
Aus ${\displaystyle {\frac {\sin \alpha x}{1-e^{\beta x}}}=-\sum _{k=1}^{\infty }e^{-k\beta x}\,\sin \alpha x}$ folgt ${\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1-e^{\beta x}}}\,dx=-\sum _{k=1}^{\infty }\int _{0}^{\infty }e^{-k\beta x}\,\sin \alpha x\,dx}$.
Und das ist ${\displaystyle -\sum _{k=1}^{\infty }{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {1}{2}}\sum _{k=-\infty }^{\infty }{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{coth}}\left({\frac {\alpha \pi }{\beta }}\right)}$.
##### 2.4
${\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1+e^{\beta x}}}\,dx={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{csch}}\left({\frac {\alpha \pi }{\beta }}\right)}$
Beweis
Aus ${\displaystyle {\frac {\sin \alpha x}{1+e^{\beta x}}}=-\sum _{k=1}^{\infty }(-1)^{k}\,e^{-k\beta x}\,\sin \alpha x}$ folgt ${\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1+e^{\beta x}}}\,dx=-\sum _{k=1}^{\infty }(-1)^{k}\int _{0}^{\infty }e^{-k\beta x}\,\sin \alpha x\,dx}$.
Und das ist ${\displaystyle -\sum _{k=1}^{\infty }(-1)^{k}\,{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {1}{2}}\sum _{k=-\infty }^{\infty }(-1)^{k}\,{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{csch}}\left({\frac {\alpha \pi }{\beta }}\right)}$.
##### 2.5
${\displaystyle \int _{0}^{\infty }e^{-ax}\,\sin bx\,dx={\frac {b}{a^{2}+b^{2}}}}$
ohne Beweis
##### 2.6
${\displaystyle \int _{0}^{\infty }e^{-ax}\,{\frac {\sin bx}{x}}\,dx=\arctan \left({\frac {b}{a}}\right)\qquad {\text{Re}}(a)\geq |{\text{Im}}(b)|\quad ,\quad {\frac {b}{a}}\neq \pm i}$
Beweis für a>b>0
Aus der Reihenentwicklung ${\displaystyle \sin bx=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,(bx)^{2k+1}}$
folgt ${\displaystyle e^{-ax}\,{\frac {\sin bx}{x}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,x^{2k}\,e^{-ax}}$.
Also ist ${\displaystyle \int _{0}^{\infty }e^{-ax}\,{\frac {\sin bx}{x}}\,dx=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,\int _{0}^{\infty }x^{2k}\,e^{-ax}\,dx}$
${\displaystyle =\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,{\frac {(2k)!}{a^{2k+1}}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}\,\left({\frac {b}{a}}\right)^{2k+1}=\arctan \left({\frac {b}{a}}\right)}$.
##### 3.1
${\displaystyle \int _{0}^{\infty }e^{-ax}\,\sin(bx)\,x^{s-1}\,dx={\frac {\Gamma (s)}{{\sqrt {a^{2}+b^{2}}}^{s}}}\,\sin \left(s\,\arctan {\frac {b}{a}}\right)\qquad a>0\,,\,b\in \mathbb {R} \,,\,{\text{Re}}(s)>0}$
ohne Beweis | 2,137 | 4,550 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 24, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-22 | latest | en | 0.22657 |
https://docs.google.com/document/d/e/2PACX-1vQujVBgJBYLNaJYqK0aM9rR0_JWFtvJWduxnIS1KmarCVERDe4t2Q61YwalxNf8SfNGUFpRATZxqjkT/pub | 1,537,658,893,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158766.53/warc/CC-MAIN-20180922221246-20180923001646-00162.warc.gz | 501,542,138 | 21,040 | ʻaoʻao o ka
Nā Ana Pili Helu - Papa 7
Unuhi ʻia mai CCSS
Domain
Cluster
Code
Ke Ana
CCSS
Ratios and Proportional Relationships
Analyze proportional relationships and use them to solve realworld and mathematical problems.
(1.1)
7.RP..A.1
Helu i ka lakio anakahi e pili i nā lakio o nā hākina, me nā lakio o ka lōʻihi, ka ʻili a me ka nui e ana ʻia ma ke anakahi like a me ke anakahi ʻokoʻa.
Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units. For example, If a person walks 1/2 mile in each 1/4 hour, compute the unit rate as the complex fraction (1/2)/(1/4) miles per hour, equivalently 2 miles per hour.
7.RP.A.2
Hoʻomaopopo a hōʻike i ka pilina lakio like o nā nui. a. Hoʻoholo inā he pilina lakio like o ʻelua nui, he laʻana: ma o ka hoʻāʻo ʻana no nā lakio kaulike ma ka pakuhi papa a i ʻole ma ke kākuhi ʻana i ka papa kuhikuhina a me ke kaulona ʻana inā he laina pololei ma ka piko pakuhi ka pakuhi. e. Hoʻomaopopo i ka helu paʻa o ka lakio like (ka lakio anakahi) ma ka pakuhi papa, ka pakuhi, ka haʻihelu, ke kiʻikuhi, a me ka wehewehe waha i ka pilina lakio like. i. Hōʻike i ka pilina lakio like ma ka haʻihelu. o. Wehewehe i ka manaʻo o ia mea he kiko (x,y) ma ka pakuhi o ka pilina lakio like ma ka pōʻaiapili, me ka maliu ʻana i nā kiko (0,0) a me (1, r) ʻoiai ʻo r ka lakio anakahi.
Recognize and represent proportional relationships between quantities. a. Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin. b. Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships. c. Represent proportional relationships by equations. For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn. d. Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate.
7.RP.A.3
Hoʻohana i ka pilina lakio like e hoʻomākalakala i nā polopolema/nane haʻi lakio a pākēneka no lākou nā kaʻina lehulehu. He mau laʻana: ke kuala nōhie/maʻalahi, ka ʻauhau, ke kūʻaiʻemi a me ke kūʻaihoʻonui, ka uku lawelawe a me ka uku komikina, ka uku, ka pākēneka hoʻonui a hōʻemi, ka pākēneka hewa.
Use proportional relationships to solve multistep ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.
Ka ʻŌnaehana Helu
The Number System
Work with radicals and integer exponents
(2.1)
7.NS.A.1
Hoʻohana a hoʻoloa i ka ʻike no ka hoʻohui ʻana a me ka hoʻolawe ʻana e hoʻohui a e hoʻolawe i nā helu rational; e hōʻike i ka hoʻonui ʻana a me ka hoʻolawe ʻana ma ke kiʻikuhi laina helu papakū a i ʻole papamoe. a. Wehewehe i ka pōʻaiapili e hui ai ʻelua nui ʻēkoʻa a loaʻa ka ʻole (0). e. Maopopo p + q he helu i kaʻawale he |q| mai p, i ka ʻaoʻao ʻiʻo a i ʻole i ka aʻoʻao ʻiʻo ʻole, aia i ka ʻiʻo o q. Hōʻike i ka huina he ʻole, ke hoʻohui i kekahi helu me kona ʻēkoʻa (ka hoʻohui huli hope). Wehewehe i nā huinanui o nā helu rational ma o ka pōʻaiapili maoli o ka nohona. i. Maopopo ka hoʻolawe ʻana i ka helu rational ma ka hoʻohui huli hope ʻana, p – q = p + (-q). Hōʻike i ke kaʻawale ma waena o ʻelua helu rational ma ka laina helu ka waiwai ʻiʻo o ko lāua koena, a hoʻohana i kēia ʻanopili ma ka pōʻaiapili maoli o ka nohona. o. Hoʻohana i ke ʻanopili o ka hana hoʻomākalakala no ke kaʻakālai i ka hoʻonui ʻana a me ka hoʻolawe ʻana i nā helu rational.
Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram. a. Describe situations in which opposite quantities combine to make 0. For example, a hydrogen atom has 0 charge because its two constituents are oppositely charged. b. Understand p + q as the number located a distance |q| from p, in the positive or negative direction depending on whether q is positive or negative. Show that a number and its opposite have a sum of 0 (are additive inverses). Interpret sums of rational numbers by describing real-world contexts. c. Understand subtraction of rational numbers as adding the additive inverse, p – q = p + (–q). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in realworld contexts. d. Apply properties of operations as strategies to add and subtract rational numbers.
7.NS.A.2
Hoʻohana a hoʻoloa i ka ʻike no ka hoʻonui ʻana a me ka puʻunaue ʻana i nā hakina e hoʻonui a puʻunaue i nā helu rational. a. Maopopo ka hoʻopili ʻia ʻana o ka hoʻonui ʻana i nā hakina a i nā helu rational ma o ka hoʻokoi ʻana i ka hoʻomau ʻia o ka hana hoʻomākalakala e hoʻokō i ke ʻanopili o ka hana hoʻomākalakala, me ke ʻanopili hoʻoili nō hoʻi, i ka loaʻa ʻana o ka hua loaʻa e laʻa me (-1)(-1) = 1 a me nā lula no ka hoʻonui ʻana i nā helu hōʻailona. Wehewehe i nā hua loaʻa o nā helu rational ma ka wehewehe ʻana i nā pōʻaiapili maoli o ka nohona. e. Maopopo ka hiki ʻana ke puʻunaue i nā helu piha, inā ʻaʻole ka helu komo he ʻole, a he helu rational nā helu puka a pau o nā helu piha (me ka ʻole o ka helu komo, he ʻole). Inā ʻo ka p a me ke q he mau helu piha, a laila –(p/q) = (- p)/q = p/(-q). Wehewehe i ka helu puka o nā helu rational ma ka wehewehe ʻana i nā pōʻaiapili maoli o ka nohona. i. Hoʻohana i nā ʻanopili o ka hana hoʻomākalakala i kaʻakālai e hoʻonui a e puʻunaue ʻana i nā helu rational. o. Hoʻololi i ka helu rational i kekimala ma ka puʻunaue lōʻihi ʻana; ʻike e pau ana ke kino kekimala o ka helu rational ma nā 0 a i ʻole e pīnaʻi ana.
Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers. a. Understand that multiplication is extended from fractions to rational numbers by requiring that operations continue to satisfy the properties of operations, particularly the distributive property, leading to products such as (–1)(–1) = 1 and the rules for multiplying signed numbers. Interpret products of rational numbers by describing real-world contexts. b. Understand that integers can be divided, provided that the divisor is not zero, and every quotient of integers (with non-zero divisor) is a rational number. If p and q are integers, then – (p/q) = (–p)/q = p/(–q). Interpret quotients of rational numbers by describing realworld contexts. c. Apply properties of operations as strategies to multiply and divide rational numbers. d. Convert a rational number to a decimal using long division; know that the decimal form of a rational number terminates in 0s or eventually repeats
7.NS.A.3
Hoʻomākalakala i nā polopelema/nane haʻi o ka nohona a me ka pili helu/makemakika ma nā hana hoʻomākalakala ʻehā i nā helu rational. (Hoʻomau ʻia nā lula no nā hakina a i nā hakina paʻakikī e ka helu ʻana i nā helu rational).
Solve real-world and mathematical problems involving the four operations with rational numbers. (Computations with rational numbers extend the rules for manipulating fractions to complex fractions.)
No Ka Haʻi a me Ka Haʻihelu
Expressions and Equations
Use properties of operations to generate equivalent expressions.
(3.1)
7.EE.A.1
Hoʻohana i nā ʻanopili o ka hana hoʻomākalakala i kaʻakālai e hoʻohui, e hoʻolawe, e heluhana, a e hoʻoloa i nā haʻihelu lālani me ke kaʻilau rational.
Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
7.EE.A.2
Maopopo ke kākau hou ʻana i ka haʻihelu ma kekahi ʻano hou ma ka polopolema/nane haʻi he mea e hoʻonui ʻike no ka polopolema/nane haʻi a me ka ʻike no ka pilina o ka nui. E noʻonoʻo i a + 0.05a = 1.05a, ʻo ia hoʻi, “e hoʻonui ma ka 5%” ua like me “hoʻonui ma ka 1.05.”
Understand that rewriting an expression in different forms in a problem context can shed light on the problem and how the quantities in it are related. For example, a + 0.05a = 1.05a means that “increase by 5%” is the same as “multiply by 1.05.”
Solve real-life and mathematical problems using numerical and algebraic expressions and equations.
(3.2)
7.EE.B.3
Hoʻomākalakala i nā polopelema/nane haʻi o ka nohona a me ka makemakika/pili helu a o nā kaʻina hana lehulehu me nā helu ʻiʻo rational a me nā helu ʻiʻo ʻole rational ma kekahi ʻano (nā helu piha, nā hakina, a me nā kekimala), ma ka hoʻohana kūpili ʻana i nā mea hana. Hoʻohana i ke ʻanopili o ka hana hoʻomākalakala i kaʻakālai e helu i nā helu ma ma nā kino like ʻole; hakuloli/hoʻololi i ke ʻano o ka helu ke pono; a ana i ka pololei o ka haʻina ma o ka helu naʻau a me nā kaʻakālai koho.
Solve multi-step real-life and mathematical problems posed with positive and negative rational numbers in any form (whole numbers, fractions, and decimals), using tools strategically. Apply properties of operations as strategies to calculate with numbers in any form; convert between forms as appropriate; and assess the reasonableness of answers using mental computation and estimation strategies. For example: If a woman making \$25 an hour gets a 10% raise, she will make an additional 1/10 of her salary an hour, or \$2.50, for a new salary of \$27.50. If you want to place a towel bar 9 3/4 inches long in the center of a door that is 27 1/2 inches wide, you will need to place the bar about 9 inches from each edge; this estimate can be used as a check on the exact computation.
7.EE.B.4
Hoʻohana i nā hualau e hōʻike i ka nui ma nā polopolema/nane haʻi o ka nohona a i ʻole ka pili helu/makemakika, a kūkulu i ka haʻihelu nōhihi/maʻalahi a me ka haʻihelu kaulike ʻole e hoʻomākalakala polopolema/nane haʻi ma o ka noʻonoʻo kūpili ʻana i ka nui. a. Hoʻomākalakala i nā polopelema huaʻōlelo/moʻolelo nane e loaʻa ka haʻihelu o ke ʻano px + q = r a me p(x + q) = r, ʻoiai p, q, a me r nā helu rational kikoʻī/pilikahi. ʻEleu ka hoʻomākalakala ʻana i kēia ʻano haʻihelu. Hoʻohālikelike i ka hāʻina hōʻailona helu me ka haʻihelu huina helu, me ka hoʻomaopopo ʻana i ke kaʻina hana o ka hana hoʻomākalakala no nā mea ʻelua. He laʻana: He 54 knm. ke anapuni o ka huinahā lōʻihi. He 6 knm. kona lōʻihi. ʻEhia kenimika kona ākea? e. Hoʻomākalakala i nā polopolema huaʻōlelo/moʻolelo nane e loaʻa ka haʻihelu kaulike ʻole o ke ʻano px + q > r a i ʻole px + q < r, ʻoiai p, q, a me r nā helu rational kikoʻī/pilikahi. Kākuhi i ka ʻōpaʻa/hui haʻina o ia kaulike ʻole a wehewehe ma ka pōʻaiapili o ka polopolema/moʻolelo nane. He laʻana: Uku ʻia ʻoe he \$50 kālā o ka pule a me \$3 kālā o kēlā me kēia kūʻai ʻana aku i kāu ʻoihana he kālepa. I kēia pule, makemake ʻoe i ka uku he \$100. E haku ʻoe i haʻihelu kaulike ʻole no ke kūʻai ʻana aku e pono ai, a e wehewehe mai i nā haʻina
Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. a. Solve word problems leading to equations of the form px + q = r and p(x + q) = r, where p, q, and r are specific rational numbers. Solve equations of these forms fluently. Compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach. For example, the perimeter of a rectangle is 54 cm. Its length is 6 cm. What is its width? b. Solve word problems leading to inequalities of the form px + q > r or px + q < r, where p, q, and r are specific rational numbers. Graph the solution set of the inequality and interpret it in the context of the problem. For example: As a salesperson, you are paid \$50 per week plus \$3 per sale. This week you want your pay to be at least \$100. Write an inequality for the number of sales you need to make, and describe the solutions.
Ke Anahonua
Geometry
Draw, construct, and describe geometrical figures and describe the relationships between them.
(4.1)
7.G.A.1
Hoʻomākalakala i nā polopolema/nane haʻi me nā kiʻi kaha pālākiō o nā kinona, me ka helu ʻana i ka lōʻihi a me ka ʻili mai ke kiʻi pālākiō mai a e hoʻolaupaʻi i ke kiʻi kaha ma kekahi pālākiō ʻē aʻe.
Solve problems involving scale drawings of geometric figures, including computing actual lengths and areas from a scale drawing and reproducing a scale drawing at a different scale.
7.G.A.2
Kaha i kiʻi (me ka lima, ka lula, ke ana huina, a me ka ʻenehana) i nā kīnona me kekahi mau lula. Kālele ma ke kūkulu ʻana i nā huina kolu mai ʻekolu huina a i ʻole ʻekolu ʻaoʻao, me ka makaʻala ʻana i ka loaʻa o nā huina kolu kūikawā, he ʻoi aku o hoʻokahi huinakolu, a i ʻole ka huinakolu ʻole.
Draw (freehand, with ruler and protractor, and with technology) geometric shapes with given conditions. Focus on constructing triangles from three measures of angles or sides, noticing when the conditions determine a unique triangle, more than one triangle, or no triangle.
7.G.A.3
Wehewehe i nā kinona papa i loaʻa i ka ʻoki ʻia ʻana o ke kīnona paʻa, e laʻa me nā māhele papa o ka ʻōpaka huinhāloa kūpono a me ka pelamika huinahā ʻākau.
Describe the two-dimensional figures that result from slicing three-dimensional figures, as in plane sections of right rectangular prisms and right rectangular pyramids.
Solve real-life and mathematical problems involving angle measure, area, surface area, and volume.
(4.2)
7.G.B.4
ʻIke i ka haʻilula no ka ʻili a me ke anapuni o ka pōʻai a hoʻohana e hoʻomākalakala polopolema/nane haʻi; hāʻawi i ka molekumu o ka pilina o ke anapuni a me ka ʻili o ka pōʻai.
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.
7.G.B.5
Hoʻohana i nā mea ʻoiaʻiʻo no nā huina hoʻopiha kaha, hoʻopiha kūpono, papakū, a pili ma ka polopolema/nane haʻi me nā kaʻina lehulehu no ke kākau ʻana a me ka hoʻomākalakala ʻana i ka haʻihelu nōhihi/maʻalahi no ka huina i ʻike ʻole ʻia ma kekahi kinona.
Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure.
7.G.B.6
Hoʻomākalakala i nā polopolema/nane haʻi o ka nohona a me ka pili helu/makemakika no ka ʻili, ka pihanahaka a me ka ʻili alo o nā kinona papa a me nā kinona paʻa e loaʻa ai ka huina kolu, ka huinahā, ka huinalehulehu, ka paʻaʻiliono, a me ka ʻōpaka kūpono.
Solve real-world and mathematical problems involving area, volume and surface area of two- and three dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms.
Ka ʻIke Pili Helu a me Ka Pāhiki
Statistics and Probability
Use random sampling to draw inferences about a population.
(5.1)
7.SP.A.1
Kaha i nā kiko, ā kaha, nā ʻāpana kaha, nā kukna, nā huina (kūpono, ʻoi, peleleu) a me nā kaha kūpono a me nā kaha moe like. A hoʻomaopopo i ia mau mea ma nā kinona papa.
Understand that statistics can be used to gain information about a population by examining a sample of the population; generalizations about a population from a sample are valid only if the sample is representative of that population. Understand that random sampling tends to produce representative samples and support valid inferences.
7.SP.A.2
Hoʻohana i ka ʻikepili/ʻike o ka hāpana pono koho/ʻohi kaulele e kuhi e pili ana i kekahi heluna kānaka nona ka hiʻohiʻona hoihoi i maopopo ʻole. Hoʻopuka i nā hāpana lehulehu (a i ʻole nā hāpana hakupuni) o ka nui like e ana i ka ʻokoʻa o nā kuhi a i ʻole nā koho. He laʻana: e koho i ka ʻawelika o ka lōʻihi huaʻōlelo o kekahi puke ma nānā ʻana i ka hāpana pono koho/ʻohi kaulele o nā huaʻōlelo o ka puke; kuhi/wānana i ka lanakila o ke koho pāloka kula ma ka nānā ʻana i ka ʻikepili/ʻike o ke anamanaʻo hāpana pono koho/ʻohi kaulele. Ana i ke kau hewa o ia koho a i ʻole wānana.
Use data from a random sample to draw inferences about a population with an unknown characteristic of interest. Generate multiple samples (or simulated samples) of the same size to gauge the variation in estimates or predictions. For example, estimate the mean word length in a book by randomly sampling words from the book; predict the winner of a school election based on randomly sampled survey data. Gauge how far off the estimate or prediction might be.
Draw informal comparative inferences about two populations.
(5.2)
7.SP.B.3
Ana mōhalu i ka nui o ka ʻunuʻunu ʻana ma ka nānā ʻana i ʻelua hoʻoili helu ʻikepili/ʻike me nā kumuloli i like, a me ke ana ʻana i ke kaʻawale o nā kikowaena ma o ka hōʻike ʻana ma ke ʻano he helu māhua o ka ana kumuloli. He laʻana: He 10 knm ʻoi aʻe ka ʻawelika o ke kiʻekiʻena o nā ʻālapa ma ke kime pōhinaʻi ma mua o nā ʻālapa ma ke kime pōwāwae, ma kahi o ka pālua ke kumuloli (ka haiahū ʻawelika holoʻokoʻa) ma kekahi kime; ma ka pakuhi kiko, ahuwale ke kaʻawale o ʻelua hoʻoili o ke kiʻekiʻena.
Informally assess the degree of visual overlap of two numerical data distributions with similar variabilities, measuring the difference between the centers by expressing it as a multiple of a measure of variability. For example, the mean height of players on the basketball team is 10 cm greater than the mean height of players on the soccer team, about twice the variability (mean absolute deviation) on either team; on a dot plot, the separation between the two distributions of heights is noticeable.
7.SP.B.4
Hoʻohana i ke ana kikowaena a me ke ana kumuloli no ka ʻikepili helu o ka hāpana pono koho/ʻohi kaulele e kuhi hoʻohālike no ʻelua heluna kānaka. He laʻana: e hoʻoholo inā ʻoi aku ka lōʻihi o nā huaʻōlelo ma ka mokuna o ka puke ʻepekema no pae papa 7 ma mua o nā huaʻōlelo ma ka mokuna o ka puke ʻepekema no ka pae papa 4.
Use measures of center and measures of variability for numerical data from random samples to draw informal comparative inferences about two populations. For example, decide whether the words in a chapter of a seventh-grade science book are generally longer than the words in a chapter of a fourth-grade science book.
#### Investigate chance processes and develop, use, and evaluate probability models.
(5.3)
7.SP.C.5
Maopopo ka pāhiki o ka hanana papaha he helu ma waena o ka 0 a me ka 1 e hōʻike ana i ka nui papaha o ke kupu ʻana o ia hanana. ʻO nā helu nui aʻe ka hōʻailona o ka papaha nui aʻe. Inā kokoke ka pahiki i ka 0, ʻaʻole paha e kupu ana, a inā kokoke ka papaha i ka ½, ʻaʻohe kupu a i ʻole he kupu paha, a inā nō kokoke ka papaha i ka 1, e kupu ana paha.
Understand that the probability of a chance event is a number between 0 and 1 that expresses the likelihood of the event occurring. Larger numbers indicate greater likelihood. A probability near 0 indicates an unlikely event, a probability around 1/2 indicates an event that is neither unlikely nor likely, and a probability near 1 indicates a likely event.
7.SP.C.6
Koho i ke kokekau i ka pāhiki o ka hanana papaha ma o ka ʻohi ʻana i ka ʻikepili/ʻike o ke kaʻina hana papaha nāna e hoʻopuka i ia hanana a ma o ke kaulona i kona alapine i ka wā lōʻihi, a wānana/kuhi i ke alapine pili i ia pāhiki. He laʻana: I ka lūlū ʻia ʻana o ka una he 600 manawa, wānana/kuhi i ka loaʻa ʻana o ka 3 a me ka 6 ma kahi o 200 manawa, akā ʻaʻole naʻe he 200 manawa kikoʻī/pilikahi.
Approximate the probability of a chance event by collecting data on the chance process that produces it and observing its long-run relative frequency, and predict the approximate relative frequency given the probability. 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.
7.SP.C.7
Kūkulu i ke kumu alakaʻi pāhiki a hoʻohana i ia mea no ka huli ʻana a me ka loaʻa ʻana o ka pāhiki o ka hanana. Hoʻohālikelike i nā pahiki o ke kumu alakaʻi a me nā alapine i ʻike maka ʻia; inā ʻaʻole launa nā pāhiki, wehewehe i ke kumu o ka launa ʻole. a. Kūkulu i ke kumu alakaʻi pāhiki makalike ma ka hoʻolilo ʻana i ka pahiki kaulike i nā hopena a pau, a hoʻohana i ia kumu alakaʻi e hoʻoholo i ka pāhiki o nā hanana. He laʻana: Inā pono koho/ʻohi kaulele wale ʻia ka haumana o kekahi papa, e huli a loaʻa ka pāhiki e koho ʻia ana ʻo Lani a e koho ʻia ana kekahi kaikamahine. e. Kūkulu i ke kumu alakaʻi pāhiki (ʻaʻole paha he makalike) ma ka nānā ʻana i nā alapine ma ka ʻikepili/ʻike i hoʻokumu ʻia e ke kaʻina hana papaha. He laʻana: e huli i ka pāhiki kokekau o ke kuʻu ʻana o ke kenikeni i hoʻohuli ʻia me ke poʻo i luna a i ʻole ke kuʻu ʻana o ke kīʻaha pepa me kona puka i lalo. Ua kaulike anei nā hopena o ka hoʻohuli kenikeni i ka papaha ke kālele ʻia nā alapine i nānā ʻia?
Develop a probability model and use it to find probabilities of events. Compare probabilities from a model to observed frequencies; if the agreement is not good, explain possible sources of the discrepancy. a. Develop a uniform probability model by assigning equal probability to all outcomes, and use the model to determine probabilities of events. For example, if a student is selected at random from a class, find the probability that Jane will be selected and the probability that a girl will be selected. b. Develop a probability model (which may not be uniform) by observing frequencies in data generated from a chance process. For example, find the approximate probability that a spinning penny will land heads up or that a tossed paper cup will land open-end down. Do the outcomes for the spinning penny appear to be equally likely based on the observed frequencies?
7.SP.C.8
Huli i ka pāhiki o nā hanana ʻano hui me ka hoʻohana ʻana i nā papa helu hoʻonohonoho pono ʻia, nā pakuhi papa, nā kiʻikuhi kumu lāʻau, a me ka hoʻomeamea ʻana. a. Maopopo ka pāhiki o ka hanana ʻano hui, e like me ka nā hanana nōhie/maʻalahi, he hakina ia o nā hopena i kahi hāpana o ke kupu ʻana o nā hanana ʻano hui a pau. e. Hōʻike i kahi hāpana o nā hanana ʻano hui ma ka papa helu hoʻonohonoho pono ʻia, nā pakuhi papa, a me nā kiʻikuhi kumu lāʻau. No ka hanana e wehewehe ʻia ma ka ʻōlelo maʻa mau (e laʻa, "ka lūlū ʻana i nā ʻeono pālua"), hoʻomaopopo i nā hopena i kahi hāpana e hoʻokupu ai i ka hanana. i. Haku/Hakulau a hoʻohana i ka hoʻomeamea e hoʻoulu i ke alapine o nā hanana ʻano hui. He laʻana, hoʻohana i nā kikohoʻe pono koho/ʻohi kaulele ma ke ʻano he mea hana hoʻomeamea e pane kokekau i ka nīnau: Inā he 40% heluna kānaka o ke ʻano koko A, he aha ka pāhiki e pono ana he ʻehā kānaka ma ka liʻiliʻi e loaʻa hoʻokahi kānaka o ke ʻano koko A?
Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation. a. Understand that, just as with simple events, the probability of a compound event is the fraction of outcomes in the sample space for which the compound event occurs. b. Represent sample spaces for compound events using methods such as organized lists, tables and tree diagrams. For an event described in everyday language (e.g., “rolling double sixes”), identify the outcomes in the sample space which compose the event. c. Design and use a simulation to generate frequencies for compound events. For example, use random digits as a simulation tool to approximate the answer to the question: If 40% of donors have type A blood, what is the probability that it will take at least 4 donors to find one with type A blood?
Nā Ana Kā Mua - Papa 7 Pili Helu
M. Peters
ʻOkakopa 2013 | 8,496 | 23,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-39 | latest | en | 0.647153 |
https://www.themathdoctors.org/finding-an-angle-with-and-without-trig/ | 1,722,660,546,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00740.warc.gz | 782,426,460 | 24,589 | # Finding an Angle With and Without Trig
### (A new question of the week)
Usually when we have a figure labeled with some lengths and angles, we can expect to find unknown angles using trigonometry. When we are expected to do this using geometry alone, we can expect that there is something special about the figure that makes it possible. But how to find that specialness? That’s what we face here.
## The problem
The problem came to us in early October from Abishek:
I’m not knowing how to start to solve the question?
The figure is nicely labeled to show the parallel lines and the length relationship between AB and CD, as well as the two known angles. Without any instructions or context, it looked like a nice trigonometry problem. I tried out both ways, then answered, giving minimal hints to a trig solution, but expecting that geometry might be the required method (because it was posted under Geometry, not Trigonometry!):
Hi, Abishek.
I can solve it using trigonometry, and find a simple, exact angle. So my first question to you is, are you allowed to use trigonometry, or does your solution have to use synthetic geometry?
My method using trig involves drawing auxiliary vertical lines through points B and D, from line AB to line CD perpendicularly. I called the lengths of these y, and found y as a function of x, then found the tangent of the unknown angle.
If you can use only geometry, which seems quite likely, I have some possible ideas to start with, though I have not completed them. One is to join the midpoint of AB to C. (Possibly it could be joined to D instead; either will result in a parallelogram.) I think there are many interesting relationships in the figure that, if proved, could lead to a solution.
Please pick some way to start, and show your work as far as you can get, so we can discuss it further.
## Solving it with trigonometry
We won’t be pursuing the trig solution in the discussion, so let’s take a look at it now:
By alternate interior angles, we find that ∠BCE = 45° and ∠BDE = 60°.
From 30-60-90 triangle ΔBDE, DE = $$\frac{y}{\sqrt{3}}$$. This is also the length of FB.
From right isosceles triangle ΔBCE (because of the 45° angle), we see that CE = BE = y, so $$y=x+\frac{y}{\sqrt{3}}$$.
Solving for y, we find that $$x=\left(1-\frac{1}{\sqrt{3}}\right)y=\frac{\sqrt{3}-1}{\sqrt{3}}y$$ so $$y= \frac{x\sqrt{3}}{\sqrt{3}-1} = \frac{3+\sqrt{3}}{2}x$$
Now, $$\text{AF}=\text{AB}-\text{FB}=2x-\frac{y}{\sqrt{3}} = \left(2-\frac{1+\sqrt{3}}{2}\right)x=\frac{3-\sqrt{3}}{2}x$$
Finally, from right ΔAFD, $$\tan(\theta)=\frac{y}{\text{AF}}=\frac{\frac{3+\sqrt{3}}{2}x}{\frac{3-\sqrt{3}}{2}x}=\frac{3+\sqrt{3}}{3-\sqrt{3}}=2+\sqrt{3}\approx 3.732$$
So $$\theta=\tan^{-1}(3.732) = 75°$$.
Getting this apparently exact and “nice” angle suggested it could be solved geometrically. (It is a common exercise in trig classes to find the exact value of $$\tan(75°) = \tan(30° + 45°)$$, so we know it is in fact exact.)
## First thoughts using geometry
Abishek replied as expected:
Yes, we should proceed in Geometry.
I have joined the midpoint of AB to C.
How do we proceed to the Next Step?
I had been working on the problem, but so far could only show the additional lines I had introduced:
I don’t yet see a purely geometrical solution. But here is my drawing, showing where I added two broken lines to use in the trigonometrical solution, and the two solid blue lines I think may be of use geometrically. I’m looking for some similar triangles that might be used to determine parts of the angle θ that we want to find.
I’ll keep looking as I have time.
The parallelograms seemed useful; I hadn’t needed them in the trigonometric solution, but they seemed like a good way to make use of the 2x. The perpendicular lines also seemed likely to be useful in this approach. And similar triangles are often a good way to find angles.
At this point, I’m just playing with ideas. Many of them will turn out to be dead ends; that is not unusual in problem-solving! And I like being able to show students this messy process behind solving problems, as well as the mere fact that I don’t see solutions immediately; too often textbooks make math look too neatly packaged, resulting in students feeling that if they don’t see the whole answer at once, they just can’t do mathematics.
## Better ideas
Meanwhile, Doctor Rick had been looking over our shoulders, so to speak, and he jumped in with some new ideas:
Hi, Abishek. I too have been trying to solve this problem. It has the “feel” of a contest problem — it surely takes some creative thinking to find a solution! I hope you realize that it isn’t a matter of following some standard set of steps, but rather about exploring, trying a lot of things to see if they lead you anywhere. Both Doctor Peterson and I (and maybe others) are doing just this.
I drew in the lines that Doctor Peterson showed you; one thing they do for us is to create two “special triangles”, a 45-45-90 (right isosceles) triangle and a 30-60-90 triangle. You might look for other triangles of these types that you could make by adding other lines. We don’t want to make the diagram too complicated, but I think that one more line will do the trick. (I too started by adding the lines from the midpoint of AB to C and D, but I wasn’t getting anywhere with that idea, so I got rid of those lines. Sometimes we have to let go of our first idea in order to make progress.)
Here’s another thought. One of the facts we know involves the ratio of lengths of two segments. How can this provide information about angles, without bringing trigonometry into the picture? I can think of only one way: If we can show that two sides of some triangle are equal in length, then we know the triangle is isosceles, so its base angles are equal.
He’s saying my solid blue lines didn’t help, but something else will, something involving forming a special triangle. His wording suggested to me that he might be closer to a solution than he said!
Now Abishek, too, showed his progress, including additions to my drawing:
I have tried modifying the picture and I got this.
I’ll have to look for more things tomorrow because it’s night over here.
I had thought of working with the triangles he labeled, too, but couldn’t find a way to use them.
## A geometrical solution
While Abishek slept, I found an answer and wrote back:
I suspect that Doctor Rick has actually solved the problem, as I now have as well. I’m going to add one more hint, namely the one additional line I used, AF:
I labeled one new length; if you label other lengths, one by one, you should be able to write an equation or two that will finish the problem.
By the way, we really could as well have called x “1”, since the unit is arbitrary; then we would be finding the actual lengths of many segments.
The new line AF is drawn perpendicular to BD, though not marked as such. (I’d also considered drawing a perpendicular to BC, but that didn’t seem to lead anywhere.) I wrote in the length of BF as x because seeing that was the key to my solution. I hoped this was a small enough hint not to spoil the fun of the final chase!
Abishek was unsure of the reason for one label on the diagram:
Then, how did we find length of BF to be x?
That was a good question; I answered,
Because AF was constructed perpendicular to BD, and angle ABF = 60°, triangle ABF is a 30-60-90 triangle (half on an equilateral triangle). That makes BF half of AB.
I included that on my diagram because, although it is “obvious”, I initially missed it; when I included that fact, I knew I could solve the problem!
I could also have labeled AF as $$\sqrt{3}x$$, due to the same triangle, but the x was the key.
This was the end of our discussion. I assume Abishek was able to finish with these hints. But we haven’t seen the end yet, so let’s do it now:
We can start as with the trig solution, finding y: $$y= \frac{3+\sqrt{3}}{2}x$$
Then, in 30-60-90 ΔBDH, $$\text{DB}=2\text{DH}=\frac{2}{\sqrt{3}}y = (1+\sqrt{3})x$$
Therefore, $$\text{DF}=\text{DB}-\text{BF}=(1+\sqrt{3})x-x=\sqrt{3}x$$
But this is equal to AF (in 30-60-90 triangle ΔABF), so ΔADF is a right isosceles triangle, and ∠FAD = 45°.
So we can conclude that ∠BAD = ∠FAD + ∠FAB = 45° + 30° = 75°. We are finished!
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1 Strand Data Date Class level Sixth Class Strand Unit Representing and Interpreting Data Title Interpreting Data about Prawns landed in Irish waters. Subject Mathematics Objective(s) The aim of the lesson plan is for the children to be able to read and interpret trend graphs. Using data collected from the Irish fishing ports (Dublin prawns), the children will compile and use simple data sets and explore and calculate averages of simple data sets. Skills Required Recall, estimation, analysis, recording. Learning objectives The child should be enabled to: Examine a data set. Read the information supplied. Plot and draw a trend graph to illustrate the information. Explore and calculate the averages of the data set given. Learning activities Talk and Discussion: Recall the language and prior knowledge of graphs pictograms, block graphs, horizontal and vertical bar charts. Introduce the concept of trend graphs. Discuss their uses. If possible examine an example of a trend graph from a newspaper, magazine etc. Discuss how data about animals that are caught in Irish waters is collected by fishermen and marine scientists. See Atlas of Commercial Fisheries around Ireland second ed. March 2014 ( ), for samples of maps using the data collected. The Marine Institute also produce an annual Stock Book which contains data about our commercial fish stocks in Ireland. This is very important work, as it tells us what the state of our fishing stocks is like and if they are being fished sustainably. The scientists collect a range of data about Ireland s fish stocks (e.g. population size, age, sex of fish etc.) so they can assess the sustainability of our
2 fish stocks and the environment. They assess the data collected and then provide advice to the Irish Government who negotiates Ireland s fishing quotas in Europe each year. Each year the fishermen are then given a Total Allowable Catch of the different species of commercial fish they can catch in Irish waters. Discuss with the students what fish they think are commercial fish. Discuss why it is important to collect data on the fish stocks. - If the data collected over a number of years shows that stock sizes are growing what advice can the scientists provide? - If a trend graphs shows that a the species is declining what sort of advice should the scientists provide? - Who might find the information / data that the scientists gather each year important? Using the Irish Nephrops (prawns) fisheries as an example, discuss how this commercial species is extremely valuable to Ireland. With landings in recent years, worth around 34m at first sale, it supports an important indigenous processing industry in Ireland. Pair Work: Divide the class into pairs. Distribute the data to each pair. Discuss the information given. Elicit from the children what must be included in a trend graph-title, scale etc. Model the drawing of a trend graph on the whiteboard. Worksheet: Allow the children to work in pairs to
3 complete the attached worksheet. Each pair is then given the opportunity to share the information and procedure with the whole class. Resources Atlas of Commercial Fisheries around Ireland second ed. March 2014 ( ). Also see information about prawns Nephrops Under Water TV Surveys on the Marine Institute website which provides information on nephrops fishing grounds in Ireland, movie footage of prawns etc. Graph paper Data set (attached on worksheet). Pencils and rulers. Colour pencils Worksheet (attached) Differentiation Paired work. Assessment Question and answer. Oral feedback from children, teacher observation. Teacher check, pupil work samples. Linkage and Integration English- Oral language: Discussion based on the findings of the class regarding the average of the data set. Opinions and reasons for their answers. Geography: Fishing in Ireland. SPHE: Healthy eating. The importance of fish in our diet and the benefits which eating fish can bring. English: Procedural Writing- Recipe
4 Sixth Class Strand: Data Strand Unit: Interpreting and Representing data Total Dublin Bay Prawns Caught in the Smalls (an area in the Irish Sea) from Dublin Bay Prawn Landings by Country (tonnes) in the Smalls - an area in the Irish Sea. Year France Republic of Ireland UK Total Source: Marine Institute Stock Book 2013 Answer the following questions: 1. Calculate the total number of tonnes of prawns landed from Calculate the average number of tonnes of prawns landed during those years of all the countries listed. 3. Calculate each countries average number of prawns landed. Which of the three countries were: a. Above the total average b. Below the total average. 4. Using the data for the Republic of Ireland, suggest a reason why the number of prawns caught in the Smalls area is greater than that of either France or the UK.
5 5. Explain why the Total Allowable Catch of Prawns that fishermen are given each year important for Ireland s commercial fishing industry? 6. Create a trend graph showing the data collected from the different countries. Explain what is happening with the landings of prawns from each country. 7. Match up the names given to the prawn from other countries: Cloicheán, дублинская креветка, Homarzec norweski, Lagostim, Cigala, Dublin Bay prawn, Norway lobster, Langoustine Countrie Name of Prawn: Irish English Spanish French Portuguese Polish Russian
6 8. The map below shows the areas where the Marine Institute has conducted surveys to collect data from the different prawn grounds in Irish waters. Can you locate the different areas, oceans and seas on the map below? To help you out, go to the Marine Institute website and look for information on Nephrops Under Water TV Surveys Areas Covered During the Marine Institute Nephrops UWTV Surveys
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### Predator Prey Lab Exercise L2
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ICES Advice on fishing opportunities, catch, and effort Greater North Sea Ecoregion Published 14 November 2017 Version 2: 6 December 2017 DOI: 10.17895/ices.pub.3528 Sole (Solea solea) in Subarea 4 (North
### BRIEFING PAPER 29 FINDINGS SERIES. Children s travel to school are we moving in the right direction?
BRIEFING PAPER 29 FINDINGS SERIES Children s travel to school are we moving in the right direction? February 2011 FINDINGS SERIES 29 BRIEFING PAPER KEY FINDINGS National surveys show that while the level | 7,478 | 32,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-18 | latest | en | 0.885097 |
https://kitsunesoftware.wordpress.com/tag/math/ | 1,618,733,664,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038469494.59/warc/CC-MAIN-20210418073623-20210418103623-00263.warc.gz | 439,739,883 | 15,176 | mathematics
# Hᵤ(2, 2) = 4 for all u>0
A bit of recreational mathematics. I’d be (pleasantly) shocked if this is novel.
2+2 = 2×2 = 2² = 4 🤔
Hᵤ is the set of hyperoperations, e.g.
H₁(a, b) = a + b
H₂(a, b) = a × b
Hu(a, b) = Hu-1(a, Hu(a, b-1)) when u≥3 & b≠0
∴ Hu(2, 2) = Hu-1(2, Hu(2, 1)) when u≥3
also Hu(2, 1) = Hu-1(2, Hu(2, 0)) when u≥3
also Hu(2, 0) = 1 when u≥3
∴ Hu(2, 1) = Hu-1(2, 1) when u≥3
∴ H3(2, 1) = H2(2, 1) = 2
∴ H3(2, 2) = H2(2, 2) ⇒ 2² = 2 × 2 = 4
also Hu(2, 1) = 2 when u≥3
∴ Hu(2, 2) = Hu-1(2, 2) when u≥3
∴ ∀ u ≥ 3 Hu(2, 2) = 4
as H₁(2, 2) = 4 and H₂(2, 2) = 4, ∀ u ≥ 1 Hu(2, 2) = 4
i.e., for any hyperoperation ∘ other than the zeroth (successor) operator, 2 ∘ 2 = 4
Standard
# Railgun notes #2
[Following previous railgun notes, which has been updated with corrections]
Force:
F = B·I·l
B = 1 tesla
I: Current = Voltage / Resistance
l: Length of armature in meters
F = 1 tesla · V/R · l
F = m · a
∴ a = (1 tesla · V/R · l) / m
Using liquid mercury, let cavity be 1cm square, consider section 1cm long:
∴ l = 0.01 m
Resistivity: 961 nΩ·m
∴ Resistance R = ((961 nΩ·m)*0.01m)/(0.01m^2) = 9.6×10^-7 Ω
Volume: 1 millilitre
∴ Mass m = ~13.56 gram = 1.356e-2 kg
∴ a = (1 tesla · V/(9.6×10^-7 Ω) · (0.01 m)) / (1.356e-2 kg)
Let target velocity = Escape velocity = 11200 m/s = 1.12e4 m/s:
Railgun length s = 1/2 · a · t^2
And v = a · t
∴ t = v / a
∴ s = 1/2 · a · (v / a)^2
∴ s = 1/2 · a · v^2 / a^2
∴ s = 1/2 · v^2 / a
∴ s = 1/2 · ((1.12e4 m/s)^2) / ((1 tesla · V/(9.6×10^-7 Ω) · (0.01 m)) / (1.356e-2 kg))
@250V: s = 0.3266 m (matches previous result)
@1V: s = 81.65 m
I = V/R = 1V / 9.6×10^-7 Ω = 1.042e6 A
P = I · V = 1V · 1.042e6 A = 1.042e6 W
Duration between rails:
t = v / a
∴ t = (1.12e4 m/s) / a
∴ t = (1.12e4 m/s) / ( (1 tesla · V/(9.6×10^-7 Ω) · (0.01 m)) / (1.356e-2 kg) )
(Different formula than before, but produces same values)
@1V: t = 0.01458 seconds
Electrical energy usage: E = P · t
@1V: E = 1.042e6 W · 0.01458 seconds = 1.519e4 joules
Kinetic energy: E = 1/2 · m · v^2 = 8.505e5 joules
Kinetic energy out shouldn’t exceed electrical energy used, so something has gone wrong.
Standard
# Railgun notes
Force on the projectile of a railgun:
F = B·I·l
B: Magnetic field
I: Current
l: Length of armature
Current = Voltage / Resistance
Resistivity of seawater:
ρ = 2.00×10^−1 (Ω·m) (because = (Ω/m-length)*(cross-sectional area))
Let cavity be 1cm square, consider section 1cm long:
Volume: 1 millilitre
mass (m): ~1 gram = 1e-3 kg
Cross-section: 1e-4 m^2
Armature length (l): 1e-2 m
Resistance: ((2.00×10^−1 Ω·m)*0.01m)/(0.01m^2) = 0.2 Ω (got that wrong first time! Along with all that followed, which is now updated…)
∴ current (I) = Voltage (V) / 0.2 Ω
Rare earth magnets can be 1 tesla without much difficulty. Assume that here.
F = 1 T · (V/0.2 Ω) · (1e-2 m)
Target velocity: 11.2 km/s = Escape velocity = 11200 m/s
v = at = 11200 m/s
∴ a = (11200 m/s) / t
s = 1/2 · a · t^2
∴ s = 1/2 · ( (11200 m/s) / t ) · t^2
= 1/2 · (11200 m/s) · t
or: t = s / (1/2 · (11200 m/s))
F = ma = (1e-3 kg) · a
∴ a = F / (1e-3 kg)
∴ t = (11200 m/s) / (F / (1e-3 kg))
= (11200 m/s) · (1e-3 kg) / F
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / F
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / ( 1 T · (V/0.2 Ω) · (1e-2 m) )
Say V = 250 volts:
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / ( 1 T · (250V/0.2 Ω) · (1e-2 m) ) = 5020m (not ~501760 meters)
Say V = 25,000 volts:
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / ( 1 T · (25000V/0.2 Ω) · (1e-2 m) ) = 50.2m (not ~5017.6 meters)
Liquid mercury instead of seawater:
Resistivity: 961 nΩ·m = 0.961e-6 Ω·m
Resistance: 9.6e-7 Ω (got this one wrong the first time, too!)
Density: 13.56 times water
F = 1 T · (V/9.6e-7 Ω) · (1e-2 m)
s = 1/2 · (11200 m/s) · (11200 m/s) · (13.56e-3 kg) / ( 1 T · (V/9.6e-7 Ω) · (1e-2 m) )
@250 volts: s = 0.3266 meters (not 3.266m as before correction)
@25kV: s = 3.266 millimetres (not 32.66 millimetres as before)
Power (DC): P = IV where I = V/R,
R = 9.6e-7 Ω
@250 volts: I = 250 / R = 250 V / 9.6e-7 Ω = 2.604e8 amperes (x10 more than before correction)
∴ P = 65.1 gigawatts (x10 than before)
@25kV: I = 25000 / R = 25000 V / 9.6e-7 Ω = 2.604e10 amperes (x10 more than before)
∴ P = 651 terawatts (x10 than before)
Duration between rails:
From t = s / (1/2 · (11200 m/s))
@250 volts:
t = 0.3266 meters / (1/2 · (11200 m/s)) = 5.8321×10^-5 seconds (x10 less than before correction)
@25kV:
t = 3.266 millimetres / (1/2 · (11200 m/s)) = 5.8321×10^-7 seconds (x10 less than before)
Electrical energy usage:
E = P · t
@250 volts:
E = 65.1 gigawatts · 5.8321×10^-5 seconds = 3.797×10^6 joules (unchanged by correction)
@25kV:
E = 651 terawatts · 5.8321×10^-7 seconds = 3.797×10^8 joules (unchanged by correction)
(For reference, 1 litre of aviation turbine fuel is around 3.5e7 joules)
Standard | 2,280 | 4,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-17 | latest | en | 0.663397 |
https://www.physicsforums.com/threads/cartesian-to-polar-cordinates.828554/ | 1,511,497,284,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807084.8/warc/CC-MAIN-20171124031941-20171124051941-00731.warc.gz | 836,794,748 | 18,649 | # Cartesian to polar cordinates
1. Aug 20, 2015
### RingNebula57
If we expres cartesian cordinates in polar coordinates we get:
x=r*cos(theta)
y=r*sin(theta)
let's differentiate those 2 eqs:
dx= dr cos(theta) -r* d(theta) * sin(theta)
dy=dr sin(theta) + r* d(theta) * cos(theta)
why isn't dx*dy= r* dr* d(theta) ( like when taking the jacobian , or when doing geometric interpretation)?
Last edited by a moderator: Aug 20, 2015
2. Aug 20, 2015
### tommyxu3
$x$ and $y$ may depend on $r$ and $\theta.$ Shouldn't it be partial derivative?
3. Aug 20, 2015
### PeroK
When you say "differentiate", what are you differentiating with respect to?
Last edited by a moderator: Aug 20, 2015
4. Aug 20, 2015
### Zondrina
I assume you are speaking in the context of integration. When transforming to polar co-ordinates, it can be shown:
$$\iint_D f(x,y) \space dA = \iint_{D'} f(r \cos(\theta), r \sin(\theta)) \space dA'$$
Where $dA' = J_{r , \theta} (x, y) \space dr d \theta$. We have to multiply the function by the Jacobian $J$ whenever an invertible transformation is used to transform an integral. It turns out:
$$J_{r , \theta} (x, y) = x_r y_{\theta} - x_{\theta} y_r = r$$
You can work this result out yourself by taking the partials of the transformation. Hence we can write:
$$\iint_D f(x,y) \space dA = \iint_{D'} f(r \cos(\theta), r \sin(\theta)) \space dA' = \iint_{D'} f(r \cos(\theta), r \sin(\theta)) \space J_{r , \theta} (x, y) \space dr d \theta = \iint_{D'} f(r \cos(\theta), r \sin(\theta)) \space r \space dr d \theta$$
Where the order of integration is still to be decided. It is very important to multiply by the volume element $J$ because you want to preserve the result of the integral (the transformation is one to one and onto).
Last edited by a moderator: Aug 20, 2015
5. Aug 20, 2015
### RingNebula57
Imagine a third variable t, and we know x,y, r, theta are functions of t. So when you write the total derivative of x to respect to t we get:
dx/dt = d(partial)x/d(partial)r * dr/dt + d(partial)x/d(partial)(theta) * d(theta)/dt, and if you multiply by dt this equation , you get the above one.
6. Aug 21, 2015
### Zondrina
I'm not sure what a total derivative has to do with this. Why don't we see how a change of variables is going to affect a double integral?
Suppose there is a rectangle $R$ in the arbitrary $uv$-plane. Suppose further the lower left corner of the rectangle is at the point $(u_0, v_0)$, and the dimensions of the rectangle are $\Delta u$ for the width and $\Delta v$ for the height respectively.
Now lets define an invertible transformation $T: (u, v) \rightarrow (x, y)$ such that the rectangle $R$ in the $uv$-plane can be mapped to a region $R'$ in the $xy$-plane. This transformation from the $uv$-plane to the $xy$-plane will be given by some $x = g(u, v)$ and $y = h(u, v)$. For example, the lower left corner of $R$ can be mapped to the boundary of $R'$ by using $T$ like so:
$$T(u_0, v_0) = (x_0, y_0)$$
$$x_0 = g(u_0, v_0), \space y_0 = h(u_0, v_0)$$
Now, define a vector function $\vec r(u, v)$ to be the position vector of the image of the point $(u, v)$:
$$\vec r(u, v) = g(u, v) \hat i + h(u, v) \hat j$$
Note the equation for the bottom side of the rectangle $R$ in the $uv$-plane is given by $v = v_0$, and the equation for the left side of the rectangle $R$ in the $uv$-plane is given by $u = u_0$. The image of the bottom side of $R$ in the $xy$-plane is given by $\vec r(u , v_0)$, and the image of the left side of $R$ in the $xy$-plane is given by $\vec r(u_0 , v)$.
The tangent vector at $(x_0, y_0)$ to the image $\vec r(u , v_0)$ is given by:
$$\vec r_u(u, v) = g_u(u_0, v_0) \hat i + h_u(u_0, v_0) \hat j = x_u \hat i + y_u \hat j$$
Similarly, the tangent vector at $(x_0, y_0)$ to the image $\vec r(u_0 , v)$ is given by:
$$\vec r_v(u, v) = g_v(u_0, v_0) \hat i + h_v(u_0, v_0) \hat j = x_v \hat i + y_v \hat j$$
We can approximate the region $R'$ in the $xy$-plane by a parallelogram determined by the secant vectors:
$$\vec a = \vec r(u_0 + \Delta u, v_0) - \vec r(u_0, v_0) ≈ \Delta u \vec r_u$$
$$\vec b = \vec r(u_0, v_0 + \Delta v) - \vec r(u_0, v_0) ≈ \Delta v \vec r_v$$
So to determine the area of $R'$, we must determine the area of the parallelogram formed by the secant vectors. So we compute:
$$\Delta A_{R'} ≈ \left| (\Delta u \vec r_u) \times (\Delta v \vec r_v) \right| = \left| \vec r_u \times \vec r_v \right| (\Delta u \Delta v)$$
Where we have pulled out $\Delta u \Delta v$ because it is constant. Computing the magnitude of the cross product we obtain:
$$\left| \vec r_u \times \vec r_v \right| = x_u y_v - x_v y_u$$
So we may write:
$$\Delta A_{R'} ≈ [x_u y_v - x_v y_u] (\Delta u \Delta v)$$
Where $x_u y_v - x_v y_u$ can be determined by evaluating $g_u(u_0, v_0), h_v(u_0, v_0), g_v(u_0, v_0)$, and $h_u(u_0, v_0)$ respectively.
Now that we have formalized all of that, divide the region $R$ in the $uv$-plane into infinitesimally small rectangles $R_{ij}$. The images of the $R_{ij}$ in the $xy$-plane are represented by the rectangles $R_{ij}'$. Applying the approximation $\Delta A_{R'}$ to each $R_{ij}'$, we can approximate the double integral of a function $f$ over $R'$ like so:
$$\iint_{R'} f(x, y) dA ≈ \sum_{i = 1}^m \sum_{j = 1}^n f(x_i, y_j) \Delta A_{R'} ≈ \sum_{i = 1}^m \sum_{j = 1}^n f(g(u_i, v_j), h(u_i, v_j)) [x_u y_v - x_v y_u] (\Delta u \Delta v)$$
Notice this looks like a typical Riemann sum. Now as $m \to \infty$ and $n \to \infty$, the double sum converges to a double integral over $R$:
$$\displaystyle \lim_{m \to \infty} \displaystyle \lim_{n \to \infty} \sum_{i = 1}^m \sum_{j = 1}^n f(g(u_i, v_j), h(u_i, v_j)) [x_u y_v - x_v y_u] (\Delta u \Delta v) = \iint_R f(g(u, v), h(u, v)) [x_u y_v - x_v y_u] \space dudv$$
Where we usually write the Jacobian $J = [x_u y_v - x_v y_u]$. So we can finally conclude:
$$\iint_{R'} f(x,y) \space dA = \iint_{R} f(g(u, v), h(u, v)) \space J \space dudv$$
This argument for an arbitrary $(u, v)$ space applies to polar $(r, \theta)$ space as well. In fact, this argument will apply for any kind of other invertible transformation $T$. | 2,124 | 6,148 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-47 | longest | en | 0.752298 |
http://jailjobs.cf/top3423-definition-of-grouping-symbols-math-is-fun.html | 1,521,337,664,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645413.2/warc/CC-MAIN-20180318013134-20180318033134-00131.warc.gz | 148,557,937 | 6,862 | definition of grouping symbols math is fun
# definition of grouping symbols math is fun
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This page contains the math-related posts from my short-lived blog Symbolism.If you know the meaning of the arrows above, you can use this to remember the category theory definitions.When used as a mathematical symbol it has a distinct code point, U2135.symbols Read articles that related to : define grouping symbols - define grouping symbols in math - definition of grouping symbols math is funsymbols This theory was supported by a team of scientists from the Laboratory of Computational Neuroscience ATR in Kyoto, Japan , who Mathematics Pronunciation Guide. American English pronunciation of mathematical terms and names. Search.I am working on a project to make equations discernable by screen readers. Your math symbol phonetics was a very big help to me. Math grouping symbols definition. Keywords for Mathematical Operations.I need the names of four grouping symbols. Ask Math Questions you want answered. 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As discussed above, the infinity symbol of Wallis () is not a number However, there are two different definitions that make good mathematical sense ofIn a recent issue of the journal Science (May 28, 2004) a group of chemists at UCLA reported the synthesis of a molecule with the Borromean topology. Also, practice lots of math problems with fun math worksheets at Splash Math.Picture graph - Definition and Meaning. Type of graph that uses symbols and pictures to represent data. - Math Symbols - The Most Valuable and Important Grouping Symbols UsedMath Symbols: . . . why math symbols are used . . . Symbols are a concise way of giving lengthy instructions related to numbers and logic. Symbols save time and space when writing. Here are the most common mathematical symbols( ) grouping symbols. 2(a3). summary of the common mathematical symbols discussed below, along with the words in English used to describe them.Math can be frustrating enough in your own language. But when learning a new languageOne method is to take short pauses before saying numbers grouped in parentheses. | 2,098 | 10,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-13 | longest | en | 0.833467 |
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Q1: You are attending a championship soccer match with your family. In your pocket only \$40 to spend on Popcorns(S) and Cokes (C), supposed to allocate such amount to maximize satisfaction of the two goods, where the price of popcorn is \$5 and the price of coke is \$4, looking at the table below:
Various levels of S and C MUs MUs/Ps MUc MUc/Pc
1 40 120
2 30 80
3 25 40
4 20 32
5 15 16
6 10 8
MUs = marginal utility of popcorn, Ps = price of popcorn
MUc = marginal utility of coke, Pc = piece of coke
1. Fill up the blanks above?
2. Find the combination of popcorns and cokes yields and the level of utility, subject to the budget constraint?
3. Calculate the total utility of the optimal consumption of the two good?
4. What is the marginal rate of substitution(MRS) in such case?
5. State the budget equation in this case?
6. Graph the budget line and an indifferent curve, showing the optimal point?
7. What does the indifferent curve represent? How it relates to the indifference map?
8. What is the level of marginal utility per dollar spent, where you get the maximization or the optimal equation of the two goods?
9. The demand curve is derived from the marginal utility concept, explain?
10. Can you classify this case as a corner solution? Explain?
Q2: Decided on open tailor shop, the following table gives the level of outputs(shirts) produced daily, with the least cost of input combinations (labor wage(w) = \$300, and cost of capital, sewing machine(r) = \$200) per day:
output Labor # Capital # Long run total cost (LTC) Long run average cost(LAC) Long run marginal cost(LMC)
100 10 7
200 12 8
300 20 10
400 30 15
500 40 22
600 50 30
700 60 42
1. Fill up the blanks above?
2. Write down the cost equation?
3. State the total production function?
4. Draw up the graph for LAC and LMC in relation to the output?
5. Graph the Isoquant curve as well as the Isocost curve, with the optimal input combination?
6. In this case, you are operating in competitive market, what are the characteristics of such market?
7. At which price you are supposed to sell the shirt?
8. Draw up the expansion path for outputs and costs in the table above?
9. What is the approximate level of output(varies discretely by 100 units in the table), where profit is maximized and you are economically and technically efficient?
10. What is the Marginal Rate of Technical Substitution (MRTS) in this case?
11. Is the cost structure in the question long-run cost or short-run cost ones, what is the difference between the two, and how can they be related?
12. There are many factors reduce costs, and shift LAC downtrend, identify three of them?
13. At which level of output is the Minimum Efficient Scale(MES)?
14. What kind of demand curve you are facing in this market?
15. Is this competitive market, what is the economic profit, and should it be different in value from the opportunity cost? Explain?
Q3: Abdullah want to expand the capacity of his restaurant, but not sure of the 2020 economic growth (GDP) of the Saudi economy. He has the probability of 40% that the economy will maintain the expected growth rate in 2019 (1.8%) and the probability of 60% it will be little higher (2.1%) as the IMF forecasted. Accordingly, the table below is:
Growth rate 2.1% 1.8%
Probability Dist. 60% 40%
Profit Profit(\$Million) Profit(\$Million)
A. Decision maintain capacity 100 persons 3 2
B. Decision expand capacity by 20% 4 1
1. Compute the expected profits for both decisions?
2. Based on the expected profit only, which decision should Abdullah make?
3. Compute the Standard Deviation for decisions A and B, facing Abdullah?
4. Which decision would Abdullah make, using the coefficient of variation?
5. If Abdullah has no idea of the probability distribution of economic growth, operating in uncertainty world. Using the information above, what decision would Abdullah make, according to each of the following rules:
A. Maximax:
B. Maximin:
C. Minimax Regret:
D. Equal Probability:
6. Which decision is riskier, and how can you classify yourself as risk lover or averter or neutral?
Compelling correspondence is essential to the achievement all things considered but since of the changing idea of the present working environments, successful correspondence turns out to be more troublesome, and because of the numerous impediments that will permit beneficiaries to acknowledge the plan of the sender It is restricted. Misguided judgments.In spite of the fact that correspondence inside the association is rarely completely open, numerous straightforward arrangements can be executed to advance the effect of these hindrances.
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This course starts with an outline of the correspondence cycle, the method of reasoning and hypothesis. In resulting modules of the course, we will look at explicit use of relational connections in close to home and expert life. These incorporate relational correspondence, bunch correspondence and dynamic, authoritative correspondence in the work environment or relational correspondence. Rule of Business Communication In request to make correspondence viable, it is important to follow a few rules and standards. Seven of them are fundamental and applicable, and these are clear, finished, brief, obliging, right, thought to be, concrete. These standards are frequently called 7C for business correspondence. The subtleties of these correspondence standards are examined underneath: Politeness Principle: When conveying, we should build up a cordial relationship with every individual who sends data to us.
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It is a unique cycle that oversees force, closeness and limits, cohesiveness and flexibility of route frameworks, and makes pictures, topics, stories, ceremonies, rules, jobs, making implications, making a feeling of family life An intelligent cycle that makes a model. This model has passed ages. Notwithstanding the view as a family and family automatic framework, one of the greatest exploration establishments in between family correspondence centers around a family correspondence model. Family correspondence model (FCP) hypothesis clarifies why families impart in their own specific manner dependent on one another ‘s psychological direction. Early FCP research established in media research is keen on how families handle broad communications data. Family correspondence was perceived as an exceptional scholastic exploration field by the National Communications Association in 1989. Family correspondence researchers were at first impacted by family research, social brain science, and relational hypothesis, before long built up the hypothesis and began research in a family framework zeroed in on a significant job. Until 2001, the primary issue of the Family Communication Research Journal, Family Communication Magazine, was given. Family correspondence is more than the field of correspondence analysts in the family. Examination on family correspondence is normally done by individuals in brain science, humanism, and family research, to give some examples models. However, as the popular family correspondence researcher Leslie Baxter stated, it is the focal point of this intelligent semantic creation measure making the grant of family correspondence special. In the field of in-home correspondence, correspondence is normally not founded on autonomous messages from one sender to one beneficiary, yet dependent on the dynamic interdependency of data shared among families It is conceptualized. The focal point of this methodology is on the shared trait of semantic development inside family frameworks. As such, producing doesn’t happen in vacuum, however it happens in a wide scope of ages and social exchange.
Standards are rules end up being followed when performing work to agree to a given objective. Hierarchical achievement relies significantly upon compelling correspondence. So as to successfully impart, it is important to follow a few standards and rules. Coming up next are rules to guarantee powerful correspondence: clearness: lucidity of data is a significant guideline of correspondence. For beneficiaries to know the message plainly, the messages ought to be sorted out in a basic language. To guarantee that beneficiaries can without much of a stretch comprehend the importance of the message, the sender needs to impart unmistakably and unhesitatingly so the beneficiary can plainly and unquestionably comprehend the data.> | 2,097 | 10,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-50 | latest | en | 0.857712 |
https://www.metabunk.org/threads/how-to-take-a-photo-of-the-curve-of-the-horizon.8859/page-3 | 1,638,946,398,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363445.41/warc/CC-MAIN-20211208053135-20211208083135-00560.warc.gz | 925,140,924 | 57,820 | # How to Take a Photo of the Curve of the Horizon
#### Z.W. Wolf
##### Senior Member.
Reminder: What we are seeing is the edge of a circle.
Stand in any one of the three circles on a basketball court. Turn around in place. From your perspective the edge of the circle in your line of sight will be the point farthest away from you. It will therefore appear to be the top of the circle. There will always be a top of the circle as you turn through 360 degrees.
The horizon on a sphere Earth is a circle all around you. The circle is defined by where your line of sight meets the surface. That's what creates the edge of the circle. There will always be a top of the circle.
As you turn 360 degrees, the edge of the circle in your line of sight will always be the part of the circle farthest away from you, and from your perspective that will always be the top of the circle.
So what does this mean?
On a huge plane surface there wouldn't be any circle. There wouldn't be anything to define the edge of the circle. (Unless someone painted one on, like the paint on the basketball court.) Therefore the horizon on a plane Earth wouldn't be a circle and wouldn't show any curve.
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#### Mechanik
##### Active Member
@Chris Rippel - Your thought experiment, or a variation of it, has tricked many globers as well as flat earthers. Stated a little differently, it almost seems to be a mathematical proof that the horizon can't be curved. i.e. A) Eye level is a flat plane through the eye of the observer B) eye level out in the distance appears as a straight line, as it is part of that flat plane. C) The horizon on a spherical ocean is a constant drop below the straight eye level line. D) A line a constant distance from another line is a parallel line. Conclusion: A line, such as the horizon, that is parallel to a straight line must also be a straight line.
This is presented as a logic problem, so let’s treat it as such. If A, B, C, and D are all true, then the conclusion must be true. However, the horizon on a spherical ocean is not a plane or a line parallel to a plane and the drop (as measured by the length of a line perpendicular to your eye level line or plane) is clearly not constant (i.e. equal) across your eye level flat plane. That makes C false and therefore the conclusion is also false.
A plane at eye level extends infinitely far to both left or right. For C to be true, the drop from your eye level plane to the surface of the water at the horizon must be constant (exactly 12 feet, or 12 inches, of 23 pixels, or whatever measurement you’re using) across your field of view or across the field of view of the camera. It’s clear from photos that placing a straight line across a photo, or across the field of vision, shows the drop is greater the farther you get from the center of your view. Therefore the drop varies from the center to right or left. The horizon on a spherical ocean is a curve and the distance of that horizon to the original plane is not constant and C is false.
We see these logic errors regularly in conspiracy theories. What seems to be a logical conclusion is reached, but one or more statements are untrue, making the conclusion false.
#### Rory
##### Senior Member.
To try and combine all the above:
A) Eye level is a flat plane through the eye of the observer - true (could add MW's "and parallel to local down")
B) Eye level out in the distance appears as a straight line, as it is part of that flat plane - true
C) The horizon on a spherical ocean is a constant drop below the straight eye level line - true (for example, eye level always at zero degrees, and a 360 degree horizon at -0.7 degrees)
D) A line a constant distance from another line is a parallel line - kind of true (see below)
Conclusion: A line, such as the horizon, that is parallel to a straight line must also be a straight line - not true
ZWW points out that the horizon forms a circle, and I've said the horizon is akin to the curved edge of a flat coin: I think this is the key. The imaginary plane at eye level is not a circle, but rather an infinite plane. So perhaps imagine this: a large horizontal square (eye level) hovering above a much smaller flat circle, such as a coin. Every point on the circle is the same distance below the square ("a constant drop") and you could say that the edge of the circle and a similar circle traced on the plane would be parallel. But it doesn't stop the edge of the circle being curved, just as it doesn't stop "eye level in the distance" being a straight line.
An actual diagram may help that. It makes sense in my head. Though I'm sure there are better ways to visualise it.
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#### Trailblazer
##### Moderator
Staff member
As you turn 360 degrees, the edge of the circle in your line of sight will always be the part of the circle farthest away from you, and from your perspective that will always be the top of the circle.
Nitpick: the circle of the horizon, by definition, is equidistant from the observer all the way round. I assume you meant “furthest away in the direction you are looking at a given time”.
#### Z.W. Wolf
##### Senior Member.
Yes, I need to fine tune that. I'm struggling to describe how it looks.
In a simple, real world example: Standing in a circle painted on the basketball court.
Along your line of sight there will always be a "top" of the circle. It's the point that separates "curving to my left" and "curving to my right. "
But if I spin in place there's always a NEW point that looks like that. There's no problem in that simple real world example.
#### Mick West
Staff member
In a simple, real world example: Standing in a circle painted on the basketball court.
I prefer the example of standing in the middle of a circus ring. The edge of the ring, at the bottom of the barrier, is a flar circle, and yet you can see a curve.
##### New Member
This is a photo I took that I believe shows the curvature of the earth.
In 2018 I observed this phenomena while holding down a beer on the deck of our rental beach house. At that time I didn't say anything to anyone because I did not want to sound like a loon. In the meantime, I Googled around and found this scholarly thread on metabunk.
When we returned to Galveston this year, I used the panoramic feature on my new iPhone 8 to take this handheld picture. The railing in the foreground helps to see the curve. A lens can make a straight line appear to be curved, but I believe the railing in the foreground would have to be curved the same way and it is not.
Now I will confess that as I pivoted the camera it may have risen slightly in the middle. Position a piece of paper above the railing so that the edge rests on the corners of the railing. A tiny sliver of the gulf appears in the middle. Flip it over and position the paper so that the edge is under the corners of the railing and the gulf definitely has a bulge in it.
The picture is almost a 180 degree view, all the way northeast up the coast to Galveston and all the way southwest down the coast to the San Luis Pass. So it is not a small layout, but about 15 miles in each direction, about a 30 mile span.
But I could be wrong, I often am.
#### Jesse3959
##### Member
I've actually thought about hiking up to a high place and tying a thick string between two trees really tight, parallel to the horizon - then hanging a few helium party balloons on it (enough to lift it if it were slack) to prove that it it's not sagging and causing an appearance of curve.
Then position the camera so the string lines up with the horizon - that way there's no argument about whether the curved horizon is from the lens or real because the string is known to be straight (or curved up from the helium balloons, but in any case not curved down..)
Why a string instead of an 8-foot straight edge? Because the string could be 50 feet long, and thus far away enough from the camera that it was in good focus along with the horizon.
I haven't had a chance to try it yet though.
#### Rory
##### Senior Member.
The string could for sure work, but I suppose it's more difficult to verify that a string is perfectly straight than it is with a solid bar, as well as to set up. Plus the solid bar has shown to work really well - especially with the option of having one above and one below the horizon.
As far the deck shot, given the amount of visible curve from a low elevation - more than from much higher elevations, and more than the math would predict - it would appear that it's being caused by the camera. Perhaps lens, but probably more to do with the panoramic aspect of it. I would imagine that the ends of the balcony are further away than the center, so that probably has a lot to with it too. But others are much more au fait with the workings of cameras than I, so now doubt they will explain it better.
#### Mick West
Staff member
As far the deck shot, given the amount of visible curve from a low elevation - more than from much higher elevations, and more than the math would predict - it would appear that it's being caused by the camera. Perhaps lens, but probably more to do with the panoramic aspect of it.
Yeah, it's the panorama. Look at the huge curve in the bottom rail. Unfortunately any curvature in this image is an illusion. There's about the same amount of ocean above the railing all the way across, relative to the height of the railing at that point.
#### Jesse3959
##### Member
The string could for sure work, but I suppose it's more difficult to verify that a string is perfectly straight than it is with a solid bar, as well as to set up. Plus the solid bar has shown to work really well - especially with the option of having one above and one below the horizon.
The string won't be perfectly straight. Because of the helium balloons, it will be ever so slightly curved up in the center, favoring the flat earthers. But I like your idea of two strings, one above and the other below the horizon.
The reason I'm interested in a string for this is because many cameras will not focus sharply at infinity and at 4 feet away at the same time. With a string, the straight edge can be much farther from the camera and still fill the horizontal field of view.
There are also some unexpected conditions relating to the effective aperture of the camera lens when especially close to an object. By moving the straight edge out these effects are reduced.
As to knowing that a straight edge is straight, I think for many people, doubters may say "Well your red stick wasn't straight." But everyone knows that a taut line has to be nearly straight - and it's buoyancy determines the direction of the curve.
(I'm not personally doubting your red ruler, but rather pointing out that a taut line is inherently straight (to the degree it is straight) while a stick can be any shape you like. It does not have an inherent straightness. A taut line does - and it is a function of it's buoyancy and tension and length.)
#### Rory
##### Senior Member.
As to knowing that a straight edge is straight, I think for many people, doubters may say "Well your red stick wasn't straight."
That's true, and though I included ways to show that the red bars were straight - rotating them; showing along their length; and measuring the distance between them - it was always more a demonstration of what 'doubters' could do for themselves. But as far as I'm aware, no globe denier has yet repeated the observation.
#### Amber Robot
##### Active Member
I would think that an ideally executed panorama would by definition not show curvature is the horizon.
#### Mick West
Staff member
I would think that an ideally executed panorama would by definition not show curvature is the horizon.
What definition are you using?
#### Amber Robot
##### Active Member
What definition are you using?
I envision an “ideal” execution of a panoramic photograph would have very little horizontal field of view that sweeps across a large angle. That would minimize distortions, rather that stitching together multiple wide angle photographs.
if you expected curvature to show in a panoramic photograph what would you expect to happen in a 360 degree panorama?
#### Jesse3959
##### Member
So we really would have to look at the method used to form the panorama.
The normal way we take a panorama is with a smart phone or an equally smart camera - it takes multiple photos as we rotate the camera then it tries its best to distort and stitch them together into a seamless scene. But the shape of the horizon is completely up to how the software distorts the pictures to make them all stitch. It's really completely useless for measuring the curve of the horizon without a known-straight reference bar in the field of view.
For a panorama to actually measure the curve, you would need a camera that records a single vertical column of pixels at a time, and the camera would be on a swivel which would allow it to turn and take in the next vertical column of pixels, one column at a time, eventually taking in the whole desired scenery.
If the swivel were perfectly level, no curve would be seen. The horizon would be a straight line, so many degrees down from level.
However, if the swivel axis were tilted slightly down so that it was exactly at right angles to the horizon in one direction, then it would produce a panorama which correctly showed the curve of the horizon.
This fellow built himself one using the guts of an old flatbed scanner: https://hackaday.com/2009/12/29/panoramic-scanner-camera/
#### Amber Robot
##### Active Member
However, if the swivel axis were tilted slightly down so that it was exactly at right angles to the horizon in one direction, then it would produce a panorama which correctly showed the curve of the horizon.
yes. If it were tilted down and spun in a plane it would be tilted up at 180.
#### Jesse3959
##### Member
yes. If it were tilted down and spun in a plane it would be tilted up at 180.
So I guess in summary, all a panorama proves is that the horizon is a circle. And that if you are at some altitude and you point a tube at the horizon in one direction, it will be pointing above the horizon in the other direction.
(And obviously the earth has to be essentially spherical for the horizon to be close enough that there's a distinct boundary, because if the edge was the horizon then we couldn't see it at all unless we were at the edge of the disk..)
#### Amber Robot
##### Active Member
So I guess in summary, all a panorama proves is that the horizon is a circle. And that if you are at some altitude and you point a tube at the horizon in one direction, it will be pointing above the horizon in the other direction.
I saw a YouTube video in which a guy did exactly that. He was on a hill that had a view of water in both directions and set up a tube and pointed it at the horizon looking one way and showed that it was above the horizon looking the other way.
#### Rory
##### Member
My attempt at this. Taken from the third floor of the Hotel Santika Premiere Belitung:
The situation at the balcony. There are glass railings that I used as a reference:
The shot. Exposure is f/22, 1/160s, ISO 3200. Taken using Canon EOS R with EF 16-35 f/4 lens, at the widest setting. I aligned the horizon with the glass railing and focused on the horizon.
Vertically stretched:
I was surprised at how easy this was, considering this was taken less than 30 m (100 ft) above sea level and I was without any preparation. I brought the correct lens by chance & had only about 10 minutes before the fog came in.
#### Mick West
Staff member
Vertically stretched:
I was surprised at how easy this was, considering this was taken less than 30 m (100 ft) above sea level and I was without any preparation. I brought the correct lens by chance & had only about 10 minutes before the fog came in.
I'm not sure about this one. You are not really high enough, and the tops of the glass panels don't form a continuous verifiably straight line. So it's hard to tell what role lens distortion has.
Did you try correcting with the lens profile in Photoshop? If you email me the original photo I can give that a go. [email protected]
##### Member
I'm not sure about this one. You are not really high enough, and the tops of the glass panels don't form a continuous verifiably straight line. So it's hard to tell what role lens distortion has.
Did you try correcting with the lens profile in Photoshop? If you email me the original photo I can give that a go. [email protected]
I had "lens aberration correction - lens distortion correction" enabled in-camera, so the camera will correct barrel distortions. doing it again in PS will only 'correct' it for the second time and reintroduce the distortions. unfortunately, I did not take raw version of the pictures.
##### Member
As a follow up to my attempt above, I have managed to get a simulated view of the scene using Walter Bislin's curvature app. The problem is that the observation is very close to the surface, and by default, the resolution of the output is too small to show the curvature. Here's a crude way I did to get a higher resolution simulation:
• Go to Walter's curvature app
• Input the height of the observation (25 m) and the focal length of my camera (16 mm)
• Maximize the browser window across my triple monitor setup (6400 px of horizontal pixels)
• Increase the zoom of the browser window to 500%
With these steps, the app gave me a simulated view with a resolution of 4530x3020.
Here is my photo & the result of Walter's simulation, vertically stretched together, side by side:
#### DavidB66
##### Active Member
At #72 above I posted a link to a photo taken from a hillside above the Bonneville Salt Flats, showing clear curvature along a very long straight road. Now a YouTuber going by the name Don't Stop Motion has gone one better by taking a video from much the same viewpoint. The resulting video is here: The Curvature of the Earth - Bonneville Salt Flats (from West Wendover) - YouTube
By zooming in and out the video gives a better idea than the still photo of the distances involved. I tried identifying surface features on Google Earth to estimate the distance of the horizon, but didn't have much luck. Overall, the view of curvature is strikingly similar to Soundly's footage from Lake Ponchartrain, except that this time it is taken over land. As I stressed in earlier comments, it is essential to be sure that the road is straight and that the relevant surface is very level. Google Earth seems to confirm this to within a few feet over a 20-mile stretch.
#### JMartJr
##### Senior Member
The problem with this simple analogy is of course that Flat Earthers will then say "So you mean the Earth is flat like a coin?" and we're back where we started! I'm trying to think of a common object that is shaped like a very flat circular dome, like the section of the Earth's surface above the horizon...
To me, that makes the whole discussion of horizon curve, and the somewhat related "horizon always rises to eye level" flat Earth claim, extremely odd. The horizon does indeed curve, and does not rise to eye level, as is demonstrably true in this thread and others ad nauseum. It seems this would be expected whether the Earth was flat or the globe that it actually is -- the difference, if any, would be in the degree of horizon curve or drop below eye level as you ascend. It hurts my head that flat apologists insist that a bit of evidence that works as well with their theory as with the correct one, mus be wrong, when if their claims are true it would argue against their hypothesis as well as against the globe model.
Coins are flat, balls are round, you get horizon curve either way. And the horizon drops as you ascend, either way.
#### Mendel
##### Senior Member.
The horizon does indeed curve, and does not rise to eye level, as is demonstrably true in this thread and others ad nauseum. It seems this would be expected whether the Earth was flat or the globe that it actually is -- the difference, if any, would be in the degree of horizon curve or drop below eye level as you ascend.
The problem is that the horizon is not where the edge of the Earth is -- you can't see all the way to Antarctica or beyond, you can't even see Europe from America. So that curved horizon can't be the edge of the Earth, it must be caused by something else, but there's no good explanation for it. It can't be caused by some limit on how you can see, because you can see things beyond the horizon if they're big enough (like wind farms or big ships).
#### Rory
##### Senior Member.
The horizon does indeed curve. It seems this would be expected whether the Earth was flat or the globe that it actually is
If the Earth was flat there wouldn't be a horizon, there'd be nothing to cause it.
#### Mendel
##### Senior Member.
If the Earth was flat there wouldn't be a horizon, there'd be nothing to cause it.
That depends on how you define "horizon". Flat Earth would have a line where the ground meets the sky. But the sun couldn't set on it while lighting other parts of Earth.
#### Rory
##### Senior Member.
Flat Earth would have a line where the ground meets the sky.
I always find it hard to imagine how things would appear if we actually lived on a plane. Wouldn't the point where, for example, sea meets sky be rather indistinguishable on a flat plane, as though fading into a mist due to atmostpheric conditions, etc?
#### FatPhil
##### Active Member
If the Earth was flat there wouldn't be a horizon, there'd be nothing to cause it.
If the earth were flat *and infinite in extent in all directions* there wouldn't be a visible horizon. There would be a theoretical horizon, but being at infinite distance would be attenuated to non-visibility (not to mention it would get in the way of the sun ducking below it).
However, none of the "circled by antarctica" flat earth models show something infinite in extent, so your argument does not address their currently favoured model.
I will confess that I generally find flat earth arguments to be so idiotic they're not even worth paying attention to - they've all been debunked a myriad times - but if we are going to waste time debunking them, we should do so with precision.
#### Mendel
##### Senior Member.
If the earth were flat *and infinite in extent in all directions* there wouldn't be a visible horizon. There would be a theoretical horizon, but being at infinite distance would be attenuated to non-visibility (not to mention it would get in the way of the sun ducking below it).
Is a far mountain not there when it's hidden by haze? We know where it is, right?
The horizon on "infinite flat world" is clearly, precisely, sharply defined. It is exactly at the eye height of the observer. And with the naked eye, it'll be indistinguishable from the horizon on the globe.
#### FatPhil
##### Active Member
Is a far mountain not there when it's hidden by haze? We know where it is, right?
I mentioned no mountain. What mountain are you talking about? What promped this question, it doesn't seem a response to what I posted. How far is this mountain? The only farness that is relevant to my argument refarging planes is farther than any finite farness - i.e. at infinity. A mountain at infinity that's never been observed - if we can talk about such a concept at all, then no we don't know where it is. Your asking the question seems to imply that objects at infinity can (a) exist; and (b) be seen. I'm a mathematician, I reject even the former until you precisely define the topology which permits "at infinity" to have an unambiguous meaning. Why have you deviated the discussion down this tangent, it seems pointless?
My mention of attentuation was to *stop* pointless deviations down this path - nothing exists "at infinity" even on an infinite plane, so it's silly to talk about such things as if they're real. Hence my couching in terms of "theoretical horizon" - it's not a horizon - the parallel planes never meet. So there's nothing to see - anything that you can see has something visible behind it, so nothing is "the" horizon. Hiding the concept of existence (which apparently you seem to have not addressed) behind the practical nature of something not even being visible in a world with particles in it (which is part of the model both sides of the argument hold) was intended to protect you from the mathematical pedantry which makes your wording look sloppy.
The horizon on "infinite flat world" is clearly, precisely, sharply defined. It is exactly at the eye height of the observer. And with the naked eye, it'll be indistinguishable from the horizon on the globe.
Do not agree at all.
The horizon on an infinite flat world is a virtual circle at infinity in the plane of the observer's eye line parallel to the surface.
The horizon on a globe is a circle parallel and below the observer's eye line.
In the former, the horizons in opposite directions are antiparallel; in the latter they are not, being rays down a cone they are at an angle of less than 180 degrees.
I don't even understand why you've proposed this falsity, as you seem to be aware of these facts, they've even been stated upthread.
#### Mendel
##### Senior Member.
@FatPhil
I mention the mountain hidden by haze because your argument seems to be that the horizon isn't there if it's hidden.
The horizon is not a physical place. It's something an observer sees; it depends on the position of the observer and where they're looking. I am describing the horizon as "the line where sky and ground/sea meet". The line does not exist, but ground and sky do.
So if you have an observer, you can say for that observer with certainty where that line is because you can tell whether they are looking at the sky or not. You can do the same thing for the "infinite plane world" in the absence of atmosphere; in fact, you can do it more easily because any sight line tilted down will hit ground; any sight line not tilted down will hit sky. So that's where the horizon is in that world.
In an infinite plane world with an atmosphere, it's just slightly more difficult because of refraction and because haze makes it harder to see objects in the distance (IR photography may come in handy). Are you arguing that adding an atmosphere to a world makes the horizon go away? like adding fog would make the mountain go away?
#### Rory
##### Senior Member.
If the Earth was flat there wouldn't be a horizon, there'd be nothing to cause it.
None of the "circled by antarctica" flat earth models show something infinite in extent, so your argument does not address their currently favoured model.
I guess to be more precise, if the Earth was a flat plane (infinite or otherwise) there wouldn't be a horizon - although, as you point out, if you were within visual range of "the edge" there'd be something different.
Whatever a person imagines to be at the edge, though, I don't think one could call it a horizon.
#### FatPhil
##### Active Member
The horizon is not a physical place. It's something an observer sees; .... The line does not exist, but ground and sky do.
OK, that's nicely abstract, I can salute that as it's run up the flagpole.
From previous posts, in the context of a sphere model, I had inferred a more concrete horizon which exists as points on the sphere's surface. Referring to the "distance to the horizon", as people commonly do, reinforces the reification of those as real points that do exist on the surface, as you can't have the distance to something that doesn't exist. Most of the time, in a real world context, I believe most people think this way, that the horizon is real points in the distance over ---> there.
Of course, the infinite plane case has a horizon with none of these properties, and the above fails. The definition of the virtual horizon as, for example, an angle (in our plane case, 0 degrees) below the viewing plane, such that above is sky and below is ground, makes the most sense (that angle being a function of azimuthal angle as you turn around your own upright axis, but in our plane case constant). (This declension definition has the property of working if you're standing on silly things like an infinite cylinder, where it varies from 0 to down to 0 to down again as you turn.)
The real/abstract thing is where the confusion lies. For the actual world we live in, I don't think the completely natural assumption that it is real points on the earth will be anything apart from the default people think of when they hear the word horizon. Thank you for clarifying that you understand the distinction.
#### FatPhil
##### Active Member
I guess to be more precise, if the Earth was a flat plane (infinite or otherwise) there wouldn't be a horizon - although, as you point out, if you were within visual range of "the edge" there'd be something different.
Whatever a person imagines to be at the edge, though, I don't think one could call it a horizon.
The image above in post #106 ( https://www.metabunk.org/threads/how-to-take-a-photo-of-the-curve-of-the-horizon.8859/post-243657 ) implies that the people who like have that model use the edge as their horizon. It works from a nomenclature perspective, even if from no other. Wrongthink has a tendency to weirdify language. ( ;-) )
#### Mendel
##### Senior Member.
The real/abstract thing is where the confusion lies. For the actual world we live in, I don't think the completely natural assumption that it is real points on the earth will be anything apart from the default people think of when they hear the word horizon. Thank you for clarifying that you understand the distinction.
I'm pretty sure I didn't understand that second sentence.
You can look at this picture and know where the horizon is, even if you don't know the distance.
If we lived on Flat Earth, on a world surrounded by ocean, the horizon would look just the same (though we could debate the sun and the clouds). I believe the "natural" assumption is to look at this picture and think "that's the horizon" without knowing how distant it is.
It's true that on the globe, the horizon line coincides with a line that is "the furthest points you can see", which is also quite hard to pin down exactly because of refraction (and haze, and sometimes inferior mirage), which is why I'm using the terms "geometric horizon" and "apparent horizon" when discussing this. On the infinite plane, you can geometrically see everything, so there is no geometric horizon on the surface; and on the finite Flat Earth, it coincides with the edge of the Earth (and we'd typically imagine those points at sea level).
Larry Niven's Ringworld has two horizons -- or none, if you don't recognize the edge of the world as horizon, and instead look for the "edge" of an infinite tube.
Last edited:
#### Rory
##### Senior Member.
The image above in post #106 ( https://www.metabunk.org/threads/how-to-take-a-photo-of-the-curve-of-the-horizon.8859/post-243657 ) implies that the people who like have that model use the edge as their horizon. It works from a nomenclature perspective, even if from no other. Wrongthink has a tendency to weirdify language. ( ;-) )
I would say I don't think anyone thinks of the edge as the horizon, or at least no one I've come across - especially because flat earthers don't believe anyone can get close to the edge. They all believe in horizons though - but getting them to think about what would cause a horizon on their hypothetical flat plane hasn't been easy, in my experience (last I heard it was mumblings about "visual limits" and "diffraction").
That image was posted by a non-flat earther to demonstrate that we always look down to the horizon whether on a ball or on a plane, so I don't think it quite relates to discussions on the horizon or "the edge".
#### FatPhil
##### Active Member
Yeah, it's the panorama. Look at the huge curve in the bottom rail. Unfortunately any curvature in this image is an illusion. There's about the same amount of ocean above the railing all the way across, relative to the height of the railing at that point.
One minor frustration is that this situation could possibly have been an interesting opportunity to demonstrate the curvature of the earth. I wonder what the horizon would have looked like if the level at which the photo had been taken was lowered by about 15cm, with the bottom of the railing above the horizon, but put centre-of-frame, thus the reference straight line. Were the earth flat, then, like the bottom of the railing, the horizon would be concave up in the photo as everything converges to the two vanishing points. However, if as you look left and right of centre it had started to fall away from the top rail, that would have been clear proof that the curvature was great enough to not just counter, but perhaps *beat* the cylindrical distortion of the panorama, at least in part of the frame. Maybe more altitude would have been required, who knows. If someone has such an opportunity again, it might be a fun one to try. (The output of my wondering is 'perhaps even w shaped', as the distortion could eventually win, but it would be setup dependent.)
I've found a wildlife viewing tower on a (alas low and flat) peninsula near me, and feel inspired to try to reproduce the 'tube and spirit-level' experiment with nothing but a long stick (maybe with some twist-ties to make 'sights'), in such a minimalistic fashion that literally anyone who can get their hands on a stick (and a peninsula, but the spirit level's not needed) can reproduce it. If I can contrive it, I will even try to do away with the stick, and make it a zero-equipment experiment. I'm not expecting clear weather to reliably have a clearly visible horizon for months, so not even I am holding my breath.
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https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.2-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese3.htm | 1,696,153,410,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510810.46/warc/CC-MAIN-20231001073649-20231001103649-00451.warc.gz | 693,242,676 | 4,938 | ### 3.3 $$\int x \sqrt{b x+c x^2} \, dx$$
Optimal. Leaf size=81 $\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{5/2}}-\frac{b (b+2 c x) \sqrt{b x+c x^2}}{8 c^2}+\frac{\left (b x+c x^2\right )^{3/2}}{3 c}$
[Out]
-(b*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(8*c^2) + (b*x + c*x^2)^(3/2)/(3*c) + (b^3*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c
*x^2]])/(8*c^(5/2))
________________________________________________________________________________________
Rubi [A] time = 0.0238282, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.267, Rules used = {640, 612, 620, 206} $\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{5/2}}-\frac{b (b+2 c x) \sqrt{b x+c x^2}}{8 c^2}+\frac{\left (b x+c x^2\right )^{3/2}}{3 c}$
Antiderivative was successfully verified.
[In]
Int[x*Sqrt[b*x + c*x^2],x]
[Out]
-(b*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(8*c^2) + (b*x + c*x^2)^(3/2)/(3*c) + (b^3*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c
*x^2]])/(8*c^(5/2))
Rule 640
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]
Rule 612
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]
Rule 620
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int x \sqrt{b x+c x^2} \, dx &=\frac{\left (b x+c x^2\right )^{3/2}}{3 c}-\frac{b \int \sqrt{b x+c x^2} \, dx}{2 c}\\ &=-\frac{b (b+2 c x) \sqrt{b x+c x^2}}{8 c^2}+\frac{\left (b x+c x^2\right )^{3/2}}{3 c}+\frac{b^3 \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{16 c^2}\\ &=-\frac{b (b+2 c x) \sqrt{b x+c x^2}}{8 c^2}+\frac{\left (b x+c x^2\right )^{3/2}}{3 c}+\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{8 c^2}\\ &=-\frac{b (b+2 c x) \sqrt{b x+c x^2}}{8 c^2}+\frac{\left (b x+c x^2\right )^{3/2}}{3 c}+\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.112991, size = 87, normalized size = 1.07 $\frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-3 b^2+2 b c x+8 c^2 x^2\right )+\frac{3 b^{5/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{24 c^{5/2}}$
Antiderivative was successfully verified.
[In]
Integrate[x*Sqrt[b*x + c*x^2],x]
[Out]
(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-3*b^2 + 2*b*c*x + 8*c^2*x^2) + (3*b^(5/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(S
qrt[x]*Sqrt[1 + (c*x)/b])))/(24*c^(5/2))
________________________________________________________________________________________
Maple [A] time = 0.053, size = 87, normalized size = 1.1 \begin{align*}{\frac{1}{3\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{bx}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{{b}^{2}}{8\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{{b}^{3}}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*(c*x^2+b*x)^(1/2),x)
[Out]
1/3*(c*x^2+b*x)^(3/2)/c-1/4*b/c*x*(c*x^2+b*x)^(1/2)-1/8*b^2/c^2*(c*x^2+b*x)^(1/2)+1/16*b^3/c^(5/2)*ln((1/2*b+c
*x)/c^(1/2)+(c*x^2+b*x)^(1/2))
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(c*x^2+b*x)^(1/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [A] time = 2.07372, size = 343, normalized size = 4.23 \begin{align*} \left [\frac{3 \, b^{3} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c\right )} \sqrt{c x^{2} + b x}}{48 \, c^{3}}, -\frac{3 \, b^{3} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c\right )} \sqrt{c x^{2} + b x}}{24 \, c^{3}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(c*x^2+b*x)^(1/2),x, algorithm="fricas")
[Out]
[1/48*(3*b^3*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(8*c^3*x^2 + 2*b*c^2*x - 3*b^2*c)*sqrt(c
*x^2 + b*x))/c^3, -1/24*(3*b^3*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (8*c^3*x^2 + 2*b*c^2*x - 3*
b^2*c)*sqrt(c*x^2 + b*x))/c^3]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{x \left (b + c x\right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(c*x**2+b*x)**(1/2),x)
[Out]
Integral(x*sqrt(x*(b + c*x)), x)
________________________________________________________________________________________
Giac [A] time = 1.31978, size = 99, normalized size = 1.22 \begin{align*} \frac{1}{24} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \, x + \frac{b}{c}\right )} x - \frac{3 \, b^{2}}{c^{2}}\right )} - \frac{b^{3} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{16 \, c^{\frac{5}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(c*x^2+b*x)^(1/2),x, algorithm="giac")
[Out]
1/24*sqrt(c*x^2 + b*x)*(2*(4*x + b/c)*x - 3*b^2/c^2) - 1/16*b^3*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqr
t(c) - b))/c^(5/2) | 3,052 | 6,403 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-40 | latest | en | 0.131964 |
http://orinanobworld.blogspot.fr/2015/02/ | 1,498,707,845,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323864.76/warc/CC-MAIN-20170629033356-20170629053356-00030.warc.gz | 318,803,170 | 34,873 | Monday, February 23, 2015
Partitioning with Binary Variables
Armed with the contents of my last two posts ("The Geometry of a Linear Program", "Branching Partitions the Feasible Region"), I think I'm ready to get to the question that motivated all this. Let me quickly enumerate a few key take-aways from those posts:
• The branch and bound method for solving an integer linear program or mixed integer linear program uses a search tree.
• At each node of the search tree, a linear program (LP) that relaxes some or all of the integrality restrictions (while adding some new restrictions based on earlier branching) is solved. (As an aside, this hold for mixed integer quadratic programs, with the change that the node relaxation will likely be a quadratic program rather than a linear program.)
• LPs have closed and convex feasible regions.
• The act of branching (separating a node into two child nodes) partitions the feasible region for the problem at the parent node into two smaller feasible regions.
Let's move on now. A common use for binary decision variables -- I suspect the most common use -- is to partition the feasible region of a problem into two or more disjoint subregions, corresponding to mutually exclusive scenarios. Consider the example in Figure 1, a mixed integer linear program (MILP) in which $x$ is integer-valued and $y$ is continuous. The green polytope is the LP relaxation; the vertical bars are the actual feasible region of the MILP.
Figure 1: Feasible region
Now suppose that there is something going on in the problem that depends on whether $x$ is 3 or less, 4-6, or 7 or more. For instance, if $x$ represents the number of workers deployed on some project, we might need a supervisor for every three workers. We need to partition into those three cases.
A general approach for partitioning a feasible region is to introduce one binary variable for each subregion/scenario, and constrain those variables to add up to 1 (so that exactly one scenario is selected). In our example, we add $z_i \in \{0,1\}$ for $i \in \{1,2,3\}$ along with the constraints\begin{eqnarray*} z_{1}+z_{2}+z_{3} & = & 1\\ x & \le & 3z_{1}+9(1-z_{1})\\ x & \le & 6z_{2}+9(1-z_{2})\\ x & \ge & 4z_{2}\\ x & \ge & 7z_{3}. \end{eqnarray*}Left to the reader as an exercise: confirm that if $z_1=1$ then $0\le x \le 3$, if $z_2=1$ then $4\le x\le 6$, and if $z_3=1$ then $7\le x\le 9$. [Update: See Rob Pratt's comment below for a tighter formulation.]
As a side note, before proceeding further, suppose that what we really want is to ensure that either $x\le 3$ or $x\ge 7$. (Don't ask why; just play along.) We do that the same way I just described, but then also constrain $z_2 = 0$. So the technique for partitioning a feasible region also works for omitting a chunk of a feasible region.
One limitation of this approach is that the subregions must be definable using linear constraints, assuming that you want your problem to remain a MILP. Another (and this is really what motivated me to do this post) is that the partitioning really takes place during the branching process. The algebra splits our feasible region into three disjoint chunks so long as $z_1,z_2,z_3$ are integer, but when we are solving node relaxations some or all of the $z$ variables will be relaxed to continuous.
Let's stick with our three-way split, while also fixing $z_2=0$, so that we force either $x\le 3$ or $x\ge 7$. The conceptual feasible region looks like Figure 2. Again, the actual feasible region for the MILP consists of two clusters of vertical line segments; the shaded polytopes are the relaxations of each scenario.
Figure 2: Two scenarios
At some node in the search tree, $z_1$ will be fixed at 1 (and $z_2$ and $z_3$ at 0), and the feasible region of the node relaxation will be the polytope on the left side of Figure 2. At some other node, $z_3$ will be fixed at 1 ($z_1$ and $z_2$ at 0), and the feasible region of the node relaxation will be the polytope on the right side of Figure 2.
What about before any of the $z$ variables are fixed (for instance, at the root node)? Since $z_1$ and $z_3$ can take fractional values, the relaxation at the root node (shown in Figure 3) is the same as what it was before we introduced the $z$ variables.
Figure 3: Root node relaxation
To see this, consider what happens if we let $(z_1, z_2, z_3) = (0.5, 0, 0.5)$. The constraints on $x$ become\begin{eqnarray*} x & \le & 3(0.5)+9(1-0.5) = 6\\ x & \le & 6(0)+9(1-0) = 9\\ x & \ge & 4(0) = 0\\ x & \ge & 7(0.5) = 3.5. \end{eqnarray*}Together with the original constraints that defined the sides of the polytope in Figure 1, this gives us the portion of the excluded region where $3.5 \le x \le 6$ (which is most of the pink region). For $x$ between 3 and 3.5 or between 6 and 7, we just need to adjust the $z_1, z_3$ split.
The key takeaway here is that the split we see in Figure 2 actually takes effect during branching, not at the root node (and not at nodes below the root but above the point where the solver branches on one of the $z$ variables). The approach of using binary variables to partition feasible regions is valid because the final solution will obey the integrality restrictions (and thus will belong to one of the disjoint scenarios). Along the road to the final solution, though, we cannot count on the partitioning holding true (and, in the case of excluded regions, we cannot count on the unwanted portion actually be excluded yet).
Saturday, February 21, 2015
Branching Partitions the Feasible Region
Yesterday's post got me started on the subject of the geometry of a linear program (LP). Today I want to review another well-known geometric aspect, this time of integer linear programs (ILPs) and mixed integer linear programs (MILPs), that perhaps slips some people's minds when they are wrestling with their models. Most solvers use some variant of the branch and bound (or, more generally, branch and cut) algorithm.
The point I want to make is in the title of this post: branching (more precisely, separating a node in the branch and bound search tree into child nodes) equates to partitioning the feasible region into disjoint chunks. The interpretation of "feasible region" in that last sentence is a bit tricky. On the one hand, the region being partitioned continues to relax the integrality restriction on at least some of the integer variables, adding points not belonging to the original feasible region. On the other hand, outside of the root node, every node problem restricts variables to a subset of the original feasible region. The problem solved at each node is a linear program (LP), sometimes called the node LP. The feasible region of the node LP at any node other than the root is neither a subset nor a superset of the original feasible region; it's a bit of both.
To clarify that (I hope), consider the two variable MILP illustrated in Figure 1, in which $x$ is a general integer variable and $y$ is a continuous variable.
Figure 1: MILP feasible region
The feasible region consists of a bunch of disjoint line segments (integer abscissas $x$, arbitrary ordinates $y$), along with an isolated point (labeled A). If we relax the integrality restriction on $x$, we get the convex polytope shown in Figure 2, which is the feasible region of the LP at the root node.
Figure 2: LP relaxation
Now suppose that we are looking for the point that maximizes $y$. The optimal solution to the root LP is labeled B in Figure 2. Let's say that $B=(x_0, y_0)$. Note that the value $x_0$ of $x$ at B is not integral.
One possible way to partition the root node is to round the value of $x$ at B up in one child node and down in the other. In effect, this adds the new constraint (cut) $x\le \left\lfloor x_{0}\right\rfloor$ to one child node (call it the left child), and the cut $x \ge \left\lceil x_{0}\right\rceil$ to the other (right) child. Figures 3 and 4 show the child nodes.
Figure 3: Left child
Figure 4: Right child
I left point B in both figures as a point of reference, but it does not belong to either feasible region.
In this particular (rather trivial) example, the maximum value of $y$ in both children occurs at an integer-feasible point (a corner where $x$ happens to take an integer value), so there will be no further branching. More generally, the algorithm might choose to split either child into either smaller chunks, etc.
One of the questions I saw not long ago asked why a particular solver would branch twice on the same variable. If that variable were binary, it would not make sense. If $z$ is a binary variable that takes a fractional value (say $z = 0.234$) in the solution of a node LP, the cuts $z \le \left\lfloor 0.234\right\rfloor$ and $z \ge \left\lceil 0.234\right\rceil$ equate to $z=0$ and $z=1$, and there is no way to further reduce the domain of $z$. For a general integer variable, such as $x$ in the illustration above, there may be reason to further subdivide the domain of $x$.
My apologies if this all seems quite basic; it is building up to the next post, where I hope to answer a not entirely trivial question.
Friday, February 20, 2015
The Geometry of a Linear Program
I frequently see questions on forums, in blog comments, or in my in-box that suggest the person asking the question is either unfamiliar with the geometry of a linear program, unsure of the significance of it, or just not connecting their question to the geometry. This post is just the starting point for addressing some of those questions. (I dislike reading or writing long posts, so I'm "chunking" my answer, and this is the first chunk.)
Ideally, the feasible region of a linear program (henceforth "LP") is a convex polytope, a geometric object that is convex, closed, bounded, and has a finite number of flat sides (intersecting at vertices or extreme points). Figure 1 shows a two dimensional example. The shaded area is the feasible region; the dots are the vertices.
Figure 1: Polytope
Being closed and bounded (and therefore, in finite dimensions, compact) ensures, via the extreme value theorem, that any continuous objective function attains its maximum and minimum over the feasible region at one or more points. Of course, the objective function of a linear or quadratic program is continuous. For a linear objective function, we can be more specific: the optimum will be attained at one or more extreme points. So we only need to check the vertices, and this in essence is what the famed simplex algorithm does.
To see why that is important, consider the following candidate for world's most trivial LP:$\begin{array}{lrcl} \textrm{minimize} & x\\ \textrm{subject to} & x & \ge & 0. \end{array}$ Its solution is, shockingly, $x=0$. Now suppose we try to use a strict inequality in the lone constraint:$\begin{array}{lrcl} \textrm{minimize} & x\\ \textrm{subject to} & x & \gt & 0. \end{array}$The lower bound of the objective function is again 0, but it is never attained, since $x=0$ is not a feasible solution. This is a consequence of the feasible region not being closed. Technically, while the objective function ($f(x)=x$) has an infimum over the feasible region, it does not have a minimum. While trivial, this is a useful example of why we never use strict inequalities (>, <) in a mathematical program.
One step more general than a polytope is a polyhedron, which retains all the characteristics of a polytope except boundedness. Figure 2 illustrates a polyhedron.
Figure 2: Polyhedron
Again, we have a finite number of flat sides (the dark line segments) intersecting in vertices (the dots), and again the region is convex and closed. The arrows indicate recession directions, directions in which we can continue forever without leaving the polyhedron. (Any direction between those two is also a recession direction.) The presence of recession directions makes the question of whether the (continuous) objective function attains a maximum or minimum a bit trickier:
• if the objective function degrades (gets bigger if minimizing, smaller if maximizing) along every recession direction, the optimal value of the objective function will be attained at one or more extreme points;
• if the objective function improves (gets smaller if minimizing, bigger if maximizing) along any recession direction, the problem is unbounded (the objective function does not have a maximum or minimum, whichever one you were hunting), which likely means you omitted or screwed up a constraint; and
• if the objective function is constant along at least one recession direction
and does not improve along any of them, the objective function achieves its optimum at one or more extreme points and along every ray starting from one of those extreme points and pointing in a recession direction where the objective stays constant.
Finally, a word about the role of convexity. Consider the feasible region in Figure 3, which is a polytope but is not convex.
Figure 3: Nonconvex polytope
Suppose that we want to find the point furthest to the right (maximize $x$, assuming that the horizontal axis represents $x$ and the vertical axis represents a second variable $y$). The optimal solution is clearly at point C. Suppose further that we currently find ourselves at point E, which is a local optimum but not a global one. For any algorithm to get from E to C, it has to move in a direction that decreases $x$ (making the objective function worse than at E, at least in the short run). That is a consequence of the fact that any line segment from E in the direction of C initially leaves the feasible region, which in turn follows from the lack of convexity. Most optimization algorithms (including the simplex algorithm) are not "smart" enough to do this; they tend to be myopic, moving only in directions that immediately improve the objective function. LPs are as tractable as they are in large part because their feasible regions are guaranteed to be convex.
Wednesday, February 18, 2015
Thunar Slow-down Fixed
My laptop is not exactly a screamer, but it's adequate for my purposes. I run Linux Mint 17 on it (Xfce desktop), which uses Thunar as its file manager. Not too long ago, I installed the RabbitVCS version control tools, including several plugins for Thunar needed to integrate the two.
Lately, Thunar has been incredibly slow to open (consistently approximately 20 seconds from click to window). The Web contains a variety of reports about Thunar being slow on first open, with several suggested fixes, but the fixes didn't apply in my case, and Thunar was slow for me every time, not just at first access. I tried opening it while running top in a terminal, but whatever was slowing down Thunar did not seem to be gobbling many CPU cycles.
So I uninstalled RabbitVCS and all associated plugins for Thunar, rebooted, and voila! Thunar now opens in under one second. As a side benefit, shutting down or rebooting the laptop, which previously suffered rather lengthy delays, is back to its normal speed.
I use RabbitVCS with the Nemo file manager on my PC (also Mint 17, but the Cinnamon desktop) with no problems, so the culprit is likely not RabbitVCS per se but rather something about the RabbitVCS - Thunar integration (?). Anyway, while I like RabbitVCS, I like having the laptop be responsive even more.
Thursday, February 12, 2015
Parsing Months in R
As part of a recent analytics project, I needed to convert strings containing (English) names of months to the corresponding cardinal values (1 for January, ..., 12 for December). The strings came from a CSV file, and were translated by R to a factor when the file was read. The factor had more than 12 levels: to the literal-minded (which includes R), "August" and "August " (the latter with a trailing space) are different months.
So I wanted a solution that was moderately robust with respect to extra spaces, capitalization, and abbreviation. A Google search turned up several solutions involving string manipulation, none of which entirely appealed to me. So I rolled my own, which I'm posting here. As usual, the code is licensed under a Creative Commons license (see the right-hand margin for details).
A few notes about the code:
• I used the lubridate package to provide a function (month()) for extracting the month index from a date object. I know that some people dislike loading packages they don't absolutely need (memory consumption, name space clashes, ...). I find the lubridate::month() function pleasantly robust, but if you want to avoid loading lubridate, I suggest you try one of the other methods posted on the Web.
• My code loads the magrittr package so that I can "pipeline" commands. If you load a package (such as dplyr) that in turn loads magrittr, you're covered. If you prefer the pipeR package, a minimal amount of tweaking should produce a version that works with pipeR. If you just want to avoid loading anything, the same logic will work; you just need to change the piping into nested function calls.
• I make no claim that this is the most efficient, most robust or most elegant solution. It just seems to work for me.
The code includes a small example of its use.
#
#
library(lubridate)
library(magrittr)
#
# Function monthIndex converts English-language string
# representations of a month name to the equivalent
# cardinal value (1 for January, ..., 12 for December).
#
# Argument:
# x a character vector, or object that can be
# coerced to a character vector
#
# Value:
# a numeric vector of the same length as x,
# containing the ordinals of the months named
# in x (NA if the entry in x cannot be deciphered)
monthIndex <-
function(x) {
x %>%
# strip any periods
gsub("\\.", "", .) %>%
# turn it into a full date string
paste0(" 1, 2001") %>%
# turn the full string into a date
as.Date("%t%B %d, %Y") %>%
# extract the month as an integer
month
}
#
# Unit test.
#
x <- c("Sep", "May", " July ", "huh?",
"august", "dec ", "Oct. ")
monthIndex(x) # 9 5 7 NA 8 12 10
Created by Pretty R at inside-R.org
Monday, February 2, 2015
Flagging a Specific Variable Value
A recent question on a web forum, one I've seen asked elsewhere, was the following: in a mathematical programming model, how does one constrain a binary variable to take the value 1 if another variable takes a specific (predefined) value, and 0 otherwise? If the binary variable is $x$, the other variable is $y$, and the target value is $m$, then what we want is \begin{eqnarray*}y = m & \implies & x = 1 \\ y \neq m & \implies & x = 0.\end{eqnarray*}I hinted at the answer to this at the end of a previous post (Indicator Implies Relation), but perhaps a bit more detail is in order.
We will need $y$ to be bounded, say $L\le y \le U$. The easy case is when $y$ is integer valued, in which case $y\neq m$ means either $y \le m-1$ or $y \ge m+1$. One approach (I don't claim it's the only one) is to introduce two new binary variables, $z_1$ and $z_2$. Essentially, $z_1$ will be an indicator for $y\le m-1$, $z_2$ will be an indicator for $y \ge m+1$, and $x$ remains an indicator for $y=m$. To do this, we add the constraints \begin{eqnarray*} y & \le & (m-1)z_1 + mx + Uz_2 \\ y & \ge & Lz_1 + mx + (m+1)z_2 \\ z_1 + x + z_2 & = & 1. \end{eqnarray*} Life gets trickier if $y$ is continuous. For any given choice of values for the discrete variables, we need the projection of the feasible region onto the subspace of continuous variables to be closed, which rules out strict inequalities. In other words, we can't say $y > m$ or $y < m$ when $x = 0$. The best we can do is pick a small positive parameter $\epsilon > 0$ and treat any value of $y$ in $(m-\epsilon, m+\epsilon)$ as if it were $m$. The resulting inequalities are as above, but with $m \pm 1$ changed to $m \pm \epsilon$: \begin{eqnarray*} y & \le & (m-\epsilon)z_1 + mx + Uz_2 \\ y & \ge & Lz_1 + mx + (m+\epsilon)z_2 \\ z_1 + x + z_2 & = & 1. \end{eqnarray*}Note that this precludes $y \in (m-\epsilon, m) \cup (m, m+\epsilon)$, i.e., the only value between $m-\epsilon$ and $m+\epsilon$ that $y$ can take is $m$.
As a practical matter, you should be careful not to pick $\epsilon$ too small. It's tempting to think in terms of the smallest positive double-precision number known to your compiler, but that could result in $x = 0$ when theoretically $y = m$ but actually the computed value of $y$ contains a bit of rounding error. | 5,125 | 20,430 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-26 | latest | en | 0.9092 |
https://essaygazebo.com/2017/09/22/solving-various-fraction-word-problems-in-math/ | 1,723,216,314,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640767846.53/warc/CC-MAIN-20240809142005-20240809172005-00010.warc.gz | 178,064,387 | 15,776 | # The Notion of Fraction Word Problems
Before we get to the point of how we can solve various types of fraction word problems, let’s see what fractions are, kinds of fractions, as well as what you can do with them.
Fraction in mathematics is a number consisting of one or more parts (shares) of a unit. Fractions are part of the field of rational numbers. By way of the recording fractions, they are divided into two formats: ordinary and decimal.
## History and Etymology of the Term
The term fraction comes from the Latin «fractura», which, in turn, is a translation of the Arabic term with the same meaning: «to break» or «break in pieces». The foundation of the theory of ordinary fractions was laid by Greek and Indian mathematicians. Through Arabs the term came to Europe, where it was already mentioned in the Fibonacci. The terms of the numerator and denominator were invented by the Greek mathematician Maximus Planudes.
Initially, European mathematicians operated only with common fractions, and in astronomy – with sexagesimal. The current designation of fractions is derived from ancient India. At first, the fractions didn’t use slashes. The application of slashes was made permanent only about 300 years ago. In Europe, the first scientist who used and distributed Indian system of account, including the method of recording fractions, was the Italian merchant and a traveler, Fibonacci. A complete theory of fractions and fraction word problems was developed in the 16th century.
Decimal fractions are first found in China around the 3rd century BC among the calculations on the counting board. In written sources decimals were represented in the traditional format, but it gradually replaced the traditional position system. Persian mathematician and astronomer Al-Kashi in his treatise «The key of arithmetic» declared himself the inventor of decimal fractions.
In Europe, the first decimals were introduced by Immanuel Bonfis around 1350, but they became widespread only after the appearance of the works of Simon Stevin «The Tenth». Stevin wrote decimal fractions in complex ways. Commas to separate the integer part began to be used in the 17th century.
Types of Fractions
• Ordinary fractions.
Ordinary or simple fraction is a record of a rational number in the form m/n, where n isn’t equal to 0. The horizontal line (which can be vertical as well) denotes the division sign, resulting in a quotient. The dividend is called the numerator of a fraction and the divisor is the denominator. There are several types for recording ordinary fractions: 1/2 (the slash is called the «solidus»); through the horizontal line, and the lowercase formula.
• Proper and improper fractions.
Proper fraction is a fraction, whose numerator module is less than the denominator module. The improper fraction is a rational number, which by the module is bigger or equal than one.
• Mixed fractions.
The fraction recorded as a whole number and a proper fraction is called a mixed fraction and is understood as the sum of this number and the fraction. Any rational number can be written as a mixed fraction. In contrast to the mixed fraction, the fraction, containing only the numerator and denominator, is called simple.
• Composite fractions.
Composite fraction is an expression, containing multiple horizontal (or inclined) features.
• Decimal fractions.
Decimal fraction is a positional record of fractions, for example, 3.24587. The part of the recording, which is before the positional comma, is the integral part of a number (fraction), and he part standing after the comma is a fractional part. Every ordinary fraction can be converted to the decimal fraction, which in this case has a finite number of digits after the decimal point, or is a periodic fraction. Generally speaking, for the positional record of a number it is possible to use not only decimal system, but also other systems (including Fibonacci).
## Actions with Fractions
In order to solve different fraction word problems, you need to know what actions you can do with fractions. So, what can you do with fractions? Actually, anything that you can do with the usual numbers: subtract, add, multiply, and divide.
## Addition and Subtraction of Fractions
In many fraction word problems you will need to add or subtract fractions. If there is the same number in both fractions’ denominators, then in order to add the fractions, it is necessary to summarize their numerators, and in order to subtract the fractions, you have to subtract their numerators (in the same order). The resulting sum or difference will be the numerator of the result; the denominator will remain the same. If there are different numbers in the denominators of fractions, you must first bring fractions to a common denominator. When adding the mixed numbers, their integers and fractional parts are added separately. When subtracting the mixed numbers, it is recommended to first convert the fractions to the improper fractions, and then subtract one from the other, and then re-cast the result, if required, to the form of the mixed number.
## Multiplying Fractions
Multiplication is also often needed when dealing with fraction word problems. Multiplying a number by a fraction means to multiply it by the numerator and split the found value by the denominator. Consequently, we have a general rule of multiplication of fractions: to multiply fractions, multiply singly their numerators and denominators and divide the first product by the second.
## Dividing Fractions
Fraction word problems can also be about dividing fractions. In order to split some number by a fraction, multiply this number by the inverse fraction. This rule stems from the definition of division.
## Solving Basic Fraction Word Problems
All students need to know how to solve basic fraction word problems, i.e. be able to find the part of a number and the number on the part. There are three main types of fraction word problems that help understand the meaning of the fraction.
The fraction word problems are always about a certain value A, which is taken as a unit («the whole»), and some part of B, which is expressed in the following fraction: m/n.
The type of fraction word problems is determined by what is unknown – A, B, or m/n. Accordingly, there are three types of fraction word problems:
• Fraction word problems for finding a part of the number, which is expressed in a fraction.
1 – A; m/n – ? In order to find the part of a number, expressed as a fraction, that number can be divided by the denominator of the fraction and multiplied by its numerator: B = A / n x m.
• Fraction word problems for finding the number by its part, expressed in a fraction.
1-?; m/n – B. In order to find the number by its part, expressed as a fraction, it is possible to divide this part by the numerator of a fraction and multiply it by the denominator: A = B / m x n.
• Fraction word problems for finding the fraction, which one number is from the other number.
1 – A; ? – B. To find a fraction, which one number is from the other number, it is possible to divide the first number by the second number: m/n = A / B.
When studying fractions, for successful solving of fraction word problems it is important for students to understand what is taken as one (whole) in each problem, for how many parts it is broken, what the value of one share is, how many parts are taken, what is the meaning of all the parts combined, what the rules for finding a fraction of the number are, as well as the rules for finding the number by the fraction, and fractions, which is composed by the number from the other number.
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Exercise $$\PageIndex{1}$$
(See Exercise 1 from "Problems on Distribution and Density Functions", m-file npr07_01.m). The class $$\{C_j: 1 \le j \le 10\}$$ is a partition. Random variable $$X$$ has values {1, 3, 2, 3, 4, 2, 1, 3, 5, 2} on $$C_1$$ through $$C_{10}$$, respectively, with probabilities 0.08, 0.13, 0.06, 0.09, 0.14, 0.11, 0.12, 0.07, 0.11, 0.09. Determine $$E[X]$$
% file npr07_01.m
% Data for Exercise 1 from "Problems on Distribution and Density Functions"
T = [1 3 2 3 4 2 1 3 5 2];
pc = 0.01*[8 13 6 9 14 11 12 7 11 9];
disp('Data are in T and pc')
npr07_01
Data are in T and pc
EX = T*pc'
EX = 2.7000
[X,PX] csort(T,pc): % Alternate using X, PX
ex = X*PX'
ex = 2.7000
Exercise $$\PageIndex{2}$$
(See Exercise 2 from "Problems on Distribution and Density Functions", m-file npr07_02.m ). A store has eight items for sale. The prices are $3.50,$5.00, $3.50,$7.50, $5.00,$5.00, $3.50, and$7.50, respectively. A customer comes in. She purchases one of the items with probabilities 0.10, 0.15, 0.15, 0.20, 0.10 0.05, 0.10 0.15. The random variable expressing the amount of her purchase may be written
$$X = 3.5I_{C_1} + 5.0 I_{C_2} + 3.5I_{C_3} + 7.5I_{C_4} + 5.0I_{C_5} + 5.0I_{C_6} + 3.5I_{C_7} + 7.5I_{C_8}$$
Determine the expection $$E[X]$$ of the value of her purchase.
% file npr07_02.m
% Data for Exercise 2 from "Problems on Distribution and Density Functions"
T = [3.5 5.0 3.5 7.5 5.0 5.0 3.5 7.5];
pc = 0.01*[10 15 15 20 10 5 10 15];
disp('Data are in T and pc')
npr07_02
Data are in T and pc
EX = T*pc'
EX = 5.3500
[X,PX] csort(T,pc)
ex = X*PX'
ex = 5.3500
Exercise $$\PageIndex{3}$$
See Exercise 12 from "Problems on Random Variables and Probabilities", and Exercise 3 from "Problems on Distribution and Density Functions," m-file npr06_12.m). The class $$\{A, B, C, D\}$$ has minterm probabilities
$$pm =$$ 0.001 * [5 7 6 8 9 14 22 33 21 32 50 75 86 129 201 302]
Determine the mathematical expection for the random variable $$X = I_A + I_B + I_C + I_D$$, which counts the number of the events which occur on a trial.
% file npr06_12.m
% Data for Exercise 12 from "Problems on Random Variables and Probabilities"
pm = 0.001*[5 7 6 8 9 14 22 33 21 32 50 75 86 129 201 302];
c = [1 1 1 1 0];
disp('Minterm probabilities in pm, coefficients in c')
npr06_12
Minterm probabilities in pm, coefficients in c
canonic
Enter row vector of coefficients c
Enter row vector of minterm probabilities pm
Use row matrices X and PX for calculations
call for XDBN to view the distribution
EX = X*PX'
EX = 2.9890
T = sum(mintable(4));
[x,px] = csort(T,pm);
ex = x*px
ex = 2.9890
Exercise $$\PageIndex{4}$$
(See Exercise 5 from "Problems on Distribution and Density Functions"). In a thunderstorm in a national park there are 127 lightning strikes. Experience shows that the probability of of a lightning strike starting a fire is about 0.0083. Determine the expected number of fires.
$$X$$ ~ binomial (127, 0.0083), $$E[X] = 127 \cdot 0.0083 = 1.0541$$
Exercise $$\PageIndex{5}$$
(See Exercise 8 from "Problems on Distribution and Density Functions"). Two coins are flipped twenty times. Let $$X$$ be the number of matches (both heads or both tails). Determine $$E[X]$$
$$X$$ ~ binomial (20, 1/2). $$E[X] = 20 \cdot 0.5 = 10$$
Exercise $$\PageIndex{6}$$
(See Exercise 12 from "Problems on Distribution and Density Functions"). A residential College plans to raise money by selling “chances” on a board. Fifty chances are sold. A player pays $10 to play; he or she wins$30 with probability $$p = 0.2$$. The profit to the College is
$$X = 50 \cdot 10 - 30N$$, where $$N$$ is the numbe of winners
Determine the expected profit $$E[X]$$.
$$N$$ ~ binomial (50, 0.2). $$E[N] = 50 \cdot 0.2 = 10$$. $$E[X] = 500 - 30E[N] = 200$$.
Exercise $$\PageIndex{7}$$
(See Exercise 19 from "Problems on Distribution and Density Functions"). The number of noise pulses arriving on a power circuit in an hour is a random quantity having Poisson (7) distribution. What is the expected number of pulses in an hour?
$$X$$ ~ Poisson (7). $$E[X] = 7$$.
Exercise $$\PageIndex{8}$$
(See Exercise 24 and Exercise 25 from "Problems on Distribution and Density Functions"). The total operating time for the units in Exercise 24 is a random variable $$T$$ ~ gamma (20, 0.0002). What is the expected operating time?
$$X$$ ~ gamma (20, 0.0002). $$E[X] = 20/0.0002 = 100,000$$.
Exercise $$\PageIndex{9}$$
(See Exercise 41 from "Problems on Distribution and Density Functions"). Random variable $$X$$ has density function
$$f_X (t) = \begin{cases} (6/5) t^2 & \text{for } 0 \le t \le 1 \\ (6/5)(2 - t) & \text{for } 1 \le t \le 2 \end{cases} = I_{[0, 1]}(t) \dfrac{6}{5} t^2 + I_{(1, 2]} (t) \dfrac{6}{5} (2 - t)$$.
What is the expected value $$E[X]$$?
$$E[X] = \int t f_X(t)\ dt = \dfrac{6}{5} \int_{0}^{1} t^3 \ dt + \dfrac{6}{5} \int_{1}^{2} (2t - t^2)\ dt = \dfrac{11}{10}$$
Exercise $$\PageIndex{10}$$
Truncated exponential. Suppose $$X$$ ~ exponential ($$\lambda$$) and $$Y = I_{[0, a]} (X) X + I_{a, \infty} (X) a$$.
a. Use the fact that
$$\int_{0}^{\infty} te^{-\lambda t} \ dt = \dfrac{1}{\lambda ^2}$$ and $$\int_{a}^{\infty} te^{-\lambda t}\ dt = \dfrac{1}{\lambda ^2} e^{-\lambda t} (1 + \lambda a)$$
to determine an expression for $$E[Y]$$.
b. Use the approximation method, with $$\lambda = 1/50$$, $$a = 30$$. Approximate the exponential at 10,000 points for $$0 \le t \le 1000$$. Compare the approximate result with the theoretical result of part (a).
$$E[Y] = \int g(t) f_X (t)\ dt = \int_{0}^{a} t \lambda e^{-\lambda t} \ dt + aP(X > a) =$$
$$\dfrac{\lambda}{\lambda ^2} [1 - e^{-\lambda a} (1 + \lambda a)] + a e^{-\lambda a} = \dfrac{1}{\lambda} (1 - e^{-\lambda a})$$
tappr
Enter matrix [a b] of x-range endpoints [0 1000]
Enter number of x approximation points 10000
Enter density as a function of t (1/50)*exp(-t/50)
Use row matrices X and PX as in the simple case
G = X.*(X<=30) + 30*(X>30);
EZ = G8PX'
EZ = 22.5594
ez = 50*(1-exp(-30/50)) %Theoretical value
ez = 22.5594
Exercise $$\PageIndex{11}$$
(See Exercise 1 from "Problems On Random Vectors and Joint Distributions", m-file npr08_01.m). Two cards are selected at random, without replacement, from a standard deck. Let $$X$$ be the number of aces and $$Y$$ be the number of spades. Under the usual assumptions, determine the joint distribution. Determine $$E[X]$$, $$E[Y]$$, $$E[X^2]$$, $$E[Y^2]$$, and $$E[XY]$$.
npr08_01
Data in Pn, P, X, Y
jcalc
Enter JOINT PROBABILITIES (as on the plane) P
Enter row marix of VALUES of X X
Enter row marix of VALUES of Y Y
Use array operations on matrices X, Y, PX, PY, t, u, and P
EX = X*PX'
EX = 0.1538
ex = total(t.*P) % Alternate
ex = 0.1538
EY = Y*PY'
EY = 0.5000
EX2 = (X.^2)*PX'
EX2 = 0.1629
EY2 = (Y.^2)*PY'
EY2 = 0.6176
EXY = total(t.*u.*P)
EXY = 0.0769
Exercise $$\PageIndex{12}$$
(See Exercise 2 from "Problems On Random Vectors and Joint Distributions", m-file npr08_02.m ). Two positions for campus jobs are open. Two sophomores, three juniors, and three seniors apply. It is decided to select two at random (each possible pair equally likely). Let $$X$$ be the number of sophomores and $$Y$$ be the number of juniors who are selected. Determine the joint distribution for $$\{X, Y\}$$ and $$E[X]$$, $$E[Y]$$, $$E[X^2]$$, $$E[Y^2]$$, and $$E[XY]$$.
npr08_02
Data are in X, Y, Pn, P
jcalc
-----------------------
EX = X*PX'
EX = 0.5000
EY = Y*PY'
EY = 0.7500
EX2 = (X.^2)*PX'
EX2 = 0.5714
EY2 = (Y.^2)*PY'
EY2 = 0.9643
EXY = total(t.*u.*P)
EXY = 0.2143
Exercise $$\PageIndex{13}$$
(See Exercise 3 from "Problems On Random Vectors and Joint Distributions", m-file npr08_03.m ). A die is rolled. Let X be the number of spots that turn up. A coin is flipped $$X$$ times. Let $$Y$$ be the number of heads that turn up. Determine the joint distribution for the pair $$\{X, Y\}$$. Assume $$P(X = k) = 1/6$$ for $$1 \le k \le 6$$ and for each $$k$$, $$P(Y = j|X = k)$$ has the binomial $$(k, 1/2)$$ distribution. Arrange the joint matrix as on the plane, with values of $$Y$$ increasing upward. Determine the expected value $$E[Y]$$
npr08_03
Data are in X, Y, P, PY
jcalc
-----------------------
EX = X*PX'
EX = 3.5000
EY = Y*PY'
EY = 1.7500
EX2 = (X.^2)*PX'
EX2 = 15.1667
EY2 = (Y.^2)*PY'
EY2 = 4.6667
EXY = total(t.*u.*P)
EXY = 7.5833
Exercise $$\PageIndex{14}$$
(See Exercise 4 from "Problems On Random Vectors and Joint Distributions", m-file npr08_04.m ). As a variation of Exercise, suppose a pair of dice is rolled instead of a single die. Determine the joint distribution for $$\{X, Y\}$$ and determine $$E[Y]$$.
npr08_04
Data are in X, Y, P
jcalc
-----------------------
EX = X*PX'
EX = 7
EY = Y*PY'
EY = 3.5000
EX2 = (X.^2)*PX'
EX2 = 54.8333
EY2 = (Y.^2)*PY'
EY2 = 15.4583
Exercise $$\PageIndex{15}$$
(See Exercise 5 from "Problems On Random Vectors and Joint Distributions", m-file npr08_05.m). Suppose a pair of dice is rolled. Let $$X$$ be the total number of spots which turn up. Roll the pair an additional $$X$$ times. Let $$Y$$ be the number of sevens that are thrown on the $$X$$ rolls. Determine the joint distribution for $$\{X,Y\}$$ and determine $$E[Y]$$
npr08_05
Data are in X, Y, P, PY
jcalc
-----------------------
EX = X*PX'
EX = 7.0000
EY = Y*PY'
EY = 1.1667
Exercise $$\PageIndex{16}$$
(See Exercise 6 from "Problems On Random Vectors and Joint Distributions", m-file npr08_06.m). The pair $$\{X,Y\}$$ has the joint distribution:
$$X =$$ [-2.3 -0.7 1.1 3.9 5.1] $$Y =$$ [1.3 2.5 4.1 5.3]
$$P = \begin{bmatrix} 0.0483 & 0.0357 & 0.0420 & 0.0399 & 0.0441 \\ 0.0437 & 0.0323 & 0.0380 & 0.0361 & 0.0399 \\ 0.0713 & 0.0527 & 0.0620 & 0.0609 & 0.0551 \\ 0.0667 & 0.0493 & 0.0580 & 0.0651 & 0.0589 \end{bmatrix}$$
Determine $$E[X]$$, $$E[Y]$$, $$E[X^2]$$, $$E[Y^2]$$ and $$E[XY]$$.
npr08_06
Data are in X, Y, P
jcalc
---------------------
EX = X*PX'
EX = 1.3696
EY = Y*PY'
EY = 3.0344
EX2 = (X.^2)*PX'
EX2 = 9.7644
EY2 = (Y.^2)*PY'
EY2 = 11.4839
EXY = total(t.*u.*P)
EXY = 4.1423
Exercise $$\PageIndex{17}$$
(See Exercise 7 from "Problems On Random Vectors and Joint Distributions", m-file npr08_07.m). The pair $$\{X, Y\}$$ has the joint distribution:
$$P(X = t, Y = u)$$
t = -3.1 -0.5 1.2 2.4 3.7 4.9 u = 7.5 0.009 0.0396 0.0594 0.0216 0.044 0.0203 4.1 0.0495 0 0.1089 0.0528 0.0363 0.0231 -2.0 0.0405 0.132 0.0891 0.0324 0.0297 0.0189 -3.8 0.051 0.0484 0.0726 0.0132 0 0.0077
Determine $$E[X]$$, $$E[Y]$$, $$E[X^2]$$, $$E[Y^2]$$ and $$E[XY]$$.
npr08_07
Data are in X, Y, P
jcalc
---------------------
EX = X*PX'
EX = 0.8590
EY = Y*PY'
EY = 1.1455
EX2 = (X.^2)*PX'
EX2 = 5.8495
EY2 = (Y.^2)*PY'
EY2 = 19.6115
EXY = total(t.*u.*P)
EXY = 3.6803
Exercise $$\PageIndex{18}$$
(See Exercise 8 from "Problems On Random Vectors and Joint Distributions", m-file npr08_08.m). The pair $$\{X, Y\}$$ has the joint distribution:
$$P(X = t, Y = u)$$
t= 1 3 5 7 9 11 13 15 17 19 u = 12 0.0156 0.0191 0.0081 0.0035 0.0091 0.007 0.0098 0.0056 0.0091 0.0049 10 0.0064 0.0204 0.0108 0.004 0.0054 0.008 0.0112 0.0064 0.0104 0.0056 9 0.0196 0.0256 0.0126 0.006 0.0156 0.012 0.0168 0.0096 0.0056 0.0084 5 0.0112 0.0182 0.0108 0.007 0.0182 0.014 0.0196 0.0012 0.0182 0.0038 3 0.006 0.026 0.0162 0.005 0.016 0.02 0.028 0.006 0.016 0.004 -1 0.0096 0.0056 0.0072 0.006 0.0256 0.012 0.0268 0.0096 0.0256 0.0084 -3 0.0044 0.0134 0.018 0.014 0.0234 0.018 0.0252 0.0244 0.0234 0.0126 -5 0.0072 0.0017 0.0063 0.0045 0.0167 0.009 0.0026 0.0172 0.0217 0.0223
Determine $$E[X]$$, $$E[Y]$$, $$E[X^2]$$, $$E[Y^2]$$ and $$E[XY]$$.
npr08_08
Data are in X, Y, P
jcalc
---------------------
EX = X*PX'
EX = 10.1000
EY = Y*PY'
EY = 3.0016
EX2 = (X.^2)*PX'
EX2 = 133.0800
EY2 = (Y.^2)*PY'
EY2 = 41.5564
EXY = total(t.*u.*P)
EXY = 22.2890
Exercise $$\PageIndex{19}$$
(See Exercise 9 from "Problems On Random Vectors and Joint Distributions", m-file npr08_09.m). Data were kept on the effect of training time on the time to perform a job on a production line. $$X$$ is the amount of training, in hours, and $$Y$$ is the time to perform the task, in minutes. The data are as follows:
$$P(X = t, Y = u)$$
t = 1 1.5 2 2.5 3 u = 5 0.039 0.011 0.005 0.001 0.001 4 0.065 0.07 0.05 0.015 0.01 3 0.031 0.061 0.137 0.051 0.033 2 0.012 0.049 0.163 0.058 0.039 1 0.003 0.009 0.045 0.025 0.017
Determine $$E[X]$$, $$E[Y]$$, $$E[X^2]$$, $$E[Y^2]$$ and $$E[XY]$$.
npr08_09
Data are in X, Y, P
jcalc
---------------------
EX = X*PX'
EX = 1.9250
EY = Y*PY'
EY = 2.8050
EX2 = (X.^2)*PX'
EX2 = 4.0375
EY2 = (Y.^2)*PY' EXY = total(t.*u.*P)
EY2 = 8.9850 EXY = 5.1410
For the joint densities in Exercise 20-32 below
a. Determine analytically $$E[X]$$, $$E[Y]$$, $$E[X^2]$$, $$E[Y^2]$$ and $$E[XY]$$.
b. Use a discrete approximation for $$E[X]$$, $$E[Y]$$, $$E[X^2]$$, $$E[Y^2]$$ and $$E[XY]$$.
Exercise $$\PageIndex{20}$$
(See Exercise 10 from "Problems On Random Vectors and Joint Distributions"). $$f_{XY}(t, u) = 1$$ for $$0 \le t \le 1$$. $$0 \le u \le 2(1-t)$$.
$$f_X(t) = 2(1 -t)$$, $$0 \le t \le 1$$, $$f_Y(u) = 1 - u/2$$, $$0 \le u \le 2$$
$$E[X] = \int_{0}^{1} 2t(1 - t)\ dt = 1/3$$, $$E[Y] = 2/3$$, $$E[X^2] = 1/6$$, $$E[Y^2] = 2/3$$
$$E[XY] = \int_{0}^{1} \int_{0}^{2(1-t)} tu\ dudt = 1/6$$
tuappr: [0 1] [0 2] 200 400 u<=2*(1-t)
EX = 0.3333 EY = 0.6667 EX2 = 0.1667 EY2 = 0.6667
EXY = 0.1667 (use t, u, P)
Exercise $$\PageIndex{21}$$
(See Exercise 11 from "Problems On Random Vectors and Joint Distribution"). $$f_{XY} (t, u) = 1/2$$ on the square with vertices at (1, 0), (2, 1) (1, 2), (0, 1).
$$f_{X} (t) = f_{Y} (t) = I_{[0, 1]} (t) t + I_{(1, 2]} (t) (2 - t)$$
$$E[X] = E[Y] = \int_{0}^{1} t^2 \ dt + \int_{1}^{t} (2t - t^2) \ dt = 1$$, $$E[X^2] = E[Y^2] = 7/6$$
$$E[XY] = (1/2) \int_{0}^{1} \int_{1 - t}^{1 + t} dt dt + (1/2) \int_{1}^{2} \int_{t - 1}^{3 - t} du dt = 1$$
tuappr: [0 2] [0 2] 200 200 0.5*(u<=min(t+1,3-t))&(u>=max(1-t,t-1))
EX = 1.0000 EY = 1.0002 EX2 = 1.1684 EY2 = 1.1687 EXY = 1.0002
Exercise $$\PageIndex{22}$$
(See Exercise 12 from "Problems On Random Vectors and Joint Distribution"). $$f_{XY} (t, u) = 4t (1 - u)$$ for $$0 \le t \le 1$$. $$0 \le u \le 1$$
$$f_X (t) = 2t$$, $$0 \le t \le 1$$, $$f_Y(u) = 2(1 - u)$$, $$0 \le u \le 1$$
$$E[X] = 2/3$$, $$E[Y] = 1/3$$, $$E[X^2] = 1/2$$, $$E[Y^2] = 1/6$$, $$E[XY] = 2/9$$
tuappr: [0 1] [0 1] 200 200 4*t.*(1-u)
EX = 0.6667 EY = 0.3333 EX2 = 0.5000 EY2 = 0.1667 EXY = 0.2222
Exercise $$\PageIndex{23}$$
(See Exercise 13 from "Problems On Random Vectors and Joint Distribution"). $$f_{XY} (t, u) = \dfrac{1}{8} (t + u)$$ for $$0 \le t \le 2$$, $$0 \le u \le 2$$
$$f_{X} (t) = f_{Y} (t) = \dfrac{1}{4} (t + 1)$$, $$0 \le t \le 2$$
$$E[X] = E[Y] = \dfrac{1}[4} \int_{0}^{2} (t^2 + t) \ dt = \dfrac{7}{6}$$, $$E[X^2] = E[Y^2] = 5/3$$
$$E[XY] = \dfrac{1}{8} \int_{0}^{2} \int_{0}^{2} (t^2u + tu^2) \ dudt = \dfrac{4}{3}$$
tuappr: [0 1] [0 1] 200 200 4*t.*(1-u)
EX = 1.1667 EY = 1.1667 EX2 = 1.6667 EY2 = 1.6667 EXY = 1.3333
Exercise $$\PageIndex{24}$$
(See Exercise 14 from "Problems On Random Vectors and Joint Distribution"). $$f_{XY} (t, u) = 4ue^{-2t}$$ for $$0 \le t, 0 \le u \le 1$$
$$f_X (t) = 2e^{-2t}$$, $$0 \le t$$, $$f_Y(u) = 2u$$, $$0 \le u \le 1$$
$$E[X] = \int_{0}^{\infty} 2te^{-2t} \ dt = \dfrac{1}{2}$$, $$E[Y] = \dfrac{2}{3}$$, $$E[X^2] = \dfrac{1}{2}$$, $$E[Y^2] = \dfrac{1}{2}$$, $$E[XY] = \dfrac{1}{3}$$
tuappr: [0 6] [0 1] 600 200 4*u.*exp(-2*t)
EX = 0.5000 EY = 0.6667 EX2 = 0.4998 EY2 = 0.5000 EXY = 0.3333
Exercise $$\PageIndex{25}$$
(See Exercise 15 from "Problems On Random Vectors and Joint Distribution"). $$f_{XY} (t, u) = \dfrac{3}{88} (2t + 3u^2)$$ for $$0 \le t \le 2$$, $$0 \le u \le 1 + t$$.
$$f_X(t) = \dfrac{3}{88} (1 + t) (1 + 4t + t^2) = \dfrac{3}{88} (1 + 5t + 5t^2 + t^3)$$, $$0 \le t \le 2$$
$$f_Y(t) = I_{[0, 1]} (u) \dfrac{3}{88} (6u^2 + 4) + I_{(1, 3]} (u) \dfrac{3}{88} (3 + 2u + 8u^2 - 3u^3)$$
$$E[X] = \dfrac{313}{220}$$, $$E[Y] = \dfrac{1429}{880}$$, $$E[X^2] = \dfrac{49}{22}$$, $$E[Y^2] = \dfrac{172}{55}$$, $$E[XY] = \dfrac{2153}{880}$$
tuappr: [0 2] [0 3] 200 300 (3/88)*(2*t + 3*u.^2).*(u<1+t)
EX = 1.4229 EY = 1.6202 EX2 = 2.2277 EY2 = 3.1141 EXY = 2.4415
Exercise $$\PageIndex{26}$$
(See Exercise 16 from "Problems On Random Vectors and Joint Distribution"). $$f_{XY} (t, u) = 12t^2 u$$ on the parallelogram with vertices
(-1, 0), (0, 0), (1, 1), (0, 1)
$$f_X(t) = I_{[-1, 0]} (t) 6t^2 (t + 1)^2 + I_{(0, 1]} (t) 6t^2 (1 - t^2)$$, $$f_Y(u) 12u^3 - 12u^2 + 4u$$, $$0 \le u \le 1$$
$$E[X] = \dfrac{2}{5}$$, $$E[Y] = \dfrac{11}{15}$$, $$E[X^2] = \dfrac{2}{5}$$, $$E[Y^2] = \dfrac{3}{5}$$, $$E[XY] = \dfrac{2}{5}$$
tuappr: [-1 1] [0 1] 400 300 12*t.^2.*u.*(u>=max(0,t)).*(u<=min(1+t,1))
EX = 0.4035 EY = 0.7342 EX2 = 0.4016 EY2 = 0.6009 EXY = 0.4021
Exercise $$\PageIndex{27}$$
(See Exercise 17 from "Problems On Random Vectors and Joint Distribution"). $$f_{XY} (t, u) = \dfrac{24}{11} tu$$ for $$0 \le t \le 2$$, $$0 \le u \le \text{min } \{1, 2-t\}$$.
$$f_X (t) = I_{[0, 1]} (t) \dfrac{12}{11}t + I_{(1, 2]} (t) \dfrac{12}{11} t (2 - t)^2$$, $$f_Y(u) = \dfrac{12}{11} u(u - 2)^2$$, $$0 \le u \le 1$$
$$E[X] = \dfrac{52}{55}$$, $$E[Y] = \dfrac{32}{55}$$, $$E[X^2] = \dfrac{57}{55}$$, $$E[Y^2] = \dfrac{2}{5}$$, $$E[XY] = \dfrac{28}{55}$$
tuappr: [0 2] [0 1] 400 200 (24/11)*t.*u.*(u<=min(1,2-t))
EX = 0.9458 EY = 0.5822 EX2 = 1.0368 EY2 = 0.4004 EXY = 0.5098
Exercise $$\PageIndex{28}$$
(See Exercise 18 from "Problems On Random Vectors and Joint Distribution"). $$f_{XY} (t, u) = \dfrac{3}{23} (t + 2u)$$ for $$0 \le t \le 2$$, $$0 \le u \le \text{max } \{2 - t, t\}$$.
$$f_X (t) = I_{[0, 1]} (t) \dfrac{6}{23} (2 - t) + I_{(1, 2]} (t) \dfrac{6}{23} t^2$$, $$f_Y(u) = I_{[0, 1]} (u) \dfrac{6}{23} (2u + 1) + I_{(1, 2]} (u) \dfrac{3}{23} (4 + 6u - 4u^2)$$
$$E[X] = \dfrac{53}{46}$$, $$E[Y] = \dfrac{22}{23}$$, $$E[X^2] = \dfrac{397}{230}$$, $$E[Y^2] = \dfrac{261}{230}$$, $$E[XY] = \dfrac{251}{230}$$
tuappr: [0 2] [0 2] 200 200 (3/23)*(t + 2*u).*(u<=max(2-t,t))
EX = 1.1518 EY = 0.9596 EX2 = 1.7251 EY2 = 1.1417 EXY = 1.0944
Exercise $$\PageIndex{29}$$
(See Exercise 19 from "Problems On Random Vectors and Joint Distribution"). $$f_{XY} (t, u) = \dfrac{12}{179} (3t^2 + u)$$, for $$0 \le t \le 2$$, $$0 \le u \le \text{min } \{2, 3 - t\}$$.
$$f_X (t) = I_{[0, 1]} (t) \dfrac{24}{179} (3t^2 + 1) + I_{(1, 2]} (t) \dfrac{6}{179} (9 - 6t + 19t^2 - 6t^3)$$
$$f_Y (u) = I_{[0, 1]} (t) \dfrac{24}{179} (4 + u) + I_{(1, 2]} (t) \dfrac{12}{179} (27 - 24u + 8u^2 - u^3)$$
$$E[X] = \dfrac{2313}{1790}$$, $$E[Y] = \dfrac{778}{895}$$, $$E[X^2] = \dfrac{1711}{895}$$, $$E[Y^2] = \dfrac{916}{895}$$, $$E[XY] = \dfrac{1811}{1790}$$
tuappr: [0 2] [0 2] 400 400 (12/179)*(3*t.^2 + u).*(u<=min(2,3-t))
EX = 1.2923 EY = 0.8695 EX2 = 1.9119 EY2 = 1.0239 EXY = 1.0122
Exercise $$\PageIndex{30}$$
(See Exercise 20 from "Problems On Random Vectors and Joint Distribution"). $$f_{XY} (t, u) = \dfrac{12}{227} (3t + 2tu)$$, for $$0 \le t \le 2$$, $$0 \le u \le \text{min } \{1 + t, 2\}$$.
$$f_X (t) = I_{[0, 1]} (t) \dfrac{12}{227} (t^3 + 5t^2 + 4t) + I_{(1, 2]} (t) \dfrac{120}{227} t$$
$$f_Y (u) = I_{[0, 1]} (t) \dfrac{24}{227} (2u + 3) + I_{(1, 2]} (u) \dfrac{6}{227} (2u + 3) (3 + 2u - u^2)$$
$$= I_{[0, 1]} (u) \dfrac{24}{227} (2u + 3) + I_{(1, 2]} (u) \dfrac{6}{227} (9 + 12u + u^2 - 2u^3)$$
$$E[X] = \dfrac{1567}{1135}$$, $$E[Y] = \dfrac{2491}{2270}$$, $$E[X^2] = \dfrac{476}{227}$$, $$E[Y^2] = \dfrac{1716}{1135}$$, $$E[XY] = \dfrac{5261}{3405}$$
tuappr: [0 2] [0 2] 400 400 (12/227)*(3*t + 2*t.*u).*(u<=min(1+t,2))
EX = 1.3805 EY = 1.0974 EX2 = 2.0967 EY2 = 1.5120 EXY = 1.5450
Exercise $$\PageIndex{31}$$
(See Exercise 21 from "Problems On Random Vectors and Joint Distribution"). $$f_{XY} (t, u) = \dfrac{2}{13} (t + 2u)$$, for $$0 \le t \le 2$$, $$0 \le u \le \text{min } \{2t, 3-t\}$$.
$$f_X (t) = I_{[0, 1]} (t) \dfrac{12}{13} t^2 + I_{(1, 2]} (t) \dfrac{6}{13} (3 - t)$$
$$f_Y(u) = I_{[0, 1]} (u) (\dfrac{4}{13} + \dfrac{8}{13} u - \dfrac{9}{52} u^2) + I_{(1, 2]} (u) (\dfrac{9}{13} + \dfrac{6}{13} u - \dfrac{51}{52} u^2)$$
$$E[X] = \dfrac{16}{13}$$, $$E[Y] = \dfrac{11}{12}$$, $$E[X^2] = \dfrac{219}{130}$$, $$E[Y^2] = \dfrac{83}{78}$$, $$E[XY] = \dfrac{431}{390}$$
tuappr: [0 2] [0 2] 400 400 (2/13)*(t + 2*u).*(u<=min(2*t,3-t))
EX = 1.2309 EY = 0.9169 EX2 = 1.6849 EY2 = 1.0647 EXY = 1.1056
Exercise $$\PageIndex{32}$$
(See Exercise 22 from "Problems On Random Vectors and Joint Distribution").
$$f_{XY} (t, u) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 2u) + I_{(1, 2]} (t) \dfrac{9}{14} t^2 u^2$$, for $$0 \le u \le 1$$.
$$f_X(t) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 1) + I_{(1, 2]} (t) \dfrac{3}{14} t^2$$, $$f_Y(u) = \dfrac{1}{8} + \dfrac{3}{4} u + \dfrac{3}{2} u^2$$ (0 \le u \le 1\)
$$E[X] = \dfrac{243}{224}$$, $$E[Y] = \dfrac{11}{16}$$, $$E[X^2] = \dfrac{107}{70}$$, $$E[Y^2] = \dfrac{127}{240}$$, $$E[XY] = \dfrac{347}{448}$$
tuappr: [0 2] [0 1] 400 200 (3/8)*(t.^2 + 2*u).*(t<=1) + (9/14)*(t.^2.*u.^2).*(t > 1)
EX = 1.0848 EY = 0.6875 EX2 = 1.5286 EY2 = 0.5292 EXY = 0.7745
Exercise $$\PageIndex{33}$$
The class $$\{X, Y, Z\}$$ of random variables is iid(independent, identically distributed) with common distribution
$$X =$$ [-5 -1 3 4 7] $$PX =$$ 0.01 * [15 20 30 25 10]
Let $$W = 3X - 4Y + 2Z$$. Determine $$E[W]$$. Do this using icalc, then repeat with icalc3 and compare results.
Use $$x$$ and $$px$$ to prevent renaming.
x = [-5 -1 3 4 7];
px = 0.01*[15 20 30 25 10];
icalc
Enter row matrix of X-values x
Enter row matrix of Y-values x
Enter X probabilities px
Enter Y probabilities px
Use array operations on matrices X, Y, PX, PY, t, u, and P
G = 3*t - 4*u
[R,PR] = csort(G,P);
icalc
Enter row matrix of X-values R
Enter row matrix of Y-values x
Enter X probabilities PR
Enter Y probabilities px
Use array operations on matrices X, Y, PX, PY, t, u, and P
H = t + 2*u;
EH = total(H.*P)
EH = 1.6500
[W,PW] = csort(H,P); % Alternate
EW = W*PW'
EW = 1.6500
icalc3 % Solution with icalc3
Enter row matrix of X-values x
Enter row matrix of Y-values x
Enter row matrix of Z-values x
Enter X probabilities px
Enter Y probabilities px
Enter Z probabilities px
Use array operations on matrices X, Y, Z,
PX, PY, PZ, t, u, v, and P
K = 3*t - 4*u + 2*v;
EK = total(K.*P)
EK = 1.6500
Exercise $$\PageIndex{34}$$
(See Exercise 5 from "Problems on Functions of Random Variables") The cultural committee of a student organization has arranged a special deal for tickets to a concert. The agreement is that the organization will purchase ten tickets at $20 each (regardless of the number of individual buyers). Additional tickets are available according to the following schedule: 11-20,$18 each; 21-30 $16 each; 31-50,$15 each; 51-100, \$13 each
If the number of purchasers is a random variable $$X$$, the total cost (in dollars) is a random quantity $$Z = g(X)$$ described by
$$g(X) = 200 + 18I_{M1} (X) (X - 10) + (16 - 18) I_{M2} (X) (X - 20) +$$
$$(15 - 16) I_{M3} (X) (X - 30) + (13 - 15) I_{M4} (X) (X - 50)$$
where $$M1 = [10, \infty)$$, $$M2 = [20, \infty)$$, $$M3 = [30, \infty)$$, $$M4 = [50, \infty)$$
Suppose $$X$$ ~ Poisson (75). Approximate the Poisson distribution by truncating at 150. Determine $$E[Z]$$ and $$E[Z^2]$$.
X = 0:150;
PX = ipoisson(75, X);
G = 200 + 18*(X - 10).*(X>=10) + (16 - 18)*(X - 20).*(X>=20) + ...
(15 - 16)*(X - 30).*(X>=30) + (13 - 15)*(X>=50);
[Z,PZ] = csort(G,PX);
EZ = Z*PZ'
EZ = 1.1650e+03
EZ2 = (Z.^2)*PZ'
EZ2 = 1/3699e+06
Exercise $$\PageIndex{35}$$
The pair $$\{X, Y\}$$ has the joint distribution (in m-file npr08_07.m):
$$P(X = t, Y = u)$$
t = -3.1 -0.5 1.2 2.4 3.7 4.9 u = 7.5 0.009 0.0396 0.0594 0.0216 0.044 0.0203 4.1 0.0495 0 0.1089 0.0528 0.0363 0.0231 -2.0 0.0405 0.132 0.0891 0.0324 0.0297 0.0189 -3.8 0.051 0.0484 0.0726 0.0132 0 0.0077
Let $$Z = g(X, Y) = 3X^2 + 2XY - Y^2)$$. Determine $$E[Z]$$ and $$E[Z^2]$$.
npr08_07
Data are in X, Y, P
jcalc
------------------
G = 3*t.^2 + 2*t.*u - u.^2;
EG = total(G.*P)
EG = 5.2975
ez2 = total(G.^2.*P)
EG2 = 1.0868e+03
[Z,PZ] = csort(G,P); % Alternate
EZ = Z*PZ'
EZ = 5.2975
EZ2 = (Z.^2)*PZ'
EZ2 = 1.0868e+03
Exercise $$\PageIndex{36}$$
For the pair $$\{X, Y\}$$ in Exercise 11.3.35, let
$$W = g(X, Y) = \begin{cases} X & \text{for } X + Y \le 4 \\ 2Y & \text{for } X+Y > 4 \end{cases} = I_M (X, Y) X + I_{M^c} (X, Y)2Y$$
Determine $$E[W]$$ and $$E[W^2]$$.
H = t.*(t+u<=4) + 2*u.*(t+u>4);
EH = total(H.*P)
EH = 4.7379
EH2 = total(H.^2.*P)
EH2 = 61.4351
[W,PW] = csort(H,P); %Alternate
EW = W*PW'
EW = 4.7379
EW2 = (W.^2)*PW'
EW2 = 61.4351
For the distribution in Exercises 37-41 below
a. Determine analytically $$E[Z]$$ and $$E[Z^2]$$
b. Use a discrete approximation to calculate the same quantities.
Exercise $$\PageIndex{37}$$
$$f_{XY} (t, u) = \dfrac{3}{88} (2t + 3u^2)$$ for $$0 \le t \le 2$$, $$0 \le u \le 1+t$$ (see Exercise 25).
$$Z = I_{[0, 1]} (X)4X + I_{(1,2]} (X)(X+Y)$$
$$E[Z] = \dfrac{3}{88} \int_{0}^{1} \int_{0}^{1 + t} 4t (2t + 3u^2)\ dudt + \dfrac{3}{88} \int_{1}^{2} \int_{0}^{1 + t} (t + u) (2t + 3u^2)\ dudt = \dfrac{5649}{1760}$$
$$E[Z^2] = \dfrac{3}{88} \int_{0}^{1} \int_{0}^{1 + t} (4t)^2 (2t + 3u^2)\ dudt + \dfrac{3}{88} \int_{1}^{2} \int_{0}^{1 + t} (t + u)^2 (2t + 3u^2)\ dudt = \dfrac{4881}{440}$$
tuappr: [0 2] [0 3] 200 300 (3/88)*(2*t+3*u.^2).*(u<=1+t)
G = 4*t.*(t<=1) + (t + u).*(t>1);
EG = total(G.*P)
EG = 3.2086
EG2 = total(G.^2.*P)
EG2 = 11.0872
Exercise $$\PageIndex{38}$$
$$f_{XY} (t, u) = \dfrac{24}{11} tu$$ for $$0 \le t \le 2$$, $$0 \le u \le \text{min } \{1, 2 - t\}$$ (see Exercise 27)
$$Z = I_M(X, Y) \dfrac{1}{2}X + I_{M^c} (X, Y) Y^2$$, $$M = \{(t, u) : u > t\}$$
$$E[Z] = \dfrac{12}{11} \int_{0}^{1} \int_{t}^{1} t^2u\ dudt + \dfrac{24}{11} \int_{0}^{1} \int_{0}^{t} tu^3\ dudt + \dfrac{24}{11} \int_{1}^{2} \int_{0}^{2 - t} tu^3\ dudt = \dfrac{16}{55}$$
$$E[Z^2] = \dfrac{6}{11} \int_{0}^{1} \int_{t}^{1} t^3u\ dudt + \dfrac{24}{11} \int_{0}^{1} \int_{0}^{t} tu^5\ dudt + \dfrac{24}{11} \int_{1}^{2} \int_{0}^{2 - t} tu^5\ dudt = \dfrac{39}{308}$$
tuappr: [0 2] [0 1] 400 200 (24/11)*t.*u.*(u<=min(1,2-t))
G = (1/2)*t.*(u>t) + u.^2.*(u<=t);
EZ = 0.2920 EZ2 = 0.1278
Exercise $$\PageIndex{39}$$
$$f_{XY} (t, u) = \dfrac{3}{23} (t + 2u)$$ for $$0 \le t \le 2$$, $$0 \le u \le \text{max } \{2 - t, t\}$$ (see Exercise 28)
$$Z = I_M (X, Y) (X + Y) + I_{M^c} (X, Y)2Y$$, $$M = \{(t, u): \text{max } (t, u) \le 1\}$$
$$E[Z] = \dfrac{3}{23} \int_{0}^{1} \int_{0}^{1} (t + u) (t + 2u) \ dudt + \dfrac{3}{23} \int_{0}^{1} \int_{1}^{2 - t} 2u (1 + 2u)\ dudt + \dfrac{3}{23} \int_{1}^{2} \int_{1}^{t} 2u (t + 2u)\ dudt = \dfrac{175}{92}$$
$$E[Z^2] = \dfrac{3}{23} \int_{0}^{1} \int_{0}^{1} (t + u)^2 (t + 2u) \ dudt + \dfrac{3}{23} \int_{0}^{1} \int_{1}^{2 - t} 4u^2 (1 + 2u)\ dudt + \dfrac{3}{23} \int_{1}^{2} \int_{1}^{t} 4u^2 (t + 2u)\ dudt =$$
tuappr: [0 2] [0 2] 400 400 (3/23)*(t+2*u).*(u<=max(2-t,t))
M = max(t,u)<=1;
G = (t+u).*M + 2*u.*(1-M);
EZ = total(G.*P)
EZ = 1.9048
EZ2 = total(G.^2.*P)
EZ2 = 4.4963
Exercise $$\PageIndex{40}$$
$$f_{XY} (t, u) = \dfrac{12}{179} (3t^2 + u)$$, for $$0 \le t \le 2$$, $$0 \le u \le \text{min } \{2, 3-t\}$$ (see Exercise 19)
$$Z = I_M (X,Y) (X + Y) + I_{M^c} (X, Y) 2Y^2$$, $$M = \{(t, u): t \le 1, u \ge 1\}$$
$$E[Z] = \dfrac{12}{179} \int_{0}^{1} \int_{1}^{2} (t + u) (3t^2 + u)\ dudt + \dfrac{12}{179} \int_{0}^{1} \int_{0}^{1} 2u^2 (3t^2 + u)\ dudt + \dfrac{12}{179} \int_{1}^{2} \int_{0}^{3 - t} 2u^2 (3t^2 + u)\ dudt = \dfrac{1422}{895}$$
$$E[Z^2] = \dfrac{12}{179} \int_{0}^{1} \int_{1}^{2} (t + u)^2 (3t^2 + u)\ dudt + \dfrac{12}{179} \int_{0}^{1} \int_{0}^{1} 4u^4 (3t^2 + u)\ dudt + \dfrac{12}{179} \int_{1}^{2} \int_{0}^{3 - t} 4u^4 (3t^2 + u)\ dudt = \dfrac{28296}{6265}$$
tuappr: [0 2] [0 2] 400 400 (12/179)*(3*t.^2 + u).*(u <= min(2,3-t))
M = (t<=1)&(u>=1);
G = (t + u).*M + 2*u.^2.*(1 - M);
EZ = total(G.*P)
EZ = 1.5898
EZ2 = total(G.^2.*P)
EZ2 = 4.5224
Exercise $$\PageIndex{41}$$
$$f_{XY} (t, u) = \dfrac{12}{227} (2t + 2tu)$$, for $$0 \le t \le 2$$, $$0 \le u \le \text{min } \{1 + t, 2\}$$ (see Exercise 30).
$$Z = I_M (X, Y) X + I_{M^c} (X, Y) XY$$, $$M = \{(t, u): u \le \text{min } (1, 2 - t)\}$$
$$E[Z] = \dfrac{12}{227} \int_{0}^{1} \int_{0}^{1} t (3t + 2tu) \ dudt + \dfrac{12}{227} \int_{1}^{2} \int_{0}^{2 - t} t(3t + 2tu)\ dudt +$$
$$\dfrac{12}{227} \int_{0}^{1} \int_{1}^{1 + t} tu(3t + 2tu)\ dudt + \dfrac{12}{227} \int_{1}^{2} \int_{2 - t}^{2} tu (3t + 2tu)\ dudt = \dfrac{5774}{3405}$$
$$E[Z^2] = \dfrac{56673}{15890}$$
tuappr: [0 2] [0 2] 400 400 (12/227)*(3*t + 2*t.*u).*(u <= min(1+t,2))
M = u <= min(1,2-t);
G = t.*M + t.*u.*(1 - M);
EZ = total(G.*P)
EZ = 1.6955
EZ2 = total(G.^2.*P)
EZ2 = 3.5659
Exercise $$\PageIndex{42}$$
The class $$\{X, Y, Z\}$$ is independent. (See Exercise 16 from "Problems on Functions of Random Variables", m-file npr10_16.m)
$$X = -2I_A + I_B + 3I_C$$. Minterm probabilities are (in the usual order)
0.255 0.025 0.375 0.045 0.108 0.012 0.162 0.018
$$Y = I_D + 3I_E + I_F - 3$$. The class $$\{D, E, F\}$$ is independent with
$$P(D) = 0.32$$ $$P(E) = 0.56$$ $$P(F) = 0.40$$
$$Z$$ has distribution
Value -1.3 1.2 2.7 3.4 5.8 Probability 0.12 0.24 0.43 0.13 0.08
$$W = X^2 + 3XY^2 - 3Z$$. Determine $$E[W]$$ and $$E[W^2]$$.
npr10_16
Data are in cx, pmx, cy, pmy, Z, PZ
[X,PX] = canonicf(cx,pmx);
[Y,PY] - canonicf(cy,pmy);
icalc3
input: X, Y, Z, PX, PY, PZ
-------------
Use array operations on matrices X, Y, Z.
PX, PY, PZ, t, u, v, and P
G = t.^2 + 3*t.*u.^2 - 3*v;
[W,PW] = csort(G,P);
EW = W*PW'
EW = -1.8673
EW2 = (W.^2)*PW'
EW2 = 426.8529
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http://www.kenthaines.com/blog?category=one+week | 1,548,208,760,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583879117.74/warc/CC-MAIN-20190123003356-20190123025356-00035.warc.gz | 318,003,090 | 22,587 | # One Week in Mr. Haines's Math Class - Friday
## Friday
### Warm-up:
I ask students to look at their worksheet from yesterday and pick one acute triangle, one right triangle, and one obtuse triangle. I then call on 3 kids at random to share their side lengths.
### Activity:
I've gone through a great deal of discovery work yesterday, so it's time for a little guided instruction. I give a small lecture about the Greeks, who were faced with a similar problem. Except the Greeks noticed something pretty cool that happens when you square the sides...
I square the sides for each triangle and ask students to look for patterns for 2 minutes. In each class, someone noticed that the right triangle's short sides added up to the long side when all sides were squared.
So we use that as a springboard to analyze acute and obtuse triangles. Pretty quickly, we see that acute triangles' short sides add to more than the longest side, whereas obtuse triangles' short sides add to less than the long side. (I keep saying, over and over, "after you square all the sides" like a broken record because I am terrified of students forgetting that step.)
So now we have a hypothesis. How do we test it? Three new triangles! I get three new triangles from my students and try out our hypothesis, which seems to work!
Time for notes in our \$1 Textbooks. We write down the Pythagorean Theorem, a diagram with the legs and hypotenuse defined and labeled, and the rule for classifying triangles using only their sides.
Finally, I give students a final triangle worksheet. In this worksheet, they are given the sides but not provided a ruler. They must use this new classification tool to classify the triangles.
If we have time, I've embedded an extension task in this worksheet. In the first problem, we had sides of 6, 8, and 11, which resulted in an obtuse triangle. The second problem, 6, 8, 8, resulted in an acute triangle. What would we need the hypotenuse to be to make a right triangle?
None
### Monday's Goal:
Continue using the Pythagorean Theorem, now finding the missing sides of various triangles.
### Resources:
Classification Worksheet
# One Week in Mr. Haines's Math Class - Thursday
## Thursday
### Warm-up:
None - we have a lot to do today!
### Activity:
I hand out two worksheets - a set of triangles that I have drawn and a worksheet where students will collect their work for the day. I also give out a ruler to each student. The instructions are simple: Using the centimeter side of the ruler, measure all three sides of each triangle. Then classify each triangle by its sides and by its angles.
For the file, check the link above or the Resources section below,
I had to hand-draw these triangles using a compass and ruler to ensure that the measurements were precise, but that's fine - I love constructing geometric figures. (In fact, I think kids should spend WAAAAY more time in geometry constructing figures of their own, but that's a side issue.)
This section of the class whips by pretty quickly, and I was able to help out any students who were struggling with using a ruler. They can all use rulers, but only if I really, truly force them.
On the back of their classification worksheet, I list a bunch of triangles by their sides and ask students to classify these triangles by their sides and angles. Pretty quickly, they realize that all the triangles are scalene, but how can you tell whether they are acute, right or obtuse? Mr. Haines? Mr. Haines? Can you come here?
At this point, I am walking from table to table distributing scratch paper and encouraging students to try to draw each triangle. There is a LOT of trial and error as students draw, then redraw, then redraw their 8, 9, 10 triangles or their 2, 8, 9 triangles. I don't worry too much about this. After all, I am giving the students a headache so they appreciate the aspirin.
Also, these are some really clever students I'm teaching! In two of my three classes, I had a student come up with the following strategy, which I will paraphrase:
"I pretend the triangle is a right triangle. So I draw the short side and the medium side with a right angle, and I try to connect them with the third side. If the third side reaches perfectly, it's a right triangle. If it's too short, the triangle is acute because the short side has to bend down to reach it. If the long side is too long, the triangle is obtuse."
Pretty cool, right?
By this point, we are edging right up against the bell, so I bring the students together, run through a quick check of their classifications, and then ask them why it took them so long. Lots of grumbling about erasing and redrawing. I mimic every announcer from every infomercial eve: "There's got to be a better way!"
The bell rings. Tomorrow, we meet Pythagoras.
None
### Tomorrow's Goal:
Introduce the Pythagorean Theorem and use it to classify triangles.
### Resources:
Triangle Worksheet (I tried very hard to get the scale of the triangles to remain after scanning and converting to a PDF. Hopefully this works for you, but it may depend on your printer. Or you could construct your own set of triangles!)
Classification Worksheet
# One Week in Mr. Haines's Math Class - Wednesday
Before we get into today's lesson, which was my favorite lesson all week, can I rant about something for a minute?
Why in the world would someone try to teach about square roots without talking about squares? I'm not referring to "squares" as in raising a number to the second power. I'm talking about "squares" as in those pointy shapes with all the sides that match each other.
You will notice that I didn't even introduce the notation or the term "square root" until my lesson on Tuesday. This was intentional. I don't want my students to get hung up on this new vocab term or this symbol that kind of looks like a long division sign. No, I just want them thinking about how to find the length of one side of a square that has an area of 73.
So already, on day 1 of this unit, I have students who are accurately estimating square roots. They just don't know that they're doing it yet. They think they're finding the missing side of a square. Once they know that concept and build a strategy to solve that sort of problem, it's not a major shift to tell students "Ok, that thing you've been doing, where you un-square a number? That's called a square root. And it looks like this check-mark-with-a-bar-next-to-it symbol you've noticed on your calculator."
Conversely, if you start this unit by projecting a slide entitled "Square Roots" and introduce all the formal vocab and symbols from the start, then students don't have anything to ground their understanding of the operation. You are asking students to use a new symbol to enact a new operation that they've never tried before. And forget about asking them to find the missing side of a square - that would be a seemingly-impossible task to those students.
Introduce the challenge first. Once students understand the challenge, then provide the notation and the vocabulary. Don't dump them both into students' laps at the same time.
Ok, rant over.
### Warm-Up:
None. We are diving into the main part of the lesson, something I have been dying to try since I saw Andrew Stadel post about it back in August.
### Activity: Movable Number Line
My number line is huuuuge. Great for class, hard to capture in a picture.
Ok, so this might have been the most awesome thing I've done all year. I made a loooooong number line out of a string that stretches almost all the way from my window to my door. I put the numbers 0 and 10 on either side of the number line.
I told kids that I would be showing the whole class a number and then calling on one person to place that number on the number line. Nobody else could talk, but we would take a poll after the student sat down:
• Thumbs Up: Perfect!
• Thumbs Sideways: Your answer is in the correct order but needs to slide either right or left
Then I held up the number 4. I chose 4 on purpose because it's incredibly familiar to students and yet a bit tricky to place. It's closer to 0 than to 10, but how much closer? Not to mention, with such a long number line, it's going to be hard to get the placement exactly right. I was expecting a lot of "Thumbs Sideways" on this first number, and I wasn't disappointed.
By the way, I use popsicle sticks with names, often called equity sticks, to choose my participants in class for this activity. I know that some teachers feel that equity sticks cause students anxiety, but I think they are worth it for this sort of activity. First of all, they strongly improve engagement. Everyone knows that they could be called up to place the next number, so they are paying attention to each number I present. Secondly, this activity does not have a clear right-or-wrong answer. In fact, I usually end up polling the class and sliding each answer ever-so-slightly in one direction or the other. Since every answer gets improved or amended, the pressure to be exactly right is lowered. Everyone is just making their best guess.
But back to the game. The first student has just placed the number 4, and we have to decide - is it perfect, or should it slide right or left? In my first class I had a student place 4 verrry close to 0. This is a great opportunity to ask students to critique the reasoning of others in a respectful way. I had lots of great comments from students, such as "If 4 was that close to 0, you wouldn't have room for 1, 2, and 3, and you'd have way too much room for the numbers bigger than 4." As the lesson went on, the justifications became more precise.
My next choice, root(49), I chose because 7 is exactly halfway between 4 and 10, and I want to see fi my students will pick up on that.
My third choice, root(20), is where the real fun starts. Now students have to use yesterday's skill of estimating square roots without any benchmarks. Where should root(20) go? Where is 5 on this number line? Where is 6? These are all questions that my students are silently asking themselves as I hold up the card with root(20) on it. At least, it sure seems that way. My students are rapt. They can't wait to find out if their popsicle stick will get pulled.
From here, my sequence was 6, root(93), root(40), root 4, root(14), root(-4). I'm sure I could have sequenced them better, but I'm not sure how. How would you sequence this activity? Let me know in the comments.
Are these placed appropriately? If you could slide one card, which one would you slide, and where?
Are these placed appropriately? If you could slide one card, which one would you slide, and where?
Anyway, I like this lesson for a few reasons:
• Student engagement was through the roof. I felt like everyone was with me in a way that almost never happens. Kids were having fun! I even had a group of students ask me in study hall the next day if they could play "the number line game" again
• Because the number line is huge, almost nobody placed their card in exactly the right spot on the first try. We always had to shift someone's card a little to the right or the left. Conversely, almost nobody put their card in the wrong order. So all the students were participating in an activity where nobody got an answer totally wrong, but nobody got an answer totally right. The pressure that students feel when coming to the front of the class was lessened in this case. Sometimes, students had a legitimate difference of opinion and we had to agree to disagree since this activity has an inherent amount of imprecision. But that's great! I'd rather my students be disagreeing and debating as long as they back up their ideas with some evidence
• By placing radicals on a number line, students are beginning to interact with radicals as objects that have a specific value. It's not just a problem to be solved. Root(14) is a number that is somewhere between 3 and 4. That approach to radicals will be useful in Algebra 1.
• I threw root(-4) in as a challenge because I thought it would spark a great debate. I also wanted to add a little bit of new information into the day's material. Yes, it's a trick question. Maybe it will feel more memorable to students because they spent 3 minutes arguing over the location of root(-4) before I admitted that there is no place for this answer on the number line. At least, not this number line.
This took about 30 minutes, which sounds crazy, but it was my first time trying a number line, and we had great discussions between every number.
Here’s a good extension. Start with numbers other than 0 and 10. Here I’ve started with 2 and 8. Where would you place root(36)? Root(90)? Root(3)?
Once we had placed all the numbers on the number line, I pulled more popsicle sticks and got students to sort the numbers on the number line into the categories "Rational" and "Irrational." This is my attempt to cement the idea of irrational numbers within our existing activity. Kids could sort the numbers perfectly, but I still don't know if they truly understand what irrational numbers are, and how they actually differ from rational numbers. Something to think about before next year.
Lastly, I got students to create a foldable for their \$1 Textbooks to help them classify triangles by their sides and angles. This is vocab that they should know, but it's going to be vital for tomorrow's activity, so it's worth the time investment to make a good foldable.
Note to self: draw triangles in the foldable so kids don't draw their own "obtuse" triangles that are clearly acute.
## Homework
IXL Activity on estimating square roots. I am lagging in my implementation of lagging homework. I'll get better at this.
## Tomorrow's Goal
Classify triangles by their sides and angles, maybe even discover the Pythagorean Theorem?
# One Week in Mr. Haines's Math Class - Tuesday
## Tuesday
### Warm-up:
I give my students a worksheet with three problem types:
• Find the area of a square
• Find the side length of a square whose area is a perfect square such as 25, 36, 49, etc,
• Estimate the side length of a square whose area is given but not a perfect square.
I like recapping every problem type from yesterday since it reminds students of our progression on Monday. It also provides an easier on-ramp for students who were absent on Monday.
The only difference today is that students are no longer allowed to use calculators. I want them developing their own strategies to estimate square roots.
### Activity:
Once we've reviewed the questions from the warm-up, I ask students to get out their \$1 Textbooks. This is my name for my class's interactive notebooks. I was skeptical of the time commitment required to do interactive notebooks well, but this year I buckled down and got it started. I realized that I don't use a textbook in my class, so it's my responsibility as a teacher to provide a resource to students so they can review the vocabulary and concepts that they are responsible for in my class. We write in our \$1 Textbooks about 2-3 days a week, and rarely more than a single slide or two of notes. But at the end of the semester, my students have a marvelous study guide for their final exam.
We write down some review material about squaring numbers, and then I give a short speech about square roots and what that term means to me. I say that trees come in all shapes and sizes, but the thing that fundamentally determines their size is their roots. Similarly, squares come in all sizes, but their size is determined by their "roots," which is the size of the base. If you want to draw a square with an area of 25 square inches, you need a square with a root of 5 inches. This is a bit of a silly analogy, but it provides some hook for this new vocab.
Once we write down our notes on squares and roots, I ask students to turn to the back of their warm-up sheet. Along the top you'll see that I ask students to calculate the squares of the numbers from 1-10 and then find or estimate a bunch of square roots. I provide the top line because I know that some students will not immediately come up with their own strategy for estimating square roots.
Almost every student can find the square roots of perfect squares because the first row of the worksheet acts as an answer key. So if a student is having trouble with estimating roots, I ask them how to find the square root of 81. Then I ask why they are having trouble with the square root of, for example, 38. When they say something like "38 isn't in this list of numbers," I ask them to find the square number that is closest to 38 but a bit smaller. Then we find the closest square number that is close to 38 but a bit too big. We use those numbers to make our estimate, which should be between 6 and 7.
Once most students have completed the worksheet, we check the work as a class. I typically project my own worksheet using a document camera. I don't love that I have to sit in the back of the room, as I can't read my students' facial expressions. But I like being able to manipulate the worksheet itself. That way my students can easily scan my sheet and their own to check their answers. Also, while I am writing estimates on the worksheet, I can find the square root using my calculator and show that on the document camera as well.
By this point, most students have noticed that all these square roots of numbers like 47 and 6 keep going to the end of the calculator. So I go into a quick math lesson about rational numbers, and how ancient Greeks thought that they were the only type of number that existed until they tried to find the area of a circle. I know this history lesson is rough (and not exactly accurate), but it taps into my students' prior knowledge of pi and helps them to believe that the square roots of certain numbers are also irrational.
Then it's time to write down some notes about rational and irrational numbers in our \$1 Textbooks before the bell! Not the best time to introduce new vocab, I know. I'll have to remedy that tomorrow.
### Homework
IXL work on finding the square roots of perfect squares. My goal is for everyone to say "Mr. Haines, I finished the homework in like 2 minutes!"
### Tomorrow's Goal:
Build fluency with estimates of square roots.
### Resources:
Squares and Roots Worksheet
# One Week In Mr. Haines's Math Class - Monday
Sometimes I read about a great lesson or idea, but I am not sure how the teacher strings all these ideas together. So I’m going to blog about an entire week of teaching and explain how I tried to meld together several different types of lessons.
### Weekly Goals:
My goals are for students to meet the Common Core standards shown below:
CCSS.MATH.CONTENT.8.NS.A.2
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 (e.g., π2). For example, by truncating the decimal expansion of √2, show that √2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations.
CCSS.MATH.CONTENT.8.NS.A.1
Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number.
CCSS.MATH.CONTENT.8.G.B.6
Explain a proof of the Pythagorean Theorem and its converse.
CCSS.MATH.CONTENT.8.G.B.7
Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.
That’s a whole lot for one week! To be fair, there is a lot in these standards that I won’t be getting to. Here are my more specific learning goals:
• Find the square root of perfect squares
• Estimate the square root of other numbers
• Understand the difference b/t rational and irrational numbers, and know that many square roots of whole numbers are irrational
• Classify triangles as acute, right, or obtuse based solely on side lengths using the Pythagorean Theorem
• Find the missing side of a right triangle using the Pythagorean Theorem
Still, it’s ambitious. And to be honest, I wasn’t sure on Monday if I would even get to the missing side of a triangle. Fortunately, I have 3 days next week to use and apply the Pythagorean theorem before exam prep starts. So if I only get through classifying triangles, I’m happy.
## Monday
### Warm-up: Number talk about the problem 14*8.
This is the only day this week that I did a warm-up that was not connected to the material. I wish I had done more warm-ups, but unfortunately as the semester exam approaches I get nervous about running out of time. My favorite method a student used was to multiply 7*8 to get 56 twice, then add 56+56=112.
### Activity:
I handed out this worksheet. The front asks students to find the area of squares. I know that geometry is my students’ weakest topic in our standards, so I try to start at the very beginning. Some students didn’t know or remember how to find the area of a square. Is this shocking to me as an 8th grade teacher? Definitely. But I just walked around the room with blank graph paper. If a student couldn’t find the area of the first few squares, I asked the students to draw each square on graph paper and count the number of boxes inside each one. Every student pretty quickly realized that they could multiply the base of the square by its height.
The bottom half of the worksheet I included non-integer side lengths to prime the students for the possibility that squares can have side lengths other than whole numbers. This will come in handy on the back of the worksheet when they have to estimate square roots. In my A period class, I didn’t allow calculators, but so many students were having trouble multiplying with decimals that I let them use their calculators. After all, fluency with the multiplication algorithm isn’t the point of this lesson. Noticing patterns in square numbers is the point of the lesson.
(My favorite part of this activity was the student who got 36 instead of 1/4 as his answer to problem 6, which had a side length of 1/2 ft. I asked him how he got 36 and he said “Well, I know that half of a foot is 6 inches, so I changed the measurement to inches and got 36 square inches. Am I allowed to do that?” High fives all around.)
Then we moved to the back of the worksheet. On the back, I provide the area of each square and students have to find the side length. I was prepared to walk around with graph paper again so students could try to draw a square with 36 boxes in it, but nobody seemed to need it. Kids blew through the first four problems.
The bottom half of the worksheet was harder. I had kids saying “I can’t find any number that multiplies to 19,” to which I replied “What’s a side length that’s just a little too short?” Kids would usually say that 4 was too short since it gets an area of 16. Then I said “What’s a side length that’s just a bit too long?” and kids would usually answer 5.
So I would say “Hmm. 4 is too short, but 5 is too long. Weird.” and walk away. By this point in the year, the kids are used to me doing stuff like this.
(Note: Initially I used A=20 instead of 19, but I had two students who thought the best estimate was 4.5 because 4*5 =20. Then they put down a side length of 6.7 for A=42 since 6*7=42. These are both decent estimates, but for the wrong reasons. I changed problem 5 to A=19 to avoid this misconception in the future.)
Some kids came up with decent estimates for all four problems on the bottom half, while others got stuck on problem 5. Kids would show me their answers and I would ask them if they got exactly 19. Invariably they would be reeeeally close but not quite. With about ten minutes left in class I said “Competition time. The table who gets the closest answer to problem 5 wins a jolly rancher.” With this added incentive kids were furiously plugging numbers into their calculators and refining their estimates.
I would always get one or two kids in each class who knew how to use the square root button, but they only shared their solution with their table since they didn’t want to jeopardize their chance of getting a Jolly Rancher. 8th graders are teenagers, but they are still kids.
With 5 minutes to go, I collected all their estimates on the board and we tried them out as a class. Usually one table had the “right” answer of 4.358898944, which gets exactly 19 on the calculator. But I told the kids that they were ever so slightly incorrect. In fact, if they had a more precise calculator, they would see that their answer was only correct when rounded. We finished the class talking about the fact that the answer seems to keep going and going into the decimals without stopping...
### Homework
Students did IXL homework on exponents, reviewing a concept from last year. I am trying to do lagging homework or “refresh” homework when I can, but I am still not good at it.
### Tomorrow's Goal:
Introduce square root notation, work on estimating square roots.
### Resources:
Squares and Roots Introductory Worksheet | 5,650 | 25,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-04 | latest | en | 0.939819 |
http://hi.gher.space/forum/viewtopic.php?f=3&t=1574&p=23354 | 1,553,507,937,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203865.15/warc/CC-MAIN-20190325092147-20190325114147-00488.warc.gz | 92,548,989 | 11,596 | ## Octagons with integral coordinates?
Higher-dimensional geometry (previously "Polyshapes").
### Octagons with integral coordinates?
In 2D, it is impossible to represent the coordinates of an equilateral triangle using only integer coordinates. However, in 3D this is possible, as a face of the alternated cube: (1,1,1), (-1,-1,1), (1,-1,-1), for example.
Similarly, a regular octagon cannot have integer coordinates in 2D; but can it have integer coordinates in 3D or higher? What about regular pentagons?
quickfur
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### Re: Octagons with integral coordinates?
A regular octagon can not have integer coordinates in any dimension. This is because an chord parallel to an edge would be of length 1+sqrt(2), and the implication here is that sqrt(2) can be expressed in integers.
On the other hand, one can get really close with the sets of points (17,0), and (12,12), with all permutations, change of sign. The area of this figure differs from the octagon by a measure of 1/169 of the unit-square.
The dream you dream alone is only a dream
the dream we dream together is reality.
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wendy
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### Re: Octagons with integral coordinates?
all regular.
bruce741
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### Re: Octagons with integral coordinates?
To explain Wendy's statement further: no integer multiple of 1 + sqrt(2) can appear as distance of two points with integer coordinates, regardless of dimension.
Proof:
Let's have one of the points in origin and the other one with coordinates [a1,a2,a3,...an].
The distance is sqrt(a1^2 + a2^2 + a3^2 + ... + an^2).
Assume this distance is a multiple of 1 + sqrt(2), k(1 + sqrt(2)) with k > 0.
We have sqrt(a1^2 + a2^2 + a3^2 + ... + an^2) = k(1 + sqrt(2)).
Since both sides are positive, we can square them:
a1^2 + a2^2 + a3^2 + ... + an^2 = k^2(1 + sqrt(2))^2
When we expand (1 + sqrt(2))^2, we get:
a1^2 + a2^2 + a3^2 + ... + an^2 = k^2(3 + 2 sqrt(2)) = 3 k^2 + 2 k^2 sqrt(2)
By further manipulation, we get
(a1^2 + a2^2 + a3^2 + ... + an^2 - 3 k^2)/(2 k^2) = sqrt(2)
This would express sqrt(2) as a ratio of integers, which is impossible.
Marek14
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### Re: Octagons with integral coordinates?
If your argument is correct, then that implies that any ratio of the form a+b*√c where a, b, c ≠ 0 cannot be expressed by two points with integer coordinates in any dimension. In the case of the triangle and hexagon (== truncated triangle), the ratio is of the form b*√(c/d), so this proof doesn't apply.
And indeed, we find that we can express equilateral triangles and hexagons with integer coordinates. And in fact, since the coordinates of any regular n-simplex can be expressed in the form b*√(c/d), given the right orientation, this means that any n-simplex and its Stott expansions/truncations thereof can have integer coordinates in some space of dimension k. And indeed, we find that k=n+1 is sufficient: all n-simplex truncates occur as facets of an (n+1)-cubic truncate, and although initially this seems to negate the proof (since k-cubic truncates can have coordinates of the form a+b*√2), it's actually not a problem because the self-duality of the n-simplex means that half of the (n+1)-cubic truncates are congruent to one of the others, and all unique n-simplex truncates occur as facets of some polytope of the form o4x3y3z3... where the first node in the CD diagram is unmarked, which in turn guarantees that their coordinates have a=0, so they reduce to b*√2, which, upon multiplication by 1/√2, become integers. A particular interesting example of this is the omnitruncated n-simplex, which is none other than the permutahedron of order (n+1). (A permutohedron is a polytope whose coordinates are all permutations of (1,2,3,4,...n), or equivalently, of (0,1,2,3,...(n-1)).)
This then leads to the next question: which of the regular polygons have the property that they can be represented by integer coordinates in a space of some dimension k? So far we have established that the triangle, square, and hexagon have this property, but the octagon does not. Are these 3 the only possibilities, or are there others?
quickfur
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### Re: Octagons with integral coordinates?
Well, the necessary (but maybe not sufficient) condition for integer polygon is that ALL distances between its vertices must be expressible as distances of integer points. In practice, an n-gon has n/2 distinct distances, rounded down. Now, any distance between two integer points must be an integer or a square of integer. No other values are permitted. For a triangle, square and hexagon, this is true, but I don't thing it holds for any other polygon.
The octagon argument I showed, I must say, is not completely correct: the k should not be an integer because the edge of the octagon doesn't have to be integer either -- k should be the distance between two adjacent vertices of the octagon. But this doesn't actually matter that much since if k is a distance between two points, we can construct a cuboid from hypercubes of edge k that will have diagonal an integer multiple of (1 + sqrt(2)), so it still holds up in the end.
Let's ask the question: why can we have integer triangles and hexagons in 3D? And one of the answers is that we can take normal cubic lattice and alternate it into tetrahedron/octahedron lattice. That still uses only integer points, and contains whole planes with triangular tilings. We can also freely rectify tilings -- but not truncate, as that can introduce noninteger points. Except that triangles can be truncated into hexagons because that divides the sides of triangle in 3 equal parts.
So you can see why not pentagons, heptagons or anything like that: there is no regular lattice, in any dimension, that would contain such shapes in the first place. Octagons and dodecagons can be probably eliminated by solving equations for truncating squares and hexagons.
Now the questions are what integer polytopes can be built from these. Prisms are of course doable. I suspect that all polytopes from simplex families are possible, and all polytopes from cube/orthoplex families that don't include octagons (and are not snub). Since demicubes and Gosset polytopes have no 4 branches, they should all exist in integer forms as well.
Moreover, since n-cube/orthoplex can be made integral in n-dimensions and n-simplex in n+1 dimensions, the same should hold for everything derived from them.
I suspect that square or hexagonal antiprism can't be made integer.
Marek14
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### Re: Octagons with integral coordinates?
I think your argument about lattices is not quite waterproof. There *are* space tessellations involving the pentagonal polytopes, for example. They aren't regular, of course, but they do exist. So the non-existence of a suitable lattice is not a necessarily what makes something impossible.
quickfur
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### Re: Octagons with integral coordinates?
It's interesting that the demicubes and Gosset polytopes can be integral... can their respective truncates be integral too? Or only a subset of them?
quickfur
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Location: The Great White North
### Re: Octagons with integral coordinates?
The pentagonal polytopes aren't impossible because there would be no possible tiling with them, but rather because this tiling can't be derived from the basic cubic tiling. There is a dearth of regular tilings; in 2D there are triangular and hexagonal tilings which can only have integral coordinates when the plane is put into 3D space and in 4D you have 16-cell and 24-cell tiling which can be both derived from tesseractic tiling. (16-cell tiling is alternated tesseractic and 24-cell tiling is the dual of 16-cell.) Basically, my argument was why triangle/hexagon can be integer, not why the other can't.
In 6 to 8 dimensions you have tilings based on Gosset polytopes.
I realized that the set of cubic/orthoplex polytopes with integer coordinates is smaller than I thought, btw. For example, you can't have integer rhombicuboctahedron because it has octagons hidden inside.
So in 3D, you have:
Cube - (1,1,1) with sign changes; (1,1,1) joined to (-1,1,1), (1,-1,1) and (1,1,-1)
Octahedron - (1,0,0) with sign changes and permutations; (1,0,0) joined to (0,1,0), (0,-1,0), (0,0,1) and (0,0,-1)
Cuboctahedron - (1,1,0) with sign changes and permutations; (1,1,0) joined to (1,0,1), (1,0,-1), (0,1,1) and (0,1,-1)
Truncated octahedron - (2,1,0) with sign changes and permutations; joined to (2,0,1), (2,0,-1) and (1,2,0)
In 4D, you have:
1000 - tesseract - (1,1,1,1) with sign changes; (1,1,1,1) joined to (-1,1,1,1), (1,-1,1,1), (1,1,-1,1) and (1,1,1,-1)
0100 - rectified tesseract - (1,1,1,0) with permutations and sign changes; (1,1,1,0) joined to (1,1,0,1), (1,1,0,-1), (1,0,1,1), (1,0,1,-1), (0,1,1,1) and (0,1,1,-1)
0010 - 24-cell - (1,1,0,0) with permutations and sign changes; (1,1,0,0) joined to (1,0,1,0), (1,0,-1,0), (1,0,0,1), (1,0,0,-1), (0,1,1,0), (0,1,-1,0), (0,1,0,1) and (0,1,0,-1)
0001 - 16-cell - (1,0,0,0) with permutations and sign changes; (1,0,0,0) joined to (0,1,0,0), (0,-1,0,0), (0,0,1,0), (0,0,-1,0), (0,0,0,1) and (0,0,0,-1)
0110 - bitruncated tesseract - (2,2,1,0) with permutations and sign changes; (2,2,1,0) joined to (2,2,0,1), (2,2,0,-1), (2,1,2,0) and (1,2,2,0); note that this contains truncated tetrahedra
0101 - rectified 24-cell - (2,1,1,0) with permutations and sign changes; (2,1,1,0) joined to (2,1,0,1), (2,1,0,-1), (2,0,1,1), (2,0,1,-1), (1,2,1,0) and (1,1,2,0)
0011 - truncated 16-cell - (2,1,0,0) with permutations and sign changes; (2,1,0,0) joined to (2,0,0,1), (2,0,0,-1), (2,0,1,0), (2,0,-1,0) and (1,2,0,0)
0111 - truncated 24-cell - (3,2,1,0) with permutations and sign changes; (3,2,1,0) joined to (3,2,0,1), (3,2,0,-1), (3,1,2,0) and (2,3,1,0)
Marek14
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### Re: Octagons with integral coordinates?
Marek14 wrote:[...]I realized that the set of cubic/orthoplex polytopes with integer coordinates is smaller than I thought, btw. For example, you can't have integer rhombicuboctahedron because it has octagons hidden inside.[...]
Of course. I don't know if you recall the uniform polytope coordinates derivation scheme from CD diagrams that I posted some time ago. Basically, given any CD diagram of an n-cubic uniform polytope, I can produce coordinates for them instantly just by reading the CD diagram. Given a diagram .4.3.3..., the single node at the end of the edge marked 4 contributes 1 to the resulting coordinates if it is marked, and every subsequent node contributes √2 to the previous accumulated value if it is marked, otherwise it's zero. So x4o3o, for example, would generate the coordinates (1,1,1) (because x4 contributes 1, then 4o3 adds 0, so the 2nd coordinate simply copies the first, and 3o also adds 0, so the 3rd coordinate also just copies the second). SImilarly, o4x3o produces (0, √2, √2), and o4x3x produces (0, √2, 2√2), etc..
If you consider what kind of coordinates are produced, you can immediately tell that anything that starts with o4... will have coordinates of the form k√2, which, since √2 becomes a common factor, you just divide away and you get integer coordinates. On the other hand, anything that starts with x4 produces coordinates of the form 1+k√2, so the only case that's going to give you integer coordinates is when k=0, which gives you only the n-cubes. Anything that has a ringed node after the x4 will give you 1+k√2 with non-zero k, of which the √2 factor cannot be eliminated, so it's impossible to get integer coordinates from it.
So this tells us that the only n-cube family uniform polytopes that can have integer coordinates are of the form o4.3.3...3. or x43o3o3o3o...3o.
Furthermore, by deleting the .4 node from the diagram, you get an (n-1)-simplex family uniform truncate. By extension, then, any n-cube truncate that has integer coordinates will also produce an (n-1)-simplex truncate with integer coordinates. Since the o4.3.3.... form of n-cube truncates allow arbitrary x or o marking of all nodes after the o4 node, this means all (n-1)-simplex truncates (including the simplex itself) can have integer coordinates. Therefore, we conclude that all uniform simplex truncates can have integer coordinates (albeit in (n+1)-space; in their "native" n-space they require radicals of the form √(a/b) with the values of a and b ranging over the square/triangular numbers, respectively, so it is not possible to eliminate them all except for n=1.).
quickfur
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### Re: Octagons with integral coordinates?
Now, it's interesting that you mention the Gosset polytopes; it makes me wonder if it's possible to derive larger regular polygons from them? Also, what subset of their truncates continue to allow integral coordinates?
A further line of thought proceeds as follows: one could argue that the reason for the apparent arbitrary preference for n-cubes and the incidental n-simplices (or, if you like, squares and triangles), is that our chosen coordinate system is unduly biased in their preference. So the next question is, suppose, instead of using regular Cartesian coordinates, we use a "skewed" coordinate system instead, say in 3D we choose 3 edges from the tetrahedron that share the same vertex and use them to establish the basis vectors of our coordinate system. Then, obviously, the tetrahedron will have integral coordinates in 3D (we wouldn't need to go to 4D to get this), under this new coordinate system. But would this allow us to also express all the other tetrahedron truncates with integer coordinates? Would the cubic truncates still have integer coordinates under this system?
Is there a choice of basis vectors that would give us integral coordinates for the pentagonal polytopes? Can all of them be expressed using a minimal basis set (i.e., only 3 vectors allowed in 3D)?
Is there a choice of (minimal) basis vectors that would give us integral coordinates for octagonal polytopes?
If so, what is the (minimal) basis set that would give us the maximal number of regular polygons with integer coordinates?
quickfur
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### Re: Octagons with integral coordinates?
The ''eutactic star" of a mirror-group is formed by unit vectors perpendicular to each of the mirror planes. The eutactic lattice is then formed by placing a parallel eutactic star at the ends of any open tips.
The interesting thing is that any wythoff mirror-edge polytope of integer edge, has its vertices on the eutactic lattice, and therefore must have chord distances which are a subset of the eutactic lattice. We can then see that if the nodes of a wythoff symbol are marked with integer branches, the distances between any two vertices must also be the square root of an integer. The only regular polygons that pass this test are Wn, where n is an integer (ie triangle, square, hexagon, horogon, W5, etc).
When there is a 4-branch or a 6-branch in the picture, you do get A+Bq or A+Bh, where q²=2, h²=3. But if all the nodes on one side of the 4-branch are marked q, then the vertices become integer again. So o4o3q and x4o3o fall on the same lattice. The Lie group people mark a four branch as q==>==x (double bar and arrow, pointing from the q to the x), since they don't admit integers from Z4 (A+Bq) or Z6 (A+Bh)
The 5-branch creates an integer system in the pentagonal numbers Z5 = A+Bf.
The tiling of octagons, or of octagrammy in 4D, produce a cover in Z4 (the span of chords of the square, or A+Bq), while the dodecagonal tiling is in Z6 (hexagon-chords, A+Bh). But all of the A+Bx, of Z4, Z5, Z6, are class 2 (infinitely dense as the decimals are, that is, mapping a 2d lattice onto the number line).
The dream you dream alone is only a dream
the dream we dream together is reality.
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wendy
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Location: Brisbane, Australia | 4,767 | 16,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-13 | latest | en | 0.850267 |
http://gedmath.blogspot.com/2014/ | 1,548,029,639,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583743003.91/warc/CC-MAIN-20190120224955-20190121010955-00103.warc.gz | 96,207,415 | 22,045 | ## Wednesday, October 08, 2014
### Finding the Least Common Denominator of Three Fractions
The least common denominator of two or more fractions is the smallest number that can be divided evenly by each of the fractions' denominators. You can determine the LCM (least common multiple) by finding multiples of the denominators of the fractions.
Find the least common denominator of the following fractions: 5/12, 7/36, and 3/8.
8, 16, 24, 36
12, 24, 36
36
The least common denominator is 36.
## Tuesday, October 07, 2014
### Least Common Multiple
Find the least common denominator of 6, 8, 12.
6, 12, 18, 24
8, 16, 24
12, 24
The least common multiple is 24.
## Monday, October 06, 2014
### Help With GED Math Problems: Finding Lowest Common Denominator for Fractions
Building the LCD or lowest common denominators for two or more fractions can be challenging. But it is an important skill for knowing how to add and subtract fractions and one that anyone studying their GED math test will need to know.
First step: Take each denominator and factor to product of prime numbers.
Second step: Build the lowest common denominator by using each factor with the greatest exponent.
What is the lowest common denominator for the following fractions: 7/12, 7/15, 19/30? Use the product of prime factor method.
12 = 2 x 2 x 3 or 2^2 x 3
15 = 3 x 5
30 = 2 x 3 x 5
Build the lowest common denominator by using each factor (i.e. 2^2) with the greatest exponents.
If I were demonstrating the concept of building lowest common denominators to students, it would go something like this, " Let's start with the denominator twelve. The denominator 12 needs at least two twos and a three. The denominator fifteen needs a three, but because we have one from the twelve... we do not need to write another one. However, the denominator twelve needs a five, so we need to add a five. The denominator thirty needs a two... which we have so we do not need to add one. It also needs a three and a five, but because we already have both, again we do not need to add. We have now build our LCD and all we need to do is multiply the factors together. So 2 x 2 x 3 x 5 = 60. The LCD of 12, 15, and 30 is 60.
LCD = 2 x 2 x 3 x 5 = 60
## Friday, October 03, 2014
### Lowest Common Denominator
Find the lowest common denominator for the following fractions: 1/2, 1/4, 1/5
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
4, 8, 12, 16, 20
5, 10, 15, 20
Because 20 is the first common multiple of 2, 4, and 5..... it is the lowest common denominator or LCD.
## Wednesday, July 30, 2014
### GED Math Test Prep: Area of Rectangle
GED Skill: Area of rectangles
You have decided to put carpet in your 10ft by 15 ft. living room. What is the area of carpet needed?
Answer: 10ft x 15ft = 150 cubic feet
### GED Math Test Prep: Simplify the equation 3x + 7y - 2z + 3 - 6x - 5z +15
Simplify the following equation.
3x + 7y - 2z + 3 - 6x - 5z +15
Step 1: Using the associative property, rearrange the terms of the equation so that "like" terms are next to each other.
3x - 6x - 2z - 5z + 7y + 3 + 15
Step 2: Combine like terms.
-3x - 7z + 7y + 18
## Monday, May 12, 2014
### Using the Product Rule with Exponents
When you multiply constants (variables) that have the same base, you add the exponents... but keep the base unchanged.
For example:
x^2c · x^3 = x^(2+3) = x^5
(x · x) (x · x · x) = x^5
"X" squared times "X" cubed equals "X" to the fifth power.
Try a few more.
1) p^5 · p^4 =
2) 2t^2 · 3t^4
3) r^2 · 2^3 · r^5
4) 3x^2 · 2x^5 · x^4
5) (p^2)(3p^4)(3p^2)
1) p^9
2) 6t^6
3) 2r^10
4) 6r^11
5) 9p^8
## Thursday, May 08, 2014
### Simplify and Solve Using the Addition Principal of Equality
4 ( 8 - 15) + (-10) = x - 7
4 ( 8 - 15) + (-10) = x - 7
32 - 60 + (-10) = x - 7
-28 + (-10) = x - 7
-38 = x - 7
-38 + 7 = x -7 + 7
-31 = x + 0
-31 = x
## Wednesday, May 07, 2014
### Solving Equations Using the Addition Principle of Equality
Can you find the error in the following problem?
5² + (4 - 8) = x + 15
25 + 4 = x + 15
29 = x + 15
29 + (-15) = x + 15 + (-15)
14 = x + 0
14 = x
## Tuesday, May 06, 2014
### Practice Solving Simple Equations Using the Addition Property of Equality
It is important to practice the addition property of equality. See below and solve five simple equations using the addition property of equality.
Practice Problem #1
x - 11 = 41
Practice Problem #2
x - 17 = -35
Practice Problem #3
84 = 40 + x
Practice Problem #4
45 = -15 + x
Practice Problem #5
-21 = -52 + x
Practice Problem #1
x - 11 = 41
x - 11 + 11 = 41 + 11
x + 0 = 52
x = 52
check
52 - 11 = 41
41 = 41
Practice Problem #2
x - 17 = -35
x - 17 + 17 = -35 + 17
x + 0 = -18
x = -18
check
-18 - 17 = -35
-35 = -35
Practice Problem #3
84 = 40 + x
84 + ( - 40) = 40 + (-40) + x
44 = 0 + x
44 = x
check
84 = 40 + 44
84 = 84
Practice Problem #4
45 = -15 + x
45 + 15 = -15 + 15 + x
60 = 0 + x
60 = x
check
45 = -15 + 60
45 = 45
Practice Problem #5
-21 = -52 + x
-21 + 52 = -52 + 52 + x
31 = 0 + x
31 = x
check
-21 = -52 + 31
-21 = -21
## Monday, May 05, 2014
### Solving Equations Using the Addition Property of Equality
The addition principle of equality states that if a = b, then a + c = b + c.
When you solve equations using this addition principle of equality, you need to use the additive inverse property. In other words, you must add the same number to both sides of an equation.
Example #1:
x - 5 = 10
x - 5 + 5 = 10 + 5 We add the opposite of (-5) to both sides of the equation.
x + 0 = 15 We simplify -5 + 5 = 0.
x = 15 The solution is x = 15
To check the answer, simply substitute 15 in for x, in the original equation and solve.
15 - 5 = 10
10 = 10
Example #2:
x + 12 = -5
x + 12 + (- 12) = -5 + (- 12) We add the opposite of (+12) to both sides of the equation.
x + 0 = -17 We simplify +12 - 12 = 0.
x = -17 The solution is x = -17
(-17) + 12 = -5
-5 = -5
## Friday, April 11, 2014
### Practice Percent Word Problem
A car which is normally priced at \$25,437 is marked down 10%. How much would Karen save if she purchased the car at the sale price?
(Spanish translation coming soon...)
## Tuesday, March 11, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. Twenty-one more than a number is 51. What is the number?
Veinte y uno más que el número es 51. ¿Cuál es el número?
2. Thirty-seven less than a number is 45. Find the number.
Treinta y siete menos que el número es 45. Encuentre el número.
1. 30
2. 82
## Monday, March 10, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 50 is 73. Find the number.
La suma del número y 50 es 73. Encuentre el número.
2. Thirty-one more than a number is 69. What is the number?
Treinta y uno más que el número es 69. ¿Cuál es el número?
3. A number decreased by 46 is 20. Find the number.
El número que está reducido por 46 es 20. Encuentre el número.
1. 23
2. 38
3. 66
## Friday, March 07, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 28 is 74. Find the number.
La suma del número y 28 es 74. Encuentre el número.
2. Thirty-nine more than a number is 72. What is the number?
Treinta y nueve más que el número es 72. ¿Cuál es el número?
3. Eighteen less than a number is 48. Find the number.
Dieciocho menos que el número es 48. Encuentre el número.
1. 46
2. 33
3. 66
## Thursday, March 06, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. A number increased by 21 is 52. Find the number.
El número que está aumentado por 21 es 52. Encuentre el número.
2. Twenty-five more than a number is 68. What is the number?
Veinte y cinco más que el número es 68. ¿Cuál es el número?
3. Forty-two more than a number is 58. What is the number?
Cuarenta y dos más que el número es 58. ¿Cuál es el número?
1. 31
2. 43
3. 16
## Wednesday, March 05, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. Twenty more than a number is 42. What is the number?
Veinte más que el número es 42. ¿Cuál es el número?
2. Forty-three more than a number is 85. What is the number?
Cuarenta y tres más que el número es 85. ¿Cuál es el número?
3. Twenty-two more than a number is 62. What is the number?
Veinte y dos más que el número es 62. ¿Cuál es el número?
1. 22
2. 42
3. 40
## Tuesday, March 04, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 26 is 42. Find the number.
La suma del número y 26 es 42. Encuentre el número.
2. Thirty more than a number is 51. What is the number?
Treinta más que el número es 51. ¿Cuál es el número?
3. Fifteen more than a number is 47. What is the number?
Quince más que el número es 47. ¿Cuál es el número?
1. 16
2. 21
3. 32
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. One-half of a number is 13. Find the number.
Una media de un número es 13. Encuentre el número.
2. A number decreased by 29 is 39. Find the number.
Un número que está reducido por 29 es 39. Encuentre el número.
3. The sum of a number and 39 is 56. Find the number.
La suma del número y 39 es 56. Encuentre el número.
1. 26
2. 68
317
## Tuesday, January 28, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. A number increased by eight is 14. Find the number.
El número que aumenta por ocho es 14. Encuentre el número.
2. Three less than a number is 2. Find the number.
Tres menos que el número es dos. Encuentre el número.
1. 6
2. 5
## Monday, January 27, 2014
### Translating Words Into Algebraic Expressions Examples: (Spanish & English)
1. Six less than a number is 9. Find the number.
Seis menos que el número es nueve. Encuentre el número.
2. Ten less than a number is 9. Find the number.
Diez menos que el número es nueve. Encuentre el número.
3. A number increased by seven is 12. Find the number.
El número que aumenta por siete es 12. Encuentre el número.
1. 15
2. 19
3. 5
## Friday, January 24, 2014
### Easy Tanslating Algebra Word Problems: (Spanish & English)
1. Seven more than a number is 11. What is the number?
Siete más que el número es 11. Encuentre el número.
2. The sum of a number and six is 16. Find the number.
La Suma del número y seis es 16. Encuentre el número.
3. A number diminished by 9 is 3. Find the number.
El número que reduce por nueve es tres. Encuentre el número.
1. 4
2. 10
3. 12
## Thursday, January 23, 2014
### Translating Words into Algebraic Expressions Simple: (Spanish & English)
1. A number diminished by 2 is 7. Find the number.
El número que reduce por dos es siete. Encuentre el número.
2. A number decreased by 7 is 8. Find the number.
El número que reduce por siete es ocho. Encuentre el número.
3. A number increased by three is 13. Find the number.
El número que aumenta por tres es 13. Encuentre el número.
1. 9
2. 15
3. 10
## Wednesday, January 22, 2014
### Translating Simple Number Word Problems: (Spanish & English)
1. Six less than a number is 5. Find the number.
Seis menos que el número es cinco. Encuentre el número.
2. Six less than a number is 7. Find the number.
Seis menos que el número es siete. Encuentre el número.
3. The sum of a number and three is 11. Find the number.
La suma del número y tres es 11. Encuentre el número.
1. 11
2. 13
3. 8
## Tuesday, January 21, 2014
### Translating Word Problems Simple: (Spanish & English)
1. One-third of a number is 1. Find the number.
Un tercer del número es uno. Encuentre el número.
2. A number increased by five is 13. Find the number.
El número que aumenta por cinco es 13. Encuentre el número.
3. One-third of a number is 2. Find the number.
Un tercer del número es dos. Encuentre el número.
1. 3
2. 8
3. 6
## Monday, January 20, 2014
### Translating Simple Algebra Word Problems: (Spanish & English)
1. Two more than a number is 8. What is the number?
Dos más que el número es ocho. ¿Cuál es el número?
2. Three more than a number is 5. What is the number?
Tres más que el número es cinco. ¿Cuál es el número?
3. A number decreased by 2 is 5. Find the number.
El número que reduce por dos es cinco. Encuentre el número.
1. 10
2. 8
3. 7
## Tuesday, January 14, 2014
### Algebra Word Problem: Setting up Problem (Spanish & English)
A total of r players came to a basketball practice. The coach divides them into four groups of t players each, but two players are left over. Which expression shows the relationship between the number of players out for basketball and the number of players in each group? | 4,178 | 12,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2019-04 | longest | en | 0.925409 |
http://www.physicsforums.com/showthread.php?t=744161 | 1,406,232,122,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997891176.68/warc/CC-MAIN-20140722025811-00061-ip-10-33-131-23.ec2.internal.warc.gz | 983,277,251 | 13,406 | # Why is Pi, Pi
by craigi
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P: 419 Here's something that has been bugging me for decades. Well I keep forgetting about it thankfully, but I've never really been able to answer it. Why is Pi actually the value that we have for it and not some other number? If the ratio of a circle's circumference to its diameter was any different, then a regular hexagon wouldn't be composed of 6 equilateral triangles, but is there a fundamental reason why this had to be the case? I know that it can be generated with various series, so if we use that as an answer then the question becomes, why those series? I've mentioned this to a few people over the years and everyone just looks at me blankly or gives some kind of hand-wavey argument. It seems that there's only me that sees it as a valid question. To phrase this a different way, why does flat geometry have 6 equilateral triangles fitting the circle? This happens to be true in our everyday experience of nature, but were it not the case and we were to live in curved space would we consider that curved space to be 'flat' and other spaces curved relative to it? If so then why do we end up with a definition of a flat space with the very precise symmetry of the 6 equilateral triangles fitting the circle?
Engineering Sci Advisor HW Helper Thanks P: 6,927 There is no obvious answer to the question "why is flat space flat" But for a "flat" space, there is nothing "magical" about the 6 equilateral triangles fitting together. The angles in any triangle add up to 180 degrees (that's one way of defining what "flat space" means). If you make a polygon by sticking triangles together, the angles inside a polygon with n sides add up to 180(n-2) degrees. The angles at the vertices of an regular polygon are all equal, so each angle is (180 - 360/n) degrees. So, you can find the values of n where regular polygons can fit together to "fill" the space around a point. The only possibilities are 6 triangles, 4 squares, and 3 hexagons. For n > 6, two polygons are not enough, and three are too many.
P: 419
Quote by AlephZero There is no obvious answer to the question "why is flat space flat" But for a "flat" space, there is nothing "magical" about the 6 equilateral triangles fitting together. The angles in any triangle add up to 180 degrees (that's one way of defining what "flat space" means). If you make a polygon by sticking triangles together, the angles inside a polygon with n sides add up to 180(n-2) degrees. The angles at the vertices of an regular polygon are all equal, so each angle is (180 - 360/n) degrees. So, you can find the values of n where regular polygons can fit together to "fill" the space around a point. The only possibilities are 6 triangles, 4 squares, and 3 hexagons. For n > 6, two polygons are not enough, and three are too many.
Sure, but my point is that there is something very special about the 6 equilaterals making a regular hexagon, and this only occurs in what we call a 'flat' space. If it's only flat following a definition that is chosen to coincide with our experience, then there is still something left to explain.
Mentor P: 21,216 Why is Pi, Pi For many thousands of years, all we knew about was the apparently flat space we lived in. The earth is large enough so that it's curvature is not immediately obvious. In a curved space you can have a triangle with three 90° angles.
P: 25
Quote by craigi Sure, but my point is that there is something very special about the 6 equilaterals making a regular hexagon, and this only occurs in what we call a 'flat' space. If it's only flat following a definition that is chosen to coincide with our experience, then there is still something left to explain.
There is nothing special about 6 equilaterals making a hexagon it is simply one result of their properties. Why are you looking for more than is there?
P: 419
Quote by mathsman1963 There is nothing special about 6 equilaterals making a hexagon it is simply one result of their properties. Why are you looking for more than is there?
Sure, but the 'is because it is' argument relies on us accepting that our definition of flatness is the only possible mathematical definition of flatness rather than an arbitrary definition chosen to match our experience of the physical world.
If Pi were slightly smaller then no number of equilaterals centred on a point would form a regular shape, smaller still and we would find that 5 of them would form a regular pentagon and so on. If this were our experience of the physical world would we define this to be 'flat' or would we still find the case of 6 equilaterals forming a regular hexagon to have a more special property worthy of the definition?
P: 25
Quote by craigi Sure, but the 'is because it is' argument relies on us accepting that our definition of flatness is the only possible mathematical definition of flatness rather than an arbitrary definition chosen to match our experience of the physical world. If Pi were slightly smaller then no number of equilaterals centred on a point would form a regular shape, smaller still and we would find that 5 of them would form a regular pentagon and so on. If this were our experience of the physical world would we define this to be 'flat' or would we still find the case of 6 equilaterals forming a regular hexagon to have a more special property worthy of the definition?
Well.... no. This has nothing to do with a definition of flatness and has nothing to do with the physical world. In Euclidean geometry 6 equilateral triangles equal a hexagon. This can be proved. That is the end of the story. If the proof does not supply you with enough of the 'why' that's too bad.
The title of the OP was why is pi, pi? and the answer is: because it is.
If pi were slightly smaller
is a nonsensical piece of speculation because its size follows logically from the properties of a circle.
Now you can define other geometries if you like and play with them to your heart's content but anything you discover in them will be defined in terms of the axioms and therefore the properties of the geometry. You will still not have a 'why' answer.
P: 419
Quote by mathsman1963 Well.... no. This has nothing to do with a definition of flatness and has nothing to do with the physical world. In Euclidean geometry 6 equilateral triangles equal a hexagon. This can be proved. That is the end of the story. If the proof does not supply you with enough of the 'why' that's too bad. The title of the OP was why is pi, pi? and the answer is: because it is. is a nonsensical piece of speculation because its size follows logically from the properties of a circle. Now you can define other geometries if you like and play with them to your heart's content but anything you discover in them will be defined in terms of the axioms and therefore the propertuies of the geometry. You will still not have a 'why' answer.
As you're aware that's just another 'is because it is' argument. Can we really do no better than that?
Perhaps if you take the axioms that lead to the Euclidian plane and derive from them, the value of pi then question becomes more formal and easier to understand.
P: 354
Quote by craigi As you're aware that's just another 'is because it is' argument. Can we really do no better than that? Perhaps if you take the axioms that lead to the Euclidian plane and derive the value of pi then then question becomes more formal and easier to understand.
That's exactly how it is estimated. Pi only applies in flat, Euclidean space. As someone already pointed out, if you measure a circle and its diameter on the surface of a sphere, you won't get a ratio of Pi.
So, if you accept Euclid's axioms, then you can derive Pi by looking at regular polygons inside and outside a circle. The more sides the polygon has, the nearer the length of all these sides will be to the length of the circumference. So, you can estimate Pi as accurately as you want by this method.
In fact, you can use this method to show that there is a common ratio between the diameter and circumference of any circle. I always think that's an overlooked point: why is there a common ratio for all circles?
Proving this using Euclidean geometry might be a good thing to try to do.
P: 25
Quote by craigi As you're aware that's just another 'is because it is' argument. Can we really do no better than that? Perhaps if you take the axioms that lead to the Euclidian plane and derive from them, the value of pi then question becomes more formal and easier to understand.
What other arguments do you expect in mathematics? The truth of a statement follows logically from axioms.
P: 60 Take the triangular number 2701 ( figurate number ) and locate in pi digits, starting after the decimal. You'll find that it starts at the 165th place after the decimal 2) 165th position 7) 166th position 0) 166th position 1) 167th position Sum the position numbers ( 165 +....167 ) = 666 = sum of the first 144 numbers in pi While 666 is the 36th triangular (T36), the consecutive triangular 703 (T37) is the 4th part of the tessellation that produces the 73rd triangular 2701 (T73) Neat little bit of pi trivia there :P I grew curious about pi after learning how Archimedes and others used geometric techniques to close in on pi, and wondered,.. if there is a relationship between shapes which can be used to approximate pi,...are there relationships between the numbers in pi and the numbers which produce shapes ( figurate numbers ) ? I have found some quite interesting things about pi which have actually led me to Hipparchus' work pertaining to astronomical calculations, which he supposedly checked against earlier tables of calculations provided by Chaldean astronomers. It may be a rather bold statement, or even premature on my part, but I am going to say that the current things taught regarding how accurately pi was known to ancient civilizations, is tripe. If you can derive approximations of pi by dropping sticks on the ground, you can certainly find other methods that are not intuitive. My main area of focus has been on the repunits { 111,222...888,999 } and their locations in pi, which incidentally have a very curious distribution,...but not surprisingly, the first repunit ( 111 ) is found at the 153rd place in pi, while 153 is the square of New Moons in one calendar year, ~ 12.369... New Moons / Year. By playing around with these repunits { 111,222...888,999 }, pi, their numbers in pi, and some other very curious relationships involving sums of series, figurates and palindromic primes ( emirp ) and using simple combinatorics, it led me to Hipparchus' work on astronomical calculations, totally unexpected, but then again, Hipparchus was thought by some to also have been rather adept at combinatorics himself, hence Schröder–Hipparchus numbers. I've got about 50 pages of work on this, still drafting it bit by bit. Anyway, random crazy guy here, thought I'd drop that on the table.
P: 66 We can ask the same question about all astronomical constants. I.E Why is the speed of light specifically 299 792 458 m/s? Your question about pi is more simple to understand, because if pi was any different, a circle wouldn't be possible. Pi CAN'T be any different than 3.34... (this can also be applied to pretty much everything else. However, pi is more simple to understand) I'm pretty sure others have answered your question pretty well. cb
P: 419
Quote by Cosmobrain We can ask the same question about all astronomical constants. I.E Why is the speed of light specifically 299 792 458 m/s? Your question about pi is more simple to understand, because if pi was any different, a circle wouldn't be possible. Pi CAN'T be any different than 3.34... (this can also be applied to pretty much everything else. However, pi is more simple to understand) I'm pretty sure others have answered your question pretty well. cb
PI isn't a constant of nature and circles do exist with a ratio of circumference to diameter other than 3.14... in other curvatures of space.
As an example in the physical world, if you could rotate fast enough around the centre of a circle and you attempted to measure Pi this way, it would take a different value because each arc of the circle would appear to contract.
You could assign it the same status as the constants of nature, but that would preclude an ontological flatness, which is what the question was designed to determine. Is there a mathematical reason why Pi takes the value it does or is it just a consequence of Euclidian axioms? It seems that it's the latter.
So then the question becomes, are the Euclidian axioms just chosen to match observation? The answer to this seems to be yes.
From a purely mathematical perspective the remainder of the question is what is the complete set of possible replacements for Euclid's 5th axiom and what if anthing is special about Euclid's choice for it.
I haven't been able to establish an argument for ontological flatness, even though I still suspect there must be one.
HW Helper P: 2,942 A lot of the "mysticism" surrounding the value of pi was worsened after Carl Sagan's book Contact. I guess a few people here have read the book? At the end of the book, Ellie is told by the aliens that there are secret messages hidden in transcendental constants like pi. Since she has no physical evidence of her "trip", she resorts to calculating pi to arbitrary precision, and ends up finding an anomaly in base-11 - a sudden long string of ones and zeros. When put on a grid, the string formed a circle of ones against a field of zeros. That gave her a warm and fuzzy feeling that the Universe had been intelligently designed, since pi was "built into the fabric of the Universe" (language very similar to that was used in the book). As much as I respect Sagan, this was an extremely nonsensical part of his book, and can easily mislead people who are already a little cranky to begin with. That scene was omitted in the movie. Thank goodness for that - usually, it's the film version that ends up mangling the better science from books.
P: 419
Quote by Curious3141 A lot of the "mysticism" surrounding the value of pi was worsened after Carl Sagan's book Contact. I guess a few people here have read the book? At the end of the book, Ellie is told by the aliens that there are secret messages hidden in transcendental constants like pi. Since she has no physical evidence of her "trip", she resorts to calculating pi to arbitrary precision, and ends up finding an anomaly in base-11 - a sudden long string of ones and zeros. When put on a grid, the string formed a circle of ones against a field of zeros. That gave her a warm and fuzzy feeling that the Universe had been intelligently designed, since pi was "built into the fabric of the Universe" (language very similar to that was used in the book). As much as I respect Sagan, this was an extremely nonsensical part of his book, and can easily mislead people who are already a little cranky to begin with. That scene was omitted in the movie. Thank goodness for that - usually, it's the film version that ends up mangling the better science from books.
For me, the question originated when considering packing structures in chemistry, as a child. My personal prejudice is that any relationship found in the digits of Pi can only be an expression of expansion series and my interest has only ever been a geometric one. I certainly don't consider that mystical or cranky, though I do confess to seeing a certain power afforded to nature by the symmetry of the regular hexagon that isn't available in other spaces. It's encouraging to find non-Euclidean geometry studied in mathematics and playing a significant role in nature.
P: 56 Everything tells you Pi must be 3.1459... because so many things are based on it. Just think about the confusion it would cause if Pi were 4.32492945!
P: 25 Can John Baez' Crackpot Index be applied to mathematics?
P: 419
Quote by mathsman1963 Can John Baez' Crackpot Index be applied to mathematics?
Technically, you'd need a scientific claim in order to apply it, but I think it's ok so long as you have some form of claim to apply it to. What do you have in mind?
You may not have answers to certain questions and you may not even like the questions being asked, but to suggest that someone is a crackpot for asking them is unnecessarily belligerent.
I'm used to having to work for answers and if your username is a fair reflection, then I expect that you are too.
Quote by z.js Everything tells you Pi must be 3.1459... because so many things are based on it. Just think about the confusion it would cause if Pi were 4.32492945!
If you lived in a world where you did measure Pi to be 4.32492945, would you find your own world more confusing than a world where it was 3.1459...? This might sound like a rhetorical question, but it's also a serious question that I genuinely don't know the answer to. It's also, to some extent, the essence of the questions that I'm asking in this thread.
Which gives me an idea - how do we define spatial complexity? Is it relative to Euclidean space or does it provide an absolute way to compare spaces? Is there one way to define the complexity of a space or multiple? Does Euclidean space have the lowest complexity or are there other spaces with the same degree of complexity? These would seem like a well formed mathematical questions, right? | 3,953 | 17,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2014-23 | longest | en | 0.965206 |
http://fromtheheartofleo.blogspot.com/2009/ | 1,529,466,899,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863411.67/warc/CC-MAIN-20180620031000-20180620051000-00495.warc.gz | 118,313,057 | 42,443 | # Cutler keeps his cool in Chicago's chill
Coming off two consecutive poor games, the Bears quarterback shook off his only interception to throw the game-winning touchdown pass in overtime.
Last update: December 29, 2009 - 6:11 AM
They got it after Cutler threw the game-winning touchdown pass in overtime in the Bears' 36-30 victory over the Vikings. Click here for more.
Here's the numbers:
36-30 (3630) = 11 prime X 330
330 corresponds to 3/30, the 89th day of non-leap years. And there are 11 players on each side of the ball in football.
3630 days prior to 20091229 brings us to Day Twenty (20) of the New Millenium. Ten (10) + Ten (10) = 20. Now hold that thought.
1) Ten (10) cycles of 3630 days prior to 20091228 brings us all the way back to 19100810 which is the eve of the Divine Birthdate in the year 1910 which factors into 191 prime X 10 which is the prime of God (764) = 191 prime X 4.
2) Ten (10) + Ten (10) = 20 cycles of 3630 days prior to 20091228 brings us back to 18110322 which is the 3/22 (322) mo/day which adds up to 131 "Prime of ONE (655)" + 191 "Prime of God (764)" = 322 = 23 "I (9) Am (14)" X 14. 23 and 14 when conjoined as 2314 is Leo (356) from 1958 which was the advent of the 23rd cycle of 89 "Prime of Leo (356)".
We can go one step further and bring in the third - "triune" - cycle of Ten (10) as follows:
36-30 (3630) = 11 prime X 330
3630 = Two teams of 11 players (11 X 11) multiplied together X Ten (10) + Ten (10) + Ten (10) = 30
And, to review and rephrase:
1) 20091229 - Ten (10) X 3630 days = 19100810 - which is the eve of the Divine Birthdate our Lord in the year 1910 which factors into 191 prime X 10 which is the prime of God (764) = 191 prime X 10
2) 19100810 -
Ten (10) X 3630 days = 18110322 - which brings us to the center-point of the numerical correspondence that confirms Heart-Soul of our Lord's Divine Birth in the year 1811 - which is prime - and on the 3/22 (322) mo/day which add up to 131 "Prime of ONE (655)" + 191 "Prime of God (764)". 1811 prime corresponds to our Lord's spiritual birthdate, 010811 - Year 1, Month 8, Day 11 - which converts into the spiritual numerical designator of 281 which comes from Day 11 = 1+1 = 2, Month 8, Year 1 = 281 prime and the 281st prime number is 1811 - including ONE (1).
3) 18110322 -
Ten (10) X 3630 days = 17111101 which is "All Saints Day" - think of the New Orleans Saints!
La Nouvelle-Orléans (New Orleans) was founded May 7, 1718, by the French Mississippi Company, under the direction of Jean-Baptiste Le Moyne de Bienville, on land inhabited by the Chitimacha.
17111101 was/is 2379 days from 17180507. 2379 is Ten (10) from 2389 which is the Leo (356th) prime number. 2379 factors into 61 prime X 39 which is the prime of Christ (389912) = 61 prime X 6392 which brings us full circle to "X", the Greek letter Chi (389):
The Greek letter "X" - which corresponds to the Roman numeral "X" which equals Ten (10) - stand for "Christos" which is where the english "Christ" originated from. Again, 2379 + 10 = 2389 which is the Leo (356th) prime number.
And if we take the year 1811 - above - it corresponds to the spiritual birth of our Lord in Year 1, Month 8, Day 11 (010811). 1811 is the 281st prime number and 281 is the 61st prime number - 61 being the prime of Christ (389912) = 61 prime X 6392. 281 is the spiritual numerical designator for 010811 for Day 11 = 1+1 = 2, Month 8, Year 1 convert into 281 prime, etc.
The New Orleans Saints became the #1 seed in the NFC playoff picture as a result of Chicago's win over the Vikings. So we had the energy of the very northern area of the Mississippi River - Minneapolis-St. Paul Minnesota - home of the Vikings - connecting through Chicago and the waters of Lake Michigan which flow down the Illinois River that connects with the Mississippi River just north of St. Louis before flowing all the way south to where it exits into the Gulf of Mexico just south of New Orleans.
Keep in mind, the Saints were founded in 1967 as an expansion team.
Here's how we connect the numberical dots...
1967 = 281 prime X 7. 1811 is the 281st prime number. The 61st prime number is 281. And 61 is the prime of Christ (389912) = 61 prime X 6392.
Helmet
Remember also that, according to Wikipedia, "New Orleans is one of five NFL teams that have yet to play in a Super Bowl. The club reached the NFC Championship Game in 2006, which they lost to the Chicago Bears 39–14."
This game was played on Sunday January 21, 2007, the eve of my late father's 82nd birthday.
#### Also, 3917 = 1957 X 2 which corresponds to the year 1957 which was the year prior to 1958, the year I was born, which was the advent of the 23rd "I (9) Am (14)" cycle of 89 "Prime of Leo (356)" years which will end in the year 2047. 1957 corresponds to SING (1957).
The Chicago Bears last game of the 2009 season will be in Detroit against the Lions on January 3rd. Don't underestimate the power and energy of the Detroit Lions. Think of the Token's famous rendition of "The Lion Sleeps Tonight" at this link.
## Monday, December 28, 2009
### Indianapolis Colts' Perfect Season Ends With A Simple Message: "God Is One"
By Brent Smith, ReutersPeyton Manning was on the bench for most of the second half as Curtis Painter, right, quarterbacked the Colts when the Jets overtook them on Sunday.
Colts quarterback Curtis Painter committed a big turnover on Sunday, but never should have been in the game. (MICHAEL CONROY The Associated Press)
# Indianapolis Colts fans right to jeer QB switch to Curtis Painter
When the probable league MVP, quarterback Peyton Manning, was lifted from Sunday's game with 5:36 left in the third quarter and the Colts up 15-10 over the grateful New York Jets, the capacity crowd booed its lungs out.
Howard Bryant of ESPN.com wrote:
Make no mistake, the Colts robbed their fans of a special moment -- an undefeated season -- that few have ever had the opportunity to witness. And the game had AFC playoff implications (the Baltimore Ravens and Denver Broncos could not be happy with Caldwell's decision, either).
The Colts are not alone -- teams in all sports outthink themselves in attempting to manage injuries -- but that's not enough of a reason to take something very special away from fans, players and the game, leaving behind only a hollow afternoon.
There was something very powerful going on with numerical-spiritual energy of this game. Let's get into the numbers to find out more:
1) The score was 15-10 when they pulled Manning which corresponds to 1510 which corresponds to God (764) Is (91) One (655) = 1510 = 151 prime X 10.
2) Manning was pulled with 5:36 min/sec left in the 3rd quarter which corresponds to 536 and ONE (1) cycle of 536 days prior to the date of yesterday's game, 20091227, brings us to 20080709 which was the 191st day of the year - 191 is the prime of God (764) = 191 prime X 4.
3) The final score was 29-15 (2915) and 2915 days from yesterday, 20091227, brings us to 20171220 which was 131 "Prime of ONE (655)" days from August 11th, the Divine Birthdate of our Lord God Jesus Christ.
4) The Colts season record is now 14-1 which is one (1) from 142 - there are 142 days between August 11th, the Divine Birthdate, and the end of each year (12/31). If the Colts win their next game their record for this year will be 15-1 (151) which is the prime of "God (764) Is (91) One (655)" = 1510 = 151 prime X 10 as seen above.
Manning was 14-21 for 192 yards when he was pulled - click here to confirm.
191 is ONE (1) from 191 which is the prime of God (764) = 191 prime X 4.
And 1421 (14-21) days prior to 20091227 brings us back to 20060205, my 48th birthday which is Day ONE (1) of my 49th year and 49 = 7 X 7.
And 1421 = 7 X 7 X 29
We since we are keying in on the energy of 7X7 (49) let's go back 49 days as follows:
20091227 - 49 days = 11/08, the Divine Conception Day, which is always 89 "Prime of Leo (356)" days from August 11th, the Divine Birthdate.
20091227 + 49 days = 2/14 (214), Valentines Day, which in non-leap years is always two (2) cycles of 89 "Prime of Leo (356)" days from August 11th, the Divine Birthdate. Think "two hearts together as one" when you think of two (2) cycles of 89 "Prime of Leo" since the astrological sign of Leo corresponds to the Human Heart, etc. Also, 214 adds up to I (9) Am (14) 191 "Prime of God" = 214.
The Indianapolis Colts were enjoying an NFL record 23 "I (9) Am (13)" consecutive wins before this loss.
Yesterday's win by the NY Jets brought back memories of Super Bowl III in Miami when the Jets beat the Baltimore Colts 16-7 on January 12, 1969 as follows:
Super Bowl III
Date January 12, 1969
Super Bowl III was the third AFL-NFL Championship Game in professional American football, but the first to officially bear the name "Super Bowl". (Although the two previous AFL-NFL Championship Games came to be known, retroactively, as "Super Bowls".) This game is regarded as one of the greatest upsets in sports history. The heavy underdog American Football League (AFL) champion New York Jets (11-3) defeated the National Football League (NFL) champion Baltimore Colts (13-1) by a score of 16–7.[1] It was the first Super Bowl victory for the AFL.
20091227 - 19690112 = 14959
14959 = 2137 prime X 7
Eight (8) cycles of 2137 prime days prior to 20091227 brings us back to 19630308 which was/is Jesus (15131) prime days from 20040810 which was/is ONE (1) day from the Divine Birthdate of our Lord. 1963 factors into 151 "God Is One" prime X 13 - see above.
The New York Jets' record back in 1969 was 11-3 (113) which corresponds to I (9) Am (14) One (655) = 678 = 113 prime X 6
The Baltimore Colts' record back in 1969 was 13-1 (131) which is the "Prime of ONE (655)".
Note: At the very beginning of this post was the picture of Manning (18) & Painter (7) which, when combined forms the number 187 or 718 if you reverse the order. First, I was born on 19580205 which was 187 days from August 11th, our Lord's birthday. Second, 718 mirrors 7/18 which is always 23 "I (9) Am (14)" + ONE (1) = 24 days from August 11th.
## Sunday, December 13, 2009
### August 11th & The Foundation For The Statue of Liberty
The website, "This Day In History", documented the following:
On August 11, 1885 -
\$100,000 raised in US for pedestal for Statue of Liberty
On another website the details of this story were revealed as follows:
"On August 11, 1885, the front page of the World proclaimed, "ONE HUNDRED THOUSAND DOLLARS!" The goal had been reached, and slightly exceeded, thanks to more than 120,000 contributions."
Statue of Liberty's Pedestal is 89 Feet Tall
The architect for Liberty's pedestal, Richard Morris Hunt, was a highly respected and popular designer of expensive homes. He designed an 89-foot-high pedestal that would sit upon a concrete foundation that would appear to grow up from within the 11-pointed, star-shaped walls of the existing Fort Wood. His fee for the project was \$1,000, which he returned to the fund to reassemble the statue.
General Charles P. Stone was the chief engineer in charge of the entire construction project, including the foundation, the pedestal and the reassembly of the statue. Liberty's foundation alone required 24,000 tons of concrete, the largest single mass ever poured at that time. It measures 52 feet, 10 inches in height. At the bottom, it is 91 feet, and at the top, it is 65 feet. The pedestal rises 89 feet above the foundation.
-------------------------------
The literal financial foundation for building of the Statue of Liberty was established on the 1891st birthday of our Lord God Jesus Christ. 1891 = 31 prime X 61 prime.
31 is the prime of The (285) Lord (3694) God (764) = 4743 = 31 prime X 153; and 61 is the prime of Christ (389912) = 61 prime X 6392.
So the title "The Lord God Christ" is numerically etched in the year 1885 which was 1891 years from 6 BC, the year our Lord was born. 1891 is 89 - the height of the foundation - bordered by two ones (1__1) which mirror 11 which mirror the 11-pointed, star-shaped walled structure of Fort Hood that the pedestal stands upon - see above.
Frédéric Auguste Bartholdi in 1880
Wikipedia wrote:
Frédéric Auguste Bartholdi
(August 2, 1834, Colmar, Haut-Rhin – October 4, 1904) was a French sculptor who is remembered mainly for designing the Statue of Liberty.
And another website wrote:
"The sculptor who designed the Statue of Liberty, Frédéric-Auguste Bartholdi, was born into a well-to-do family in Colmar, France on August 2, 1834."
18340802 is Leo (356) days from August 11, 1834, the Divine Birth of our Lord God Jesus Christ. In perfect correspondence with this "Leonic" energy, prior to designing the Statue of Liberty, one of Frederic-Auguste Bartholdi's famous works was the Lion of Belfort:
The Lion of Belfort
"[The Lion of Belfort] was finished in 1880 and is entirely made of red sandstone. The blocks it is made from were individually sculpted then moved under Belfort castle to be assembled. The colossal work is 22 meters long and 11 meters high and dominates the local landscape.
The lion symbolizes the heroic French resistance during the Siege of Belfort, a 103 days long Prussian assault (from December 1870 to February 1871). The city was protected from 40,000 Prussians by merely 17,000 men (only 3,500 were from the military) led by Colonel Denfert-Rochereau.
Instead of facing Prussia to the east as was intended, it was turned the other way because of German protests."
Click here to read view the PDF file of this article which was published on 12/12/1894 - 89 "Prime of Leo (356)" days from 9/14 (914) mo/day which factors into 457 prime X 2 which is the 89th prime number.
Let's get back to nuts and bolts of the Statue of Liberty:
click image to enlarge
There are 354 steps inside the statue and its pedestal, with 25 windows in the crown which comprise the jewels beneath the seven rays of the diadem. The keystone which the statue holds in her left hand reads, in Roman numerals, "July 4, 1776" the day of the adoption of the Declaration of Independence.
Note: 354 is two (2) - the ONLY female prime number - from Leo (356).
The Statue of Liberty was engineered to withstand heavy winds. Winds of 50 miles per hour (80 km/h) cause the Statue to sway 3 inches (76 mm) and the torch to sway 5 inches (130 mm). This allows the Statue to move rather than break in high wind load conditions.
Feature[49] Customary Metric
Height from base to torch 151 ft 1 in 46 m
Foundation of pedestal (ground) to tip of torch 305 ft 1 in 93 m
Heel to top of head 111 ft 1 in 34 m
Length of hand 16 ft 5 in 5 m
Index finger 8 ft 1 in 2.44 m
Circumference at second joint 3 ft 6 in 1.07 m
Head from chin to cranium 17 ft 3 in 5.26 m
Head thickness from ear to ear 10 ft 0 in 3.05 m
Distance across the eye 2 ft 6 in 0.76 m
Length of nose 4 ft 6 in 1.48 m
Right arm length 42 ft 0 in 12.8 m
Right arm greatest thickness 12 ft 0 in 3.66 m
Thickness of waist 35 ft 0 in 10.67 m
Width of mouth 3 ft 0 in 0.91 m
Tablet, length 23 ft 7 in 7.19 m
Tablet, width 13 ft 7 in 4.14 m
Tablet, thickness 2 ft 0 in 0.61 m
Height of granite pedestal 89 ft 0 in 27.13 m
Height of foundation 65 ft 0 in 19.81 m
Weight of copper used in Statue[50] 60,000 pounds 27.22 metric tonnes
Weight of steel used in Statue 250,000 pounds 113.4 metric tonnes
Total weight used in Statue 450,000 pounds 204.1 metric tonnes
Thickness of copper sheeting 3/32 of an inch 2.4 mm
The following information provided by the US Park Serviceat their excellent site: http://www.cr.nps.gov/phad/stlilcs.html IDLCS: 40500Name: Fort Wood WallsAlso Known as: Star FortBuilt: 1808 - 1811Altered: 1844Altered: 1886Altered: 1930 - 1965
The Statue of Liberty was built within the walls of Fort Hood which was completed in 1811 - see above from this link.
1811 is the 281st prime number.
281 is the numerical designator for our Lord's spiritual - ie. "corrected" - date of birth: 010811. Day 11 = 1+1 = 2, Month 8 and Year 1 combine to form 281 prime. 1811 prime mirrors 010811.
The year 1811 not only factors into the completion of Fort Hood, but it was the year that the intellectual creator of the Statue of Liberty, Édouard René Lefèbvre de Laboulaye, was born. 1811 is also 23 "I (9) Am (14)" years from 1834 the year that Frédéric Auguste Bartholdi, the designer of the Statue of Liberty, was born.
According to Wikipedia:
Édouard René Lefèbvre de Laboulaye (January 18, 1811 in Paris - May 25, 1883[1] in Paris) was a French jurist, poet, and author.
Nevertheless, he is most remembered as the intellectual creator of the Statue of Liberty in New York Harbor, and the lesser known twins in Paris, France and the Luxembourg Garden.
Always a careful observer of the politics of the United States, and an admirer of its constitution, he wrote a three-volume work on the political history of the United States, and published it in Paris during the height of the politically repressed Second Empire. During the Civil War, he was a zealous advocate of the Union cause, publishing histories of the cultural connections of the two nations, while the United States was in the throes of its Civil War (1862 and 1863). At the war's conclusion, in 1865 he had the idea of presenting a statue representing liberty as a gift to the United States, a symbol for ideas suppressed by Napoleon III. The sculptor Frederic-Auguste Bartholdi, one of Laboulaye's friends, turned the idea into reality.
-----------------------------
Édouard René Lefèbvre de Laboulaye's numerical age (NA) at the time of his passing/ascension was 26425 days as follows:
18830525 b. - 18110118 d. = 26425 = 151 prime X 175
151 is the prime of "God (764) Is (91) One (655)" = 1510 = 151 prime X 10
And 151 is the height from base to torch (Bartholdi's design): 151' 1" (46.50m).
Finally, let's examine the date the cornerstone was set into place as follows:
Date the cornerstone was laid on Bedloe's Island:
5 August 1884
Source of Granite for the Pedestal: Leete's Island, Connecticut
Official accepting Statue on behalf of US:
President Grover Cleveland
Date of Acceptance by President: October 28, 1886
Part of Acceptance Statement by President Cleveland:
"We will not forget that liberty here made her home;
nor shall her chosen altar be neglected".
18861028 was the 815th day from/including the laying of the cornerstone on 18840805. The end of the 23rd "I (9) Am (14)" cycles of 815 days from/including 18840805 brings us to 19351201 which was/is EXACTLY "23 (9) Am (14)" + 89 "Prime of Leo (356)" days from August 11th in the year 1935 which was/is 23 "I Am" years from 1958 which was the advent year for the 23rd cycle of 89 "Prime of Leo (356)" year cycle that will end in the year 2047.
Note again that the Leo (356th) prime number is 2389 which is composed of 23 "I Am" plus 89 "Prime of Leo".
The Leo (356th) Prime 2389 = "I Am Leo!"
### "Bob" Heft, Designer of the 50-star United States flag passes 12/12/09
Robert G. Heft, designer of the 50-star US Flag
Wikipedia wrote:
Robert G. "Bob" Heft (1942 - December 12, 2009), born in Saginaw, Michigan, was a designer of the 50-star flag, and one of the proposed designs for a 51-star flag for the United States of America. He spent his childhood in Lancaster, Ohio, where he created the flag as a school project.
He designed the current U.S. flag in 1958 while living with his grandparents. He was 17 years old at the time and did the flag design as a class project. He unstitched the blue field from a family 48-star flag, sewed in a new field, and used iron-on white fabric to add 100 hand-cut stars, 50 on each side of the blue canton.[1]
He originally received a B- for the project. After discussing the grade with his high school teacher, Stanley Pratt, it was agreed that if the flag was accepted by Congress, the grade would be reconsidered. Heft's flag design was chosen and adopted by presidential proclamation after Alaska and before Hawaii was admitted into the union in 1959. According to Heft, his teacher did keep to their agreement and changed his grade to an A for the project.
=================
On a separate website I found evidence of the exact date that Bob designed the 50-star flag as follows:
#### I designed the 50 star flag on April 18, 1958, two years before it became official (July 4th, 1960 at 12:01 p.m. EST). It even preceded the 49 star flag by nearly 1 ½ years. It was out of sequence. Two weeks after making the 50 star flag, I thought that I would go even further and made the 51 star flag....just in case. I'm still waiting for that to happen (Ha-Ha)
--------------------
Now let's dig into the numbers as follows:
The Universal Numerical Date (UND) for 19580418 was/is 717222 which factors into 10867 prime X 66.
66 is 23 "I (9) Am (14)" from 89 which is the prime of Leo (356) = 89 prime X 4. So let's continue to dig....
1958 was the year I was born. But, more important, 1958 was the advent of the 23rd "I Am" cycle of 89 "Prime of Leo (356)" year cycle that will end in the year 2047 = 23 X 89. Note that 23X89 mirrors 2389 which is the Leo (356th) prime number.
19580418 was the completion of the 73rd day from the day I was born - 73 = 23 "I (9) Am (14)" plus 50 as in "50 stars" on the flag. And if we move backward two (2) X 50 = 100 days prior to 19580418 we will arrive at the 23rd "I Am" birthday of "The King" - Elvis Presley.
19580418 was/is actually the 108th day of the year which corresponds to the 1/08 mo/day which was "The King's" - Elvis Presley's - 23rd "I Am" birthday. And 108 days from 19580418 brings us to 19580804 which was/is two (2) female prime X 89 "Prime of Leo (356)" days from 19580207 which was the 26th birthday of Dean Fagerstrom, author of "The Book of Anglion" from which this Divine Science of Numerical Correspondences comes from.
Let's get back to 10867 which is the prime of 717222 which is the UND for 19580418 and which factors into 10867 prime X 66...
One obvious leading question is: what's the date for the end of the 89th "Prime of Leo (356)" cycle of 10867 prime days?
10867 prime X 89 days = 967163 UND which is August 11th, the Divine Birthdate of our Lord God Jesus Christ in the year 2642.
Robert Heft passed/ascended yesterday, 20091212, which was exactly 18866 days from 19580418, the day he was called to design the 50-state flag. 18866 factors into 9433 prime X 2.
9433 prime days from 19580418 brings us to 19840214 which was/is Valentines Day - which in non-leap years is always two (2) X 89 "Prime of Leo (356)" days from August 11th, the Divine Birthdate. 19840214 was Aaron Russo's - a great modern American patriot - 41st birthday.
The 2/14 (214) mo/day adds up to 23 "I (9) Am (14) + 191 which is the prime of God (764) = 191 prime X 4, etc.
26420811 is 250,013 days from the day I was born, 19580205.
33270214, Valentines Day, is 250,013 days from the Divine Birthdate, 26420811
Two (2) cycles of 89 "Prime of Leo (356)" days from 33270214 brings us to August 11th, the Divine Birthdate of our Lord God Jesus Christ.
Our Lord picked a very powerful day for Bob Heft to pass/ascend as the 12/12 mo/day is ALWAYS 89 "Prime of Leo (356)" days from the 9/14 (914) mo/day which factors into 457 X 2 which is the 89th prime number. In addition the UND for 20091212 was/is 736088 which factors into 911 prime X 808 which evokes the massive energies that threatened - and continue to threaten - this great American Constitutional Republic.
PS. Just saw this piece of information shared in today's article from the Columbus Ohio Dispatch as follows:
Robert G. Heft: 1942-2009 - Creator of 50-star flag was teen in Lancaster
Sunday, December 13, 2009 3:33 AM
THE COLUMBUS DISPATCH
"Heft, who was 16 at the time, crafted a new flag from an old 48-star flag and \$2.87 worth of blue cloth and white iron-on material."
\$2.87 (287) + 1 cent = \$2.88 (288) and 288 days prior to 19580418, the day Robert Heft designed the 50-state flag, brings us to July 4th, 1957 which was the 181st anniversary of our Declaration of Independence. 181 is a prime number that corresponds to We (55) Are (195) One (655) = 905 = 181 prime X 5 which mirrors perfectly the power of ONE (1) extra cent to get us to \$2.88 (288) days, etc.
Note: the time that this article was posted was 3:33 (333) am which is 23 "I (9) Am (14)" from Leo (356).
Finally: OHIO (6896) is composed of 89, the prime of Leo (356), in between two sixes "6__6" - 66 being 23 "I Am" from 89 which is the prime of Leo (356), etc.
## Friday, December 11, 2009
### Poppa Leo's Pizza
click image to enlarge
Contact Poppa Leo's Pizza
Phone: (517) 351-1811 Web: View Website
Poppa Leo's Pizza is closed - as the phone number(s) are not in service. But the energy behind the numbers lives on to glorify the Divine Birth of our Lord God Jesus Christ as follows:
1811, the last four digits, corresponds to Year 1, Month 8, Day 11 which is our Lord's spiritual - ie. corrected - birthyear. The correct way of displaying this date is 010811. This date converts into the numerical designator of 281 prime as Day 11 = 1+1 = 2, Month 8 and Year 1 when combined for 281 prime. And the 281st prime number happens to be 1811, etc.
Now let's add up the number 517 + 351 + 1811 = 2679
If we move back 2679 days prior to today, 20091211, we will arrive at 20020811, the Divine Birthdate of our Lord God Jesus Christ.
Now let's move onto the address as follows:
Poppa Leo's Pizza
5214 S Martin Luther King J
Lansing, MI 48911
23 "I (9) Am (14)" cycles of 5214 days prior to today, 20091211, brings us to 16810810 which means that Day ONE (1) of this period is August 11th, the Divine Birthdate of our Lord.
## Tuesday, December 8, 2009
### Season # 8 Biggest Loser Final
The exact time here in the midwest when Danny won the 8th Season's 1st prize was 8:59 pm. 859 days prior to 20091208 brings us back to 20070802 which was/is Leo (356) days from August 11, 2009 the Divine Birthdate of our Lord God Jesus Christ.
### The "Key" To Fatima
I believe there's going to be a spiritual revelation that is backed by simple-numerical roots which reveal and glorify our Father, the One Lord God the Savior Jesus Christ.
Islam (91314) = 89 prime X 1026
Taken another way, Islam (91314) = 513 X 178 (89 prime X 2). And 178 is composed of two (2) - the only female prime number - cycles of 89 which is the prime of Leo (356) = 89 prime X 4 which is the astrological sign for the Divine Birthmonth of August whose ruling planet is the Sun and which corresponds to the Human Heart.
I was awakened this morning at 5:13 am and was called to note the following:
20091208 + 513 days = 20110505 which is 89 "Prime of Leo (356)" days from 2/5, my 53rd birthday.
20091208 - 513 days = 20080713 which is the 7/13 (713) mo/day which factors into 23 "I (9) Am (14)" + 31 "The Lord God" (The (285) Lord (3694) God (764) = 31 prime X 153).
513 is a non-prime prime which is ONE (1) from 514 which corresponds to the english word for "Key" as follows:
514 = 257 prime X 2 and 257 = KEY
5/14 (514) factors into Key (257) prime X 2 and May 14th is always 89 "Prime of Islam" days from August 11th, the Divine Birthdate of our Lord.
Also, the "Key (257th) Day" in non-leap years is always the 9/14 (914) mo/day which factors into 457 prime X 2 which is the 89th prime number.
Fatima began one day from 5/14 - on May 13th - the 5/13 (513) mo/day - in the year 1917. Sister Lucia, the last of the 3 remaining children of Fatima passed/ascended on 2/13 in the year 2005, exactly 89 "Prime of Islam" days from 5/13 which was Day ONE (1) of the 89th year from/including May 13, 1917, the beginning of Fatima.
In addition to being the 89th year since Fatima, the year 2005 was highly revelatory for another reason as it factors into 401 prime X 5 which is the prime of the word Rosary (961197) = 401 prime X 2397. Also, 2005 was decreed to be "The Year of the Eucharist" by the late Pope John Paul II - who, too, passed/ascended that year - and the Divine Birthdate that year, August 11th, fell on a Thursday which is the day that the "Five Luminous Mysteries" - also referrred to as "the Mysteries of Light" - are to be prayed since being added to the Rosary prayer by Pope John Paul II during the 2002-2003 "Year of the Rosary", etc.
The year 1917 is "Key" for it was 1923 years from 6 BC which was the year of the Divine Birthdate of our Lord God Jesus Christ on August 11th and 1923 factors into 641 prime X 3 which is the prime of the word Rose (9615) = 641 prime X 15, etc.
Fatima and Islam
http://www.medjugorje-online.com/apparitions/fatima.php
Far back in history, the Muslims once occupied Portugal. The village of Fatima was given the Islamic name of well-loved Princess Fatima who lived in a nearby castle. Princess Fatima died at an early age, just after converting to Catholicism. Like most Muslim-born girls, this princess had been named after the beloved daughter of Islam's founder. Mohammed once said of his daughter, "Fatima is the most blessed woman in heaven after the Virgin Mary."
In fact, the Blessed Virgin Mary is mentioned more than thirty times in the Koran. Muslims believe in Mary's Immaculate Conception and the Virgin Birth of Christ. In the Koran, when Mary is born, her mother says: "And I consecrate her with all of her posterity under Thy protection, O Lord, against Satan!" Angels are pictured as accompanying the Blessed Mother and saying: "Oh Mary, God has chosen you and purified you, and elected you above all the women of the earth."
Nearly half a century ago, Bishop Fulton J. Sheen wrote the following. "I believe that the Blessed Virgin chose to be known as Our Lady of Fatima as a pledge and a sign of hope to the Moslem people, and as an assurance that they, who show her so much respect, will one day accept her divine Son, too. Mary is the portal for Moslems to accept Christ."
======================
This video was uploaded to youtube on the 155th anniversary of the Dogma of The Immaculate Conception of our Lord God Jesus Christ in the womb of the Most Blessed Virgin Mary in 1854. See http://en.wikipedia.org/wiki/Immaculate_Conception
And, just like 4 years ago in 2005 on this day it's snowing in the Chicagoland area. On 20051208, the 151st anniversary of the Dogma of the Immaculate Conception becoming the official Doctrine of the universal Catholic Church, there was a tragic accident here in Chicago with Southwest Airlines Flight # 1248 as follows:
http://en.wikipedia.org/wiki/Southwest_Airlines_Flight_1248
"On December 8, 2005, the airplane slid off a runway at Chicago-Midway while landing in a snowstorm and crashed into automobile traffic, killing Joshua Woods, a six-year-old boy in a car."
1248 = 156 X 8
And today is Day ONE (1) of the 156th anniversary of the Doctrine of the Immaculate Conception.
"At around 7:15 p.m. CST, the pilot attempted a landing with nearly eight inches of snow on the ground in the area. Airport officials stated that the runway was cleared of snow prior to the time of landing. The latest reported weather had the wind from between east and east-southeast (100°) at 11 knots (20 km/h)."
7:15 pm is the 1155th minute of the day which is ONE (1) conjoined with 155 which mirrors the 155th anniversary of the Doctrine of the Immaculate Conception today. 1155 days from 20051208 brings us to 20090205, my 51st birthday which was ONE (1) day from the end of the 14427th cycle of 51 days since the birth of our Lord God Jesus Christ on August 11th in 6 BC.
The other interesting thing to note about 20051208 is that it was the 151st anniversary of the Dogma of the Immaculate Conception which is the prime of God (764) Is (91) One (655) = 1510 = 151 prime X 10. And 20051208 was also the "Release Date" in the UK of the The Chronicles of Narnia: The Lion, the Witch and the Wardrobe:
http://en.wikipedia.org/wiki/The_Chronicles_of_Narnia:_The_Lion,_the_Witch_and_the_Wardrobe
Note also that Narnia (519591) = 151 "God Is One" prime X 3441
Pretty Amazing!!
ONE (1) final note: On 20051208 I was the reader of both readings of the Mass of the Immaculate Conception at the Sheil Catholic Center on the campus of Northwestern University.
PS. It is interesting to note the following from wikipedia:
"The feast of the Immaculate Conception, celebrated on 8 December, was established as a universal feast in 1476 by Pope Sixtus IV. He did not define the doctrine as a dogma, thus leaving Roman Catholics free to believe in it or not without being accused of heresy; this freedom was reiterated by the Council of Trent. The existence of the feast was a strong indication of the Church's belief in the Immaculate Conception, even before its 19th century definition as a dogma."
The year 1476 was/is 378 years from the year 1854 which was the year 1854 which was year that "the Immaculate Conception was solemnly defined as a dogma by Pope Pius IX in his constitution Ineffabilis Deus on 8 December 1854."
And if you do the math you will note that 18541208 was/is day ONE (1) of the 379th prime year from/including 1476 which is the result of adding 23 "I (9) Am (14)" prime + Leo (356) = 379 prime. It's interesting to also note that yesterday was the day I broke my personal record for the "Stairmaster" exercise machine when I stepped 379 "I Am Leo" prime floors in 46 minutes. 378 was my former "record" which is ONE (1) from 379, etc. And this post is # 152 which is ONE (1) from 151 which corresponds to God (764) Is (91) One (655) = 1510 = 151 prime X 10, etc.
## Monday, December 7, 2009
### The "Lion From The Tribe of Judah" Revelation
David Justice, Colorado State Cordinator of We The People Congress manifested this amazing numerical revelation At EXACTLY 12:07 pm on November 17, 2009 while standing in front of Jon McNaughton's "One Nation Under God" painting outside the New Orleans Ballroom inside Pheasant Run Resort on the occasion of Continental Congress 2009.
12:07 pm corresponds to 1207
20091117 + 1207 days = 20130308
March 8th is known as "The Jesus Foundation Day" for in most years it is always Jesus (15131) prime days from August 11th, the Divine Birthdate 41 years into the future.
http://cc2009.us/
http://mcnaughtonart.com/
Fast forward to today, 12/07 (1207) - which mirrors 12:07 pm - my youtube channel reveals 89 total views on this video with 89 being the prime of Leo (356) = 89 prime X 4.
click image to enlarge
One of my most popular recent videos is on this same topic:
David Justice, Colorado State Coordinator for We The People Congress provides some commentary on the significance of the presence of having Jon McNaughton's painting, "One Nation Under God", displayed outside of the New Orleans Ballroom where Continental Congress 2009 is meeting for their deliberations.
## Sunday, December 6, 2009
### The Four Boxes Of Light Revelation
While in the process of finishing up the task of putting up our indoor and outdoor Christmas lights I purchased 4 boxes of "50 White Lights" for a grand total of \$11.51 which corresponds to 1151 prime which is the 191st prime number.
Today, 20091206 - the Feast of St. Nicholas Day - is the 1151st day from/including 20061013 which was the 89th anniversary of the great "Miracle Of The Sun" at Fatima in 1917. 89 is the prime of Leo (356) = 89 prime X 4.
See http://en.wikipedia.org/wiki/The_Mira...
The cost of each of the 4 boxes was \$2.49 and 249 days ago was/is the 4/01 (401) mo/day in the year 2009 - 401 is the prime of Rosary (961197) = 401 prime X 2397.
The time on the receipt was 4:12 pm - 412 days prior to today, 20091206, brings us to 20081020 which was the 189th anniversary of the birth of "The Bab" the forerunner of Baha'ulla, the Founder of the Bahai Faith - see http://www.bahai.org
And 412 days from today brings us to 20110122 which will be my father's 86th birthday.
This came at the very end of a very powerful week which began on 20091130, Monday, when I watched "The 13th Day" the new movie about Fatima.
## Wednesday, November 18, 2009
### Jon McNaughton's "One Nation Under God" Painting
David Justice Comments On Jon McNaughton's "One Nation Under God" Painting
David Justice, Colorado State Coordinator for We The People Congress provides some commentary on the significance of the presence of having Jon McNaughton's painting, "One Nation Under God", displayed outside of the New Orleans Ballroom where Continental Congress 2009 is meeting for their deliberations.
http://cc2009.us/
# The "Lion From The Tribe of Judah" Revelation
David Justice, Colorado State Cordinator of We The People Congress manifested this amazing numerical revelation At EXACTLY 12:07 pm on November 17, 2009 while standing in front of Jon McNaughton's "One Nation Under God" painting outside the New Orleans Ballroom inside Pheasant Run Resort on the occasion of Continental Congress 2009.
12:07 pm corresponds to 1207
20091117 + 1207 days = 20130308
March 8th is known as "The Jesus Foundation Day" for in most years it is always Jesus (15131) prime days from August 11th, the Divine Birthdate 41 years into the future.
Finally, the phrase "One Nation Under God" works out as follows:
One (655) Nation (512965) Under (35459) God (764) = 549843 = 26183 prime X 21
Two (2) cycles of 26183 "One Nation Under God" days prior to yesterday, 20091117, brings us to 18660704, the 90th anniversary of our Declaration of Independence in 1776.
Two (2) cycles of 26183 "One Nation Under God" days from yesterday, 20091117, brings us to 21530402 which is 131 "Prime of ONE (655)" days from August 11th, the Divine Birth of our Lord God Jesus Christ.
Two (2) is the ONLY female prime number.
Now when we apply this 26183 "One Nation Under God" prime to the day Bob Schulz was born, 19391117, an interesting historical aspect manifests as follows:
19391117 - Two (2) cycles of 26183 "One Nation Under God" prime days brings us back to 17960702 which was/is the 20th anniversary of the original date of the legal separation of the 13 American from England as documented by Wikipedia as follows:
During the American Revolution, the legal separation of the American colonies from Great Britain occurred on July 2, 1776, when the Second Continental Congress voted to approve a resolution of independence that had been proposed in June by Richard Henry Lee of Virginia.[4] After voting for independence, Congress turned its attention to the Declaration of Independence, a statement explaining this decision, which had been prepared by a Committee of Five, with Thomas Jefferson as its principal author. Congress debated and revised the Declaration, finally approving it on July 4. A day earlier, John Adams had written to his wife Abigail:
“ The second day of July, 1776, will be the most memorable epoch in the history of America. I am apt to believe that it will be celebrated by succeeding generations as the great anniversary festival. It ought to be commemorated as the day of deliverance, by solemn acts of devotion to God Almighty. It ought to be solemnized with pomp and parade, with shows, games, sports, guns, bells, bonfires, and illuminations, from one end of this continent to the other, from this time forward forever more.[5] ”
Adams' prediction was off by two days. From the outset, Americans celebrated independence on July 4, the date shown on the much-publicized Declaration of Independence, rather than on July 2, the date the resolution of independence was approved in a closed session of Congress.[6]
Here are pictures of some of the attendees of CC2009 who agreed to stand in front of this painting in support of the Christian heritage that founded our great nation.
According to Wikipedia:
### Addition of the words "under God"
"Under God" was officially incorporated into the Pledge of Allegiance in 1954. The man to first initiate the addition of "under God" to the Pledge was Louis A. Bowman (1872-1959). The National Society of the Daughters of the American Revolution gave him an Award of Merit as the originator of this idea.[7][8] He spent his adult life in the Chicago area and was Chaplain of the Illinois Society of the Sons of the American Revolution. At a meeting on February 12, 1948, Lincoln's Birthday, he led the Society in swearing the Pledge with two words added, "under God." He stated that the words came from Lincoln's Gettysburg Address. He repeated his revised Pledge at other meetings.[8]
First, as we saw above, One (655) Nation (512965) Under (35459) God (764) = 549843 = 26183 prime X 21.
Second, 20091117 - 19480212 = 22559 days
26183 "One Nation Under God" prime - 22559 = 3624 = 151 prime X 24
151 is the prime of God (764) Is (91) One (655) = 1510 = 151 prime X 5
1150 "God Is One" days prior to 20091117 brings us to 20050929 which is the 9/29 (929) mo/day which is the prime of "Free From Fear" as follows:
Free (6955) From (6964) Fear (6519) = 20438 = 929 prime X 22
More will follow....
# 8A football:Kicks start Loyola's heart in win over Mount Carmel
http://www.chicagotribune.com/sports/highschool/football/chi-15-prep-1foot-loyola-mt-carmelnov15,0,3289509.story Loyola Academy kicker Leo Sheridan (#37) is congratulated by teammate Chancellor Carter (85) after a one of his five successful field goals. (Chris Sweda, Chicago Tribune / November 14, 2009)
Dear [Father, Mother, etc.] Sheridans:
Congratulations on an INCREDIBLE GAME!!
Your son, Leo, was AMAZING!! The Smarts were there watching this historic game unfold. When we saw Leo boot that 53 yard field goal in practice before the game - and with the wind blowing north - I/we knew it was going to be a perfect day for Leo's leg to shine. The warm sunny weather didn't hurt either. What a beautiful day! And to think that just the night before our oldest son, Frederick, had the great honor to attend and be driven to and from the Weddington's pre-game varsity dinner in Lake Forest by your son, Leo!
I think I may have mentioned to you, Leo, at one of those progressive dinners that one of my hobbies is numbers - with a special focus on prime numbers. Well the prime number behind Leo (356) is one of my all-time favorites and, if you'll forgive me for sharing this stuff, it fits perfectly with what happened yesterday as follows:
Using basic "reduced gematria" which applies to all the languages of the word, but most powerfully to the english language which is considered by this "Divine Science of Numerical Correspondences" to be the Amazon river-like-capstone "Language of the Divine Providence":
abcdefghi = 123456789
jklmnopqr = 123456789
stuvwxyz = 12345678
Leo (356) = 89 prime X 4
Leo is the astrological sign that corresponds to the Human Heart and is ruled by the Sun. There's actually a mathematical proof of the birth of our Lord God Jesus Christ (on the 11th day of August, 6 BC) that's embedded in the prime number system. It was taught to me years ago by an elderly gentleman by the name of Dean Fagerstrom whose whose an incredible guy who has had some pretty amazing things happen to him throughout his life. Dean documented this Divine Science of Numerical Correspondences in an unpublished book, "The Book of Anglion", which I have read and studied in detail.
And it was very fitting that yesterday, while Leo was booting his five field goals - Leo is the 5th astrological sign - Northwestern was beating Illinois 21-16 to bring home the first "Land of Lincoln Trophy" - see this link. The 89th Meridian(Longitude) West runs straight through the Heart of this "Land of Lincoln" on a north-south basis. If you reverse 21-16 you get 16-21 (1621) which is a prime number that corresponds to The (285) King (2957) = 3242 = 1621 prime X 2. Our state capital, Springfield sits on the 89th Meridian West and it's EXACTLY 89 miles from St. Louis. And, just a reminder, Leo (356) = 89 prime X 4.
Let's look at the Illinois Lottery's Midday "Pick 3" and "Pick 4" yesterday - see this link.
First, the Midday Pick 3 was 112 which is composed of 23 prime + 89 prime = 112. If you let your fingers do the walking through the prime number table you will discover that the Leo (356th) prime number - include the number one (1) in the count - is the fusion of those two prime numbers into 2389. 23 corresponds to I (9) Am (14) = 23 prime which is our Lord's original name and title given to Moses on Mt. Sinai - ie. "I Am....the Lord Thy God..." Exodus 20:2. And, I was mentioned twice above, 89 is the prime of Leo (356) = 89 prime X 4
Note that when you combine I (9) and Am (14) you get 914 which factors into 457 prime X 2. 457 is the 89th prime number. Keep this in mind when I share the following:
The Midday Pick 4 yesterday was 5052. When you add 5052 days to yesterday you arrive at 20230914 which is the 9/14 (914) mo/day in the 23 "I (9) Am (14)" year. And 9/14 (914) is the ONLY day on the calendar that factors into 457 which is the 89th prime number. When you add 457 - the 89th prime number - to yesterday, 20091114, you will arrive at 20110214 - Valentines Day - which is ALWAYS two (2) - the only female prime number - times 89 "Prime of Leo (356)" days from August 11th, the Divine Birthdate of our Lord. It's be the grace of our Lord Divine Providence that Valentines Day - in non-leap years - is always two (2) X 89 "Prime of Leo (356)" = 178 days from August 11th, His Divine Birth - ie. the sign of Leo which corresponds with the Heart whose ruling planet is the Sun as in "two Hearts together as one" etc.
Check out the Evening Pick 3 of 397....
Yesterday was 397 days from 20081013 which was the 91st anniversary of the Miracle of The Sun at Fatima in 1917.
And if that wasn't enough numerical fireworks, check out the Evening Pick 4 of 6441....
Yesterday was 6441 days from July 4, 2027 which was the 251st anniversary of our great Declaration of Independence in 1776.
Here's ONE simple-yet-amazing numerical FACT that connects your son's # 37 with 89, the prime of Leo (356) and the Divine Birthdate, August 11th, as follows:
20091114 - # 37 X 89 "Prime of Leo (356)" (3293) days = November 8, 2000. November 8th is ALWAYS 89 "Prime of Leo (356)" days from August 11th. But there's something even more powerful going on here for in this Divine Science of Numerical Correspondences, November 8th is documented - mathematically - as the Divine Conception Day of our Lord God Jesus Christ in the womb of the Most Blessed Virgin Mary in 7 BC. 11/08 (1108) actually factors into 277 prime X 4 which is the prime of "Divine (4949555) Love (3645)" = 498600 = 277 prime X 1800. The 277th day in non-leap years from/including November 8th is always August 11th, the Divine Birthdate of our Lord.
November 8th holds special meaning for our family as my wife, Maribeth, was born on this day in 1959 EXACTLY 641 prime days after the day I was born, 19580205. 641 is the prime of Rose (9615) = 641 prime X 15. Anytime you add or subtract 641 days to the calendar you will move backward/forward 89 "Prime of Leo (356)" days. My wife's birthday, 11/08, is 89 days from my birthday, 2/5, and her birthday is 89 days from August 11th, the Divine Birthdate.
Now let's go back 89 cycles of Leo (356) days prior to yesterday and see where we land as follows:
20091114 minus 89 X Leo (356) (31684) days = 19230215 which is EXACTLY ONE (1) day from Valentines Day in the year 1923 which factors into 641 "Prime of Rose (9615)" X 3. The year 1923 is very special since it corresponds to the number of years between 1917, the Year of Fatima, and 6 BC, the Year of our Lord's Birth. The advent of the 89th year since Fatima was the year 2005 which was also known as the "Year of the Eucharist" which is when Sister Lucia - the last of the three remaining children who witnessed Fatima - and Pope John Paul II passed away. Sister Lucia passed/ascended into Heaven on 20050213 which was 89 "Prime of Leo (356)" days from 5/13 which was Day ONE (1) of the 89th year since Fatima began in 1917.
And thanks to Pope John Paul II's "Year of The Rosary October 2002-2003" he added the Five Luminous Mysteries and by the time the "Year of the Eucharist 2005" rolled around it was amazingly fitting that the Divine Birthdate, August 11th, fell on a Thursday which is the ONLY day when we are supposed to pray the Five Luminous Mysteries. The other very powerful aspect to the year 2005 - being the 89th year since Fatima (1917) - is that it factored into 401 prime X 5 which is the prime of the word Rosary (961197) = 401 prime X 2397.
I could go on and on.....
It was very cool that #33 Daveed Carter, Chance's brother - blocked that punt. I swear to God but just prior to that blocked punt I yelled out at the top of my lungs "BLOCK THE PUNT!! BLOCK THE PUNT!!" And sure enough, there was Daveed flying over the line bocking the punt which rolled the football into the arms of #56 Jim Ford - I played with Jim's father, the late "Jimbo Ford" very briefly at Northwestern as a "walk-on" quarterback in my freshman year (1976). When you add 23 "I (9) Am (14)" to Daveed's # 33 you get Jim Ford, Jr's #56 which is EXACTLY 1/2 of 112 which is composed of 23 "I (9) Am (14)" + 89 "Prime of Leo (356)" etc.
And if you want go really go deeper on all of this, don't every overlook the "Power and Energy of Sweetness" - ie. Walter Payton - for 112 (23 "I Am" + 89 "Prime of Leo") days brings us back to 20090725 which was day ONE (1) of Walter's # 56th birthday, Walter Payton's numerical age (NA) on the day he passed/ascended, 19991101, was 16535 days, which was/is 191 prime days from 16726 which was the EXACT number of yards he rushed to set the All-Time NFL Rushing Record which lasted for years until Emmit Smith of the Dallas Cowboys broke it in 2002.
Born July 25, 1954(1954-07-25)
I better leave well-enough alone!!
It was just an AMAZING game! The energy, the weather, the sun and the great crowd all made for a memory which will be remembered for a very long, long time.
Please note that the Ramblers are now 11-1 (111) which is ONE (1) - as in one kick - from 112 = 23 "I (9) Am (14)" + 89 "Prime of Leo (356)". We will be rooting for another two (2) wins for a 2009 Illinois State Championship Crown which would make the final record 13-1 (131) which is the prime of ONE (655) = 131 prime X 5.
ENOUGH!!
The Smarts will be at the game this afternoon watching your other son, Jimmy, play in the Catholic League Championship game. What a weekend of football for you guys and all of us!!
Peace and God Bless,
Fred Smart | 13,733 | 50,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-26 | latest | en | 0.929803 |
https://aunl.org/document/4230c4/finding-perimeter-amp-area--common-core-sheets | 1,631,839,949,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053918.46/warc/CC-MAIN-20210916234514-20210917024514-00384.warc.gz | 158,163,905 | 10,872 | # Finding Perimeter & Area - Common Core Sheets
Finding Perimeter & Area. Find the perimeter and area of each figure. Each figure is in inches (in). Not to scale. 1. Page 2. Math. Name: www.
## Finding Perimeter & Area - Common Core Sheets
Finding Perimeter & Area. Find the perimeter and area of each figure. Each figure is in inches (in). Not to scale. 1. Page 2. Math. Name: www.
## Finding Perimeter and Area - Common Core Sheets
5. 26 in. 6. 10 ft. 2. 7. 4 cm. 8. 32 ft. 9. 5 ft. 10. 28 in. Solve the problems. Finding Perimeter and Area. Math www.CommonCoreSheets.com. Name: Answers. 1 ...
## Finding Angles By Degrees - Common Core Sheets
Determine if the angle described is 'acute', 'obtuse','right' or 'straight'. Finding Angles By Degrees. Math www.CommonCoreSheets.com. Name: Answers. 1 ...
## Finding Volume (word) - Common Core Sheets
2) What is the volume of a figure with that is 6 inches wide, 3 inches tall and 4 inches long? 3) A cube has ... Solve each problem. Finding Volume (word). Math.
## FINDING AREA, PERIMETER, AND CIRCUMFERENCE
A square is a kind of rectangle, and the perimeter of any rectangle is the sum of its ... The circumference of any circle equals two times its radius multiplied by pi.
## Perimeter and area Student information sheet Perimeter Area
Nuffield Free-Standing Mathematics Activity Perimeter and area Student sheets. Copiable page 1 of 4 ... In this activity you will find the perimeter and area of rectangles and shapes made from ... Why are the units m2? Student worksheet.
## Round - Common Core Sheets
18) Round 770 to the nearest hundred. 19) Round 212 to the nearest ten. 20) Round 92,240 to the nearest ten thousand. 1.
## Math - Common Core Sheets
The 4 in the tens place is ______ the value of the 4 in the hundreds place. 3) 54.45. The 5 in the tens place is ______ the value of the 5 in the hundredths place.
## Subtracting Across Zero - Common Core Sheets
Use subtraction to solve the following problems. Subtracting Across Zero. Math www.CommonCoreSheets.com. Name: Answers. 1 ...
## 8's Multiplication Maze - Common Core Sheets
Math. Name: 1 www.CommonCoreSheets.com. 8's Multiplication Maze. Determine the multiples of 8 to find your way through the maze. Remember the numbers ...
## Social Studies - Common Core Sheets
7) Ferdinand Magellan and his crew sail around the world. (1519) D. 8) Tristan De Luna explorers ... Placing Events on a Timeline. Determine which letter best ...
## Interpreting Histograms - Common Core Sheets
6) How many students sent between 8 and 9 texts? 7) How many students are represented in this histogram? 8) If a student sent 7 texts which bar would they be ...
## Understanding Ratios - Common Core Sheets
21. 5. 80. 6. 10 : 7. 7. 24. 8. 81. 9. 3 : 2. 10. 16. Solve each problem. Understanding Ratios. Math www.CommonCoreSheets.com. Name: Answers. 1. Answer Key ...
## Rounding Decimals - Common Core Sheets
5) Round to the nearest whole number. 6.805. 7. 6) Round to the nearest tenth. 9.725. 9.7. 7) Round to the nearest hundredth. 118.380. 118.38. 8) Round to the ...
## Using Unit Prices - Common Core Sheets
Online the you could buy 6 books for \$32.16. Which place has a lower unit price? book store = \$5.45 , online = \$5.36. 2) At the store beef jerky was \$73.70 for 5 ...
## Basic Probability - Common Core Sheets
11) If you were to select 1 shape at random from the array, what shape do you have the greatest probability of selecting? 12) Which shape has a 37.5% chance (6 ...
## Identifying Factors - Common Core Sheets
1 , 2 , 7 , 14 , 49 , 98. 14) 86. 1 , 2 , 43 , 86. 15) 29. 1 , 29. List the factors for each of the numbers. Identifying Factors. Math www.CommonCoreSheets.com. Name ...
## Two Step Problems - Common Core Sheets
1) A restaurant owner bought 4 boxes of disposable cups for \$142, with each box containing 3,507 cups. If he wanted to divvy up the cups among his 7.
## Mind Matter - Common Core Sheets
See if you can figure out what these tricky brain teasers are trying to say. S. I. T. Chair. You're Under Arrest. Sit Down. High Chair. Sand. Mind. Matter. Sand Box.
## 1. World Explorers - Common Core Sheets
7) Ferdinand Magellan and his crew sail around the world. (1519) ... Determine which letter best represents the location the event should occur on the timeline.
## Simplifying Expressions - Common Core Sheets
Simplify the expressions shown. Simplifying Expressions. Math www.CommonCoreSheets.com. Name: Answers. 1. Page 2. Answer Key. 1-10 95 89 84 79 74 68 ...
## Identifying Shapes - Common Core Sheets
11-15 27 20 13 7 0. 1) Emily's family is building a pool in the shape of a nonagon. How many sides will the pool have? 2) Ned sketched a logo into the shape of ...
## Ordering Decimal - Common Core Sheets
Ex) A. 53.783. B. 53.65. C. 53. D. 53.063. 1) A. 66.087. B. 67. C. 66.87. D. 66.6. 2) A. 7.67. B. 7. C. 7.72. D. 7.38. 3) A. 38.48. B. 38.84. C. 38.126. D. 38.18.
## Reducing Fractions - Common Core Sheets
Ex) 10. = 1. 40. 4. 1). 8. = 1. 64. 8. 2) 40. = 5. 64. 8. 3) 50. = 5. 60. 6. 4) 18. = 2. 27. 3 ... 5⁄6. 19. 1⁄8. 20. 2⁄3. Reducing Fractions. Reduce each fraction as much as ...
## Understanding Union and Intersect - Common Core Sheets
10. (A∪C)-B. 11. A∪(B-C). 12. B∪A. Determine the shaded region of each diagram. Understanding Union and Intersect. Math www.CommonCoreSheets.com.
## Converting Days to Weeks - Common Core Sheets
Determine what number makes each conversion true. Converting Days to Weeks. Math www.CommonCoreSheets.com. Name: Answers. 1 ...
## Multiplication Word Problems - Common Core Sheets
Math www.CommonCoreSheets.com. Name: Answers. 1-10 90 80 70 60 50 40 30 20 ... 4. 5. 6. 7. 8. 9. 10. Multiplication Word Problems. Solve each problem. 1 ...
## Identifying Combination Amounts - Common Core Sheets
1) At Victor's Pizza Palace you can order one type of pizza and one drink for \$9. Their menu is shown below. Pizza. Drink ... name, a color and a mascot. The.
## Determining Factors and Multiples - Common Core Sheets
Math www.CommonCoreSheets.com. Name: Answers. 1-10 92 83 75 67 58 50 42 33 25 17. 11-12 8 0. 1) Which number is a factor of 20, but not a multiple of 2?
## Identifying Types of Triangles - Common Core Sheets
R S. Determine if each triangle is acute(A), obtuse(O) or right(R) and if it is an equilateral(E), isosceles(I) or scalene(S). Identifying Types of Triangles. Math www.
## Rotating on a Coordinate Plane - Common Core Sheets
4b. (2,7). 4c. (8,8). 4d. (10,4). 4. Graph. Rotate each shape as described. Rotating on a Coordinate Plane. Math www.CommonCoreSheets.com. Name: Answers.
## Creating a Box Plot on a Numberline - Common Core Sheets
20. 25. 30. 35. 40. 45. 50. 55. 60. 65. 70. 75. 80. 85. 90. 4) 9 , 13 , 19 , 16 , 14 , 19 Min=9 Max=19 q1=13 Q2=15 q3=19. 0. 5. 10. 15. 20. 25. 5) 57 , 41 , 40 , 67 , 66 ...
## Reading An Analog Clock - Common Core Sheets
12. Reading An Analog Clock. Determine the time shown on the clock. 1. Page 2. Math www.CommonCoreSheets.com. Name: Answers. Answer Key. 7). 8). 9).
## Solving with Squared and Cubed - Common Core Sheets
100 x. 2 = 100 x = 100. 20) x. 2. = 121 x. 2 = 121 x = 121. 1. 2. 2. 3. 3. 4. 4. 5. 5. 6. 6. 7. 7. 8. 8. 9. 9. 10. 10. 1. 11. ... Solving with Squared and Cubed. Math www.
## Division Word Problems - Common Core Sheets
If they wanted to split all the tickets so each friend got the same amount, how many ... many flowers would be in the last vase that isn't full? 878 ÷ 9 = 97 r5 ... 5) A movie theater needed nine hundred eighty popcorn buckets. If each package has ...
## Division Drills (Mixed) - Common Core Sheets
Fill in the blanks for each problem. Division Drills (Mixed). Math www.CommonCoreSheets.com. Name: 1. Page 2. Answer Key. 24 ÷ 8 = 3. 70 ÷ 7 = 10. 36 ÷ 9 =. | 2,476 | 7,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-39 | latest | en | 0.801665 |
http://nrich.maths.org/public/leg.php?code=72&cl=3&cldcmpid=4926 | 1,503,393,595,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110573.77/warc/CC-MAIN-20170822085147-20170822105147-00718.warc.gz | 314,842,405 | 10,163 | # Search by Topic
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https://www.physicsforums.com/threads/calculating-flow-rate-when-emptying-liquid-from-a-closed-top-tank.1051772/ | 1,726,321,534,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.38/warc/CC-MAIN-20240914125424-20240914155424-00158.warc.gz | 882,170,177 | 24,482 | # Calculating flow rate when emptying liquid from a closed-top tank
• labuch
In summary: P_{atm}=\frac{P_1}{\gamma}+z_1P_{atm}=\frac{P_1}{\gamma}+\frac{P_{atm}}{2g}\frac{z_1}{\gamma}P_{atm}=\frac{P_1}{\gamma}+\frac{P_{atm}}{2g}\frac{z_1}{\gamma}P_{atm}=\frac{P_1}{\gamma}+\frac{P_{at
labuch
Homework Statement
Hello,
Here is my problem: "We have a closed tank of large size which contains a liquid topped by air at a pressure equal to the atmospheric pressure. If the tank is closed, as the water flows out, the volume of air above the free surface increases, so the pressure decreases (law of perfect gases). We seek to find the pressure variation (𝛥P) at the surface of the water in the tank as a function of the z axis (whose origin is at B) and to give the expression of v (the flow velocity) as a function of z."
Relevant Equations
I know that:
-according to the perfect gas equation P=nRT/V so that pressure is proportional to volume, so that P(t=0)*V(t=0)=P(t=1)*V(t=1)
-That the initial pressure at the surface of the water is P =Patm
-by applying bernoulli between A and B, we have classical V0=sqrt(2gh)
I can imagine the experiment: the pressure at the surface will drop while the tank is emptying (if no air bubbles enter through the evacuation of course, otherwise it restores the pressure atm at the surface). The flow speed decreases as a function of time until the external pressure maintains a liquid level above the drain (thus stopping the flow). I can't put this resonance in mathematical form (𝛥P as a function of z and v as a function of z)
thank you in advance for those who will help me,
Sincerely
here is a simple visualisation of the probleme
Hello @labuch ,
labuch said:
-by applying bernoulli between A and B, we have classical V0=sqrt(2gh)
No, you don't. That expression is based on pressure terms canceling, and in your case they don't.
But you can build an appropriate model along the same lines and end up with a differential equation. I don't expect that to have an analytical solution, but perhaps I can be proven wrong.
Interesting problem; not homework I suppose ?
##\ ##
Lnewqban and labuch
BvU said:
Hello @labuch ,
No, you don't. That expression is based on pressure terms canceling, and in your case they don't.
But you can build an appropriate model along the same lines and end up with a differential equation. I don't expect that to have an analytical solution, but perhaps I can be proven wrong.
Interesting problem; not homework I suppose ?
##\ ##
hello, thank you for your response, Yes you are right I thought that this expression could be true at the very beginning of the draining that's why I called it V0 and I had the feeling that it was there that I had to dig. This homework is actually a problem around which I have to create a practical work. My problem is that I don't know how to develop the expression to solve the problem mathematically
labuch said:
here is a simple visualisation of the problemeView attachment 325111
Bernoulli's is not valid for this problem as the flow is not steady (its time varying).
I propose to begin with the "Momentum Equation" (Reynolds Transport Theorem - Newtons Second for a control volume) to develop your differential equation
$$\sum \boldsymbol F = \frac{d}{dt}\int_{cv} \rho \boldsymbol v ~dV\llap{-} + \int_{cs} \rho \boldsymbol v \boldsymbol V \cdot d \boldsymbol A$$
Although its application is probably going to be cumbersome and not obvious... the answer is going to be in there somewhere
Last edited:
erobz said:
Bernoulli's is not valid for this problem as the flow is not steady (its time varying).
I propose to begin with the "Momentum Equation" (Reynolds Transport Theorem - Newtons Second for a control volume) to develop your differential equation
$$\sum \boldsymbol F = \frac{d}{dt}\int_{cv} \rho \boldsymbol v ~dV\llap{-} + \int_{cs} \rho \boldsymbol v \boldsymbol V \cdot d \boldsymbol A$$
Although its application is probably going to be cumbersome and not obvious... the answer is going to be in there somewhere
Hello,
I thought of another solution because I could not use your formula. I think that we can probably assimilate the tank to a tank of right section of constant area S. We can note z=f(t) the height of air in the tank and Zo the initial value. The law of perfect gases applied by considering the temperature as fixed leads to :
Patm.zo=P.z
The velocity of the fluid at the upper free surface of the liquid is (dz/dt).
If we note "s" the cross-sectional area of the outlet orifice, the conservation of flow gives: S.(dz/dt)=s.Vs where Vs is the outlet velocity at the bottom of the tank. Taking into account these two equations, Bernoulli's theorem will give you the differential equation verified by z=f(t). However I can't get to the final equation...
Bernoulli's is not supposed to be valid for non-steady flow, but then again in my fluids text does use the solution ## v = \sqrt{2gh}## for the exit velocity in a draining tank example problem...like we find in Torricelli's Law derived from Bernoulli's principle (so I'm not sure what exactly is affected and to what extent under this assumption).
Under that assumption (which also neglects viscosity), let 1 be the upper surface of the fluid, and 2 be the position of the fluid jet.
$$\frac{P_1}{\gamma} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{\gamma} + \frac{v_2^2}{2g} + z_2$$
##z_2## is the datum
##P_2## is 0 gage pressure (atmospheric)
## v_1 \ll v_2## the kinetic head in the tank is negligible in comparison to the jet at 2.
##A_1## is the cross section of the tank
##A_2## is the cross section of the jet
##H## is from ##z_2## to the top of the tank
##z_{1_o}## is the initial height of fluid in the tank measured from ##z_2##
$$\frac{P_1}{\gamma} + z_1 = \frac{v_2^2}{2g}$$
Let the absolute pressure above the tank surface 1 be ##P(z_1) = P_1(z_1) + P_{atm}##
Applying the ideal gas law we are looking for the pressure of the gas as a function of ##z_1##
$$P_{atm} A_1 ( H - z_{1_o}) = P(z_1)A_1( H - z_1)$$
Can you take it from there?
##P_1## will be a function of ##z_1##, and ##v_2## will be a function of ##\dot z_1##.
Last edited:
member 731016
While we are on the subject can anyone explain the consequences of applying Bernoulli's to non-steady flow?
Here is the derivation in my textbook. The local acceleration (i.e. the variation of velocity with time at a specific point on a path line - ## \frac{\partial V}{\partial t}## in this problem will clearly not be zero, which negates everything after (4.15).
member 731016
Based on the figure in post #2, here is my take on this.
Ideal Gas Law: $$P_g(H-h)=P_{atm}(H-h_0)$$where ##P_g## is the absolute pressure of the gas in the head space.
Bernoulli Equation: $$P_g+\rho g h=P_{atm}+\rho g h_B+\frac{1}{2}\rho v^2$$
Combining these two equations gives: $$\frac{1}{2}\rho v^2=\rho g(h-h_B)-\left(\frac{h_0-h}{H-h}\right)p_{atm}$$
Finally, the overall mass balance is $$A_{tank}\frac{dh}{dt}=-A_{exit}v$$
member 731016, BvU, erobz and 1 other person
I found this example of Unsteady Bernoulli's. Can someone reconstruct an interpretation from the example? There is obviously a change in the flow energy per unit volume (a deviation from Bernoulli's for steady flow), but where/what is it?
## 1. How do you calculate the flow rate of liquid emptying from a closed-top tank?
The flow rate of liquid emptying from a closed-top tank can be calculated using Torricelli's Law, which states that the speed of efflux, v, of a fluid under gravity through an orifice is given by $$v = \sqrt{2gh}$$, where g is the acceleration due to gravity and h is the height of the fluid column above the orifice. The flow rate Q can then be found using $$Q = A \cdot v$$, where A is the cross-sectional area of the orifice.
## 2. What factors affect the flow rate when emptying a closed-top tank?
The flow rate when emptying a closed-top tank is affected by the height of the liquid column above the orifice, the size and shape of the orifice, the viscosity of the liquid, and the atmospheric pressure inside the tank. Additionally, any resistance or friction in the piping system can also impact the flow rate.
## 3. How does the height of the liquid column influence the flow rate?
The height of the liquid column directly influences the flow rate because it determines the pressure at the orifice. According to Torricelli's Law, the flow rate is proportional to the square root of the height of the liquid column. As the height decreases, the pressure and thus the flow rate also decrease.
## 4. Can the flow rate be constant when emptying a closed-top tank?
No, the flow rate is not constant when emptying a closed-top tank. As the liquid level drops, the height of the liquid column decreases, which in turn reduces the pressure and the flow rate. Therefore, the flow rate decreases over time as the tank empties.
## 5. How can you maintain a steady flow rate when emptying a closed-top tank?
To maintain a steady flow rate when emptying a closed-top tank, you can use a pump to control the discharge rate or install a flow control valve. Another method is to use a pressurized system where the pressure is regulated to compensate for the decreasing liquid level, ensuring a consistent flow rate.
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3K | 2,548 | 9,461 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-38 | latest | en | 0.909978 |
https://u.osu.edu/mathcircle/ | 1,506,067,048,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688926.38/warc/CC-MAIN-20170922074554-20170922094554-00168.warc.gz | 715,999,384 | 18,661 | # Math Circle: 2017 – 2018
This Math Circle is for middle school (and high school) students who enjoy thinking about mathematical ideas, problems, and puzzles. We hope to get students involved in investigating some mathematical questions, talking about their ideas, coming up with guesses and suggestions, and arriving at answers and explanations.
Meetings will be held 2:00 – 4:00 PM on occasional Sundays, in a room on the OSU campus. Our first choice is room 154 in the Math Tower
FLOORS: For info on floor tilings in the OSU Math Tower see: Tessellation Project .
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• September 24, 2017: Recursive Sequences.
Room: 154 Math Tower
Fibonacci numbers provide an example of a sequence where each term is a specific linear combination of earlier terms. What other sequences are there following similar rules?
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Here are topics of meetings held during 2015-2016:
• April 23, 2017: Integer Equations.
Room: 154 Math Tower
Does the equation 38x – 15y = 1 have a solution where x and y are positive integers?
How many positive integer solutions are there to: 45x + 23y = 3000 ? We will talk about such problems, discuss an algorithm for solving them, and relate that process to modular arithmetic.
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• March 26, 2017: Counting and Probability.
Room: 154 Math Tower
In twenty coin flips, how likely is “three heads in a row”? What’s the probability of getting a pair in a five card hand? If a stick is cut in two places, how likely is it that the three pieces can form a triangle?
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• February 12, 2017: Area.
Room: 154 Math Tower
Can you find the area of a plane polygon if you are given its vertices? For instance, if A = (1, 2), B = (5, -1), and C = (6, 8) are points in the plane, find the area of triangle ABC. These ideas lead to investigations of polygons with lattice point vertices. (A “lattice point” is a point that has integer coordinates).
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• January 22, 2017: Simple Doodles.
Room: 154 Math Tower
Draw a closed curve (with some self-crossings) on a piece of paper.
What mathematical questions come to mind when you contemplate that picture?
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• December 4, 2016: Congruence by scissors.
Room: 154 Math Tower
Two polygons A and B in the plane are called “scissors-equivalent” if A be cut apart (with some straight cuts) and the pieces rearranged to make B. In that case, the areas of A and B must be equal.
Is every polygon scissors-equivalent to a square?
If so, then any two polygons of equal area must be scissors-equivalent.
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• November 13, 2016: Differences.
Room: 154 Math Tower
If f is a sequence of numbers, form its “difference-sequence” Δf by taking differences of successive terms. For instance, if s is the sequence of squares 0, 1, 4, 9, 16, 25, . . . then Δs is 1, 3, 5, 7, 9, . . . We will work out some examples and investigate different patterns.
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• October 16, 2016: Infinity.
Room: 154 Math Tower
We will talk about infinite processes, with the first focus on cardinality (counting). We might also discuss infinite decimals, sums of infinite series, etc.
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Here are topics of meetings held during 2015-2016:
• April 10, 2016: Polynomials.
Room: 154 Math Tower
Algebra has developed a lot since the cubic formula was discovered in the 1540s. We will discuss polynomials and some of their properties, especially those connected with roots and factors. Those ideas motivate the study of infinite series, as illustrated by some of Euler’s work dating from the 1700s.
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• March 20, 2016: Moebius bands, and surfaces.
Room: 154 Math Tower
We will discuss Moebius strips, make some paper models, and draw more abstract models on the board. Those pictures lead us to think about closed surfaces like the torus and Klein’s bottle. Hexaflexagons provide interesting models of Moebius bands. We will fold a few hexaflexagons and then investigate some of their mathematical properties.
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• February 14, 2016: Modular Arithmetic.
Room: 154 Math Tower
Impose a new rule on the integers stating: 12 equals 0. To avoid confusion we use the symbol ≡ (three parallel dashes) instead of = for equality in this new system. From that new rule we deduce that 13 ≡ 1, 20 ≡ 8, and -3 ≡ 9.
How can we tell whether given integers a and b are equal here? Is 1841 ≡ 2093 ?
Can you solve equations like 5x ≡ 3 and 3x ≡ 5 ?
This new system of “integers modulo 12” contains exactly 12 different elements. By posing standard problems in this new system and its relatives, we obtain insights into algebra.
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• January 17, 2016: Game Theory.
Room: 154 Math Tower
Two-person games and mixed strategies.
We expect to have a second meeting on this topic the following Sunday, Jan 24.
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• November 15, 2015: Pascal’s Triangle.
Leader: Matisse Peppet (a high school student)
Room: 232 Cockins Hall
Pascal’s Triangle is an array of numbers built recursively using an addition rule. The numbers have many properties involving combinatorics, factorial formulas, and expansion of powers of (1 + x). We will discuss various aspects and generalizations of that triangle of numbers.
The triangle’s name refers to Blaise Pascal, a French mathematician who wrote about this “arithmetical triangle’ in the 1650s. However many other scholars investigated those patterns in earlier centuries.
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• October 25, 2015: Centers of Mass.
Where can two people sit on a seesaw and balance each other even though they have different weights? If weights of 3 and 5 pounds are placed one meter apart on a rigid wire, where is their equilibrium point, the spot where a fulcrum will balance them? Where is the center of mass of three several weights on a line? Or in the plane, or 3-dimensions?
Archimedes (who died in 212 BCE) wrote about those ideas, and his work is still worth investigating today.
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• September 27, 2015: Combinatorial Games.
Two players, named L and R, take turns playing a game (a finite game with complete information). A game G is “positive” if it is a win for L (no matter who goes first). A game is “negative” if it is a win for R. Games can be added and subtracted, and we consider the algebra of games.
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• August 30, 2015: 2:00 – 4:00 PM: Pythagorean Theorem and Pythagorean Triples.
Hundreds of different proofs of the Pythagorean Theorem have been published. Which one is the simplest? Which is the most elementary? Which of the proofs do you like the best?
Sometimes a right triangle has sides of integer length. For instance, the lengths 3, 4, 5 belong to a right triangle because 3^2 + 4^2 = 5^2, (that is: 9 + 16 = 25). How can we generate more examples of this type?
We seek whole numbers a, b, c satisfying a^2 + b^2 = c^2.
How is this question is related to lists of rational points on the unit circle?
DIRECTIONS.
The Math Tower (231 W. 18th Ave) is on the OSU campus. It is the tallest of several connected buildings For driving and parking directions, click the link on the top left of this page. Room 154 is the first door to your left after you enter the building.
Teachers and parents are welcome to join in.
Feel free to invite other middle and high school students who would enjoy this session. To help us organize, please send an email if you might attend this Math Circle meeting.
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Here are topics of meetings held during 2014-2015:
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• May 3, 2015: 2:00 – 4:00 PM: Permutations and symmetry.
A cube C is symmetric in several ways. A quarter-turn about a face-center moves C back to its original position. Similarly, a (1/3)-turn about an axis connecting opposite corners also returns C to the same place. Those rotations are examples of “symmetry operations” of C.
How many different symmetry operations does C have?
A permutation of set A = {1, 2, 3} is a mixing of the elements. For instance, one permutation switches symbols 1 and 2 while leaving 3 fixed. Another one sends symbol 1 to 2, sends 2 to 3, and sends 3 to 1.
How many different permutations does A have?
Those two examples can be viewed with the same lens: an object (cube C or set A) and its symmetry group. We will discuss several examples, and talk about compositions of symmetry operations.
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• April 19, 2015: 2:00 – 4:00 PM: Exploding dots.
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• April 5, 2015 (Easter !! ): 2:00 – 4:00 PM: Measure, dimension, and fractal sets: Part 2.
This session is a continuation of the previous one, building on ideas of cardinality and measure.
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• March 22, 2015: 2:00 – 4:00 PM: Measure, dimension, and fractal sets.
A key example is the “Cantor set” built from the interval [0,1] on the number line by “removing middle thirds”. Here’s the procedure:
Step 1. Remove the open interval (1/3, 2/3). This leaves two separated intervals [0, 1/3], [2/3, 1].
Step 2. Remove the middle third of each of those two remaining intervals (i.e., remove (1/9, 2/9) and (7/9, 8/9) ). This leaves 4 smaller intervals, each of length 1/9.
Repeat this process:
Step n. Remove the middle third of each of the intervals left at the end of Step n – 1.
The collection of points remaining after all (infinitely many) steps is the Cantor set S.
Question: How big is S ?
The answer is a bit tricky. If we compute the total length of our set after n steps, and take the “limit” as n gets large, we conclude that S must have length 0 (or “measure zero”). On the other hand, S has infinitely many points.
For a better answer, we investigate how to measure sizes of subsets of the number line. This is closely related to the idea of dimension. A point, line, and plane have dimensions 0, 1, and 2. Our set S has a “dimension” that is between 0 and 1. The exact dimension of S (its Hausdorff dimension) equals
Dim(S) = log(2)/log(3) = .630929753571457 (approximately).
If we measure S according to its correct dimension, it turns out to have “length” equal to 1!
[Historical Note: The dimension of the middle-third Cantor set was first computed by Hausdorff himself in 1919, and appeared in the same paper in which he introduced what is now called Hausdorff dimension. However, the calculation of the Hausdorff length of S was found 82 years later, in 2001.]
We will explore different ways of computing the Hausdorff dimension of “fractal” sets (like the Cantor set). Participants should be familiar with some of the geometry of the line and plane. It’s also useful to know how to add up a geometric series. We will develop the rest of the ideas as we go along.
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• March 1, 2015: 2:00 – 4:00 PM: Sequences and Differences.
The sequence s: 0, 1, 4, 9, 16, 2.5, 36, . . . is made by squaring the counting numbers. This infinite sequence is described by a simple rule s(n) = n^2. For instance, s(3) = 9 and s(5) = 25. That compact formula contains the same information as the long sequence of numbers.
Given a sequence f, build its difference-sequence Δf by subtracting successive terms
Δf(n) = f(n+1) – f(n).
The square sequence s above has difference-sequence Δs: 1, 3, 5, 7, 9, 11, . . . Repeat the process to find the second-difference sequence is constant: ΔΔs: 2, 2, 2, 2, 2, . . . That constant sequence is pretty simple, and ΔΔΔs is all zeros.
Perform similar difference calculations for the cube sequence c: 0, 1, 8, 27, 64, 125, . . . Is ΔΔc a constant sequence?
We will investigate these types of questions, and the converse idea: Given a sequence of numbers, can we construct a polynomial formula that describes it?
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• January 18, 2015: 2:00 – 4:00 PM: The Fourth Dimension.
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• November 9, 2014: 2:00 – 4:00 PM: Non-Standard Digits.
We use “positional notation in base ten” (the Hindu-Arabic number system) with standard digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Every whole number is represented uniquely as a string of those digits. You learned to count with this notation.
Maybe you have seen some other bases, like binary (base two) and hex (base sixteen). Can you do standard arithmetic in other bases? For instance, find (26) times (23) in base seven, without converting any numbers to base ten.
Sticking to base ten, let’s eliminate the digit 0, replacing it with the new digit X, where X represents “ten”. Is every whole number represented uniquely in this “X system” using digits {1, 2, 3, 4, 5, 6, 7, 8, 9, X}? For instance, the number after 79 (seventy-nine) is 7X (seventy-ten).
We will explore this X-system for number names, and some variations.
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• October 19, 2014: 2:00 – 4:00 PM: Groups and Symmetry.
Geometric figures exhibit various types of symmetry. For instance, a square rotated by ninety degrees looks the same as before. There seem to be eight different symmetries of that square (rotations and reflections). The set of all those symmetrical motions forms a “group” of eight elements. We will examine some small symmetry groups and discuss ways to represent similar groups.
Note: This topic requires knowledge of high school math. It’s not intended for students in middle or elementary school.
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• September 21, 2014, 2:00 – 4:00 PM: Tiling with dominoes and trominoes.
We investigate some old tiling puzzles. An 8-by-8 checkerboard can easily be tiled by dominoes. (Each domino covers two squares). A board with two opposite corners removed cannot be tiled by dominoes. [Why?]
If some other two squares were removed, could the resulting board be tiled by dominoes?
If one square is removed from the board, can the remaining 63 squares be tiled by straight trominoes?
Here are topics of meetings held during 2013-2014:
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• June 8, 2014 from 2:00 to 4:00 PM: Algebraic Numbers.
A complex number s is “algebraic” if it satisfies g(s) = 0, for some nonzero polynomial g(X) with integer coefficients. The “degree” of s is the smallest degree of a nonzero polynomial that kills s. For instance, rational numbers are algebraic of degree 1, while any irrational square root is algebraic of degree 2.
Can you prove that the cube root of 5 is algebraic of degree 3 ?
Is \sqrt{2} + \sqrt{3} algebraic? If so, what is its degree?
If s and t are algebraic, must s + t be algebraic?
cos(0) and cos(60) are algebraic of degree 1, while cos(30) and cos(45) have degree 2.
Are cos(20) and cos(72) algebraic?
Euclidean geometry provides construction tools: straight-edge and compass. Starting with a segment of length 1 we can use those tools to construct segments of other lengths, like 3, 1/6, and \sqrt{5}. In the 1800s mathematicians proved:
Every constructible length is an algebraic number whose degree is a 2-power.
This Theorem implies the impossibility of solving some famous ancient problems: trisecting a given angle, duplicating a cube, and constructing a regular heptagon.
• May 18, 2014 from 2:00 to 4:00 PM: Finding areas.
How can we measure the area of a region D drawn on paper? We might draw a grid of small squares on that paper and estimate area by counting the number of squares that are inside D. If D is a polygon with lattice point vertices (i.e. they have integer coordinates), “Pick’s Theorem” says that the area of D can be found by counting the number of lattice points in the interior and on the boundary.
For general D, is there a mechanical tool that can calculate the area of D after we use the tool’s pointer to trace around the boundary of D?
• April 13, 2014 from 2:00 to 4:00 PM: Farey Fractions.
The n-th Farey sequence F(n) is the list of fractions between 0 and 1 having denominator at most n. Those fractions are expressed in lowest terms, and listed in order of size. For instance, the sequence F(5) starts out: ( 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, . . . ). Students will explore those sequences and investigate their properties. Then we will try to explain (prove) why those observations hold true.
• March 9, 2014 from 2:00 to 4:00 PM: What is Pi?
For any circle, C = 2 pi R or A = pi R^2. These formulas for circumference and area are familiar, but lead to hard questions. If the number pi is defined using that circumference formula, why must that area formula hold true?
If we draw a circle of the size of Ohio on the earth’s surface and measure its area and circumference, will we get an accurate calculation of pi ?
Other shapes: Are there definitions for pi and R so that those two formulas will still work for a square? For a triangle? What shape has the largest value for pi? The smallest pi?
• February 9, 2014 from 1:30 to 3:30 PM: Fractions and repeating decimals.
The Division Algorithm comes in two flavors, producing either: (1) a quotient and remainder; or (2) a decimal. For instance, 18 divided by 7 is either: (1) quotient 2 and remainder 4; or (2) 2.57142 . . . Why do those algorithms make sense? Why does every fraction have a decimal expression that eventually repeats itself? What does this have to do with modular arithmetic?
• January 12, 2014 from 2:00 to 4:00 PM: Flexagons and Moebius bands.
Participants will fold flexagons and investigate some of their mathematical properties. This may lead to discussion of Moebius strips, and some other surfaces that arise from gluing the edges of a rectangular region.
• December 1, 2013 from 2:00 to 4:00 PM: Graphs.
We discuss the graphs that are made of vertices and edges. Think of a vertex as a “dot” and an edge as a path between two vertices.
Which graphs can be drawn on paper, tracing every edge without repetition and without lifting pencil from paper? Can this be done with a path that doesn’t cross itself?
Suppose a closed curve is drawn in the plane, perhaps with several “simple” self-crossings. Is there a way to color each of the resulting regions red or blue, so that adjacent regions have different colors? What does that situation have to do with graphs?
• November 10, 2013 from 2:00 to 4:00 PM: Complex numbers.
Represent a complex number a + bi as the point (a, b) in the plane. For numbers z = a + bi and w = c + di, their sum z + w is represented geometrically as the “vector sum”: the parallelogram diagonal. The product zw also has geometric meaning, involving polar coordinates. These ideas provide a powerful connection between algebra and plane geometry.
• October 13, 2013 from 2:00 to 4:00 PM: How can a square be covered by squares?
This question has different interpretations, depending on how terms are defined. We will discuss some possibilities and work toward answers.
• September 8, 2013 from 2:00 to 4:00 PM: Fibonacci numbers and related sequences.
The sequence of Fibonacci numbers is a popular elementary topic, involving interesting properties and generalizations. You may have worked with such sequences before, so we will discuss some of their less familiar properties. I hope some participants will bring calculators to this session.
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(assuming some calculus background)
• March 23, 2014 from 2:00 to 4:00 PM: Topology of Closed Surfaces.
A square is made of thin, flexible rubber, and each edge is assigned a “directed-label”. The label is one of the letters a or b, and the direction is indicated by an arrow along that edge. Assume that each letter appears exactly twice. Now let’s glue together the two “a” edges, matching their arrows. Similarly glue the two “b” edges. This produces a surface with no edges. Different surfaces arise, depending on how the directed-labels were chosen. For example, we could get a Torus, or a Klein Bottle.
How many “essentially different” surfaces are possible?
What surfaces can arise if we put directed-labels on edges of a hexagon,
using letters a, b, c ? Can every surface be build by gluing edges of some polygon with directed-labels?
• February 23, 2014 from 2:00 to 4:00 PM: Generating functions.
A sequence {a_n} of numbers can be studied by analyzing the associated power series: \sum a_n x^n.
For instance, the constant sequence 1, 1, 1, 1, … corresponds to the function 1 + x + x^2 x^3 + . . . = 1/(1 – x).
What function corresponds to the sequence of Fibonacci numbers?
What do “partial fractions” have to do with this topic?
• January 26, 2014 from 2:00 to 4:00 PM: Cyclotomic polynomials.
This topic arises from questions about “roots of unity” and their algebraic properties. Those properties also relate to some interesting questions involving linear algebra, geometry, and combinatorics.
Here are topics of meetings held during 2012-2013:
• May 19, 2013 from 1:00 to 3:00 PM: Non-standard digits
Led by Daniel Shapiro
Standard base ten notation uses the first ten numbers as the digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Could a different collection of numbers be used as digits, instead of those ten?
If so, is every number expressed uniquely in that system? Do the usual algorithms for addition and multiplication still work?
• April 28, 2013: Cutting and pasting
Led by Daniel Shapiro
Call two plane figures “equi-decomposable” if there is a way to break one figure into several pieces, and rearrange those pieces into the other figure.
Is every polygon equi-decomposable with some square? Is there a 3-dimensional version of this theory?
• April 7, 2013: Aspects of counting
Led by Brad Findell and Daniel Shapiro
There are many ways to count the number of things in a collection. We will discuss one or two aspects.
The group of students might split in two, with one discussion aimed at younger students and a more advanced topic for high school students.
• February 24, 2013: What’s the deal with zero
Why are students not allowed to “divide by zero”? Is there some good reason for that prohibition?
• November 18, 2012: What do means mean
A “mean” of two numbers is some sort of average, right? But different contexts require different kinds of averages. We will explore several different means and the relationships among them.
• October 21, 2012: Tangent Lines
Led by Bart Snapp
When is a line “tangent” to a given curve? We will draw lots of tangent lines to a given curve and explore some of the patterns geometrically and algebraically.
Participants are invited to bring a ruler (and pencil) to this session.
• September 23, 2012: What’s the Difference
From a given row of numbers, construct a new row by taking differences of consecutive numbers. For instance, if the first row is 0, 1, 4, 12, 16 the row of differences is 1, 3, 8, 4.
If the initial row is the list of squares, 0, 1, 4, 9, 16, … , then the derived row is 1, 3, 5, 7, … . Repeating the process with that row yields the second derived row 2, 2, 2, … . We investigate various sorts of sequences using such differences.
The Four Numbers Game is a related situation but using “absolute differences” in a cyclic fashion, as described in: 4 Numbers Game [pdf]
Here are topics of meetings held during 2011-2012:
• May 6: Triangulations.
Many polygonal figures can be built from triangular pieces. Can *every* polygon in the plane be broken into triangles? We will investigate triangulated figures in different ways, leading to results in geometry and number theory.
• April 15: Paper Folding Constructions.
Led by Bart Snapp.
In high school geometry, students learn to construct different objects using two “Eudlidean tools”: a straight-edge and a compass. For instance, given three points A, B, C, can you construct the circle that passes through them? Given a circle and a point P outside that circle, can you construct a line through P and tangent to that circle?
Today we will use “paper folding” instead of those other tools. You are invited to bring tracing paper, a ruler, and a marker, to help work out some of those constructions.
• March 25: Extremal Geometry.
Led by Professor Matthew Kahle.
We’ll begin with some problems involving length, and then move on to some involving area.
• Warmup problem #1: Given ten points in a unit square (side length one), show that some pair of points is no more than distance 1/2 apart.
• Warmup problem #2: Given nine points in a triangle of area one, show that some three of the points must form a triangle of area 1/4 or less. If you solve that — can you do better than nine? What is the smallest number of points for which this is true?
• March 11: Roots of Unity.
Complex numbers can be represented as points in the plane, and there are geometric ways to look at their addition and multiplication. A complex “root of unity” is a number z having some power equal to 1. That is: zn = 1, for some positive integer n. We will discuss some of their properties and uses.
• February 26: The Fourth Dimension and Beyond.
Visitor Chris Altomare will lead this session.
232 Cockins Hall.
• February 5: Polyhedra and Symmetry.
How many rotations of space carry a given cube to itself? What other geometric figures are as symmetric as a cube?
• January 15, 2012: Combinatorial Games.
The two players, L and R, play a finite game with no hidden information. A game G that is a win for L (no matter who goes first) is called a “positive” game. A game that is a win for R is “negative”. If G is a win for the second player it is a “zero game”. Games can be added and subtracted, and we consider the algebra of games.
• December 18: Tilings.
An 8-by-8 checkerboard can easily be tiled by dominoes (where each domino covers two squares). A simple argument proves that a checkerboard with two opposite corners removed cannot be tiled by dominoes. So if two squares are removed from the board, when is there a domino tiling.
• December 4: Infinities.
Even if I don’t know how to count, I can detect whether two sets have the same size by matching their elements. This idea allows us to compare sizes of infinite sets.
• November 20: Bases.
We all express positive integers in base ten and own calculators that use that notation. But can you calculate with base B expressions, for another number B? Even if those are familiar, can you find (1/3) in base 5? Or (3/4) in base 7? Which “base B decimals” terminate?
This session will meet in 232 Cockins Hall.
• October 23: Vectors.
Some ways that vectors apply to elementary geometry.
This session will meet in room 154 (first floor).
• October 2: Mass point geometry.
Led by visitor Max Warshauer from Texas State University.
For further information on this topic see the article by Tom Rike posted at the San Jose Math Circle.
Meet in room 724.
Here are topics of meetings held during 2010-2011:
October 24: Walking the dog. Comparing lengths of paths.
December 5: Taxicab geometry.
January 30: Symmetry: Tilings and polyhedra.
March 6: Mobius strips: Ideas of topology.
April 3: Impossible scores.
April 17: Hyperbolic geometry: Five squares at each corner? | 7,200 | 27,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-39 | longest | en | 0.837587 |
https://opentextbc.ca/introstatopenstax/chapter/the-standard-normal-distribution/ | 1,579,550,502,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250599789.45/warc/CC-MAIN-20200120195035-20200120224035-00543.warc.gz | 594,063,406 | 27,283 | The Normal Distribution
# The Standard Normal Distribution
The standard normal distribution is a normal distribution of standardized values called z-scores. A z-score is measured in units of the standard deviation. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:
x = μ + (z)(σ) = 5 + (3)(2) = 11
The z-score is three.
The mean for the standard normal distribution is zero, and the standard deviation is one. The transformation z = produces the distribution Z ~ N(0, 1). The value x in the given equation comes from a normal distribution with mean μ and standard deviation σ.
### Z-Scores
If X is a normally distributed random variable and X ~ N(μ, σ), then the z-score is:
The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. Values of x that are larger than the mean have positive z-scores, and values of x that are smaller than the mean have negative z-scores. If x equals the mean, then x has a z-score of zero.
Suppose X ~ N(5, 6). This says that X is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then:
This means that x = 17 is two standard deviations (2σ) above or to the right of the mean μ = 5.
Notice that: 5 + (2)(6) = 17 (The pattern is μ + = x)
Now suppose x = 1. Then: z = = = –0.67 (rounded to two decimal places)
This means that x = 1 is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5. Notice that: 5 + (–0.67)(6) is approximately equal to one (This has the pattern μ + (–0.67)σ = 1)
Summarizing, when z is positive, x is above or to the right of μ and when z is negative, x is to the left of or below μ. Or, when z is positive, x is greater than μ, and when z is negative x is less than μ.
Try It
What is the z-score of x, when x = 1 and X ~ N(12,3)?
Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let X = the amount of weight lost (in pounds) by a person in a month. Use a standard deviation of two pounds. X ~ N(5, 2). Fill in the blanks.
a. Suppose a person lost ten pounds in a month. The z-score when x = 10 pounds is z = 2.5 (verify). This z-score tells you that x = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
a. This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean five.
b. Suppose a person gained three pounds (a negative weight loss). Then z = __________. This z-score tells you that x = –3 is ________ standard deviations to the __________ (right or left) of the mean.
b. z = –4. This z-score tells you that x = –3 is four standard deviations to the left of the mean.
c. Suppose the random variables X and Y have the following normal distributions: X ~ N(5, 6) and Y ~ N(2, 1). If x = 17, then z = 2. (This was previously shown.) If y = 4, what is z?
c. z = = = 2 where µ = 2 and σ = 1.
The z-score for y = 4 is z = 2. This means that four is z = 2 standard deviations to the right of the mean. Therefore, x = 17 and y = 4 are both two (of their own) standard deviations to the right of their respective means.
The z-score allows us to compare data that are scaled differently. To understand the concept, suppose X ~ N(5, 6) represents weight gains for one group of people who are trying to gain weight in a six week period and Y ~ N(2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since x = 17 and y = 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means.
Try It
Fill in the blanks.
Jerome averages 16 points a game with a standard deviation of four points. X ~ N(16,4). Suppose Jerome scores ten points in a game. The z–score when x = 10 is –1.5. This score tells you that x = 10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).
The Empirical RuleIf X is a random variable and has a normal distribution with mean µ and standard deviation σ, then the Empirical Rule states the following:
• About 68% of the x values lie between –1σ and +1σ of the mean µ (within one standard deviation of the mean).
• About 95% of the x values lie between –2σ and +2σ of the mean µ (within two standard deviations of the mean).
• About 99.7% of the x values lie between –3σ and +3σ of the mean µ (within three standard deviations of the mean). Notice that almost all the x values lie within three standard deviations of the mean.
• The z-scores for +1σ and –1σ are +1 and –1, respectively.
• The z-scores for +2σ and –2σ are +2 and –2, respectively.
• The z-scores for +3σ and –3σ are +3 and –3 respectively.
The empirical rule is also known as the 68-95-99.7 rule.
The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then X ~ N(170, 6.28).
a. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010. The z-score when x = 168 cm is z = _______. This z-score tells you that x = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
a. –0.32, 0.32, left, 170
b. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a z-score of z = 1.27. What is the male’s height? The z-score (z = 1.27) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.
b. 177.98 cm, 1.27, right
Try It
Use the information in (Figure) to answer the following questions.
1. Suppose a 15 to 18-year-old male from Chile was 176 cm tall from 2009 to 2010. The z-score when x = 176 cm is z = _______. This z-score tells you that x = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
2. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a z-score of z = –2. What is the male’s height? The z-score (z = –2) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.
From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males from 1984 to 1985. Then Y ~ N(172.36, 6.34).
The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then X ~ N(170, 6.28).
Find the z-scores for x = 160.58 cm and y = 162.85 cm. Interpret each z-score. What can you say about x = 160.58 cm and y = 162.85 cm as they compare to their respective means and standard deviations?
The z-score for x = -160.58 is z = –1.5.
The z-score for y = 162.85 is z = –1.5.
Both x = 160.58 and y = 162.85 deviate the same number of standard deviations from their respective means and in the same direction.
Try It
In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean µ = 496 and a standard deviation σ = 114. Let X = a SAT exam verbal section score in 2012. Then X ~ N(496, 114).
Find the z-scores for x1 = 325 and x2 = 366.21. Interpret each z-score. What can you say about x1 = 325 and x2 = 366.21 as they compare to their respective means and standard deviations?
Suppose x has a normal distribution with mean 50 and standard deviation 6.
• About 68% of the x values lie within one standard deviation of the mean. Therefore, about 68% of the x values lie between –1σ = (–1)(6) = –6 and 1σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation from the mean 50. The z-scores are –1 and +1 for 44 and 56, respectively.
• About 95% of the x values lie within two standard deviations of the mean. Therefore, about 95% of the x values lie between –2σ = (–2)(6) = –12 and 2σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations from the mean 50. The z-scores are –2 and +2 for 38 and 62, respectively.
• About 99.7% of the x values lie within three standard deviations of the mean. Therefore, about 95% of the x values lie between –3σ = (–3)(6) = –18 and 3σ = (3)(6) = 18 from the mean 50. The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations of the mean 50. The z-scores are –3 and +3 for 32 and 68, respectively.
Try It
Suppose X has a normal distribution with mean 25 and standard deviation five. Between what values of x do 68% of the values lie?
From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males in 1984 to 1985. Then Y ~ N(172.36, 6.34).
1. About 68% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
2. About 95% of the y values lie between what two values? These values are ________________. The z-scores are ________________ respectively.
3. About 99.7% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
1. About 68% of the values lie between 166.02 cm and 178.7 cm. The z-scores are –1 and 1.
2. About 95% of the values lie between 159.68 cm and 185.04 cm. The z-scores are –2 and 2.
3. About 99.7% of the values lie between 153.34 cm and 191.38 cm. The z-scores are –3 and 3.
Try It
The scores on a college entrance exam have an approximate normal distribution with mean, µ = 52 points and a standard deviation, σ = 11 points.
1. About 68% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
2. About 95% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
3. About 99.7% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
### References
“Blood Pressure of Males and Females.” StatCruch, 2013. Available online at http://www.statcrunch.com/5.0/viewreport.php?reportid=11960 (accessed May 14, 2013).
“The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of z-scores.” London School of Hygiene and Tropical Medicine, 2009. Available online at http://conflict.lshtm.ac.uk/page_125.htm (accessed May 14, 2013).
“2012 College-Bound Seniors Total Group Profile Report.” CollegeBoard, 2012. Available online at http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf (accessed May 14, 2013).
“Digest of Education Statistics: ACT score average and standard deviations by sex and race/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.” National Center for Education Statistics. Available online at http://nces.ed.gov/programs/digest/d09/tables/dt09_147.asp (accessed May 14, 2013).
Data from the San Jose Mercury News.
Data from The World Almanac and Book of Facts.
“List of stadiums by capacity.” Wikipedia. Available online at https://en.wikipedia.org/wiki/List_of_stadiums_by_capacity (accessed May 14, 2013).
Data from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013).
### Chapter Review
A z-score is a standardized value. Its distribution is the standard normal, Z ~ N(0, 1). The mean of the z-scores is zero and the standard deviation is one. If z is the z-score for a value x from the normal distribution N(µ, σ) then z tells you how many standard deviations x is above (greater than) or below (less than) µ.
### Formula Review
z = a standardized value (z-score)
mean = 0; standard deviation = 1
To find the kth percentile of X when the z-scores is known:
k = μ + (z)σ
z-score: z =
Z = the random variable for z-scores
A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable X in words. X = ____________.
ounces of water in a bottle
A normal distribution has a mean of 61 and a standard deviation of 15. What is the median?
<!– <solution id=”fs-idm63161360″> 61 –>
X ~ N(1, 2)
σ = _______
2
A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable X in words. X = ______________.
<!– <solution id=”fs-idp15866016″> diameter of a rubber ball –>
X ~ N(–4, 1)
What is the median?
–4
X ~ N(3, 5)
σ = _______
<!– <solution id=”fs-idm120454704″> 5 –>
X ~ N(–2, 1)
μ = _______
–2
What does a z-score measure?
<!– <solution id=”fs-idm57247936″> The number of standard deviations a value is from the mean. –>
What does standardizing a normal distribution do to the mean?
The mean becomes zero.
Is X ~ N(0, 1) a standardized normal distribution? Why or why not?
<!– <solution id=”fs-idp48272224″> Yes because the mean is zero, and the standard deviation is one. –>
What is the z-score of x = 12, if it is two standard deviations to the right of the mean?
z = 2
What is the z-score of x = 9, if it is 1.5 standard deviations to the left of the mean?
<!– <solution id=”fs-idp6573504″> z = –1.5 –>
What is the z-score of x = –2, if it is 2.78 standard deviations to the right of the mean?
z = 2.78
What is the z-score of x = 7, if it is 0.133 standard deviations to the left of the mean?
<!– <solution id=”fs-idp38051616″> z = –0.133 –>
Suppose X ~ N(2, 6). What value of x has a z-score of three?
x = 20
Suppose X ~ N(8, 1). What value of x has a z-score of –2.25?
<!– <solution id=”fs-idp19875360″> x = 5.75 –>
Suppose X ~ N(9, 5). What value of x has a z-score of –0.5?
x = 6.5
Suppose X ~ N(2, 3). What value of x has a z-score of –0.67?
<!– <solution id=”fs-idp31923120″> x = –0.01 –>
Suppose X ~ N(4, 2). What value of x is 1.5 standard deviations to the left of the mean?
x = 1
Suppose X ~ N(4, 2). What value of x is two standard deviations to the right of the mean?
<!– <solution id=”fs-idm47755696″> x = 8 –>
Suppose X ~ N(8, 9). What value of x is 0.67 standard deviations to the left of the mean?
x = 1.97
Suppose X ~ N(–1, 2). What is the z-score of x = 2?
<!– <solution id=”fs-idm119642576″> z = 1.5 –>
Suppose X ~ N(12, 6). What is the z-score of x = 2?
z = –1.67
Suppose X ~ N(9, 3). What is the z-score of x = 9?
<!– <solution id=”fs-idm56843792″> z = 0 –>
Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the z-score of x = 5.5?
z ≈ –0.33
In a normal distribution, x = 5 and z = –1.25. This tells you that x = 5 is ____ standard deviations to the ____ (right or left) of the mean.
<!– <solution id=”fs-idm62539280″> 1.25, left –>
In a normal distribution, x = 3 and z = 0.67. This tells you that x = 3 is ____ standard deviations to the ____ (right or left) of the mean.
0.67, right
In a normal distribution, x = –2 and z = 6. This tells you that x = –2 is ____ standard deviations to the ____ (right or left) of the mean.
<!– <solution id=”fs-idm102069296″> six, right –>
In a normal distribution, x = –5 and z = –3.14. This tells you that x = –5 is ____ standard deviations to the ____ (right or left) of the mean.
3.14, left
In a normal distribution, x = 6 and z = –1.7. This tells you that x = 6 is ____ standard deviations to the ____ (right or left) of the mean.
<!– <solution id=”fs-idm98531584″> 1.7, left –>
About what percent of x values from a normal distribution lie within one standard deviation (left and right) of the mean of that distribution?
About what percent of the x values from a normal distribution lie within two standard deviations (left and right) of the mean of that distribution?
<!– <solution id=”fs-idm47813520″> about 95.45% –>
About what percent of x values lie between the second and third standard deviations (both sides)?
Suppose X ~ N(15, 3). Between what x values does 68.27% of the data lie? The range of x values is centered at the mean of the distribution (i.e., 15).
<!– <solution id=”fs-idm33513040″> between 12 and 18 –>
Suppose X ~ N(–3, 1). Between what x values does 95.45% of the data lie? The range of x values is centered at the mean of the distribution(i.e., –3).
between –5 and –1
Suppose X ~ N(–3, 1). Between what x values does 34.14% of the data lie?
<!– <solution id=”fs-idm68579024″> between –4 and –3 or between –3 and –2 –>
About what percent of x values lie between the mean and three standard deviations?
About what percent of x values lie between the mean and one standard deviation?
<!– <solution id=”fs-idp3556992″> about 34.14% –>
About what percent of x values lie between the first and second standard deviations from the mean (both sides)?
About what percent of x values lie betwween the first and third standard deviations(both sides)?
<!– <solution id=”fs-idp56159920″> about 34.46% –>
Use the following information to answer the next two exercises: The life of Sunshine CD players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts.
Define the random variable X in words. X = _______________.
The lifetime of a Sunshine CD player measured in years.
X ~ _____(_____,_____)
<!– <solution id=”fs-idp89883200″> X ~ N(4.1, 1.3) –>
### Homework
Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
What is the median recovery time?
1. 2.7
2. 5.3
3. 7.4
4. 2.1
<!– <solution id=”fs-idm68359392″> b –>
What is the z-score for a patient who takes ten days to recover?
1. 1.5
2. 0.2
3. 2.2
4. 7.3
c
The length of time to find it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true?
1. The data cannot follow the uniform distribution.
2. The data cannot follow the exponential distribution..
3. The data cannot follow the normal distribution.
1. I only
2. II only
3. III only
4. I, II, and III
<!– <solution id=”fs-idp16013392″> b –>
The heights of the 430 National Basketball Association players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with mean, µ = 79 inches and a standard deviation, σ = 3.89 inches. For each of the following heights, calculate the z-score and interpret it using complete sentences.
1. 77 inches
2. 85 inches
3. If an NBA player reported his height had a z-score of 3.5, would you believe him? Explain your answer.
1. Use the z-score formula. z = –0.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average.
2. Use the z-score formula. z = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average.
3. Height = 79 + 3.5(3.89) = 92.615 inches, which is taller than 7 feet, 8 inches. There are very few NBA players this tall so the answer is no, not likely.
The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. Systolic blood pressure for males follows a normal distribution.
1. Calculate the z-scores for the male systolic blood pressures 100 and 150 millimeters.
2. If a male friend of yours said he thought his systolic blood pressure was 2.5 standard deviations below the mean, but that he believed his blood pressure was between 100 and 150 millimeters, what would you say to him?
<!– <solution id=”eip-774″> Use the z-score formula. 100 – 125 14 ≈ –1.8 and 100 – 125 14 ≈ 1.8 I would tell him that 2.5 standard deviations below the mean would give him a blood pressure reading of 90, which is below the range of 100 to 150. –>
Kyle’s doctor told him that the z-score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. If X = a systolic blood pressure score then X ~ N (125, 14).
1. Kyle’s systolic blood pressure is 175.
2. Kyle’s systolic blood pressure is 1.75 times the average blood pressure of men his age.
3. Kyle’s systolic blood pressure is 1.75 above the average systolic blood pressure of men his age.
4. Kyles’s systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men.
2. Calculate Kyle’s blood pressure.
1. iv
2. Kyle’s blood pressure is equal to 125 + (1.75)(14) = 149.5.
Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean µ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ N(10.2, 0.8). Calculate the z-scores that correspond to the following weights and interpret them.
1. 11 kg
2. 7.9 kg
3. 12.2 kg
<!– <solution id=”eip-182″> 11 – 10.2 0.8 = 1 A child who weighs 11 kg is one standard deviation above the mean of 10.2 kg. 7.9 – 10.2 0.8 = –2.875 A child who weighs 7.9 kg is 2.875 standard deviations below the mean of 10.2 kg. 12.2 – 10.2 0.8 = 2.5 A child who weighs 12.2 kg is 2.5 standard deviation above the mean of 10.2 kg. –>
In 2005, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ = 520 and standard deviation σ = 115.
1. Calculate the z-score for an SAT score of 720. Interpret it using a complete sentence.
2. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score?
3. For 2012, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternate to the SAT and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test they took?
Let X = an SAT math score and Y = an ACT math score.
1. X = 720 = 1.74 The exam score of 720 is 1.74 standard deviations above the mean of 520.
2. z = 1.5
The math SAT score is 520 + 1.5(115) ≈ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520.
3. = ≈ 1.59, the z-score for the SAT. = ≈ 1.70, the z-scores for the ACT. With respect to the test they took, the person who took the ACT did better (has the higher z-score).
<!– <para id=”fs-idp207361600″>Use the following information to answer the next three exercises: X ~ U(3, 13) –>
### Glossary
Standard Normal Distribution
a continuous random variable (RV) X ~ N(0, 1); when X follows the standard normal distribution, it is often noted as Z ~ N(0, 1).
z-score
the linear transformation of the form z = ; if this transformation is applied to any normal distribution X ~ N(μ, σ) the result is the standard normal distribution Z ~ N(0,1). If this transformation is applied to any specific value x of the RV with mean μ and standard deviation σ, the result is called the z-score of x. The z-score allows us to compare data that are normally distributed but scaled differently. | 6,928 | 24,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2020-05 | latest | en | 0.917764 |
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A005097 (Odd primes - 1)/2. 142
1, 2, 3, 5, 6, 8, 9, 11, 14, 15, 18, 20, 21, 23, 26, 29, 30, 33, 35, 36, 39, 41, 44, 48, 50, 51, 53, 54, 56, 63, 65, 68, 69, 74, 75, 78, 81, 83, 86, 89, 90, 95, 96, 98, 99, 105, 111, 113, 114, 116, 119, 120, 125, 128, 131, 134, 135, 138, 140, 141, 146, 153, 155, 156 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Or, numbers k such that 2k+1 is prime. Also numbers not of the form 2xy + x + y. - Jose Brox (tautocrona(AT)terra.es), Dec 29 2005 This sequence arises if you factor the product of a large number of the first odd numbers into the form 3^n(3)5^n(5)7^n(7)11^n(11)... Then n(3)/n(5) = 2, n(3)/n(7) = 3, n(3)/n(11) = 5, ... . - Andrzej Staruszkiewicz (astar(AT)th.if.uj.edu.pl), May 31 2007 Kohen shows: A king invites n couples to sit around a round table with 2n+1 seats. For each couple, the king decides a prescribed distance d between 1 and n which the two spouses have to be seated from each other (distance d means that they are separated by exactly d-1 chairs). We will show that there is a solution for every choice of the distances if and only if 2n+1 is a prime number [i.e., iff n is in A005097], using a theorem known as Combinatorial Nullstellensatz. - Jonathan Vos Post, Jun 14 2010 Starting from 6, positions at which new primes are seen for Goldbach partitions. E.g., 31 is first seen at 34 from 31+3, so position = 1 + (34-6)/2 = 15. - Bill McEachen, Jul 05 2010 Perfect error-correcting Lee codes of word length n over Z: it is conjectured that these always exist when 2n+1 is a prime, as mentioned in Horak. - Jonathan Vos Post, Sep 19 2011 Also solutions to: A000010(2*n+1) = n * A000005(2*n+1). - Enrique Pérez Herrero, Jun 07 2012 A193773(a(n)) = 1. - Reinhard Zumkeller, Jan 02 2013 I conjecture that the set of pairwise sums of terms of this sequence (A005097) is the set of integers greater than 1, i.e.: 1+1=2, 1+2=3, ..., 5+5=10, ... (This is equivalent to Goldbach's conjecture: every even integer greater than or equal to 6 can be expressed as the sum of two odd primes.) - Lear Young, May 20 2014 See conjecture and comments from Richard R. Forberg, in Links section below, on the relationship of this sequence to rules on values of c that allow both p^q+c and p^q-c to be prime, for an infinite number of primes p. - Richard R. Forberg, Jul 13 2016 The sequence represents the minimum number Ng of gears which are needed to draw a complete graph of order p using a Spirograph(R), where p is an odd prime. The resulting graph consists of Ng hypotrochoids whose respective nodes coincide. If the teethed ring has a circumference p then Ng = (p-1)/2. Examples: A complete graph of order three can be drawn with a Spirograph(R) using a ring with 3n teeth and one gear with n teeth. n is an arbitrary number, only related to the geometry of the gears. A complete graph of order 5 can be drawn using a ring with diameter 5 and 2 gears with diameters 1 and 2 respectively. A complete graph of order 7 can be drawn using a ring with diameter 7 and 3 gears with diameters 1, 2 and 3 respectively. - Bob Andriesse, Mar 31 2017 LINKS T. D. Noe, Table of n, a(n) for n = 1..1000 Richard R. Forberg, Comments on A005097 Peter Horak and Bader F. AlBdaiwi, Diameter Perfect Lee Codes, arXiv:1109.3475 [cs.IT], 2011-2012. Daniel Kohen and Ivan Sadofschi, A New Approach on the Seating Couples Problem, arXiv:1006.2571 [math.CO], 2010. Dhananjay P. Mehendale, On Hamilton Decompositions, arXiv:0806.0251 [math.GM], 2008. Eric Weisstein's World of Mathematics, Legendre Symbol FORMULA a(n) = A006093(n)/2 = A000010(A000040(n+1))/2. a(n) = (prime(n+1)^2-1)/(2*sigma(prime(n+1))) = (A000040(n+1)^2-1)/(2*A000203(A000040(n+1))). - Gary Detlefs, May 02 2012 a(n) = (A065091(n) - 1) / 2. - Reinhard Zumkeller, Jan 02 2013 a(n) ~ n*log(n)/2. - Ilya Gutkovskiy, Jul 11 2016 a(n) = A294507(n) (mod prime(n+1)). - Jonathan Sondow, Nov 04 2017 a(n) = A130290(n+1). - Chai Wah Wu, Jun 04 2022 MAPLE with(numtheory): p:=n-> ithprime(n):seq((p(n+1)^2-1)/(2*sigma(p(n+1))), n= 1..64) # Gary Detlefs, May 02 2012 MATHEMATICA Table[p=Prime[n]; (p-1)/2, {n, 2, 22}] (* Vladimir Joseph Stephan Orlovsky, Apr 29 2008 *) (Prime[Range[2, 70]]-1)/2 (* Harvey P. Dale, Jul 11 2020 *) PROG (PARI) forprime(p=3, 1e4, print1(p>>1", ")) \\ Charles R Greathouse IV, Jun 16 2011 (Haskell) a005097 = (`div` 2) . a065091 -- Reinhard Zumkeller, Jan 02 2013 (Magma) [n: n in [1..160] |IsPrime(2*n+1)]; // Vincenzo Librandi, Feb 16 2015 (Python) from sympy import prime def A005097(n): return prime(n+1)//2 # Chai Wah Wu, Jun 04 2022 CROSSREFS Complement of A047845. Cf. A000040, A006005, A006093. A130290 is an essentially identical sequence. Cf. A005384 (subsequence of primes), A266400 (their indices in this sequence). Numbers n such that 2n+k is prime: this seq(k=1), A067076 (k=3), A089038 (k=5), A105760 (k=7), A155722 (k=9), A101448 (k=11), A153081 (k=13), A089559 (k=15), A173059 (k=17), A153143 (k=19). Numbers n such that 2n-k is prime: A006254 (k=1), A098090 (k=3), A089253 (k=5), A089192 (k=7), A097069 (k=9), A097338 (k=11), A097363 (k=13), A097480 (k=15), A098605 (k=17), A097932 (k=19). Cf. also A266409, A294507. Sequence in context: A347912 A082583 A274332 * A102781 A130290 A139791 Adjacent sequences: A005094 A005095 A005096 * A005098 A005099 A005100 KEYWORD nonn,easy AUTHOR N. J. A. Sloane STATUS approved
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https://exceljet.net/formulas/lookup-lowest-monday-tide | 1,726,224,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00867.warc.gz | 214,736,296 | 15,885 | ## Summary
To find the lowest tide on a Monday, given a set of data with many days of high and low tides, you can use an array formula based on the IF and MIN functions. In the example shown, the formula in I6 is:
``````{=MIN(IF(day=I5,IF(tide="L",pred)))}
``````
which returns the lowest Monday tide in the data, -0.64
To retrieve the date of the lowest Monday tide, the formula in I7 is:
``````{=INDEX(date,MATCH(1,(day=I5)*(tide="L")*(pred=I6),0))}
``````
Where the worksheet contains the following named ranges: date (B5:B124), day (C5:C124), time (D5:D124), pred (E5:E124), tide (F5:F124).
Both are array formulas and must be entered with control + shift + enter.
Data from tidesandcurrents.noaa.gov for Santa Cruz, California.
## Explanation
At a high level, this example is about finding a minimum value based on multiple criteria. To do that, we are using the MIN function together with two nested IF functions:
``````{=MIN(IF(day=I5,IF(tide="L",pred)))}
``````
working from the inside out, the first IF checks if the day is "Mon", based on the value in I5:
``````IF(day=I5 // is day "Mon"
``````
If the result is TRUE, we run another IF:
``````IF(tide="L",pred) // if tide is "L" return prediction
``````
In other words, if the day is "Mon", we check if the tide is "L". If so, we return the predicted tide level, using the named range pred.
Notice we do not provide a "value if false" for either IF. That means if either logical test is FALSE, the outer IF will return FALSE. For more information on nested IFs, see this article.
It's important to understand that the data set includes 120 rows, so each of the named ranges in the formula contain 120 values. This is what makes this an array formula – we are processing many values at once. After both IFs are evaluated, the outer IF will return an array that contains 120 values like this:
``````{FALSE;FALSE;FALSE;FALSE;FALSE;3.27;FALSE;0.3;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;2.02;FALSE;0.17;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;3.04;FALSE;-0.55;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;1.96;FALSE;-0.64;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;3;FALSE;-0.02;FALSE;FALSE;FALSE;FALSE}
``````
The key thing to notice here is only values associated with Monday and low tide survive the trip through the nested IFs. The other values have been replaced with FALSE. In other words, we are using the double IF structure to "throw away" values we aren't interested in.
The array above is returned directly to the MIN function. The MIN function automatically ignores the FALSE values, and returns the minimum value of those that remain, -0.64.
This is an array formulas and must be entered with control + shift + enter.
### Minimum with MINIFS
If you have Office 365 or Excel 2019, you can use the MINIFS function to get the lowest Monday tide like this:
``````=MINIFS(pred,day,"Mon",tide,"L")
``````
The result is the same, and this formula does not require control + shift + enter.
### Get the date
Once you find the minimum Monday tide level, you will undoubtedly want to know the date and time. This can be done with an INDEX and MATCH formula. The formula in I7 is:
``````{=INDEX(date,MATCH(1,(day=I5)*(tide="L")*(pred=I6),0))}
``````
Working from the inside out, we need to first locate the position of the lowest Monday tide with the MATCH function:
``````MATCH(1,(day=I5)*(tide="L")*(pred=I6),0))
``````
Here, we run through the same conditional tests we applied above to restrict processing to Monday low tides only. However, we apply one more test to restrict results to the minimum value now in I6, and we use a slightly simpler syntax based on boolean logic to apply criteria. We have three separate expressions, each testing one condition:
``````(day=I5)* // day is "Mon"
(tide="L")* // tide is "L"
(pred=I6) // prediction is min value
``````
Each of these expressions runs on 120 values and returns an array of 120 TRUE FALSE results. When these arrays are multiplied by one another, the TRUE FALSE values are coerced to 1s and 0s. The result is a single array like this:
``````{0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;1;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0}
``````
Because there is only one value in the entire data set that meets all three conditions, there is only a single 1 in the array.
Now you can see why we have configured the MATCH function to look for the number 1 in exact match mode. MATCH locates the 1, and returns a position of 88 directly to the INDEX function. We can now rewrite the formula like this:
``````=INDEX(date,88) // returns 23-Dec-19
``````
The INDEX function then returns the 88th value in the named range date, which is 23-Dec-19. This is the date that corresponds to the lowest Monday tide level.
This is an array formulas and must be entered with control + shift + enter.
### Get the time
The formula to retrieve the time of the lowest Monday tide is almost the same as the formula to get the date. The only difference is that the named range time is provided to INDEX instead of date. The formula in I8 is:
``````{=INDEX(time,MATCH(1,(day=I5)*(tide="L")*(pred=I6),0))}
``````
In other respects the behavior of the formula is the same, so we end up with a similar result:
``````=INDEX(time,88) // returns 2:44 PM
``````
As before, INDEX returns the 88th item in the array, which is 2:44 PM.
This is an array formulas and must be entered with control + shift + enter.
Note: in the event of a tie (two Monday low tides with the same value), the INDEX and MATCH formulas above will return the first match.
### Date and time with XLOOKUP
With the XLOOKUP function, you can simplify the formulas used to get the date and time associated with the lowest tide:
``````=XLOOKUP(1,(day=I5)*(tide="L")*(pred=I6),date) // get date
=XLOOKUP(1,(day=I5)*(tide="L")*(pred=I6),time) // get time
``````
This is an example that nicely shows off XLOOKUP's flexibility. We can use exactly the same logic from the INDEX and MATCH formulas above, in a simple and elegant formula.
Dynamic Array Formulas are available in Office 365 only.
Author
### Dave Bruns
Hi - I'm Dave Bruns, and I run Exceljet with my wife, Lisa. Our goal is to help you work faster in Excel. We create short videos, and clear examples of formulas, functions, pivot tables, conditional formatting, and charts. | 2,050 | 6,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-38 | latest | en | 0.825686 |
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Section 5.6 #40: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t^2).$$ Solution: Since $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t)=\dfrac{1}{\sqrt{1-x^2}},$$ we, using the chain rule, compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t^2) &= \dfrac{1}{\sqrt{1-(t^2)^2}} \dfrac{\mathrm{d}}{\mathrm{d}t} t^2 \\ &= \dfrac{2t}{\sqrt{1-t^4}}. \blacksquare \end{array}$$
Section 5.6 #50: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \log(t^2+4) - \dfrac{1}{2} \arctan \left( \dfrac{t}{2} \right) \right].$$ Solution: Since $$\dfrac{\mathrm{d}}{\mathrm{d}t} \log(t) = \dfrac{1}{t}$$ and $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arctan(t) = \dfrac{1}{1+t^2},$$ we use the chain rule to compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \log(t^2+4) - \dfrac{1}{2} \arctan \left( \dfrac{t}{2} \right) \right] &= \dfrac{1}{t^2+4} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ t^2+4 \right] - \left( \dfrac{1}{2} \right) \dfrac{1}{1+\left(\frac{t}{2}\right)^2} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \dfrac{t}{2} \right] \\ &= \dfrac{2t}{t^2+4} -\dfrac{1}{4(1+\frac{t^2}{4})} \\ &= \dfrac{2t}{t^2+4} - \dfrac{1}{4+t^2}. \\ &= \dfrac{2t-1}{t^2+4}. \blacksquare \end{array}$$
Section 5.7 #17: Calculate $$\displaystyle\int \dfrac{x-3}{x^2+1} \mathrm{d}x.$$ Solution: Since $$\displaystyle\int f(x) + g(x) \mathrm{d}x = \displaystyle\int f(x) \mathrm{d}x + \displaystyle\int g(x) \mathrm{d}x,$$ we compute $$\begin{array}{ll} \displaystyle\int \dfrac{x-3}{x^2+1} \mathrm{d}x &= \displaystyle\int \dfrac{x}{x^2+1} \mathrm{d}x - 3 \displaystyle\int \dfrac{1}{x^2+1} \mathrm{d}x \\ &\stackrel{u=x^2+1, \frac{1}{2}\mathrm{d}u= x\mathrm{d}x}{=} \dfrac{1}{2} \displaystyle\int \dfrac{1}{u} \mathrm{d}u - 3 \arctan(x) + C \\ &= \dfrac{1}{2} \log(u) - 3 \arctan(x) + C \\ &= \dfrac{1}{2} \log(x^2+1) - 3\arctan(x) + C. \blacksquare \end{array}$$
Section 5.7 #28: Calculate $$\displaystyle\int_{\ln(2)}^{\ln(4)} \dfrac{e^{-x}}{\sqrt{1-e^{-2x}}} \mathrm{d}x.$$ Solution: Note that $$\sqrt{1-e^{-2x}} = \sqrt{1-(e^{-x})^2}.$$ Therefore calculate $$\begin{array}{ll} \displaystyle\int_{\ln(2)}^{\ln(4)} \dfrac{e^{-x}}{\sqrt{1-e^{-2x}}} \mathrm{d}x &\stackrel{u=e^{-x},-\mathrm{d}u=e^{-x}\mathrm{d}x}{=} -\displaystyle\int_{\frac{1}{2}}^{\frac{1}{4}} \dfrac{1}{\sqrt{1-u^2}} \mathrm{d}u \\ &= - \arcsin(u) \Bigg|_{\frac{1}{2}}^{\frac{1}{4}} \\ &=-\left[ \arcsin\left( \dfrac{1}{4} \right) - \arcsin \left( \dfrac{1}{2} \right) \right] \\ &= \arcsin \left( \dfrac{1}{2} \right) - \arcsin \left( \dfrac{1}{4} \right) \\ &\approx 0.2709. \blacksquare \end{array}$$
Section 5.7 #29: Calculate $$\displaystyle\int_{\frac{\pi}{2}}^{\pi} \dfrac{\sin(x)}{1+\cos^2(x)} \mathrm{d}x.$$ Solution: Let $u=\cos(x)$ so that $-\mathrm{d}u=\sin(x) \mathrm{d}x$. Now if $x=\dfrac{\pi}{2}$ then $u=\cos\left(\dfrac{\pi}{2}\right)=0$ and if $x=\pi$ then $u=cos(\pi)=-1$. Now calculate $$\begin{array}{ll} \displaystyle\int_{\frac{\pi}{2}}^{\pi} \dfrac{\sin(x)}{1+\cos^2(x)} \mathrm{d}x &= \displaystyle\int_{0}^{-1} \dfrac{1}{1+u^2} \mathrm{d}u \\ &= \arctan(u) \Bigg|_{0}^{-1} \\ &= \arctan(0)-\arctan(-1) \\ &= 0- \left( - \dfrac{\pi}{4} \right) \\ &= \dfrac{\pi}{4}. \blacksquare \end{array}$$
Section 5.8 #28: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right).$$ Solution: Since $\tanh(x)=\dfrac{\sinh(x)}{\cosh(x)}$, we may proceed with the quotient rule or we simply use the formula $\dfrac{\mathrm{d}}{\mathrm{d}x} \tanh(x) = \mathrm{sech}^2(x)$. We will do the latter: compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right) &= \dfrac{1}{\tanh(\frac{x}{2})} \dfrac{\mathrm{d}}{\mathrm{d}x} \tanh \left( \dfrac{x}{2} \right) \\ &= \dfrac{1}{\tanh(\frac{x}{2})} \mathrm{sech}^2 \left( \dfrac{x}{2} \right) \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{x}{2} \right] \\ &= \dfrac{1}{2} \dfrac{\coth(\frac{x}{2})}{\sinh(\frac{x}{2})} \dfrac{1}{\cosh^2(\frac{x}{2})} \\ &= \dfrac{1}{2\sinh(\frac{x}{2})\cosh(\frac{x}{2})}. \end{array}$$ If one notes that $2\sinh\left(\dfrac{x}{2} \right)\cosh \left( \dfrac{x}{2} \right) = \sinh(x)$, we may write $$\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right) = \dfrac{1}{\sinh(x)} = \mathrm{csch}(x). \blacksquare$$
Section 5.8 #30: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}x} \left[ x \cosh(x)-\sinh(x) \right].$$ Solution: Since $\cosh'(x)=\sinh(x)$ and $\sinh'(x)=\cosh(x)$, we use the product rule to compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} [ x \cosh(x) - \sinh(x) ] &= [\cosh(x) + x \sinh(x) ] - \cosh(x) \\ &= x \sinh(x). \blacksquare \end{array}$$
Section 5.8 #43: Calculate $$\displaystyle\int \cosh(2x) \mathrm{d}x.$$ Solution: Let $u=2x$ so that $\dfrac{1}{2} \mathrm{d}u=\mathrm{d}x$. Then, compute $$\begin{array}{ll} \displaystyle\int \cosh(2x) \mathrm{d}x &= \dfrac{1}{2} \displaystyle\int \cosh(u) \mathrm{d}u \\ &= \dfrac{1}{2} \sinh(u) + C \\ &= \dfrac{1}{2} \sinh(2x) + C. \blacksquare \end{array}$$
Section 5.8 #55: Calculate $$\displaystyle\int_0^{\ln(2)} \tanh(x) \mathrm{d}x.$$ Solution: Since $\tanh(x) = \dfrac{\sinh(x)}{\cosh(x)}$, let $u=\cosh(x)$ so that $\mathrm{d}u=\sinh(x)$. Also whenever $x=0$ we have $$u=\cosh(0)=\dfrac{e^0+e^{-0}}{2}=1,$$ and if $x=\ln(2)$ we have $$u=\cosh(\ln(2))=\dfrac{e^{\ln(2)}+e^{-\ln(2)}}{2}=\dfrac{2+\frac{1}{2}}{2}=\dfrac{5}{4}.$$ So compute $$\begin{array}{ll} \displaystyle\int_0^{\ln(2)} \tanh(x) \mathrm{d}x &= \displaystyle\int_0^{\ln(2)} \dfrac{\sinh(x)}{\cosh(x)} \mathrm{dx} \\ &= \displaystyle\int_1^{\frac{5}{4}} \dfrac{1}{u} \mathrm{d}u \\ &= \log(u) \Bigg|_1^{\frac{5}{4}} \\ &= \log \left( \dfrac{5}{4} \right) - \log(1) \\ &= \log\left( \dfrac{5}{4} \right). \blacksquare \end{array}$$
Section 5.8 #75: Calculate $$\displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x.$$ Solution: Note that $$\dfrac{1}{3-9x^2} = \dfrac{1}{(\sqrt{3})^2-(3x)^2}.$$ So let $u=3x$ so that $\dfrac{1}{3} \mathrm{d}u = \mathrm{d}x$. Since $$\displaystyle\int \dfrac{1}{a^2-x^2} \mathrm{d}x = \dfrac{1}{a} \arctan\left( \dfrac{x}{a} \right) + C,$$ we now calculate $$\begin{array}{ll} \displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x &= \displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x \\ &= \displaystyle\int \dfrac{1}{\sqrt{3}^2-(3x)^2} \mathrm{d}x \\ &= \dfrac{1}{3} \displaystyle\int \dfrac{1}{\sqrt{3}^2-u^2} \mathrm{d}u \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh} \left( \dfrac{u}{\sqrt{3}} \right) + C \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh}\left( \dfrac{3x}{\sqrt{3}} \right) + C \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh} \left( \sqrt{3}x \right) + C. \blacksquare \end{array}$$
Section 5.8 #86: Calculate $$\displaystyle\int_0^1 \dfrac{1}{\sqrt{25x^2+1}} \mathrm{d}x.$$ Solution: Note that $$\dfrac{1}{\sqrt{25x^2+1}} = \dfrac{1}{\sqrt{(5x)^2+1}}.$$ So letting $u=5x$ implies $\dfrac{1}{5} \mathrm{d}u = \mathrm{d}x$ and if $x=0$ then $u=0$ and if $x=1$ then $u=5$. Therefore compute $$\begin{array}{ll} \displaystyle\int_0^1 \dfrac{1}{\sqrt{25x^2+1}} \mathrm{d}x &= \displaystyle\int_0^1 \dfrac{1}{\sqrt{(5x)^2+1}} \mathrm{d}x \\ &= \dfrac{1}{5} \displaystyle\int_0^5 \dfrac{1}{\sqrt{u^2+1}} \mathrm{d}u \\ &= \dfrac{1}{5} \mathrm{arcsinh}(u) \Bigg|_0^5 \\ &= \dfrac{1}{5} \left[ \mathrm{arcsinh}(5) - \mathrm{arcsinh}(0) \right] \\ &= \dfrac{\mathrm{arcsinh}(5)}{5}. \blacksquare \\ \end{array}$$ | 3,421 | 7,381 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-23 | latest | en | 0.326043 |
https://zerooneseo.com/matt-lieberman-bxcyvm/6710e7-mean-deviation-formula-for-ungrouped-data | 1,632,434,220,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057447.52/warc/CC-MAIN-20210923195546-20210923225546-00172.warc.gz | 1,115,351,624 | 8,682 | Now, first of all we have to figure out the median class and for that we will apply the following formula … Here is the simplified formula of mean for your consideration. mean absolute deviation, mean absolute deviation formula, mean absolute deviation calculator, mean absolute deviation calculator, mean absolute deviation for ungrouped data We can evaluate the variance of a set of data from the mean that is, how far the observations deviate from the mean. Need for Variance and Standard Deviation. Find the variance and standard deviation. So, standard deviation of the given data is 4.69. 3.2 Measures of Dispersion for Ungrouped Data 95 H17011 Two Observations Calculating the population variance and standard deviation for ungrouped data. Median of an Ungrouped Data Set The median refers to the middle data point of an ordered data set at the 50% percentile.If a data set has an odd number of observations, then the median is the middle value. 2) Calculate mean by formula. Formula for Mean Deviation: (ungrouped data) MD Where, μ = mean x n x = each value N = number of values 6. (a) H.M. for Ungrouped data . Calculate the mean deviation about the mean of the set of first n natural numbers when n is an even number. Therefore in order to keep all the values positive, absolute deviations are used to find the mean of deviations called mean absolute deviation. 22,25,24,23,24,20. 7. Below are the numerical examples with step by step guide solution on variance and standard deviation for ungrouped data. The average mean of this set is 16.4 now, out this mean in the formula of standard deviation as shown below. MATH 2. Namaste to all Friends, This Video Lecture Series presented By VEDAM Institute of Mathematics is Useful to all students of Class 11th & 12th of … Skip navigation Sign in. Also Check: Standard Deviation Formula Variance Formula Example Question. 5. Covers frequency distribution tables with ungrouped data. Range and Mean Deviation. However, we can further implement this analytical claim of statistics, by measuring the scattering and dispersion of data around these measures of central tendency. It is also referred to as mean absolute deviation. Example Find the standard deviation of 4, 9, 11, 12, 17, 5, 8, 12, 14 The mean of the data is 10.222 Subtracting the mean individually from each of the numbers given and square the result. Example 1 : Find the coefficient of variation of 24, 26, 33, 37, 29, 31. Loading... Close. The calculation of this entity can be found by using a formula called, standard deviation formula, which is used by mathematicians or statisticians. Click Create Assignment to assign this modality to your LMS. MEMORY METER. Formula for mean deviation for ungrouped data or an individual series is given from BMA 2202 at Mount Kenya University Mean deviation or average deviation is the arithmetic mean of the absolute deviations. variance and standard deviation, variance calculator, variance formula, variance examples, standard deviation, standard deviation formula, standard deviation calculator, standard … But a major problem is that mean deviation ignores the signs of deviation, otherwise they would add up to zero.To overcome this limitation variance and standard deviation came into the picture. Mean absolute deviation from mean for grouped and ungrouped data can be calculated by using the formulas: Similarly mean absolute deviation from median can be calculated by using the formulas: Mean and Standard Deviation Formula. Harmonic Mean (H.M.) Harmonic Mean is defined as the reciprocal of the arithmetic mean of reciprocals of the observations. For a given series of data, statistics aims at analysis and drawing conclusions.The various measures of central tendency – mean, median and mode represent the values in a series. 3) Calculate standard deviation in two steps Let x 1, x 2, ..., x n be the n observations then the harmonic mean is defined as . A man travels from Jaipur to Agra by a car and takes 4 hours to cover the whole distance. Mean deviation in statistics is defined as the mean of the distances of each value from their mean. H9018 x, H9018 x. For that, let us arrange the given data in ascending order. Examples: Median Formula. Watch Queue Queue. The steps to calculate mean & standard deviation are: 1) Process the data. ... Ungrouped Data to Find the Mean. 2) Find the distance of each value from that mean. Solution •For grouped data, class mode (or, modal class) is the class with the highest frequency. If you are interested in calculating this function manually by your hands then the best way of doing this is to prepare a table by the data set you are having like, for example, we are having 5,12,16,21,28 data. Two of our most-viewed posts deal with Mode and Median of Grouped Data: how to calculate these statistics for data that is supplied in the form of frequencies for classes of data (bins), rather than the individual data values.Here we’ll complete that topic with a look at the less troublesome cases of Mean and Standard Deviation, including some issues that arise in the grouping process itself. While computing standard deviation, arranging data in ascending order is not mandatory. 3) Find the mean of those distances. The age (in years) of 6 randomly selected students from a class are. Search. In Mathematical terms, sample mean formula is given as: $\bar{x}$= 1/n $\sum_{i=1}^{n}x$ Solution : Let us use actual mean method to find standard deviation. If it has an even number of observations, the median is the average of the two middle values. Standard deviation for ungrouped data. Variance and standard deviation (ungrouped data) Introduction In this leaflet we introduce variance and standard deviation as measures of spread. MA TH Definition: The MEAN DEVIATION, or average deviation, is the dispersion of the data about the mean of these data. WELCOMEMATH 3. MA TH Definition: The MEAN DEVIATION is a measure of variation that makes use of all the scores of the distribution. Mean Deviation tells us how far, on average, all values are from the middle. This video explains how to find out the mean & mean deviation about mean for ungrouped frequency data !! This indicates … Because the data in this example are on earnings of all employees of this company, we use the population formula to … The data in a tabular form which indicates the frequency is known as a frequency distribution. 1 mo 12. In applying this method first of all we compute the arithmetic mean of the given data, either ungrouped or grouped. Mean Deviation -The mean of the distances of each value from their mean. Standard Deviation Formula for Ungrouped Data: Standard deviation is normally represented by the symbol known as sigma or letter ‘σ’. This video is unavailable. Δ =L + i. Δ + Δ. Mode – Grouped Data Mean Formula. To calculate mean deviation for ungrouped data, we have to find absolute deviation for every observation of central tendency and evaluate the mean about the given observation. Finding weighted mean by hand or using the TI calculator % Progress . Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number. For ungrouped data, sort and tabulate the data in a table. Here below you can check out the median formula for the grouped data. Find the standard deviation of the first n natural numbers. ! MA TH MEAN DEVIATION (for ungrouped data) 4. 4. Three steps on finding the mean: 1) Find the mean of all values. Then we take the deviation from the actual mean. The following formulas are applied: We have studied mean deviation as a good measure of dispersion. Example 7 Calculate the mean deviation about median for the following data : Median = + 2 − ×ℎ Where, = lower limits of median class N = sum of frequencies = frequency of median class C = Cumulative frequency of class before median class Here, = 20, N = 6. •To find mode for grouped data, use the following formula: ⎛⎞ ⎜⎟ ⎝⎠ Mode. Standard deviation for ungrouped data. Question: Find the variance for the following set of data representing trees heights in feet: 3, 21, 98, 203, 17, 9 Solution: Step 1: Add up the numbers in your given data set. This method is already defined above. 6. Mean Deviation is used to determine how far the data values are dispersed from the mean value. Learn the definition, formula, and solved examples at BYJU’S. Sample mean is represented by the symbol $\bar{x}$. 5. Step 2: Square your answer: 351 × 351 = 123201 …and divide by the number of items. •Mode is the value that has the highest frequency in a data set. Solution : First let us find the standard deviation for the given data. 3 + 21 + 98 + 203 + 17 + 9 = 351. The sample mean is the average and is calculated as the addition of all the observed outcomes from the sample divided by the total number of events. skewness meaning, skewness in statistics, skewness formula, skewness coefficient, pearson coefficient of skewness for ungrouped data Variance and Standard deviation of ungrouped data Example 1. For grouped data, obtain the mid-value of each intervals. Mean Deviation 1. The data has to be arranged in ascending order and then the quartile formula for ungrouped data is applied to calculate the values of Q 1,, Q 2 ,and Q 3 Solved Examples For Quartile Formula Below you can see solved examples for quartile formulas which will clear your concepts of quartile formulas … Here is an example (using the same data as on the Standard Deviation page): Example: You and your friends have just measured the heights of your dogs (in millimeters): Mean Deviation for Grouped Data Calculator. The standard deviation is given by the formula: s means 'standard deviation' m is the mean N is the number of values. 19.5.3. Question 2 : Calculate the standard deviation of the following data x 3 8 13 18 23 f 7 10 15 10 8 . Example 5.11. Know how to calculate the Mean Deviation for Grouped Data and Continuous Frequency Distribution. The set of first n natural numbers when n is an odd.! Randomly selected students from a class are takes 4 hours to cover the whole.. 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Calculating the population variance and standard deviation as a frequency distribution the middle the dispersion of the distances of value! | 4,298 | 18,792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-39 | latest | en | 0.851462 |
https://howkgtolbs.com/convert/62.42-kg-to-lbs | 1,619,126,568,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039604430.92/warc/CC-MAIN-20210422191215-20210422221215-00525.warc.gz | 422,891,648 | 12,241 | # 62.42 kg to lbs - 62.42 kilograms to pounds
Do you want to learn how much is 62.42 kg equal to lbs and how to convert 62.42 kg to lbs? Here you go. In this article you will find everything about kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to emphasize that whole this article is dedicated to a specific number of kilograms - this is one kilogram. So if you want to know more about 62.42 kg to pound conversion - read on.
Before we get to the more practical part - that is 62.42 kg how much lbs conversion - we are going to tell you few theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 62.42 kg to lbs? 62.42 kilograms it is equal 137.6125439404 pounds, so 62.42 kg is equal 137.6125439404 lbs.
## 62.42 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, known also as International System of Units (in short form SI).
From time to time the kilogram is written as kilogramme. The symbol of the kilogram is kg.
First definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. This definition was simply but totally impractical to use.
Later, in 1889 the kilogram was described by the International Prototype of the Kilogram (in abbreviated form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was in use until 2019, when it was substituted by another definition.
Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is 0.001 tonne. It is also divided into 100 decagrams and 1000 grams.
## 62.42 kilogram to pounds
You know a little about kilogram, so now we can move on to the pound. The pound is also a unit of mass. It is needed to underline that there are more than one kind of pound. What are we talking about? For example, there are also pound-force. In this article we want to centre only on pound-mass.
The pound is used in the Imperial and United States customary systems of measurements. Of course, this unit is used also in another systems. The symbol of the pound is lb or “.
The international avoirdupois pound has no descriptive definition. It is defined as 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 62.42 kg?
62.42 kilogram is equal to 137.6125439404 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 62.42 kg in lbs
The most theoretical section is already behind us. In this part we want to tell you how much is 62.42 kg to lbs. Now you learned that 62.42 kg = x lbs. So it is high time to get the answer. Just look:
62.42 kilogram = 137.6125439404 pounds.
This is a correct outcome of how much 62.42 kg to pound. It is possible to also round off the result. After rounding off your result will be as following: 62.42 kg = 137.324 lbs.
You learned 62.42 kg is how many lbs, so look how many kg 62.42 lbs: 62.42 pound = 0.45359237 kilograms.
Obviously, this time you may also round it off. After it your result will be exactly: 62.42 lb = 0.45 kgs.
We also want to show you 62.42 kg to how many pounds and 62.42 pound how many kg outcomes in tables. See:
We are going to begin with a table for how much is 62.42 kg equal to pound.
### 62.42 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
62.42 137.6125439404 137.3240
Now see a chart for how many kilograms 62.42 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
62.42 0.45359237 0.45
Now you learned how many 62.42 kg to lbs and how many kilograms 62.42 pound, so we can go to the 62.42 kg to lbs formula.
### 62.42 kg to pounds
To convert 62.42 kg to us lbs a formula is needed. We are going to show you a formula in two different versions. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 137.6125439404 result in pounds
The first formula will give you the most exact result. Sometimes even the smallest difference could be significant. So if you need an accurate outcome - this version of a formula will be the best solution to know how many pounds are equivalent to 62.42 kilogram.
So let’s move on to the another version of a formula, which also enables conversions to learn how much 62.42 kilogram in pounds.
The shorter version of a formula is as following, see:
Number of kilograms * 2.2 = the outcome in pounds
As you can see, this formula is simpler. It could be better solution if you need to make a conversion of 62.42 kilogram to pounds in fast way, for example, during shopping. Just remember that final result will be not so exact.
Now we are going to learn you how to use these two versions of a formula in practice. But before we are going to make a conversion of 62.42 kg to lbs we are going to show you easier way to know 62.42 kg to how many lbs without any effort.
### 62.42 kg to lbs converter
An easier way to check what is 62.42 kilogram equal to in pounds is to use 62.42 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Calculator is based on first formula which we gave you above. Due to 62.42 kg pound calculator you can quickly convert 62.42 kg to lbs. Just enter amount of kilograms which you need to convert and click ‘convert’ button. You will get the result in a second.
So let’s try to convert 62.42 kg into lbs using 62.42 kg vs pound converter. We entered 62.42 as an amount of kilograms. This is the result: 62.42 kilogram = 137.6125439404 pounds.
As you can see, our 62.42 kg vs lbs calculator is user friendly.
Now we can move on to our main topic - how to convert 62.42 kilograms to pounds on your own.
#### 62.42 kg to lbs conversion
We will start 62.42 kilogram equals to how many pounds conversion with the first version of a formula to get the most exact result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 137.6125439404 the result in pounds
So what have you do to check how many pounds equal to 62.42 kilogram? Just multiply number of kilograms, in this case 62.42, by 2.20462262. It gives 137.6125439404. So 62.42 kilogram is equal 137.6125439404.
You can also round it off, for example, to two decimal places. It is 2.20. So 62.42 kilogram = 137.3240 pounds.
It is time for an example from everyday life. Let’s convert 62.42 kg gold in pounds. So 62.42 kg equal to how many lbs? And again - multiply 62.42 by 2.20462262. It gives 137.6125439404. So equivalent of 62.42 kilograms to pounds, if it comes to gold, is 137.6125439404.
In this case it is also possible to round off the result. It is the result after rounding off, in this case to one decimal place - 62.42 kilogram 137.324 pounds.
Now let’s move on to examples calculated with short formula.
#### How many 62.42 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 137.324 the result in pounds
So 62.42 kg equal to how much lbs? And again, you have to multiply number of kilogram, in this case 62.42, by 2.2. Let’s see: 62.42 * 2.2 = 137.324. So 62.42 kilogram is 2.2 pounds.
Make another calculation using this version of a formula. Now convert something from everyday life, for instance, 62.42 kg to lbs weight of strawberries.
So calculate - 62.42 kilogram of strawberries * 2.2 = 137.324 pounds of strawberries. So 62.42 kg to pound mass is 137.324.
If you know how much is 62.42 kilogram weight in pounds and are able to convert it using two different versions of a formula, let’s move on. Now we are going to show you these results in charts.
#### Convert 62.42 kilogram to pounds
We know that outcomes presented in tables are so much clearer for most of you. We understand it, so we gathered all these outcomes in charts for your convenience. Due to this you can quickly make a comparison 62.42 kg equivalent to lbs results.
Let’s begin with a 62.42 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
62.42 137.6125439404 137.3240
And now let’s see 62.42 kg equal pound chart for the second formula:
Kilograms Pounds
62.42 137.324
As you can see, after rounding off, if it comes to how much 62.42 kilogram equals pounds, the outcomes are the same. The bigger number the more considerable difference. Remember it when you want to make bigger number than 62.42 kilograms pounds conversion.
#### How many kilograms 62.42 pound
Now you know how to calculate 62.42 kilograms how much pounds but we want to show you something more. Are you interested what it is? What do you say about 62.42 kilogram to pounds and ounces calculation?
We want to show you how you can convert it step by step. Start. How much is 62.42 kg in lbs and oz?
First things first - you need to multiply amount of kilograms, in this case 62.42, by 2.20462262. So 62.42 * 2.20462262 = 137.6125439404. One kilogram is exactly 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To convert how much 62.42 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So your outcome is 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then your result is equal 2 pounds and 33 ounces.
As you see, conversion 62.42 kilogram in pounds and ounces quite easy.
The last conversion which we are going to show you is conversion of 62.42 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To calculate it you need another formula. Before we show you this formula, see:
• 62.42 kilograms meters = 7.23301385 foot pounds,
• 62.42 foot pounds = 0.13825495 kilograms meters.
Now let’s see a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 62.42 foot pounds to kilograms meters you need to multiply 62.42 by 0.13825495. It gives 0.13825495. So 62.42 foot pounds is 0.13825495 kilogram meters.
You can also round off this result, for instance, to two decimal places. Then 62.42 foot pounds will be equal 0.14 kilogram meters.
We hope that this calculation was as easy as 62.42 kilogram into pounds conversions.
We showed you not only how to do a calculation 62.42 kilogram to metric pounds but also two another conversions - to check how many 62.42 kg in pounds and ounces and how many 62.42 foot pounds to kilograms meters.
We showed you also other solution to do 62.42 kilogram how many pounds conversions, it is using 62.42 kg en pound converter. This will be the best solution for those of you who do not like converting on your own at all or need to make @baseAmountStr kg how lbs calculations in quicker way.
We hope that now all of you can make 62.42 kilogram equal to how many pounds conversion - on your own or using our 62.42 kgs to pounds converter.
It is time to make your move! Let’s calculate 62.42 kilogram mass to pounds in the way you like.
Do you want to make other than 62.42 kilogram as pounds conversion? For example, for 5 kilograms? Check our other articles! We guarantee that conversions for other amounts of kilograms are so simply as for 62.42 kilogram equal many pounds.
### How much is 62.42 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 62.42 kg in pounds , we prepared one more section. Here you can see all you need to know about how much is 62.42 kg equal to lbs and how to convert 62.42 kg to lbs . Let’s see.
What is the kilogram to pound conversion? It is a mathematical operation based on multiplying 2 numbers. How does 62.42 kg to pound conversion formula look? . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
See the result of the conversion of 62.42 kilogram to pounds. The exact answer is 137.6125439404 lb.
It is also possible to calculate how much 62.42 kilogram is equal to pounds with another, easier type of the formula. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So this time, 62.42 kg equal to how much lbs ? The result is 137.6125439404 lbs.
How to convert 62.42 kg to lbs quicker and easier? It is possible to use the 62.42 kg to lbs converter , which will do all calculations for you and you will get a correct result .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,581 | 13,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-17 | longest | en | 0.951484 |
https://q4interview.com/quantitative-aptitude/decimal-fraction/easy/1/1 | 1,603,140,005,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107866404.1/warc/CC-MAIN-20201019203523-20201019233523-00166.warc.gz | 484,269,798 | 36,864 | Get Off-Campus Placement Jobs Info !!!
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# Decimal Fraction Questions
Home > Quantitative Aptitude > Decimal Fraction > General Questions
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Choose the correct option.
A$$\frac{13}{20}$$
B$$\frac{11}{20}$$
C13
D11
Explanation:
8 new n/w,20 is already employed
Total n/w = 20+8 = 28 n/w
so that 75% came from same university then $$28 \times \frac{75}{100} = 21$$
we all know 8 n/w is recently hired then 21-8 = 13
=> the fraction is = $$\frac{13}{20}$$
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Arrange the fractions $$\frac{3}{5},\frac{4}{7},\frac{8}{9} and \frac{9}{11}$$ in their descending order.
A$$\frac{2}{9}>\frac{6}{11}>\frac{4}{5}>\frac{3}{7}$$.
B$$\frac{8}{9}>\frac{6}{11}>\frac{4}{5}>\frac{3}{7}$$.
C$$\frac{8}{9}>\frac{9}{11}>\frac{4}{5}>\frac{3}{7}$$.
D$$\frac{8}{9}>\frac{9}{11}>\frac{3}{5}>\frac{4}{7}$$.
Explanation:
clearly, $$\frac{3}{5}=0.6,\frac{4}{7}=0.571,\frac{8}{9}=0.88,\frac{9}{11}=0.818$$.
now, $$0.88>0.818>0.6>0.571$$.
so,$$\frac{8}{9}>\frac{9}{11}>\frac{3}{5}>\frac{4}{7}$$.
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Choose the correct option.
what value will replace the question mark in the following equations? (i) $$5172.49+378.352+?=9318.678$$
(ii) $$?-7328.96=5169.38$$
A12496.43,3768.798
B3767.836, 12498.34
C1249.34,3767.836
D4394.12,3767.836
Explanation:
(i) $$5172.49+378.352+x=9318.678$$
then, $$x=9318.678-(5172+378.352)=9318.678-5550.842$$
=3767.836
(ii) $$?-7328.96=5169.38$$
let $$x-7328.96=51.69.38$$
then,$$x=5169.38+7328.96=12498.34$$
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Choose the correct option.
Evaluate:
(i) 35+.07
(ii) 2.5 +0.0005
(iii) 136.09+43.9
A(i) 500 (ii) 3.1 (iii) 5000
B(i)500 (ii)5000 (iii)3.1
C(i) 3.1 (ii)5000 (iii)500
D(i) 5000 (ii) 500 (iii) 3.1`
Explanation:
(i) $$\frac{35}{.07}=\frac{35\times100}{.07\times100}=\frac{3500}{7}=500$$
(ii) $$\frac{2.5}{0.0005}=\frac{2.5\times10000}{.0005\times10000}=\frac{25000}{5}=5000$$
(iii) $$\frac{136.09}{43.9}=\frac{136.09\times10}{43.9\times10}=\frac{1360.9}{439}=3.1$$
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Choose the correct option.
Simplify: $$\frac{0.05\times0.05\times0.05+0.04\times0.04\times0.04}{0.05\times0.05-0.05\times0.04+0.04\times0.04}$$
A0.9
B0.06
C0.09
D0.009
Explanation:
Given expression = $$\left(\frac{a^3+b^3}{a^2-ab+b^2}\right)$$, where a=0.05, b=0.04
=(a+b)=(0.05+0.04)=0.09
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Choose the correct option.
Which of the following has fraction in ascending order?
A$$\frac{2}{5},\frac{3}{5},\frac{1}{3},\frac{4}{7},\frac{5}{6},\frac{6}{7}$$
B$$\frac{1}{3},\frac{2}{5},\frac{4}{7},\frac{3}{5},\frac{5}{6},\frac{6}{7}$$
C$$\frac{1}{3},\frac{2}{5},\frac{3}{5},\frac{4}{7},\frac{5}{6},\frac{6}{7}$$
D$$\frac{1}{3},\frac{2}{5},\frac{3}{5},\frac{5}{6},\frac{4}{7},\frac{6}{7}$$
Explanation:
Converting each of the given fractions into decimal form, we get
$$\frac{1}{3}=0.33,\frac{2}{5}=0.4,\frac{4}{7}=0.57,\frac{3}{5}=0.6,\frac{5}{6}=,0.82,\frac{6}{7}=0.857$$
Clearly $$0.33<0.4<0.57<0.6<0.82<0.857.$$
so, $$\frac{1}{3},\frac{2}{5},\frac{4}{7},\frac{3}{5},\frac{5}{6},\frac{6}{7}$$
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Choose the correct option.
Which of the following has fraction in ascending order?
A$$\frac{8}{9},\frac{9}{11},\frac{7}{9},\frac{2}{3},\frac{3}{5}$$
B$$\frac{2}{3},\frac{3}{5},\frac{7}{9},\frac{9}{11},\frac{8}{9}$$
C$$\frac{3}{5},\frac{2}{3},\frac{9}{11},\frac{7}{9},\frac{8}{9}$$
D$$\frac{3}{5},\frac{2}{3},\frac{7}{9},\frac{9}{11},\frac{8}{9}$$
Explanation:
Converting each of the given fractions into decimal form, we get
$$\frac{2}{3}=0.66,\frac{3}{5}=0.6,\frac{7}{9}=0.77,\frac{9}{11}=0.81,\frac{8}{9}=0.88$$
Clearly $$0.6<0.66<0.77<0.81<0.88.$$
so, $$\frac{3}{5}<\frac{2}{3}<\frac{7}{9}<\frac{9}{11}<\frac{8}{9}$$
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Choose the correct option.
Which of the following are descending order of their value?
A$$\frac{11}{17},\frac{7}{11},\frac{5}{9},\frac{8}{15}$$
B$$\frac{5}{9},\frac{7}{11},\frac{8}{15},\frac{11}{17}$$
C$$\frac{5}{9},\frac{8}{15},\frac{11}{17},\frac{7}{11}$$
D$$\frac{11}{17},\frac{7}{11},\frac{8}{15},\frac{5}{9}$$
Explanation:
Converting each of the given fractions into decimal form, we get
$$\frac{5}{9}=0.55,\frac{7}{11}=0.63,\frac{8}{15}=0.533,\frac{11}{17}=0.647.$$
Clearly $$0.647>0.63>0.55>0.533<0.88.$$
so, $$\frac{11}{17}>\frac{7}{11}>\frac{5}{9}>\frac{8}{15}$$
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Choose the correct option.
What is the diffrence between the biggest and the smallest fraction among $$\frac{2}{3},\frac{3}{4},\frac{4}{5} and \frac{5}{6}$$ ?
A$$\frac{1}{30}$$
B$$\frac{1}{6}$$
C$$\frac{1}{12}$$
D$$\frac{1}{20}$$
Explanation:
Converting each of the given fractions into decimal form, we get
$$\frac{2}{3}=0.66,\frac{3}{4}=0.75,\frac{4}{5}=0.8,\frac{5}{6}=0.8333,\frac{8}{9}=0.88$$
Clearly $$0.8333>0.8>0.75>0.66$$
so, $$\frac{5}{6}>\frac{4}{5}>\frac{3}{4}>\frac{2}{3}$$
therefore, required diffrence =$$\left(\frac{5}{6}-\frac{2}{3}\right)=\frac{1}{6}$$
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Choose the correct option.
Which of the following fraction is greater then $$\frac{3}{4}$$ and less then $$\frac{5}{6}$$ ?
A$$\frac{9}{10}$$
B$$\frac{1}{2}$$
C$$\frac{2}{3}$$
D$$\frac{4}{5}$$
Explanation:
$$\frac{3}{4}=0.75,\frac{5}{6}=0.8333,\frac{1}{2}=0.5,\frac{2}{3}=0.66,\frac{4}{5}=0.8,\frac{9}{10}=0.9$$
Clearly, 0.8 lies between 0.75 and 0.8333
therefore, $$\frac{4}{5}$$ lies between $$\frac{3}{4} and\frac{5}{6}$$
Workspace
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# On Monday, a person mailed 8 packages weighing an average
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On Monday, a person mailed 8 packages weighing an average [#permalink]
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13 Aug 2012, 06:57
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On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of $$12\frac{3}{8}$$ pounds, and on Tuesday, 4 packages weighing an average of $$15\frac{1}{4}$$ pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?
(A) $$13\frac{1}{3}$$
(B) $$13\frac{13}{16}$$
(C) $$15\frac{1}{2}$$
(D) $$15\frac{15}{16}$$
(E) $$16\frac{1}{2}$$
Practice Questions
Question: 16
Page: 154
Difficulty: 600
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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13 Aug 2012, 06:57
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SOLUTION
On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of $$12\frac{3}{8}$$ pounds, and on Tuesday, 4 packages weighing an average of $$15\frac{1}{4}$$ pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?
(A) $$13\frac{1}{3}$$
(B) $$13\frac{13}{16}$$
(C) $$15\frac{1}{2}$$
(D) $$15\frac{15}{16}$$
(E) $$16\frac{1}{2}$$
The total weight of 8 packages is $$8*12\frac{3}{8}=99$$ pounds;
The total weight of 4 packages is $$4*15\frac{1}{4}=61$$ pounds;
The average weight of all 12 packages is $$\frac{total \ weight}{# \ of \ packages}=\frac{99+61}{12}=13\frac{1}{3}$$.
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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07 Sep 2012, 11:47
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1
Since the final average cannot be greater than $$15\frac{1}{4}$$, answers C, D and E are out.
We can use the property of weighted averages.
$$15\frac{1}{4}=15\frac{2}{8}$$, the distance between the two initial averages is almost 3.
Since the number of packages are in a ratio of 8:4 = 2:1, the differences between the final average and the initial averages are in a ratio 1:2.
So, the distance between $$12\frac{3}{8}$$ and the final average is almost 1, close to $$12\frac{3}{8}+1\approx{13}\frac{1}{4}$$.
The final answer should be close to $$13\frac{1}{4}$$.
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##### General Discussion
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Joined: 02 Sep 2009
Posts: 60627
Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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07 Sep 2012, 10:15
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Gmbrox wrote:
There must be something easy that i just don't get for me :
8*12(3/8) = 36 as you simplify the 8 between them. Therefore how do you manage to arrive at 99?
I guess it must be something different of spelling or something?
Thanks a lot for your help !
It's not 12 multiplied by 3/8. it's $$12\frac{3}{8}=\frac{12*8+3}{8}=\frac{99}{8}$$ (the same way as $$1\frac{1}{2}=\frac{3}{2}$$).
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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15 Nov 2014, 02:55
2
I used ratio of packages, which is 2:1.
converted both to the same fractions, so
15 1/4 = 15 2/8
2x(12 3/8) + 1x( 15 2/8) = 24+15+ 6/8+2/8 = 39 and 8/8, 8/8 is also obviously 1. Could also be together 40 but that's not easily divisible with three and you know you're left with a remainder. Instead just:
39/3 + 1/3 = 13 and 1/3
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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20 Nov 2014, 03:08
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Total weight on Monday $$= 8*12 + 8 * \frac{3}{8} = 96 + 3$$
Total weight on Tuesday $$= 4*15 + 4 * \frac{1}{4} = 60 + 1$$
Average of all days $$= \frac{96 + 60 + 4}{12} = 8 + 5 + \frac{4}{12} = 13\frac{1}{3}$$
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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20 May 2016, 04:29
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1
Bunuel wrote:
SOLUTION
On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of $$12\frac{3}{8}$$ pounds, and on Tuesday, 4 packages weighing an average of $$15\frac{1}{4}$$ pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?
(A) $$13\frac{1}{3}$$
(B) $$13\frac{13}{16}$$
(C) $$15\frac{1}{2}$$
(D) $$15\frac{15}{16}$$
(E) $$16\frac{1}{2}$$
Solution:
To solve this question we can use the weighted average equation.
Weighted Average = (Sum of Weighted Terms) / (Total Number of Items)
We'll first determine the sum (numerator). We see that on the first day we had 8 items that averaged 12 3/8 pounds. We don't know the weights of the individual packages, but we can determine that the sum of all 8 packages is:
Sum of first day's packages = 8 x 12 3/8 = 99 pounds
Similarly, the sum of the second day's packages is:
Sum of second day's packages = 4 x 15 ¼ = 61
We now can use the weighted average equation to find the average weight of the 12 packages:
Weighted Average = (99 + 61) / 12
Weighted Average = 160 /12
Weighted Average = 13 1/3
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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13 Aug 2012, 08:46
a person mailed 8 packages=99 pounds (total of all 8 packages)
and on Tuesday, 4 packages=61 pounds (total of all 4 packages)
total weight =99+61 =160 pounds
the average weight =160/(8+4) =option a
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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16 Aug 2012, 11:10
Bunuel wrote:
RESERVED FOR A SOLUTION.
Bunuel, had an off-topic request for you: Could you please post questions from non-OG sources as well? I'm not sure if that might breach a copyright arrangement bsaed on the source you use, and of course your comments on other's questions are supremely valuable for those of us subscribed to your daily updates - but if you could include occasional 700+ non-OG questions, would be much appreciated by your "followers"
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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07 Sep 2012, 10:09
There must be something easy that i just don't get for me :
8*12(3/8) = 36 as you simplify the 8 between them. Therefore how do you manage to arrive at 99?
I guess it must be something different of spelling or something?
Thanks a lot for your help !
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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07 Sep 2012, 10:20
Thank you a lot bunuel, Ok after reviewing the official book, i now got it, it is a mix number, it does not exists in france so that's why. If anyone has difficulties to understand like me :
12(3/8) = 12+3/8 and not 12x(3/8).
Can be confusing.
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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25 Aug 2014, 00:54
I used weighted-averages to solve this as follows:
Let 12.375 be the lowest weight, therefore, (15.25 - 12.375) = 2.875, the difference in weights between the light and heavy packages.
Hence, the amount we need to add to 12.375 is:
(8x0 + 4x2.875)/(8+12) = 11.50 / 12 ~ 12/12 = 1 (Need to round final answer down a bit since I rounded numerator up)
Therefore, the average weight of all packages is approximately 12.375 + 1 = 13.375, since I need to round the numerator down, ans = 13.333, or A
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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20 Jul 2016, 12:55
The total weight of packages on Monday = 12 $$\frac{3}{8}$$ * 8 = 99 pounds
The total weight of packages on tuesday = 15 $$\frac{1}{4}$$ * 4 = 61 pounds
Average = Total weight / Number of packages = $$\frac{(99+61)}{12}$$ = 13 $$\frac{1}{3}$$ pounds
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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20 Jul 2016, 15:25
(2*12 3/8)+(1*15 1/4)=40 pounds
40/3=13 1/3 pounds average
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On Monday, a person mailed 8 packages weighing an average [#permalink]
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15 Dec 2016, 15:15
This Question is giving us mean of two data and asking us for the combined average.
Using $$Mean = Sum/#$$
Sum(8) = $$\frac{99}{8}*8=99$$
Sum(4)=$$\frac{61}{4}*4=61$$
Hence combined average =$$\frac{99+61}{12}= \frac{40}{3}$$
Hence A
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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28 Jul 2017, 16:02
Using the weighted average formula =
weight of x (x) + weight of y (y) = weight of x and y (x+y)
In this situation, x =8, weighting of x = 12 3/8; y = 4, weighting of y = 15 1/4
substitute in the formula to find
99 + 61 = weight of x and y (8 +4)
99 + 61 = weight of x and y (12)
(99+61)/12 = weight of x and y
= 13 1/3
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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11 Oct 2019, 03:16
Bunuel wrote:
On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of $$12\frac{3}{8}$$ pounds, and on Tuesday, 4 packages weighing an average of $$15\frac{1}{4}$$ pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?
(A) $$13\frac{1}{3}$$
(B) $$13\frac{13}{16}$$
(C) $$15\frac{1}{2}$$
(D) $$15\frac{15}{16}$$
(E) $$16\frac{1}{2}$$
Practice Questions
Question: 16
Page: 154
Difficulty: 600
This is how i solved
the ratio of Packets on Mon : Tuesday = 8 : 4
avg of monday = 12 3/8 or 99/8 and Tuesday = 15 1/4 or 61 /4
We have a formula for weighted avg ie
8 /12 * 99/8 + 4/12 * 61 /4
=99 /12 + 61 /12 = 160 /12 = 13 1/3 ans choice A
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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24 Nov 2019, 17:14
I know this question without this fraction and I’m complete lost WHERE this fraction come frommm?
On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of pounds, and on Tuesday, 4 packages weighing an average of pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?
(A) 13 1/3
(B) 13 13/16
(C) 15 1/2
(D) 15 15/16
(E) 16 1/2
I could find the result of entire number but I still don’t get where this fraction come from!
I have to use the weighted average calculation right?
Posted from my mobile device
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On Monday, a person mailed 8 packages weighing an average [#permalink]
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Updated on: 05 Dec 2019, 10:14
I think the fastest approach is to combine the balancing method and approximation.
You have 4 packages with weight of 15 pounds - approximately - and 4 packages with weight of 12 - approximately - pounds. The average of the two weights is (15+12)/2=13.5 {you will get the same if you calculate (4*15+4*12)/8 }
But then, you still have 4 packages with weight close to 12 pounds so the average will be definitely less than 13.5. Only A fits.
Originally posted by rencsee on 05 Dec 2019, 09:23.
Last edited by rencsee on 05 Dec 2019, 10:14, edited 1 time in total.
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
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05 Dec 2019, 10:10
letsamf wrote:
I know this question without this fraction and I’m complete lost WHERE this fraction come frommm?
On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of pounds, and on Tuesday, 4 packages weighing an average of pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?
(A) 13 1/3
(B) 13 13/16
(C) 15 1/2
(D) 15 15/16
(E) 16 1/2
I could find the result of entire number but I still don’t get where this fraction come from!
I have to use the weighted average calculation right?
Posted from my mobile device
Hey,
You need to know the common fractions in and out. Haven’t found the original post but pls see attached a picture from my notes. A lot of question will translate easy when you know the fractions and their decimal equivalent and reversely as well.
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4D1A0A49-DC08-4EFE-8ABD-AA1797E92D1F.jpeg [ 2.45 MiB | Viewed 732 times ]
Re: On Monday, a person mailed 8 packages weighing an average [#permalink] 05 Dec 2019, 10:10
Display posts from previous: Sort by | 4,556 | 14,490 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-05 | latest | en | 0.854166 |
http://www.crauswords.com/collatz/collatz.html | 1,686,245,180,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655092.36/warc/CC-MAIN-20230608172023-20230608202023-00461.warc.gz | 56,453,372 | 26,418 | The Collatz Conjecture (also known as the 3n + 1 conjecture, the Ulam conjecture, Kakutani's problem, the Thwaites conjecture, Hasse's algorithm, the Syracuse problem and the Hailstone problem), concerns itself with the properties of the series of numbers which is generated when you start from any integer greater than zero, and repeat the following steps:-
• If the current number is even, divide by 2 to generate the next member.
• If the current number is odd, multiply by 3 and add 1 to generate the next member.
Examples.
• The Collatz series for 1:
`1 4 2 1`
• The Collatz series for 3:
`3 10 5 16 8 4 2 1`
• The Collatz series for 7:
`7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1`
• The Collatz Series for 27:
``` 27 82 41 124 62 31 94 47 142 71 214 107 322 161
484 242 121 364 182 91 274 137 412 206 103 310 155 466
233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336
668 334 167 502 251 754 377 1132 566 283 850 425 1276 638
319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822
911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866
433 1300 650 325 976 488 244 122 61 184 92 46 23 70
35 106 53 160 80 40 20 10 5 16 8 4 2 1```
Note that in all of these examples, the numbers in the series vary up and down for a time, but finally decay to a value of 1. Over time, researchers have tested all odd numbers up to 1020, as well as a great many much larger numbers and in every case studied to date, the series continues until it reaches 1. This explains the title Collatz Conjecture, the Conjecture being that ALL numbers will ultimately suffer this fate. Ever since 1937, a proof of this conjecture has been lacking. In the end, the contents of this article may not provide the long anticipated proof, but it will provide the most convincing yet simple to understand evidence you will ever see in support of its truth.
The Collatz Series for 1, 3, and 7 are quite easy to corelate with the operation of the Collatz process, but as the series becomes longer, and especially as the numbers within it become bigger, it becomes more difficult to gain a clear picture of what is happening. Imagine how it would look if we were dealing with numbers of 100 or more digits! This article introduces a more concise method of presenting the series. Each number is replaced by a letter O or E depending on whether the number is odd or even, and the resulting string of Os and Es is referred to as the Signature of the number. The Signature for 27 then has the following appearance:-
OEOEEOEOEOEOEOEEOEEOEOEEOEOEOEEOEOEOEOEEOEEEOEOEOEEOEOEE OEOEOEOEOEOEEEOEOEOEOEEEEOEEOEEOEEEEOEEEOEOEOEEEEEOEEEEO
A further change in the presentation is to insert a space immediately before each O in the series. This breaks the series up into fragments called Syllables. The number of Syllables in a Signature is a very important factor in the analysis of the Collatz process as you will see in subsequent sections of this article.
OE OEE OE OE OE OE OEE OEE OE OEE OE OE OEE OE OE OE OEE OEEE OE OE OEE OE OEE OE OE OE OE OE OEEE OE OE OE OEEEE OEE OEE OEEEE OEEE OE OE OEEEEE OEEEEO
It should be noted that all Signatures in this article begin with O, which implies that only odd numbers are of interest. This is because application of the Collatz process to an even number immediately reveals an underlying (and smaller) odd number.
Signature Categories
As a matter of interest only, Signatures fall into two categories, namely Final and Partial:-
• Final Signature.
If you apply the Collatz process repetitively to a number, adding the letters E or O as appropriate to the Signature string as you go until you encounter the digit 1, then the Signature string which results will be a Final Signature.
• Partial Signature.
A Partial Signature results when the Collatz process is interrupted before a digit 1 has been encountered. Alternatively, you can create a Partial Signature using a text editor by typing a string consisting only of Os and Es. The first and last letters must both be O, and there must not be any examples of consecutive Os. It is most unlikely that a Signature created in this way will be of the Final variety.
Important Notes.
• The Signature of a series is a string of letters which correspond to the numbers of the series, with an O for each odd number and an E for each even number.
• The members of most Collatz series go up and down in what appears to be a totally random fashion in much the same way as a Hailstone rises and falls in a storm cell until it grows to a such a size and weight that it has no alternative but to fall to the ground. Because of this, the numbers considered here are often referred to as Hailstone Numbers
• Most commentaries on the Collatz conjecture warn of the unpredictable and haphazard behavior of the number series it produces. This is undoubtedly a fact, but by the time you have finished studying this article, you will see that some of the behaviors are reassuringly regular and very predictable and that the Normal or Gaussian Distribution of statistics is a remarkably accurate predictor for some of this behavior.
• The first of the above series appears to be unique. It is the only one which returns to its starting value.
• The number 10 appears in both the second and third examples. This will happen to any number which is half of some even number, whilst also being one more than three times some odd number.
• Theoretically, there are two ways in which a number could fail to return to 1, and thereby fail the conjecture. It could continue its up and down behavior forever, or it could enter a loop, and circulate around that loop forever. In either event, an infinitely long Signature would ensue, and this would quickly become apparent to the Rule-of-8 process (see later) if ever it should happen to encounter such a number.
• Signatures can vary greatly in length. The longest Signature for any number less than 1017 has 2091 letters. But after you have studied the Tutorial on Signatures and Syllables you will be able to design and build numbers which have vastly longer signatures than this. There is no upper limit to the length of a Signature.
• The algorithm for calculating a Signature from a Number is defined quite simply by the two rules stated at the beginning of this page. The algorithm for performing the reverse operation is considerably more intricate, and although an understanding of its operation is not essential, a complete listing of it is provided in Appendix 2.
• It will be very much to your advantage if you have a working copy of the Collatz software on your computer as you study this section of the article, so before you proceed any further, please refer to Appendix 4. and complete the Download and Installation instructions. Try to have the program showing on the screen at the same time as you read the contents of this tutorial.
• Calculate a Signature from a Number. Type the number directly into the Number field, and click the Number to Signature button. The Signature will appear in the Signature field, and some explanatory notes will appear in the Collatz Results field.
• Calculate a Number from a Signature. Type the signature directly into the Signature field, and click the Signature to Number button. The Number will appear in the Number field, and the Collatz Results field will be populated with an abbreviated description of the algorithm used to calculate the number. Note especially the format of the number display. It represents an infinite series of numbers, all of which have signatures which begin with a common set of characters.
• How to design a number which has very special characteristics. For example say you wanted a number which would start its Collatz series with ten consecutive odd numbers, followed by ten consecutive even numbers, followed by another ten odd numbers. All you need to do is to create a signature with ten consecutive OE syllables followed by a syllable containing ten consecutive Es, followed again by ten consecutive OE syllables. In other words, the complete signature would be
OEOEOEOEOEOEOEOEOEOE
EEEEEEEEEE
OEOEOEOEOEOEOEOEOEOE
Type it into the Signature field, and click the Signature to Number button. You should receive the number 451306495. Use a calculator to extract the Collatz series for this number, and you will find that the series begins with the odd and even numbers appearing in exactly the order you specified.
• Here is a number for you to ponder. It is quite ordinary except for the fact that it has 3,011 digits. An interesting experiment for you to try, is to copy the array of digits into the clipboard of your computer and then paste them into the Number window of the Collatz program. Clicking the Number to Signature button will start the calculation of the Signature. Depending on your computer it may take quite a bit of time (think in terms of minutes rather than seconds), but the interesting result will be well worth the wait. You will find that the first 20,000 characters of the signature will consist of 10,000 OE Syllables!
```19950631168807583848837421626835850838234968318861924548520089498529438830221946631919961
68403619459789933112942320912427155649134941378111759378593209632395785573004679379452676
52465512660598955205500869181933115425086084606181046855090748660896248880904898948380092
53941633257850621568309473902556912388065225096643874441046759871626985453222868538161694
31577562964076283688076073222853509164147618395638145896946389941084096053626782106462142
73333940365255656495306031426802349694003359343166514592977732796657756061725820314079941
98179607378245683762280037302885487251900834464581454650557929601414833921615734588139257
09537976911927780082695773567444412306201875783632550272832378927071037380286639303142813
32414016241956716905740614196543423246388012488561473052074319922596117962501309928602417
08340807605932320161268492288496255841312844061536738951487114256315111089745514203313820
20293164095759646475601040584584156607204496286701651506192063100418642227590867090057460
64178569519114560550682512504060075198422618980592371180544447880729063952425483392219827
07404473162376760846613033778706039803413197133493654622700563169937455508241780972810983
29131440357187752476850985727693792643322159939987688666080836883783802764328277517227365
75727447841122943897338108616074232532919748131201976041782819656974758981645312584341359
59862784130128185406283476649088690521047580882615823961985770122407044330583075869039319
60460340497315658320867210591330090375282341553974539439771525745529051021231094732161075
34748257407752739863482984983407569379556466386218745694992790165721037013644331358172143
11791398222983845847334440270964182851005072927748364550578634501100852987812389473928699
54083434615880704395911898581514577917714361969872813145948378320208147498217185801138907
12282509058268174362205774759214176537156877256149045829049924610286300815355833081301019
87675856234343538955409175623400844887526162643568648833519463720377293240094456246923254
35040067802727383775537640672689863624103749141096671855705075909810024678988017827192595
33812824219540283027594084489550146766683896979968862416363133763939033734558014076367418
77711055384225739499110186468219696581651485130494222369947714763069155468217682876200362
77725772378136533161119681128079266948188720129864366076855163986053460229787155751794738
52463694469230878942659482170080511203223654962881690357391213683383935917564187338505109
70271613915439590991598154654417336311656936031122249937969999226781732358023111862644575
29913575817500819983923628461524988108896023224436217377161808635701546848405862232979285
38756234865564405369626220189635710288123615675125433383032700290976686505685571575055167
27518899194129711337690149916181315171544007728650573189557450920330185304847113818315407
32405331903846208403642176370391155063978900074285367219628090347797453332046836879586858
02379522186291200807428195513179481576244482985184615097048880272747215746881315947504097
32115080498190455803416826949787141316063210686391511681774304792596709375```
This means that if you were to perform the Collatz process on this number, then each time you multiplied by 3, added 1 and divided by 2 you would get another odd number, and this would continue for an amazing 10,000 times. Thereafter, the pattern would cease, and the normal mixture of odd and even numbers would return, until the number 1 is finally encountered as it always is.
Some additional results will appear in the Collatz Results window as shown below:-
The initial number has 3011 digits.
The number of Signature letters is 134405
The number of Signature Syllables is 48126
The biggest number encountered in the Collatz series is 3262700 - - - - 4400000
This Number contains 4772 digits
• Try some signatures of your own choosing. You should be able to create numbers with some very unlikely characteristics. Relax about the size of the numbers involved. The program should have no problem whatever handling numbers with at least a few thousand digits. There is no end to the interesting games you can play in this way.
• In case you are wondering... Well I hope you are wondering about how one finds a number of over 3000 digits which begins with a string of 10,000 OE syllables. Clearly, this is not the result of any sort of a search routine. It is done by crafting a signature which consists of 10,000 OEs and using the Signature / Number conversion program to obtain the desired number.
• Towards an infinitely long signature It has already been mentioned that an exception to the Collatz Conjecture demands the existence of an infinitely long Signature, so you may be thinking that as long as you can isolate a Number which produces a Signature designed to your specification, you might be able to isolate one which generates an infinitely long signature. The facts however are that although you can create signatures of any given length, infinitely long signatures will apparently always remain beyond your grasp.
In articles which discuss the Collatz conjecture, you will often find comments to the effect that some numbers should continually increase rather than eventually decrease to 1, based on the fact that odd numbers are multiplied by 3 (and incremented by 1), but even numbers are divided only by 2. Viewing the problem in that way is simplistic and misleading. It is better to break the Signature up into Syllables as defined in the introduction.
Every Syllable begins with OE, which implies:-
• Multiplication by 3.
• Division by 2.
Following the division by 2, our number may be odd or even with equal probability.
This implies that:-
It will terminate at OE or extend beyond OE, with equal probability of 1/2.
If it extends beyond OE
It will terminate at OEE or extend beyond OEE, with equal probability of 1/4.
If it extends beyond OEE
It will terminate at OEEE or extend beyond OEEE, with equal probability of 1/8.
And so on...
Clearly, the probability decreases by a factor of 2 for each E added to the Syllable, so longer Syllables are progressively less likely. However they do have a greater impact on the Collatz process due to the greater number of divisions by 2.
The reasoning presented above is captured in tabular form in the following:-.
Signature Syllables [D]ivisions by 2 [P]robability [D]x[P] [D]x[P] (normalised) OE 1 1 / 2 1 / 2 32768 / 65536 OEE 2 1 / 4 2 / 4 32768 / 65536 OEEE 3 1 / 8 3 / 8 24576 / 65536 OEEEE 4 1 / 16 4 / 16 16384 / 65536 OEEEEE 5 1 / 32 5 / 32 10240 / 65536 OEEEEEE 6 1 / 64 6 / 64 6144 / 65536 OEEEEEEE 7 1 / 128 7 / 128 3584 / 65536 OEEEEEEEE 8 1 / 256 8 / 256 2048 / 65536 OEEEEEEEEE 9 1 / 512 9 / 512 1152 / 65536 OEEEEEEEEEE 10 1 / 1024 10 / 1024 640 / 65536 OEEEEEEEEEEE 11 1 / 2048 11 / 2048 352 / 65536 OEEEEEEEEEEEE 12 1 / 4096 12 / 4096 192 / 65536 OEEEEEEEEEEEEE 13 1 / 8192 13 / 8192 104 / 65536 OEEEEEEEEEEEEEE 14 1 / 16384 14 / 16384 56 / 65536 OEEEEEEEEEEEEEEE 15 1 / 32768 15 / 32768 30 / 65536 OEEEEEEEEEEEEEEEE 16 1 / 65536 16 / 65536 16 / 65536
About the contents of this table.
• Signature Syllables.
A list of possible Signature Syllables which contain varying numbers of Es from 1 up to 16. Syllables longer than this will of course be encountered when very large numbers are submitted to the Collatz process. Numbers having thousands of digits will be met and processed in later sections of this article.
• [D]ivisions by 2.
This is simply the number of Es contained within the Syllable.
• [P]robability.
The probability of a random Signature Syllable being of this type. As discussed previously, each Syllable type has a probability of half that of the previous Syllable type. The sum of the probabilities in this column will approach a value of 1 as the list is extended.
• [D]x[P].
The product of the number of divisions by 2 and the probability of this Syllable being generated. The resulting number provides a measure of the overall likelihood of achieving a division by 2 by means of this Syllable when a Syllable is generated.
• [D]x[P] (normalized)
The numbers in this column have exactly the same values as the numbers in the previous column, but they have been normalized so that each number has a denominator of 65536. When we see a number such as 6144 / 65536, it tells us that when 65536 Syllables of a Signature are generated, 6144 of the divisions by 2 will be generated by Syllables which have the form OEEEEEE.
Adding all of the items in column 5 gives us the sum 131054 / 65536 which equals 1.9997. This is the average number of divisions by 2 generated by each Signature Syllable. The fact that this number is so close to 2 is significant. In fact, adding additional lines to the table would move it even closer to 2. Summing up then, each Signature Syllable provides one multiplication by 3 (and an addition of 1) as well as an average of two divisions by 2. Reducing this thought to the simplest possible form implies that on average, calculating one additional Signature Syllable for the number multiplies that number by a factor of 3/4. This relationship is highly significant, and is worthy of a title. I propose that it should be called The rule of 3/4.
To see some experimental evidence which supports this rule, refer to Result 2: Profile of the Signature Syllable data of 100 digit numbers.
A very interesting circumstance arises when we calculate a series of 8 consecutive Signature Syllables. On each of the eight occasions the subject number will be multiplied by a factor which, in the long run, will average out at 3/4. What actually happens is encapsulated in the following mathematical statement.
38 / 48 = .10011
In short it gives us a division by a number very close to 10. This in turn translates to a reduction of one in the number of digits in the number being processed. To a first approximation then, an n digit number will generate n*8 Signature Syllables on its journey to the expected concluding value of 1. This is the basis for what I call The Rule of 8 which will be demonstrated quite soon. When you study this topic, I believe you will be pleasantly surprised at how closely numbers right across the vast number spectrum obey this rule.
Three important caveats.
This Rule of 8 is only an approximation (although a remarkably precise one), and as a result small departures from it will be caused by the following:-
• The 3 in the mathematical statement is always accompanied by the addition of 1. This is not expected to cause a big departure in the operation of the Rule, but it is always present, and the departure is always in the same direction.
• The 4 is the average calculated in the Signature Syllable analysis discussed previously. On any given Syllable calculation it will in fact be some power of 2, but averaged over a large number of calculations the probabilities involved will dictate that the outcome will be a division by very close to 4.
• The mathematical statement above doesn't give us exactly one tenth, although it is, fortuitously, remarkably close to that figure. As a result, we are entitled to be quietly confident that the Rule of 8 will be closely observed.
It will be very much to your advantage if you have a working copy of the Collatz software on your computer as you study this section of the article, so before you proceed any further, please refer to Appendix 4 and complete the Download and Installation instructions.
On the page dealing with Signature / Syllable Analysis. mention was made of the fact that as the Collatz process operates on a number, it will reduce the length of that number by roughly one digit for each group of 8 Signature Syllables which it calculates for that number. For a single number, this is of course just an approximation, but the Rule-of-8 demonstration discussed here allows you to execute the Collatz process on a large group of numbers all of the same length. When the results are averaged over a group of some (or many) thousands of numbers, a quite remarkable result emerges as you will soon see.
The demonstration is part of the Rule-of-8 program, and is best described by reference to the typical output shown in the following graphic.
As the demonstration runs using the defaults supplied, the program applies the collatz process to one million 30 digit numbers, and in so doing produces a set of one million signatures for those numbers. The important information here is not the content of the signature, but its length. The program maintains a list of signature lengths, and as the signature length of each number is determined, the list item for that length is incremented by one. When all one million numbers have been processed this list can be used to display the histogram you see in the graphic. Naturally, as the program runs, it provides an indication of progress by redrawing the histogram at regular intervals. This can be quite spectacular to watch, and is recommended for your entertainment.
It is hoped that the following dot points will add meaning to what you see in the above graphic:-
• The X axis of this histogram is calibrated in terms of signature lengths which range between 64 and 654. These figures are also included in tabular form above the histogram.
• The Y axis is calibrated between 0 and 9937, indicating that the longest signature achieved by any of the one million 30 digit number was in fact achieved by a total 9937 of them.
• A downward pointing red arrow below the x axis points to the location of the average value of Signature Syllables predicted by the Rule of 8, and an upward pointing blue arrow labeled Average points to the location of the current average for the numbers processed so far. It is quite entertaining to watch this indicator as the program runs. Initially it wanders up and down but quickly settles down to a number very close to 240 as predicted by the Rule of 8.
• Although 64 is the shortest signature length encountered in this particular demonstration, it would be quite wrong to assume that 64 is the shortest possible signature for 30 digit numbers. In fact, there is always at least one, and often two numbers which will result in a signature having only one syllable. In the case of 30 digit numbers, the numbers 422550200076076467165567735125 and 105637550019019116791391933781 will collapse to 1 with a single Syllable Signature. These numbers look impressively large, but considerably less so when expressed as (2100 - 1) / 3 and (298 - 1) / 3.
• Similarly, although 654 was the longest signature encountered in the demonstration, it is almost certainly not the longest possible signature. Unlike the shortest signature, finding the longest signature seems not to be a trivial matter. This could be fertile ground for people who can't resist a mathematical challenge.
• The shape of this histogram will strike a chord with anyone who has more than a passing interest in the subjects of probability and statistics. The graph is immediately recognizable as a bell curve, or "normal" distribution with some obvious differences. The main difference is that it is very far from being the smooth curve normally expected. It seems that certain Signature Syllable count values are favored by the Collatz process while others are disadvantaged. Why this is so may be another interesting question for additional research in the future.
• Another point worth noting is the way in which the right extreme of the curve seems to extend out much further than the left extreme. For the time being we will just assume this is because there is an absolute lower limit of 1 for the numbers which can appear in the left portion this histogram, while the right portion might possibly extend indefinitely, as would be the case if an exception to the Collatz Conjecture were to be encountered.
A more detailed look at the Rule of 8 results.
Greater insights into the Rule of 8 can be obtained by running a much more ambitious test. The next graphic shows the result of running the program with one million ninety digit numbers. This means that we are testing numbers in the vicinity of 1090 which is approximately the number of atoms in ten billion universes. This is obviously a very big number, but successful trials have been performed using numbers having 1,000 digits and even up to 10,000 digits. Regardless of the enormity of the numbers being tested, the results always conform very closely with the following description.
The Collatz curve for ninety digit numbers.
• The Normal Distribution curve is again firmly in evidence with the great majority of data points in this histogram crowded very compactly around the value of 720, which is the value suggested by the Rule of 8.
• The curve of the histogram thins out very substantially as it approaches both the left and the right extremes of the graphic.
A closer look at the extremes.
The tapering of the histogram at the extremes finally results in little more than a single pixel being displayed. Naturally we would like to know what is actually happening there. This is taken care of by printing the bars of the histogram in two passes. The first pass prints only the short bars ... the ones which represent signature lengths which were achieved by 10 or less numbers. These bars are stretched so that the longest of them occupy the entire height of the graph. Also they are dawn using a distinctive colour to distinguish them from the rest of the graph. The second pass is drawn using black, and the scaling is organized so that the longest bar occupies the entire height of the graph.
The extreme left of the curve.
• The thinning of the curve mentioned above continues to the left, to the extent that the last of the results shrink to only a single pixel.
• The shortest recorded Signature has a Syllable length of 364. This is most certainly not the shortest possible signature. It is mentioned elsewhere in this article that, for numbers having a given number of digits (90 in this case), there will always be at least one example of a signature length of just one syllable. The likelihood of encountering such a number in a run of the Rule of 8 program on 90 digit numbers is as good as zero. It is the same as the likelihood of selecting one particular atom out of 1090 atoms or, to put it another way, one particular atom out of all the atoms in 10 billion copies of the observable universe. I venture to suggest that this is not very likely.
• The stretched signatures are drawn using red, and clearly show the continued reduction in frequency.
• The Y axis of the histogram is calibrated in terms of frequency. The most frequently encountered signature length was 719, and was encountered 5598 times.
The extreme right of the curve.
• The thinning of the curve also continues to the right, with the last Signature having a Syllable length of 1295. This is very likely not the longest signature possible for numbers having 90 digits. Finding longer signatures than this is not such a simple matter as finding the shortest possible signature. More research is required to settle this question.
• In the Introduction to the Collatz problem it was mentioned that any number which fails the Collatz conjecture will be characterized by an infinite length signature. The rapid thinning out of longer signatures here doesn't bode well for the prospects of finding an infinitely long one which is required to disprove the Collatz conjecture.
• In order to obtain a more detailed view of the extreme right hand end of the Normal Distribution Curve for Numbers having exactly 100 digits, the program was modified to provide an output of the Signature Length, and the Number which produced that Signature for any cases in which the Signature Length was greater than 1400.
The following paragraph contains a list of forty 100 digit numbers, all of which yield signatures having lengths of more than 1400 syllables. It may appear from this that they are rather plentiful, and we shouldn't be surprised about this since they are selected randomly from a collection of 10100 100 digit numbers. Although they are quite plentiful, they are fiendishly hard to find. You can find them using the Rule of 8 function of the Collatz program if you use the option function of that program to request a run of say 10 million 100 digit numbers, and request that any number which produces a signature of over 1400 syllables be displayed in the Collatz Report section of the screen. Typically, a run of 10 million numbers will produce two or three examples of signature lengths in excess of 1400. On the other hand it is not unusual to find no such examples at all.
Patience is a virtue.
1402 : 6842757495773267586887806151688807436727743186126349007201127828468961393836806410856342616825894625
1403 : 2180801148686324953439462939778882626303731853654111339462437156761016480513749135384422020443688379
1406 : 2678311937747823955428233908126798043266098510223891544662249648949944961907723760460734814766910333
1407 : 3823718535532971207517904536989829309675030479388977545081310923773973862966462501763453945111079159
7622170809671124244743551875404553634800556574478143656625786196519428409570986806220345366335963493
1408 : 4820216256912984513189605847117956512837519700474367403251546673836856799328899781794800128254180253
2506365190376979278615060456758297938410847350707104358935686789427934817493845788432634124228325519
1409 : 6060257294094424836930858444496460904397254379087841820258975656997038526771140399033637812395623249
1410 : 4686973975765635616838998517650308280055079164945267339841401598957946602817094876336812022569995391
9374073467186332319437831785343186847740210300395202900173031197209535622358506974005191163376365815
1412 : 3732465449176743594224707380191993481984089791756450392918583200376536728650534763667263757341603203
7154264313251634209772562532969336146398156934218428580194249436324043572067810466236232627824661951
1414 : 3659640113614688382853914953076440177360139842033070807131266614715044469239717140998327646571334199
1415 : 8995089998187912595026462086517046267339660140522371736764191610482759529859248935170505760178449767
1417 : 4043559559806976359551087111494822836155102541571017178364399858575755907192497135991522484779844111
1419 : 7131190755481947195107934275281318476355170881457672715019446987921830936264922416160406395592394129
6771146834247137694732961971255838525322742172319874644186854495776487064109198768529832871573782059
7856528451007070433975984903163448123727975214790085824317098724864610725742702890346177760390827071
1420 : 9309794760299155642628929637431924875904944644155461315035051583979900098373752828867489265705838575
2334693199694858085441325298114605202857326117892976383782195097461485535727656770536332159567954119
1423 : 2830287823686069734541803020779532604253613699343910521381354929520209221476192189132279145735299901
1424 : 7438269916262190780119924450170867585002538104826307458335788281293123333618941445385862130534123371
7697272880142611622999325454520993661086991386431588693804692081387250383617266411700217074288672211
1427 : 4518015638758504762454896952014173200458087802783061434847829915557389908484527774750244585945495709
1428 : 2987351939502758017968763002262795094030303442986575672495787377421193383546038031368008177191767481
1429 : 8092646192449084550390721847739445404832588938966339379318593425997002227885473491409252483725167387
1432 : 9123695420409529171114400592587784174324640783510261967307613794037332553017495609149175866329395983
1436 : 3706105287219384670033366475152581904264055815421809436915465735952194777186386907699933009131046089
1447 : 5517046498299205737900588092793379349801960574321693647391635231648009790605397094833200279046740291
1449 : 2484469091877287715642028352428745726373577826592312905854489492569789134629962447426019564086370415
1454 : 5366017916635901776062517542587112469857321682017825015324268581233285357636552379577372694849721765
1455 : 6739376114482765766373616813230128205294275852107565774722510864761543950500853450866031186171124015
6793689273525437379018565305878252468658627531172429794461157003568686480121401099229595020365282249
1458 : 8098659249748649346462339440788399673519597757016037607323163102355838625354820422426232514262620585
1459 : 2735671485420203313402614296821931151400352116648886211126918787671682198087559807028241063050176535
1471 : 1305049607908580325169640257013784663287258107157267305227843138837988758167021876344271894149757407
1480 : 2114906264399229703377091102898189426947788654789037024114887199131871912693467009113804621622760153
1483 : 9922131373936600124796146073324036161679659546472207392368908204560210671900600300025651470071977871
1485 : 9192617866236964186795541944302165469471124072578069540398093355392057707815135005706757256462122995
1501 : 1925325427301238693436974716463324231743360748827222059797465213996993658480666367192272714798797665
7200774491333522318966513865183887923813564533295043070788032954829439786486921553433708195977833007
1504 : 8981501245210358466941063069963587676629346197118428970148956094519761258248051502546615028160501531
1564 : 7987221001644153042341078535366145131744118543385943691991226902014405396825535417707018213937349053
Even a cursory examination of these forty signatures provides some revealing facts about the Collatz sequence:-
• The first 20 signatures are associated with signature lengths which range between 1402 and 1420, a span of only 19 lengths.
• The range of lengths occupied by the second group of 20 signatures extends from 1420 up to 1504, a span of 84. This demonstrate clearly that the density of signatures is decreasing quite rapidly.
• In the first group of signatures there are six examples of signature lengths which play host to multiple signatures, while in the second group of 20 there are only two such examples.
• The highest signature length discovered was 1504. There are very likely longer signature lengths than this, but searching for them would be a very time consuming and probably unrewarding task.
• It would be foolish to insist that there are no examples of 100 digit numbers with signature lengths greater than 1504, but it is surely unlikely that there exists a 100 (or less) digit number with an infinitely long signature.
• Which brings us to the crucial point, (mentioned in several other places in this report), that an exception to the Collatz Conjecture demands the existence of a finite number which has an infinitely long signature.
• The above results apply to numbers having "only" 100 digits. I have used the Rule of 8 program to perform the same process on 1000 digit numbers, and also to a lesser extent on 10,000 digit numbers, and the result is the same in all cases.
How maximum Signature Length changes with Digits per Number.
The following graphic image serves an an illustration for the data contained in the table which immediately follows it. It is a graph of the signature length values of 100,000 thirty digit numbers.
The key items of data are the center point of the curve as suggested by the Rule of 8 and indicated by the values on the x axis of the graph, and the length of the longest signature for each number length. The ratio between these two values is recorded in the fourth column of the table, and it is this ratio which provides a very telling story about the operation of the Collatz series. You will observe that the value of the ratio decreases as the number of digits in the subject numbers increases. Not only does it show that the prospects of finding an infinitely long signature are not very good at the outset, but those prospects become even less hopeful as the number of digits in the numbers being processed increases from 30 in the first line of the table up to the maximum 10,000 digits in the final entry.
Anyone armed with this information who continues a search for an exception to the Collatz Conjecture really is embarking on a forlorn quest!
Number of
Digits
Center of curve
as determined by
Rule of 8
Longest
Signature
Length
Ratio of
Signature Length
to Center of Curve
30 240 587 2.45
40 320 697 2.18
50 400 788 1.92
60 480 895 1.86
70 560 984 1.76
80 640 1135 1.77
90 720 1184 1.64
100 800 1285 1.61
150 1200 1792 1.49
200 1600 2250 1.41
1000 8000 9303 1.16
10000 80000 83324 1.04
Some illuminating results generated by the Rule of 8 function.
Result 1: Accuracy of the Rule of 8 function.
The following table contains the results of a series of tests performed by the Rule of 8 function. The columns contain the following data:-
• The Number Length is the number of digits in the numbers used for each test in the series. A minimum of one thousand numbers were processed in each test.
• Note that in every case, the Average Number of Syllables per Signature is very close to the (8 times the Number Length) figure predicted by the Rule of 8, and shown in square brackets.
• The Highest Number of Syllables per Signature is to be interpreted as the highest number encountered on this run. There will probably be a few higher numbers but these can safely be categorized as outliers.
• The Lowest Number of Syllables per Signature is not recorded. This is because there will always be at least one number for any given number length which will have a Signature with only one Syllable.
For example:-
• 2 digit numbers :- 21 and 85
• 3 digit numbers :- 341
• 4 digit numbers :- 1365 and 5461
Number
Length
Average number
of Syllables
per Signature
Highest number
of Syllables
per Signature
Number
Length
Average number
of Syllables
per Signature
Highest number
of Syllables
per Signature
14 [16] 43 200 1602 [1600] 2201
22 [24] 65 300 2397 [2400] 3250
31 [32] 96 400 3200 [3200] 3963
38 [40] 129 500 4001 [4000] 4857
47 [48] 176 600 4804 [4800] 5553
55 [56] 203 700 5600 [5600] 6401
63 [64] 218 800 6397 [6400] 7458
71 [72] 265 900 7200 [7200] 8213
10 78 [80] 272 1000 8000 [8000] 9183
20 158 [160] 406 2000 16001 [16000] 17429
30 238 [240] 589 3000 24017 [24000] 26084
40 319 [320] 636 4000 32038 [32000] 33767
50 398 [400] 723 5000 40044 [40000] 42091
60 479 [480] 955 6000 48015 [48000] 50528
70 559 [560] 909 7000 55984 [56000] 58473
80 638 [640] 1121 8000 64013 [64000] 66737
90 718 [720] 1118 9000 72002 [72000] 74303
100 799 [800] 1237 10000 80047 [80000] 83017
In the topic Signature / Syllable Analysis it was confidently predicted that the Rule of 8 would be closely observed as the Collatz Process is applied to numbers in general. This table demonstrates that the prediction was indeed justified.
The progressive results provided by the Rule of 8 function include a profile of the entire set of Signature Syllables generated during a batch processing run. The following table was generated during a typical run using one million 100 digit numbers. The Rule of 8 predicts that the total number of Signature Syllables should be 8 x 100 x 1,000,000 = 800,000,000. The figure that is reported by the program is 799,108,217. This adds quite significantly to the confidence we can feel in the validity of the Rule.
Of equal interest are the numbers of each of the Syllable types. Note that almost exactly half of all the Syllables are of type OE, and that all of the subsequent types progressively decrease by a factor of close to 2. This justifies the equal probability assumptions made in the topic Signature / Syllable Analysis.
Behavior 1.
At first sight, it is quite surprising to study the Colltz sequence for the number 27, and observe that it reaches a high point of 9232 which is over 340 times greater than its starting point. Perhaps it is natural to wonder how this compares to numbers in general, and especially to much larger numbers. A small adjustment was made to the Rule of 8 function of the Collatz program so that it could create a list of the biggest numbers encountered during a typical run. Each time a number bigger than the last big number on the list was encountered, it was added to the end of that list. The following list resulted when a sequence of one million randomly chosen 30 digit numbers were put to the test.
0 521864777009781972171399170839 1565594331029345916514197512518 [1]
0 521864777009781972171399170839 2348391496544018874771296268778 [1]
0 521864777009781972171399170839 3522587244816028312156944403168 [1]
1 323420776575166478500093050537 3683964783176505669165122403790 [1]
1 323420776575166478500093050537 5525947174764758503747683605686 [1]
1 323420776575166478500093050537 8288920762147137755621525408530 [1]
1 323420776575166478500093050537 12433381143220706633432288112796 [2]
1 323420776575166478500093050537 13987553786123294962611324126898 [2]
1 323420776575166478500093050537 20981330679184942443916986190348 [2]
1 323420776575166478500093050537 23603997014083060249406609464144 [2]
57 916855648438295922460724587307 24472957408552401667321569784324 [2]
87 621623765988457188136100529945 28709147380980211756617196305502 [2]
87 621623765988457188136100529945 43063721071470317634925794458254 [2]
87 621623765988457188136100529945 64595581607205476452388691687382 [2]
87 621623765988457188136100529945 96893372410808214678583037531074 [2]
87 621623765988457188136100529945 145340058616212322017874556296612 [3]
87 621623765988457188136100529945 183946011686143720053872485312906 [3]
87 621623765988457188136100529945 275919017529215580080808727969360 [3]
357 126481844367082466405697580063 384123720446478525524194192690150 [3]
357 126481844367082466405697580063 576185580669717788286291289035226 [3]
357 126481844367082466405697580063 864278371004576682429436933552840 [3]
997 930817431324349698729704998367 1242071656860735704986768744595224 [4]
3759 833314741525573657655193087439 1687177372739226871344116090500816 [4]
4790 802879304996710367566694233327 2002334859286487864932913145951352 [4]
4790 802879304996710367566694233327 2534205056284461204055718200344688 [4]
12498 724912411722761965880444096681 3478991944441262688411679665259144 [4]
13198 288842135708910560684485601487 4995984046651086558655411738847260 [4]
13198 288842135708910560684485601487 5620482052482472378487338206203170 [4]
13198 288842135708910560684485601487 8430723078723708567731007309304756 [4]
18895 337595270660394052978947869065 9122282670150866104209365939507014 [4]
18895 337595270660394052978947869065 13683424005226299156314048909260522 [5]
18895 337595270660394052978947869065 20525136007839448734471073363890784 [5]
46838 877683494233739625091193455399 23082073760820664932033630627369574 [5]
46838 877683494233739625091193455399 34623110641230997398050445941054362 [5]
46838 877683494233739625091193455399 51934665961846496097075668911581544 [5]
56900 808023609957127685989776095391 58102268660421757049325755468362498 [5]
56900 808023609957127685989776095391 87153402990632635573988633202543748 [5]
121196 305282207143663775670173201499 123597536981034324928986406162487428 [6]
121196 305282207143663775670173201499 160725630931863703306657628708270410 [6]
121196 305282207143663775670173201499 241088446397795554959986443062405616 [6]
121196 305282207143663775670173201499 257451382945302186864008960438223202 [6]
121196 305282207143663775670173201499 386177074417953280296013440657334804 [6]
159448 620544853285226825440776099267 557672052424846300330533042985763488 [6]
• Column 1: The number of numbers which had been processed before processing was begun on this one.
• Column 2: The thirty digit number currently being processed.
• Column 3: The value of the current number in the Collatz series.
• Column 4: The number of additional digits in the Collatz series number, compared to the number being processed.
As you can see, in this case the Collatz series can contain numbers which are over a million times larger than the number which started the series.
What follows is the basis of an algorithm which can be used to construct a program which accepts a complete Collatz Signature and computes the number which generated it. Note that this algorithm works equally well for both partial and final signatures.
The Signature used as the input in this description will be OEOEOEEEEEOEEEEO.
We begin with two equal odd numbers called Α (the Greek letter alpha) and Ω (the Greek letter omega). Both will be initialized to the odd number 2n+1 before processing begins. Ω will change its value as processing proceeds, as dictated by the contents of the Signature which is listed vertically in the first column. The changes to the value of Ω are effected by carefully controlled changes to the value of n. Whatever changes are made to the value of n in Ω will also be made to the value of n in Α. The result of all this will be that at the completion of the algorithm, Α will contain the required number.
Ω Α Ω 2n+1 2n+1 O 2n+1 This is odd so Multiply by 3 and add 1. 6n+4 2n+1 E 6n+4 This is even so Divide by 2. 3n+2 2n+1 O 3n+2 For this to be odd, n must be odd, so set n to 2n+1. 6n+5 4n+3 6n+5 is odd so Multiply by 3 and add 1. 18n+16 4n+3 E 18n+16 This is even so Divide by 2. 9n+8 4n+3 O 9n+8 For this to be odd, n must be odd, so set n to 2n+1. 18n+17 8n+7 18n+17 is odd so Multiply by 3 and add 1. 54n+52 8n+7 E 54n+52 This is even so Divide by 2. 27n+26 8n+7 E 27n+26 For this to be even, n must be even, so set n to 2n. 54n+26 16n+7 54n+26 is even so Divide by 2. 27n+13 16n+7 E 27n+13 For this to be even, n must be odd, so set n to 2n+1. 54n+40 32n+23 54n+40 is even so Divide by 2. 27n+20 32n+23 E 27n+20 For this to be even, n must be even, so set n to 2n. 54n+20 64n+23 54n+20 is even so Divide by 2. 27n+10 64n+23 E 27n+10 For this to be even, n must be even, so set n to 2n. 54n+10 128n+23 54n+10 is even so Divide by 2. 27n+5 128n+23 O 27n+5 For this to be odd, n must be even, so set n to 2n. 54n+5 256n+23 54n+5 is odd so Multiply by 3 and add 1. 162n+16 256n+23 E 162n+16 This is even so Divide by 2. 81n+8 256n+23 E 81n+8 For this to be even, n must be even, so set n to 2n. 162n+8 512n+23 162n+8 is even so Divide by 2. 81n+4 512n+23 E 81n+4 For this to be even, n must be even, so set n to 2n. 162n+4 1024n+23 162n+4 is even so Divide by 2. 81n+2 1024n+23 E 81n+2 For this to be even, n must be even, so set n to 2n. 162n+2 2048n+23 162n+2 is even so Divide by 2. 81n+1 2048n+23 O 81n+1 For this to be odd, n must be even, so set n to 2n. 162n+1 4096n+23 162n+1 is odd as required.
So, numbers having the given Signature will have the form 4096n+23
``` When n=0, the initial number is 23.
O E O E O E E E E E O E E E E O
23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
When n=1, the initial number is 4096 + 23 = 4119.
O E O E O E E E E E O E
4119 12358 6179 18538 9269 27808 13904 6952 3476 1738 869 2608
E E E O
1304 652 326 163```
The algorithm provides us with an infinite series of numbers all of which have a Signature beginning with OEOEOEEEEEOEEEEO. Explore the Signatures which appear when you substitute other values of n. When you do you will find that the first 16 characters of the Signature will correspond exactly to the Signature of 23. Thereafter, all bets are off as to how the Signatures will progress except for the fact that they will all terminate with an O at number 1.
It has been mentioned several times in this report that it is an easy matter to derive large numbers which have quite short signatures. However the inverse operation of finding the longest possible signature for numbers which have a (relatively) small number of digits is far from simple. This appendix will describe a method of obtaining relatively long examples.
Consider the following diagram of a number map. Virtually everyone who knows anything about the Collatz conjecture could start at 1442 and calculate their way down to the 1. It is simply a matter of applying the Collatz process to generate the familiar sequence of numbers, but how do you get back to 1442 if you start from, for example, 5. The difficulty is that at each number you have a choice of two operations. You can always multiply by 2, but if the current number is 1 modulo 3, you can also subtract 1 and divide by 3. How do you decide which choice to make?
The aim of the process is to get the longest possible signature, and to achieve this you must take the divide by 3 option as often as possible.
^^^^
1442
^^^^ 721
1624 - 541 - 1082 - 2164 >
812
406
^^^ 203
916 - 305 - 610 >
458
^^^ 229
130 - 43 - 86 - 172 - 344 - 688 >
^^^ 65
148 - 49 - 98 - 196 >
74
^^ 37
22 - 7 - 14 - 28 - 112 >
^^ 11
52 - 17 - 34 >
26
^^ 13
16 - 5 - 10 - 20 - 40 >
4
2
1
The Collatz program which supports the principles discussed in this article was generated over a period of some years while the principles themselves were emerging from my research. During this time, a program called Crossword Express was also being developed, and at the time it seemed appropriate to incorporate the much shorter Collatz program into the main body of Crossword Express. As a result, if you go the extra mile to download and install this program you will receive a significant bonus of an extensive puzzle generation platform which I believe surpasses the performance of all competing products, and it won't cost you a cent.
The entire package is written in the Java programming language, and so is able to be installed and run on both Apple and Windows computers. There are some differences in the steps involved between these two systems, and so what follows are two separate paragraphs which provide all of the details you will need to achieve a satisfactory result.
• Unzip the file by means of a double click on the icon of the .zip file. This will give you a folder called CrosswordExpress. You can now safely delete the CrosswordExpress.zip file.
• Move the CrosswordExpress folder to the location on your hard drive where you would like it to be stored permanently. The Applications folder would be a good choice, but the decision is up to you.
• Double click the CrosswordExpress folder in the new location you chose for it, right click the Crossword-Express.jar file, and select Open With followed by Jar Launcher.app. This is a once only operation which in effect tells the computer that the application being opened is a trusted application.
• Double click the CrosswordExpress folder in the new location you chose for it, then right click the Crossword-Express.jar file, and select Make Alias. This will give you a file called Crossword-Express.jar alias.
• Drag the Crossword-Express.jar alias file to an easily accessible location, so that you can start Crossword Express and make use of the Collatz program whenever the need arises. The Desktop would be a good choice, but once again, the choice is yours.
• This is where the recommended procedure for Apple and Windows diverge. Thirty years of experience with Crossword Express has revealed that there are many unzip utilities available for use with Windows, and many of these have a faithful band of regular users. Consequently, I make no firm recommendation on the subject of unzipping this file, except to say that when you are finished, the Desktop must contain a new folder called CrosswordExpress. You can now safely delete the CrosswordExpress.zip file.
• Move the CrosswordExpress folder to the location on your hard drive where you would like it to be stored permanently. The Programs folder would be a good choice, but the decision is up to you.
• Double click the CrosswordExpress folder in the new location you chose for it, right click the Crossword-Express.jar file, and select Open With followed by Java (TM) Platform SE binary. This is a once only operation which in effect tells the computer that the application being opened is a trusted application.
• Double click the CrosswordExpress folder in the new location you chose for it, then right click the Crossword-Express.jar file, and select Create shortcut. This will give you a file called Crossword-Express.jar-Shortcut.
• Drag the Crossword-Express.jar-Shortcut file to an easily accessible location, so that you can start Crossword Express and make use of the Collatz program whenever the need arises. The choice is yours.
Possible Installation Difficulties.
The executable file of Crossword Express is CrosswordExpress.jar Such files (having a .jar extension) will run on any computer which has a Java Runtime Environment (JRE) installed. If it fails to run on your computer, it simply means that your computer has not yet been equipped with a JRE. This is easily remedied:-
• For Apple
Select Settings / Java / Update The resulting dialog will tell you the version of Java you have installed. If this is not the recommended version, an Update button will appear, and you can use this to download the correct version. For detailed assistance on the subject, visit https://java.com/en/download/help/mac_install.xml
• For Windows.
Select Start / Control Panel / Java / Update. Clicking the Update Now button will either tell you that you have the correct version already installed, or it will automatically install the correct version. For detailed assistance on the subject, visit https://java.com/en/download/faq/java_win64bit.xml
And here is the payoff for your care and attention.
To start the Collatz program, select Collatz from the list of puzzle options, and then click and GO. This will present you with a window similar to this:-
The window that you see will be larger than this, but all of the components of the window are resizeable using simple mouse operations.
When you are studying Number / Signature Conversions, you will be working with the Number window, the Signature window, and the associated green arrow buttons.
When you are studying The Rule-of-8 you will be using the Rule-of-8 Demonstration menu option. | 15,136 | 56,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-23 | latest | en | 0.783855 |
http://sqlml.azurewebsites.net/2017/08/19/machine-learning-algorithms-part-2/ | 1,600,914,360,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400213006.47/warc/CC-MAIN-20200924002749-20200924032749-00023.warc.gz | 123,608,969 | 10,181 | # Machine Learning Algorithms – Part 2
This post describes some generative machine learning algorithms.
# Gaussian Discriminant Analysis
Gaussian Discriminant Analysis is a supervised classification algorithm. In this algorithm we suppose the p(x|y) is multivariate Gaussian with parameter mean $$\overrightarrow{u}$$ and covariance $$\Sigma : P(x|y) \sim N(\overrightarrow{u}, \Sigma)$$.
If we suppose that y is Bernoulli distributed, then there are two distributions that can be defined, one for p(x|y=0) and one for p(x|y=1). The separator between these two distributions could be used to predict the values of y.
$$P(x|y=0) = \frac{1}{(2\pi)^{\frac{n}{2}}|\Sigma|^{\frac{1}{2}}} exp(-\frac{1}{2}(x – \mu_0)^T\Sigma^{-1}(x – \mu_0))$$
$$P(x|y=1) = \frac{1}{(2\pi)^{\frac{n}{2}}|\Sigma|^{\frac{1}{2}}} exp(-\frac{1}{2}(x – \mu_1)^T\Sigma^{-1}(x – \mu_1))$$
$$P(y) = \phi^y.(1-\phi)^{1-y}$$
We need to find $$\phi, \mu_0, \mu_1, \Sigma$$ that maximize the log-likelihood function:
$$l(\phi, \mu_0, \mu_1, \Sigma) = log(\prod_{i=1}^{m} P(y^{(i)}|x^{(i)};\phi, \mu_0, \mu_1, \Sigma))$$
Using the Naive Bayes rule, we can transform the equation to:
$$= log(\prod_{i=1}^{m} P(x^{(i)} | y^{(i)};\mu_0, \mu_1, \Sigma) * P(y^{(i)};\phi))) – log(\prod_{i=1}^{m} P(x^{(i)}))$$
To maximize l, we need to maximize: $$log(\prod_{i=1}^{m} P(x^{(i)} | y^{(i)};\mu_0, \mu_1, \Sigma) * P(y^{(i)};\phi)))$$
For new data, to predict y, we need to calculate $$P(y=1|x) = \frac{P(x|y=1) * P(y=1)}{P(x|y=0)P(y=0) + P(x|y=1)P(y=1)}$$ (P(y=0) and P(y=1) can be easily calculated from training data).
# EM Algorithm
EM algorithm is an unsupervised clustering algorithm in which we assign a label to each training example. The clustering applied is soft clustering, which means that for each point we assign a probability for each label.
For a point x, we calculate the probability for each label.
There are two main steps in this algorithm:
E-Step: Using the current estimate of parameters, calculate if points are more likely to come from the red distribution or the blue distribution.
M-Step: Re-estimate the parameters.
Given a training set $$S = \{x^{(i)}\}_{i=1}^m$$. For each $$x^{(i)}$$, there is hidden label $$z^{(i)}$$ assigned to it.
Z is a multinomial distribution with $$P(z = j)=?_j$$.
k the number of distinct values of Z (number of classes).
For each $$x^{(i)}$$, we would like to know the value $$z^{(i)}$$ that is assigned to it. In other words, find $$l \in [1, k]$$ that maximizes the probability $$P(z^{(i)} = l | x^{(i)})$$
Log-likelihood function to maximize: $$l(?) = log(\prod_{i=1}^m P(x^{(i)}; ?)) = \sum_{i=1}^m log(P(x^{(i)}; ?))$$
$$= \sum_{i=1}^m log(\sum_{l=1}^k P(x^{(i)}|z^{(i)} = l; ?) * P(z^{(i)} = l))$$ $$= \sum_{i=1}^m log(\sum_{l=1}^k P(x^{(i)}, z^{(i)} = l; ?))$$
Let’s define $$Q_i$$ as a probability distribution over Z: ($$Q_i(z^{(i)} = l)$$ > = 0 and $$\sum_{l=1}^k Q_i(z^{(i)} = l) = 1$$)
$$l(?) = \sum_{i=1}^m log(\sum_{l=1}^k Q_i(z^{(i)} = l) \frac{P(x^{(i)}, z^{(i)} = l; ?)}{Q_i(z^{(i)} = l)})$$ $$= \sum_{i=1}^m log(E_{z^{(i)} \sim Q_i}[\frac{P(x^{(i)}, z^{(i)}; ?)}{Q_i(z^{(i)})}])$$
Log is concave function. Based on Jensen’s inequality log(E[X]) >= E[log(X)].
Then:
$$l(?) >= \sum_{i=1}^m E_{z^{(i)} \sim Q_i}[log(\frac{P(x^{(i)}, z^{(i)}; ?)}{Q_i(z^{(i)})})]$$ $$= \sum_{i=1}^m \sum_{l=1}^k Q_i(z^{(i)}=l) log(\frac{P(x^{(i)}, z^{(i)}=l; ?)}{Q_i(z^{(i)}=l)})$$
If we set $$Q_i(z^{(i)}) = P(z^{(i)}| x^{(i)}; ?)$$ then: $$\frac{P(x^{(i)}, z^{(i)}; ?)}{Q(z^{(i)})}$$ $$= \frac{P(z^{(i)}| x^{(i)}; ?) * P(x^{(i)}; ?)}{P(z^{(i)}| x^{(i)}; ?)}$$ $$= P(x^{(i)}; ?)$$ = Constant (over z). In that case log(E[X]) = E[log(X)] and for the current estimate of ?, $$l(?) = \sum_{i=1}^m \sum_{l=1}^k Q_i(z^{(i)}=l) log(\frac{P(x^{(i)}, z^{(i)}=l; ?)}{Q_i(z^{(i)}=l)})$$ (E-Step)
EM (Expectation-Maximization) algorithm for density estimation:
1-Initiliaze the parameters ?
2-For each i, set $$Q_i(z^{(i)}) = P(z^{(i)}| x^{(i)}; ?)$$ (E-Step)
3-Update ? (M-Step)
$$? := arg\ \underset{?}{max} \sum_{i=1}^m \sum_{l=1}^k Q_i(z^{(i)}=l) log(\frac{P(x^{(i)}, z^{(i)}=l; ?)}{Q_i(z^{(i)}=l)})$$
Mixture of Gaussians
Given a value $$z^{(i)}=j$$, If x is distributed normally $$N(?_j, ?_j)$$, then:
$$Q_i(z^{(i)} = j) = P(z^{(i)} = j| x^{(i)}; ?)$$ $$= \frac{P(x^{(i)} | z^{(i)}=j) * P(z^{(i)} = j)}{\sum_{l=1}^k P(x^{(i)} | z^{(i)}=l) * P(z^{(i)}=l)}$$ $$= \frac{\frac{1}{{(2?)}^{\frac{n}{2}} |?_j|^\frac{1}{2}} exp(-\frac{1}{2} (x^{(i)}-?_j)^T {?_j}^{-1} (x^{(i)}-?_j)) * ?_j}{\sum_{l=1}^k \frac{1}{{(2?)}^{\frac{n}{2}} |?_l|^\frac{1}{2}} exp(-\frac{1}{2} (x^{(i)}-?_l)^T {?_l}^{-1} (x^{(i)}-?_l)) * ?_l}$$
Using the method of Lagrange Multipliers, we can find $$?_j, ?_j, ?_j$$ that maximize: $$\sum_{i=1}^m \sum_{l=1}^k Q_i(z^{(i)}=l) log(\frac{\frac{1}{{(2?)}^{\frac{n}{2}} |?_j|^\frac{1}{2}} exp(-\frac{1}{2} (x^{(i)}-?_j)^T {?_j}^{-1} (x^{(i)}-?_j)) * ?_j}{Q_i(z^{(i)}=l)})$$ with the constraint $$\sum_{j=1}^k ?_j = 1$$. We also need to set to zero the partial derivatives with respect to $$?_j$$ and $$?_j$$ and solve the equations to find the new values of these parameters. Repeat the operations until convergence.
# Naive Bayes Classifier
Naive Bayes classifier is a supervised classification algorithm. In this algorithm, we try to calculate P(y|x), but this probability can be calculated using the naive bayes rule:
P(y|x) = P(x|y).P(y)/P(x)
If x is the vector $$[x_1,x_2,…,x_n]$$, then:
$$P(x|y) = P(x_1,x_2,…,x_n|y) = P(x_1|y) P(x_2|y, x_1) ….P(x_n|y, x_{n-1},…,x_1)$$ $$P(x) = P(x1,x2,…,x_n)$$
If we suppose that $$x_i$$ are conditionally independent (Naive Bayes assumption), then:
$$P(x|y) = P(x_1|y) * P(x_2|y) * … * P(x_n|y)$$ $$= \prod_{i=1}^n P(x_i|y)$$ $$P(x) = \prod_{i=1}^n P(x_i)$$
If $$\exists$$ i such as $$P(x_i|y)=0$$, then P(x|y) = 0. To avoid that case we need to use Laplace Smoothing when calculating probabilities.
We suppose that $$x_i \in \{0,1\}$$ (one-hot encoding: [1,0,1,…,0]) and y $$\in \{0,1\}$$.
We define $$\phi_{i|y=1}, \phi_{i|y=0}, \phi_y$$ such as:
$$P(x_i=1|y=1) = \phi_{i|y=1} \\ P(x_i=0|y=1) = 1 – \phi_{i|y=1}$$ $$P(x_i=1|y=0) = \phi_{i|y=0} \\ P(x_i=0|y=0) = 1 – \phi_{i|y=0}$$ $$P(y) = \phi_y^y.(1-\phi_y)^{1-y}$$
We need to find $$\phi_y, \phi_{i|y=0}, \phi_{i|y=1}$$ that maximize the log-likelihood function:
$$l(\phi, \phi_{i|y=0}, \phi_{i|y=1}) = log(\prod_{i=1}^{m} \prod_{j=1}^n P(x_j^{(i)}|y^{(i)}) P(y^{(i)})/P(x^{(i)}))$$
Or maximize the following expression:
$$log(\prod_{i=1}^{m} \prod_{j=1}^n P(x_j^{(i)}|y^{(i)}) P(y^{(i)}))$$ | 2,812 | 6,562 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-40 | longest | en | 0.698627 |
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