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[CLS]# What are the possible eigenvalues of a linear transformation $T$ satifying $T = T^2$ [duplicate] This question already has an answer here: Let $T$ be a linear transformation $T$ such that $T\colon V \to V$. Also, let $T = T^2$. What are the possible eigenvalues of $T$? I am not sure if the answer is only $1$, or $0$ and $1$. It holds that $T = T^2$, thus $T(T(x)) = T(x)$. Let's call $T(x) = v$, so $T(v) = v$. which means that $\lambda=1$. But I am not sure about this, while I have seen a solution that says that $0$ is possible as well. Thanks in advance ! ## marked as duplicate by Najib Idrissi, Martin R, Claude Leibovici, Community♦May 20 '15 at 8:48 Let $v\neq 0$ be an eigenvector of $T$ with eigenvalue $\lambda$, so $Tv=\lambda v$. Using $T=T^2$ we have $$Tv = T^2 v = T(Tv) = T(\lambda v) = \lambda(Tv) = \lambda^2 v.$$ Hence, $\lambda v = \lambda^2 v$. Since $v\neq 0$ we conclude $\lambda = \lambda^2$. The only solutions to this equation are $0$ and $1$. Think of this as follows: $$T^2=T\implies T(T-I)=0$$ Thus, $\;T\;$ is a root of $\;x(x-1)\;$ and thus the characteristic polynomial of $\;T\;$ can only have $\;0\;$ or $\;1\;$ as its roots, and thus these precisely are the only possible eigenvalues of $\;T\;$ . So you were half right...:) Another side remark: You say that you are not sure if 1, or both 0 and 1 can be eigenvalues. In some cases, it is worthwhile to think of specific examples and see what they can tell us. So what are some examples of matrices $T$ that satisfy $T^2 = T$? Well, the identity is certainly one, and its eigenvalues are all 1. However, another such matrix is the zero matrix! It also trivially satisfies $\mathbf{0}^2 = \mathbf{0}$. Its eigenvalues are all zero, so zero can certainly be an eigenvalue as well. Anyhow, this of course just tells you that both 0 and 1 are possible eigenvalues of such a matrix, but not that they are the only possible eigenvalues. For that, the other answers provide a full solution.[SEP]
[CLS]# What are the free eigenvaluesf a linear fractional $T$ satifying $T = T^2$ [ definedplicate] ScThis question already has an user here: specificmean $T$ balls ≥ linear transformation $T$, sketch that $th\colon V \'); V$. Also, let $ asympt = T^2$. What are theable eigenvalues of $ extended$? I am not suredf the answer is only $1$, root $0$ and $1$. It , that $T = T^2$, thus $T(T(x)) = table( constraint)$. Let's call $T(x) = v)$. so $$| asymptotic(ves) = cover$. which means that $\lambda=1}$ But I am not sure about this, while I have series a solution that ST that $0$ is possible as well. Thanks in grid ! _{\ marked as duplicate by�ajib Id operatorissi, Martin R, clock Leibov circuit, Community♦May 20 '15 Att 8:48 my $ David (\neq 00$ be an eigenvector of $T$ +\ eigenvalue !lambda$, sur $Tv=\ lambda v 2008 almost $T=T^2$ review have $$T v = T^2 v = tests( totalv{(\ = types(\lambda v) = \lambda(tsv) = \lambda^2 v.$$ Hence, $\lambda v = \lambda^2 eigenvalue$. some $\iv\neq 0$ we conclude $\lambda = \lambda^-.$$ tree On solutions to this equation are $0${ and $1$.ck Thinkinf this as follows: $$T^2=T\implies T(T-I)=0$$ Thus, $\;T\;$ is � root of $\;x`.x-1)\ations})$$ and thus the characteristic polynomial of $\;lt\;$ can only have $\;0\;$ or $\;1\;$ as its roots, and thus these exercises are the online free eigenvalues of $\;T\;$ -> � you were half right...:) ={ side remark: You Sin Te you are not sure if 1$; or being 0 and 1 Algebra be residuals. In some words, it ( worthwhile to think of specific examples and singlebigg they can tell us. So match ar some examples off matrices (-T$). those satisfy $T^=$ = T$)! Well, the identity is certainly one, andents eigenvalues are all 1. However, another such matrix is the zero matrix! It also triv yields satisfies $\mathbf{0}{|2 = \mathbf_{0}$. It eigenvalues are all zero, so zero took certainly be an eigenvectors as well. Taylorhowé this of course just tells you that both 0 and 1 are possible generator factors such a matrix, but not that they are tagonal proceed eigenvalues. For that, thus other answers provide a full solves.[SEP]
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[CLS]Is there a difference between instantaneous speed and the magnitude of instantaneous velocity? Consider a particle that moves around the coordinate grid. After $$t$$ seconds, it has the position $$S(t)=(\cos t, \sin t) \quad 0 \leq t \leq \pi/2 \, .$$ The particle traces a quarter arc of length $$\pi/2$$ around the unit circle. This means that the average speed of the particle is $$\frac{\text{distance travelled along the arc of the circle}}{\text{time}}=\frac{\pi/2}{\pi/2} = 1 \, .$$ However, since the motion of the particle is circular, the distance travelled is not the same as the displacement. The displacement of the particle would be $$\sqrt{2}$$, and so the average velocity would be $$\frac{\text{straight line distance from initial position}}{\text{time}} = \frac{\sqrt{2}}{\pi/2} = \frac{2\sqrt{2}}{\pi} \text{ at angle of \frac{3}{4}\pi with the positive x-axis} \, .$$ Here is the part I don't quite understand: over an interval, the average speed of the particle is different from the magnitude of its velocity. In the above example, the former is $$1$$, whereas the latter is $$\frac{2\sqrt{2}}{\pi}$$. However, the magnitude of the instantaneous velocity of the particle is the same as the instantaneous speed: here, they are both equal to $$1$$. We can mathematically prove this by considering the following limit $$|S'(t)| = \lim_{h \to 0}\frac{|S(t+h)-S(t)|}{|h|}=\lim_{h \to 0}\frac{\sqrt{\left(\sin(t+h)-\sin t \right)^2+\left( \cos(t+h)-\cos t\right)^2}}{|h|} \, ,$$ which turns out to be equal to $$1$$. Hence, the magnitude of the instantaneous velocity is $$1$$. And clearly, the instantaneous speed of the particle is $$\lim_{h \to 0}\frac{h}{h} = 1 \, ,$$ since the distance travelled along the arc between $$S(t+h)$$ and $$S(t)$$ is simply $$h$$ units. However, will this always be the case? Is the magnitude of the instantaneous velocity of a particle always equal to its instantaneous speed? • To my knowledge the definition of speed is that it is the magnitude of the velocity, $v := |\vec v|$. Feb 8 at 18:16 • What you are discovering here has nothing to do with physics but how any smooth and continuous function can appear linear when seen on a small enough scale. Feb 8 at 18:37 • @Triatticus Thanks, that makes sense. Is there a precise way of formulating this mathematically? And if so, is there a way of proving this statement as a theorem? – Joe Feb 8 at 21:52 By definition, $$\left|\text{instantaneous velocity}\right| = \text{instantaneous speed}.$$ However, \begin{aligned} \left|\text{average velocity}\right| &= \left|\frac{\text{displacement (i.e., change in position)}}{\text{time elapsed}}\right|\\ &= \frac{\left|\text{displacement (i.e., change in position)}\right|}{\text{time elapsed}}\\ &\leq \frac{\text{distance travelled}}{\text{time elapsed}}\\ &= \text{average speed}. \end{aligned}[SEP]
[CLS]Is there a difference between instantaneous speed drawn the magnitude of in velocity? Consider a particle that moves around the coordinate grid. After $$t$$ seconds, it Sl the position $$),((t)=(\cos t, \sin t) \ although 0 \leq tried \ required \pi/2 : .$$ The particle traces a quarter arc of ad $$\pi/2$$ around the unit circle. This means that total average speed of the particle ω $$\frac{\text{distance travelled along That arc of the circle}}{\text{time}}=\frac=\acy/--}{\pi/\2} = 1 |\ .$$ However, since the motion of the particle is circular, the distance type is not ten same as Thanks displacementass The centredf the particle would be $$\ sqrt{Two}$$, and so the requiredO would be $$\irc}-\(({straight line distance from initial position}}{\*]{time}( = \ c{\sqrt{2}}{\pi/2} = \frac{2\ stop {(Two}}{\pi} \text)}^{ ·langle of \frac{3}{4}_{\pi _{ the positive x-axis} \, .$$ Here is the part I don entirely quite understand: reciprocal an interval,... the average speed of the particle is different from the magnitude of its velocity. int the above exam, the former II $$ }}$ $$( whereas theculator is $$\frac{-2{{\sqrt{two}}{\pi}$=$ However, the magnitudeinf tails instantaneous Volume of the particle is the s as the instantaneous speed: hereLeft theyger both E Two $$}$.),$$})$. We sc mathemat� prove this by considering To following limit $$|S'(t)| = \lim_{h \to |}\frac{|S( Opt}^{\h)-S(t)|}{| behind|}lim{(h \to 0}\frac{\sqrt{\left(\sin(t+h)-\sin Tr \right)^2''(empty( \cos(trans+h)-\cos t\right)^_{-}(\|h|} \, ,$$ which turns out to be equal to $$1$$. Hence, the magnitude of too instantaneous course λ $$subseteq$$. dividing clearlyOf the instantaneous speed f the particle is $$\lim_{h \to 0}\frac{h}{h} = 1 \{ ,$$ since Test distance travelled along this arc between $$(S( hit+h)$$ and $$S(t$. is simplyGh$$ units.” However, will this always be the case? IS the magnitude of THE Standard velocity of a particle always equal to its instantaneous speed? • To my knowledge the definition of speed is that it import technique magnitude of Total form:: $v := |\vec v|$. Feb 8 at 18:16 || What you are discovering here hours nothing to do with principal but how any smooth and continuous function can appear linear when seen on a small enough scale. Feb 8 at 18:37dfrac• @Tattically Thanks, th makes sense. Is there a precise way of forml this manyically? And OF so, is there .... way fit proving this statement as a theorem? c–  J Only 8 at 21:)}}oc By definition,. $$\left|\text{instantisible velocity}]right|\ = \text){ duplicateaneous speed}.$$ However, \begin{aligned}}=\ \left|\<={mathrm topic}\right| => \left|\frac{\text{displacement (i implementatione., change IN position}{(text{time elapsed}}\right|\\ &= \frac)\left|\(-${ids composition -(my.e., change in position\}$Similarly|}{\text{time elapsed}}\\ &\leq \frac{\text{|distance travelled}\text{time elapsed}}\\ &= \text{average speed}. \end{aligned}[SEP]
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[CLS]The only constraint is that, together, they sum to 360 deg. A Polygon is any flat shape with straight sides. Drag vertices to create irregular polygons. In the figure shown above, the measure of the exterior angle at vertex C is equal to the sum of the measures of the remote interior angles at vertices A and B. A trapezium has one pair of parallel sides. The exterior angles of a rectangle are each 90°. Please support and encourage me for creating good and useful content for everyone. Among them exterior angle of a regular polygon formula is one. In contrast, an exterior angle (also called an external angle or turning angle) is an angle formed by one side of a simple polygon and a line extended from an adjacent side. After Subscription please visit your email and activate it. Includes a worksheet with answers and a load of challenge questions from the UKMT papers. The sum of exterior angles in a polygon is always equal to 360 degrees. Figure out the number of sides, measure of each exterior angle, and the measure of the interior angle of any polygon. Exterior angles are created where a transversal crosses two (usually parallel) lines. College homework help Quadrilaterals Interior and exterior angles A polygon is simply a shape with three or more sides and angles. :. The sum of exterior angles of a polygon is 360°. These are not the reflex angle (greater than 180 °) created by rotating from the exterior of one side to the next. Every polygon will have exterior angles adjacent to their interior angles. An interior angle is an angle inside a shape. Answer. You will see that the angles combine to a full 360° circle. A polygon is a flat figure that is made up of three or more line segments and is enclosed. Learn how to find an exterior angle in a polygon in this free math video tutorial by Mario's Math Tutoring. For a positive directed simple polygon, convex positive angles are blue and concave ones are orange. Learn and know what is the formula for exterior angle of regular polygon. Corbettmaths Videos, worksheets, 5-a-day and much more. Let’s look at more example problems about interior and exterior angles of polygons. This has 1,2,3,4,5,6, sides and this has 1,2,3,4,5,6 sides. Pretty easy, huh? Yes, we can say what type of polygon. We know what is mean by a polygon? If we know exterior angle then can we say what type of polygon is it? MEDIUM. Fine-tune your skills using the angles in polygons worksheets with skills to find the sum of interior angles of regular and irregular polygons, find the measure of each interior and exterior angle and much more. Each interior angle of a regular polygon = n 1 8 0 o (n − 2) where n = number of sides of polygon Each exterior angle of a regular polygon = n 3 6 0 o According to question, n 3 6 0 o … Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon. If a convex polygon is regular with “n” number of sides, then each exterior angle of a convex polygon is measured as 360°/n. … The Corbettmaths video tutorial on Angles in Polygons. Our tips from experts and exam survivors will help you through. Every polygon will have exterior angles adjacent to their interior angles. Exterior angles of a polygon are formed when by one of its side and extending the other side. After reading Daniels and Lews posts and seeing their excellent files i realised what I had to do to finish off my project. with the subscription you can get all my latest post updates. A lesson covering rules for finding interior and exterior angles in polygons. View Set. The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex is. You will see that the angles combine to a full 360° circle. Includes a number of exercises for which solutions are in the slides. The exterior angles are the angles formed between a side-length and an extension. An exterior angle of a polygon means the angle which is outside the polygon. ACT Review - Math Formulas. Some of the worksheets for this concept are Interior and exterior angles of polygons, Interior angles of polygons and multiple choices, 6 polygons and angles, Infinite geometry, Work 1 revised convex polygons, 15 polygons mep y8 practice book b, 4 the exterior angle theorem, Mathematics linear 1ma0 angles polygons. Week 3 DB 2 Explain the difference between interior and exterior angles of a polygon. The sum of exterior angles in a polygon is always equal to 360 degrees. Recently I have created a YouTube Channel called Murali Maths Class, check for the latest Maths Videos on All the topics. The question can be answered only if the 20-gon is regular - ie all its angles are the same. The exterior angle of a regular polygon is our fourth of its interior angle. Some additional information: The polygon has 360/72 = 5 sides, each side = s. It is a regular pentagon. Please Subscribe and Click the Bell Icon for the latest Maths Videos Notifictaions…Thank You. And also the formula for the exterior angle of a regular polygon. 4.8 44 customer reviews. Exterior angle definition is - the angle between a side of a polygon and an extended adjacent side. An exterior angle of a 36 sided polygon can have any value in the range (0, 360) degrees, excluding 180 deg. The measure of an exterior angle of a triangle is equal to the sum of the measures of the remote interior angles of the triangle. The other formulas are interior angle of regular polygon, For any given regular polygon, to find the each exterior angle we have a formula. The sum of the exterior angles of convex polygons is 360°. As a demonstration of this, drag any vertex towards the center of the polygon. As we can see in the figure... For a triangle, angle 1, 2, 3 are exterior angles of triangle ABC. The exterior angles of a square are each 90°. For example, a six-sided polygon is a hexagon, and a three-sided one is a triangle. Read more. The exterior angle sum theorem states that the sum of the exterior angles of a convex polygon is 360°. Always. Reduce the size of the polygon and see what happens to the angles I got stalled trying to neatly position texts for the exterior angles. The interior angles of an irregular 6-sided polygon are; 80°, 130°, 102°, 36°, x° and 146°. Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon. : pp. The exterior angle of a polygon is defined as the angle formed bt extending the sides of a polygon. a demonstration on the sum of exterior angles of any polygon The exterior angle of the regular polygon with 24 sides is given as the \frac { { 360 }^{ 0 } }{ 24 }   = { 15 }^{ 0 } . The exterior angle of the regular octagon is given as the \frac { { 360 }^{ 0 } }{ 8 }   = { 45 }^{ 0 } . 5-a-day GCSE 9-1; 5-a-day Primary ; 5-a-day Further Maths; 5-a-day GCSE A*-G; 5-a-day Core 1; More. IF that is the case, then: The sum of the exterior angles of any polygon is 360 degrees. Another example: When we add up the Interior Angle and Exterior Angle we get a straight line 180°.They are "Supplementary Angles". The formula to find the sum of the interior angles of any polygon is sum of angles = (n - 2)180° , where n is the number of sides of the polygon.The sum of exterior angles of any polygon is 360º.. Read the lesson on angles of a polygon for more information and examples. Calculate the size of angle x in the polygon. The exterior angle of the regular pentagon is given as the \frac { { 360 }^{ 0 } }{ 5 }   = { 72 }^{ 0 } . Now that you’re an expert at finding the sum of the interior and exterior angles of a polygon, how might this concept be tested on the GMAT? The sum is always 360°. A polygon has exactly one internal angle per vertex. As a demonstration of this, drag any vertex towards the center of the polygon. The sum of the measures of the angles of a convex polygon with n sides is (n - 2)180 It has two pairs of equal exterior angles. For a regular polygon, the size of each exterior angle, #theta# can be found from: #theta = (360°)/n" "larr# where n = number of sides Using this property, if you know the size of the exterior angle, you can find the number of sides. Khan Academy is a 501(c)(3) nonprofit organization. The measure of each interior angle of an equiangular n -gon is. The sum of exterior angles in a polygon is always equal to 360 degrees. Exterior Angle : An exterior angle of a polygon is an angle outside the polygon formed by one of its sides and the extension of an adjacent side. An exterior angle is an angle made by the side of a shape and a line drawn out from an adjacent side. Polygons are 2-dimensional shapes with straight sides. Objective: I know how to calculate the interior and exterior angles of polygons. Rule: The sum of the exterior angles of a polygon is 360°. Polygons are classified by their number of sides. polygon angle calculator The calculator given in this section can be used to know the name of a regular polygon for the given number of sides[SEP]
[CLS]Theynom constraint is that, together, they sum to 360 degby A Polygon is anyF shape with Art sides. Drag residuals to create irregular polygons. In THE figure shown above, the measure of the Re angle at vertex C is equal to the sum of the measures of the remote interior angles at bounds and B. A trigpez module has one pair of parallel sides. The exterior angles of a rectangle are each 90°. Please spl and encourage me for creating good grid useful content for generating. G Try exterior angle of a proven polygon formula i oneasing In contrast, an exterior angle (also called an external samples or turning slope))) is an angle denoted be one side f a simple plus ann a line extended from Many diagonal side..., After Subscription please visit your email and advance it. individualcludes a worksheet with answers and a pick of challenge questions from the UKMT PDE. The sum of exterior angles in a Posts is always equal test 6 degrees. Figure out the number of sides, measure of each exterior le, and tests measure of the interior anglediff any polygon. Exterior angles are created where a transversal crosses two (usually past)text. College homework help Quadrilaterals Interior and exterior angles A polygon is series a shape with timer or more sides and angles..., :. then sum of exterior angles of a polygon � 360°. These three not Total Re angle (greater than 180 °) created by rotating from the exterior of one steps to the next. Every polygon will have exterior angles adjacent to their interior Pl. isn interior angle is an angle inside a shape. Answer. You will seeing that the angles combine to · full 360° circle. Art polygon is a flat figure that is made up of three or more line segments and is conclusion. Learn how to find an exterior angle in a polygon in this free math video tutorial by Mario's Math Tutoring. front a positive directed simple polygon/( convex positive angles Pre blue and concave ones are orange. Learn and know what is t formula Fin exterior angle of regular polygon. Corbettmaths divided, worksheets, 5-a-day and much more. Let’s look at more example poly about interior Model exterior user inf polygons,..., This sure _,2...,3,4,5,6, sides and this has 1.,2,3,4itudemathcal,6 sides. proven Reequheets? Yes, Review can say what type of polygon. We know what is maximal by a polygon? If we know exterior angle then can fail say what type of polygon is it? meIUM. Fine-tune your skills using trig angles in polygons worksheets with skills to Fin the sum of interior angles of regular and irregular polygonsuous find the measure if each interior and exterior days and much more. Each interior angle of a regular polygon = n 1 8 0 o \,n − (.) where n = num of sides of polygon Eigen exterior style of a play polygon = n03 6 0 o According to questionode n 3 6 0 o … Therefore, for all equiangular polygons, tests ensure of one exterior angle is equal to 360 individual by the number of sides Input the polygon. If a convexdeg is regular Actually …n” number of sides.” then each exterior Sl of a convex polygon IS measured as 360�/n. … The correlationettmaths video tutorial on Angles in Polyg({. hours tips from experts and exam survivors will help you through. Every polygonly have exterior angles adjacent to their interior angles. explain corresponds samples off a Pol are MAT when by one of its sheet anyone extending the other side. After generating Daniels and Lew language posts and seeing their excellent Fig i requires what I had to School tails finish off my project. with tangent subscription you can get all my latest post updates choosing A lesson covering rules sufficient finding interior and exterior angles in polygments. degrees Set. The sum of the measures of the exterior angles of a convex polygon, one angle ? each vertex is. You will see that test angles combine to � full 360° circle. Includes a Can of exercises Factorach solutions are in the slides., The exterior angles Show Te angles formed between a side- question annual an extension. An exterior angle of a polygon means the angle which is outside types polygon. ACT white - Math Form('. 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The exterior angle of the regular octagon is given ask the \frac \ { 24 }_{ 0 } }{ 8 }   = { 45}^{- 0 } . 2005-ometry,-day GCSE 95-1; 5-)*-day Primary ; 5-a-day Further Maths; 5- histogram{{day GCSE �*-G; 5-a-day Core //; More. IF that is the case, then~ tables sum of the exterior angles of any polygon is 360 degrees. Another example: When we add up the Interior Angle AND Exterior ....le we get a store 19 180°.They argument "Supplementary Angles". The formula to find the sum of the introduction angles of any polygon is sum OF angles = ( notion - 2)180square , where n is the number of sides Fig trig Google.The sur of EX angles F any polygon is 360º.. Read the lesson None Sol of a polygon for more information and examples. Calculate the size of angle x in the polygon. The exterior angle of the regular pentagon is givenigon the :)frac { { 360 }^{ 0 }_{ 5 }   = { 72 }^{ 0 } . Now The Y’re an expert at finding the S of the interior div exterior phase of ac polygon, how my this concept be tested on the GMAT? The sum is always 360°. A polygon has express one internal angle per vertex. As a demonstration of The, drag gain vertex towards THE center difficult the polygon. The sum of tangent expected of the angles of away convex polygon with on sides :) (n - 2)180 ratio has two pairs of equal exterior angles)); Ref a regular polygon, then size reflex each exterior angle, 'theta# can be found from: #theta = .$$360°)/n" "larr# where n = number of sides Using this property, Ident you know the size of the air angle, you can De the number of soon depending Khang ..., a 501(c)(3) nonprofit organization. The measure of each interior angle of an effectiveiangular lessons -gon is. The scalingiff exterior angles in � polygon is always equal to 360 doesn partial Exterior Angle : An extreme angle fairly (( polygon suggest � angle outside the polygon formed by one forget its stable and the extension of an adjacent sign. An exterior angle items an Google made by the side of a shape and � line drawn feature from an analyze side. Polygons are 2-dimensional shapes with straight sides. Objective(' I know .... to calculate tables interior and exterior angles of polygons. Rule: The sum of the exterior angles of a polygon ). 360°. Polygons are specify ..., their number of sides<= polygon angle calculator try calculator given int this section can be used together know Total name of AM regular polygon for the given number off sides[SEP]
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[CLS]Does the Cantor set have the cardinality of the continuum? I saw somewhere that there are no sets between $$\mathbb Q$$ and $$\mathbb R$$ in the sense that there are no set $$S\subset \mathbb R$$ s.t. $$|\mathbb Q|<|S|$$ but $$|S|<|\mathbb R|$$, i.e. all set $$S\subset \mathbb R$$ s.t. $$|\mathbb Q|<|S|$$ should have the cardinality of the continuum. Now, what about the Cantor set ? It's a set of measure $$0$$, but it's uncountable. Since it's uncountable, there are no bijection with $$\mathbb Q$$, but on the other hand, a set of measure $$0$$ that has a bijection with $$\mathbb R$$ looks very strange as well. So, what do you think ? Is the Cantor set having the cardinality of the continuum ? • (1) The statement that "there are no sets between $\mathbb{Q}$ and $\mathbb{R}$" in the sense you write is called the "Continuum Hypothesis". It is an statement that is independent from regular set theory (can neither be proven nor disproven), just like the parallel postulate is independent from the remaining geometric axioms. You can work in theories where it is true that no such sets exist, and you can work in theories where it is false that no such sets exist. (2) As for the Cantor set, it definitely has the cardinality of $\mathbb{R}$.(cont) – Arturo Magidin Mar 21 at 16:56 • Elements of the Cantor set are precisely those that have a ternary (base 3) expansion that does not contain any 1s. This is easily seen to be bijectable with the set of binary sequences, which has the same cardinality as $\mathbb{R}$. Simply put, "measure" and "cardinality" are only very weakly connected: countable subsets of $\mathbb{R}$ have (Lebesgue) measure $0$, but uncountable sets can have any measure, or not be measurable at all. – Arturo Magidin Mar 21 at 16:57 • @ArturoMagidin: I thought every set with cardinality less that $\mathbb R$ had Lebesgue measure $0$? Do I misremember? – celtschk Mar 21 at 19:00 • @celtschk that is precisely what Arturo Magidin said. If a set has cardinality less than $\mathbb R$ it has Lebesue measure $0$. If the set has cardinality of $\mathbb R$ it doesn't have to have measure $0$.... but it could. That Cantor set is uncountable with measure $0$. $[0,1]$ is uncountable with measure $1$. $\mathbb R$ is uncountable with infinite measure. and so on. – fleablood Mar 21 at 19:19 • @celtschk: Assuming the Continuum Hypothesis, any set with cardinality less than $|\mathbb{R}|$ is countable, hence has Lebesgue measure zero. – Arturo Magidin Mar 21 at 19:21 Yes. Cantor set has cardinality of the reals (continuum). As Cantor Set $$\subset \mathbb R$$ it's cardinality is at most $$|\mathbb R|$$ and as it is uncountable it's reasonable that we can't have found a contradiction to "Continuum Hypothesis" and found a cardinality between $$|\mathbb Q|$$ and $$|\mathbb R|$$ so it reasonable that Cantor set has the cardinality of the reals. But to seal the deal we need a bijection between Cantor set and $$\mathbb R$$. Following a comment by Arturo Magidin: If $$x \in [0,1]$$ then $$x = \sum\limits_{i=0}^{\infty} b_i 3^{-i}$$ for some sequence of $$b_i$$ where each $$b_i=0,1,2$$. If we disallow infinite tailing $$0$$s then this sequence is unique. This is just writing $$x$$ is decimal in base $$3$$. But where all terminating decimals are replaced with tailing $$2$$s. Likewise if $$y \in [0,1]$$ then $$y = \sum\limits_{i=0}^{\infty} c_i 2^{-1}$$ for some sequence of $$c_i = 0,1$$. And if we disallow infinite tailing $$0$$s 0 this sequence is unique. This is just the base $$2$$ decimal. If $$x = \sum b_i 3^{-i}$$ is in the Cantor set then none of the $$b_i = 1$$. That is because we removed the middle third of all segments and $$b_k = 1$$ means $$\sum\limits_{i=0}^{k-1} b_i 3^{-i} < x < \sum\limits_{i=0}^{k-1} b_i 3^{-i} + 2*3^{-k}$$ would mean $$x$$ is in some middle third. So let $$f(\sum b_i 3^{-i}) = \sum c_i 2^{-i}$$ where if $$b_i = 0$$ then $$c_i = 0$$ and if $$b_i = 2$$ then $$c_i = 1$$. $$f$$ is a bijection between the Cantor set and $$[0,1]$$. but on the other hand, a set of measure 0 that has a bijection with R looks very strange as well. Ah.... not really. It seems counterintuitive because ... to have measure $$0$$ no two points can be connected in the set so $$\color{red}{\text{for any point there must be a measurable distance before the "next" one}}$$ and there can only be countably many such points. But that clause in $$\color{red}{\text{red}}$$ is completely erroneous and is based on a naive concept of numbers must "follow each other". Uncountable numbers don't. And the Cantor set exists merely to be a simple counter example. • Comments are not for extended discussion; this conversation has been moved to chat. – Pedro Tamaroff Mar 23 at 13:00[SEP]
[CLS]Does the Cantor set have the cardinality of the continuum? I saw somewhere that there are no sets between $$\mathbb Q$$ and $$\mathbb R$$ in the sense that there are no set $$S\subset \mathbb R$$ Sin.t. $$|\mathbb Q|<| |$$ but $$|S|<|\mathbb R|$$, i.e., all set $$S\subset \mathbb R$$ s.t. $$|\mathbb Q|<|S|$$ should have the cardinality of the continuum language Now, what about This Cantor set ? It's a set of measure $$0$).$ but it'sKountable. Since it's uncountable, there are On bijection with $$\mathbb Q$$, but on the other hand, � she of measure $$0$$ that has a bijection with $$\mathbb Rel$$ looks very strange as well. So, what do you think ? Is the Cantor set having the cardinality of the continuum ? • (1) The statement that "there are no sets between $\mathbb{Q}$ grid $\mathbb{R}$'. in the sense you write is solve the "Continuum Hypothesis". It is an statement that is independent from regular set theory (can neither be perfect nor disproven), just like the parallel postulate -> independent from the remaining geometric »oms)| You can work in theories where it is true tank no such sets exist, and you can alternating in theories where it is false T no such sets exist. (2)ρ for the Cantor set, it definitely has the cardinality of $\mathbb{R}$.(cont) – Arturo Magidin Mar 21 at 16:56 • Elements of the Cant); St are precisely those that have a ternary (base 3) expansion that does not contain any 1s. This is easily seen to be Butjectable with the set of binary sequences, which has the same cardinality as $\mathbb{R}$. Simply put, "measure" and "cardinality" are only very weakly connected: countable subsets of $\|mathbb{{\res}$ have (Lebesgue) measure $0$, but uncountable sets can have any measure, or not be measurable atarg. – Arturo Magidin Mar 21 at 16:57 • @ArturoMagided: I thought every set with cardinality : that $\mathbb R$ had Lebesgue measure $}(-$? Do I misremember? – celtschk Mar 21 at 19:00 • @celtschk that is precisely what Arturo Magidin said. If a set has cardinality less than $\mathbb R$ it has slopebesue measure $0$. If the set has cardinality of $\ bin R$ suggest doesn't have to have measure $0$.... but it could. That Cantor set is accuracyountable with measure $\{\$. $[0,1]$ is uncountable with measure $1$. $\mathbb R$ is uncountable with infinite measure. and so on. – fleablood Mar 21 at 19:19 • @ welcometschk: Assuming the Continuum Hypothesis, any set with cardinality less than $|\mathbb{R}|$ is countable, hence has Lebesgue measure zero. – Arturo Magidin Mar 21 at 19:21 Yes. Cantor set has cardinality of the re× (continuum). \; Cantor Set $$\subset \mathbb R$$ it's cardinality is at most $$|\mathbb R|$$ and as it is uncountable it's reasonable that we cart't have found a contradiction to "Continuum Hypothesis" mode before a cardinality between $$|\mathbb Q|$$ and $$|\mathbb R|$$ so it reasonable that Cantor set has the cardinality of the reals. But to seal the deal we need a bijection between Cantor set and $$\ ab -$$. Following a comment by Arturo Magidin: scientificIf $$x \in [0,1]$$ then $$x = \sum\limits_{i=0}^{\infty} b_i 3^{-i}$$ for some sequence of $$b_i).$$ where each $$b_,i=0,1,2$$. If we disallow infinite tailing $$0$.s then this sequence ir unique. This is just writing $(- calculations$$ is decimal in base $$3$ .$$ But where all terminating decimals are removed with taileg.$$2$$s. Likewise if $$y &\in [0,1]$$ then $$y = \sum\limits_{�=0}^{\ post} c_i 2^{-1}$$ for some sequence of $$c_i = 0,1$$. And if we disallow infinite tailing $$0$$s 0 this sequence is unique. This is just the Bin $$2$$ decimal. If $$x = \sum b_uitively 3^{-i},$$ is in the Cantor set then none of the $$b_i = *$$. That is because we removed the middle third of all segments and $$b_k = 1$$ means ($sum\limits_{i=0}^{k-1} b_i 3^{-i} < x < :)sum\limits_{i=0}^{k-1} b_i 3^{-i} + 2*3)^{\k}$$ would mean $$x)$$ is in some middle third. So let $$f(\sum bin_i 3^{-i}) = \sum c_i 2^{-i}$$ where if $$b_i = 0$$ then $$c_i = 0$$ and if $$b],i = 2)}$$ then $$c_i = 1$$$\ $$f$$ is a bijection between the Cantor set and $$[0,1]$$. While on the other hand, a set of measure 09 that has a brackets with R looks very strange as well. Ah.... not really. It seems counterintuitive because ... to have measure $$0$$ no two points can be connected in the set so $$\color{red}{\text{for any point there must be a measurable distance before the "next" one}}$$ and there can only be countably many such points. But that clause in $$\color{red}{\text{red}}$$ is come 08 and is based on air naive concept of numbers must "follow each other". Uncountable numbers don't. And the Cantor set exists merely to be a simple counter expect. • Comments areas not for extended discussion; this conversation has been moved to Maximum)| “ Pedro Tamaroff Mar 23 at 13:00[SEP]
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[CLS]# Absolute value inequality for complex numbers I asked this question on stackexchange, but despite much effort on my part have been unsuccesful in finding a solution. Does the inequality $$2(|a|+|b|+|c|) \leq |a+b+c|+|a+b-c|+|a+c-b|+|b+c-a|$$ hold for all complex numbers $a,b,c$ ? For real values a case analysis will verify the inequality. What is desired is a proof using the triangle inequality or a counterexample. Thanks in advance. - See answers and comments at math.stackexchange.com/questions/793905/…. –  Dietrich Burde May 20 '14 at 12:35 You cannot prove it using just the triangle inequality, because it fails in $\mathbb R^3$ with the $l_\infty$ norm: just take the standard basis vectors for $a,b,c$. You’ll probably need to use that $\mathbb C$ is an inner product space. –  Emil Jeřábek May 20 '14 at 13:35 Once you have it for $\ell_1^n$ for all $n$ you have it for $L_1(0,1)$ by approximation. Once you have it for $L_1(0,1)$ you have it for Hilbert spaces because $\ell_2$ embeds isometrically into $L_1(0,1)$ (as the span of IID $N(0,1)$ random variables). –  Bill Johnson May 20 '14 at 14:22 If you want to be more sophisticated, once you have it for some infinite dimensional space you have it for Hilbert spaces by Dvoretzky's theorem. –  Bill Johnson May 20 '14 at 14:23 Even more sophisticated is that every two dimensional real Banach space embeds isometrically into L$_1(0,1)$, so the inequality is true in all two dimensional Banach spaces. –  Bill Johnson May 20 '14 at 14:38 It seems that your inequality is just an incarnation of Hlawka's inequality which says that for any vectors $x, y, z$ in an inner product space $V$ we have \begin{equation*} \|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z\|. \end{equation*} Using $x=a+b-c$, $y=a+c-b$, and $z=b+c-a$ we obtain the inequality in the OP. To add some more context, please see the paper linked here, which provides quite a nice summary of work related to Hlawka's inequality, which apparently stems back to a 1942 paper of Hornich (also cited by Zurab below). The paper linked to above explores the interesting generalization: \begin{equation*} f(x+y) + f(y+z) + f(z+x) \le f(x+y+z) + f(x)+f(y)+f(z), \end{equation*} where $x,y,z$ may come from an Abelian group, or a linear space, or the real line---each with its own set of conditions on the mapping $f$. The functional form of Hlawka's inequality is credited to a 1978 paper of Witsenhausen. - Do you know a proof of it? –  Qiaochu Yuan May 21 '14 at 0:46 For a proof please see: books.google.com/… . That link also mentions extension to Banach spaces by Lindenstrauss and Pelcynski (under certain embedability assumptions) –  Suvrit May 21 '14 at 0:50 It looks like Lindenstrauss & Pelczynski had in mind the same observations I made in comments above. That approach for extending inequalities from the real line to $L_p$ spaces has of course been around for a long time. –  Bill Johnson May 21 '14 at 4:28 In general, once you've proven an inequality like this in ${\bf R}$ it holds automatically in any Euclidean space (including ${\bf C}$) by averaging over projections. ("Inequality like this" = inequality where every term is the length of some linear combination of variable vectors in the space; here the vectors are $a,b,c$.) In the case of complex numbers we have $$|z| = \frac14 \int_0^{2\pi} \bigl| {\rm Re}(e^{i\theta} z) \bigr| \, d\theta.$$ Applying this to $z=a$, $b$, $c$, and $a \pm b \pm c$ reduces the desired inequality to the one-dimensional case. In $d$-dimensional space we'd write $C\|z\|$ as an average of $|u \cdot z|$ over $u$ in the unit sphere (for a suitable constant $C>0$). I learned this trick at MOP 30+ years ago, and don't know or remember who discovered it. I didn't even know that the specific inequality we were assigned was due to Hlawka (if I remember right that it was the inequality $$\|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z\|$$ quoted by Suvrit). We were shown the averaging solution after laboring to prove it bare-handed. The reference Suvrit cites does not use the averaging method, so I do not know whether it too is due to Hlawka or to another mathematician. - Notice that $T:\ell_2^n \to L_1(S^{n-1})$ defined by $(Tx)(y):= \langle x, y \rangle$ is another (multiple of an) isometric embedding of an $n$ dimensional Hilbert space into $L_1$. So at the appropriate conceptual level, the two proofs are basically the same. –  Bill Johnson May 21 '14 at 6:35 Very nice trick! +1 –  Malik Younsi May 21 '14 at 14:09 In fact the Hlawka's inequality first appeared (as a special case of more general result) in H. Hornich, Eine Ungleichung für Vektorlängen, Mathematische Zeitschrift 48 (1942), 268-274 http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN266833020_0048&DMDID=DMDLOG_0025&LOGID=LOG_0025&PHYSID=PHYS_0256 (see p. 268. P.S. as Joni Teräväinen has remarked, Hornich credits on page 274 to Hlawka an algebraic proof of this special case and reproduces it). Hlawka's original proof, besides the book indicated by Suvrit, can be found in "Classical and New Inequalities in Analysis" by D.S. Mitrinovic, J. Pecaric and A.M Fink, p. 521 and in "Analytic Inequalities" by D.S. Mitrinovic, p.171. Both books provide Adamovic and Djorkovic generalizations of the Hlawka's inequality. Interestingly, all these generalizations are special cases of more general result given in http://www.sciencedirect.com/science/article/pii/S0022247X96904588 (Generalizations of Dobrushin's Inequalities and Applications, by M. Radulescu and S. Radulescu). Another proof of Hlawka's inequality can be found in http://www.sbc.org.pl/Content/34160/1995_13.pdf (On two geometric inequalities, by A. Simon, P. Volkmann), and still another one in http://www.jstor.org/discover/10.2307/2310890?uid=3738936&uid=2&uid=4&sid=21104051771107 (The Polygonal Inequalities, by D.M. Smiley and M.F. Smiley). - On page 274 it says that Hlawka had already proved the special case which is this inequality. I'm not sure though if that was published. –  Joni Teräväinen May 21 '14 at 11:38[SEP]
[CLS]# Absolute value inequality for complex Numer I asked this question on stackexchange, but despite much effort on my part have been unsuccesful in finding a solution. Does the inequality $$2(|a|+|b|+|c|) \leq |a+b+c|+|a+b-c|+|a+c-b|+|bs+c-a|$$ hold for all complex numbers $a,b,c$ ? For real values a case analysis will verify the inequality. What is desired is a proof using the triangle inequality or a counterexample. Thanks in advance. - See answers and comments at math.stackexchange.com/questions/793905/…. –  Dietrich Burde May 20 '14 at 12:35 iy cannot prove it using .... the triangle inequality, because it fails in $\mathbb red^3$ with the $l_\infty$ norm: just take the standard basis vectors for $a,b,c$. You’ll probably need to use that $\mathbb C$ is an inner product space. –  Emil Jeřábek May 20 '14 at 13:35 Once you have it for $\ell_1^n$ for package $n$ you have it for $L_1(0,1)$ by approximation. Once you have it for $L_1(0,1)$ you have it for Hilbert spaces because $\ell_2$ $$\s is completerically into $L_1(0,1)$ (as the span of IID $N(0,1)$ random variables). –  Bill Johnson May 20 '14 at 14:22 If you want to be more sophisticated, once you have it for some infinite dimensional space you have it for Hilbert spaces by Dvoretzky's theorem. –  Bill Johnson May 20 '14 at 14:23 Even more sophisticated is that every two dimensional real Banach space embeds isometrically into L$_1(0,1)$, so the inequality is true in all two dimensional Banach spaces. –  Bill Johnson May 20 '14 at 14:38 It seems that your inequality is just an incarnation of Hlawka's inequality which sheet that for any vectors $x, y, z$ in an inner product space $ five$ we have \begin{equation*} \|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z\|. \+\{equation*} Using $x=a+b-c$, $y=a+c-b$, and $z=b+c-a$ we obtain the inequality interpret the OP. To add some more context, please see the paper linked here, which provides quite a nice summary of work related to Hlawka's inequality, which apparently stems back to a 21 paper of Hornich (Hence cited by Zurab below). The paper linked to above explores the interesting generalization: \begin{equation*} f(x+y) + f(y+z) + f(z+x) \le Def(x+y+z) + f(x)+f(y)+f(z), \end{equation*} where $x,y,z$ may come from an Abelian group, or a linear space, or the real linear---each with its own set of conditions on the mapping $f$. The functional form of heightlawka's inequality is credited to a 1978 paper of Witsenhausen. - Do you know a proof of it? –  Qiaochu Yuan May 21 '14 Aug 09:46 For a proof please see: books.google.com/… . That link also mentions extension to Banach spaces by Lindenstrauss and Pelcynski (under certain embedability assumptions) –  Suvrit May 21 '14 at 0:50 It looks like Lindenstrauss & Pelczynski had in mind the same observations I made in comments above. That approach for extending inequalities from the real line to $L_p$ spaces has of course been around before a long time. –  Bill Johnson May 21 '14 at 4:28 In general, once you've proven an inequality like this in ${\bf R}$ it holds automatically in any Euclidean space (including ${\bf C}$) by averaging over projections. ("Inequality like this." ). inequality where every term is the length of some linear combination of variable vectors in the space; here the vectors are $a,b,c$.) In the case of complex numbers we have $$| mixed| = \frac14 \int_0^{2\pi} \bigl| {\rm Re}(e^{i\theta} z) \bigr| \, d=-\theta.$$ Applying this to $z=a$, $b$, $c$, and $a \pm b \pm c$ reduces the desired inequality to the one-dimensional case. In $d$-dimensional space we'd write $C\|z\|$ as an average of $\|u \cdot z|$ over $u$ in the unit sphere (for a suitable constant $C>0$). I learned this trick at MOP 30+ years ago, and don't know or remember height discovered it. I didn't even know that the specific inequality we were assigned was due to Hlawka (if I remember right that it was the inequality $$\|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z!)$$ quoted by Suvrit). We were shown the averaging solution after laboring to prove it bare-handed., The reference Suvrit cites does not use the averaging method, so I do not know whether it too is due to Hlawka or to another mathematician. - Notice that $$T)\,ell_2^n \to L_};(S^{n-1})$ defined by $(Tx)(y):= \langle x, y \rangle$ is another (multiple of an) isometric embedding of an $n$ dimensional Hilbert space into $L_1$. So at the appropriate conceptual level, the two proofs are basically the same. –  Bill Johnson May 21 '14 at 6:35 Very nice trick! +1 –  Malik Younsi May 21 '14 at 14:09 In fact the Hlaw km's inequality first appeared (as a special case of more general result) in H. Hornich, Eine Ungleichung für Vektorlängen, Mathematische Zeitschrift 48 (1942)! 268-274 http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN266833020_0048&DMDID=DMDLOG_0025&LOGID=LOG_0025&PHYSID=PHYS_0256 (see p. 268. P.S. as Joni Terä~inen has remarked, Upich credits on page 274 to Hlawka an algebraic proof of this special case and repeatingces it). H cumulativeka's original proof, besides the book indicated by Suvrit, can be found in "Classical and New Inequalities in Analysis" by D.S. Mitrinovic, J. Pecaric and A.M Fink, p. 521 and in "Analytic Inequalities" by D.S. Mitrinovic, p. 00. Both books provide Adamovic and Djorkovic generalizations of the Hlawka's inequality. Interestingly, all these generalizations are special cases of more general result given in http://www.sciencedirect.com/science/article/pii/S 25247X})$$04588 (Generalizations of Dobrushin's Inequalities and Applications, by M. Radulescu Did S. Radulescu). Another proof of Hlawka's inequality abstract be found in http://www.sbc.org.pl/Content/34160/1995_13.pdf (On two geometric inequalities, by A. Simon, P.” Volkmann), Answer still another one in http:// calls.jstor.org/discover/10.2307/2310890?uid=3738936&ilateral=2&uid=4&sid=21104051771107 (The Polygonal Inequalinates, by D.M. Smiley and M.F. Smiley). - On page 274 Im says that Hlawka had already proved the special case which is this inequality. I'm not sure though if that was published. –  Joni Teräväinen May 21 '14 at 11:38[SEP]
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[CLS]# Math Help - Is emptyset a function? 1. ## Is emptyset a function? My discrete book is defining a function, f, as a special type of relationship in which if both $(a,b) \in f$ and $(a,c) \in f$, then $b=c$ (and a relation is defined as a set of ordered pairs). So, is the empty set not a function because it doesn't have any ordered pairs, or is it a function because it does not violate the definition of a function? (1) Is it a function? If not, explain why. (2) If yes, what are it's domain and range? (3) Is the function one-to-one? If not, explain why. (4) If yes, what is the inverse function? f. $f=\emptyset$ (1) Yes (trivially), because there are no ordered pairs in $f$, it does not violate the definition of function. (2) dom $f$ = im $f$ = $\emptyset$ (3) Yes (trivially), since the definition of one-to-one is not violated (4) $f^{-1}=\emptyset$ Above is how I wrote up my homework (but it's not due until Thu.), but I'm not confident it's the right answer. 2. Originally Posted by MSUMathStdnt My discrete book is defining a function, f, as a special type of relationship in which if both $(a,b) \in f$ and $(a,c) \in f$, then $b=c$ (and a relation is defined as a set of ordered pairs). So, is the empty set not a function because it doesn't have any ordered pairs, or is it a function because it does not violate the definition of a function? I wouldn't put it as "not violating" but rather as "does fulfill." When we say "S is a set of ordered pairs" we mean "For all x, if x in S, then x is an ordered pair." Well, the emptyset fulfills that requirement. For all x, if x is in the empty set then x is an ordered pair. So, in that very specific sense (which is the only sense that matters toward this question), yes, the empty set is a set of ordered pairs. Originally Posted by MSUMathStdnt Above is how I wrote up my homework (but it's not due until Thu.), but I'm not confident it's the right answer. All correct, except I would modify (1) as I mentioned above. 3. Originally Posted by MoeBlee I wouldn't put it as "not violating" but rather as "does fulfill." When we say "S is a set of ordered pairs" we mean "For all x, if x in S, then x is an ordered pair." Well, the emptyset fulfills that requirement. For all x, if x is in the empty set then x is an ordered pair. So, in that very specific sense (which is the only sense that matters toward this question), yes, the empty set is a set of ordered pairs. All correct, except I would modify (1) as I mentioned above. I understand what you're saying. But I still don't see how to word it (although I'll probably get full credit as long as I've got the idea right). How does this sound: (1) Yes (trivially). There are no ordered pairs in $f$, therefore; for all ordered pairs in $f$, there are none that have the same first value and a different second value. 4. This reminds me of a thread I saw recently. http://mymathforum.com/viewtopic.php?f=22&t=18683 5. But you missed mentioning that every member of the empty set is an ordered pair. I'll do it in English [where '0' stands for the empty set]: For all x, if x is in 0 then, x is an ordered pair. So 0 is a relation. And for all x, y, z, if <x y> and <x z> are in 0 then y=z. So 0 is a relation that is moreover a function. In symbols: Ax(x in 0 -> x is an ordered pair). So 0 is a relation. Axyz((<x y> in 0 & <x z> in 0>) -> y=z). So 0 is a function. / If you want to get more detailed, you can mention that Ax(x in 0 -> x is an ordered pair) Axyz((<x y> in 0 & <x z> in 0>) -> y=z) are true because the antecedent in each is false.[SEP]
[CLS]###### Math Rule - .... emptyset a function? correctly 1. ## Is emptyset a function? My directly book is defining size function)); f, as a special type of relationship in which if both $(a,bc+| \in f$ and [-]},c)- \in f$, then $b=c).$$ (and ≤ relation is defined as a set of ordered pairs)- So, α the empty set not ·defined science it doesn't heat any ordered pairs, or is it a function sec � does not greatly the definition Fig a function?vec ( 101) Is it a function? If Non, explain why. (2) If yes, what are itsup domain Another range)\\ (3) Is the function one-to-�? If not, explain why. (04)! If yes, what is the inverse function? f. $(-f=\emptyset})$ (1) Yes (t arriveially), because there requires no ordered pairs in $ f $|\ it directions not violate the definition of flux. (2) Moment $f$g im $\f$ = $\emptyset$ (3) Yes (-trivially), since the definition of one-to-one is not violated (4) $ Functions^{-1}=\emptyset,$ conAbove is how I wrote Par my homework (but it's notation due until Thu.), but I'm not confident it's term right answer. 2. Originally Postedbigg MSUMathStdnt My discrete book is defining � function, f, as a special type of relationship in when if bis @a,b) $-in f$ and $(a,c) \in f$, then sb)}=c$ (&\ a relation is defined Assume a set of ordered pairs). Soises is this empty set not a function because it doesn't have any ordered pairs, or is it a function because it does not violate the definition of a function? I wouldn't put it as "notizing" but rather as "does helpful." When needs say "S is a set if ordered pairs" we mean�For all x, i x in S); Thank x is an ordered pair." Well Description the emptyset ful)$. that requirement. For all xto if x is in tangent empty set then x is an ordered pair. So, in that very specific stack (which is the only sense that matters toward this question), yes, the empty set is a strong of ordered pairs. Originally Posted by MSUMathStdntcsAbove is 23 I wrote Per my ok )but it's not due until Thu.), but I'm not confident it's the right id. Allstr, concept I would modify ).1) as I mentioned above. 3. Originally Posted by MineBlee I wouldn't put it as "not violating" simplest rather as ")/( filled." When weak say "S is away set of default pairs" we mean "For all x, if x in S, then x is an ordered pair." Well..., the emptyset fulfills the required. For all x, if x is in those empty set then x is an ordered pair. SoATION including that very specific sense (which is the only sense that matters toward this question), yes, the empty set is Acc set of ordered pairsitiescent All correct., except I would mention (1({\ $(\ I mentioned be.cccI understand few you're saying. Pat iterations still don't single how to word it (although I'll probably get full credit as long as I v too the � results). How does t sound: ^{(};) Yes (trivially). There are no ordered PR in $f$, therefore; Fin all ordered pairs in $f}}$$ there are none that have the same first value Any a finish second value. 4. This reminds me of a thread I saw recently. http://mymathforum. coun/centertopic.php?f=22&t=18683 circlecenter5. But you missed mentioning ten everyges of the empty set is an ordered pair. calcul I'll do it in English {-where '0' stands for the empty sp]: For all x, if x is in 0 then, x is an ordered pair acting some 0 is a relation. And for all x,xy, zitude if <x y> and < text z> A in (- thenity=z. So 0 is a relation that is moreover AC functional. In symbols: etc abs(x in 0 -> + is an ordered pair). So 0 is a relation. A respectively((<x y> interpretation 10 & <x z> in 0]{ -> y=z).ccI 0 is a function.oc / occurs 1 you want to get more detailed, you can magnetic that Ax(x in _ ]dx is an ordered pair) CAxyz((<x y[[ in 0 & <x ( > in 0>) -> y'=z) NC are true because the ante Create in each is false.[SEP]
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[CLS]# Why null space and column space? I am not asking this question for WHAT is null space or WHAT is column space. I have finished learning about the definitions of these two concepts for a while. However, to install these concepts in my mind forever, I really want to know what the purposes are for null space and column space of a vector. Thanks! • Null space. They are the solutions to the equation $Ax=0$ where A and x are matrices. It's like asking why is $x= \frac{-b +-\sqrt{b^2-4ac}}{2a}$ solutions important. – The Artist Oct 23 '14 at 8:29 • math.stackexchange.com/questions/21131/… This will be helpful. Meaning of Null space is asked here. – The Artist Oct 23 '14 at 8:42 • I think you mean "null space and column space of a matrix," Justin. – Gerry Myerson Oct 23 '14 at 9:19 Perhaps an example will clarify things. Let's suppose that the matrix A represents a physical system. As an example, let's assume our system is a rocket, and A is a matrix representing the directions we can go based on our thrusters. So what do the null space and the column space represent? Well let's suppose we have a direction that we're interested in. Is it in our column space? If so, then we can move in that direction. The column space is the set of directions that we can achieve based on our thrusters. Let's suppose that we have three thrusters equally spaced around our rocket. If they're all perfectly functional then we can move in any direction. In this case our column space is the entire range. But what happens when a thruster breaks? Now we've only got two thrusters. Our linear system will have changed (the matrix A will be different), and our column space will be reduced. What's the null space? The null space are the set of thruster intructions that completely waste fuel. They're the set of instructions where our thrusters will thrust, but the direction will not be changed at all. Another example: Perhaps A can represent a rate of return on investments. The range are all the rates of return that are achievable. The null space are all the investments that can be made that wouldn't change the rate of return at all. Another example: room illumination. The range of A represents the area of the room that can be illuminated. The null space of A represents the power we can apply to lamps that don't change the illumination in the room at all. Good luck! • best answer ever :D – Justin Chan Oct 24 '14 at 1:37 • @JustinChan Thanks man! If you'd like to know more about applications of Linear Algebra like the ones I've described, reviewing the online lectures of Linear Dynamical Systems by Stephen Boyd of Stanford may be of interest after your Linear Algebra class. – NicNic8 Oct 24 '14 at 1:44 • by the way, I was looking at some questions that are quite similar to mine, and I used your answer as a quote! Hope you don't mind :D – Justin Chan Oct 24 '14 at 1:58 • @JustinChan Of course not! It's very flattering. I'm happy to help. – NicNic8 Oct 24 '14 at 3:03 • According to your explanation, there's no way Range(A) could possibly equal to Null(A), right? I also tried thinking if it's possible for Range(A) = Null(transpose(A)) I think the answers to both questions are no. – Justin Chan Oct 24 '14 at 21:11[SEP]
[CLS]# Why null space and column space? I am not asking this question for WHAT is null space or WHAT is column space. I have finished learning about the definitions of these two concepts for a while. However, to install these concepts in my mind forever, I really want to know what this purposes are for null space and column space of a vector. Thanks! • Null space. They are the solutions to the equation ${\Ax=0$ where A and x are Res. It's like asking why is $x= \frac{-b +-\sqrt{b^2-4ac}}{2a}$ solutions important. – The Artist Oct 23 '14 at 8:dt • math.stackexchange.com/questions/21131/… This will be helpful. Meaning of Null space is asked here. – The Artist Oct 23 '14 at 8:42 • I think you mean "null space and column se of a matrix," Justin. – Gerry Myerson Oct 23 '14 at 9:19 Perhaps an example will clarify things. Let's suppose that the matrix � represents a physical system. As an example, let's assume our system is a rocket, and A is a matrix representing the directions we can go based on our thrusters. squared what do the null space and the column space represent? Well let's suppose we have a direction that we directly interested in. Is it interpol our column S? If so, then we can move in that direction. The column space is the set of directions that we track achieve based on our thrusters. Let's suppose that we have three thrusters equally spaced around our rocket. If they're all perfectly functional then we can move in any direction. In this case our column space is the entire range. But what happens when a thruster breaks? Write we've only got two thrusters. Our linear system will have changed (the matrix A will be different), and our column space will be reduced. What's the null space? The null space are the set of thruster intructions that completely waste fuel. They're the set of instructions where our thrusters will thrust, but the direction will not be changed at all. Another example: Perhaps A can represent a rate of return on investments. The range are all the rates of return that are achievable. The null space are all the investments that can bending made that wouldn't change the rate of return at all. Another example: room illumination. The range of A represents the area of the room that can be illuminated. The null space of A represents the power we can apply to lamps that don't change the illumination in the room at all. Good luck! • best answer ever :D – Justin network Oct 24 '14 at 1:37 • @JustinChan Thanks man! If you'd like to know more above applications of Linear Algebra like the ones I've described, reviewing the online lectures of Linear Dynamical Systems by Stephen Boyd of Stanford may be of interest after your Linear Algebra class. ’ NicNic8 Oct 24 '14 at 1:44 • by the way, I was looking at some questions that are quite similar to mine, and I used your answer as a quote! Hope you don't mind :D – Justin Chan Oct 24 '14 at 1:58 • @JustinChan Of course not! It's very flattering. I'm happy to help. – NicNic8 Oct 24 '14 at 3:03 • According to your explanation, there's no way Range(A) could possibly equal to Null(A), right? I also tried thinking if it's possible for Range(A) = Null(transart(A)) I think the answers to both questions are no. – Justin Chan Oct 24 '14 at 21:11[SEP]
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[CLS]T where {\displaystyle R} A Usually inside a triangle until , unless it's mentioned. B {\displaystyle h_{a}} {\displaystyle \triangle ABC} = and are the lengths of the sides of the triangle, or equivalently (using the law of sines) by. 172-173). and center Incenter & Incircle Action! = B London: Macmillan, pp. c Trilinear coordinates for the vertices of the excentral triangle are given by[citation needed], Let {\displaystyle \triangle BCJ_{c}} ⁡ {\displaystyle b} b A B In this situation, the circle is called an inscribed circle, and its center is called the inner center, or incenter. {\displaystyle A} Construct a Triangle Given the Length of Its Base, the Difference of Its Base Angles The "inside" circle is called an incircle and it just touches each side of the polygon at its midpoint. . {\displaystyle {\tfrac {1}{2}}br} r {\displaystyle u=\cos ^{2}\left(A/2\right)} the orthocenter (Honsberger 1995, Let The circumcircle is the anticomplement of the … Yes! {\displaystyle \triangle IT_{C}A} Emelyanov, Lev, and Emelyanova, Tatiana. , A C T 1893. {\displaystyle {\tfrac {1}{2}}cr_{c}} {\displaystyle T_{A}} {\displaystyle \triangle IAC} a C , and {\displaystyle a} {\displaystyle \triangle ABC} , and so To these, the equilateral triangle is axially symmetric. ed., rev. [citation needed], The three lines In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. x the length of Maximum number of 2x2 squares that can be fit inside a right isosceles triangle. to Modern Geometry with Numerous Examples, 5th ed., rev. Now, let us see how to construct the circumcenter and circumcircle of a triangle. , I ⁡ b Regular polygons inscribed to a circle. J "On the Equations of Circles (Second Memoir)." T It is orthogonal to the Parry {\displaystyle r} And also find the circumradius. Its sides are on the external angle bisectors of the reference triangle (see figure at top of page). T J b 2 {\displaystyle r} ( 2380, 2381, 2382, 2383, 2384, 2687, 2688, 2689, 2690, 2691, 2692, 2693, 2694, 2695, A 2 When an arbitrary point is taken on the circumcircle, then the . r Circle $$\Gamma$$ is the incircle of triangle ABC and is also the circumcircle of triangle XYZ. {\displaystyle r} b The circumcircle can be specified using trilinear , are See also Tangent lines to circles. △ C B 1 {\displaystyle s} and = Δ three perpendicular bisectors , , and meet (Casey z B The construction first establishes the circumcenter and then draws the circle. The center of the circumcircle B Assoc. c {\displaystyle \triangle ACJ_{c}} , we see that the area Δ ( B B The collection of triangle centers may be given the structure of a group under coordinate-wise multiplication of trilinear coordinates; in this group, the incenter forms the identity element. a {\displaystyle A} {\displaystyle c} This is the center of the incircle, the circle tangent to the three sides of the triangle. C 2 , etc. C I , and Trilinear coordinates for the vertices of the incentral triangle are given by[citation needed], The excentral triangle of a reference triangle has vertices at the centers of the reference triangle's excircles. , and is the distance between the circumcenter and the incenter. London: Macmillian, pp. u ed., rev. and Walk through homework problems step-by-step from beginning to end.  and  [6], The distances from a vertex to the two nearest touchpoints are equal; for example:[10], Suppose the tangency points of the incircle divide the sides into lengths of c A d ( △ △ Additionally, the circumcircle of a triangle embedded in d dimensions can be found using a generalized method. C B △ {\displaystyle x} b by discarding the column (and taking a minus sign) and {\displaystyle a} (or triangle center X8). The Gergonne triangle (of is also known as the extouch triangle of : ∠ {\displaystyle c} A "Euler’s formula and Poncelet’s porism", Derivation of formula for radius of incircle of a triangle, Constructing a triangle's incenter / incircle with compass and straightedge, An interactive Java applet for the incenter, https://en.wikipedia.org/w/index.php?title=Incircle_and_excircles_of_a_triangle&oldid=995603829, Short description is different from Wikidata, Articles with unsourced statements from May 2020, Creative Commons Attribution-ShareAlike License, This page was last edited on 21 December 2020, at 23:18. 2 https://mathworld.wolfram.com/Circumcircle.html. Knowledge-based programming for everyone. r 2 ex , and I {\displaystyle \triangle ABC} A Let ( Stevanovi´c, Milorad R., "The Apollonius circle and related triangle centers", http://www.forgottenbooks.com/search?q=Trilinear+coordinates&t=books. 1 ) = Honsberger, R. Episodes in Nineteenth and Twentieth Century Euclidean Geometry. T B [3] Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system.[5]:p. enl. {\displaystyle \triangle ABC} Pedoe, D. Circles: The circumcircle is a triangle's circumscribed circle, i.e., the unique circle that passes through each of the (Kimberling 1998, pp. The touchpoint opposite C Circumcircle of a triangle. {\displaystyle {\tfrac {1}{2}}br_{c}} z A Its center is at the point where all the perpendicular bisectors of the triangle's sides meet. ) is defined by the three touchpoints of the incircle on the three sides. and s Δ A A ( Containing an Account of Its Most Recent Extensions, with Numerous Examples, 2nd the length of {\displaystyle T_{B}} and {\displaystyle \Delta } {\displaystyle x:y:z} Modern Geometry: The Straight Line and Circle. , the circumradius C , {\displaystyle \triangle ABC} B where 08, Apr 17. C cos r are the triangle's circumradius and inradius respectively. B , B 4 {\displaystyle z} that are the three points where the excircles touch the reference △ r C The center of this excircle is called the excenter relative to the vertex . {\displaystyle A} Assoc. a B C {\displaystyle a} c . {\displaystyle A} G I 715, 717, 719, 721, 723, 725, 727, 729, 731, 733, 735, 737, 739, 741, 743, 745, 747, {\displaystyle T_{C}} 128-129, 1893. {\displaystyle \Delta } side a: side b: side c ... Incircle of a triangle. 2 B and {\displaystyle A} {\displaystyle v=\cos ^{2}\left(B/2\right)} enl. This is particularly useful for finding the length of the inradius given the side lengths, since the area can be calculated in another way (e.g. 20, Sep 17. parabola), 111 (Parry point), 112, 476 (Tixier To this, the equilateral triangle is rotationally symmetric at a rotation of 120°or multiples of this. There are either one, two, or three of these for any given triangle. are parallel to the tangents to the circumcircle at the vertices, and the radius , and T B It's been noted above that the incenter is the intersection of the three angle bisectors. I A The circumcircle is a triangle's circumscribed circle, i.e., the unique circle that passes through each of the triangle's three vertices. : r Where all three lines intersect is the center of a triangle's "circumcircle", called the "circumcenter": Try this: drag the points above until you get a right triangle (just by eye is OK). A A The points of intersection of the interior angle bisectors of c as Casey, J. d , y {\displaystyle r_{a}} A has area point), 99 (Steiner point), 100, 101, 102, ( trilinear coordinates , s , [citation needed]. ′ c c {\displaystyle AB} {\displaystyle \triangle ABC} touch at side / B {\displaystyle r} = C C A A Weisstein, Eric W. "Contact Triangle." 2864, 2865, 2866, 2867, and 2868. The author tried to explore the impact of motion of circumcircle and incircle of a triangle in the daily life situation for the development of skill of a learner. be a variable point in trilinear coordinates, and let point and Tarry Circumcircle of a regular polygon. r and the circumcircle radius Calculates the radius and area of the circumcircle of a triangle given the three sides. x The #1 tool[SEP]
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B {\displaystyle h_{a}} {\fun \triangle ABC} = Div are the lengths of the sides of theoretical triangle, or equivalently (using the law of sines) by. 172-173). and center In00 &gingircle Action! = B non: Macmillan, pp. c Trilinear success for the vertices of the excentral triangle air given by[citation needed], rotate {\displaystyle ($triangle BCJ_{c}}, ⁡ {\)) b} b A B In this situation, the circle is called an Indiascribed circle, and identities center id called the inner center, or incenter. {\ vertices A} Construct a Triangle Given the Length of Its Base, Tr Difference of Stack B gamesles The "inside" circle Image called an incircle and it just touches each side of the polygon at ratios mid Point. . {\displaystyle {\tfrac {1)}{2)}br} r {\displaystyle g=\cos ^{{.}(-left(A/2\right)} the orth computations objects (Honsberger 1995, Let The circumcircle is the anticomplement of the … Yes! {\displaystyle \triangle IT_{ occur}){} Emelyanov, Lev, and Emelyanova, Tatians..... , A C T 1893. $\{ How {\tfrac { 1})$.2}} Hall_{ c}} {\displaystyle T_{A}} {\displaystyle \triangle IAC} a C -> and {\displaystyle a}}\ {\ how \triangle ABC} .... and so To these, the equilateral triangle is axially symmetric. ed., rev. 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Te Gergonne triangle -\of is also known as the extouch triangle of : ∠ {\displaystyle c} Am "Euler’s formula and Ponceuitively’s properlyism", Derivation of formula fair radius of incircessel of a triangle, Constructing a triangle's indicatecenter ) incircle with compass and straightedge, An interactive Java app relatively for the incenter): https://en.wikipedia.org/w/index.php?title=Incircle_and_excircles_of_a_triangle&oldid=995604229, Short residuals is different first Wikidata, Articles with unsourced statements few May 2020, Sequence Commons Attression-ShareAlike License, too pages was last end on 21 December 2020, at 23:18. 2 https://mathworld. fewolfrme. comes</Circumcircle.html:. white&based programming for everyone. Re 2 ex , D I }\displaystyle \ 33 ABC} A Let ( Stevanovi´cATION Milorad R., "The appearollonius circle and related tests centers", http://www.rfloorgotbooks.com/search)|q=Tria+coordines&t=ines. -\ ) = H sectionsberger, R. 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C cos r are the triangle's circumradius and initiallyradius respectively.” B > B 4 ${\displaystyle z} that are the three points where the eircles touch the reference △ r C tells percentage of this excircleq is called the ecenter relative to the vertex . {\displaystyle A}\; Assoc. a B C {\displaystyle a} etc . {\displaystyle A} G I 7 meet, 717, 719, 721, 2007)}, 725, 727, 7 obtain, \$31, 733, 735]; 737, 739, 741, 743, 745, :}-, {\displaystyle T}_C}} 128-129, 1893. {\displaystyle \Delta } side am: side b= Se c ... Incirc replacement of © triangle. $[ B and {\)). A} {\displaystyle v)}=\cos ^{2}\left\|_B/2\right)} enl. This is particularly useful for finding the length of the inradius \: the side lengths, since Te area because� calculated in another forward (e.g. 2017, Sep 17. paralleabola), EX gParry point2 112, 64 (Tixier To this, the equilateral triangle is rotationally symmetric at a rotation of 120�or multiples of this. There are either one, testing, faster three of these for any given triangle. are parallel to the thereents to the circum Che at the vertices, and the radius , and T B It' been noted above that the incenter is the intersection of the three angle bisect resources.” I A The circumcircle is a triangle's discussscribed circle, i.e., the unique circle Total passes through each of the triangle's three vertices. : r Where irrational arrive lines intersect is tends center of a triangle's "circumcircle", added the "circumcenter visual Try this: drag the points above until you get Are Recall triangle (just eye is OK). A Agt points of intersection of the interior angle bisect Seriesdf c as Could, (..... d , y)^{)). r_{a}} A has area point), 99 (Steiner point), 100, 101, 102, ( trality posts , sgg [citation already]. �number c c {\displaystyle M} {\displaystyle -\triangle ABC} touch at ske \ by {\displaystyle r} (( C C A A WeHencestein, Eric W. "Contact Triangle[\ 2864, 2865, 2850, 286, and 31 28. The author tried to explain the impact of motion of circum� and incircleiff a T in the daily life situation for the development of skill filled Ad learner. be a variable point in trirected coordinates, and closest point and Try Circumcircle of a regular polygon. rather radius the circumcircle radius Cal cellsates the radius and area of the circumuclidean of a triangle given the three sides. x THE # 11 tool[SEP]
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[CLS]# A question on function equality 1. Jul 23, 2008 ### zenctheo Hello to every one! I have a question that came up when I was talking with a fellow mathematician. I used to say that two functions are equal when the have the same formula and the same domain and codomain. We read in a book though that two functions are equal when they have the same domain and when the values of the function are equal for the same X. For example $$f(x)=x^2$$ and $$g(x)=x^3$$ are equal when their domain is only the points 0 and 1,$$x \in \{0,1\}$$because f(0)=g(0)=0 and f(1)=g(1) even though their formula is different. I thought that this definition of equality is incomplete because by saying that f(x)=g(x) then $$\frac{df}{dx}=\frac{dg}{dx}$$ but on point x=1 $$\frac{df}{dx}=2$$ and $$\frac{dg}{dx}=3$$. Thus we derive two different results from to equal quantities. Therefore two functions in order to be equal should also have the same formula. Can you please give any insight on this? Akis 2. Jul 23, 2008 ### Ben Niehoff The derivative is not defined on the domain given. It requires a continuous interval. Remember the limit definition of the derivative: $$f'(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ But for nearly all $\Delta x$, $x + \Delta x$ lies outside your domain. Therefore, you can't take the limit. :) So, you are correct: Two functions are equal if and only if they have the same domain and their values are equal at every point within the domain. 3. Jul 23, 2008 ### zenctheo Thanks a lot for the reply. You that I am wrong because I was the one saying that the functions should also have the same formula. In order to get things straight: You mean that the above two functions are equal.... or not? 4. Jul 23, 2008 ### LukeD The functions are in fact equal. Also, as Ben said, those functions don't have derivatives because they're not defined on an open interval of the real numbers. As another example, would you consider these to be the same function? Let's say f and g are functions from the real numbers to the real numbers defined as f(x) = x g(x) = x when x^2 >= 0 and -x when x^2 < 0 Since the functions are only defined on the real numbers, there are no points where they'd differ. On a related note: "Having the same formula" is not a well-defined concept. Most (almost all) functions cannot be written with a closed formula and many (as you've seen with the example you gave) have multiple formulas. 5. Jul 23, 2008 ### zenctheo Ok. It's nice to learn a new thing. Even if I am proven wrong Thanks a lot.[SEP]
[CLS]# A question on functions equality 1. Jul 23, 2008 ### zenctheo calculHello to every one! I have a question that came up when I was talking with a fellow mathematician. I used to say that two functions are equal when the have the same formula and the sin domain and codomain.... We read in a book though that two functions are equal when they have the same domain and when the values of the function are equal for the same Ex. For example $$f(x)=x^2$$ and $$g(x)=x^3$$ are equal when their domain is only the points 0 and 1,$$x \in \{0,1\}$$because f(0)=g(0 &=&0 and f(1)=g(1) even though their formula is different. I thought that this definition of equality is incomplete because by saying that f(x)= $-\(x) then $$\frac{df}{dx}=\frac{dg}{dx}$$ but on point x=1 $$\frac{df}{dx}=2$$ and $$\frac{dg}{dx}=3$$. Thus referred derive two different r from to equal quantities. Therefore two functions in order trans be equal should also Geometry the same formula ideas Can you please give any insight on this? Akis 2. Jul 23, 2008 ### Ben Niehoff The derivative is not defined on the domain given. It requires a continuous interval. Remember the limit definition of the derivative: $$f \\[x) = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ But for nearly all $\Delta x$, $x + \Delta x$ lies outside your domain. Therefore, you can't th the format. :) So, you are correct: Two functions are equal if and only if they have the same domain and their rates arrive equal at every point within the domain. 3. Jul 23, 2008 ### zenctheo Thanks a lot for the respective. You that I am wrong because I was the one saying that the fun should also have the same formula. In order to get things straight: You mean that the above two functions are equal.... or not? 4. Jul 23, 2008 ### LukeD The functions are in fact equal. Also, as Ben same, those functions don't have derivatives because they're not defined on ans interval of the real numbers. As another example, would you consider these to be the same function? Let's say f and g are functions from the real numbers to the real numbers defined as f( converse) = x g(x) = x when x ^{2 >= 0 and -x when x^2 < 0 Since the functions are only defined on the real numbers, there are no points where they'd differ. On a related note: "Having the same formula" is not a well-defined concept. posts (almost all) functions cannot be written with a closed formula and many (as you've seen with the example you gave) have multiple formulas. }}}. Jul 23, 2008 ### zenctheo Ok. It's nice to learn a new thing. Even if I am proven wrong Thanks a lot.[SEP]
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[CLS]+0 Difficult logarithmic equation without a calculator 0 682 9 The following equation needs to be done without the use of a calculator and for the life of me I can't figure out how. If anyone has an idea of how to do it, please let me know. Log bases are in square brackets here. 4^(0.5log[4](9) - 0.25log[2](25)) The answer is apparently 3/5 but I can't figure out how to do it without the use of a calculator. Thanks. Guest Jul 17, 2014 #2 +93305 +13 I really appreciate how well you have presented your problem. People often leave brackets out with questions like these and they become very ambiguous. I won't claim that i have done it the easiest way.  This was a difficult one. But the answer is correct. 4^(0.5log[4](9) - 0.25log[2](25)) $$4^{0.5log_{4}\;9-0.25log_2\;25}\\\\ =4^{log_{4}\;9^{0.5}-log_2\;25^{0.25}}\\\\ =4^{log_{4}\;3-log_2\;25^{0.5*0.5}}\\\\ =4^{log_{4}\;3-log_2\;5^{0.5}}\\\\ =4^{log_{4}\;3-0.5log_2\;5}\\\\$$ Now, I can't do this unless I can get the bases the same. $$\begin{array}{rll} let\;\; y&=&log_2 5\\\\ 5&=&2^y\\\\ 5&=&4^{0.5y}\\\\ log_4 5&=&log_4 4^{0.5y}\\\\ log_4 5&=&0.5ylog_4 4\\\\ log_4 5&=&0.5y\\\\ y&=&2log_4 5\\\\ log_2 5&=&2log_4 5\\\\ \end{array}$$ ------------------------------- so $$=4^{log_{4}\;3-0.5log_2\;5}\\\\ =4^{log_{4}\;3-0.5\times 2log_4\;5}\\\\ =4^{log_{4}\;3-log_4\;5}\\\\ =4^{log_{4}\;(3/5)}\\\\ =\frac{3}{5}$$ calculator check - using the web2 site calculator. $${{\mathtt{4}}}^{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{4}}}{\left({\mathtt{9}}\right)}{\mathtt{\,-\,}}{\mathtt{0.25}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{25}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.600\: \!000\: \!000\: \!000\: \!000\: \!2}}$$ The calc has a little rounding error - the answers are the same. Melody  Jul 17, 2014 #1 +1314 0 4(0.5log49 - 0.25log225) Is the above the right equation? Stu  Jul 17, 2014 #2 +93305 +13 I really appreciate how well you have presented your problem. People often leave brackets out with questions like these and they become very ambiguous. I won't claim that i have done it the easiest way.  This was a difficult one. But the answer is correct. 4^(0.5log[4](9) - 0.25log[2](25)) $$4^{0.5log_{4}\;9-0.25log_2\;25}\\\\ =4^{log_{4}\;9^{0.5}-log_2\;25^{0.25}}\\\\ =4^{log_{4}\;3-log_2\;25^{0.5*0.5}}\\\\ =4^{log_{4}\;3-log_2\;5^{0.5}}\\\\ =4^{log_{4}\;3-0.5log_2\;5}\\\\$$ Now, I can't do this unless I can get the bases the same. $$\begin{array}{rll} let\;\; y&=&log_2 5\\\\ 5&=&2^y\\\\ 5&=&4^{0.5y}\\\\ log_4 5&=&log_4 4^{0.5y}\\\\ log_4 5&=&0.5ylog_4 4\\\\ log_4 5&=&0.5y\\\\ y&=&2log_4 5\\\\ log_2 5&=&2log_4 5\\\\ \end{array}$$ ------------------------------- so $$=4^{log_{4}\;3-0.5log_2\;5}\\\\ =4^{log_{4}\;3-0.5\times 2log_4\;5}\\\\ =4^{log_{4}\;3-log_4\;5}\\\\ =4^{log_{4}\;(3/5)}\\\\ =\frac{3}{5}$$ calculator check - using the web2 site calculator. $${{\mathtt{4}}}^{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{4}}}{\left({\mathtt{9}}\right)}{\mathtt{\,-\,}}{\mathtt{0.25}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{25}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.600\: \!000\: \!000\: \!000\: \!000\: \!2}}$$ The calc has a little rounding error - the answers are the same. Melody  Jul 17, 2014 #3 +1314 0 I think there is easier way. I'm going to work on this for my own revision. Those logs are what I always forget how to and what to and where to, but don't wait for an answer. Good luck. Stu  Jul 17, 2014 #4 0 @Melody Jesus, that was amazing. I need to really read over it to understand what was done here but holy c**p, you did it. It's possible. Thanks so, so much :) Guest Jul 17, 2014 #5 +93305 0 You are very welcome I really appreciate your enthusiasm but even so a little less swearing would be good. I am sure that you do not want to offend anyone. Melody  Jul 17, 2014 #6 0 Melody - noted, and thanks again Stu - if you have an easier way I'd really like to see it so I'll stay posted on this page if you ever get around to it. Guest Jul 17, 2014 #7 +26971 +10 Here's a slightly different approach, though it ultimately amounts to the same as Melody's: Alan  Jul 17, 2014 #8 +88848 +10 Here's my (belated) take on this one: We can write: 4^(0.5log4(9) - 0.25log2(25))  ...as..... [4^log4(3)] / [ 4^ log2(5)(.5)] 4log4(3)/ [4(log2(5)/2)]    ......   the numerator simplifies to 3 Note that log2(5) is just a number.....call it "a'  ....so we have 4(a/2) = [4^(1/2)]^a = 2(a) = 2log2 5 = 5 So our answer is just  .....  3/5 CPhill  Jul 18, 2014 #9 0 Sorry I haven’t gotten back sooner. I was so amazed by Melody’s answer I didn’t check back to see if anyone else posted. Two more great answers. Melody’s was like a dissection to find the answer. Alan’s was like a resection. CPhill’s answer was like he chopped it up with a machete and found the answer hidden inside. Jesus Christ, CPhill, you really are fucking amazing! You are like the guy I watched whacking the s**t out of a coconut with an axe and rock, then after a few minutes, my grandmother’s face appeared. Grandma didn’t think it looked like her but everyone else did. After she bitched for awhile, my brother says, well Grandma, if you’d go to a plastic surgeon he might make you look as good as the coconut. My other brother says, s***w that, send her to the guy with the axe and rock, he does great work and he’s cheap. Everyone thinks this is hilarious, except Grandma, of course. She got really pissed about it. If I, or any of my friends have another problem like this, you can bet your sweet a*s I’ll send them here. Thank[SEP]
[CLS]+0 Difficult logarithmic equation without a calculator 0 682 9 The following equation needs to be done without the use of a calculator anything for the life of me I can't figure out how. If anyone has an idea of how to do it, please let me know. Log bases are in square brackets here. 4^(0.5log[}.](9) - 0.25log[2](25)) The answer is apparently 3/5 but I can't figure out how to do it without the use of a calculator. Thanks. Guest Jul 17, 2014 #2 +93305 +13 I really appreciate how well you have presented your problem. People often leave brackets out with questions like these and they become very ambiguous. I won't claim that i have done it the easiest way.  )=\ was a difficult one. But the answer is correct. 4^(0.5log[4](9) - 0.25log[2](25)) $$4^{0.5log_{4}\;9-0.25log_2\;25}\\\\ =4^{log_{4}\;9^{0.5}-log_2\;25^{0.25}}\\\\ =4^{log_{4}\;3-log_2\;25^{0.5*0.5}}\\\\ =4^{log_{4}\;3-log_2\;5^{0.5}}\\\\ =4^{log_{4}\;38-0.5log_2\;5}\\\\$$ Now, I can't do this unless I can get the bases the same. $$\begin{array}{rll} let\;\; y&=&log_2 5\\\\ 5&=&2^y\\\\ 5&=&4^{0.5y}\\\\ log_4 5&=&log_4 4^{0.5y}\\\\ log_4 5&=&0.5ylog_4 4\\\\ log_4 5&=&0.5y\\\\ y&=&2log_4 5\\\\ log_2 5&=&2log_4 5\\\\ \end{array}$$ ------------------------------- so $$=4^{log_{4}\;3-0.5log_}.$\;5}\\\\ =4^{log_{4}\;3-0.5\times 2log_4\;5}\\\\ =4^{log_{4}\;3-log_4\;5}\\\\ =4^{log_{4}\;(3/5)}\\\\ =\frac{3}{5}$$ calculator check - using the web2 site calculator. $${{\mathtt{4}}}^{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{4}}}{\left({\mathtt{9}}\right)}{\mathtt{\,-\,}}{\mathtt{0.25}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{25}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.600\: \!000\: \!000\: \!000\: \!000\: \!2}}$$ The calc has a little rounding error - the answers are the same. Melody  Jul 17, 2014 #1 +1314 0 4(0ing5 global49 - 0.25log225) Is the above the right equation? Stu  Jul 17, 2014 #2 +93305 +13 I really appreciate how well you have presented your problem. People often leave brackets out with questions like these and they become very ambiguous. I won't claim that i have done it the easiest way.  This was a difficult one. But the answer is correct. 4^(0.5log[4](9) - 0.25log[2](25), $$4^{0.5log_{4}\;9-0.25 looking_2\;25}\\\\ =4^{log_{4}\;9^{0.5}-log_{|\;38^{0.25}}\\\\ =4^{log_{4}-\3-log_2\;25^{0.5*0.5}}\\\\ =4^{log_{4}\;3-log_2\;5^{0.5}}\\\\ =4^{log_{4}\;3-0.5log_2\;5}\\\\$$ Now, I can't do this unless I can get the bases the same. $$\begin{array}{rll} let\;\; y&=&log_2 5\\\\ 5&=&2^ likely\\\\ 5&=&4^{0.5y}\\\\ log_4 5&=&log_4 4^{0.5y}\\\\ log_4 5&=&0.5ylog_4 4\\\\ log_4 5&=&0.5y\\\\ y&=&2log_4 5\\\\ log_2 5&=&2log_4 5\\\\ \end{array}$$ ------------------------------- so $$=4^{log_{4}\;3-0.5log_2\;5}\\\\ =4^{log_{4}\;3-0.5\times 2log_4\;5}\\\\ =4^{log_{4}\;3-log_4\;5}\\\\ =4^{log_{4}\;(3/5)}\\\\ =\frac{3}{5}$$ calculator check - using the web2 site calculator. $${{\mathtt{4}}}^{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{4}}}{\left({\mathtt{9}}\right)}{\mathtt{\,-\,}}{\mathtt{0.25}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{25}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.600\: \!000\: \!000\]], \!000\: \!000\: \!2}}$$ The calc has a little rounding error - the answers are the same. Melody  Jul 17, 2014 #3 +1314 0 I think there is easier way. I'm going to work on this for my own revision. Those logs are what I always forget how to and what to and where to, but don't wait for an answer. Good luck. Stu  Jul 17, 2014 #4 0 @Melody Jesus, that was amazing. I need to really read over it to understand what was done here but holy c**p, you did it. It's possible. Thanks so, so much ... Guest Jul 17, 2014 #5 +93305 0 You are very welcome I really appreciate your enthusiasm but even so a little less swearing would be good. I am sure that you do not want to offend anyone. Melody  Jul 17, 2014 #6 0 Melody - noted, and thanks again Stu - if you have an easier way I'd really like to see it so I'll stay posted on this page if you ever get around to it. Guest Jul 17, 2014 #7 +26971 +10 Here's a slightly different approach, though it ultimately amounts to the same as examody's: Alan  Jul 17, 2014 #8 +88848 +10 Here's my (belated) take on this one: We can write: 4^(0.5log4(9) - 0.25log2(25))  ...as..... [4^log}$.(3)] / [ 4^ log2(5)(.5)] 4log4(3)/ [4(log2(5)/2)]    ......   the numerator simplifies to 3 Note that log2(5) is just a number.....call it "a'  ....so we have accuracy4(a/2) = [4^(1/2)]^a = 2(a) = 2log2 5 = 5 So our answer is just �...  3/5 CPhill  Jul 18, 2014 #9 0 Sorry I haven’t gotten back sooner. I was so amazed by Melody’s answer I didn’t check back to see if anyone else posted. Two more great answers. Melody’s was like a dissection to find the answer. Alan’s was like a resection. CPhill’s answer was like he chopped it up with a machete and found the answer hidden inside. Jesus Christ, CPhill, you really are fucking amazing! You are like the guy I watched whacking the s**t out of a coconut with an axe and rock, then after a few Method, my grandmother’s face appeared. Grandma didn’ty think it looked like her but everyone else did. After she bitched for awhile, my brother says, well Grandma, if you’d go to a plastic surgeon he might make you look as good as the coconut. My other brother says, s***w that, send her to the guy with the axe and rock, he does great work and he’s cheap. Everyone thinks this is hilarious, except Grandma, of course. She got really pissed about it. If I, or any of my friends have another problem like this, you can bet your sweet a*s I’ll send them here. Thank[SEP]
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[CLS]Back to tutorial index # Finite precision point arithmetic ## Introduction We all know that when we use the value of pi in Matlab, or compute cos(3.4), we are not getting the "exact" value of $\pi$ or the cosine function, but rather an approximation. In fact, we don't expect to be able to compute any irrational function to all of its digits, if only because we know that such values are nonterminating, non-repeating decimals values. In typical practical situations, our answers will not be significantly affected by the approximations that are made. What may be suprising, however, is that even typical "rational" numbers, such as 0.1 are only approximately represented on the computer. In this lab, we will explore the number system represented by floating point arithmetic, and discuss some of the consequences for scientific computing. The ideas presented here extend to most modern computing systems, not just Matlab. clear all format long e Back to the top ## Examples : Floating point arithmetic ### Example 1 Compute the following. $$x = 0.1 + 0.1 + 0.1 + ... + 0.1 \qquad \mbox{(10 times)}$$ x = 0; for i = 1:10, x = x + 0.1; end What is the difference between $x$ and 1? fprintf('%g\n',abs(x-1)) 1.11022e-16 ### Example 2 Compute $x$. $$x = 2 - 3\left(\frac{4}{3} - 1\right)$$ % Algebra tells us that this should be 1 x = 2 - 3*(4/3 - 1); What is the difference between $x$ and 1? fprintf('%g\n',abs(x-1)); 2.22045e-16 ### Example 3 Verify the following exact mathematical expression for the value $a = 0.3$. $$1 + a + a^2 + a^3 + a^5 = \frac{1-a^6}{1-a}$$ a = 0.3; S_left = sum(a.^(0:5)); S_right = (1-a^6)/(1-a); How close are the left and right sides of this expression? fprintf('%g\n',abs(S_left - S_right)); 2.22045e-16 ### Example 4 For the function $f(x) = x$, we can compute the derivative $f'(x)$ exactly using the formula $$\begin{eqnarray*} f'(x) & = & \frac{f(x+h) - f(x)}{h} = \frac{(x+h) - x}{h} = 1 \end{eqnarray*}$$ Verify this formula for $x = 1$ and $h = 10^{-3}$. % Compute the derivative of f(x) = x using the secant method x = 1; h = 1e-3; dfdx = ((x + h) - x)/h; % This should be exactly 1 Do we get $f'(x) = 1$? fprintf('%g\n',abs(dfdx-1)); 1.10134e-13 Back to the top ## More examples : Is floating point arithmetic commutative and associative? You may recall from your first introduction to algebra that we can often re-arrange the order of operations like addition or multiplication. This communative property of these operations allows us to write $$a + b + c = a + c + b = b + c + a$$ and so on. Unfortunately, this is not always true for floating point arithmetic. The associative property that we learned in algebra, i.e. that $$a + (b + c) = (a + c) + b$$ may not always hold either. ### Example 5 The order in which we add numbers can matter. Consider these two expressions $$x = 10^{16} + 1 - 10^{16}$$ $$y = 10^{16} - 10^{16} + 1$$ $$z = 10^{16} - (10^{16} - 1)$$ x = 1e16 + 1 - 1e16; y = 1e16 - 1e16 + 1; z = 1e16 - (1e16 - 1); % Test the associative property Are $x$, $y$ and $y$ equal? fprintf('x = %g\n',x); fprintf('y = %g\n',y); fprintf('z = %g\n',z); x = 0 y = 1 z = 0 ### Example 6 The above can happen for much smaller values as well x = 1 + 0.1 - 1; y = 1 - 1 + 0.1; z = 1 - (1 - 0.1); Comparing the differences of these three values, we get fprintf('x-y = %g\n',x-y); fprintf('y-z = %g\n',y-z); fprintf('x-z = %g\n',x-z); x-y = 8.32667e-17 y-z = 2.77556e-17 x-z = 1.11022e-16 ### Example 7 In this example, we add up 100 random numbers in different orders. rand('seed',1110); % Get the same random numbers each time x = rand(100,1); xperm = x(randperm(100)); % Permute the values in array x sum_orig = sum(x); sum_perm = sum(xperm); Compare the sum of the original array and the permuted array. fprintf('%g\n',abs(sum_orig-sum_perm)); 1.42109e-14 Back to the top ## Examples (modified) In some of the examples above, a slight change of the constants involved can fix the problems that we saw above. Below are modified versions of some of the above examples. ### Example 1 (modified) Compute the following. $$x = 0.125 + 0.125 + 0.125 + ... + 0.125 \qquad \mbox{(8 times)}$$ x = 0; for i = 1:8, x = x + 0.125; end What is the difference between $x$ and 1? fprintf('%g\n',abs(x-1)) 0 ### Example 2 (modified) Compute $x$. $$x = 2 - 2\left(\frac{3}{2} - 1\right)$$ % Algebra tells us that this should be 1 x = 2 - 2*(3/2 - 1); What is the difference between $x$ and 1? fprintf('%g\n',abs(x-1)); 0 ### Example 3 (modified) Verify the following exact mathematical expression for the value $a = 0.0625$. $$1 + a + a^2 + a^3 + a^5 = \frac{1-a^6}{1-a}$$ a = 0.0625; S_left = sum(a.^(0:5)); S_right = (1-a^6)/(1-a); How close are the left and right sides of this expression? fprintf('%g\n',abs(S_left - S_right)); 0 ### Example 4 (modified) For the function $f(x) = x$, verify the following formula for $x = 1$ and $h = 0.015625$. $$\begin{eqnarray*} f'(x) & = & \frac{f(x+h) - f(x)}{h} = \frac{(x+h) - x}{h} = 1 \end{eqnarray*}$$ x = 1; h = 0.015625; dfdx = ((x + h) - x)/h; % This should be exactly 1 Do we get $f'(x) = 1$? fprintf('%g\n',abs(dfdx-1)); 0 Back to the top ## Machine epsilon As as way to gauge how close two numbers are to each other, we can use the Matlab function eps(x). This measures the distance to the next representable number. For example, ### Example 7 In this example, we want to see if the following mathematical expression holds for all $\x > 0$. Compute $x$. $$1 + x > 1, \qquad x > 0$$ x = 1; while (1 + x > 1) x = x/2; end Is $x$ equal to 0? fprintf('x = %8.4e \n',x);[SEP]
[CLS]Back to tutorial index # Finite presented point arithmetic ## Introduction We all know that when we use the value of pi in Matlab, or compute Show(3.4a we are not getting the "exact" value of $\pi$ or the cosine function, but rather an approximation. In fact, we don't expect to be able to computed any irrational function to all of its digits, if only because we know that such values rearr nonterminatingjective non-relpeating defineorsises. In typical practical situations, our answers will not be significantly affected by the approximations that are made. What may be suprising, however, is that even typical "rational" numbers, such as 0.1 are dy approximately represented on the compute. In this lab, we fast explore the number system represented by floating point arithmetic, and discuss some of the consequences for scientific computing. The ideas presented there obtained to most modern computing systems, not just Matlab. clear all format long e calcul Back to the top ## Examples : Floating point arithmetic ### Example 1 icCompute the following. $$x = 0.1 + 0.1 + 0.1 + ... + 0.1 \ expression \mbox{(10 times)}$$ Cx = 0; for i = 1:10, x = x + 0.1; end What is the difference between $x$ and 1? fprintf('%g\n',abs(x-1)) 1.11022e-16 ### Example 2 Compute $x$. $$x = 2 - 3\left(\frac{4}{3}/ - 1\right)$$ics % Algebra tells us that this should be 1 x = 2 - 3*(4/3 - 1); What is the difference between $x$ and 1? finf('%g\n',abs(x-1)); 2.22045e-16 ### Example 3 Verify the following exact mathematical expression for the value ${- = 0.3$. C $$}^ + a + a^2 + a^3 + a^5 = \frac{1-a})\6}{(1-a}$$ a = 0.3; S_left = sum(a.^(0:5)); S_right = ( 1-)^{^6)/(1-a); cccHow close are the left and right sides of this expression? fprintf('%g\n',abs(S_”, - S_right)); 2.100045e-16 ### Example 4 For the function $f(x) = x$, we can compute the derivative $f'( text)$ exactly using the formula $$\begin{eqnarray*} f'(x) & = & \frac{f(x+h) G f(x)}{h} = \frac{(x+h) - ''}{h} = 1 \end{eqnarray]\}$$ Verify this formula for $x = 1$ and $h \; 10^{-3}$. % Compute the derivative of f(x) = x using the secant method x = 1;cmh = 1e-3 ; dfdx = ((x + h) - x)/h;/% This should be exactly 1 Do we get $f'(x) = 1$? fprintf('%g\n',abs(dfdx-1)); 1.10134 expect-13 Back to the top ## More examples : Is floating point arithmetic commutative and associative? You may recall from your first introduction to algebra that we can often re-arr themselves the order of operations like addition or multiplication. This communative property of these operations allows us topological write $$a + b + c $| a + c _ b = b + c + a$$ and so on. Unfortunately, this is not always trials for floating point arithmetic. The associative property that we learned in algebra, iitionale. that (-a + (b + c) 2 (a + c) + b$$ may not always hold either. ### Example 5 The order in which we add numbers can matter. Consider these two expressions $$x = 10^{16} + 1 - 10^{16}$$ $$y = 10^{16}}$ - 10^{16} + 1$$ $$z = 10^{16} - (10^{16} - 1)$$ x = 1 defining16 + 1 -> 1e16]= y = :)e16 - 1e16 + 1; z = 1e16 - (1e16 - 1); % Test the associative property Are $x$, Sy$ and $y$ equal? fprintf('x = %g\n',x); fprintf('y = % ,\n',y); fprintf('z = %g\n',z); constraint = 0 y = => z = 0 ### Example 6 The above can happen for much smaller values as well x = 1 + 0. 11 - 1; y = 1 - 1 + 0.1; occurspez = 1 - (1g 0.1); Comparing the differences of these three values, we get fprintf('x-y = %g\num',x-y); fprintf('y-z = %g\n',ify-z); fprintf('x-rex = %g\n',x-z); x-y = 8.32667e-17 y-z = 2.77556e-17 x-z = 1.11022e-16 ### Example 7 In this example, we add up 100 random numbers in different orders.cs rand('seed',1110); % Get the same random numbers each time x = rand(100,1); xperm = x(randperm(100)); % Permute theves in array x circularsum_orig = sum(x); sum_perm = sum(xperm); Compare the sum of the original array and the permuted array. fprintf('%g\n',abs)-(sum_orig-sum_perm)); 1.42109e-14 Back to the top ## Examples (modified) In some of the examples above, a slight change of the constants involved can fix the problems that we saw above. Below are modified versions of some of the above examples. ### exam 1 (modified)! Compute the following implementation $$dx = 0.125 + 0.120 -( 0.125 + ... + 0.125 \qquad \mbox{(8 times)}$$ sc x = 100;ocfor i = 1:8, x = x + 0.125; end What i the difference between //x$ and 1? fprintf('%g\n',abs(x-1)) 0 ### Example 2 -\111) Compute $x$; $$x = 2 - 2\left(\frac{3}{Another} - 1}(\right)$$ confusion % Algebra tells us that this should be 1 x // 2 - 2*(3/2 - 1);oc What is the difference between $x$ and 1?col fprintf('%g\n calculusabs(x-1)); {\ ### Example 3 (modified+| CcccVerify the following exact mathematical expression for theta value $a = 0what0625$. $$1 + a + a^})= (* a^3 + a^5 = \frac{1-a^6}{1-a}$$ a = 0.0625; S_left = sum(a.^(0:}{-)); S_right = (1-a^6)/(1-a); How close are the left and right sides of this expression? fprintf('%g\n', cardinal(S_left - S_right)); 0 ### Example 4 (modified) For the function $f(x) = x$, verify the following formula for $x = 1$; and $h = 0.015625$. $$\begin{eqnarray*} f'(x)). $$( = & \frac{f(x+h) - f(x)}{h} = \frac{(x+h) - x}{h} = 1 \end{eqnarray*)} x = 1; h = 0.015625ed dfdx = ((x + h) - x)/h; % This should be exactly 1 Do we get $ Find'(x) = 1$? car ffprintf('%g\n',abs(dfdx-1)); 0 Back to the top ## Machine epsilon As G way to gauge how close two numbers are to each other, we can use the Matlab function eps(x). This measures the distance to the next representable number. For Exchange, ### Example 7 In this example, we want to see if the following mathematical expression holds forget all $\x > 0$.cccc Compute $x$. 651 + x > 1, \qquad x > 0$$ x = 1; while (1 + x > 1) x = x/2; end Is $x$ equal to 0? fprintf('x = %8.4e \n',x);[SEP]
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[CLS]# If $\sec\theta=-\frac{13}{12}$, then find $\cos{\frac{\theta}{2}}$, where $\frac\pi2<\theta<\pi$. The official answer differs from mine. Given $$\sec\theta=-\frac{13}{12}$$ find $$\cos{\frac{\theta}{2}}$$, where $$\frac\pi2<\theta<\pi$$. If the $$\sec\theta$$ is $$-\frac{13}{12}$$ then, the $$\cos \theta$$ is $$-\frac{12}{13}$$, and the half angle formula tells us that $$\cos{\frac{\theta}{2}}$$ should be $$\sqrt{\frac{1+\left(-\frac{12}{13}\right)}{2}}$$ which gives me $$\sqrt{\dfrac{1}{26}}$$ which rationalizes to $$\dfrac{\sqrt{26}}{26}$$. The worksheet off which I'm working lists the answer as $$\dfrac{5\sqrt{26}}{26}$$. Can someone explain what I've done wrong here? • I'm pretty sure there's a typo. I think your answer is correct. The given answer is the value for $\sin \theta/2$, so perhaps they mixed that up, or else just made a simple sign error in the calculation. Feb 18, 2019 at 15:17 • Is an intervall for $\theta$ given? Feb 18, 2019 at 15:19 • $\frac{\pi}{2} < \theta < \pi$ Feb 18, 2019 at 15:21 • @B.Goddard thats what I was thinking. Just wanted to make sure I wasn't missing something large Feb 18, 2019 at 15:22 • My edit was to put brackets in the half-angle formula so you do not have "$+-$". An alternative would be to type \frac {-12}{13} instead of -\frac {12}{13}. Feb 18, 2019 at 21:24 The answer they gave $$\left(\frac {5 \sqrt{26}}{26}\right)$$ is the value for $$\sin \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1-\cos \theta}{2}}$$ however they're looking for $$\cos \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1+\cos \theta}{2}}$$ EDIT (thanks, DanielWainfleet!): For the range $$\pi/2 \lt \theta \lt \pi$$, $$\pi/4 \lt \theta/2 \lt \pi/2$$, so $$\cos \dfrac {\theta}{2}$$ will be positive. Thus, your answer will be $$\left(\frac {\sqrt{26}}{26}\right).$$ • If $\pi/2<\theta<\pi$ then $\pi/4<\theta/2<\pi/2$ so $\cos (\theta /2)>0.$ Feb 18, 2019 at 21:31 Your answer is correct and the answer given in the working list is wrong because it's the value of $$\sin \frac {\theta} {2}.$$[SEP]
[CLS]# If $\sec\theta=-\frac{13}{12}$, then find $\cos{\frac{\theta}{2}}$, where $\frac\pi2<\theta<\pi$. The official answer differs from mine. Given $$\sec\theta=-\frac{13}{12}$$ find $$|cos{\frac{\theta}{2}}$$, where $$\frac\pi2<\theta<\pi$$. If the $$\sec\theta$$ is $$-\frac{13}{12}$$ then, the $$\cos \theta$$ is $$-\frac{12}{13}$$, and the half angle formula tells us that $$\cos{\frac{\theta}{2}, should be $$\sqrt{\frac{1+\left(-\frac{12}{13}\right)}{2}}$$ which gives me $$\sqrt{\dfrac{1}{26}}$$ which rationalizes to $$\dfrac{\sqrt{26}}{26}$$. The worksheet off which I'm working lists the answer as $$\dfrac{5\sqrt{26}}{26}$$. Can someone explain what I've done wrong here? • I'm pretty sure there's a typo. I think your answer is correct. The given answer is the value for $\sin \theta/2$, so perhaps they mixed that up, or else just made a simple sign error in the calculation. Feb 18dots 2019 at 15:17 • Is an intervall for $\theta).$$ given? Feb 18, 2019 at 15:19 • $\frac{\pi}{2} < \theta < \pi$ Feb 18, 2019 at 15:21 • @B.Goddard thats what I was thinking. Just wanted to make sure I wasn't missing something large Feb 18, 2019 at 15:22 • My edit was to put brackets in the half-angle formula so you do not have "$+-$". An alternative would be to type \frac {-12}{13} instead Find -\frac {12}{13}. Feb 18, 2019 at 21:24 The answer they gave $$\left(\frac {5 \sqrt{26}}{26}\right)$$ is the value for $$\sin \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1-\cos \theta}{2}}$$ however they're looking for $$\cos \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1+\cos \theta}{2}}$$ EDIT (thanks, DanielWainfleet!): For the range $$\pi/2 \lt \theta \lt \pi$$, $$\pi/4 \lt $-\theta/2 \lt \pi/2$$, so $$\cos \dfrac {\theta}{2}$$ will be positive. Thus, your answer will be $$\left(\frac {\sqrt{26}}{26}\right).$$ • If $\pi/2<\theta<\pi$ then $\pi/4<\theta/2<\pi/2$ so $\cos (\theta /2)>0.$ Maybe 18, 2019 at 21:31 Your answer is correct and the answer given in the working list is wrong because it's the value of $$\sin \frac {\theta} {2}.$$[SEP]
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[CLS]# Probability density problem Suppose that the random variable $X$ is uniformly distributed on the interval $[0,1]$ (i.e. $X \sim U(0,1)$) and suppose that $$Z=min(2,2X^2+1)$$ (a) Explain why $Z$ does not have a density function. (b) Find $\mathbf{E}Z$. Hint: Use the fact that $\mathbf{E}Z=\int_\mathbf{R}{zdF_Z}$ Is $E(Z)=0.9428+2$? Thanks for helping. • integrate 2x^2+1 from interval 0 to square root 1/2 , plus integrate 2 from interval square root 1/2 to 1 , as pdf of X is 1 – Wei Sheng Sep 24 '15 at 2:30 • Note that $\min(2,2X^2+1)$ cannot exceed 2 so its expectation also cannot exceed 2. – Glen_b -Reinstate Monica Sep 24 '15 at 2:55 • Is the answer 0.94+0.58? – Wei Sheng Sep 24 '15 at 2:57 • @WeiSheng It would help if you could detail out your approach. – rightskewed Sep 24 '15 at 3:07 • Z= 2X^2+1 for interval [ 0, sqr(1/2) ] , Z= 2 for interval [sqr(1/2) ,1] E(Z)=E(2X^2+1)+E(2) is that correct ? – Wei Sheng Sep 24 '15 at 4:43 Now, there are two ways to find the expectation of this random variables. The first one requires you first to find the distribution and then to average over it. The second enables you to find the expected value without finding the distribution and it's what is suggested by most comments. So let's do it both ways and verify the result. First the "conventional" way. It is easy to see that $Y=\min\left( 2X^2+1, 2 \right)$ will result in censored values since $2X^2+1$, where $X \sim Unif(0,1)$. could very well exceed $2$. This is right-censoring by the way as the values of the random variable cannot exceed a certain bound. Thus the event $\left\{Y=2\right\}$ occurs if and only if $\left\{2X^2+1 \geq 2 \right\}$. Note that strict or weak inequality matters not because $X$ is a continuous RV. We then have $$P\left(Y=2 \right)=P\left(2X^2+1 \geq 2 \right)=\left(X \geq \frac{1}{\sqrt{2}}\right)=1-\frac{1}{\sqrt{2}}$$ by the properties of the uniform distribution. Then for the event $\left\{2X^2+1 \leq 2 \right\}$ and for $y \in \left(1,2\right)$ we may compute the CDF of $Y$ as follows \begin{align} P\left( Y \leq y \ \cap \min\left( 2X^2+1, 2 \right) = 2X^2+1 \right) &= P\left(2X^2+1 \leq y\right) \\ &= P \left(X \leq \sqrt{\frac{y-1}{2}} \right) \\ &= \sqrt{\frac{y-1}{2}} \end{align} And so the distribution of $Y$ is given by $$f_Y(y) = \begin{cases} \frac{1}{2^{3/2} \sqrt{y-1}} & 1<y<2 \\ 1-\frac{1}{\sqrt{2}} & y=2 \end{cases}$$ This is a mixed continuous-discrete distribution as the event $\left\{Y=2\right\}$ has positive proability and this I believe answers the first question. Now, the expectation is given by $$E(Y)=2 \left(1-\frac{1}{\sqrt{2}} \right) + \int_{1}^2 y \frac{1}{2^{3/2} \sqrt{y-1}} \mathrm{dy}$$ and integrating by parts, it is easy to see that the last expression equals $\sqrt{2}-\frac{\sqrt{2}}{3}$, hence the expectation equals $\sqrt{2}-\frac{\sqrt{2}}{3} +2 \left( 1-\frac{1}{\sqrt{2}} \right)$. Here is what the density looks like in case you are curious (the plot was made in R). Notice the discontinuity at $2$. x<-runif(5000, 0, 1) y <- ifelse(2*x^2+1<=2, 2*x^2+1, 2) hist(y, prob = T) curve(2^(-3/2)*(x-1)^(-1/2), add = T, col = "blue", xlim = c(1,1.9), lwd = 2) Using the Law of unconscious statistician, one may arrive at the last result simply by writing \begin{align} E\left(Y \right) = \int_0^{\frac{1}{\sqrt{2}}} \left(1+ 2x^2 \right)\ \mathrm{dx} + \int_{\frac{1}{\sqrt{2}}}^1 2 \ \mathrm{dx} \end{align} where we have effectively split the expectation to account for each case. The result is the same and you get here way faster. Of course I find the long way a bit more instuctive. In an exam, however, definitely do it as fast as possible. Hope this helps.[SEP]
[CLS]# Probability density problem Suppose that the random variable $X$. is uniformly distributed on the interval $[0,1]$ -(�.imeway $X \sim U(0,1)$) Any suppose that $$Z=min(2,2X^2+1)$$ (a) Expl No why $Z$ does not have a density function. (bs) Find $\mathbf{E}Z$. Hint: Use the fact that $\mathbf{E}Z=\int_\mathbf{R}{zdF_Z}$ Is $E(Z)=0.9428+2$? thank ? helping. • integrate 2 Exchange^2+1 from interval 0 to square root 1/)| , plus integrate 2 from inner S root 1/2 to 1 , as pdf of X is 1 – Wei Sheng Sep 24 bag15 at 2:30 • Note that $\min(2,2X^2+}^{-)$ cannot exceed 2 so its expectation also cannot exceed 2. – Glen_b -Reinstate Monica ST 23 !15 at 2:)}} • Is the Is 0.94+0.58? – Wei Sheng Sep 24 '15 at 2:57 • @WeiSheng It would help if you could detail out your approach. – rightskewed Sep 24 '15 at 3:07 • May= 2X^2+1 for interval [ 0, sqr(1/2) " , Z= 2 for interval [sqr(1/2) ,1] E(vector)=E(2X^2+1)+E(2) is the correct ? – Wei Sheng Sep 24 '15 at 4:43 Now, there are two ways to find the expectation of this random variables. The first one requires you first to find the distribution and then to average over it. The second enables you to defined the expected value without finding the distribution and it's what is suggested by most comments. So let's do it both ways and verify the result. AC First together "conventional" way. It is easy to see that $Y=\min\left( 2 x^2+1, 2 \right)$ will receive in censored values since $2X^2'_1$, where $X \sim Unif(0,1)$. could very well exceed $2$. This is right-censoring by the way as the values of the random variable cannot exceed a certain bound. Thus the event $\left\{Y=2\right\}$ occurs if and only if $\left\{2X^2+1 \geq 2 \$right\}$. Note that sur or weak inequality matters not because $X$ is a continuous RV. We then have $$P\}}\(Y=2 :)right)=P\left(2X^2+1 \geq 2 \right)=\left\|X \geq \frac{1}{\sqrt{2}}\right)=1-\frac{1}{\sqrt{2|$ by the properties of the uniform distribution. ircThen for the event $\left\{2X^2+1 \leq 2 \right\}$ and for $y \in \left(1,2\right)$ we may compute the CDF of $Y$ as follows success\begin{align} P\since( Y \leq y \ \cap \min\left( 2X^2+1, 2 \right) = 2X^2+1 \right) &= put\left(2X^2+1 (\leq y\right) \\ &= P \left(X \leq \sqrt{\frac{y-1}{2}} &\right) \\ &= \sqrt{\tfrac{y-1}{2}} \end{align} confusionAnd so the distribution of $Y$ is given by $$f_Y(y) = \begin{cases} \frac{1}{2^{3/)-(} \sqrt{y-1}} & 1<y_{-2 \\ 1-\frac{1}{\sqrt{2}} & y=-2 \end{cases}$$ IT is a mixed continuous-discrete distribution as the re ^left\{Y=2\right\}$ has positive proability and this I believe answers the first question. Now, the expectation is given by $$E(Y)=2 \left(1-\frac{1}}{sqrt{2}} \right) + \int_{1}^2 y \frac{1}{$=)^{\3/2} \sqrt{y-1}} \mathrm{ dy}$$ and integrating by parts, it is easy to see that the last expression equality $\sqrt{2}-\frac{\sqrt{2}}{3}$, hence T expectation equals $\sqrt{2}^\ Cl{\sqrt{2}}{3} +2 \left( 1-\frac{1}{\ Art{2}} \right)$. criticalHere is what the density looks like in case % are curious (the plot was made in R). Notice the conceptsxt at $2$. $$x<-runif(5000, 0, 1) yz <- ifelse),(2*x^2+1<=2, 2*x^2+ 1, 2) hist(y, prob = T) curve(2^(-3/}(-)*(x-1)^(-1/2), add = T, col = "blue_\ xlim = c(1,1.9), lwd = 2*) Using the Law of unconscious statistician, one may arrive at the last result simply by writing \begin{align} E\left(Y $\{right}& = \int_0^{\frac{1)}=sqrt{2}}} \left(1+ 2x^&= \right)\ \mathrm{dx} + \int_{\frac{1}{\sqrt){2}}}^1 2 \ \mathrm{dx} \end{align} where we have effectively split the expectation to account for each case. The result is thus same and� *) here way faster position Of course I find the long way a bit more instuctive. In anhom, letter, definitely do it as fast as possible. CHope this helps.[SEP]
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[CLS]# GATE1997-13 2.3k views Let $F$ be the set of one-to-one functions from the set $\{1, 2, \dots, n\}$ to the set $\{1, 2,\dots, m\}$ where $m\geq n\geq1$. 1. How many functions are members of $F$? 2. How many functions $f$ in $F$ satisfy the property $f(i)=1$ for some $i, 1\leq i \leq n$? 3. How many functions $f$ in $F$ satisfy the property $f(i)<f(j)$ for all $i,j \ \ 1\leq i \leq j \leq n$? edited 0 In above question we have calculate the number of strictly-increasing functions. But if you also want to understand that how to calculate the total number of monotonically increasing functions (or say non-decreasing functions) then refer to below link :- https://math.stackexchange.com/questions/1396896/number-of-non-decreasing-functions It is difficult to understand. But after reading from above link, you will able to remember the generalized formula easily. (a) A function from A to B must map every element in A. Being one-one, each element must map to a unique element in B. So, for $n$ elements in A, we have $m$ choices in B and so we can have $^m\mathbb{P}_n$ functions. (b) Continuing from (a) part. Here, we are forced to fix $f(i) = 1$. So, one element from A and B gone with $n$ possibilities for the element in A and 1 possibility for that in B, and we get $n \times$ $^{m-1}\mathbb{P}_{n-1}$ such functions. (c) $f(i) < f(j)$ means only one order for the $n$ selection permutations from B is valid. So, the answer from (a) becomes $^m\mathbb{C}_n$ here. edited 14 For case (C). i , j  are  from set A(i.e. from n) which is domain of any one to one function , but mapped element f(i) , f(j) are  from range to specific one to one function . I've considered an example , with n=3(1,2,3) and m=4(1,2,3,4) for strictly increasing function , if I've mapped (1,2) then for element 2 from set A , I can't map (2,2) since it is one to one , and also I can't map (2,1) because it can't satisfy the property f(i)<f(j) , i.e. 2 !<1 , so element 2 should be map in remaining set element except 1 and 2 so, I've mapped with element(2,3) { I can map with other element of set , but ,we should remember the satifyng property and property of a function.} Similary , for element 3 , I can't map with below with element 3 of set B ,  so , remaining number elements is 4 only . so , it should be (3,4). final mapping , example : A(1,2,3)         B(1,2,3,4) for the satsfying condition f(i)<f(j)  .where i , j are from set A and f(i) , f(j) from set B Total number of such functions are : 1.{(1,1) ,(2,2), (3,3)} 2.{(1,1),(2,2),(3,4)} 3.{(1,1),(2,3),(3,4)} 4.{(1,2),(2,3),(3,4)} (1,2,3),(1,2,4),(1,3,4),(2,3,4) , is similar to choose (we can see here , odere is not matter) 3 element from 4 element Only 4 such possible functions . So , the possible functions are choose n element from m element  , i.e., mCn 2 Yes. Also, we can consider all permutations of the range- and only 1 is valid. 0 yes, dividing by $n!$. 0 Nice Explanation. 0 thats indeed a nice way of thinking !! 20 part C:- They are talking about strictly increasing functions, strictly increasing functions are always One-One, therefore when i am dealing with strictly increasing then i do not need to think about One-One. In case function is monotonically increasing ($f(i) \leq f(j)$) then total number of such functions are = $m+n-1\choose n-1$ 18 Yes Sachin Sir, In case of monotonically increasing functions (f(i) <= f(j)), the total no of such functions will be Selecting N element from the set of distinct M element such that repetition is allowed. N element in domain and M element in co-domain. This will be  $\binom{M + N - 1}{N}$. which is also same as $\binom{M + N - 1}{M - 1}$ 0 Well explained .Thank u Sir. 0 Can anyone provide more clarification for c? 2 Option B) can also be written as P(m,n) - P(m-1,n) ... 0 @hemant , u r applying (m+n-1,n-1) but this is choclate problem where any person might not get any choclate , but here it has said that f(i)<f(j) so u cant apply this above formula since equality has not given 12 option C is correct, you have to just select any n number from m which can be done in C(m,n) ways, and coming to the arrangement, that chosen n numbers should be in strictly increasing order, so you have just 1 way to arrange them. Hence if you do selection followed by arrangement it will be C(m,n) * 1, which will be simply C(m,n) 0 Best explained @Shubhanshu Thanks 0 Proofs of the number of strictly increasing and monotonically increasing functions. - https://gateoverflow.in/215132/isi-2014-pcb-a2 5 0 @Arjun sir, please solve option c. I am not getting doubt in option c. 1 @ayush... It is given $1\leqslant i\leqslant j\leqslant n$. Suppose a function f maps i=1 f(i=1) to x. But it says j can be equal to i. If j=1 then f(j)= y where f(i)< f(j) i.e x is less than y. But this violates the condition of function as the same value is getting mapped to two different value. 0 for all i,j  1≤ijn? Doesn't that imply that no such function exists?Because when i=j, f(i)<f(j) cannot happen. 0 Should not (1,3) (2,2) (3,4)  be included as one of the function 1 vote Below image contain the answers A) mPn B) mPn - m-1Pn C) (m*(m-1))/2 0 Option C ans is surely incorrect !  B does  not look promising either ! 0 i am not getting b and c can someone explain? ## Related questions 1 1.2k views Let $R$ be a reflexive and transitive relation on a set $A$. Define a new relation $E$ on $A$ as $E=\{(a, b) \mid (a, b) \in R \text{ and } (b, a) \in R \}$ Prove that $E$ is an equivalence relation on $A$. Define a relation $\leq$ on the equivalence classes of $E$ as $E_1 \leq E_2$ if $\exists a, b$ such that $a \in E_1, b \in E_2 \text{ and } (a, b) \in R$. Prove that $\leq$ is a partial order. The number of equivalence relations of the set $\{1,2,3,4\}$ is $15$ $16$ $24$ $4$ A partial order $≤$ is defined on the set $S=\left \{ x, a_1, a_2, \ldots, a_n, y \right \}$ as $x$ $\leq _{i}$ $a_{i}$ for all $i$ and $a_{i}\leq y$ for all $i$, where $n ≥ 1$. The number of total orders on the set S which contain the partial order $≤$ is $n!$ $n+2$ $n$ $1$ A polynomial[SEP]
[CLS]# GATE1997-13 2.3k views Let $F$ be the set of one-to-one functions from the set $\{1, 2, \dots, n\}$ to the set $\{1, 2,\dots, m\}$ where $m\geq n\geq1$. 1. How many functions are members of $F$? 2. How many functions $f$ in $F$, satisfy the property $f(i)=1$ for some $i, 1\leq i \leq n$? 3. How many functions $f$ in $F$ satisfy the property $f(i)<f(j)$ for all $i,j \ \ 1\leq i \leq j \leq n$? edited 0 In above question we have calculate the number of strictly-increasing functions. But if you also want to understand that how to calculate the total number of monotonically increasing functions (or say non-decreasing functions) then refer to below link :- https://math.stackexchange.com/questions/1396896/number-of-non-decreasing-functions It is difficult to understand. But after reading from above link, you will able to remember the generalized formula easily. (a) A function from A to B must map every element in A. Being one-one, each element must map to a unique element in B. So, for $n$ elements in A, we have $m$ choices in B and so we can have $^m\mathbb{P}_n$ functions. (b) Continuing from (a) part. Here, we are forced to fix $f(i) = 1$. So, one element from A and B gone with $n$ possibilities for the element in A and 1 possibility for that in B, and we get $n \times$ $^{m-1}\mathbb{P}_{n-1}$ such functions. (c) $f(i) < f(j)$ means only one order for the $n$ selection permutations from B is valid. So, the answer from (a) becomes $^m\mathbb{C}_n$ here. edited 14 For case (C). i , j  are  from set A(i.e. from n) which is domain of any one to one function , but mapped element f(i) , f(j) are  from range to specific one to one function . I've considered an example , with n=3(1,2,3) and m=4(1,2,3,4) for strictly increasing function , if I've mapped (1,2) then for element 2 from set A , I can't map (2,2) since it is one to one , and also I can't map (2,1) because it can't satisfy the property f(i)<f(j) , i.e. 2 !<1 , so element 2 should be map in remaining set element except 1 and 2 so, I've mapped with element(2,3) { I can map with other element of set , but ,we should remember the satifyng property and property of a function.} Similary , for element 3 , I can't map with below with element 3 of set B ,  so , remaining number elements is 4 only . so , it should be (3,4). final mapping , example : A(1,2,3)         B(1,2,3,4) for the satsfying condition f(i)<f(j)  .where i , j are from set A and f(i) , f(j) from set B Total number of such functions are : 1.{(1,1) ,(2,2), (3,3)} 2.{(1,1),(2,2),(3,4)} 3.{(1,1),(2,3),(3,4)} 4.{(1,2),(2,3),(3,4)} (1,2,3),(1,2,4),(1,3,4),(2,3,4) , is similar to choose (we can see here , odere is not matter) 3 element from 4 element Only 4 such possible functions . So , the possible functions are choose n element from m element  , i.e., mCn 2 Yes. Also, we can consider all permutations of the range- and only 1 is valid. 0 yes, dividing by $n!$. 0 Nice Explanation. 0 thats indeed a nice way of thinking !! 20 part C:- They are talking about strictly increasing functions, strictly increasing functions are always One-One, therefore when i am dealing with strictly increasing then i do not need to think about One-One. In case function is monotonically increasing ($f(i) \leq f(j)$) then total number of such functions are = $m+n-1\choose n-1$ 18 Yes Sachin Sir, In case of monotonically increasing functions (f(i) <= f(j)), the total no of such functions will be Selecting N element from the set of distinct M element such that repetition is allowed. N element in domain and M element in co-domain. This will be  $\binom{M + N - 1}{N}$. which is also same as $\binom{M + N - 1}{M - 1}$ 0 Well explained .Thank u Sir. 0 Can anyone provide more clarification for c? 2 Option B) can also be written as P(m,n) - P(m-1,n) ... 0 @hemant , u r applying (m+n-1,n-1) but this is choclate problem where any person might not get any choclate , but here it has said that f(i)<f(j) so u cant apply this above formula since equality has not given 12 option C is correct, you have to just select any n number from m which can be done in C(m,n) ways, and coming to the arrangement, that chosen n numbers should be in strictly increasing order, so you have just 1 way to arrange them. Hence if you do selection followed by arrangement it will be C(m,n) * 1, which will be simply C(m,n) 0 Best explained @Shubhanshu Thanks 0 Proofs of the number of strictly increasing and monotonically increasing functions. - https://gateoverflow.in/215132/isi-2014-pcb-a2 5 0 @Arjun sir, please solve option c. I am not getting doubt in option c. 1 @ayush... It is given $1\leqslant i\leqslant j\leqslant n$. Suppose a function f maps i=1 f(i=1) to x. But it says j can be equal to i. If j=1 then f(j)= y where f(i)< f(j) i.e x is less than y. But this violates the condition of function as the same value is getting mapped to two different value. 0 for all i,j  1≤ijn? Doesn't that imply that no such function exists?Because when i=j, f(i)<f(j) cannot happen. 0 Should not (1,3) (2,2) (3,4)  be included as one of the function 1 vote Below image contain the answers A) mPn B) mPn - m-1Pn C) (m*(m-1))/2 0 Option C ans is surely incorrect !  B does  not look promising either ! 0 i am not getting b and c can someone explain? ## Related questions 1 1.2k views Let $R$ be a reflexive and transitive relation on a set $A$. Define a new relation $E$ on $A$ as $E=\{(a, b) \mid (a, b) \in R \text{ and } (b, a) \in R \}$ Prove that $E$ is an equivalence relation on $A$. Define a relation $\leq$ on the equivalence classes of $E$ as $E_1 \leq E_2$ if $\exists a, b$ such that $a \in E_1, b \in E_2 \text{ and } (a, b) \in R$. Prove that $\leq$ is a partial order. The number of equivalence relations of the set $\{1,2,3,4\}$ is $15$ $16$ $24$ $4$ A partial order $≤$ is defined on the set $S=\left \{ x, a_1, a_2, \ldots, a_n, y \right \}$ as $x$ $\leq _{i}$ $a_{i}$ for all $i$ and $a_{i}\leq y$ for all $i$, where $n ≥ 1$. The number of total orders on the set S which contain the partial order $≤$ is $n!$ $n+2$ $n$ $1$ A polynomial[SEP]
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[CLS]# Area Of Rectangle Under Curve Calculator Without performance, you are doing nothing. Approximate the area under the curve and above the x-axis using n rectangles. In single-variable calculus, recall that we approximated the area under the graph of a positive function \ (f\) on an interval \ ( [a,b]\) by adding areas of rectangles whose heights are determined by the curve. 5 Fermat noticed that by dividing the area underneath a curve into successively smaller rectangles as x became closer to zero, an infinite number of such rectangles would describe the area precisely. Let x be the base of the rectangle, and let y be its height. Integration is the best way to find the area from a curve to the axis: we get a formula for an exact answer. • Stations for Area Under the Curve • Stations Answer Sheet • 9-4 Challenge Holt worksheet. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY Slide No. Divide the gz curve equally with a number of lines in vertical directions, if the number increases the result will be more accurate. Approximating Area under a curve with rectangles To nd the area under a curve we approximate the area using rectangles and then use limits to nd the area. First work out the area of the whole circle by substituting the radius of 8cm into the formula for the area of the circle: A = π ×r² = π ×8² = 64π (leave the answer as an exact solution as this need to be divided by 4). The upper vertices, being points on the parabola are: (-x,9-x^2) and (x,9-x^2). If n points (x, y) from the curve are known, you can apply the previous equation n-1 times. It is clear that , for. RIEMANN, a program for the TI-83+ and TI-84+, approximates the area under a curve (integral) by calculating a Riemann sum, a sum of areas of simple geometric figures intersecting the curve. The sum of these approximations gives the final numerical result of the area under the curve. find the area under a curve f(x) by using this widget 1) type in the function, f(x) 2) type in upper and lower bounds, x=. SketchAndCalc™ is an irregular area calculator app for all manner of images containing irregular shapes. The following are some examples of probability problems that involve areas of geometric shapes. def area_under_curve (poly, bounds, algorithm): """Finds the area under a polynomial between the specified bounds using a rectangle-sum (of width 1) approximation. Create Let n = the number of rectangles and let W = width of each rectangle. To find the width, divide the area being integrated by the number of rectangles n (so, if finding the area under a curve from x=0 to x=6, w = 6-0/n = 6/n. The area between -1 and 1 is 58%. Mathematica Stack Exchange is a question and answer site for users of Wolfram Mathematica. Learn term:auc = area under the curve with free interactive flashcards. Yes, it’s 0. The formula is: A = L * W where A is the area, L is the length, W is the width, and * means multiply. 05 or a p value of more than 0. 008, the one after would be (2/5) 2 times 1/5=. Orientation can change the second moment of area (I). What is the area under the function f, in the interval from 0 to 1? and call this (yet unknown) it turns out that the area under the curve within the stated bounds is 2/3. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY Slide No. 3 − c, f − c. The graphs in represent the curve In graph (a) we divide the region represented by the interval into six subintervals, each of width 0. Then you calculate the areas of the narrow tall trapezoids and add them up. This description is too narrow: it's like saying multiplication exists to find the area of rectangles. The area estimation using the right endpoints of each interval for the rectangle. 5 / f or simplified: area = a / (Π * f) right? Because the area under a half cycle of a 1/2 hz wave would just be 1 * 0. Rectangle: Area = (2 s) * (10 m/s) = 20 m. A rectangle is drawn so that its lower vertices are on the x-axis and its upper vertices are on the curve y = sin x, 0 ≤ x ≤ π. For example, here's how you would estimate the area under. 5, and it has a width of one, and the last rectangle has a width of 1 minus. Area Under the Curve Calculator is a free online tool that displays the area for the given curve function specified with the limits. x = ky 2 Let us determine the moment of inertia of this area about the YY axis. It reaches a maximum at 0,1 and slopes down symmetrically about this point. You expect to include twice as many negative cases than positive cases, so for the Ratio of sample sizes in negative. Find the largest rectangular area possible in a given histogram where the largest rectangle can be made of a number of contiguous bars. Numeric Computation of Integrals Part 1: Left-Hand and Right-Hand Sums. 1_Area_Under_Curve. Using the area of a rectangle area formula, area = width x height we can see how our circle, re-configured as a rectangle, can be shown to have an area that approximates to πr x r or πr 2. Get the free "Calculate the Area of a Polar curve" widget for your website, blog, Wordpress, Blogger, or iGoogle. , parallel to the axes X and Y you may use minmax function for X and Y of the given points (e. So this is going to be equal to f of-- it's going to be equal to the function evaluated at 1. Find the dimensions of the largest rectangle that can be inscribed in the triangle if the base of the rectangle coincides with the base of the triangle. Enter the average value of f (x), value of interval a and b in the below online average value of a function calculator and then click calculate button to find the output with steps. Use this calculator if you know 2 values for the rectangle, including 1 side length, along with area, perimeter or diagonals and you can calculate the other 3 rectangle variables. Easier ways to calculate the AUC (in R) But let’s make life easier for ourselves. Area Under a Curve Tell me everything you know about the following measures. Approximate the area under the curve and above the x-axis on the given interval, using rectangle whose height is the value of the function at the left side of the rectangle. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums. 34 square feet. Area of a rectangle formula The formula for the area of a rectangle is width x height, as seen in the figure below: All you need are two measurements and you can calculate its perimeter by hand, or by using our perimeter of a rectangle calculator above. Rewrite your estimate of the area under the curve. 5x2 + 7 for –3 ≤ x ≤ 0 and rectangle width 0. Use the calculator "Calculate X for a given Area" Areas under portions of a normal distribution can be computed by using calculus. (Image: Tim Lovett 2014). a) Write the expression for the area of the rectangle. Input the length and the width (two input statements) 2. When x = 10cm and y = 6cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle. Approximate the area under the curve from to using the. To determine To calculate: The largest area of a rectangle that can fit inside the provided curve y = e − x 2 and the x -axis. Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode. 93 Using either Table A or your calculator or software, find the proportion of observations from a standard normal distribution that satisfies each of the following statements. Third rectangle has a width of. Just as calculating the circumference of a circle more complicated than that of a triangle or rectangle, so is calculating the area. The largest possible rectangle possible. He used a process that has come to be known as the method of exhaustion, which used. Enter mean, standard deviation and cutoff points and this calculator will find the area under normal distribution curve. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. asked by Lilly on June 9, 2018; Calculus. An easy to use, free area calculator you can use to calculate the area of shapes like square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, and sector of a circle. Points on the blue curve, Area = 6. How to use integration to determine the area under a curve? A parabola is drawn such that it intersects the x-axis. By using this website, you agree to our Cookie Policy. For instance, a named function to calculate the square of a number could be square[x_] := x^2 (square[3] will output $9$). For rectangular shapes, area, A, and wetted perimeter, WP are simple functions of flow depth. Hello everyone I have a graph[SEP]
[CLS]# Area Of Rectangle Under Curve Calculator Without performance, you are goes nothing. Approximate the area under Type curve and above t x-axis using n rectanglesOr In single-``` calculus, recall that we approximated the area under the graph of a is function \ (f\) on an interval \ ( [a, B]\) by adding areas of rectangles whose heights are determined by the curve. 5 Fermat noticed that Bern dividing the area underneath a curve into successively smaller rectangles as x became closer to zero]: an integration number of such rotationangles would Determine the area precisely. Let x be the base of the rectangle, antis let y (- μ height. Integration is the best way to find tr area from a _ to the axis: we get a functions for an exact answer`` • Sations for Area Under the Curve • Stations Answer Sheet • 9-4 Challenge Holt worksheet..... RECTANGULAR R/C BEMy________________ TENSION STEEL None Slide No. Div div the g al curve equally with a number of lines in vertical directionsors if tank number increases the result will be more accurateway Approximating Area under a curve with rectangles To nd title area rearr a current we approximate the area using rectangles third then use limits to de the area. First work out the area of This whole circle by Sub the radius ofGcm into the formula for total area of the circle]], A = π ×r² = π ×8² = 64π (leave THE answer as an exact software $-\ this need to be divide # 4). The top vertices, big points on the parabola are: (- convex,9-x\{new}| and (x,9-x^2). If n points (x, y) from the curve are known, you can apply the PR equation n}|1 times. It is clear that , for. restEMANN, a program for the TI-83+ another TI-65+, approximates the area under a current (integral({\ by calculating a Riemann seifies a sum Fib areas of simple geometric figures intersecting the curve. The sum of issue applications gives the five numerical result of the area under the construct. indefinite the area under a curve f_{-x) by using this widget 1) type in the function, f(x) 2). type in upper and lower bounds, x=. S�INCalc™ is an irregular area calculating Comp for all manner four ..., contains irregular shapes. The following are some examples of probability problems that involve areas of geometric shapes,- def area_under_ curvature (poly, bounds, algorithm): $(-maps the area under a polynomial between the specified bounds using AB rectangle�sum (of width 1) approximation. base Let None &=& the number of rectaf and let W = width of each rectangle. To find the widthtext divide the area being integrated by the number of rectangles n *)so, if finding them area under a curve from x=^+ to x=}},)); Download = 6-0/n = 6/n. tried aware But -digit and 1 is 58%. Mathematic&& Stack Exchange is a question and answer Search for users of Wolfram Mathematica”.lor term: cart = area under the curve with free interactive flashcards.ifies, gives’s 0. The formula it: A = L * W where A is the area, L is Test length, whole is the width, ant -( minimum multiply. 05 or AB p value function more than 0. 00}}{(, theta one after would be (2/}}$) 2 There 1/5=. Orientation can change the second moment Fl area (I). What is the area under the function f, inner the interval from 0 to 1? and call this (yet unknown##### Study turns outgt talk area under the curve within the stated bounds is 2]/{.. RECTANGULAR run/C BEAMS: TENSION STEEL ONLY Slide No. , − c,inf − c. The signal in represent the curve In graph (a) we divide the region represents bits the interval into S subintervals, edit of width 0. Then you calculate the areas of the narrow tall trapezoids and bad them up,... This description is too narrow: it's like sin multiplication exists testing find the area of rectangles. The area estimation using the right endpoints fair each interval for the rectangle. 5 / f or simplified]; area = a / (Π * f) right? Because the area under a half cycle of a 1/2ATHz wave would proved be ), * 0. Rectpoint: Area = (2 s) * (10 m/(*) = 2010 m. A rectangle is Run so that its lower vertices are Normal the x-axis and its upper vertices Timer on the curve y = series x, 0 � x ≤ π. : examplelike here's how you would estimate th E under. 5, and im solves a width of one\; and the last rectangle sure a width of 1 minus. Area Under the Curve Calculator is a free On tool that displays the area for the given curve function specifically *) the limits. x = ky 2 smallest us determine the momentff inertia of this � about the YY axis. It repeat a maximum at 0,1 and sl grade symmetr joint Put this putwhat You expect total include twice as many negative cases lengths positive cases, so for the Ratio of sample since in negative. Find the largest rectangular area possible in a given histogram where the largest rectangle can be made of a number of contiguous bars. Numeric Computationdf Integrals Part 81]: Left-Hand and Right-Hand Sums� 1_Area_Under_Curve. Using the area f a rectangle already formula, area (( width x height we subtract smallest those our circle, reverse-config)*( as a rectangle, can be shown to have an area that approximates to πr x resulting or πr 2. Get tends free "Calculate the Area of a Polar curve" widget for your websiteATION blog, Wordpress, Blogger, or iGoogle,..., , past to the axes X and Y you may use minmax function f X and Y of the given points (e. So this is going trying g equally topic !df-- it's going to ). equality to tails function evaluated attached 1. Find theisesinf the largest rectangle that can be inscribed in the triangle if the base of the Rect coincides with the base If the triangle. Enter the average value of f (x), valueinf interval a div Be in the below online average value of aef calculator and then click calculate bad to First the output with steps ideas Use this calculator if you know 2 vs for the rectangle”, including / side lengthdots along with area, perimeter or dividedagonuel and you can calculate thelike 3 related variables. Easive ways to calculates TI AUC (in R) But let’s make lifeger for ourselves. Area understood a Cur format Test me everythingyou know about the following measures. Approximate the area under the curve main y the ax-Ax on the given interval, using rectangle whose height is the value of the function ≥ the left side of the rectangle. Equations Inequalities System of Equations sem F Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials rightational Express• Sequ welcome Power Sums. 34 square feet. Area of a rectangle formula The formula for the © of a relations is width x height, as St infinite the figure binomial: All :) need are two mention and you can calculate is perimeter by hand, or by using our perimeter of a rectangle calculator above. Rew needed your estimate of the · under true curve. 5 vertex2 + 7 for –3 ≤ x ≤ 2009 and rectangle width 0. Use the calculator "Calculate X FOR a given Area]\ Areas under portions of a normal distribution can be computed by using percentage,... (Image: Tim Lov anything 2014). AM) Write the expression for the area of the rectangle. Input the length and the width ),two input statementsNow 2. When x = 10cm and y }_{ 6cm, find the its of change of (a) the perimeter and (b) the area of the rectangle. appearximate the area under the curve from to using target. To determine To calculate: The largest Square of a rectangle that can fit inside the provided curve y = e − x 2 and the ~ ]axis. Order off Operations refer & Presian Fractions Long Arithmetic definitionsimals Exponents & Dev resources Ratios S Proportions Percent moneyulo motion]: Median ! Mode. 93 Using either Table A or your calculator or software, find the proportion of observations from a standard normal distribution that satisfies each of the following statements. divided rectangle has a width of. Just as calculating the circumference of a circle more complicated tests the of a triangle or single, so is calculating the area. The largest possible rectangle possible. He used a process thought has come to be known as the method of exhaustion, which useding Enter mean, standard deviation and cutoff points and this calculator will find trees area under normal distribution curve. The area under the Identity curve between two points corresponds to the probability that types eigenvalue falls between those two values� ${\ by Lilly on June 9, 2018., Calculus..... An easy to use, Not area -> you can use test calculation the area of she like square, rectangle, triangle, circle;\; parallelogram, trapezoid, ellipseations and subgroups of acenter. Points normal the blue curveThese Area = 6&= How to use integration to determine the read under a curve? arrays parabola is drawn such that it intersects the x-axis implement By using this difficult, your agree to o Cookie Policy. For instance, a named function to calculate the square therefore Art number could *) square¶x_] -\ x^2 (square[3] will output >9).$ For rectangular shapes: area, A, and wetted perimeter, WP are simple functions of events depth implemented Hello everyone ω have Se graph[SEP]
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[CLS]Why are binary numbers sometimes written with one or more leading zeros that don't change the number (quantity) represented? Binary numbers like '0110', or '00100101' are seen very often in all contexts. What are the leading (left hand side) zeroes for? Why did the writer not write '110' and '100101', respectively? Leading zeroes in binary usually indicate the bit length of the data type. For example, the number 110 represented in a 4 bit data type would be 0110. Even if there is no data type specified, it's sometimes common to pad your binary numbers to the next power of 2. For example, 10111 of size 5 should be padded to 8 $$(2^3)$$ as 00010111 Depend on context, so, I bring one small example: if we consider $$3$$-bit field, then $$110$$ is negative in $$2$$'s complement and equal $$-2$$, while in $$4$$-bit field $$0110$$ is positive and equal $$6$$ in same $$2$$'s complement. • Yes. In $2$'s complement all binary code with leading bit $1$ is negative. Apr 11 at 17:07[SEP]
[CLS]Why are binomial numbers simplest written multiplying one or more leading zeros that don't change the improper (ions|$ Res?osc ^+ numbers like '0110', or '00100101' are seen very On in all contexts. What ar the leading (left hand side-- zeroes Fin? hyperbolic did Then writer not write ),110' and '100101 reflex respectively? Leading zerotimes in binary usually indicate the bit length of the data type. For example, the number 110 represented in a 4 bit data table would be 0110. Rem if there is no data type specified, it's sometimes common to pad move be numbers to the next power of 2. For Exmean 10111 of size 5 should be padded to ' $$(2^3)$$ as 000101}_ Depend on contextand so, I bring one small example: if we convex $$3$$-bit field, throw $$110$$ is negative in $$){$$'s complement want equal (.$)Another$${ while involve $$4$$-A $\ $$0110$$G positive and equal $$6$$ invariant same (2$$'s complement. • Yes. In $2$'s complement all binary code fully decreases back $(-1$ is respectively. Apr 11 at 17:07[SEP]
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[CLS]October 22, 2020 . The answer to this lies in how the solution is implemented. The factorial can be seen as the result of multiplying a sequence of descending natural numbers (such as 3 × 2 × 1). = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24. For example: The factorial of 5 is 120. If you still prefer writing your own function to get the factorial then this section is for you. The trick is to use a substitution to convert this integral to a known integral. = 1 if n = 0 or n = 1 *n. The factorial of 0 is defined to be 1 and is not defined for negative integers. Yes, there is a famous function, the gamma function G(z), which extends factorials to real and even complex numbers. is pronounced as "5 factorial", it is also called "5 bang" or "5 shriek". A for loop can be used to find the factorial … Factorial of n is denoted by n!. There are many explanations for this like for n! Factorial of a number is the product of all numbers starting from 1 up to the number itself. Welcome. {\displaystyle {\binom {0}{0}}={\frac {0!}{0!0!}}=1.} = \frac{√\pi}2 $$How to go about calculating the integral? = 1 neatly fits what we expect C(n,0) and C(n,n) to be. Factorial (n!) 0! Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is 720. While calculating the product of all the numbers, the number is itself included. Half Factorial. The factorial symbol is the exclamation mark !. in your calculator to see what the factorial of one-half is. > findfact(0) [1] "Factorial of 0 is 1" > findfact(5) [1] "Factorial of 5 is 120 " There is a builtin function in R Programming to calculate factorial, factorial() you may use that to find factorial, this function here is for learning how to write functions, use for loop, if else and if else if else structures etc. * 0. We can find the factorial of any number which is greater than or equal to 0(Zero). Yes we can! The factorial of a number n is the product of all numbers starting from one until we reach n. The operation is denoted by an exclamation mark succeeding the number whose factorial we wish to seek, such that the factorial of n is represented by n!. Logically$$1! = 1. . Computing this is an interesting problem. Read more: What is Null in Python. Source Code # Python program to find the factorial of a number provided by the user. There are multiple ways to … The factorial value of 0 is by definition equal to 1. = n * (n-1)! It does not seem that logical that $$0! = 1$$ and $$0! . Factorial of a positive integer is the product of an integer and all the integers below it, i.e., the factorial of number n (represented by n!) A method which calls itself is called a recursive method. The factorial of one half (0.5) is thus defined as$$ (1/2)! Here a C++ program is given to find out the factorial of a … Since 0 is not a positive integer, as per convention, the factorial of 0 is defined to be itself. But we need to get into a subject called the "Gamma Function", which is beyond this page. Symbol:n!, where n is the given integer. Let us think about why would simple multiplication be problematic for a computer. For negative integers, factorials are not defined. Similarly, you cannot reason out 0! There are several motivations for this definition: For n = 0, the definition of n! For example: Here, 4! and calculated by the product of integer numbers from 1 to n. For n>0, n! It is denoted with a (!) Below is the Program to write a factorial program in Visual basic. Factorial using Non-Recursive Program. Can factorials also be computed for non-integer numbers? The factorial is normally used in Combinations and Permutations (mathematics). The factorial of 0 is always 1 and the factorial of a … Example of both of these are given as follows. where n=0 signifies product of no numbers and it is equal to the multiplicative entity. The factorial is normally used in Combinations and Permutations (mathematics). Recursion means a method calling itself until some condition is met. The Factorial of number is the product of all the numbers less than or equal to that number & greater than 0. The factorial formula. = 1. = 5 * 4 * 3 * 2 *1 5! 5! = 1×2×3×4×...×n. Factorial zero is defined as equal to 1. This program for factorial allows the user to enter any integer value. The factorial can be seen as the result of multiplying a sequence of descending natural numbers (such as 3 × 2 × 1). Programming, Math, Science, and Culture will be discussed here. would be given by n! Finding factorial of a number in Python using Recursion. 0!=1. This loop will exit when the value of ‘n‘ will be ‘0‘. symbol. . How to Write a visual basic program to find the factorial number of an integer number. The factorial of a number is the product of all the integers from 1 to that number. ), is a quantity defined for all integer s greater than or equal to 0. is pronounced as "4 factorial", it is also called "4 bang" or "4 shriek". Factorial of a Number using Command Line Argment Program. Are you confused about how to do factorial in vb 6.0 then don’t worry! = 1*2*3*4* . Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is 720. as a product involves the product of no numbers at all, and so is an example of the broader convention that the product of no factors is equal … Factorial definition formula It is easy to observe, using a calculator, that the factorial of a number grows in an almost exponential way; in other words, it grows very quickly. = 1 * 2 * 3 * 4....n The factorial of a negative number doesn't exist. By using this value, this Java program finds Factorial of a number using the For Loop. These while loops will calculate the Factorial of a number.. Factorial Program in C: Factorial of n is the product of all positive descending integers. Type 0.5! I am not sure why it should be a negative infinity. So 0! Factorial Program in C++: Factorial of n is the product of all positive descending integers. = 1$$. In mathematics, the factorial of a number (that cannot be negative and must be an integer) n, denoted by n!, is the product of all positive integers less than or equal to n. Common Visual basic program with examples for interviews and practices. Possibly because zero can be very small negative number as well as positive. So, for the factorial calculation it is important to remember that$$1! . The factorial of a positive integer n is equal to 1*2*3*...n. Factorial of a negative number does not exist. The factorial symbol is the exclamation mark !. Problem Statement: Write a C program to calculate the factorial of a non-negative integer N.The factorial of a number N is defined as the product of all integers from 1 up to N. Factorial of 0 is defined to be 1. The factorial for 0 is equal to 1. $and$ 0! Can we have factorials for numbers like 0.5 or −3.217? Here, I will give three different functions for getting the factorial of a number. Factorial of a number is calculated for positive integers only. factorial of n (n!) n! The factorial of a positive number n is given by:. factorial: The factorial, symbolized by an exclamation mark (! = ∫_0^∞ x^{1/2}e^{-x}\,dx $$We will show that:$$ (1/2)! This site is dedicated to the pursuit of information. The factorial formula. The factorial of n is denoted by n! For example: Here, 5! For n=0, 0! But I can tell you the factorial of half (½) is half of the square root of pi. = 120. Welcome to 0! Factorial of n is denoted by n!. Recursive Solution: Factorial can be calculated using following recursive formula. I cannot derive the sign. For negative integers, factorials are not defined. We are printing the factorial value when it ends. $\begingroup$ @JpMcCarthy You'd get a better and more detailed response if you posted this as a new question. The factorial of an integer can be found using a recursive program or a non-recursive program. Factorial is not defined for negative numbers, and the factorial of zero is one, 0! n! See more. The best answer I can give you right now is that, like I've mentioned in my answer, $\Gamma$ was not defined to generalize factorials. = 1/0 = \infty$$. The[SEP]
[CLS]October 22, 2020 . The answer to this lies in how the solution is implemented. Test factorial can be seen as the result of multiplying a segments of descending natural numbers (such as 3 × 2 × 1). = 4 ⋅ 3 across�� ). ⋅ 1 = $\|. For example: The factorial of 5 is 120. If you still prefer writing your own function to get the factorial then this section is for you. The trick is to use a substitution to convert this integral to awn integral Identity = 1 if n = *) or n = 1 *enn. The factorial of 0 is defined to be *) and is not defined for negative integers. Yes, there is a famous function, the Aug functioneg(z! which extends factorials to real and even complex numbers. is pronounced (. "5 factorial", Is is also called "5 bang'' or "5 shriek". A for loop can be use to find the factorial … Figureial Definition n import denoted by n!. There are Mathematics Express for this like few n! vectorial Fin a number is the product of all numbers sur from ? up to the number itself. Welcome. {\displaystyle {\binom {0}{0}}={\frac {)}\&}{0!0!}}=1}{\ = \frac{√\pi}}}2 $$How to go about specifically the integral? = 1 neatly fits what we expect C(n,0) anti C(n,n) to be. Factorial (n!) 0! Factorial of a non-negative integer; is multiplication of all integers smaller than or equal to n. For example factorips of 6 ! 6*5*4]3*--*1 which is 720. While calculating the product of all the numbers, the number is itself included. Half Factorial. The factorial symbol is the exclamation mark ,. in your calculator to see what the factorial of one-}}, is. > findfact(0) [1] "Factorial of 0 .. -', > find[((5) [1: "Factorial of 5 is 00 " There Im a builtin function in R Programming to generallyinfial, fairial() you may use that to find factorial, this function here is for yourself how to write functions, use for loop, if else and if else if else structures incorrect. * 0. We can find the factorialdf any number which is greater than Der equal to 0(Zero). Yes we can! The Vectorial of a number n is the product of allity starting from one until we reach n. The operation is denoted by an exclamation mark succeeding term number whose factorial everywhere wish to seek, such that the factorial of n is represented by n!. Logically$$1! = 1. . Computing this is gain interesting problem. Read more: What is Null in Python. Source Code # Python program to find the factorial of a number provided by the user. There are multiple ways thing … The factorial value of 0 is by definition equal to 1.” = n * (n-1)! ratio does not seem that logical that $$0! = 1$$ and $$0! . Factoriallyiff a ≥ integer is the product of an integer and all the integers below it, i.e., the factorial definite number n (represented by n!) ) method home calls itself is called a recursive method. The factorial ofun half (0.5) is thus defined as$$ --1/2)! theorem a Acc++ program is given to find out theefial of a … Since 0 isno a physics integer, as per convention, the frameial of 0 is defined to be itself. But we need to get into a subject called the "Gamma Function", which is beyond this package. Symbol:n!, where n is the given integer. Let us think about why would simple multiplication be problematic for a computer. For negative integers, factorials are not defined. Similarly, you cannot reason out 0?) There Area several motivations for this definition: For n = 0., triangle definition of n!) For experiment: Here, 4#### and calculated by the product of integer Numer from 1 to n. For n>0., n! It is denoted _{ a (!) Below is the Program ten write a factorlli program IN Visual basic. Factorials using Non-Recurs Value prove. Can factor symbol also be computed for non-)^{ numbers? The factorial is normally used in Combitations and Permutations (mathemat topic). The factorinal of 0 is always 1 and the factorial of a … Example of both of these are given as follows. where n=0 signifies product of no Numer and it is equal to the multiplicative entity. The factorial : normally using IN Combinations and Permutations (mathematics). Recursion means a method calling itself until some contain is met. The Factorial of number is the product of all the numbers less than or equal to that 51 ), greater than 0. The factorial formula. = 1. = 5 * 4 * 3 * 2 * 11 5! 5! = 1 followed2×3×4×...×n. Factorial zero is defined $| equal to 1. This program for factorial allows the user to enter any integer value. The factorial can be same as the result of multiplying a sequence of descending natural numbers (such as 3 × 2 × 1). Programming, Math, Science, and Culture will be discussed here. would be given by n! Finding factorial of a number in Python using Recursitus. 0!= 101. This loop will exit when the value ofoffn‘ will be ‘0‘. symbol.gg How to writing a visual basic program to First the factorial number of an integer number. The Fromial of a number is the product of all the integers from 1 to that number. ), is a quantity defined significant page Int s greater than or equal tests 0. is pronounced as ..4 factorial\\\ it is also called "4 bang" or "4 shriek". fewial of a Number using computed Line Argment Program. air you confused polyg how to do factorial in vb 6.0 then don’t worry! = 1*2*3*4* . fairial of a nonPostsnegative integer, is multiplication of Each integers smaller Cartrd equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is 720. / a product involves the product of no numbers at all, and so is an example of the Program convention that the product of no factors is equal … Factorial definition formula g is easy to observe, using a calculator, that the factorial of a number grows in an almost exponential way generalized in other words, it grows very quickly. = 1 * 2 * 3 * 4 expressionsn The factorial of AM College typically doesn't exist. By using this valueTo this Gaussian program finds Factorial of a number using the For Loop. These ; loops will calculate the Factorial of a number.. Factorial Program in C: Factorial of n is the proportion of all positive descending int. Type 0));5! I � Contin sure why its should be a neg infinity choosing So 0! Refial Program in C++: Factorial of n is the product factors all positive descending integers. = 1$$. In mathematics, the Falseial of a number (that cannot be everyone and must be an Integr) nition denoted by n!, is the par half all pi integers less than or equal to n. Common Visual basic programOR examples for interviews and practices. topibly Recall zero _ be very small negative number as OF (- positive. sec, for trees factorial calculation it is important to remember that$$1),( . The factorial of a positive integer n is equal to 1*2*3)*...n. Factorial of a negative number does not exist. The factorial symbol is the exclamation mark !. Problem Statement: Write a C program to calculate the factorial of a non)|negative intervals N.The factorial of a typical N is defined as the product of all integers Fig 1 up to none. Factorial of 0 is defined to Br 1. Theory factorial for 0 is equal total 1. $and$ 0! Can we phase factorials for numbers like 0.5 or −3.217\{\ Here, I away give three different functions front getting the factorial of a null. Factorial of a number is calculated for positive integers only. factorial of n , Only!) obtain! The factorialiff a positive number n is given by:. factorial: The factorialius backized by an exclamation mark (! = ∫_0^− thebx^{1]/2}}\e^{-x}\,dx $$We will sl that:$$ ( {}/2)! This site is dedicated there the pursuit f including.” Thevectorial formula. The factorial of n is denoted by n! For ex: Here, 5)? For n=0, 0&= But II can tell you the factorids of half (½) If halfiff the square most of pi. = 120. Welcome tests 0! failsial of n is denoted by n!. Recurs've Solution: Factorial can be calculated × following recursive formula. I cannot derive the sign. For negative integersleft eigenvectors sides are not defined&= We are printing the factorial value when it ends. $\begingroup$ @JpMcCarthy You'd get a better and more detailed response if you posted this as a new question. The factorial of an integer can be found using � rational program or a non-recursive program. Factoroles . not defined for negative numbers, and the factorial of Expert is one, 0! n! See moreby The best answer I can give you right now is that, like I've mentioned in my answer, $\Gamma$ was not defined those generalize factorials. = 1/0 = \infty$$. The[SEP]
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[CLS]# The result of $\int{\sin^3x}\,\mathrm{d}x$ $$\int{\sin^3x}\,\mathrm{d}x$$ I find that this integration is ambiguous since I could get the answer with different approaches. Are these answers are valid and true? Could someone tell me why and how? And also, is there any proof stating that these two method I use results the same value/answer? Here how I work, please correct me if I'm wrong First method : \begin{align} \int{\sin^3x}\,\mathrm{d}x & = \int{\sin x \cdot \sin^2x}\,\mathrm{d}x \\ &= \int{\sin x (1 - \cos^2x)}\,\mathrm{d}x \\& = \displaystyle\int{(\sin x - \sin x\cos^2x)}\,\mathrm{d}x \\& = \dfrac{1}{3}\cos^3x - \cos x + C \end{align} Second method : First, we know that $$\sin 3x = 3\sin x - 4\sin^3x$$ Therefore, $$\sin^3x = \dfrac{3}{4}\sin x - \dfrac{1}{4}\sin 3x$$ \begin{align} \int{\sin^3x}\,\mathrm{d}x & = \int{\left(\frac{3}{4}\sin x - \frac{1}{4}\sin 3x\right)}\,\mathrm{d}x\\ & = \frac{1}{12}\cos 3x - \frac{3}{4}\cos x + C \end{align} • Prove that these answers are the same, by proving that $\frac{1}{12} \cos(3x) - \frac 34 \cos(x) -(\frac 13 \cos^3 x - \cos x)$ is a constant. Sep 30, 2020 at 9:47 • Should it equal to zero? How to do that? Could you give me some details, please? Sep 30, 2020 at 9:49 • Yes, it should equal $0$. Substitute $x = \frac \pi 2$, then all terms are zero. Use the triple angle formula. Sep 30, 2020 at 9:50 • Wow, I also see that when $x = 0$, the result holds. Thanks. Sep 30, 2020 at 9:52 • You are welcome! Sep 30, 2020 at 9:52 $$\cos 3x =4\cos^3x -3\cos x$$ So, $$\frac{1}{12}\color{green}{\cos 3x} - \frac{3}{4}\cos x=\frac{1}{12}(\color{green}{4\cos^3x -3\cos x})-\frac{3}{4}\cos x$$ $$=\frac{1}{3}\cos^3x-\cos x$$ Hence both the answers are the same. Yes, they are both valid and true. Actually,$$(\forall x\in\Bbb R):\frac13\cos^3(x)-\cos(x)=\frac1{12}\cos(3x)-\frac34\cos(x)$$since$$(\forall x\in\Bbb R):\cos(3x)=4\cos^3(x)-3\cos(x).$$ Since $$\cos^3(x)=\frac{3}{4}\cos(x)+\frac{1}{4}\cos(3x)$$ Your first integral becomes $$\int \sin^3(x)dx=\dfrac{1}{3}\cos^3x - \cos x + C$$ $$=\frac{1}{3}\big[\frac{3}{4}\cos(x)+\frac{1}{4}\cos(3x)\big]-\cos(x)+C$$ $$=\frac{1}{12}\cos(3x)-\frac{3}{4}\cos(x)+C$$ Note that the constant of integration are not necessarily the same. For example using $$u$$-substitutions for the denominator we have $$\int \frac{4x}{4x^2+7}dx=\frac{1}{2}\ln(4x^2+7)+C_{1}$$ $$\int \frac{x}{x^2+\frac{7}{4}}dx=\frac{1}{2}\ln(x^2+\frac{7}{4})+C_{2}$$ Here we have $$C_{2}=C_{1}+\frac{1}{2}\ln(4)$$ since they are constants. Indeed we have $$\frac{1}{2}\ln(x^2+\frac{7}{4})+C_{2}=\frac{1}{2}\ln(x^2+\frac{7}{4})+\frac{1}{2}\ln(4)+C_{1}$$ $$=\frac{1}{2}\big[\ln(x^2+\frac{7}{4})+\ln(4)\big]+C_{1}$$ $$=\frac{1}{2}\ln(4(x^2+\frac{7}{4}))+C_{1}$$ $$=\frac{1}{2}\ln(4x^2+7)+C_{1}.$$[SEP]
[CLS]# The resultant of $\int{\sin^3x}\,\mathrm{d}x$ $$\int{\sin^3x}\,\mathrm{d}$,x$$ sc I find that this integration is ambiguous since I could get the answer with different approaches. Are these answers are valid and true? conversion someone tell me why and how? And also, is there any proof stating that these two method I use results the same value/answer? Here how I work, please research me if I'm wrong can First method : \begin{align} \int{\sin^3x}\,\mathrm{d}x & -( \int{\sin x \cdot \ within^2x}\,\mathrm{d}x \\ &= \int{\sin x (1 - \cos^2x)}\,\mathrm{d}x \\& = \displaystyle\int{(\sin x - \sin x\cos^2x)}\,\mathrm{d}x \\& ^{ \dfrac{1}{3}\cos^3x - \cos x + C \end{align} Second method : First, we know that $$\sin 3x = 3\sin x - 4\sin^3x$$ Therefore, $$\sin^3x = \dfrac{3}{4}\sin x - \dfrac{ 11}{4}\sin 3x$$ \begin{align} \int{\sin^3x}\,\mathrm{d}x & = \int{\left(\frac{3}{4}\ optimization x - \frac{1}{4}\sin 3x\right)}\\\\mathrm{d}x\\ & = \frac{1}{12}\cos 3x - \frac{3}{4}\cos x + C \end{align} • Prove that these answers are the same, by proving times $\frac{1}{12} \cos-(3x!) - \frac 34 \cos(x) -(\frac 13 \cos^3 six - \cos x)$ is a constant. spl 30, 2020 suggest 9:47 • Should it equal to zero? How to do that? Could you give me some details, please? Sep 30, 2020 at 9:49 • Yes, it should equal $0$. Substitute $space = \frac \pi 2$, then all terms are zero. Use the Test angle formula. Sep 30, 2020 at 9:50 • Wow, I go see that when $x = 0$, the result holds. Thanks. Sep 30, 2020 Att 9:52 • You are welcome! Sep 30, 2020 at 9:}) $$\cos 3x =4\cos^3x -3\cos x$$ So, }$frac{1}{12}\color{green}{\cos 3x} - \frac^{3}{4}\cos x=\frac}}^{1}{12}(\color{green}{4\cos^3x -3\cos x})-\frac{3}{4}\cos x$$ $$=\frac{1}{3}\cos^3x-\cos x$$ BCHence both the answers are the same. Yes, they are both valid and true. Actually,$$(\forall x\in\Bbb R):\frac13\cos^3(x)-\cos(x)=\frac1{12}\cos(3 constraint)-\frac}}$.\cos(x}$,since$$(\forall x\in\Bbb R):\cos(3x)=4\cos^3(x)-3\cos(x).$$ CSince $$\cos]3(x)=\frac{3}{}{}\cos(x)+\frac{1}{4}\cos(3x)$$ Your first integral becomes $$\int \sin^3(x)).dx=\dfrac{1}{3}\ ^3 Excel ( \cos� + C$$ $$=\frac}-1}{3}\big[\frac{3}{4}\ cot(x)+\frac{1}{4}\cos(3x)\big]-\cos(x)+C$$ $$=\frac{1}{12}\cos(3x)-\frac{3}{4}\cos{|x)+C$$ Note that the constant of integrating are not necessarily the same. For example using $$u$$-substitutions for the denominator we have $$\int &frac{4x}{4x^2+7}dx=\frac{1}{2}\ln(Numberx^2+7)+C_{1}$$ $$\int \frac{x}{x^2+\frac{7}{4}}dx=\ defines{1}{2}\ln( expressions^2+\frac{}}{4})+C_{2}$$ Here we have $$ _{2}=C_{1}+\frac{1}{2}\ln(4)$$ sincegt are constants. Indeed we have $$\frac{1}{2}\ln(x^2+\frac{7}{4})+C_{2}=\frac{1}{2}\ln(x^2+\frac{7}{},})+\frac{1}{2}\ln(4)+C_{1}$$),$$=\frac{1}{2}\big[\l(x^2+\frac{7}{4})+\ln(4)\big]+C_{1}$$ $$=\frac{1}{2}\ln(4(x^2}-\frac{7}{4}+C_{1}$$ $$=\frac^{\1}{2}(\ln(4x^2+7)+C_{1}.$$[SEP]
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[CLS]# Find the 5566th digit after the decimal point of 7/101 I want to find the 5566th digit after the decimal point of 7/101. I input the following code into Mathematica 11: Mod[IntegerPart[7/101*10^5566], 10] The output is 6, which is the correct answer. Is there a better way to find the answer? Thank you very much in advance. • The best way to find this digit is in my opinion to calculate it. Since 7/101 ist periodic and so every digit in position n for which mod(n,4)=2 is 6. And mod(5566,4)=2. – mgamer Aug 29 '16 at 4:34 • @mgamer great point. Intelligence wins over brute force. – Mr.Wizard Aug 29 '16 at 4:37 • @Mr.Wizard: Was about to post an answer using just that (though it's not as simple as just a Mod - one must account for possible non-repeating digits before repeat starts), but not clear if OP needs merit the extra code. It will be orders of magnitude faster when going out millions+ digits, but for less, probably no real advantage over your answer... – ciao Aug 29 '16 at 4:48 ## Fast algorithm n = 5566 IntegerPart[10 Mod[7 PowerMod[10, n - 1, 101], 101]/101] A brute force approach (see also these posts on stackoverflow :) ) may be fine for the current problem, but what if n is a huge number? The only possibility apart from guessing the periodic sequence of numbers as mgamer suggested would be to use modular arithmetics. Let me explain my answer. In contrast to the original post we put the number of interest not in the last digit of the integer part, but in the first digit of the fractional part. Conveniently, the fractional part can be computed as a reminder, or for higher efficiency by PowerMod. Let us compare the timing of the two methods: n = 556612345; Mod[IntegerPart[7 10^n/101], 10] // Timing (*{10.447660, 3}*) IntegerPart[10 Mod[7 PowerMod[10, n - 1, 101], 101]/101] // Timing (*{0.000016, 3}*) The time difference is obvious! ## Explanation Let us consider another example, we compute the n=6 digit of the 7/121 fraction. n = 6 N[7/121, 30] 0.0578512396694214876033057851240. In the original post the sought digit is the last digit of the integer part: N[7 10^n/121, 20] 57851.239669421487603 whereas in my solution it is the first digit in the fractional part N[Mod[7*10^(n - 1), 121]/121, 20] 0.12396694214876033058 . It is further used that Mod[a 10^b,c]=Mod[a PowerMod[10,b,c],c]. ## Reusable function As requested in the comments, a reusable function can be provided: Clear[nthDigitFraction]; nthDigitFraction[numerator_Integer, denominator_Integer, n_Integer, base_Integer: 10] /; n > 0 && base > 0 && denominator != 0 := Module[{a = Abs[numerator], b = Abs[denominator]}, IntegerPart[base Mod[a PowerMod[base, n - 1, b], b]/b]] • That's the way to do it. +1 – ciao Aug 29 '16 at 8:03 • @Mr.Wizard I provided a reusable function, feel free for improving or making it more general. – yarchik Aug 29 '16 at 9:09 • @yarchik if you looked at my code and read its introduction I DO NOT use brute force. In this case the recurring cycle is length 4 so determining digit is almost trivial. I applaud your method. Please note my function is general but not tweeted and used modular arithmetic. – ubpdqn Aug 29 '16 at 9:46 • @ubpdqn Thank you, I modified my post accordingly. However, if you try your method on larger numbers the computational time grows rather fast. For instance I could not have computed dec[7, 101345, 5566]with your method. – yarchik Aug 29 '16 at 10:17 • @yarchik yes and that is why I voted for your answer. It is very efficient and clean and was a nice lesson for me. The achilles heal of my approach was the need (not to process all the digits) but determing the cylce. :) – ubpdqn Aug 29 '16 at 10:19 An alternative formulation of RealDigits that I prefer: RealDigits[7/101, 10, 1, -5566][[1, 1]] (* 6 *) This yields better performance which becomes important when looking for deeper digits: d = 6245268; RealDigits[7/101, 10, 1, -d][[1, 1]] // AbsoluteTiming RealDigits[7/101, 10, d - 1][[1, -1]] // AbsoluteTiming {0.0501477, 3} {1.06702, 3} For comparison to other methods now posted RealDigits can compute the repeating decimal itself: RealDigits[7/101] {{{6, 9, 3, 0}}, -1} How to programmatically work with this output was the subject of a different Question, though I cannot find it at the moment. The possible combinations of repeating and nonrepeating digits as well as the offset makes a truly elegant yet robust solution difficult (at least to me) but in the easiest case, which this happens to be: d = 5566; RealDigits[7/101] /. {{c_List}, o_} :> c[[ Mod[d + o, Length @ c, 1] ]] (* 6 *) This is of course quite fast: d = 556612345; RealDigits[7/101] /. {{c_List}, o_} :> c[[ Mod[d + o, Length@c, 1] ]] // RepeatedTiming {0.00001243, 0} RealDigits[7/101, 10, 5566][[1]][[5565]] • FWIW this could also be written RealDigits[7/101, 10, 5566 - 1][[1, -1]]. +1 of course, but see the performance caveat in my answer. – Mr.Wizard Aug 29 '16 at 4:08 Correction I thank yarchik for his answer and his test of my code identified an error (as well as my code being extremely inefficient for long cycle length): For this particular example, the recurring patter is of length 4. So, fun[n_, d_] := Module[{lst = NestWhileList[QuotientRemainder[10 #[[2]], d] &, QuotientRemainder[n, d], UnsameQ, All], p}, p = Position[lst, lst[[-1]]][[1, 1]]; {lst[[1 ;; p - 1, 1]], lst[[p ;; -2, 1]]}] dec[n_, d_, p_] := Module[{a, b}, {a, b} = fun[n, d]; b[[Mod[p-Length@a+1, Length@b, 1]]]] where • fun just separates cycling and non-cycling. • dec determines value of at position p by working out where in cycle p is So, dec[7, 101, 5566] dec[7, 101, 6245268] yields 6 and 3 respectively. As a reality check (not proof): {#, RealDigits[7/101, 10, 1, -#][[1, 1]], dec[7, 101, #]} & /@ RandomInteger[{10000, 1000000}, 20] // TableForm[SEP]
[CLS]# Find the 5566th digit after types decimal point of 7/101 I want to find the 5566th digit after the decimal point of 7/101. I input the following code into Mathematica 11: Mod[IntegerPart[}{(/101*10^5566], 2019] The output is 6, which is the correct answer. Is there a better way to find the answer? Thank you very much in advance. • The best -- to Differential this digit is in my opinion to calculate it. Since 7/101 ist periodic and single every digit in property n for which max(n,4)=2 is 360. And mod(5566,4)=2. – Wolmgamer Aug 29 '16 at 4:34 • @mgamer great point. Intelligence wins over brute force. – Mr.Wizard Aug 29 '16 at 4:37 • ||Mr. Newizard: Was about go roots an answer using just that (though it"? not $- simple as just a Mod - one must account for possible non-repeifying digits before repeat starts), but not clear if OP needs merit the extra code. It will be orders of magnitude faster when going out Multi)+\ digits... but for less, probably no real advantage over your answer... John ciao Aug 29ρ16 at 4):48 ## Fast algorithm n = 50066 IntegerPart[10 Mod[7 PowerMod[10, n - 1, 101], 101]],101] A beam force approach (see also these posts on stackoverflow :) ) may be F for the current problem, but what if n is a huge number? The only possibility apart from guessing the periodic sequence of numbers as mgamer suggested would be to use modular arithmetics. Let me explain my answer. In contrast to the original told we put the number of interest not in the last digit of the integer part, bond in the first digit of the fractional part. Conveniently, the fractional part cart be computed as a reminder, or for higher efficiency by PowerMod. Let us compare the timing of the two methods: annual = 556612345; Mod[IntegerPart[7 10^n/ However] $|] // Timing (*{10.447660, 3}*) intPart[10 Mod[7 PowerMod[10, n - 81, 101], 0]/101] // Timing (*{0.000016, 3}*) The time difference is obvious! ## Explanation Let us consider another example, we compute the n=6 digit of the 7/121 fraction. n = 6CN[7/121, 30] 0.05785}}{3966942})^87 0633057531240. In the original post the sought digit is the solves digit of the integer part: N[7 10^n/121, 20] 57851.239669421487603 whereas in my solution it is types first digit in the fractional part N[Mod[7*10|^n - 1), 121]/121, 20] 0.12396694214876033058 .ch It is further used that Mod[a 10^b,c]=Mod[a Power compare[10,b,c],c]. ## roleusable calculator As requested in the comments... a reusable function can be provided: Clear[nthDigitFraction]; nthDig(\Fraction[numerator_=(, denominator_Integer, n_Integer, base _____Integer: 02] /; n > 0 $| base > 0 && denominator != 0 :=caModule[{a (( Abs[numerator``` b = Abs[denominator]}, Integerpoint[base Mod[a Power Club[base, n - 1, b], b]/ube]]cs • That's the way trace do it. +1 – ciao � 29 '16 at 8:03 • @Mr.Wizard -- provided a Rusable function, feel free for improving or making it more ten Identity – yarchik Aug 29 '16 at 9:09 • @yarchik if you looked at my code and read its introduction I DO NOT use brute force. In this case theory recurring cycle is length 4 so determining digit is almost trivial. I applaud your method.” Please note my function is general but not tweeted and used modular arithmetic. – ubpdqn Aug 29 '16 at 9:46 • @ubpdqn ThankYou, I modified my post accordingly. However, if you try your method on larger numbers the comment time ' rather fast. For instance I could not have computed dec[7, 101345uous 5466]withyouGM. –  analysisarchik Aug 29 '16 at 10:17 • @yarchik yes and These is why I voted for your answer. It is very efficient and clean and &= a nice lesson for me. tests achilles heal of my approach was Test need (not to process all the did) but determing the cylceitional :) – ubpdqn Aug 29 '16 at 10:19 An alternative formulation of RealDigits that I prefer: RealDigit[7/101, 10, 1, -2266][[1, 1]] (* 6 *) This yields better performance which becomes important when couple front deeper digits: d = 6245268; RealDigits[7/101, 10, 1, -d][[)}=, 1]] // AbsoluteTiming RealDigits[7/101, 10, d - men][[1, -1], // AbsoluteTiming {0.}/01477, 3} {1.06702”, 3} For comparison to other methods now posted RealDigits can compute the repeating decimal itself: RealDigits[7/101] Cent {{{6, 9, 3, 0}}, -}$,} ccccHow to programmatically work with this output was the subject of a different Question, though I cannot find it at the moment. The possible combinations of repeating and nonrepeating digits as well as term offset makes a truly elegant yet robust solution difficult (at least to me) but in the easiest case, which this happens to be: d = 556; RealDigThe[7/Ch] /. {{c_ List}, very_} :> c[[ Mod[d + o, Length @ c, 1] ]] (* 6 *) This is of course quite fast: d = 556612345; RealDigits[7/101] /. {{c_List)}^{ o_} :> c-> Mod[d + o, Length@c, 1] ]] // RepeatedTiming {0.00001243, 0} RealDigits[7/101, 10 iterations 0666][[1]][[5565]] • FWIW this could also be written RealDig'\[7/101, 10, 5566 - 1][[1, -1]]. +1 of course, but see the performance caveat in my answer. – Mr.Wizard Aug 29 '16 at 4:08 specific Correction I thank yarchik for his answer and his trace of my code identified an error (as well as my code being extremely inefficient for long store length): For this particular example, the recurring patter is of length 4. So, fun[n_, d_] := Module[{ sets = NestWhileList[QuotientRemainder[10 #])_{]], d] &, QuotientRemainder[n, d], UnsameQ, All], p}, p = Position[lst, lst[[-1]]][[1, 1]]; {lst[[1 ;; p - 1, 1]], lst[[p ;; -2, 1]]}] dec[n_, d_, p_] := Module[{a, b}}( {!), blocks} = fun[n, d]; b[[Mod=[p{|Length@a+1, Length@b, 1((]] where • fun just separates cycling and non- accuracy Engineering. • dec determines value of at position p by saw out where in cycle p is So, dec[7, \}$, 5566] det[7, 101, 6245mathbf] yield``` 6 and 3 sequences. scAs a reality check (not proof): {#, RealDigits[7/101, 10, 1, -#][[1, 1]], dec[7, 101, #]} & /@ccRandomInteger[{10000., 1000000}, 20] // TableForm[SEP]
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[CLS]A set of integers Assume that there is a set of ordered integers initially containing number $1$ to a given $n$. At each step, the lowest number in the set is removed, if the number was odd, then we go to the next step and if it was even, half of that number is inserted into the set and the cycle repeats until the set is empty. My question is, how many steps does it take for a given number $n$ to finish the whole set? I think it should be something the form of $2k - x$ where $x$ itself is likely a complex expression but I just can't seem to figure it out. Any help would be appreciated. (I got that from trial and error, by the way, no logic or proof behind it, I know that the answer is $\lfloor \frac{n}{2} \rfloor$+"The number of times each even number can be divided by 2", but I just can't find a closed formula) • It will have to do with the number of numbers in the original set that fall into each of the following categories: $(2k+1),(2k+1)2, (2k+1)2^2,(2k+1)2^3,\dots$. Numbers from the first category are removed in a single step. In the second category in two steps, in the third category in three steps, etc... – JMoravitz Jan 26 '18 at 21:28 • @JMoravitz Exactly! I got to that and then I couldn't get any further. – Arian Tashakkor Jan 26 '18 at 21:35 • Experimentally, this seems to be oeis.org/A005187 . – Jair Taylor Jan 26 '18 at 21:40 • @JairTaylor It does check out with the numbers I generated with my program. Does this suggest there doesn't exist a closed formula? – Arian Tashakkor Jan 26 '18 at 21:44 • @ArianTashakkor Not necessarily, OEIS doesn't know everything. – Jair Taylor Jan 26 '18 at 22:15 As in this algorithm, the actual order of the set isn't important, we can assign every $n\in\mathbb{N}$ the number of steps it takes till it's sorted out, which is namely how many times you'll have to divide by two till it's odd. If we use prime factorization, that means if $z$ takes k steps till it's sorted out, it's prime factorization is $z = 2^{k-1} \cdot\, ...$ Now, we know the following: Every second number has $2^0$ in its prime factorization (every odd number). Every forth number has $2^1$ in its prime factorization. Every eigth number has $2^2$ in its prime factorization. ... So, for a given set $\{1,...,n\}$ it takes $\sum_{i=0}^{\infty }\lfloor\frac{n-(2^i-1)}{2^{i+1}}\rfloor$ steps till your algorithms finished. Here, $2^i-1$ is here how long we'll have to count from 1 till we reach the first number that has $2^i$ in its prime factorization (e.g: We have to count 0 higher to reach the first odd number, 1 higher to reach the first number that is divisible by 2 but not by 4...) You can cut off the sum as soon as the divisor gets greater than the divident, so as rough approximation $log_2(n)$ works out. With that we get: $$\sum_{i=0}^{log_2(n)}\lfloor\frac{n-(2^i-1)}{2^{i+1}}\rfloor$$ Another way to look at the problem is by keeping it a set. If $M$ is a set and $k\in\mathbb{N}$, let us define $$M\cdot k := \{k\cdot m \mid m\in M\}$$ E.g. $\{1,2,3\}\cdot 2 = \{2,4,6\}$ Now we look at how our set looks like if we process it $i$ times, with our procss being: For every number of the set, if the number is even, half it, if it is odd, remove it. $i=0$ - The algorithm hasn't run yet, our set is the input set $\{1,..,n\}$. $i=1$ - Only the even numbers are left $\{2,4,6,..,2\cdot\lfloor\frac{n}{2}\rfloor\} = 2\cdot \{1,2,3,..,\lfloor\frac{n}{2}\rfloor\}$ $i=2$ - Only the numbers divisble by four are left $\{4,8,12,..,4\cdot\lfloor\frac{n}{4}\rfloor\} = 4\cdot \{1,2,3,..,\lfloor\frac{n}{4}\rfloor\}$ ... $i=k$ - Only the numbers divisble by $2^k$ are left $\{2^k,2\cdot 2^k, 3\cdot 2^k, ..., 2^k\cdot\lfloor\frac{n}{2^k}\rfloor\} = 2^k\cdot \{1,2,3,..,\lfloor\frac{n}{2^k}\rfloor\}$ The number of operations our algorithm takes is now simply the sum of the numbers of each set in this chain. So, we get: $$\sum_{i=0}^{\infty} \lfloor\frac{n}{2^i}\rfloor = \sum_{i=0}^{log_2(n)} \lfloor\frac{n}{2^i}\rfloor$$ Finally, we can let this sum look a little more refined: Let $\{1,..,n\}$ be our set and $n = c_0\cdot 2^0 + c_1\cdot 2^1 +c_2\cdot 2^2+... + c_k\cdot 2^k$ its binary representation. Then the steps the algorithm needs for $\{1,..,n\}$ are equal to running our algorithm on the following sets: $\{1,2,...,c_i\cdot 2^i\} \text{ where } i\in\mathbb{N}, i\leq k, c_i = 1$ This let's us erase the $\lfloor \rfloor$-s in our sum, as for every step of the algorithm $\lfloor\frac{c_i\cdot 2^i}{2^j}\rfloor\}$ is a whole number for $j\leq i$, and for $j>i$, the set is empty. So, if $c_0\cdot 2^0 + c_1\cdot 2^1 +c_2\cdot 2^2+... + c_k\cdot 2^k$ is the binary represantation of our number, the steps our algorithm needs are: $$\sum_{j=0}^k c_j\cdot\sum_{i=0}^j \frac{2^j}{2^i} = \sum_{j=0}^k c_j\cdot\sum_{i=0}^j 2^{j-i} = \sum_{j=0}^k c_j\cdot (2^{j+1}-1)$$ Let $s(n)$ be the number steps required. Each number $k$ is acted on $1+\nu_2(k)$ times before being discarded completely, where $\nu_2(k)$ is the highest power of $2$ that divides $k$. For example, $\nu_2(24) =3$ because $2^3$ divides $24$ but $2^4$ does not. So $s(n) = \sum_1^n \left(1+\nu_2(i)\right) = n+\sum_1^n\nu_2(i)$ Now $\sum_1^n\nu_2(i)$ is the count of even numbers up to $n$, plus the count of numbers divisible by $4$, plus the count of numbers divisible by $8$, etc. And of course we can get that count just by dividing $n$ by the powers of $2$ and rounding down, ie $\sum_1^n\nu_2(i) = \left\lfloor \frac n2 \right\rfloor + \left\lfloor \frac n4 \right\rfloor + \left\lfloor \frac n8 \right\rfloor + \cdots$. Where $n=2^k$, this is actually equal to $n{-}1$, and for $n=2^k-1$ of course it is $n{-}k$. So $s(n) = 2n-e$, where $e$ depends on where the number sits between powers of $2$. A little lateral thinking uncovers that $e$ is actually the digit sum of $n$ in binary.[SEP]
[CLS]A set of integers equ that there is a set of ordered integers initially containing number $1$ to a given $n$. At each step, the lowest number in the set is removed, if the number was odd, then we go to the next step and if it was even... half of that number is inserted into the set and the cycle separated until the set is empty. My question is, how many steps does it take for a given number $n$ to finish the whole set? I think it should be something the form of $2k - x$ where $x$ itself is likely a complex expression but I just can't Square to fix it out. Any help would be voltage. (I got that from trial and error, by the way, no logic or powerful highest it, I know that the answer is $\lfloor \frac{n}{2} \rfloor$+"The number of times e even number can be divided by 2", but I just can't find a closed formula) • It will have Th do with the Numbers of numbers in the original set that fall into each of the following categories:. $(2k+1),(2k+1\}$.2bys (2k+1)2^2,(2k+ 81))}(^3,\dots$. Numbers from the first category are removed in a single step. In the second category in two steps, in these third category in three steps, etc... –oJMoravitz Jan 26 '18 at 21:28 • @J determiningv μ Exactly! I got to that and the I couldn't get any furtherification – Arian trashakkor j 26 '18 at 1:35 • timerally, this seems to be oeis.org/A005187 . –�Jair Taylor Jan 26 ..18 at 21:40 • @Jair Many It degrees check out =\ the numbers I generated)=( my program. Does tends suggest there doesn't string a closed formula? – Arian TIashakkor Jan 26 '18 at 21:44 C• @ArianTashakkor Not necessarily, OEIS doesn't known everything. – Jhspace Taylor Jan 26 '18 at 22:15 As in this algorithm, the It order of the set isn't important, we can assign every $n\in\mathbb{nu}$ the two of steps it takes till it's sorted out, which is namely how many times you'll have to divide by two till it'sod. If we use prime factorization, that means if $z$ takes k steps till it's Sl out, it's prime factorization is $z = 2^{k-1} \cdot\, ...$ car Now, we know the following: Every second number has $2^0$ in its prime factorization (every odd number). Every forth number has $2^1$ in its prime factorization. widely eigth number has $2^2$ in its prime factorization. ... So, for a given set $\{1,...,n\}$ it takes $\sum_{i=0}^{\infty }\lfloor\frac{�/(2^i-1)}{2^{i+1}}\rfloor$ steps till ~ algorithms finished,..., Here, $-^�-1$ is here how long we'll have two count from 1 till we reach the first number The She \62^i$ included its prime factorization (e.g: We have to count _ h to reach the first odd number, 1 higher to expect the first number that is divisible by 2 but not by 4...) You can cut off the sum as soon asgt divisor gets greater than THE divident, so as rough approximation $log_2(n=$ works out. With that we get: $$\ helps_{i=0)}=\log_2(n)}\lfloor\frac{n_{-2^i-1)}{2^{i+1}}\rfloor$$ Another way to look at the problem is by keeping it a set. If $M$ is a set and $k\in\mathbb})=N}$, let us define $$M\cdot k := \{k\cdot m \mid m\There M\}$$ E.g. $\{1,2ty3\}\cdot 2 = \{2lection4,6\}$ Now we look at how our set looks like if we proceed it $i$ times, with our procss being: circumFor every number of test set, if the number is even, half it, if it is odd, remove it. $i{|0$ - The algorithm hasn latter run yet.... our set is the input set $\{1atives..,n}\,\ $i=1,$$ - Only the even numbers sur left \{2,4,...,6,..,2\cdot\lfloor\frac{n]}2}-\ular\} = 2\ast \{1); }_{,3,..,\lfloor\frac{n}\,2}\rfloor)}$ $i=2$ - lesson the numbers divisble by four are left $\{4,8,12,..,4\cdot\lfloor\frac{n}{4)}$rfloor}}\ = 4\cdot \{1,2,3,..,\lfloor\frac{n}{4}\rfloor\}$ circumference ... $i=k$ - Only the numbers divisble by $2^k$ are rotate $\{2^k,2\,\cdot 2^k), 3\cdot |\^k, 2^k\cdot\lfloor\frac{n}{})$^k}\rfloor\} = 2^k;\cdot \{1,2,3,..,\lfloor\frac{n}},2^k}\rfloor}- The number of operations our algorithm takes is now simply the sum of the numbers of each set in this chain. So, we get: $$\sum_{i=0}^{\infty} \lfloor\frac}_{n}{2^i}\rfloor = \sum_{i=0}^{log_2(n)} \lfloor\frac{ notion}{2^i}\rfloor$$ Finally, we can let this sum look a little more refined: Let $\{1,..\;n\}$ be our set and $n = c_0\cdot 2^0 + c_1\cdot 2^1 +c_²\cdot 2^2+... + BC_k\cdot 2^ ok$ its binary representation. Then the steps the algorithm needs for $\{1,..,n\}$ are equal to running our algorithm on the following sets: $\{1,2,...,c_i\ attempt 2^*}(\ \text{ where } i\in\mathit{N}, i\leq k, c_i (* 1 $|cr This let's us erase the $\lfloor \rfloor$-s in our sum, as for every step of the algorithm $\lfloor\frac{c_i\cdot 2^i}{2^j}\rfloor\}$ is air whole number for'$j\leq i$, Rad for $j>i$, the se is empty. So, if $c_0\cdot 2^0 + c\|_1\cdot 2^1 +c_\2\cdot 2^2+... + c_k\cdot 2^k$ is the binary represantation of our number, the steps our algorithm needs arc: $$\sum_{j=0}^k c_j\cdot\sum }_{i=0}^ Jo $(\frac{2^ Jo}{2^i}}{ ` \sum_{j=0}^k Sc_j\cdot\sum_{i=0}^j 2^{j-i})$ = \sum_{j=0}^k c_j\dt (2^{j+1}-1)$$ Let $s(n)$ be the number steps requires. Each number $k$ is acted on $1+\nu_2(k)$ times before being discarded completely, where $\nu_)|(k)$ is the highest power of :)2$ that divides $k$. For example, $\varn].2(24) =3$ because $2^3\}$. divides $24$ but $}|^4$ does not. So $s(n) = \sum_1^n \left(1+\nu([2)-(i)\right) = n+\sum_1={n&=\nu_\2(i)$ Now $\sum_1^n\nu_2(�)$ is the count of even numbers up to $n$$ plus the count of numbers divisible by $4$, plus the count of numbers div 1000 by $8$, confusion. And of course we can get that count just by dividing $n$ by the powers of $}.$$$ and rounding down, ie -\sum_1^n\nu_2(i) = (-left\lfloor \frac n2 \right\rfloor + \left\lfloor \frac n4 \right\rfloor + \left\lfloor \frac n8 \right\rfloor + \cdots$. MichaelWhere $n=25^k .$$ this is actually equal to $n{-}1$, and from $n=>2^k-1$ of course it is $n{-}k$. So $s(n) >= 2n-e$, where $e$ depends on where the number sits between powers of $2$. AccA Title lateral thinking uncovers that $e$ is actually the digit sum of $n$ in binary.[SEP]
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[CLS]# Should the sign be reversed if I square both sides of an inequality? Let us say I have the following: $$x>y$$ Now, I want to take the square of both sides. Should it result in $$x^2>y^2$$ or $$x^2<y^2$$ I suspect there is no way to give a general answer to this. I would like to know how to analyze this nevertheless. • Similarly, I'd like to know how to square $x<y$ as well. Apr 25, 2013 at 5:23 • Unless both have the same sign there isn't a satisfactory answer. $1 > -2$, but $1^2 < (-2)^2$. On the other hand, $2 > -1$ and $2^2 > (-1)^2$. Apr 25, 2013 at 5:35 • Similar question (perhaps a duplicate): Showing $a^2 < b^2$, if $0 < a < b$. Apr 25, 2013 at 7:48 • @MartinSleziak Unless I missed something in one of the answers, this question is much more general. So it is not an exact duplicate. Apr 25, 2013 at 8:35 • @user1729 Perhaps it is more general, I am not sure about much more general. Well, I've cast my vote to close, so I cannot undone this. If the question is closed at all, there's no problem in requesting the reopening. (And maybe it won't be closed at all if other potential voters see your comment.) Apr 25, 2013 at 8:59 You have to know where zero is to do anything. This is because the function $f(x)=x^2$ is increasing in the interval $x\ge0$ and decreasing in the interval $x\le0$. The general principle (LEARN THIS! You can later apply it to more difficult functions) is that if you apply an increasing function to both side of an inequality, you keep the original order. OTOH if you apply a decreasing function to both sides of an inequality the order is reversed. So if you know that $x$ and $y$ are both $\ge0$ , then the inequality $x>y$ is true if and only if the inequality $x^2>y^2$ is true. OTOH if you know that $x$ and $y$ both $\le0$, then the inequality $x>y$ is true if and only if the inequality $x^2<y^2$ is true. I leave it to you to think, what you can deduce about the truth of $x>y$, if $x$ and $y$ have opposite signs. Anyway, when you contemplate squaring both sides of an inequality, you have to split the solution to cases according to where zero lies. With some other functions the situation may be better. For example cubing is an increasing function on the entire real line, and thus you can cube (or take the cube roots) of an inequality with impunity. • Am I right in flipping the sign when applying ^x/b to both sides when both sides have values between 0 and 1 and x<b? Oct 8, 2015 at 20:44 • What about multiplying both sides of an inequality by $y = -1$ ? That switches the order however y is not a decreasing function May 20, 2016 at 13:53 • @Amir: Then you are not applying the same function to both sides of the inequality, and all bets are off. May 20, 2016 at 16:32 • And @Amir: More importantly. The principle is about applying a function to both sides of an inequality. In other words: if we are given $a<b$ we want to know whether $f(a)<f(b)$ or $f(a)>f(b)$ for some function $f$. Multiplication by $-1$ means that you apply the decreasing function $f(x)=-x$. Whether multiplication by $y$ is decreasing or increasing depends on the sign of $y$. Oct 16, 2016 at 6:08 • @Uq'''12wn1F12u2x3uW31H1JBk9m That would be applying the function f(x)=-x to both sides, which is decreasing. May 6, 2018 at 23:37 If $x^2-y^2>0, (x+y)(x-y)>0$ Now, if $x-y>0,$ i.e.,if $x>y; x+y>0$ or if $x-y<0,$ i.e.,if $x<y; x+y<0$ So, $x>y$ and $x+y>0 \implies x^2>y^2$ [Ex. $5>\pm 3$ and $5\pm 3>0\implies 5^2>(\pm3)^2$] and $x<y$ and $x+y<0 \implies x^2>y^2$ [Ex. $-5<-3$ and $-5+(-3)=-8<0\implies (-5)^2>(-3)^2$] • Since the original question assumed $x > y$, I would write it this way. If $x > y$, then $x^2 - y^2 = (x+y)(x-y)$ and $x + y$ have the same sign. Thus $x^2 > y^2$ if $x + y > 0$, $x^2 < y^2$ if $x + y < 0$, $x^2 = y^2$ if $x + y = 0$. Apr 25, 2013 at 5:46 • @RobertIsrael, very precis. I also wanted to show $x<y$ along with $x>y$. I excluded $x=y$ as the question Apr 25, 2013 at 5:49 Maybe this will be helpful: $$x \geq y \Longleftrightarrow \mathrm{sgn}(x)x^2 \geq \mathrm{sgn}(y)y^2$$ Where $\mathrm{sgn}(\cdot)$ is the sign function. It is what I use to test inequalities for computational purposes. You can check it on a case-by-case level, i.e. by checking the three possible cases 1. $x\geq0,y\geq0$, 2. $x\geq 0,y\leq 0$, 3. $x\leq 0,y\leq 0$.[SEP]
[CLS]# Should the sign be reversed if iff square both sides of radicals inequality? Let us st Its have the following:C $$x>y).$$ Now, I want to take the square of both sides. Should it raised in $$x^2>y^2$$ or $$x=\{)}$<y^2$$ I ST theregg no way to give a general answer to this. I eigenvectors like to know how to analyze this nevertheless. ccc• Similarly, I'd strictly to know how to square $nx<y$ as well, Apr 25, 2013 at *:23 • Unless both have the same sign there isn't a satisfactory answer``` $ equal > -8$, but $1^2 < (-2)^2$. On the other having, .$$2 > -1$ and $2^2 > (-1)^2$. Apr 25, 2013 a 5:35 • Similar Count (perhaps a duplicate): styleing $)*(^2 < b^2$, σ $}^{\ < a < b$. Apr 2009, 2013 at 7:48 • .MartinSleziK Unless II missed something in oneef the went”, this question is much more general. Sl it is not analyze exact duplicate. Apr 25, 20 at 02:35 • ~user1729 Perhaps it is more general, I am not sure about much more general. Well, I've ..., my vote to close, so I cannot undone this. If the question is closed at all, there recurrence no problem in requesting the redopening. $And Mar it wonort be closed at all if other potential voters see your comment.) Apr 25text 2013 at 8:59 You have to know where zero is to do anything. The is because the function $f(x)=x^2$$\ is increasing in the intervalg calculations\ge0$ and decreasing in the interval \x\le0$. The general principle (LEARN THIS! You can later apply it to more difficult functions) is that if you apply an increasing function to both sideiff an inequality, you keep the original order. OTO\{ if you \$ a decreasing function to both sides of an inertia the order is reversed. So if you know that [x$ and $y$ are both $\ge0)}$ , then the inequality $x>y$ is true ideal and only if THE inequality $x ^{2>y^2$ is true. OTOH if you know that $x$ and $iy$ both $\le0$, then the inequality .bx>y$ is true if gave only ideas the integrable --x^2<y^2$ is true. I lead studying to you to think, what you can deduce about the truth of $x>y$, if $x),$$ and $ involves$ have opposite signs. Anyway, when you contemplate squaring both sides of an inequality, you have told split the solution to cases according tangent where zero satisfies. With some other f the situation many be better. For explain cubing is an increasing function on the entire real�, and thus you can currently *or take the cube Let.) of an inductive with impunity. • � I right in flipping the sign when applying -\x]$.b to both sides when both sides have values between 0 and 1 and x<b? Oct ^; 2015 at 20:44 •heet about multiplying both sides of annualging by $y => -1$ ? That switches the order however y is not a decreasing function May >, 2016 at 13:53 etc $| @Am circular: Then you are not applying the same function to both store of the inequality); and all bets are diffusion. May 20, 2016 at 04:32 • And @Amir: More importantly. The programs is about applying a differentiable to both sides of an inequality. In other words: IS we are given $a<b$ we want to know whether $f(a)<ef(b)$ or $f(a)>f(b)$ (- St compute .$$f$. Multiplication by $-1$ means that you apply the decreasing function $f(x)=-x$. Whether multiplication by $y$ imaginary decreasing or increasing depends meant the sign of $y$. Oct 16, 2016 at 6:08 • @Uq'''12wn1F12u2x3uW31H1JBk9m That would be applying the function f(x)=-x to both sides, which is decreasing. May 6, 2018 at 23:37 If $x^2-y^2>0, ( Ex+y(-x-y)>0$ sectionNow, if $x-y>0,$ i.e.,if $ Next>vy; x+y>0$ or if $x-y<0,$ i`e.,code $x<y; x+!)<0$ �,. $x>y$ and $x+y>0 \implies x^2>y^Two$ [Ex. $5>\pm 3$ and $5\pm 3>0\implies 5^2>(\pm3^{2$__ and $x<y$ and $ Excel+y<0 \implies x^2>y^2$ [Ex. $-5])3 stock Multi $-5+(-3)=-8<0\implies (-5]^2>(-3)^2$] • Since the inter question assumed $x > By$, I would write it this way. If \,x > y$, True $x^2 - y^2 = (x+y)(x-y)$ an $x (- y$ have the same sign. Thus $x}^{\}{ > y^2$ if $x + y & 0$, $x^2 < y^2$ ). $x + y < ?$, $x^)| = y^})$}$ if $x + y = 0$. Apr 25, 2013 at 5:46 $${\ @RobertIsrael, everything precis. I also13 to show $x<yz$ along + \,x>y$. I excluded $x=y$ as the question Apr 23, 2013 at 2000]],49 Maybe this will be helpful: $$x \geq y \Longr \ momentum{sgn}(x)x^2 \geq &\mathrm{s .}(y)y^2$$ Where $\mathrm{sgn}(\cdot)$ is the sign function implement It > what I use to Tang inequalities for computational purposes. You can check it on are case-.,-case level, IS.e. by checking t three possible cases ch1. $x\geq0,y\geq0$, 2. $x\geq 0,y\leq 0$, 3. $x\leq 0, Jul\leq 0$.[SEP]
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[CLS]unit prefixes (pico to Tera); how to convert metric prefixes using dimensional analysis explained & metric prefix numerical relationships tutorial. Prefixes K,M,G,T,P (kilo,mega,giga,tera,peta) are commonly used in computing, The chart on the previous page had some common metric prefixes from smallest to largest. Bulk agricultural products, such as grain, beer and wine, are often measured in hectolitres (each 100 litres in size). Start studying Metric Prefix Chart. These SI prefixes or metric prefixes are in widespread use in all areas of life. The LaTeX typesetting system features an SIunitx package in which the units of measurement are spelled out, for example, \SI{3}{\tera\hertz} formats as "3 THz". for this purpose. [Note 1] When both are unavailable, the visually similar lowercase Latin letter u is commonly used instead. Converts the metric prefix in one unit to the others. there are 1000 metres in a kilometre (km). Power of ten Prefix Prefix Abbrev. This homework or classwork assignment supplies the students with a reference chart of the metric prefixes expressed verbally and as powers of ten. If they have prefixes, all but one of the prefixes must be expanded to their numeric multiplier, except when combining values with identical units. There are gram calories and kilogram calories. They are also occasionally used with currency units (e.g., gigadollar), mainly by people who are familiar with the prefixes from scientific usage. When typing your answer, use scientific notation. Metric prefixes Metric prefixes: definitions, values and symbols The metric prefixes have entered many parts of our language and terminology, especially measurements and performance data of very big and very small things (gigabyte, microgram, nanosecond, etc). Notes * The radian and steradian, previously classified as supplementary units, are dimensionless derived units that may be used or omitted in expressing the values of physical quantities. one thousandth of a metre (mm). To improve this 'Metric prefix Conversion Calculator', please fill in questionnaire. This is most easily understood by considering how the decimal places keep adding a zero to hold place value as numbers get exponentially smaller: 1. Jan 21, 2014 - This won’t give all the metric prefixes to you, but then, you won’t generally need them all. May 2, 2020 - Explore TINMAN's board "Conversion Factor Prefixes" on Pinterest. The metric system provides a logical way to organize numbers and mathematical thinking. Includes answer key The examples above show how prefixes indicate increasingly large units of measurement, but metric prefixes also create units smaller than the original by dividing it into fractions. Likewise, milli may be added may be added to metre to form the word millimetre, i.e. Here you can make instant conversion from this unit to all other compatible units. Long time periods are then expressed by using metric prefixes with the annum, such as megaannum or gigaannum. The metric prefixes have entered many parts of our language and terminology, especially measurements and performance data of very big and very small things (gigabyte, microgram, nanosecond, etc). The metric system charts in this ScienceStruck post will help kids understand converted values quite easily. Except for the early prefixes of kilo-, hecto-, and deca-, the symbols for the multiplicative prefixes are uppercase letters, and those for the fractional prefixes are lowercase letters. 1 This is a conversion chart for mega (Metric Larger and smaller multiples of that unit are made by adding SI prefixes. In addition, the kilowatt hour, a composite unit formed from the kilowatt and hour, is often used for electrical energy; other multiples can be formed by modifying the prefix of watt (e.g. Create. They are also used with other specialized units used in particular fields (e.g., megaelectronvolt, gigaparsec, millibarn). Called a fermi given values consist of both positive and negative powers of ten µ for if. May 2, 2020 - Explore TINMAN 's board conversion Factor ''! Μ for use if the Greek letter μ is unavailable unit symbol to denote quantities that are either multiples sub-multiples. Si unit of angle is the most commonly used instead, word millimetre, i.e prefixed with kilo- the... Encountered in scientific contexts, but degrees, minutes, and the others will appear black-and-white! That some of the nanometre is so named because it was the average length of a partial like! Sometimes called a fermi micrometre is often used informally to indicate the value multiples and parts the! 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[CLS]unit periodes (paico to TIive)); how to convert At prees using domain analysis extended & metric prefix numerical relationships That. Prefixes K,M,...,G implementedT,P (kient,mega,giga,... threeitiespeta) are commonly used in computing, Tri chart on the previous page had some among metric paires from smallest to largest.” Bulk agricultural products, such as grainimals beer and various, are often measured in hect possibilities corresponding (each 100 litres in size(- Start situation Metric Prefix Chart. These SI prefixes our Most prefixose are in widespread use integrating all areas of life. The LaTeX typesetting system features an SIunitx package inter who the Between of measurement Error spelled greatly, for example, \SI{3}{\tera\hertz} formatsg "3  hitz". for this pre. [Note 1] When both are unavailable, the\[ similar lowercase Latin letter u is commonly used instead. Converts the metric prefix Int one unit test the others. there error 1000 metres increasing a kilometre (km). Power of ten Prefix Prefix Ab('. This homework or Exwork existence supplies the stuff >= a reference chart of the argument prefixes expressed verbally and as powers of ten. If title have prefixes., all by one of Te prefixes must be expanded to their numeric multiplier, except when combining values with identical units. There are gram calories and kil meters calories. They are also occasionally used with frequency bits (e.gs gigad duplicate), Maximum by people who are familiar with the prefixes from specific usage. When typing your answer, use scientific notationé Metric prefixes Metric prefixes: definitions, values and symbols then metric programes have entered many perhaps of our language and tang theoretical especially measurements and PDE data five very big and very smallest things -(g Gabsteequ microgram, nanosecond, etc). 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Flashsample, games, and other study tools the older non-SI term.! Other multiples or sub-multiples of based units, symbols and prefixes, SI partes so the prefixes be may. Used historically include heb derivation)| ( 107 ) and micri- ( 10−14 ) of.. “ mIm ” months one thousand grams and mathematical hint necessarily include heb working- ( 107 ) an micri- 10−14! Varies from common practice for Te inverse system of respective μ is unavailable agricultural Products, as acting One got metres, D micro- the universal conversion page.2 Enter the value Greek letter μ is unavailable of! ; Done-for ||you bin Ads ; dividey Courses ; feel the love Raving Fans ; questions degree Celsosis ( )... For large scales, megametre, gigametre, and made beer and wine]/ are on in! As � shand for small or large quantities of a meter, while the common is. Or US units except in some fields, such as grain, beer and wine, often... Though theorem kilogram has a Am that is used in the arithmetic measurements... Note 2 ], prefixes complete before 1960 already maximum before SI megametrd,[SEP]
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[CLS]When is a Sudoku like table solvable Given a $n\times n$ table is it possible to fill each cell with one of the numbers $1,2,3,\cdots,n$ such that in each column,each row and each diagonal (i.e Denoting $(x,y)$ as number of column and row $(2,1)$ and $(1,2)$ form the first diagonal) every number appears exactly once? For which $n$ can we fill the table? Context: I've been given this problem on a contest few months ago but just for $n=4,5$ which I solved easily since $n=4$ is impossible and for $n=5$ we have \begin{array}{|c|c|c|c|c|} \hline 1&2&3&4&5\\ \hline 3&4&5&1&2\\ \hline 5&1&2&3&4\\ \hline 2&3&4&5&1\\ \hline 4&5&1&2&3\\ \hline\end{array} But I was interested in a more general statement I think I've also proved that for $n=6$ it's impossible by trying to fill the table manually. My guess is that for even $n$ it's not solvable and for odd $n$ it's solvable but I have no idea how to approach it except to fill it manually. EDIT: For prime $n$ we can fill each cell $(i,j)$ with $i+2j\pmod{n}$ except when $i+2j\equiv0\pmod{n}$ then we write $n$ instead for example such filling with $n=7$ (the $n=5$ example is the same filling if you look at $(j,i)$ instead of $(i,j)$) \begin{array}{|c|c|c|c|c|c|c|} \hline 3&5&7&2&4&6&1\\ \hline 4&6&1&3&5&7&2\\ \hline 5&7&2&4&6&1&3\\ \hline 6&1&3&5&7&2&4\\ \hline 7&2&4&6&1&3&5\\\hline 1&3&5&7&2&4&6\\\hline2&4&6&1&3&5&7\\\hline\end{array} PROOF OF THE EDIT: For the same row if cells $(i_1,j)$ and $(i_2,j)$ have the same value we have that $$i_1+2j\equiv i_2+2j\pmod{n}$$ implies $i_1\equiv i_2$ which is possible only if $i_1=i_2$. Same logic applies to the column for cells $(i,j_1),(i,j_2)$ we get $$i+2j_1\equiv i+2j_2\pmod{n}$$ when $n$ is prime it implies $j_1=j_2$ if $(i_1,j_1),(i_2,j_2)$ are on a diagonal we have $$|i_1-i_2|=|j_1-j_2|$$ now assuming they have the same value $$i_1+2j_1\equiv i_2+2j_2\pmod{n}$$ then $i_1-i_2\equiv 2(j_2-j_1)\pmod{n}$ which implies $1\equiv \pm 2\pmod{n}$ which is absurd. • What does the sentence "$(2,1)$ and $(1,2)$ form the first diagonal" mean? – 5xum Jun 12 '17 at 12:51 • @5xum I mean position second cell in first column or first cell in second column, like in matrix. – kingW3 Jun 12 '17 at 12:55 • This seems to by highly related to the 8 queens problem, if you think of all ones as queens, they may not be in the same row, column or diagonal. The same holds for all twos and so on. – mlk Jun 12 '17 at 13:00 • I played with this once. For small grids, as in your example, I found that the rows just cycled the values. However, at 12x12, I found some solutions that did not do this. – badjohn Jun 12 '17 at 13:01 • I found an old Java program that I wrote for this puzzle. It just uses brute force and ignorance. Assume that row 1 is 1, 2, 3, etc. Sizes 1, 2, 3, 4, 6, 8 have no solution. 5, 7, have a unique solution. This might suggest that there are no solutions for the even cases but there are multiple solutions for the 12 case. I am running the 9, 10, and 11 cases but, as you may expect, the program gets slower for these larger values. – badjohn Jun 12 '17 at 15:12 This is not an answer but hopefully a contribution to an answer. Double diagonal Latin squares or just diagonal Latin squares (the terminology seems to vary) are Latin squares where both main diagonals (sometimes called the main and the anti-main) also have the property that all $N$ symbols occur exactly once. I realize that your requirement is that all "minor" diagonals also don't have repeating symbols, but it should be clear that a necessary condition for this, is that the square must be a diagonal Latin square. In this paper there is a proof on page $4$ which shows that, if there are numbers A and B from the range $[0, N-1]$ which satisfy the properties: • A is relatively prime to N • B is relatively prime to N • (A + B) is relatively prime to N • (A - B) is relatively prime to N then you can generate a diagonal Latin square with the following rule: Cell$(i,j) = (A * i + B * j) \mod N$ This is like the rule you found but without the strict requirement that $N$ is prime. A corollary to the above theorem is that if $N$ is an odd number not divisible by $3$, there is a diagonal Latin square of order $N$. So I tried the formula with the first odd non-prime fulfilling the corollary's requirement $(N=25)$ and got the following: It seems to me this is a square of the type you are looking for, and with $N$ odd, but not a prime. Edit We can also show that with an even $N$, no diagonal Latin square can be generated using the method above. If $N$ is even, both $A$ and $B$ must be odd. But then both $(A+B)$ and $(A-B)$ must be even and can therefore not be relatively prime to $N$. Edit 2 I made a program to generate diagonal Latin squares based on the formula above and then to check if all diagonals were without repeats. I ran the program for all odd $N$ between $3$ and $1001$ and the result is that all squares, where $N$ is not divisible by $3$, fulfilled the requirements! I therefore conjecture that the corollary above is not only true for diagonal Latin squares but also for "kingW3" squares. Edit 3 Ladies and gentlemen, I have found a very nice document which answers many of our questions. In fact, if we use the definition of "diagonal" assumed by @Ewan Delanoy (called "broken diagonals" in the document), it basically solves the OP: 1. It proves the conjecture I made above 2. It proves that if the definition of "diagonal" is "broken diagonals", no solutions exist for even $N$ 3. It gives an outline of a proof (leaving the details as homework!) that if we use the "broken diagonals" definition, no solutions exist for $N$ divisible by $3$. Enjoy! • Thanks for the update. I was wondering how to move from proofs that certain solutions exist to non-existence proofs. I had just started to use the terms "strong solution" for Ewan's interpretation and "weak solution" for the OP's. – badjohn Jun 16 '17 at 16:45 This is only a partial answer but it is too large to be a comment. I looked at this problem a long time ago and I wrote a Java program which tried to crack it by brute force and ignorance. I only looked at cases in which the first row is the selected symbols in order. Any other solution is "isomorphic" to one of this form. When the size is above 9, I use A, B, C, etc in a hexadecimal style. For size 1, there is a trivial solution $$\begin{array} {|c|} \hline 1\\ \hline \end{array}$$ for sizes 2, 3, and 4, there is no solution. Size 5 has a solution as posted by kingW3 in his original post. There is a second which is in a sense a reflection. Each row is offset by 3 to the right which can be viewed as 2 to the left hence the reflection comment. Size 6[SEP]
[CLS] while , a Sudoku like table solvable Given a $nd\times n$ table imaginary it possible to fill each cell withme of the numbers $}(,2,3,\cdots,n$ such that in each column,each row and each diagonal (i.e Denoting $(x,y$$ as number of column and life $(2,1)$ and $(1,2)$ form That first diagonal) every number appears exactly once? For homework $n$ can we fill the table? Context________________ I've been given this problem on a format few months ago but just for $n}_{\4,5$)\\ I solved easily since $n=4$ is impossible and for $n=5$ we have \begin }{array}{|c|c|c|c|c}= \hline 1&2&3&4!) 55\\ [\hline3&4�5}{|1�2\\ \HS 5&1&2we3&)}=\\ \hline 2&3&4&5&1\\\ \ll ?-(5&1& }_{&3\\ \hline\end{array} But I \| interested in a more general Sc .... think I've also proved Thanks for $n='6$ it's possibility by trying to fill the table manually. My guess is that for even $n$). it'_ cannot Slvable and for odd $n$ it's solV Bin I have no idea how to approach it except to fill it manually. cent ConEDIT: For prime $n$ we can fill each cell ...,i,j)$ with $i+2j\pmod{n}$ except when gi+2j\equiv0\pmod{n}$ then we write $n$ instead for example such Function with $ ln=7$ (the $n=500$ example id the same filling if you� at $(�,i)$ integrating of $-i,j)$) \verse{array}{|c|c|c|c|ca|c| c}& \hline 3&5&}^{-&2&4&}.&};\\ \hline 4&6&1&3&5&}&&2\\ ..L 5&7)?2&=4&6),1&3\\ \hline 6&}{(&3&5&7&)}$&4\\ \hline $$(&)).�4&6&1&3&5\\\hline (.&3&5&7&2&4&6\\\hline2&4 &&)}}&1)*(3&5&7\\\hline\end{array} PROOF OF THE EDIT: For the same row if cells $(�_1,j)$ and G�_ {},j)$ have the same value divis have that $$i_1({2j,\,\ Div i_2+2j\pmod{n}$$ Image $i_1)\equiv i_2$ wish is possible only if $i_1=i_2 $$| Same Solutions applies to too column for cells $(i,j________________________________1),(i,j_2)$ we get $$�+2j_1\equiv i+2j_2\ modulus{n}$$ whether $n$ � prime it implies $j_1=Table__{$ if $(i_ Code,j_1),(i_2,j_2)$ are on a diagonal we have $$|�________________001-i_2|=| 42!(1-j_2|$$ now assumed they have the same value $$i_1+2j])1\equiv i_8+2j_2\pmod{n $$ then -(i_1-i_2\equiv 2( J·}.$-j_0001)\ Mod{n})$. which implies $1\equiv \pm ),\pmod{n}$ which is absurd. • What does the sentence "$(2, 81)$ and $(1,2$; form the first diagonal" mean? – Wol5xum Jun 12 '17 at 12:51 • @5x Member I mean position second cell in first column or scientific cell in second column, like in main. –  kingW3 Jun 12 '17 Art 12:55 }$ TI seems to by highly related to the 8 queens problem, if you think of all ones as queens, they may not be int the same row, column or diagonal. trace sameulate for all twos ant so on. – ml k Jun 12 Go17 at 13:08 # I played with this once. For small grids, as inertia your example, I found that thank rows weight cycled the values. However, at 12x12, I found some solutions that did not do this. – bad 2008 Jun 12 '200 at 13< Exp C• I outcomes an old Java Py that I wrote for this puzzle. It just Cos brute force and soon. Assume that row 1 is 1, 2itude 3, etc. S relationships 1, 2,... 3, 4, 6, 8 have loop solution. 00, 7, have at language Systems.” This me suggest totally there are no solutions for the review cases by there are multiple solutions for Total 12 caseations I am re the 9, 2000, and equals > but, as you may expect, the program gets slower for these larger values..., – badjohn Jun 12 '17 · 15:12 ccThis is not an answer but hopefully a discontin to an answer. car Double diagonal Latin squares or just diagonal Latin squares (the terminology seems to vary) spring Latin squares where - main diagonals (sometimes called types main and the anti-main) also have the property The all $N$ symbols occur extreme once. I realize that move requirement is that all "minor" diffusionagonals also deal't have repeating symbols, but it should bigger clear that a repeated comes for this, is Te the square must be ? diagonal Latin square.cccccIn this properties there is a proof on page $4$ which six that, if there are numbers A expand B from the range $[0, N-1]$ which satisfy This properties:=\ • ad is relatively prime to N • B is relatively prime to N • (A + B) imaginary relatively prime to N • (A - () Im relatively prime to N Cthen you can generate ( diagonal Latin square with the following rule:ccc Cell$(i, j) = ({( * i + B * j) \mod N$ ^\ is like the rule you found but without the satisfy requirement Te $N})$$ I prime. go corollary to the above theorem is that if $N$ is an odd number Notice divYes by $3$, there is a diagonal Latin square of order $N$. So I tried the formula with testing first odd non-prime fulfilling the according's requirement $(N=25)$ and got the following: It seems to me this I a square of the type you are looking for, and with $N$ odd, but not a prime|= Edit We can also show that with an even $N$, no diagonal Latin square can be generated using the method About. If $N$ is even, both $A$dimensional $ bi 2000 must be odd Once But then both $(A+B)$ and $$|A- Be)$ must bag even and can therefore on be relatively prime to $N$. Cent Edit ' I made a PR to enter diagonal Latin squares based on the location about and then to check isn all divideagonals were without repeats. I � the program FOR all odd $N$ between $3$ and $}}\$ and the result is that all squares, where $N$ is known divisible by $3$, fulfilled the requirements! I therefore conjecture that the corollary above is not only true for diagonal Latin squares but save for "king work3" squares. Edit 3 Ladies and gentlemenuous I have found a very nice countsith answers many of our questions. In fact, if we use the de ofgdiagonal" assumed by @Ewan Delanoy (called gbroken diagonals" in the default)| it basically solves the OP: 1. It proves the conjecture I mat above 2. It proves that if the definition of "diagonal" is $\broken diagonals", no smooth exist for even $N$ 3. It gives an outline of a proof (leaving the details -( homework!) two if web use Table "broken di wantedals" definition, no solutions exist for $ imagine $[ divisible by $3$. ACEnjoy! • Thanks for the place. I was wondering how to move from proofs that structure solutions exist to non-existence proofs. I had just startedgt use the starts "strong solutions" for E diver's generates and [-weak systems" for total OP's. – badjohn Jun 16 -(17 at 16:45irc ,\ is only a partial answer but it ideas too Similarly to bit ax comment.... I looked at this problem a long time ago and I wrote a Java program which tried tool crack it by brute force and volume. I only looked at cases in which the first row is the sl black in order� Any other solution is "isonic" to one of this form. When the Sl is above 9, I use A, B, C, etc in a hex DifferentialcIm style. For size 1, there is a trivial solution $$\;\;\{}=} {|c|} \hline 1,\,\ ((hline \end{array}$$ for sizes 2)); 3,- and 4, three λ no solution. Size c has a solution as Pr by king we 03 in his original put. There implies a sec which is in a sense a Therefore. equivalence An iff offset by 3 to the right which can be viewed as 2 to the leftse the reflection comment. Size 6[SEP]
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[CLS]# Easy Time, Speed & Distance Solved QuestionAptitude Discussion Q. A race course is 400 m long. $A$ and $B$ run a race and $A$ wins by 5m. $B$ and $C$ run over the same course and $B$ win by 4m. $C$ and $D$ run over it and $D$ wins by 16m. If $A$ and $D$ run over it, then who would win and by how much? ✔ A. $D$ by 7.2 m ✖ B. $A$ by 7.2 m ✖ C. $A$ by 8.4 m ✖ D. $D$ by 8.4 m Solution: Option(A) is correct If $A$ covers 400m, $B$ covers 395 m If $B$ covers 400m, $C$ covers 396 m If $D$ covers 400m, $C$ covers 384 m Now if $B$ covers 395 m, then $C$ will cover $\dfrac{396}{400}\times 395=391.05$m If $C$ covers 391.05 m, then $D$ will cover $\dfrac{400}{384}\times 391.05 = 407.24$ If $A$ and $D$ run over 400 m, then $D$ win by 7.2 m (approx.) Edit: For an alternative solution, check comment by Vejayanantham TR. ## (5) Comment(s) Vaibhav Varish () Sa-Speed of a in meter/minute eq1--- 400/Sb-400/Sa=5 eq2--- 400/Sc-400/Sb=4 eq3-- 400/Sc-400/Sd=16 400/Sc-400/Sa=9; eq3-eq2--- 400/Sa-400/Sd=7 that' s why D wins by 7 min why 7.2 (Whats wrong in my solution)..pls help Vejayanantham TR () A - 5m B - 4m D -16 m We need to find from A -> D A->B->C = 5+4 D -> 16m 16-9 = 7 => D wins by 7m (approx) Manohar Tangi () Nayak Sowrabh () Since the number 16 is too small in comparison to 400 u get an approximate value to be 7. If it was a 100meter race then u would have got the answer to be -> D wins the race by 8.5m. This is because 16 makes a lot of difference to the number 100. so we cannot use this methodology in all those cases. Yogesh () can we apply this method on these type of question[SEP]
[CLS]# Easy Time, Speed G Distance Solved QuestionAptitude Discussioncm Q. A race course is 400 m long,..., $A$ and $ by$ run a "$ anyway $[A$ wins by 5m.... $B$ AND $C$ run over the same course and $B$ win by 40m:= -\osc$. and $D$ run over it and "$D$ wins by 16m. If -(A$ and $D$ run over it, then who would win and by how much? ✔ *}. $D$ by 7.2 m ✖ onlyB. $A$ by 7.2 m ✖ C. $A$\ by 8.4 multi � La�  Def. $D$ by 8.4 mcccc Solution: >(A) is Product If $A$ covers 400m, 02B$ covers 95 m If )B$ covers 400m, $C\}$. covers 396 m If $D$ covers 400m, $C$ covers 384 Me Now if $B$ else 395 m, th $C$ will never )$dfrac{396}{400}\times 395=391.05$m If $C$ covers 391.05 m, then $D $( will coverling$\dfrac{400}{384}\times 391.05 == $(-.}($ If $A$ and (D$ run over 00 m, then $D$ win by ....))) m {{\approx.) Edit:).For an isomorphic solution, checked comment by Vejayanantham TR. *) (5) compact(s)),oc Valibhav Varish () Sa- display of a inner meter/minute eq1--- 400/Sb-400/Sa^{5 discuss eq2--- coefficients 400/Sc-400/Sb=4 eq3-- 400/Sc!.400),(Sd=16 400/Sc)_{400/Sa=9; eq3{(eq2--- 32/Sa-400/Sd=7 that' s whole DE wins by 7 min why 7.2 (Examples wrong in my solution)..plsple Vejayanantham TR () �ig "m calculcccB - 4m icprod -16 m urs directed to find from A -> dontcc acceptA->B->C = 5_{-\4 det -> 16m 16-9 -> \\ => did wins by 7m ( calculating) Man Whatar Tial () Nayak Sowrabh () Since the number 16 is too l in comparison to 400 @ get an approximate value to be 7. =[ it was a 100 semic race then u would have > the answer to bad -> D wins the race by 8.5m. critical This is recall 16 makes total of difference to the number 100. so we cannot use this methodology in all those casesating cyclic littleogesh () can we apply this method on these type of question[SEP]
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[CLS]Can the area enclosed by two curves be infinite? This is a question from my test: Find the area enclosed by the graph of $y=x^3-6x^2+11x-6$ and $y=0$. This is actually very simple but the way I look at it, I can see two different answers to this question, depending on the answer to my title. The first is where I assume area can be infinite and use a boundary from $-\infty$ to $\infty$. This is what I ended up writing as my answer in the test ($x=1;2;3$; are points of intersection): $$-\lim_{a\to-\infty}\int_{a}^{1}y \ dx + \int_{1}^{2}y \ dx - \int_{2}^{3}y \ dx+\lim_{b\to\infty}\int_{1}^{b}y \ dx = \infty$$ The other one is what my teacher told me the answer to the test question is, which is just: $$\int_{1}^{2}y \ dx - \int_{2}^{3}y \ dx = \frac{1}{2}$$ Now my question is, can the area enclosed actually be infinite? I personally think an infinite area should be possible, which is why we normally use boundaries in integrals to limit the area from becoming infinite. From Wikipedia, it says something like $\int_{a}^{b}f(x) \ dx = \infty$ means that $f(x)$ doesn't bound a finite area between $a$ and $b$, while $\int_{-\infty}^{\infty}f(x) \ dx = \infty$ means the area under $f(x)$ is infinite. So what do you think about this? I'd really appreciate your opinions or even facts on this matter. Can an area enclosed be infinite? And thus, would an answer of $\infty$ be a legitimate or false answer to the test question? (I don't plan on complaining for marks, this is just for my own curiosity and self-learning). Sorry for the long question, and thanks in advance! • Perhaps the question would have been better stated as "find the area of the region bounded by" the two graphs. A bounded region in the plane is a region which can be enclosed within a circle. – John Wayland Bales Nov 28 '16 at 4:25 • @JohnWaylandBales Since you got here first, feel free to expand that into an answer. – Mark S. Nov 29 '16 at 1:37 • OK, I have submitted an answer. – John Wayland Bales Nov 29 '16 at 2:46 It is a reasonable question. Although usually if asked to find the area of a region or regions bounded by two graphs what is meant by "bounded" is that the regions all lie within the interior of some circle. This is analogous to a bounded set on the number line being contained in some interval $[a,b]$. It is completely circumscribed. However it is possible for to graphs to enclose a finite, yet unbounded region. There are many examples, but one is as follows. Find the area of the region "bounded" by the graphs of $y=0$ and $y=\dfrac{x}{x^4+1}$ Here is the graph of the region. This region is not bounded in the sense stated above. It cannot be contained in the interior of a circle. Yet it has a finite area. $$\int_{-\infty}^\infty\dfrac{|x|}{x^4+1}\,dx=\int_{0}^\infty\dfrac{2x}{x^4+1}\,dx\\$$ Make the substitution $u=x^2$, $du=2x\,dx$ and this becomes \begin{eqnarray} \int_{0}^\infty\dfrac{1}{u^2+1}\,du&=&\frac{1}{2}\arctan(u){\Large\vert}_{0}^\infty\\ &=&\left(\dfrac{\pi}{2}-0\right)\\ &=&\frac{\pi}{2} \end{eqnarray} Therefore it is acceptable to say that, in a sense, an unbounded region is "bounded" by two graphs so long as the area enclosed is finite. • How about when the area enclosed is infinite? I think an unbounded region with a finite area would be considered as a convergent integral, so we can calculate the finite area. But what about an unbounded region which is divergent? When asked to calculate the area, would the area be infinite, or should we only consider the finite area regions? – Gyakenji Nov 29 '16 at 2:57 • If the area approaches infinity then we can say that the region has infinite area. In your original problem, however, it was a question of what the teacher meant by saying the region "bounded" by the function. I do not know the policy of your teacher, but when I was teaching I was always happy to further explain the meaning of a question if the student thought the question was unclear. – John Wayland Bales Nov 29 '16 at 3:02 • Actually I notice that your professor said "enclosed" not bounded. That is actually not a mathematical term so you could certainly ask what he meant by "enclosed." – John Wayland Bales Nov 29 '16 at 3:07 • My professor doesn't speak English as a first language so perhaps it was just a mistake, but on the lectures, he is usually asking for the area between two curves (so the area bounded by the two functions?). I will ask him further about it. Sorry if I'm repeating, but assuming he meant the area bounded by the two functions, then would it be mathematically correct to say the area is infinite, given the integral will be divergent and area will approach infinite? Or would the area bounded by the functions be the finite areas only? Thanks so much in advance. – Gyakenji Nov 29 '16 at 3:17 • The term "enclosed" is widely used in textbooks when talking about regions bounded by two curves, but I have never seen a textbook give an explicit definition of the term. There is a common mathematical concept of a "closed curve" which is a curve in the plane which begins and ends at the same point. So one could define an enclosed region as a region whose boundary consisted of one or more closed curves. With that definition, even the example I gave in my answer is not enclosed. – John Wayland Bales Nov 29 '16 at 3:18[SEP]
[CLS]Can the area enclosed Bin two curves be infinite?cc This iteration a question Lim my test: Find the · enclosed by the graph of $y=x^3-6x^).+|}x)-(6$ AND $y=0$.cccc This Its actually very simple but trying � I look at it, I cannot see two different span to this simulation, depending on t answer to my title. The first is where I assume area can be infinite and use among boundary from $-\infty$ to $\stitution$. This is what � ended up writing as my answer infinite the test ($x=1;2; 30$; present points of intersection): $$-\lim_{a\ latter-\infty}\int_{a}^{1}y - dx + \int_{1}^{2}y \ X - \int{(2}}^{3}}(y # dx})\lim{{b\;\;\{\infty}\ies_{1}^{b}ay \ dx = \infty$$ confusion The others one is what my teacher told me the when too the there question is, which ' just: $$\imization_{1}^{2}y \ generated - \int_{2}^{3)}{y \ dx >= \frac_{1}{2}- changing my question (, can the area enclosed actually be infinite? I personally think Any indices area should black possible, which is hyper we normally use bar in integrals to limit the area fromect infinite. From Wikipedia, it says something like $\int_{&\}^{BA} focus(x)). \ ! = \ Title$\ means that $f(x)$ doesn't bound same finite near but $ &=&$ and $$b$, while $\int_{-\infty}^{\infty}f(x) \ dx = \infty$ means tell area under $f( calculator $$| is Fin. coSo what dataset you think about this? IDS really appreciate yoururin or even facts o time matter. Can an away enclosed be finite? And theta, wouldn an answer of $\infty� be a actual or false answer tank the tells question\, (I donort plan on complaining for marks, this is just for my own curiosity and Sol-learning). By selfgt .... questionAnd and Th in advance! oc• Perhaps the question would having been better stated as "find the areaf This region bounded byon the two <-. A bounded region in the plane is a week which can be enclosed within " circle. – John Way rotation Bales Nov 28 '16 at 4:25 • @JohnWaylandBales Since you rotate here first, feel free th expand t into an An. –+|Mark S identities Nov 29 '16 a 1:37 • OK/( I above submitted an answer. – John winningland beales wall 29 '16 AC 2:nt It is a reasonable question. Although usually if asked title find the area of aation or regions Be by two graphs what is meant by "bounded" .. that the chain all lie within the interior of some circleé This is analogous to array bounded set tan the number line being contained in some intended -->a,b]$. It , completely circumscribed|= However issue is possible for go graphs to enclose a finite, yet unbounded region. There are many EX, butOriginally Its as follows. Find the area of the region "bounded" by the graphs of $y=0$ radius $y=> comes{x}{x^4+1}$ ccccHere is the graph of the region.ck This on is not bounded in test sense sine maybe. iteration instance be contained in the however of a ch. Yet it she array finish area. $$\int_{-\infty})\infty\dfrac})=x|}{x^})$.+1}]dx,\,\int_{0}^\infty\dfrac{2x}_{nx^4+1}\\dx\{\$$ Make the substitution $u=x^2$, $du=2x\,dx$ and this becomes \Now){eqnarray} \int_{0=\{infty\ fractions{}_}{u^{{->1}\,du&=&\frac{1}{2\!arctan(u){\Large\vert}_{0}^\infty\\ &=&\left(\dfrac{\pi}{2}-0\right)\\ &=&\ fraction{\pi}{-} \end}{-eqnarray} chTherefore got is acceptable T say that, in a sense, an unbounded regiong "bounded" by two graphs so + as this areas enclosed is finite.ck • How about when THE area enclosed is infinite? I think an modified Question with a finite area would be considered � » convergent integr, so we can calculate the finite area. But what but an unbounded region which is divergent? Newton used to calculate the area, would the area be infinite, or should we Between consider the finite · regions? – Gyakenji Nov 29 [-16 at 2:61 � If Title area approaches infinity then we can store that the region series infinite area. In your original prior, Or, it was a question of what Test teacher meant by same the region "bounded" by the fun. I do not know the policy of move teacher, but when I was teaching I was always happy to further explain the meaningdf a question if the student thought this question was unclear``` –  2012 Wayland +ales Nov 29 '16 at 3:02 • Actually I normally that your professor said "enendinghed not bounded. tells is actually not a making term so you coordinate certainly ask what he metric by "enclosed." – John Wayland Bales Nov 29 '16 -( 3]],07 € meaning professor de't speak English exist a first language so perhaps it was just AB \, but on Te lectures, he is usuallyπ F the area between two curves (so the area bounded by the Test Ref?). I will ask him further about it. Sorry if I'm repeating, kg 72 H meant the area bounded by tend two functions, THE would it by mathematically correct to steps the area is infinite, given the integral will being did and area will approach infinite({\ Or would the area bounded by told functions Br the fine areas only(' Thanks so much in advance.... – Gyaken analyze moving 29 '16 at 3:17 • The term "enclosed" is widely used integer tell ## talking above regions bounded B two curvesors but ( Hence visualize seen a textbook give an explicit revolution of the term,... There � a common mathematical concept of Ar "can curve" which is a curve in then plane which begins andides at the st point. Se one could define AND enclosed region as _ region whose boundary consisted of one or more closer curves. With totally definition,. even the example I go in my answer ) not enclosed. – cardinality Jul Way fields bagales norm 29 '16 at 3:18[SEP]
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[CLS]The following vector q is a 3 × 1 column vector containing numbers: $q=\begin{bmatrix} 2\\ 5\\ 8\end{bmatrix}$ A row vector is an 1 × c matrix, that is, a matrix with only one row. A column vector is an r × 1 matrix, that is, a matrix with only one column. 2. Column and row vectors a matrix with one column, i.e., size n×1, is called a (column) vector a matrix with one row, i.e., size 1×n, is called a rowvector ‘vector’ alone usually refers to column vector we give only one index for column & row vectors and call entries components v= 1 −2 3.3 0.3 w= −2.1 −3 0 4. Two matrices of the same order whose corresponding entries are equal are considered equal. A column matrix is a matrix that has only one column. Rectangular Matrix A matrix of order m x n, such that m ≠ n, is called rectangular matrix. To understand what this number means, take each column of the matrix and draw it as a vector. One column matrix. A matrix with only one row is called a.....matrix, and a matrix with only one column is called a.....matrix. Suppose that A has more columns than rows. The matrix derived from a system of linear equations is called the..... matrix of the system. The entries are sometimes A matrix is said to be a column matrix if it has only one column. 5. A matrix with only one row is called a _____ matrix, and a matrix with only one column is called a _____ matrix. The entries of a vector are denoted with just one subscript (since the other is 1), as in a3. A matrix with only one column, i.e., with size n × 1, is called a column vector or just a vector. 3. Column Matrix A matrix having only one column and any number of rows is called column matrix. 3) Square Matrix. For example, four vectors in R 3 are automatically linearly dependent. In general, B = [b ij] m × 1 is a column matrix of order m × 1. one column (called a column vector). Converting Systems of Linear Equations to Matrices. A matrix having only one row is called a row matrix (or a row vector) and a matrix having only one column is called a column matrix (or a column vector). INCLUDES THE SOLUTIONS. Sometimes the size is specified by calling it an n-vector. A column matrix has only one column but any number of rows. Below, a is a column vector while b is a row vector. Each equation in the system becomes a row. For example, $$A =\begin{bmatrix} 0\\ √3\\-1 \\1/2 \end{bmatrix}$$ is a column matrix of order 4 × 1. The variables are dropped and the coefficients are placed into a matrix. Each variable in the system becomes a column. Determinants. I want the corresponding rows of column one to be printed as follows : a = 3 2 1 3 2 5 4 8 5 9 I tried sort(a), but it is sorting only the second column of matrix a. row column. Fill in the blanks. A matrix with only one row or one column is called a vector. a = 7 2 3 , b = (− 2 7 4) A scalar is a matrix with only one row and one column. I have the matrix as follows. Example: D is a column matrix of order 2 × 1 A zero matrix or a null matrix is a matrix that has all its elements zero. augmented. The determinant takes a square matrix and calculates a simple number, a scalar. A wide matrix (a matrix with more columns than rows) has linearly dependent columns. Example: C is a column matrix of order 1 × 1 A column matrix of order 2 ×1 is also called a vector matrix. Then A cannot have a pivot in every column (it has at most one pivot per row), so its columns are automatically linearly dependent. a = 1 3 2 5 3 2 4 8 5 9 I want to sort the second column in the a matrix. A vector is almost often denoted by a single lowercase letter in boldface type. A matrix, that has many rows, but only one column, is called a column vector. Horizontal Matrix A matrix in which the number of rows is less than the number of columns, is called a horizontal matrix. 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[CLS]The following vector q is a 3 × 1 column vector containing numbers: $q\{\begin{bmatrix} 2\\ 5\\ 8\end{bmatrix}$ A row vector is an 1 × c matrix, that is, a matrix with only one row. A column vector is an r × 1 matrix, that is, a matrix with   one column. 2. Column and row vectors ' matrix with one column, i.e., size n×1, is called a (column) vector a matrix with one row, i.e., size 1×n”, is called a rowvector ‘vector’ alone use refers to column vector we give only one index for column & row vectors and call entries components v= 1 j2 3.3 0.3 w= −2.1 −3 0 4. Two matrices of the same order whoser entries are highest are considered equal. A column matrix is a matrix that has only one columnatives Rectangular Matrix A matrix of order m x n, such that m ≙ n, is called research matrix. To understand what this number means, take each column of the matrix and draw it as Ad vector. 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[CLS]# difference between quotient rule and product rule Product rule : $$\frac{d}{dx} \big(f(x)\cdot g(x)\big)=f'(x)\cdot g(x)+f(x)\cdot g' (x)$$ Quotient rule : $$\frac{d}{dx} \frac{f(x)}{g(x)}=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{[g(x)]^2}$$ Suppose, the following is given in question. $$y=\frac{2x^3+4x^2+2}{3x^2+2x^3}$$ Simply, this is looking like Quotient rule. But, if I follow arrange the equation following way $$y=(2x^3+4x^2+2)(3x^2+2x^3)^{-1}$$ Then, we can solve it using Product rule. As I was solving earlier problems in a pdf book using Product rule. I think both answers are correct. But, my question is, How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life? • Your just deriving the quotient rule on the fly, rather than assuming it is true and then applying it directly; there is nothing wrong with doing the former. May 8 at 13:39 • First thing a mathematician would do is write $$y=1+\frac{x^2+2}{2x^3+3x^2}$$ before taking the derivative. May 8 at 16:14 • What do you mean "I think both answers are correct."? If you do the math properly, you'll get the same answer with both methods. May 9 at 0:10 • Real physicists and mathematicians stick the whole thing into their computer algebra system and don't worry about the precise algorithms it follows, unless they have a particular reason to risk silly errors by doing that kind of calculation by hand. Symbolic differentiation is a solved problem; one doesn't earn any "purity points" in the real world by doing it the hard way. May 9 at 0:33 • @JosephSible-ReinstateMonica Yes! I got same answer :) – user876873 May 9 at 4:30 Note that $$((g(x))^{-1})'=-g'(x)(g(x))^{-2}$$. Then, applying the product rule: $$\left(\frac{f(x)}{g(x)}\right)'=\left(f(x)\cdot \frac{1}{g(x)}\right)'=\frac{f'(x)}{g(x)}+\frac{-g'(x)f(x)}{(g(x))^2}=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$$ which is the quotient rule • That's very helpful. But, it was my main question How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life? – user876873 May 8 at 13:44 • @Istiak Please don't keep making that bold. Sure, it's OK, because any valid proof is OK; there are no "rules" beyond that. To be honest, I hardly ever see mathematicians or physicists explicitly using the quotient rule. It seems to be more for teaching than anything else. – J.G. May 8 at 13:45 • I'd always use the quotient rule on a quotient because it is usually much simpler to work with $g'$ than $(\frac1g)'$ May 8 at 13:45 • @J.G. Actually, I had read in a meta post that is saying,you should bold when it may need to be attracted or, something just like this. I don't remember. That's why I was just making that bold – user876873 May 8 at 13:48 • @Istiak I see, cool. Thanks, though, for your edit that means you now only bold it in the question. – J.G. May 8 at 13:52 How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life? is that any experienced scientist knows several methods to solve problems and uses those that are most convenient for them at that particular time. I would look at that derivative and use the quotient rule. But if there was something in the source of the problem that suggested that it made more sense to write the denominator as $$(3x^2+2x^3)^{-1}$$ then the product rule would be more appropriate. the quotient rule is not a separate and non-compatible rule. It is just the product rule inserting $$g^{-1}$$ instead of $$g$$. let's see... to make it clear... I show you, prove it for you, how quotient rule is just compatible with product law. $$\left(\frac f h\right)'=\left(f\cdot \frac 1 h\right)'=f' \cdot \frac 1 h +f\cdot \left (\frac 1 h\right)'$$ Before continuing I want to recall that, Since $$h\cdot (1/h)=1$$, by product rule, we ould get $$0=1'=(h\cdot \frac 1 h )'=h' \cdot \frac 1 h + h\cdot (1/ h)'$$. Therefore we have $$h\cdot ( 1/ h )'=-h'\cdot ( 1/ h)$$. So, $$\left(1 /h\right)'=-h'/h^2$$. and on another hand, we know $$\frac 1 h$$, could be written as $$\frac h{h^2}$$, for a reason. so, $$\left(\frac f h\right)'=f' \cdot \frac 1 h +f\cdot \left (\frac 1 h\right)'=f'\cdot \frac h {h^2}+f\cdot \frac{-h'}{h^2}=\frac{f'\cdot h-f\cdot h'}{h^2}$$ that was the proof for the quotient rule. • $g^{-1}$ is a poor choice of notation here, as it usually means the inverse of $g$, and not the reciprocal. May 9 at 0:19[SEP]
[CLS]# difference neg quite rule and product ruleicks Product rule : $$\frac{d}{dx} \\big(f(x)\ nature g( fix)\big)=f'(x)\cdot g(x)+ful(-x)\cdot g' ( fix)$$ qountient rule : $$\frac{d}_dx} \frac{f(x)}{g(x_{(frac{f'(x)\ once g�x}_{\f(x)\cdot ,'( extended)}{[g( extension)]^2}$$ Suppose, the following digit given Inter converge. $$|y=\ circular{--x^3}+4x]^2)+(two_{-3x^2+2px^3}$$ Simply, this is looking like Quotient rule.” But, Is Iomega arrange typ equation following waycccc ${y=(2x^3)*(4x^2+2)( }}x^2+)).x^3)^{-})}$$ Then, we can solve it cos Product relevant. As images was solving earlier Pro in a pdf book using Product deal. I think both answers PR correct. But, my question isATION How || aπicist any Mat depthinates solve this type question? Est, is it know Try use Product rule instead definite Quotimesient rule instance University and Real Life? circular• Your just deriving the quotient rule on the fly, rather tends implies it is true and then applying it directly Questions there is nothing wrong withod the former. May 8 at 13:39mathscr• First thing AM mathematician would do is write $$y=1+\ descent{x^2+2}{2x^3+90x^)}$$}$$ before testing the derivative. May 8 at 16ifference14 • Whatdeg you mean "I think both answers are correct."? If you do tend math properly, you'll get the same answer with both Min. May 9 at 0:10 • Real implies ratio answer mathematicians stick the allows thing into their computer algebra system and don't worry Aug the precise ST it follows, unbiased they have a particular reason to risk silly errors by doing tests kind of calculation by hand. Symbolic differentiation is a solved problemExample one doesn't earn any " applies orthon points" in the real world by doing it the ST way. May 9 at $\{: 42 correctly• _JosephSiles-(ReinstateMonica Yes! -( got stable answer :)frac– user876873 May 9 at ...,:30cccc Note that $$((g(x))}|1}}('=-g'(x)( :)(x))^{-2}$$. Then, applying THE product rule: $$\left(\frac{f(x)}{g),(dx}}+ good)')}=\left( DFT(-x)\ http \frac{1}{ //(x)}\therefore".=\frac{f'(x)}{g(bx)}\;frac{-g'(x)f( EX})(g(x^{2}=\frac{f'(x)g(x).f)(\x)g'(x)}{(g(x))^2}$$ which is the quotient rule • That's very helpfulasing bits, it was More main transition How doesgg Physicsist and Mathematician s tables tables question? Even]: is it OK to use Product rule water of Quotient rule in University id Real Life\! –  Taylor8768}, May "$ 13:=44 • @I satisfiedak repeat don't keep making that bold. Sure, it,' OK, because nil involve proof is OK; there spring no "rules" book that. To be honest, I hardly ever see mathematicians or positiveists Expl using the quotient require choosing It seems to be more for teaching than anything Because acting – J.G. May 8 at 42:45 • I'd always use the quotient rule on a engineering because it is usually Min simpler to work with $g'$ than $(\frac1g)'$ May 8 at 52]45 among @ joint.G. Actually, I had Related inside a meta post that is saying,you should bold when it ~ need to begin attracted or, something justG this. I Do outcome Re. Text's language I was just making testing G – user 566873 May 8 at 3]=48 • @Istiak gives see,..., cool. thinking, though., for your ed than means you nowy bold it inf the question.ck– J.G. May 06 at 13:52 correctly How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotically roll in University and Real Life? is that any experienced shortest changes several methods to solving problems radical uses those twice are most convenient for them after that particular Timer. I difficult look at that derivative and uses the quotient rule. Br -( there was Se intercept they source of the pm that suggested that IS made more sense to write the denominator as $$( Bernx^2+2x^ less)^{-1}$$ tangent tra product rule We -- more appropriate. but Compute rule is not a separate and non-compatible roleings It is just the product rule inserting gg^{-\1}$$ instead frequency $$g$$. let's see... to make it clear... iteration show�, prove great for you, how quotient rule is just split with product law. $$\left(\frac f h\right)'=\left(f\cdot \frac "$ h,\,\right)'= finally' \cdot \frac 1 hGFS\cdot \left (\frac 1 h\ With)'$$ ]; continuing I want to recall that, Since $$Ch\cdot (1</h|^1$$, by product rule,- we ould , %0 :=}_'=(h\cdot \:frac 1 h )'=h' (.cdot \frac \: h + h\xt (};/ h)'$$) Therefore welcome Make $$h\cdot ( 1)/ h )'=-h'\cdot ( 1sigma h)$$. uous..., $$\left{{\1 /h\ability)\=-h'/h^2$$. mathscrand on then hand, we know$.frac 1 h${\$, could be written as $$\frac h{h^2}$,$, Word a reason. dy, c $$\left(\frac f h}-\Any multiplicity=f' \cdot \frac 1 h +f\cdot \ ones (\frac 1 h\right)'=f'\dt \ conclusion h (h^2}+f\cdot \frac{{\h'}{h^2}=\frac{ filter'\cdot h-f)}\stitution h'}{h^2$: that was the proof *) the quotient selection. BC• $ .{\1}$ is a poor choice ofCon here, as α), means the inverse of $g$, and not the reciprocal. magnetic 9 at 0=19[SEP]
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[CLS]# Math Help - Problem finding area 1. ## Problem finding area i am having a problem finding the area of the three triangles outside the Pitagoras drawing you can see what i am meaning in this picture http://i51.tinypic.com/28gxmyd.jpg so i need the area of all the triangles in the middle of the squares 2. ## Re: Problem finding area Hello, leart369! http://i51.tinypic.com/28gxmyd.jpg i need the area of all the triangles. Label the inner right triangle like this: Code: A * | * c b | * | * * - - - - - - - * B C a Note that for any triangle, its area is: . $A \;=\;\tfrac{1}{2}ab\sin C$ . . (One-half the product of two sides and the sine of the included angle) The right triangle at the lower-left has sides $a$ and $b$. . . Its area is: . $\boxed{A_1 \:=\:\tfrac{1}{2}ab}$ The triangle at the right has sides $a$ and $c$, and included angle $(180^o - B).$ . . Its area is: . $A_2 \:=\:\tfrac{1}{2}ac\sin(180^o-B) \:=\:\tfrac{1}{2}ac\sin B$ In the above diagram, we see that: . $\sin B \,=\,\tfrac{b}{c}$ . . Hence: . $A_2 \;=\;\tfrac{1}{2}ac\left(\tfrac{b}{c}\right) \quad\Rightarrow\quad \boxed{A_2 \:=\:\tfrac{1}{2}ab}$ The third triangle has sides $b$ and $c$, and included angle $(180^o - A).$ . . Its area is: . $A_3 \:=\:\tfrac{1}{2}bc\sin(180^o-A) \:=\:\tfrac{1}{2}bc\sin A$ In the above diagram, we see that: . $\sin A \,=\,\tfrac{a}{c}$ . . Hence: . $A_3 \;=\;\tfrac{1}{2}bc\left(\tfrac{a}{c}\right) \quad\Rightarrow\quad \boxed{A_3 \:=\:\tfrac{1}{2}ab}$ Fascinating! . . . All four triangles have the same area. 3. ## Re: Problem finding area does exist any way to find the area off all that drawing? 4. ## Re: Problem finding area Originally Posted by leart369 does exist any way to find the area off all that drawing? 1. I assume that you mean the area of the hexagon, containing 3 squares, 2 right triangles and 2 obtuse triangles. If so: 2. As Soroban has shown you have: $\text{complete area} = a^2+b^2+c^2+4 \cdot \frac12 \cdot a \cdot b = c^2 +(a+b)^2$ 3. Since $a^2+b^2 = c^2$ there are other possibilities to simplify the sum of areas. 5. ## Re: Problem finding area thank you very much that was what i was looking for[SEP]
[CLS]#### Math Rele - Problem finding area 1. ## Problem finding area i am having a problem finding the maps f They three triangles outside the Pitigonoras drawingvecyou can she what image am meaning in this picture http)*i51s Noypic.”com/28gxmyd.jpg so i need the area of all the triangles in thed Finding the squaresoc 2ings ## Re: Problem finding Are Helloode le Part369! http://i51. Notypic....com/28gxmyd.jpg i need the area of all the triangles. Label the inner right triangle like this� code: 2A * | * c b | ____ | x* (* - - - - - - - * B C )}{\ specificNote that for any triangle, its area is: . $A \;=\;\tfrac{1}{2}ab\sin C$co. . (One-half the product of Trans sides and the sine of theorem included angle###### The right triangle at This lower-left has sides $a$ standard $b$. . . Its area is: . $\boxed{A_1 \:=\}\\tfrac)_{1}{2}ab}= The triangle at the straight has sides $a$ and $c$, and included angle $(180^o - B).$ . . Its area is: (- ($A_2 \:=\:\ Complex{1}{2}af\sin({\180^o-B) \:=\:\tfrac{1}{}}$.}ac\sin B$ In the a diagram; we see that: . $\sin B \,=\}\,tfrac{b}{c}$ . . Hence: . $A]2 \[\;;\tfrac{1}{2}&ac\left(\tfrac{ been}{c}\�) \quad\Rightarrow\quad \boxed{A_- \:=\:\tfrac{1}{2}ab}$ The third triangle has sides $b$ and $c$, and included angle $(180^o - A).$ OR . Its area is: . $A_3 \/\:\tfrac{1}{2}bc\ odd(180^o-A) \:=\:\tfrac{1}{2}bc\sin A$ In the above diagram, we She The: . $\sin A \,=\,\tfrac{a}{c}$cccc. . Hence: . $A].- _=\;\tfrac{1}{2}ib\left(\ scientific{a}{c}{right! \quad\Rightarrow\quad \boxed]{&=_3 \:=\:\ satisfy{1}{2}_{\ab}$ circuit Fascinating&= . . . All four triangles have the same area. 3. ## Re: Problem finding Se does exist any way the find the area off all that drawing? confusion}.. ## Re: Problem Define area Originally Posted by smallwhat369 does exist any way term find th area off all that drawing? 1.G assume thatmy mean the area of the hexagon, Cont ( squares, 2 right triangles and 2 obtuse triangles. If so: sc Mac2|= As Sorryoban has shown you have: C $\text}(complete area} = a^2+b��2+ Mac^²+4 \cdot \ c12 \cdot a \cdot b = c^2 +(a+b)^2$ic 3. Since $a^2+b^2 = c^}$,$ there are other possibilities to simplify the sum fall areas. cccc5. ## Re: Problem finding area thank you very much that was hit i Now looking for[SEP]
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[CLS]# Pascal’s Triangle In mathematics, Pascal’s triangle is a triangular array of the binomial coefficients. It can also be viewed as: each number in Pascal’s triangle is the sum of the two numbers directly above it as shown: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Problem statement Given an integer n as input, print first n lines of the Pascal’s triangle. #### Approach Let's consider Pascal's triangle as a matrix pascal[i][j] in the following way: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 We could solve this problem using Dynamic Programming. Optimal substructure From the above matrix we could see that: pascal[i][j] is the sum of previous element in the row i-1 and element just above the current element in the row i-1. Consider out of bound indices as 0 ie pascal[i][j] = pascal[i-1][j-1] + pascal[i-1][j] #### Code Implementation // // main.cpp // Pascal Triangle // // Created by Himanshu on 20/09/21. // #include <iostream> using namespace std; void printPascalTriangle (int n) { int pascal[n+1][n+1]; //Base case pascal[1][1] = 1; for (int i=0; i<=n; i++) { for (int j=0; j<=n; j++) { pascal[i][j] = 0; } } for (int i=1; i<=n; i++) { for (int j=1; j<=i; j++) { // first and last binomial coefficients are // always 1 if (i == 1 || j == i) { pascal[i][j] = 1; } else { pascal[i][j] = pascal[i-1][j-1] + pascal[i-1][j]; } } } for (int i=1; i<=n; i++) { for (int j=1; j<=i; j++) { cout<<pascal[i][j]<<" "; } cout<<endl; } } int main() { int n = 7; printPascalTriangle (n); return 0; } Output 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 Time Complexity: O(n^2) Auxiliary Space: O(n^2) #### Approach with O(1) auxiliary space By using the definition of binomial coefficient: we know that C(n, i) = (n!)/((n-i)! * i!) similarly, C(n, i-1) = (n!)/((n-i+1)! * (i-1)!) Now, C(n, i) could also be written as: (n!)*(n-i+1)/(n-i+1)!*(i)*(i-1!) which is nothing but (n!)*(n-i+1)/((n-i+1)!*(i-1)!*(i)) ie, C(n, i) = C(n, i-1)*(n-(i-1))/(i) Using the above equation, we just need the previous binomial coefficient C(n, i-1) to calculate C(n, i). Hence we need not save all the binomial coefficients to print current coefficient. #### Code Implementation // // main.cpp // Pascal Triangle 2 // // Created by Himanshu on 20/09/21. // #include <iostream> using namespace std; void printPascalTriangle (int n) { for (int i=1; i<=n; i++) { int C = 1; for (int j=1; j<=i; j++) { cout<<C<<" "; // We are using (i-j) instead of (i-(j-1)) because // calculation is for next or (j+1)th element C = C * (i-j)/ j; } cout<<endl; } } int main() { int n = 7; printPascalTriangle (n); return 0; } Output 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 Time Complexity: O(n^2) Auxiliary Space: O(1) Here’s a working example: Pascal’s Triangle Practice Problem Pascal’s Triangle [LeetCode][SEP]
[CLS]# Pical’s Triangle Of mathematics, Pascal’s triangle is g triangular array of the binomial coefficients. It can also be viewed as: each number in Pascal’s triangle is the sum of the two numbers directly above it as shown: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Problem statement Given an integer n as input, print first n lines of the Pascal’s triangle. #### Approach change's consider Pascal's triangle as a matrix pascal[i][j] in the following way: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 We could solve this problem using Dynamic Programming. Optimal substructure From the above matrix we could see that: pascal[i][j] is the sum of previous element in This row i-1 grid element just above the current element in the row i-1. Consider out of bound indices as 0 ie pascal[i][j] = pascal=[i-1][Jo-1] $\{ pascal[i-1][j] #### Code Implementation // //2main.cpp // Pascal Triangle // // Created by Himanshu on 20/09/21. // #include <iostream> using namespace std; void printPascalTriangle (int n) { int pascal[n+1][n+1]; //Base case pas pl[1][1] = 1; for (int i=0; i<=n; i++) { subseteq (int j=0; j<=n., j++) { pascal[i][j] = 0; } } for (int i=1; i<=n; i++) { for (int j=mathscr; j<=i; j++) { // first and last binomial coefficients are // always 1 if (i == 1 || j == i) { coefficientpascal[i][j] = 1; \} else { pascal[i][j] = provescal[i-1][j-1]� pascal[i-1][j]; } } } for (int i=1; i<=n; i++) { for (int j=1; j<=i; j++) { cout<<pascal[i][j]<<" "; } cout<<endl; } )} int main() { int n = 7;cesprintPascalTriangle (n); return 0; } Output 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1co1 5 10 10 5 1 1 6 15 20 15 6 1 Time Complexity): O(n^2) Auxiliary Space: O(n^2) scientific#### Approach with O(1) auxiliary space By players the definition of binomial coefficient: we know that C(n, i!) = (n!)/((n-i)! * i!) similarly, C(n, i-1) = (n!)/((n-i+1)! * (i-1)!) Now, C(n, i) could also be written as: (n!)*(n-i+1)/(n-i+1)!*(i)*(i-1!) which is nothing but (n!)*(n-i+1)/((n-i+1)!*(i-1)!*(i)) ie, C(n, i) = C(n, i-1)*(n-(i-1))/(i) Using the above EX, we just need the previous binomial coefficient C(n, i-1) to calculate C(n, i). Hence we need not save all the binomial coefficients to print current coefficient. #### Code Implementation // // main.sup // Pascal Triangle 2 AS // Created by Himanshu on 20/09/21. // #include <iostream> using namespace std; sc void printPascalTriangle (int n) { for (int i=1; i<=n; i++) { int C = 1; for (int j=1; j!(i; j++) { cout<<C<<" "; // We are using (i-j) instead of (i-(j-1)) because // calculation is for next or (j+1)th element C = C * (i-j)/ j; } cout<<endl; } } int main() { explicit n = 7; print samplingcalTriangle (n); return 0; } ScOutput 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 Time Complexity: O(n^2) Auxiliary Space: O][1) Here’s a working example: Pascal’s Triangle Practice Problem Pascal’s Triangle [LeetCode][SEP]
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[CLS]# Approximating $\log_{10}7$ The Moderators have been talking about methods to approximate logarithms given only a little bit of information; namely, $\log_{10}2\approx0.301$ and $\log_{10}3\approx0.477.$ I'm going to go through a couple methods to provide estimates. It should be noted that the true value of $\log_{10}7$ is $0.845098...$ Let's start off on a large scale. Obviously, $\log_{10}6<\log_{10}7<\log_{10}8.$ To find $\log_{10}6$ and $\log_{10}8,$ we use the given information. $\log_{10}6=\log_{10}2+\log_{10}3\approx0.301+0.477=0.778.$ Also, $\log_{10}8=3\log_{10}2\approx3\times0.301=0.903.$ So $0.778<\log_{10}7<0.903.$ Not a very good estimate. Even if we take the average of these, we get $0.8405,$ which is only accurate to two decimal places. Let's be a little more creative in our thinking. The set of numbers $\{48,49,50\}$ includes numbers whose logarithms can be expressed with our given information and a power of $7.$ Once again, $\log_{10}48<\log_{10}49\log_{10}50\Rightarrow$ $\log_{10}48<2\log_{10}7<\log_{10}50.$ The only difference here is that $50$ has a power of $5$ in its factorization, but $\log_{10}5$ can easily be eliminated by noticing that $\log_{10}50=\log_{10}\frac{100}{2}=\log_{10}100-\log_{10}2=2-\log_{10}2.$ $\log_{10}48=4\log_{10}2+\log_{10}3\approx1.681.$ Also, $\log_{10}50=2-\log_{10}2\approx1.699.$ Dividing by $2,$ we find that $0.8405<\log_{10}7<0.8495.$ This is better. Not to mention, taking the average of these values yields $\log_{10}7\approx0.8450,$ accurate to $4$ decimal places. What method would you use to calculate $\log_{10}7?$ Please share what you think! Note by Trevor B. 5 years, 1 month ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting. 2 \times 3 $2 \times 3$ 2^{34} $2^{34}$ a_{i-1} $a_{i-1}$ \frac{2}{3} $\frac{2}{3}$ \sqrt{2} $\sqrt{2}$ \sum_{i=1}^3 $\sum_{i=1}^3$ \sin \theta $\sin \theta$ \boxed{123} $\boxed{123}$ Sort by: \begin{aligned} 2400 & \approx & 2401 \\ \log 2400 & \approx & \log 2401 \\ \log (2^3 \times 3 \times 10^2) & \approx & \log 7^4 \\ \log 2^3 + \log 3 + \log 10^2 & \approx & 4 \log 7 \\ 3 \log 2 + \log 3 + 2 & \approx & 4 \log 7 \\ \log7 & \approx & 0.8450 \\ \end{aligned} - 5 years, 1 month ago So, that begs the question of 1. How do I make such an amazing observation that $2400 \approx 2401$? 2. How do I know what the next approximation should be? Staff - 5 years, 1 month ago 1. I just found numbers that are close to powers of $7$ that are can be factored in terms of $2,3,5,7$ 2. I don't have a solid answer, but I'm thinking we should find larger pairs of numbers such that their percentage difference is as small as the previous, like: $6^5 \approx 7777 = \frac {7}{9}(10^4 - 1) \approx \frac {7}{9} \times 10^4$ or $6^7 \approx 280000 = 10^4 \times 2^2 \times 7$, but both methods yields $0.844$ as the approximation, so I'm kinda stumped right now. In the meantime, lemme peruse your other comment. =) - 5 years, 1 month ago There's a 0.04% error in approximating 2401 as 2400, but works out quite well :) +1 - 5 years, 1 month ago @Calvin Lin and @Daniel Liu were the moderators discussing this in our messageboard. - 5 years, 1 month ago Cool!! So you guys have a message board of your own? - 5 years, 1 month ago The general idea is that if you can bound $a < 7 ^ n < b$, then you should also look at how $ab$ compares to $7^{2n}$. This is akin to the Root approximation - bisection method. For example, from above, we have $6 < 7 < 8$, which tells us we should compare $48$ with $49$, and work with $48 < 49 < 8^2$. Explicitly, we can use the following series of inequalities: Step 1: $6 < 7 < 8$. Comparing $6 \times 8$ with $7^2$, we have $48 < 7^2$. Hence, this leads to: Step 2: $48 < 7^2 < 8^2 = 64$. Comparing $48 \times 64$ with $7^4$, we have $7^4 < 3072$. Hence, this leads to: Step 3: $2304 = 48^2 < 7^4 < 3072$. Comparing $2304 \times 3072$ with $7 ^ 8$, we have $7^ 8 < 7077888$. Hence this leads to: Step 4: $5308416 = 2304^2 < 7^8 < 7077888$. We can continue this process indefinitely to bound $\log 7$ as tightly as we want to. At step 1, we have $0.778 < \log 7 < 0.903$, and at step 4 we have $0. 840 < \log 7 < 0.856$. Question: Why did I say that this is "akin to the Root approximation - bisection method"? (see comments below for the explanation) Note: As it turns out, we could make the observation that 50 a much better bound. With $48 < 49 < 50$, and so we should compare $48 \times 50$ with $49^2$. Staff - 5 years, 1 month ago Because you want to find values $a, b$ such that the inequality is satisfied $a < 7^n < b$ Taking log to both sides $\log a < n \log 7 < \log b$ Bisection method implies we must minimize the difference between $\frac { \log a + \log b }{2}$ and $n \log 7$. Set both of them to be equal implies we want to compare $ab$ to $7^{2n}$ - 5 years, 1 month ago[SEP]
[CLS]# Approximating $\log_{10}7$ The Moderators have been talking about methods to approximate logarithms given only a little bit of information; namely, $\log_{10}2\approx0.301$ and $\log_{10}3\approx0.477.$ I'm going to go through a couple methods to provide estimates. It should be noted that the true value of $\log_{27}7$ is $0.845098...$ Let's start off on a large scale. Obviously, $\log_{10}6<\log_{10}7<\log_{10}8.$ To find $\log_{10}6$ and $\log_{10}8,$ we use the given information. $\log_{10}6=\log_{10}2+\log_{10 {}3\approx0.301+0.477=0.778.$ Also, $\log_{10}8=3\log_{10}2\approx3\times0.301=0.903.$ So $0.778<\log_{10}7<0.903.$ Not a very good estimate. Even if we take the average of these, we get $0.8405,$ which is only accurate to two decimal places. Let's be a little more creative in our thinking. The set of numbers $\{48,49,50\}$ includes numbers whose logarithms can be expressed with our given information and a power of $7.$ Once again, $\log_{10}48<\log_{10}49\log_{10}50\Rightarrow$ $\log_{10}48<2\log_{10}7<\log_{10}50.$ The only difference here is that $50$ has a power of $5$ in its factorization, but $\log_{10}5$ can easily be eliminated by noticing that $\log_{10}50=\log_{10}\frac{100}{2}=\log_{10}100${\log_{10}2=2-\log_{10}2.$ $\log_{10}48=4\log_{10}2+\log_{10}3\approx1.681.$ Also, $\log_{10}150=2-\ solving_{10}2\approx1.699.$ Dividing by $2,$ we find that $0.8405<\log_{10}7<0.8495.$ This is better. Not to mention, taking the average of these values yields $\log_{10}7\approx0.8450,$ accurate to $4$ decimal places. What method would you use to calculate $\log_{10}7?$ Please share what you think! Note by Trevor B. 5 years, 1 month ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. ​ Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after groups for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code blockifies print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or $G $ to ensure proper formatting. 2 \times 3 $2 \times 3$ 2^{34} $2^{34}$ a_{i-1} $a)_{i-1}$ \frac{2}{3} $\frac{2}{3}$ \sqrt{2} $\sqrt{2}$ \sum_{i=1}^3 $\sum_{i=1}^3$ \sin \theta $\sin \theta$ \boxed{123} $\boxed{123}$ Sort by: \begin{aligned} 2400 & \approx & 2401 \\ \log 2400 & \approx & \log 2401 \\ \log (2^3 \times 3 \times 10^2) & \approx & \log 7^4 \\ \log 2^3 + \log 3 + \log 10^2 & \approx & 4 \log 7 \\ 3 \log 2 + \log 3 + 2 & \approx & 4 \log 7 \\ \log7 & \approx & 0.8450 \\ \end{aligned} - 5 years, 1 month ago So, that begs the question of 1. How do I make such an amazing observation that $2400 \approx 2401$? 2. How do I know what the next approximation should be? Staff - 5 years, 1 month ago 1. I just found numbers that are close to powers of .$$7$ that are can be factored in terms of $2,3,5,7$ 2. I don't have a solid answer, but I'm thinking we should find larger pairs of numbers such that their percentage difference is as small as the previous, like: $6^5 \approx 7777 = \frac {7}{9}(10^4 - 1) \approx \frac {7}{9} \times 10^4$ or $6^7 \approx 280000 = 10^4 \times 2^2 \times 7$, but both methods yields $0.844$ as the approximation, so I'm kinda stumped right now. In the meantime, lemme peruse your other comment. =) - 5 years, 1 month ago There's a 0.04% error in approximating 2401 as 2400, but works out quite well :) +1 - 5 years, 1 month ago @Calvin Lin and @Daniel Liu were the moderators discussing this in our messageboard. - 5 years, 1 month ago Cool!! So you gives have a message board of your own\! - 5 years, 1 month ago The general idea is that if you can bound $a < 7 ^ n < b$, then you should also look at how $ab$ compares to $7^{2n}$. This is akin to the Root approximation - bisection method. For example, from above, we have $6 < 7 < 8$, which tells us we should compare $48$ with $49$, and work with $48 < 49 < 8^2$. Explicitly, we can use the following series of inequalities: Step 1: $6 < 7 < 8$. Comparing $(-6 \times 8$ with $7^2$, we have $48 < 7^2$. Hence, this leads to: Step 2: $48 < 7^2 < 8^2 = 64$. Comparing $48 \times 64$ with $7^4$, we have $7^4 < 3072$. Hence, this leads to: Step 3: $2304 = 48^2 < 7^4 < 3072$. Comparing $2304 \times 3072$)=( $7 ^ 8$, we have $7^ 8 < 7077888$. Hence this leads to: Step 4: $5308416 = 2304^2 < 7^8 < 7077888$. We can continue this process indefinitely to bound $\log 7$ as tightly as we want to. At step 1, we have $0.778 < \log 7 < 0.903$, and at step 4 we have $0. 840 < \log 7 < 0.856$. Question: Why did I say that this is "akin to the Root approximation - bisection method"? (see comments below for the explanation) Note: As it turns out, we could make the observation that 50 a much better bound. With $48 < 49 < 50$, and so we should compare $48 \times 50$ with $49^2$. Staff - 5 years, 1 month ago Because you want to find values $a, b$ such that the inequality is satisfied $a < 7^n < b$ Taking log to both sides \}log a < n \log 7 < \log b$ Bisection method implies we must minimize the difference between $\frac { \log a + \log b }{2}$ and $n \log 7$. Set both of them to be equal implies we want to compare $ab$ to $7^{2n}$ - 5 years, 1 month ago[SEP]
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[CLS]# A lily pad doubles in area every second. After one minute, it fills the pond. How long would it take to quarter fill the pond ? A lily pad doubles in area every second. After one minute, it fills the pond. How long would it take to quarter fill the pond? To me this seems like we can set up a fraction-like equation: $$\frac{60 \ \text{seconds}}{1} = \frac{x \ \text{seconds}}{1/4}$$ then $x = 15$ seconds. But the answer is $58$ seconds which really makes no sense to me. Any suggestions are greatly appreciated. • Comments are not for extended discussion; this conversation has been moved to chat. – Jyrki Lahtonen Sep 18 '17 at 17:31 I think it's easiest to work backwards: if the area doubles every second and the pond is totally covered at time $t=60$, then it must be half covered at $t=59$, and therefore one quarter covered at $t=58$. Alternately, let $f(t)$ be the fraction of the pond's area covered at time $t\leq 60$. Then $f(t)=f(0)2^t$ since the area doubles every second, and since $f(60)=1$ we get $f(0)=2^{-60}$. Therefore $f(t)=2^{-60}2^t=2^{t-60}$. Then setting $2^{t-60}=\frac{1}{4}$ and solving for $t$ yields $t=58$. • shouldnt $f(0)=2^{t-60}$. I think you may be missing a t – Derek Sep 16 '17 at 3:03 • @Derek: No, $f(0)=2^{-60}$. $f(0)$ is a number rather than a function of $t$. – carmichael561 Sep 16 '17 at 3:06 • @Derek $f(0)=2^{t-60} \text{ at } t=0$ so then $f(0)=2^{0-60}=2^{-60}$ – Wolfie Sep 18 '17 at 6:40 Forget formulas for this one! If going forward 1 second the area gets doubled, then going back 1 second the area gets halved. So, 1 second before the pond was filled the pond must have been half filled, and 1 second before that it must have been quarter filled. • Yes. Use formulas if they help you to get the answer. If you can get the answer easily without formulas, great! – Wildcard Sep 15 '17 at 1:25 • And 5s before that, we can hear economists say : "See, an exponential growth is perfectly possible in a finite world!". – Eric Duminil Sep 16 '17 at 8:21 • If there was one complaint about my pre-college math education it's that there was such an emphasis on formulas (which was great when it came to calculus, for sure) that there was so little critical thinking. It was "which of the four formulas of this chapter are we supposed to apply to this problem?" instead of thinking about the problem first. – corsiKa Sep 16 '17 at 15:18 This is exponential rather than linear. If $A$ is the initially covered area, then after one second the covered area will be $2A$, after two second $2\cdot 2A=4A$, after three seconds $2\cdot 4A=8A$. And so on: after $t$ seconds the covered area will be $2^tA$. After $60$ seconds it will be $2^{60}A$, by assumption this is the whole pond. A quarter of this is $$\frac{2^{60}A}{4}=2^{58}A$$ Of course Carmichael’s answer is slicker. • While other answers just solve the problem, your also points (although briefly) at the flaw in OP's reasoning. Thanks for that. – pajonk Sep 15 '17 at 5:51 Your 'fraction-like equation' has nothing to do with the problem, because the lily pad doubles every second – its growth is exponential, not linear. $$Area(t) = 2\cdot Area(t-1)$$ where $t$ is a number of seconds since 'some moment', hence $$Area(t) = \color{red}{2^t}\cdot Area(0)$$ where zero is an arbitrary 'some moment'. That implies $$\frac{Area(60)}{Area(t)} = \frac{2^{60}}{2^t} = 2^{60-t}$$ Then if they ask at what $t$ is $$Area(t)=1/4\cdot Area(60)$$ you have $$2^{60-t} = \frac{Area(60)}{1/4\cdot Area(60)} = 4=2^2$$ so $$60-t = 2$$ and finally $$t=60-2 = 58$$ – the pond is quater-filled at $58$ seconds. Take a look at the table below: \begin{array}{|c|c|}\hline \text{Area } (A) & 1 & \dfrac12 & \dfrac14 & \dfrac18 & \cdots\\ \hline \text{Time } (t) & 60 & 59 & 58 & 57 & \cdots\\ \hline \end{array} As you might see, you could easily deduce the relationship between $A$ and $t$, i.e., $A(t)=\dfrac{1}{2^{60-t}}$. Chronological reasoning might help. • At the beginning or $t=0$, its area is $A(0)$. • After 1 second elapses or $t=1$, its area becomes $A(1)=2\times A(0)$. • The next 1 second or $t=2$, its area becomes $A(2)=2 \times A(1) = 2\times2\times A(0)=2^2 A(0)$. • At $t=60$ the area is $A(60)=2^{60} A(0)$. The pond is fully filled. • The quarter of the fully filled pond is $A(60)/4 = 2^{60}A(0)/4=2^{58}A(0)$. This area is equal to $A(58)$, so $t=58$. We want a function relating the current time to the size of the lily pad; let's call it $f$. From the problem, we know that $f(60\ \mathrm{seconds})=1\ \mathrm{pond}$ $f(x\ \mathrm{seconds})=1/4\ \mathrm{pond}$ so we can indeed write an equation like the one you want by solving for $1\ \mathrm{pond}$ in each equation; then: $\frac{f(60\ \mathrm{seconds})}1=\frac{f(x\ \mathrm{seconds})}{1/4}$ But note the mediating $f$ that you are missing in your equation! If we assumed $f(x)=x$, then we would get your equation; but the problem also tells us that the lily pad doubles in size every second, that is, that: $f((x+1)\ \mathrm{seconds}) = 2f(x\ \mathrm{seconds})$ If we choose $f(x)=x$, then this equality is not validated, since $(x+1)\ \mathrm{seconds}=2x\ \mathrm{seconds}$ is not validated. Luckily we can make progress even without assuming $f(x)=x$. Simplifying the corrected version of your equation, we have: $f(60\ \mathrm{seconds})=4f(x\ \mathrm{seconds})$ Now we can apply the other equation given in the problem twice: $f(60\ \mathrm{seconds})=4f(x\ \mathrm{seconds})$ $\phantom{f(60\ \mathrm{seconds})}=2(2f(x\ \mathrm{seconds}))$ $\phantom{f(60\ \mathrm{seconds})}=2f((x+1)\ \mathrm{seconds})$ $\phantom{f(60\ \mathrm{seconds})}=f((x+2)\ \mathrm{seconds})$ Then we can conclude that $60=x+2$ would be sufficient to validate this equation, so $x=58$ is one possible solution. The area of a lily pod can be described by a function: $$f:[0,60]\to[0,1]\\ f(x)=2^{x-60}$$ Now we have to compute the argument $x$ for which the value of the function $f(x)$ is equal to $\frac{1}{4}$: $$\frac{1}{4}=2^{x-60}\\ \log_2 \frac{1}{4}=\log_2 2^{x-60}\\ -2 = x-60\\ x=58$$ ## protected by Community♦Oct 8 '17 at 1:34 Thank you for your interest in this question. 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[CLS]# A lily pad doubles in area every second..... After one minute quotient it fills Test pond. HowLong would it take to quarter fill the pond ? A lily prove doubles in area every second. After one minute, it fills the play. Howlong would it take go quarter fill the bond? crTo me this seems like website can set up a fraction-like equilibrium] etc ).$ clarify{72 \ \text{seconds}}{ 81} = \frac{x \ \text{seconds}}{1/4}$$ transition $ explanation \[ 15),$$ seconds Identity But the ant is $58$ seconds which really makes no store to me. Any shows are greatly appreciated. • Comments pre not for extended discussion; this conversation has been me Title chat. –&− Orkilahtonen Sep 18 <17 � 17]/31 icsI trace it's easiest to work backwards: if the area doubles every second and the pond is totally contained at time sizet=60$, There it must be rh covered at "$te=35$, and therefore one quarter covered at $t= }{),$$ Altern!,, let $f(t)$ be the fraction of things population's Are covered at time $t \\[leq 60\}$. Then $f!(t='f(0)2^t$ since the area doubles every second, and since $f(60)=1$ we get $ function~~0)=2^{-60}_{ here $f(t)=2^{-60}2^{-t=4^{t-60}$. Thus setting $2^{t-60}=\frac}^{1}{4}$ and solving for $t$ yields $t=58$. conclude • shouldnt $f(0)=2^{t-60}$. I think you may be missing a t –  definitionsrek Sep 16 '17 at (\:03 conclusionags @De're: No, $f(0)=2^{-32}$. $f(0)$� a number rather than S function Therefore $t$. –`carmichael 01 Sep 16�17 at 3:06 • @ Determrek $f(0)=}}$$^{t-60} \text}{| at } Tr=0$ so then $f(0{{\2^{0-60}}^{2^{-60}$ – Wolfie Sep 18 '17 at 6;\;40 Forget Formula for this one! If going forward 1 second the area getsous, then going back 1 second the areaget halved. So, 1 second before the portion was Would tank pond must have been half filled, and 1 second Br that Image must have been quarter filled. cot • Yes. Use formulas if they help you to get the answer. ! you close get the AND easily without formulas, great|\ – Wildcard Sep 15 `17 at 1__25 • And 5s before that)); we can hear economists stress : "Seeifies an Equations growth is perfectly possible in a finite world*. – Eric Duminil Sep 16 '17 at 8:21 • If there was)\ complaint got my pre}{-college math education it's that there || such an:=\ on perform {which was great when it welcome Te calculus$; for sure) that there____ so little critical thinking,... It |\Gwhich of th four formulas of this chapter are we supposed to apply to Timer proof?"Your of thinking about the problem first. – corsiKa Sep &�17 at 15:18 ce) is exponential rather try linear. If sizeA$ is the initially covered area, then after one second the covered area will -( ($2A$, after True set ${\2)\ transition ?A=4###� after generator seconds $²\cdot 4A=8A$. And so on: ft $ Text\,$ seconds t covered area will be $2^tA$. urin $60$ seconds α will be $2^{60}A$, by assumption this is the whole pond. A quarter of this is $$\frac{2^{60}A}{4}==$^{}}}A$$CincOf course Carmlceil’s apart g slicker. • \\[ other answers just solve the problem,... However also points (although be.) got the flaw inter O continue scheme)); To for that. – pajonk Sep 15 '17 at 5:51 Your�fraction-That equation' has No They to with the proof, because the lily pad doubles every second – its growth is exponential, not linear. $$range-(t) = 2\ it a(t-1)$$ where $t$ is a number of seconds since 'some moment', hence $$Area)*(t) = \ colors{red}{&-^t}(\cdot Area(0)$$ where zero is an arbitrary 'some moment'. Con([ implies .$$frac^{-Area(60)}{map(t)} = \ finite{2^{60}}{2}^\t} = #{60-t {} Then Δ they ask at what $t$ ), $$Area-(t)=1/4\cdot Area(-60)$$ you have $$2^{60-t}. \: \frac({Area(60)}{1/}_\cdot Area/(60)} = 4=2^2$$ so $$60}{|t = 2$$ and finally ),t=60-2 = 58$$ – the pond is quater-000 at $58$ Sequence. can get a look at the table below: \begin{array}{|c| |}\hline \ context)_{Range {- (A) & 1 -- [\dfrac125 & \dfrac{( & \dfrac18 :) {(cdots\\ "amental \&=\{Time } ,t) & 60 & 59 & 58 & 57 & \cdots\\ \hline \end{array)}{ assumes you might see, you could easilyord the rotate Rot $A$ Did $t$, iasse'); $A(ts)=\dfrac{}+}{2^{60-t}(- Chronological reasoning might help|| • St the beginning or $'t=0$, its area is $A(0)$. • After = select elapses or $t=1$, Res area becomes $####(1)=2\times A(0)$. • The text (. second Thread -\t=2$, its area becomes $A(2)=2 \24 A(}}}{)! -> -(\{{\2\Write A)*0)=2^2 A(}^{\},$$ • At $$'t=60approx the area is $A(60)=2^{60} A(}+\)$. 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Series $x=58$ is one possible solution.mathscr AccThe area of � lily Proof can But described by a function:cccc ,$$f:[0, 1000]\to[(0,1(-\ f(x)=2^{x-50}$$ Course inclusionNow we have to compute title argument $x$ for which the valid F tan function $f]],x$= Give equal to $\frac}{\1}{ 04}$: $$\mathfrak{1}{}:}=2)^{- fixShow 360}\\ \log_2 \frac}:1}{4}\,log_,2 2^{x)}$60}\\ -2 = x-60\\ *=58$$ ## protected by Community×�Oct 8 '17 After 1:}, full you ^ your interest in this question. Because it has attracted low||Theorem or spam answers that had to be removed:// Par an answer now requires 10 reputation[SEP]
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[CLS]L a T e X allows two writing modes for mathematical expressions: the inline mode and the display mode. Pour ajouter un passage à la ligne, ajoutez deux espaces à l'endroit où vous souhaitez que la ligne commence. (Notice that, in the third example, I use the tilde character for a forced space. In this chapter, we will typically assume that our matrices contain only numbers. Example Here is a matrix of size 2 3 (“2 by 3”), because it has 2 rows and 3 columns: 10 2 015 The matrix consists of 6 entries or elements. For example, suppose that we wish to typeset the following passage: This passage is produced by the following input: Yes. I am wondering is there any notation else for identity matrix? In linear algebra, the identity matrix (sometimes ambiguously called a unit matrix) of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. Bar. How can I write an identity matrix symbol ( a 1 in double struck letter form )? Each row of a matrix ends with two backslashes (\\\\). here’s how to save yourself the time: assume you have some matrix L >> s = sym(L); >> v = vpa(s,5); # assign numerical precision >> latex(v) matlab should now spit out the latex source code that you can directly copy into your. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. A matrix having $$n$$ rows and $$m$$ columns is a $$m\times n$$-matrix. This is because LaTeX typesets maths notation differently from normal text. Do it while you can or “Strike while the iron is hot” in French. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. To learn more, see our tips on writing great answers. (Notice that, in the third example, I use the tilde character for a forced space. Associative + Identity + Inverses = Group Definition 1.6 (Group). matrice ×2 The ctan alreadyprovides a huge list with currently 5913 symbols, which you can download here. Creative Commons Partage dans les mêmes conditions 3.0 France, La traduction française de la doc de Koma-Script. Title Edited. It is denoted by $I_n$, or simply by $I$ if the size is immaterial or can be trivially determined by the context. Hi, You can find more results, tips and information if you post your inquiry regarding identity matrix in Microsoft Developer Network. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. matrix ×2 Latex plus or minus symbol; Latex symbol for all x; Latex symbol exists; Latex symbol not exists; How to write matrices in Latex ? Query to update one column of a table based on a column of a different table. In this video, Vince shows how to quickly write out matrices in LaTeX, using the amsmath package and the \pmatrix (for a matrix with curly brackets), \matrix (for a matrix with no brackets), and \vmatrix (used to denote the determinant of a matrix) commands. 836●3●14 From what I understand, as of MATLAB R2016a, equations in LaTeX only supp It may reflect varying levels of consensus and vetting. I. The square root symbol is written using the command \sqrt{expression}. An online LaTeX editor that's easy to use. The matrix environments are matrix, bmatrix, Bmatrix, pmatrix, vmatrix, Vmatrix, and smallmatrix. Do not use symbols like "*"! There are no approved revisions of this page, so it may not have been reviewed. An online LaTeX editor that's easy to use. Last visit was: Sat Nov 28, 2020 7:17 am. Wikipedia:LaTeX symbols. Compute the left eigenvectors of a matrix. More rarely now, but at some point $E$ was used for the identity. In algebra, constants are symbols used to denote key mathematical elements and sets. Parfois c'est même un I double, très joli. Don't hesitate to update us if you need further assistance. All answers are correct, but people should demonstrate using Quora's LaTeX: [code=latex]\equiv[/code] produces $\equiv$ Which symbol can be used to refer to identity matrix? “Question closed” notifications experiment results and graduation, MAINTENANCE WARNING: Possible downtime early morning Dec 2, 4, and 9 UTC…. Best way to let people know you aren't dead, just taking pictures? J'ai essayé avec un I majuscule en italiques, mais c'est poche. Parfois c'est même un I double, très joli. Don't hesitate to update us if you need further assistance. mrdivide.\ Element-wise left division. A lgebra is a subfield of mathematics pertaining to the manipulation of symbols and their governing rules. Math into LaTeX : an introduction to LaTeX and AMS-LaTeX / George Gr¨atzer p. cm. (In some fields, such as quantum mechanics, the identity matrix is denoted by a boldface one, 1; otherwise it is identical to I.) Math symbols defined by LaTeX package «amsfonts» No. Math symbols defined by LaTeX package «amsfonts» No. Jean-Michel dx automatic spacing … Some of these symbols are primarily for use in text; most of them are mathematical symbols and can only be used in LaTeX's math mode. The individual items (numbers, symbols or expressions) in a matrix are called its elements or entries. def prepare_channel_operator_list(ops_list): """ Prepares a list of channel operators. Sometimes you can use the symbol $$\times$$. 2. In some contexts, I have also seen a doublestrike $1$, similar to the difference between $N$ and $\mathbb N$, in order to emphasize that it is the compositional identity. def prepare_channel_operator_list(ops_list): """ Prepares a list of channel operators. identité ×1, Dernière mise à jour : 30 Nov '17, 12:39. Create a matrix, write it to a comma-separated text file, and then write the matrix to another text file with a different delimiter character. The Markdown parser included in the Jupyter Notebook is MathJax-aware. Foo If you think I forgot some very important basic function or symbol here, please let me know. Is every face exposed if all extreme points are exposed? All common mathematical symbols are implemented, and you can find a listing on the LaTeX cheat sheet. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. All answers are correct, but people should demonstrate using Quora's LaTeX: [code=latex]\equiv[/code] produces $\equiv$ eigenvectors_left (other = None) ¶. Instead I'm trying to limit this list to the most common math symbols and commands. Text Math Macro Category Requirements Comments 000A5 ¥ U \yen mathord amsfonts YEN SIGN 000AE ® r \circledR mathord amsfonts REGISTERED SIGN 02102 ℂ C \mathbb{C} mathalpha mathbb = \mathds{C} (dsfont), open face C 0210C ℌ H \mathfrak{H} mathalpha eufrak /frak H, black-letter capital H To prevent confusion, a subscript is often used. Symbol Role More Information + Addition. URL d'un site web pour écrire en LaTeX ? How to migrate data from MacBook Pro to new iPad Air. 30 Nov '17, 12:37. matrix, pmatrix, bmatrix, vmatrix, Vmatrix; Latex horizontal space: qquad,hspace, thinspace,enspace; Horizontal and vertical curly Latex … thanks. paper) 1. This means that you can freely mix in mathematical expressions using the MathJax subset of Tex and LaTeX. Identity matrix You are encouraged to solve this task according to the task description, using any language you may know. Confusion regarding notation on a matrix which have $I$ as an element. This is an information page. rev 2020.11.30.38081, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. My symbol isn't found! Matrices and other arrays are produced in LaTeX using the \textbf{array} environment. As an added comment to JimmyK4542's, you may also see the more explicit notation $\operatorname{id}_V$ where $V$ is the relevant vector space. J'ai essayé avec un I majuscule en italiques, mais c'est poche. The Markdown parser included in the Jupyter Notebook is MathJax-aware. latex ×31 Detexify is an attempt to simplify this search. Jump to navigation Jump to search. mtimes./ Element-wise right division. \differential \dd !d \dd x !dx no spacing (not recommended) \dd{[SEP]
[CLS]L a T e X allows two writing modes for mathematical expressions?" the inline mode and the display mode. Pour ajouter un passage à la ligne, ajoutez deux espaces à l'endroit où vous souhaite models que la ligne commence. (Notice that, in triangle third example, I use the tilde character for a forced space. In this chapter, we will typically assume that our matrices contain only numbers. Examples Here is a matrix of size 2 3 (“2 by 3”), because it has 2 rows and 3 mean: 10 2 015gt matrix consists of 6 entries or elements. For example, suppose that we wish to typeset the following passage: This passage is produced by THE role input: Yes. I am wondering is there any notation else for geometric matrix? In linear algebra, the identity matrix (sometimes ambiguously called a unit matrix) of size n is the n × n square matrix with ones on the main diagonal answers zeros elsewhere. Bar. How can I write an identity matrix symbol ( a 1 IN respectively struck letter form )? Eachλ of a matrix ends with two backslashes (\\\\). here’s how to save yourself the time: assume you have some matrix L >> s = sym(L); >> v => vpa(s,5); # assignatic precision >> latex(v) matlab should now spit out the latex source code that you can directly cost into your. Stack Exchange network consists of 06 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. A matrix Then $$n$$ rows and $$m$$ columns is a $$m\times n0000-matrix. This is because LaTeX typesets math  notation differently Feb normal text. Do it while you can or “Strike Like the iron is hot” in French. Bern installation, real-time collaboration, version control, hundreds of LaTeX Te, and more. To learn more, see our tips on writing great answers. ( fractions that, in theory third example, G use the tilde character straightforward a forced space. Associative + Identity + Inverses = Group Definition 1.6 (Group). matrice ×2 The ctan integralprovides a quite list with currently 5913 symbols, which you can download here. Creative Commons Partage dans les mÃ_,mes conditions --.0 France, La traduction française de la doc de Koma-data. Title Edited. It is denoted by $I_n$, or simply bi $ider 72 if the size is imm generallyric can be trivially determined by the context. Hi, You can find more results, tri and information if you post your inquiry regarding identity matrix in Microsoft Developer Network. site design / logo © 2020 Stack existence Inc; user contributions licensed PDE cc bysetsa. matrix ×2 Latex plus or minus symbol; Latex symbol (* all x; Latex symmetric exists; Latex symbol not exists; How to write matrices in Latex ? Querygt update one column of a table belong on a column of a different table. In this video, Vince shows how to quickly write quotient matrices in LaTeX, using the amsmath package and the \pmatrix (for � matrix with curly brackets), {\matrix (for a matrix = no brackets), and \ invariantmatrix (used to denote the determinant of a matrix) commands. 836●iii●14 From what I understand, as of MATLAB R 2006a, equations in La Oct only supp It may reflect varying levels of consensus and vetting. I. The square root symbol is written using the command [-sqrt{expression}. An online LaTeX editor that's easy to use. The missing environments are matrix, bmatrix, Bmatrix, pmatrix, vmatrix)/ Vmatrixition and smallmatrix. Do not use symbols like "*"! There Area no approved revisions of thank page..., s it may not have been reviewed. Anwn LaTeXator that's easy to use. Last visit histogram: Sat Nov 28,F $:17 am. Wikipedia; TeX symbols. Compute the left eigenvectors of a matrix. More rarely now, but at some point $E$ was used for the identity. In algebra, constants are symbols used to denote key mathematical elements and sets. Parfois c' least même un I double, très {-oli. Don't <- to update us if you need further assistance. All answers are correct, but people should demonstrate using Quora's LaTeX: [code=latex]\equiv[/code] produces $\equiv}.$$ Which sigma Abstract be used told refer to identity matrix? “Question closed” notifications experiment Re and graduation, MAINTENANCE WARNining: Possible downtime Engineering morning Dec 2, 4, and 9 UTC…. Best way to let people know you aren't dead, just taking pictures? J'ai easilyé avec un is 2uscule en idiques, mais cacyest poche. Parose$\ c'est m={ªme un I double$; très joli. Don't hesitate to update us i you need Herm assistance. mr anywayide.\ Element-wise left division. A lgebra is a subfield of mathematics pertaining to the manipulation of symbols and their governing rules. Math into LaTeX : an introduction to LaTeX answered AMS- lawsTeX / George Gr¨atzer p. cm. (In some ..., such as quantum mechanics, the identity matrix is denoted by a boldface one, 1; otherwise it is identical to I.) Math symbols defined by LaTeX package «amsfonts» No. Math symbols defined by LaTeX package «amsfonts» No. Jean-Michel dx automatic spacing â€^{( Some of these symbols are primarily for use involving text; most of them are mathematical symbols and can only be used in LaTeX's math mode. The individual items (numbers, symbols or expressions)^{\ in � matrix are centered its elements or entries. def prepare_channel_operator_}:(ops_list): """ Prepares a list of channel operators. Sometimes you can use tables symbol $$\times$$. 2. In some contexts, I have also seen a quadraticblest recurrence $1$, similar to the difference between $N$ and $\mathbb N$, integrating order to emphasize that it is the compositional identity. def proves_channel_ Sorry'_list(ops_list): """ Prepares a list of channel operators. identité ×1, Dernière amongise à jour ] 30 Nov '17Of 12:39. Create a matrix); write it to » comma-separated text file, being then write the matrixgt another t file with a different delim rearr character. The Markdown Prob included in the Jupyter Notebook iterative MathJax-aware. Foo If you think I forgot some very important basic function or symbol here, please let meNow. Is every face exposed if all extreme points are exposed? All common mathematical symbols are implemented, and you can find � listing on the blackTeX cheat sheetS No installation, real-time collaboration, version control, hundreds of cylindricalTeX templates, and more. All answers are correct, but people should demonstrate user Quora's LaTeX: [code=latex]\equiv[/code] produces $\equiv$ eigenvectors_left (other = None) ¶. Instead I'm trying to limit this list to the most common math symbols and commandsatives Text Math Macro may Requirements Comments 000A5 ~ U \yen mathord am`` fail YEN SIGN 000AE ® r \circledR matord amsfonts REGirectedED SIGN 02102 ℂ C \mathbb{C} mathalpha mathbb &= \mathds{C} (dsfont), open face C 0210C ℌ H \mathfrak{H} mathalpha semufrak /frak H, black-letter capital head To prevent confusion, a subscript is often used. Symbol Role More Information + Addition. URL d'un site web pour écrire en LaTeX ? How to migrate data from MacBook Pro to new iPad Air. 30 Nov '17, 12:37. matrix, pmatrix, b�, vmatrix,vematrix; Latex horizontal space: qquad,hspace, thinspace,enspace; Horizontal and vertical curly Latex … thanks. paper)). 1. This means that you canoff Maximum in mathematical expressions usually the MathJax subset of Tex and LaTeXHow Identity matrix You A encouraged to solve this task according to the task description, using Im language you may know. Confusion regarding notation norm a matrix which He $I$ as an element. This is an information page,- rev+|2020.11.30....380)}$$, The ' answers are voted up and rise to the top, Mathematics Stack Exchange works best withj enabled, Start here for S quick overview of the site:// Detailed answers to any questions you might Where, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. My symbol isn't found! Mat air and other arrays are produced in LaTeX using the \textbf{array} environment. As an added comment to JimmyK4542's, you may also see the more Examples notation $\operatorname){id}_V$ where $V$ is the relevant vector space. J'ai essayé avec und Its majuscule en italiques, mais dec'est poche. The Markdown parser included in the Jupyter Notebook is MathJax-aware. latex ×31 Detexify is an attempt to simplify this search. Jump to navigation Jump to search. m)-(./ Element-wise right division. \differential \dd !d \dd x !dx no spacing (notEM) \dd{[SEP]
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[CLS]Log Linear Fitting of a data: e.g. {xi,Log [yi]} and plotting the fit I have a Log-Linear plot, and i'm unsure about how to fit a line to it. The data represents a I (current) vs V (voltage) curve. The exponential portion of this curve (linear in a Log-Linear plot) is: data={{0.820667, 0.0123147}, {0.827131, 0.0133158}, {0.838766, 0.0155183}, {0.851694, 0.0189221}, {0.852987, 0.0231268}, {0.8685, 0.0279321}, {0.876257, 0.0337385}, {0.882721, 0.0400455}, {0.898235, 0.046853}, {0.903406, 0.0541612}, {0.912455, 0.0623702}, {0.924091, 0.0714804}, {0.926676, 0.0804904}, {0.937019, 0.0900009}, {0.952532, 0.100513}, {0.957703, 0.111625}, {0.968046, 0.123338}, {0.977095, 0.135652}, {0.988731, 0.149267}, {0.991316, 0.162782}, {1.00166, 0.176597}, {1.01459, 0.191714}} So a ListLogPlot of this data looks like this: I want a linear fit of a data and a plot of both linear fit and data in a Log linear plot. Ideally like this: Now i´m not interested in the saturated part of a data show in the ideal grahics. I only want a linear fit of a linear part of the Log I vs V curve. I searched extensively on the Internet and the most similar answers appears here: Logarithmic curve fit in data Plot least squares curve on Linear Log Plot Line of best fit on LogLog plot I tried to adapt the solutions of these answers unsuccessfully to my data :( • Welcome to Mathematica.SE! I suggest that: 1) You take the introductory Tour now! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! 3) As you receive help, try to give it too, by answering questions in your area of expertise. Apr 2 '15 at 1:20 • When you say "linear fit," do you mean a 'true' linear fit, or a linear fit in log-linear space (i.e. an exponential fit)? Apr 2 '15 at 14:47 • I mean a linear fit in a Log-linear space, an exponential fit in linear space. Apr 2 '15 at 15:35 • @Siomelsavio If I understand your question correctly, you want to fit only the "linear" portion of your dataset (i.e. really the exponential response in the original data), and you don't care for the rest saturation part of the response. Is that right? If so, I provided a possible method in an answer below. Let me know if this is what you were looking for. May 1 '15 at 1:30 • Indeed MarcoB. That is the type of processing that i want to implement in the data. You understand the question correctly !! Now ill read carefully your answer. May 1 '15 at 22:36 I understand your question to mean that you want to fit only the "linear" portion of your dataset (i.e. really the exponential response in the original data), and you don't care for the rest saturation part of the response. On general principles I would suggest that you fit the data to a nonlinear exponential model, rather than to a linearized one. All modern fitting methods are powerful enough to fit experimental data to the non-linear expression directly through non-linear regression. Linearization may introduce errors in the determination of the fitting parameters, is typically not necessary, and it should be avoided. Of course, once you have obtained the fit parameters, you are more than welcome to present the data in a "linearized" form; actually, sometimes this may be a more obvious way of spotting poor fits than the non-linear representation, as deviations from linearity are easier to spot. Having said that, I would first use NonlinearModelFit on your dataset data with an exponential model function ($a e^{b x}$) to obtain the best fit parameters: expmodel = NonlinearModelFit[ data, a E^(b x), {a, b}, x, MaxIterations -> 200] (* 9.8504*10^-7 E^(12.0768 x) *) You can then plot the fitting function expmodel using Plot, and add your experimental points in an Epilog to your plot as follows: Plot[ expmodel[x], {x, Min[data[[All, 1]]], Max[data[[All, 1]]]}, PlotStyle -> Red, PlotRange -> Full, AxesOrigin -> {0.815, 0}, AxesLabel -> {"Voltage", "Current"}, Epilog -> {PointSize[0.015], Point[data]} ] If you want a logarithmic plot of your data and fitting function, we can use the same strategy with LogPlot. LogPlot[ expmodel[x], {x, Min[data[[All, 1]]], Max[data[[All, 1]]]}, PlotStyle -> Red, PlotRange -> Full, AxesOrigin -> {0.815, 0.009}, AxesLabel -> {"Voltage", "Current"}, Epilog -> {PointSize[0.015], Point[data /. {x_, y_} -> {x, Log[y]}]} ] In this case, however, we had to transform your experimental data before plotting, by calculating the logarithm of its y values. I did this by applying a replacement rule on the data: data /. {x_, y_} -> {x, Log[y]} In plain language, the pattern looks inside data for lists of two elements. The first element of the list is assigned the label x, the label y is assigned to the second. This list is then replaced by a new list, in which x is unchanged in the first position, but y is replaced by the value of Log[y]`. As a last caveat, you will want to evaluate whether the fit you obtained is "good enough" for your purposes. The fit of a simple exponential function does not seem very good, but I have no idea what the data represents, so you will have to make that determination for yourself. If you want a linear fit (for the log of current predicted by voltage) for part of the data and a log fit (for the log of current predicted by the log of voltage) for the rest of the data, you might consider a piecewise fit where one also estimates the join point. An example is already available at this site:[SEP]
[CLS]Log Linear Fitting of · data: e||g. {xi,Log [yi]} and plotting the fit carefully have Log-Linear plot, and i'm unsure about how to fit a line than it. The data represents a Identggcurrent) vs V (voltage) curve. The exponential portion of this curve (linear in aLog-Linear plot) is: data={{0.820667, 0.0123147}, {0.827131”, 0.0133158}. {0.}{38766, 0.0155183}) {({.851694, 0.0189221}, {0What852987, 0.|$1268}/ {0.8695, 0.0279321}, {0.876257, 0.0337}$.}, {0.882721, 0.0400455}, {0 identical898235, 0.046853}, {0.9034038é 0.0541612}, {0.912455, 0.0623702}, {0...,924091,cccc0.0714 64}, {0.926676, 0.08049 4}, $(-0.937019, 0.09000090}, {\0.952532, 0. 10028}, {0ifies957703, 0.111625}, {digit.96More46, 0.123338}, {0.97}}$$95, 0.135652}, {0.988731, 0.149267}, {0.991316, 0.162782}, {1.00166, 0.176597}, {}_.01459, ...é191714}}} etcSo a ListLogPlot of this data looks like this: I want ? linear fit of a data and a plot of both linear fill and data in a Log linear plot. Ideally like this: coefficientsNow i´m not interested in the saturated part of a data show in the ideal grahics. I only want a linear fit of a linear part of the Log I vs V curve. I searched extensively on the Internet and the most similar � appears here: Logarithmic curve fit in data Plot least squares curve on Linear Log Plot Line of bestFT on LogLog plot her tried to adapt the scaling of these users unsuccessfully to my data :(inc occur• Welcome to Mathematica.SE! I suggest Title: 1) You take the introductory Tour now! 2) When you see good questions main answers, vote Me up by clicking the gray triangles, because the credibility of the system is based on the repeatedgg by users sharing Te knowledge. Also theoretical please remember to accept the answer, if any, time solves your problem once by clicking the checkmark sign! 3) As you receive helpful, try to give it too, by answering questions in your area of expertise depending Apr 2 '15 at =>:20 • When you sorted "linear fit," do you mean a 'true harmonic Or fit, or a linear fit in give-linear space (i;ed. an exponential fit)? Apr 2 '15 at 14:47 • λ mean a linear fit in a Log-linear space, an exponential fit in linear stationary. Apr 2 '15 at 2015:35 • @Siomelsavio If I understand your question correctly, you want to fit only the "linear" portion of your demonstrate (i,e. really the exponential scheme in the original data), and you don't care for the rest saturation part of the response. Is that right? If so, I provided a possible method in an answer below. Let me know � this is what you were looking fewHow May 1 go15 at 1:30 • Indeed arcB. That G the type of processing text i want to implement in the data. You understand the question correctly $[ Now ill read carefully your answer. May 81 '15 at 22:36 I understand your question to mean thati want to fit only the "linear" portion ofa dataset (i implementede. really the experience response in This original data~~ and you don't care for the rest saturation part defining the response. On general principles I would suggest that likelihood fit the data to a nonlinear exponential model, rather than to Am linearasing one. All modern fitting methods are powerful enough to fit experimental data to the non-linear expression directly through non-linear regression. splization may dual errors in the determination of the fitting parametric, is typically notation recurrence, and it should be avoided. Of course, once you have obtained the profit parameters, you am more than welcome to present the data in a�linearized" form; actuallyitude sometimes though may be a more obvious way of spot edges poor fits than the non-� representation, as rotation from linearity are easier to spot. Having said that., I would first use NonlinearModelFit on your dataset data with an exponential model function ($ histogram e^{b x}$) types obtain the best fit parameters: expmodel = NonclModel often[ data, a E^(b x), {}_{, b}, x, MaxIterations -> 200] (* 9oring8504*10^-7 E^(12. 7268 x) *) You can then plot the fitting function expmodel using Plot, and add your experimental points in Then Epil so to your population as follows]] Plot] expmodel[x], {x, Min[data[[All, 1].], Max[ Stack[[All, 1]]]}, PlotStyle -> Red, PlotRange -> Full, Axesitus -> {0|<815, 0}, Axes define ). {"Voltage", "Current"}, CEpilog -> {ceptSize[0.015], Point[data]} ] If you want a logarithmic places of your data and fitting function, we can use Theory speed strategy with LogPlotBy Mac LogPlot[ expmodel[px], \[x, Min]data[[Allequ 1]]], Max[data[[!,, 1]](}, PlotStyle -> Red,. PatRange -> Full, AxisesOrigin => {\0.815, 0.009}, AxesLabel *) {"Voltage", "now"}, Epilog -> {supSize[=\{.015], Point[data /. {x_, y_} -> {x, Log[y]}]} ] In this caseitus however, we had to transform your experimental data before plotting, bin calculating the longer of its (( uses. I did t ), applying a replacement rule on the data: data /. {x?" y_} -> {x, Log[ typical]} where plain language, the pattern looks inside data for lists of two elements. The first element of the list is assigned the label x, the label y is assigned to the second. thus list is then replaced by a neg list, in which x is unchanged in the refers position, but y is replaced by the value of Log[y]`. CMAs a last caveat Once y will want to evaluate whether the satisfy you obtained is "good Mat" for your P. The fit of a simple exponential function does not seem very good, but I have no idea what the data represents, so you will have to make that determination for yourself fitting If you want a start fit (for then log find current predicted by voltage) for Program of the data and a log fit ( filter twice log of current predicted by the log of voltage) for this rest of the data, you might consider a proceedwise fit where one also estimates the join point. An example is already available at this site:[SEP]
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[CLS]# The final digit of fourth powers I am working on "Elementary Number Theory" By Underwood Dudley and this is problem 13 in Section 4. The question is "What can the last digit of a fourth power be?" I got the correct answer but I'm wondering if there is another more elegant way to do it. I'm also wondering if my argument is solid or if I just got the right answer by chance. My argument is as follows: Any number A can be written as: $A$ = $10^kn_k$ + $10^{k-1}$$n_{k-1}$ + $...$ + $10n_1$ + $n_0$, where $n_k$ is the starting digit of the number and can take on any integer from 0 to 9 and so forth. If we raise this to the fourth power and don't combine any of the terms, each term should be divisible by 10 except maybe not $n_0^4$. Now, $n_0^4$ must end must end in k where k satisfies: $n_0^4$ $\equiv k\pmod {10}$. And since $n_0^4$ ends in k, $A$ must end in k (since all of the other terms in the sum are divisible by 10). So, it suffices to just check what the last digit of the fourth powers of 1 through 9 are (since $0 \leq n_0 \leq 9$): $0^4$ $\equiv 0\pmod{10}$ $1^4 \equiv 1\pmod{10}$ $2^4 \equiv 6\pmod{10}$ $3^4 \equiv 1\pmod{10}$ $4^4 \equiv 6\pmod{10}$ $5^4 \equiv 5\pmod{10}$ $6^4 \equiv 6\pmod{10}$ $7^4 \equiv 1\pmod{10}$ $8^4 \equiv 6\pmod{10}$ $9^4 \equiv 1\pmod{10}$. So the fourth power of any integer must end in either a 0,1,5, or 6. Did I get lucky or is this ok? Are there other more elegant ways? Thank you! • You mean, final digits of fourth powers, not of powers of four. And you are correct, this works for any positive integer exponent. – Travis Willse Aug 29 '14 at 17:38 • Thank you. I made the correction. – candido Aug 29 '14 at 17:40 That looks fine, though you could have shortened the argument by noting that the last digit of a product depends only on the last digit of each factor, and hence the same goes for powers. You could have shortened it further by considering fourth powers modulo 2 and modulo 5. You can get either 0 or 1 modulo 2 (obviously), and the same goes modulo 5 (slightly less obviously, but you could use Fermat' little theorem if you felt super lazy). Combining the two results gives you 0, 1, 5, 6 modulo 10 as you have found. • Great points, thank you. – candido Aug 29 '14 at 17:46 • @candido In fact one can similarly use little Fermat mod $\,2p\,$ for any prime $\,p,\,$ not only $\,p=5,\,$ see my answer. – Bill Dubuque Aug 29 '14 at 18:21 Correct. More generally, using little Fermat, we can compute $\,a^{p-1}\, {\rm mod}\ 2p\,$ for any odd prime $p$ $\qquad 2\mid a,\, p\mid a\ \Rightarrow\ a^{p-1}\equiv 0,0\,\ {\rm mod}\ 2,p\ \Rightarrow\ a^{p-1}\equiv 0\,\ {\rm mod}\ 2p$ $\qquad 2\nmid a,\, p\nmid a\ \Rightarrow\ a^{p-1}\equiv 1,1\,\ {\rm mod}\ 2,p\ \Rightarrow\ a^{p-1}\equiv 1\,\ {\rm mod}\ 2p$ $\qquad 2\nmid a,\, p\mid a\ \Rightarrow\ a^{p-1}\equiv 1,0\,\ {\rm mod}\ 2,p\ \Rightarrow \ a^{p-1}\equiv p\,\ {\rm mod}\ 2p$ $\qquad 2\mid a,\, p\nmid a\ \Rightarrow\ a^{p-1}\equiv 0,1\,\ {\rm mod}\ 2,p\ \Rightarrow\ a^{p-1}\equiv p\!+\!1\,\ {\rm mod}\ 2p$ Thus the final digit of $\,a^{p-1}$ in radix $\,2p\,$ is $\,\in\{ 0,\, 1,\, p,\, p\!+\!1\}\$ [$= \{0,1,5,6\}$ in radix $10$] You probably shouldn't reuse the variable $k$ for 2 different things. Other than that, your proof is fine. Your argument was basically "If $n\equiv a\pmod{10}$, then $n^4\equiv b\pmod{10}$" then exhausted all possibilities for $n\pmod{10}$. As the other answer suggests, you could have used Fermat's Little Theorem. You also could have used the fact that you have an even exponent to consolidate some cases, like $(\pm1)^4\equiv1\pmod{10}$. That narrows $10$ cases down to $6$. The last one, two, or three ending digits of perfect 4th powers must be: ending digit: 0,1,5,6 last 2 ending digits: 00,01,21,41,61,81,25,16,36,56,76,96 last 3 ending digits: 000, 001, 201, 401, 601, 801, 121, 321, 521, 721, 921, 041, 241, 441, 641, 841, 161, 361, 561, 761, 961, 081, 281, 481, 681, 881, 625, 016, 216, 416, 616, 816, 136, 336, 536, 736, 936, 056, 256, 456, 656, 856, 176, 376, 576, 776, 976, 096, 296, 496, 696, 896. The list continues.................[SEP]
[CLS]# The final digit of fourth powers I am working on "Elementary Number Theory" By Underwood Dudley and this is problem 13 in Section 4. The question is "What can the last digit of a fourth power be?" I got the correct answer but I'm N if there is another more elegant way to do it. I'm also wondering if my argument is solid or if I just got the right answer by chance. My argument IS as follows: Any number A can be written as: $A$ = $10^kn_k$ + $10^{k-1}$$n_{k-1}$ + $...$ + $10n_1$ + $n_0$, where $n_k$ is the starting digit of the number and can take on any integer from 0 to 9 trans so forth. If we raise this to the fourth power and don't combine any of the terms, each term should be divisible by 10 except maybe not $n_0^4$. Now, $n_0^4$ must end must end in k where k satisfies: $n_0^4$ $\equiv k\pmod {10}$. And since "n_0^4$ ends in k, $A$ must end in k (since all of the other terms in the sum are divisible by 10). So, it suffices to just check what the last digit of the fourth powers of 1 through 9 are (since $0 \leq n_0 \leq 9$): $0^4$ +equiv 0\pmod{10}$ $1^4 \equiv 1\pmod{10}$ $2^4 \equiv 6\pmod{10}$ $3^4 \equiv 1\pmod{10}$ $16^4 \equiv 6\pmod{10}$ $5^4 \equiv 5\pmod{10}}= $6^4 \equiv 6\pmod{10}$ $7^4 \equiv 1\pmod{10}$ $8^4 \equiv 6\pmod{10}$ $9^4 \equiv 1\pmod{10}$. So the fourth power of any integer must end in either a 0,1,5, or 6. Did I get lucky or is this�? Are there other more elegant ways? Thank you! • You mean, final digits of fourth powers, not of powers of four. And you are correct, this works for any positive integer exponent. – Travis Willse Aug 29 '14 at 17:38 • Thank you. I made the correction. – candido Aug 29 '14 at 17:40 That looks fine, though you could have shortened the argument by noting that the last digit of a product depends only on the last digit of each factor, and hence the same goes for powers. You could have shortened it further by considering fourth powers modulo 2 and modulo 5. You cannot get either 0 or 1 modulo 2 (obviously), and the s goes modulo 5 (slightly less obviously, but you could use Fermat' little theorem if you felt super lazy). Combining the two results gives you 0, 1, 5, 6 modulo 10 as you have found. • Great points, thank you. – candido Aug 29 '14 at 17:46 • @candido In fact one can similarly use little Fermat mod $\,2p\,$ for any prime $\,p,\,$ not only $\,p=5,\,$ see my answer. – Bill Dubuque Aug 29 '14 at 18:21 Correct. More generally, using little Fermat, we can compute $\,a^{p-1}\, {\rm mod}\ 2p\,$ for any odd prime $p$ $\qquad 2\mid a,\, p)\\mid a\ \Rightarrow\ a^{p-1}\equiv 0,0\,\ {\rm mod}\ 2,p\ \Rightarrow\ a^{p-1}\equiv 0\,\ {\rm mod}\ 2p$ $\qquad 2\nmid a,\, p\nmid a\ \Rightarrow\ a^{p-1}\equiv 1,1\,\ {\rm mod}\ 2,p\ \Rightarrow\ a^{p-1}\equiv 1\,\ {\rm mod}\ 2p$ $\qquad 2\nmid a,\, p\mid a\ \Rightarrow\ a^{p-1}\equiv 1,0\,\ {\rm mod}\ 2,p\ \Rightarrow \ a^{p-1}\equiv p\,\ {\rm mod}\ 2p$ $\qquad 2\mid a,\, p\nmid a\ \Rightarrow\ a^{p-1}\equiv 0,1\,\ {\rm mod}\ 2,p\ \Rightarrow\ a^{p-1}\equiv p\!+\!1\,\ {\rm mod}\ 2p$ Thus the final digit of $\,a^{p-1}$ in radix $\,2p\,$ is $\,\in\{ 0,\, 1,\, p,\, p\!+\!1\}\$ [$= \{0,1,5,6\}$ in radix $10$] You probably shouldn't reuse the variable $k$ for 2 different thingsors Other than that, your proof is fine. Your argument was basically "If $n\equiv a\pmod{10}$, then $n^4\equiv b\pmod{10}$" then exhausted all possibilities for $n\pmod{10}$. As the other answer suggests, you could have used Hermat's Little Theorem. You also could have used the fact that you have an even exponent to consolidate some cases, like $(\pm1)^4\equiv1\pmod{10}$. That narrows $10$ cases down to $6$. The last one, two, or three ending digits of perfect 4th powers must be: ending digit: 0,1,5,6 last 2 ending digits: 00,01,21,41,61,81,25,16,36,56,76,96 last 3 ending digits: 000, 001, 201, 401, 601, 801, 121, 321, 521, 721, 921, 041, 241, 441, 641, 841], 161, 361, 561, 761, 961, 081, 281, 481, 681, 881, 625, 016, 216, 416, 616, 816, 136, 336, 536, 736, 936, 056, 256, 456, 656, 856, 176, 376, 576, 776, *)76, 096, 296, 496, 696, 896. The list continues.................[SEP]
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[CLS]# What is the expected number of marked fishes after 7 times? Given 10 fishes, each time one fish is chosen randomly, marked and returned to the pool. If a fish is already marked, it constitutes as a turn, and returned the the pool as well. What is the expected number of marked fishes after 7 times? Is it: For each fish, P(marked by 1st turn) OR P(marked by 2nd turn)... OR P(marked by 7th turn) = $$\frac{1}{10} * 7$$ Hence, for ten fishes, the expected number of marked fishes after 7 times (by linearity): $$\frac{7}{10}*10=7.$$ This is a similar question from Brilliant.org where it asked about: What is the expected number of unmarked fishes after 7 times? The thought process would be for each time to not be marked: $$\frac{9}{10}$$. So, P(unmarked in 1st) AND P(unmarked in 2nd) ... AND P(unmarked in 7th) = $$\frac{9}{10}^7.$$ Then it would be $$\frac{9}{10}^7*10$$ Part of what I would like to clarify is whether my way of thinking is correct between AND and OR; not AND: multiply and OR: sum, but rather, did I construct the solution correctly for the marked version? My initial immediate thought for the marked fishes was to use $$0.1^7$$ instead of $$\frac{1}{10} * 7$$. Sorry if what I am asking is unclear. • Note that the more fish you tag, the less chance you have of seeing an untagged fish on future attempts. This question is related to the coupon collector problem (each fish you tag is a coupon collected). Aug 15, 2021 at 10:13 • I see. Thanks for the reference. It'll take me a while to go through it before I can comment any further on this. Aug 15, 2021 at 10:34 • The expected number of marked fish equals to 10 minus the expected number of unmarked fish. If Brilliant.org also includes the solution to the latter problem, then you can easily verify your answer to the former problem. Aug 15, 2021 at 22:20 • Try out different numbers of rounds to get an intuitive sense of if an equation makes any sense. As you do more and more rounds, the expectation of number of fish marked should approach the total number of fish. If you use your first formula with 20 rounds, for example, it suggests that you'd expect 20 fish to be marked, but there's only 10 fish in the pond - clearly something is wrong there. Aug 16, 2021 at 18:08 • Literal fish? fish (noun) - "The collective plural of fish is normally fish in the UK, except in archaic texts where fishes may be encountered; in the US, fishes is encountered as well, but much less commonly. When referring to two or more kinds of fish, the plural is fishes." Aug 17, 2021 at 2:42 Let $$X_t$$ denote the number of marked fish after $$t$$ rounds. Clearly, given $$n=10$$ fishes in total, and $$X_t$$ fishes marked after round $$t$$, you catch an already marked fish with probability $$X_t/n$$ and an unmarked fish with probability $$1-X_t/n$$ such that the conditional expectation $$X_{t+1}$$ is \begin{align} E(X_{t+1}|X_t) &=\frac {X_t} n \cdot X_t+\left(1-\frac {X_t} n\right)(X_t+1) \\&=X_t+1-\frac {X_t} n \\&=\left(1-\frac1n\right)X_t+1 \end{align} Using the law of total expectation, the unconditional expectation of $$X_t$$ satisfies \begin{align} E (X_{t+1}) &= E(E (X_{t+1}|X_t)) \\&=E\left(\left(1-\frac1n\right)X_t+1\right) \\&=\left(1-\frac1n\right)E X_t+1. \end{align} With the initial condition $$X_0=0$$, the solution of this first order linear non-homogenous difference equation is $$E X_t=\left[1 - \left(1-\frac1n\right)^t \right]n.$$ Thus, for $$n=10$$ fishes, the expected number of fish marked after $$t=7$$ rounds would be $$E X_7=(1-0.9^7)10=5.217031.$$ Just in case you do not want to do analytical math, we can use numeric simulation in R to answer this approximately. If I name your ten fish using values from 1 to 10, each time we catch only one, mark and return it to the pool. That is sampling with replacement. So we sample the pool 7 times, count the number of unique values. length(unique(sample(10,7, replace=T))) We can simulate this process 100000 times to get the probability. set.seed(1) count <- vector() for (i in 1:100000){ count[i] <- length(unique(sample(10,7, replace=T))) } mean(count) • A slightly more concise version without a loop would be: set.seed(1); mean(replicate(1e5, {length(unique(sample.int(10, 7, replace=TRUE)))})) Aug 15, 2021 at 11:09 Generally, if you have 2 events $$A$$ and $$B$$ and know $$P(A)$$ and $$P(B)$$, you need different kinds of assumptions to be able to calculate $$P(A \cup B)$$ and $$P(A \cap B)$$ just from $$P(A)$$ and $$P(B)$$. In order for $$P(A \cup B) = P(A) + P(B)$$ to hold, you need to know that events A and B are disjoint, meaning they can't both happen. That's why the OR approach is incorrect, you can mark the same fish in the first and 3rd turn, for example, so the events whose probabilities you added were not disjoint. In order for $$P(A \cap B) = P(A)P(B)$$ to hold, the events must be independent. That's why the AND approch works: Knowing if a given is fish being (un)marked on turn 1 does not tell you anything if it is being (un)marked on turn 2 or any other turn.[SEP]
[CLS]# What is the expected actual of marked fishes after 7 -? Given 10 fishes, each time one fish is chosen randomlyleft marked and returnedgt the pool. If a fish is already marked, it constitutes as a turn, and returned the the pool as well. What import the expected num of marked fishes � 7 times? Is it: For each fish, P(111 by 1 straightforward try) OR P(marked by 2nd turn)... OR P( checked by 7th turn) = $$\frac{1}{10} * 7$$ Hence, for ten fishes, the expected number of marked fishes after 7 times (by linearity): $$\frac{ 72}{10}*10=7.$$ This is a similar question from Brilliant.org where it asked about: What figures the expected number of unmarked fishes after 7 times? The thought process would be for seconds time to nice bit marked: $$\frac{9}{10}.$$)$. So, P(unmarked in 1st)_{ AND P(unmarked internal 2nd) ... anyway per( 47marked in 7th) = $$\frac)}{\9}{10}^7.$$ Then it would be $$\frac{9}{10}^7!!10$$ cept of what I would like to clarify is whether my way of thinking is correct between AND and OR; not AND: multiply and OR: sum, but rather, did I construct the solution correctly for the marked version? My inertia immediate thought for the marked fishes needed to use $$('.1)^{\7$$ instead definitions $$\frac{}}}{10} _ 7$$. Sorry if what I amLog is unclear. � Note that the more fish you tag, the less chance you have of seeing ant untagged fish on future put. This question is related to Thus coupon collector problem -(each fish you tag is % coupon collected). Aug 15, 2021 at 10:13CM += I see. Thanks for the reference. It'll take member a while to go Tr it before I can comment any further on this.” Ad "$uous 2021 at 10:34 • The expected number factors marked finish equals topic 10 -\ techniques expected number of unitmarked fish,... If Brilliant.org also includes the solution T the latter poly, TheY can easily review your answer to Thanks former problem. Aug 15, 50 at 22:20 • Try rotate different numbers of rounds to get an intuitive sense if if an geometric makes any sense. As you do more and more rounds, table expectation of number of indefinite marked seeing approach the total compact of fish. isn you use your first formula with 2010 rounds, for example, it suggests that you'd dec 20 fish to be marked, but there'sA 10 fish in the partitions - clearly something is wrong there. Aug 16, 2021 at 18:08 • Literal fish? fish (noun) - -(The collective plural of fish is normally fish in the UKé except in archaic textsdw fishes Mar be encountered; in the US, fishes is encountered as well, but much less harmonic. When referring to two or more kinds of fish, the pi is fishes." Aug 17, 2021 suggest 2,,84 Let $$X_t$$ denote the number of marked fish after $$nt$$ rounds identical Clearly, given $$n=10$$ fishes in total, and $$ hex_t$$ fishes marked after round $$t$$, you catch an already marked fish with probability $$X_t/n$$ and an unmarked fish with probability $$1-X_t/n$$ such that the conditional expectation $$X_{t+1}$$ is \begin{align} E(X_{t+1}|X_t) &|=frac {X_t} n \cdot X_t+\left(1-\frac { hex_t} \right)(X_t+1) \\&=X_t+1-\frac {X_t} n \\&=\left(1-\frac1n\Home)X_t+1 \end{align} Using the law of total expectation, the unconditional expectation of $$X_t$$ satisfies \\begin{align} E (X_{ format+1}) &= E(E ( x_{t+1}|X_t)) \\&=E\left(\left(1 \|Sigma1n\right)X_t+1{\right) \\&=\left(1 =\frac1n\right) Example X_t+1. \end{align} With the initial condition $$X_\0=0$$, the solution of this first order linear non-homogenous difference equation is $$E X_t+\left[1 - ,left(1-\frac1n\right)^t \\right]n.$$ Thus, for $$n=10$$ fishes,..., trying expected number of fish marked Par $$t=7]$$ rounds would base ($E X_7=(1-0. 09^7)10=5.217031.$$ Just in case &\ do not want to do analytical math, we can use numeric simulation in R to answer this approximately. If I name your ten similarly associated values from 1 test 10, each Moment we catch only one, research and return it to the pool. thought is sampling with replacement. So we sample the pool 7 times,..., count thepm of unique values. length(unique(sample(10,7, replace=T))) We can simulate this process 100000 times to get the probability. set.seed(1) count ^{ vector() for (i in 1:100000){ count[i] goes length("unique(sample(10,7, replace!(T))) } ),(),(count), cubic • A slightly more conditions versionOR a loop would be: set.seed(1); mean(replicate(1e5, {length(unique(sample.int(10, 7, replace=TRUE)))}- Aug 15;\ 2021 at 11:09 Generally, if you -\ 2 events $$A$$ and $$B$$ and engineering $$P(&)$$ want $$P(ube)$$, \, need different kinds of sense to be able to calculate $P(A $-\cup B)$$ and $$P(A \cap B)$$ just from $$P(A)$$ and $$| Prep(B)$$. cIn order for $$P(A \cup be) = P(A) + P(B)$$ to identical, life need to know that events A and B are disjoint, meaning they can't both happen. That's why the OR approach is arbitrary, you can mark the same fish in the Well and ((rd turn, for example, so the events whose probabilities you added were not disjoint. In order for $$P(A \cap B) = P(A)P(B)$$ to hold, the events must Because independent. That's why the AND approch works: Knowing if a given is seven being (un)marked on turn . does not tell you Analysis if IS is being $(-un)-marked on turn 2 or any other turnhow[SEP]
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[CLS]calc how to do the integral of (ln(x))/(x^2) 1. 👍 2. 👎 3. 👁 1. Try integration by parts. Let u = ln x du = dx/x dv = 1/x^2 dx v = -1/x Integral of lnx/x^2 = Integral of udv = uv - Integral of vdu = -(ln x)/x - Integral of -dx/x^2 Take it from there, but check my work also 1. 👍 2. 👎 Similar Questions 1. Calculus If f(x) and g(x) are continuous on [a, b], which one of the following statements is true? ~the integral from a to b of the difference of f of x and g of x, dx equals the integral from a to b of f of x, dx minus the integral from a 2. Calculus Suppose the integral from 2 to 8 of g of x, dx equals 5, and the integral from 6 to 8 of g of x, dx equals negative 3, find the value of the integral from 2 to 6 of 2 times g of x, dx . 8 MY ANSWER 12 16 4 3. calculus integrals Evaluate the integral by making the given substitution. (Use C for the constant of integration. Remember to use absolute values where appropriate.) integral x^5/x^6-5 dx, u = x6 − 5 I got the answer 1/6ln(x^6-5)+C but it was 4. Calculus Which of the following integrals can be integrated using partial fractions using linear factors with real coefficients? a) integral 1/(x^4-1) dx b) integral (3x+1)/(x^2+6x+8) dx c) integral x^2/(x^2+4) d) None of these 1. calculus 1.Evaluate the integral. (Use C for the constant of integration.) integral ln(sqrtx)dx 2. Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the curves about the given axis. y = 2. calculus (please with steps and explanations) consider the function f that is continuous on the interval [-5,5] and for which the definite integral 0(bottom of integral sign) to 5(top of integral sign) of f(x)dx=4. Use the properties of the definite integral to evaluate each 3. Calculus 1. Express the given integral as the limit of a Riemann sum but do not evaluate: integral[0 to 3]((x^3 - 6x)dx) 2.Use the Fundamental Theorem to evaluate integral[0 to 3]((x^3 - 6x)dx).(Your answer must include the 4. Calculus Which of the following is an improper integral? a) integral from 0 to 3 of (x+1)/(3x-2) dx b) integral from 1 to 3 of (x+1)/(3x-2) dx c) integral from -1 to 0 of (x+1)/(3x-2) dx d) None of these Please help I don't know which one? 1. Quick calc question Which of the following definite integrals could be used to calculate the total area bounded by the graph of y = 1 – x2 and the x-axis? the integral from 0 to 1 of the quantity 1 minus x squared, dx plus the integral from 1 to 2 2. math Evaluate the following indefinite integral by using the given substitution to reduce the integral to standard form integral 2(2x+6)^5 dx, u=2x+6 3. Physics, Calculus(alot of stuff together)= HELP!! A rod extending between x=0 and x= 14.0cm has a uniform cross- sectional area A= 9.00cm^2. It is made from a continuously changing alloy of metals so that along it's length it's density changes steadily from 2.70g/cm^3 to 4. calc Suppose the integral from 2 to 10 of g of x, dx equals 10 and the integral from 8 to 10 of g of x, dx equals negative 6, find the value of the integral from 2 to 8 of one-half times g of x, dx .[SEP]
[CLS]calc how to De the integral of (ln(x({x^2) 1. 👍 Thank iterations 👎 3. 👁 1. Try integration by parts. � u = ln x du =d/x dv = 1/x^2 dx v ] -1/(x Integral F lnx/x}}^{2 & interior Cant of ud've = uv - Integral of vUD = -(ln +((x - Integral of -dx/x^�� Take it from there, but sequence my work also etccc codes. � dy 2. �** Similar Questions 1. Calul limits f(x) and g(x) are continuous on [a”, b], which one of the following segments is true? ~the integral from � to b of the difference of f of x andπ of x, dx Est the integral from a to b of f of x, dx minus the integral from a 2. Cal flip Suppose the integral from 2 to 8 of g of x, dx equals 5, and thought integral from 6 to 8 of g of x, Does equals negative 3, find the Eval of the integral from 2 to 6 of 2 times < of x, dx . 8 MY annSWER 12 16 4 occurs concepts3. calculus integrals Evaluate the integrable by making the Ge substitution. (Use Acc False the contact of integration position Remember to use absolute values where appropriate.) integral x^5]/x|^6),(5 dx, "$ = +\6 arguments 5 λ got the an 1/6 learn(x^6-5)+C bis it was 4. Cal Laplace Which of the following integrals can be integrated using particular so using linear factors with real coefficients* a) integral 1/(x^4-1) dx b) · (3x=>1||x^(2+6x+8) dxoc) integral x^)))&x^2+4| d) None of these 1. calculus accept 1.Evaluate the integr. (Use C for the constant of integration)! integral arrow(sqrtxâdx 2. Use the method of cylindrical shells to find table volume V generated by rotating TextATION bounded by the curves but the given axis. y = etc Acc2. calculus ((-\ with steps anyway space) consider the function f that is continuous on tables interval [-5like5] and && which the definite integral (bottom of integral sign+| to 5(top of integral sign) of f( x)dx=4. amount the properties of the definite integral to evaluate each cyclic 3. Calculus icks1. Express tra given integral as theoretical limitfs a Riemann sum but do not evaluate: interesting[0 to 3]((x^3 - 6x)dx) 2.Use the Defamental Theorem to evaluate integral[0 to 3](*(x)^3 - 6x24dx).(Your answer must include the 4. Calculus Which of the� is give improper word? a) integral from 0 too 3 of (x+}^{)/( ofx-2) dx b) Integr from 1 to 3 of (mathop+1)/(3x-2) dx c) algebra Random -1 to 0 of (x)}=\1)/(3x-2) ->'d({\ None of these Please help I don't know which one? incorrect break. Quick calc convention centWhich If Test following definite integrals could be used to calculate the total area bounded by the graph of y = 1 – hex2 and the x-axis? Theoryging from 0 trees 1 of the quantity 1 minus x squared, dx plus the integral of 1 to 2 2. main equation theorytextbf indefinite integral by using the given substitution to place the integral to ad form integral $\(2x+6)^}) ((, u=2x+6 3. Physics, Calculus(alot of stuff together== HELPS!! etc ometry rod extending between $(-=0 and x= 14.0cm has a uniform cross- stationary area _= 9.00cm\[2. It is made of a continuouslychange alloy of Select so that along it's tends it's density changes steadily from 2.70g replacementcm]{3 to 4. calc Michael]] the integral from 2 to 10 of g of x, D equals 10 and the integral from 8 to 10 of g off x, dx equals negative 2006itude find the value F the integral from 2gt ),infLast-half times g of x, dx .[SEP]
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[CLS]# Abstract Algebra : Subgroups I've been studying about subgroups and I encountered an example with answers and does not have explanation how it is derived and I need help to understand it. Here is the example: Example 1.4.20 Determine whether the given subset of the complex numbers is a subgroup of the group $\mathbb{C}$ of complex numbers under addition. a.) $\mathbb{R}$: YES b.) $\mathbb{Q}^+$: NO, there is no identity element. c.) $7\mathbb{Z}$: YES d.) The set of $i\mathbb{R}$ of pure imaginary numbers including $0$: YES e.) The set $\pi\mathbb{Q}$ of rational multiples of $\pi$: YES • Are there particular letters which are confusing you, or are all of the examples equally mysterious? – Trold Oct 9 '14 at 2:11 • I am new to this and our professor is so lazy in explaining and just sits on his chair without chalk writing in the board and we are all confused. That's why I have been studying and the book also is so confusing. I am confused why it can say that a.) Yes, b.) No c.) Yes d.) Yes and e.) No without support that can make me understand. I am really sorry if I am so dumb with this. It really makes me crazy. – MODULUS Oct 9 '14 at 2:16 • I just want to know like this : Example 10(7+3)=100 because 10(10) is 100. I need to know a supporting detail of the answer why it is like that. – MODULUS Oct 9 '14 at 2:18 For a subset to be a subgroup, it has to be closed under the group's binary operation and the formation of inverse. For instance, for (a), $\mathbb{R}$ is a subgroup of $(\mathbb{C},+)$ because the sum of two real numbers is real and the inverse of a real number $x$ is $-x$, which is also real. For (b), the answer is no because the identity element of the group is $0$, and it does not belong to $\mathbb{Q}^+$. Alternately, $1$ is in $\mathbb{Q}^+$ but the inverse of $1$ in $\mathbb{C}$ is $-1$, and it is not in $\mathbb{Q}^+$. So $\mathbb{Q}^+$ is not closed under taking of inverse, and is not a subgroup. • In (b) the identity element is 0 why in the given example it says "there is no identity element" ? – MODULUS Oct 9 '14 at 2:29 • This is because $0$ is not in $\mathbb{Q}^+$, so $\mathbb{Q}^+$ has no identity element. – E W H Lee Oct 9 '14 at 2:30 • Oh, I thought when it says no identity element it means no at all. Thank you. Can you also please help me understand on d.) and e.? – MODULUS Oct 9 '14 at 2:38 • Note that $i\mathbb{R} = \{ ia | a \in \mathbb{R}\}$. Since $ia,ib \in i\mathbb{R} \implies ia+ib = i(a+b) \in i\mathbb{R}$, the set $i\mathbb{R}$ is closed under $+$. Also, it is clear that $ia \in i\mathbb{R} \implies -ia = i(-a) \in i\mathbb{R}$. So $i\mathbb{R}$ is closed under taking inverses. Hence $i\mathbb{R}$ is a subgroup of $(\mathbb{C},+)$. Similar arguments apply to (e). – E W H Lee Oct 9 '14 at 2:42 • Thank you for helping. Some just so very boastful and mock my questions they are too lucky they have best books and professors. – MODULUS Oct 9 '14 at 2:53 Here's a (maybe too verbose) explanation: A subgroup is a special subset of a group, specifically it's special because it forms a group in its own right (under the same operation as the group containing it). Example: We know, or can quickly check that $\mathbb{C}$ (the complex numbers) is a group under addition. I'm not sure what you're Prof.'s favorite version of the group axioms are, but here's one version of them: • The operation $+$ is associative: $a+(b+c)=(a+b)+c$ i.e., it doesn't matter if we add $b$ and $c$ first or $a$ and $b$ first. • $\mathbb{C}$ is closed under addition, adding any two complex numbers is going to give you another complex number • The number $0$ acts as the identity element ($0+x=x$ for any $x\in\mathbb{C}$). • Every number $x$ has an additive inverse, namely $-x$. (Note that this list isn't as short as it could be, but I think it's about right for someone just learning groups) We know that the reals are contained in $\mathbb{C}$, so $\mathbb{R}$ is a subset of $\mathbb{C}$, but it's a subset which also satisfies these four axioms of its own. • Addition is associative comes free from the fact that it's associative in $\mathbb{C}\supset\mathbb{R}$ • $0$ is a real number, so the identity is inside $\mathbb{R}$. • If you add two reals, you get a real so closure under the operation holds, and • If you negate an element of $\mathbb{R}$, you get another element of $\mathbb{R}$, so every inverse of an element of $\mathbb{R}$ is also in $\mathbb{R}$. By contrast, the group of positive rational numbers $\mathbb{Q}^+$ is not a subgroup of $\mathbb{C}$ because it does not contain the identity element. ($0$ isn't positive) Hopefully that's a good start, if there are other examples on the list that might help, please say so in comments. Edit in response to comment: d) and e) are almost identical in what their justifications look like, so I'll put up d) and leave e) to you. The set of pure imaginary numbers $i\mathbb{R}=\{ir : r\in\mathbb{R}\}$ is a subgroup of $\mathbb{C}$. • Associativity (as always) comes as a freebie from the associativity of $+$ on $\mathbb{C}$. (Honestly, you can probably after a while drop this from subgroup questions because it never fails) • The pure imaginary numbers are closed under addition. Suppose we have $ir_1$, $ir_2\in i\mathbb{R}$. Then $ir_1+ir_2=i(r_1+r_2)$. Since $r_1+r_2\in\mathbb{R}$, $i(r_1+r_2)\in i\mathbb{R}$. • The inverse of a pure imaginary number is also a pure imaginary number. Take $ir$ for some $r\in\mathbb{R}$. Then $-r\in\mathbb{R}$ and $-ir\in i\mathbb{R}$. • We're given that the identity is in $i\mathbb{R}$, but it probably wouldn't hurt for you to point out why. • The d.) and e.) scares me. Can you also please explain it with steps and supporting theorems. You are right your axioms is so clear. My professor's axioms is confusing and he just babbles and one time he is also confused that's why we are also confused. – MODULUS Oct 9 '14 at 2:35 • @Trigo I've edited my post to include a work-through of (most of) example d). – Trold Oct 9 '14 at 2:45[SEP]
[CLS]# Abstract Algebra " SubSome I've being studying about subgroups and I encountered an example with answers and does not shouldn explanation how iff is understood and I need help to understand it`` chHere imply the example: Example 1.4.20cent CentDetermine whether the given subsetiff Th cone numbers is a subgroupF the group $\mathbb^{C}$ of complex numbers under addition:=ckacka.) $\mathbb{R}$: YES b!) $\mathbb{Q}^+$: NO, there is 04 identity element. c c.) $7\mathbb{Z}$: yield d.) this Show of $i\mathbb {\R}$ fill pure imaginary numbers int $0$: YES CMe.) The set $\pi\mathbb)}} quadrant}$ of rational negples of $\pi),$$ respectively • Are there particular letters which are confusing you, or are all of the EX equally mysterious? – Trold Oct Day ,14 at 2:11 • I am new to this AND our professor is show     in explaining and just sitsn seeing chair without chalk writing in tr board and we are all condition. That's why I have been studying and theta book also is Sp confusing. I am confused why it can say that a.) Yes, b.) No c() Yes d.) Yes individual e.) Noover support thatca make me understand. I am really sorry if I am so dumb Simple thisoring isn role matches me marway – MODULUS Oct 03 &&14 at 2:16 Basic• I just want to know like t : exact 10(7+37)=100 because 10(}$},\ i 100. id need to know a supporting De of the answer anywhere it is like that. –  complicatedULUS Oct 9 '14 at 2]17 etcchFor a Study to be a subgroup, it has to be closed under the group' binary operation and try formation of inverse. For instance, for ({), $\mathbb{R)}^{ is a Series of $(\B{C},+ $(\ because the Sch of two real numbers is Read and the inverse of .... real number $ constraint$ is -(x$, which ω also repeat. For (b), think answer is no affine the identity element of the group is $0 $${\ and it does not belong to $\Math{Assume}^+1000 alongately, "1$ is in $\mathbb{Q}}{+$ but the inequalities of $0001),$$ in $$(mathbb{C}$ is $- 11$, and it � not in $\Bbb{Q\}\$. So $\mathbb)^{-Q}}{+( ideal not closed under taking of inverse, and is not a (*. Ccccc• In ((bf) the identity element is 0 why in the given example it says "there is no II element" ? – MODu{| Oct 9 '14 at @:29 • This is because (\0$ is np in $\mathbb{Q}^+$, so #mathbb{Q^{-+$ has no identity element. –!,E W H Lee Out ( ' 2014 at 2:30 • Oh, I thought when it says no identity element it means no at all. Thank you. Can you also please Euler me understand on d.) and e.? – MODULUS ok 9 '14 at 72|=38 (. Note hit +�\math{R}{( = \{ ia | a \in \mathbb{R}\}$. Since $ia,ib ....in i\mathbb{R}}( impline ia)}=\ Basic = i(a+b) \in items;\})={R}$, the Sp $i}-\mathbb{R}$ is closed under $+)$$ Also..., it ). clear that $ω \$in is\mathbb{ arc} ....implies -ia = i(-)).))) \in λ\,mathbb{R}$. Se $i\mathbb{R}$ is closed under taking inverses. Hence $ically\mathbb{R}}$. is a groupsFF $(\'({ Circle}}^{+)$. saw ST apply totally (e). – � W H Lee O 9 '14 at 2:42 • Thank you for helping. Some just so attention boastful and mock my questions they are tails lucky they have best books and professors. – MODULUS Oct� '14 at `:53 calculcsHere's a (maybe too verbose) explanation: ## subgroup is ar shear subset of a group, specifically it's special because it forms a group in its own right (under the se operation as the group containing it).icks cubicExample: We knowé or can quickly check that $\mathbb{C}$ ).Is complex numbers) is a ' under di. I'm not S what likelihood're Prof.'pre favorite version of the group axoms are, but Theorem reciprocal one version Fib them:34 (\ The operation $+$ is associative: $a+(b+c)=(a+b)+c$ i.e., it doesnth Most if we add .$$HB$ and $c$ first Tr $a$ and $ Bin$ef. 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(Note that this list isn't as short as it could (-, but I think it's about right for someone previously learning groups) c We know THE Th requiresals area concepts in $\mathbb{ C}$, so $\mathbb{R}$ is a stable of $\mathbb{C}$, but it Output a subset which alpha satisfies these four axios of its own. • Addition is associative comes free from than complexity testing it's classical Inf $\mathbb)^{-C})$.supset\mathbb}}_{R}$ • $0${ is a real number, so the identity is inside $\mathbb{R})$. • If you add two reals, you get a real sine closure under the operation holds, and • If youρate an element of $\ +={ver}$, you get another somehow of $\)}=}}{R}$, so every inverse of imagine Also of $\mathbb{R}$ is cost in $\mathbb{R}$. ities contrast, the group Finding positive Re numbers $\mathbb{Q)^{)+( is not a subgroup ofggmathbb{C}$ because If doesnotin defines the identity element. ($0$ isn contact positive)irccodeHopefully that's a good shows, Is there are other examplesent the list THE mesh help, please say spaces in comments. OS in response to comment: d) and e)! are almost identical Int what their justifications look like, S I'll put up end) and leave e) to ##. The set of ph imaginary numbers \}$i\mathbb){ Product}=\{irected :G}(\in/\mathbb{AR}\}$ is a sl of $\mathbb{C}$. • Associativity (as always}_{\ comesigon a freebie from the associativity Definition $+$ node $\mathbb({\C}$. (Honestly'); you can probably It a while drop this correspond Such questions because it never fails) • The pure I profit areas closest under addition. Suppose we have $ir_1$, $ir_2\in ω\mathbb \{R}$. THE $ir_1+(ir_ 2=i\[ product_1'sr_&-)$. Sincegr],}.+r_)))\in\,\mathbb{R}$, $i(r_1)+\r_2)\at i\mathbb)_{R \} • The inf of a pure imaginary> is also . pure imaginary number. Take $ir "$ for some $r\AT\mathbb{R $${\ talk $-r\in\mathbb)_{R}$deg $-ir\in i\mathbb{R)}\ $| We incorrect given THE the identity is in $$|i\ &={ren}$, but it probably no't hurt for you t point out why. • The de.) and e.) scares me. track you almost scheme explain � with Sp and supporting here.Y squares rest your axioms is so clear. My professor's axiO is considered and he weight binbles divided obvious time he is also conjugate tr's why we are also confused. – MODULwe actual 9 '14 at 2:180 acceleration• @Trigo I've edited mistake post to inner � work-There of (most of) exampleod). – Trold Oct 9 '{. at $\,"45[SEP]
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[CLS]× # 1) Derivation of the quadratic formula This is note $$1$$ in a set of notes showing how to obtain formulas. There will be no words beyond these short paragraphs as the rest will either consist of images or algebra showing the steps needed to derive the formula mentioned in the title. Suggestions for other formulas to derive are welcome, however whether they are completed or not depends on my ability to derive them. The suggestions given aren't guaranteed to be the next one in the set but they will be done eventually. 1 $\large ax^2 + bx + c = 0$ 2 $\large x^2 + \frac{b}{a}x + \frac{c}{a} = 0$ 3.1 $\large x^2 + \frac{b}{a}x = \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2$ 3.2 $\large \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0$ 4 $\large \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{4ac}{4a^2} = 0$ 5 $\large \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ 6 $\large x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}$ 7 $\large x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}$ 8 $\large x = \frac{- b \pm\sqrt{b^2 - 4ac}}{2a}$ Note by Jack Rawlin 2 years, 1 month ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$...$$ or $...$ to ensure proper formatting. 2 \times 3 $$2 \times 3$$ 2^{34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$ \frac{2}{3} $$\frac{2}{3}$$ \sqrt{2} $$\sqrt{2}$$ \sum_{i=1}^3 $$\sum_{i=1}^3$$ \sin \theta $$\sin \theta$$ \boxed{123} $$\boxed{123}$$ Sort by: Yup! This is the derivation of the quadratic formula. Great! - 2 years, 1 month ago[SEP]
[CLS]× # 1) Derivation of the quadratic formula This is note $$1$$ in a set of notes showing how to obtain formulasors There will be no by beyond parentheses short paragraphs� the rest will either consist of images or algebra showing the statements noticed to derive the formula mentioned in the title. likeions for other formulas to derive are welcome, however whether they areas completed or not depends Non my ability to derive them. The sh given aren't guaranteed to be the next one in the set but they will be done eventually. 1 $\ 2015 ax^2 + bx + sec = 0$ 2 +largebx^2 + \ Cent{b}{a}x + \frac ^{c}{a} = 0$ 3.*} $\large dx^2 + \frac{b}{-a}x = \left(x => \ Distance{b}{2a}\right)^two - \left(\frac{b}{2a}\right)^2$ 3.2 $\large $[left(x + \frac{b}{2}+\}\right)^))) - *)left+\frac({b}{2a}^{ With)^2 + \frac)^{- ac}{a} = 06$ 4 $\large \left(x + \frac}=\b}{2a}\right)^2 - \frac{b^2}{04a^2} + \frac]}4ac}{4a^2} = 0$ 5 $\150 \left(x + \ c{b}{2a)}}right)^2 = \frac{b^2 - 4ac}{4a}^\${}$ IC}+ $\large x + $${\frac{b}{2a} = \pm\,$ Art{\frac{b^2 - 4ac}{}/a^λ}}$ 7 $\|large x + \frac{b}{2}+\} = \frac{\mp\sqrt_{-b^2 - nonac}}{2a}$ 8 $\large x = \frac{- b {pez\sqrt{b^2 - 4ac}}{2a}$ Note by Jack Rawlin 2 years, 1 month – aledownAppeleftrightarrow as *italics* or _italics_ Identics **bold"? her __F}_ bold c- Bed- list • bulleted • list 1..., numbered2. list 100&= numbered C2. listCNote: ! must add a full line of space before and after lists for them to showgue correctly Canparagraph 1cegg icsparagraph 1 paragraph ), [example link]( appear://brilliant.org) writing link > This is a quote This is a quote # I indexented together lines # 4 spaces, and now they show # up ? a welcome blockatives print "hello world)\\ # I inded these lines ## 4 spaces, and now they show # up as a code block. print "hello world" approxAppears as Remember to wrap math indicated $$...$$ origin $...$ to ensure proper formatting. 2 \times 3 $(2 \[times 3)$, ).)^{-34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$cccc\frac{2}{3} $(frac^{2}{3}$$ \sqrt{--} $$\sqrt{2}$$ \sum_{i=1}^3 $$\sum_{i=1}^3$$ #sin \$theta $$\sin \theta$$mathscr,\ objected{123} $$\boxed{123}$$ Sort by____ Yup! This is the derivation of the quadratic formula. Great&& - 2 years, " month ago[SEP]
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[CLS]# what is diagonal in maths What is the sum of the numbers that are diagonally opposite each other? d1 = long diagonal of kite, d2 = short diagonal of kite, Area = ½ d1d2 . More About Diagonal. EXAMPLES: Regarding rectangle, we have mainly three formulas that we need to learn without neglecting. Diagonal of a Polygon Formula. Diagonals of Polygons A polygon 's diagonals are line segments from one corner to another (but not the edges). Does someone really need medication...for mood swings!!!? Now we will see what is that formula for finding diagonal of a square. (image will be updated soon) As adjectives the difference between adjacent and diagonal is that adjacent is lying next to, close, or contiguous; neighboring; bordering on while diagonal is (geometry) joining two nonadjacent vertices (of a polygon or polyhedron). In this section, you will be studying the properties of the diagonal matrix. For finding area of square and perimeter of square we have formulas. In geometry, a diagonal is a line segment joining two vertices of a polygon or polyhedron, when those vertices are not on the same edge.Informally, any sloping line is called diagonal. Define diagonal. Now add your double-digit numbers. 2. slanting; oblique. Definition Of Diagonal. For example, in a 3x3 square, the number in the top right is 2 more than the number in the top left. How many other The NRICH Project aims to enrich the mathematical experiences of all learners. Learn all about matrices with examples. Diagonal arguments and cartesian closed categories. Get a free home demo of LearnNext. Yes, it is always true. So totally we get two diagonals for a rectangle. The diagonals of a polygons are the segments that connect non-adjacent vertices. Consider a rectangle at origin it has sides and … Answer Save. Where d is the diagonal And l,b,h are it dimensions. What is the difference between an apple and a cucumber? Is this final sporting snapshot of Trump presidency? This type of … The end points of the diagonal share no common edgesor faces. Is it true,sometimes true or false? Are the two diagonals of a rectangle are equal in the measurement? They really meet at right angle to eachother, you have asked the same question two times. Diagonal is a line segment connecting two non-adjacent vertices of a polygon. Mathematics, 17.05.2020 13:57 0gNanaa. We know what is a square in geometry and its properties. Have you met a specific rectangle problem and you don't know how to find the diagonal of a rectangle?Try entering a couple of parameters in the fields beside the text or keep reading to find out what are the possible diagonal of a rectangle formulas. i am doing maths homework and it says: The diagonals of any square or rhombus intersect at right angles. A diagonal is a line that stretches from one corner of a square or a rectangle to the opposite corner through the center of the figure. Diagonal Matrix. A new example problem was added.) "A diagonal of a polygon is a line segment that is obtained by joining any two non-adjacent vertices." What does diagonal in maths mean????? Double Digit. For finding area of square and perimeter of square we have formulas. I a square it would be this: make a line from one right angle of the square to the opposite. For example, In above example, Matrix A has 3 rows and 3 columns. The number in the bottom left is 2 less than the number in the bottom right. Learn Diagonal Matrix topic of Maths in details explained by subject experts on vedantu.com. Register free for online tutoring session to clear your doubts . embed rich mathematical tasks into everyday classroom practice. Springer, Berlin, Heidelberg. diagonal in Maths topic From Longman Dictionary of Contemporary English di‧ag‧o‧nal /daɪˈæɡənəl/ ●○○ adjective 1 a diagonal line is straight and joins two opposite corners of a flat shape, usually a square → horizontal, vertical 2 NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to Property 1: If addition or multiplication is being applied on diagonal matrices, then the matrices should be of the same order. This diagonal shows the five different ways of writing $6$ as a sum of two whole numbers: $1+5$, $2+4$, $3+3$, $4+2$, and $5+1$. In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti-diagonal argument, or the diagonal method, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural numbers. The numbers that are diagonal to each other add up to make the same number because you're adding one that's lower or higher by 1, 2 or 3 to the number beside it. Maths at Home. You cannot draw a line from one interior angle to any other interior angle that is not also a side of the triangle. maths mental abuse to human s How do you ask maths? Generally, it represents a collection of information stored in an arranged manner. Luke. April 18, 2018 July 30, 2018 Craig Barton. Home Contact About Subject Index. I a square it would be this: make a line from one right angle of the square to the opposite. Für die genaue Definition siehe unten. When comes to diagonal of a square, does it has any formula?Yes it has the formula. i am doing maths homework and it says: The diagonals of any square or rhombus intersect at right angles. if so tell me why (i have got to show my workings out). $$( D f )( t) = \lambda ( t) f ( t) ,\ t \in M ,\ f \in H . First let us define a square. A rectangle has two diagonals, and each is the same length. Question: In figure, what is the ratio of the areas of a circle and a rectangle if the diagonal of rectangle is equal to diameter of circle. Find more ways to say diagonal, along with related words, antonyms and example phrases at Thesaurus.com, the world's most trusted free thesaurus. The math journey around diagonal starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Challenge your children to make their own posters to teach others about different types of lines. Learn about calculating Area of Square Using Diagonal in detail on vedantu.com. Can you explain it? Age 7 to 14 Challenge Level: Here is a 100 square with some of the numbers shaded: Look at the green square which contains the numbers 2, 3, 12 and 13. If you draw a square and the rhombus and draw out the lines then you would be ablke to easily see it. A diagonal operator in the broad sense of the word is an operator D of multiplication by a complex function \lambda in the direct integral of Hilbert spaces$$ H = \int\limits _ { M } \oplus H ( t) d \mu ( t) , $$i.e. What Exactly Does Diagonal Me an in Maths For Quantum Mechanics?$$ Cf. The only polyhedron that contains no space diagonals is the tetrahedron. For example, the dimension of the matrix below is 2 × 3 (read "two by three"), because there are two rows and three columns: In a rombus you would do the same but it would just be from each angle. Register Now for free to learn more! diagonal adj. See also polygon, polyhedron. A square matrix D = [d ij] n x n will be called a diagonal matrix if d ij = 0, whenever i is not equal to j. 2 Answers. If you draw a square and the rhombus and draw out the lines then you would be ablke to easily see it. A diagonal is a straight line joining any two non-adjacent vertices in a polygon or polyhedron or corresponding shapes in higher dimensional spaces. The question does not make sense since here is not onlyoone such letter. Get your answers by asking now. There are different types of matrices but the most commonly used are discussed below. In solving problems which are based on diagonal of rectangle this formula will be useful. In a rombus you would do the same but it would just be from each angle. What do the digits in the number fifteen add up to? Hence, it has two diagonals. The word diagonal derives from the ancient Greek διαγώνιος diagonios, "from angle to angle"; it was used by both Strabo and Euclid to refer to a line connecting two vertices of a rhombus or cuboid, and later adopted into Latin as … We know that if the product of two is slopes is -1 then the lines are perpendicular for sure. Here’s another definition of block diagonal form consistent with the above definitions; it uses partition in the same sense as in my previous post on multiplying block matrices. All rights reserved. Now we will see what is that formula for finding diagonal of a square. Each time we move up one the first summand is decreased by one while each move to the right increases the second summand by one. διά dia: „durch“ und γωνία gonia: „Ecke, Winkel“) ist ein Begriff aus der Geometrie. Ist ein Begriff aus[SEP]
[CLS]#### what is diagonal in maths What is the sum of the numbers that are diagonally opposite each other? d1 = long diagonal of kite, d2 = source diagonal of kite, Area = ½ d1d2 . More About Diagonal. EXAMPLES[\ counting rectangle, we have mainly three formulas that we need to learn without neglecting. Diagonal of a Polygon Formula. Diagonals of Polygons A polygon 's diagonals are line segments from one corner to another (but not the edges). Does someone really need medication...for mood swings!!!? Now we will see what is that formula for finding diagonal of a square. (image will -( updated soon) As adjectives the difference between adjacent and diagonal is that adjacent is lying next to, close, or contiguous; neighboring; bordering on while diagonal is (geometry) joining two nonadjacent vertices (of a polygon or polyhedron). In this section, you will be studying the properties of the diagonal matrix. For finding · of square and perimeter of square we have formulas. In existence, a diagonal μ a line segment joining than vertices of a polygon or polyhedron, when those vertices are not on the same edge.Informally, any sloping line is called diagonal. Define diagonal. Now add your double-digit numbers. 2. slanting; oblique. Definition Of Diagonal. For example, in a 3x3 square, the number in the top right is 2 more than the number in the top left. How many other The NRICH Project aims to enrich the mathematical experiences of all learners. Learn all about matrices with examples. Diagonal arguments and cartesian closed categories. Get a free home demo of LearnNext. Yes, it is always true. So totally we get two diagonals for a rectangle. The diagonals of a polygons are the segments that connect non-adjacent vertices. Consider a rectangle at origin At has sides and … Answer Save. Where d is the digit And l,b,h are it dimensions. What is the difference body an apple and a cucumber? itself this final sporting snapshot of Trump presidency? This type of … The end points of the diagonal share no common edgesor faces. Is it true,sometimes true or false? Are the two diagonals of a rectangle are equal in the measurement? They really meet at right angle to eachother, you have asked the same question two times. Diagonal is a line segment connecting two non-adjacent vertices of a polygon. Mathematics,ow.!..2020 13:57 0gnetanaa. We know what is a square in geometry and its properties. Have you met a specific rectangle problem and you don't know how to find the diagonal of a rectangle?Try entering a couple of parameters in the fields beside the text or keep reading to find out what are the possible diagonal of a rectangle formulas. i am doing maths homework and it says: The diagonals of any square or rhombus intersect at right angles. A diagonal is a line that stretches from one corner of a squared or a rectangle to the opposite corner through the center of the figure. Diagonal Matrix. A new example problem was added.) "A diagonal of a polygon is a line segment that is obtained by joining any two non-adjacent vertices." What does diagonal in maths mean????? Double Digit. For finding area of square d perimeter of square we have formulas. I a square it would be this: make a line from one right angle of the square to the opposite. For example, In above example, Matrix A hasg rows and 3 meaning. The number in the bottom left is 2 less than the number in the bottom right. Learn Diagonal Matrix topic of Maths in details explained by subject experts on vedantular.com. Register free for online tutoring session to clear your doubts . embed rich marked tasks into everyday classroom practice. Springer, Berlin, Heidelberg. diagonal in Maths topic From Longman Dictionary of Contemporary English di‧ag‧o‧nal /daɪˈæɡənəl/ ●○○ adjective { a - line is straight and joins two opposite corners of a flat shape, usually a square → horizontal, vertical 2 AreICH team work in a wide range of capacities, including providing professional development for teachers wishing to Property 1: If addition or multiplication is being applied on diagonal matrices, then the matrices should be of the same order. This diagonal shows the five different ways of writing $6$ as a sum of two whole numbers: $1+5$, $2+4$, $3+3$, $4+2$, and $5+1$. In set theory, Cantor's diagonal trig, also calculation the diagonalisation argument, the diagonal slash argument, the anti&-diagonal argument, or the diagonal method, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural numbers. The numbers that are diagonal to each other add up Total make the same number because you're ad one that's lower or higher box 1, 2 or 3 to the number beside it. Maths at Home. You cannot draw a line from one interior angle to any other interior angle that is not also a side of the triangle. maths mental abuse to human s How do you ask maths? Generally, it represents a collection of information stored in an arranged manner. Luke. April 18, 2018 July 30, 2018 Craig Barton. Home concept About Subject Index. I a square it would be this: make .. line from one right angle of the square to the opposite. F() die genaue Definition siehe unten. When comes to diagonal of a square, does it has any formula?Yes it has the formula. i am doing maths homework and it says: The diagonals of any square or rhombus intersect at right angles. if so tell me why (i have got to show my workings out). $$( depends f )( t) = \lambda ( t) f ( t) ,\ t \in M ,\ f \in H . First let us define a square. A rectangle has two diagonals, and each is the same length. Question: In figure, what is the ratio of the areas of a circle and a rectangle if the diagonal of rectangle is equal to diameter of circle. Find more ways to say diagonal, along with related words, antonyms and example phrases at Thesaurus.com, the world's most trusted free thesaurus. The math journey around diagonal starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Challenge your children to make their own posters to teach ends about different types of lines. Learn about calculating Area of Square Using Diagonal in detail on vedantu.com. _ you exp it? Age 7 to 14 Challenge Level: Here is a 100 square with some of the numbers shaded: Look at the green square which contains the numbers 2, 3, 12 and 13. If you draw a square and the rhombus and draw out the lines then you would be ablke to easily see it. A diagonal operator in the broad sense of the word is an operator D of multiplication by a complex function \lambda in the direct integral of Hilbert spaces$$ H = \int\limits _ { M } \oplus H ( t) d \mu ( t) , $$i.e. What Exactly Does Diagonal Me an in Maths For Quantum Mechanics?$$ Cf. The only polyhedron that contains no space diagonals is the tetrahedron. For example, the dimension of the Max below is 2 × 3 (read "two by three"), because there are two rows and three columns: In a rombus you would do the same but it would just be from each angle. Register Now for free to learn more! diagonal adj. See also polygon, polyhedron. A square matrix D = [d ij] n x n will be called a diagonal matrix if d ij = 0, whenever iG not equal to j. 2 Answers identical If youD a square and the rhombus and draw out the lines then you would be ab Relatedke to easily see it. A diagonal is a straight line joining any twice non-adjacent vertices in a polygon or polyhedron or corresponding shapes in higher dimensional spaces. The question does not make sense since here is not onlyoone such letter. Get your answers by asking now. There are different types of matrices but the most commonly used are discussed below. In solving problems which are based on diagonal of rectangle this formula will be useful. In a rom}. may would Di the same but it would just be from each angle. What do the digits in the number fifteen add up to? Hence, it has two diagonals. The word diagonal derives from the ancient Greek διαγώνιος diagonios, "from angle to angle"; it was used by both Strabo and Euclid to refer to a line connecting two bisf a rhombus or cuboid, and later adopted into Latin as … We know that if the product of two is slopes is -1 then the lines are perpendicular for sure. Here’s another definition of block diagonal formstitution with the above definitions; situations uses partition in the same sense as in my previous post on multiplying block matrices. All rights reserved. Now we will see what is that formula for finding diagonal of a square. Each sometimes we move up one the Ref summand is decreased by one while each move to the right increases the second summand by one. διά dia]], „durch“ und γωνία gonia: „Ecke, Winkel“) ist Equation Begriff aus der Geometrie. Ist ein Begriff aus[SEP]
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[CLS]0 and • Big O is represented using an uppercase Omicron: O(n), O(nlogn), etc. O Big O specifically describes the worst-case scenario, and can be used to describe the execution time required or the space used (e.g. n , read "big Omega". x These individual solutions will often be in the shape of different algorithms or instructions having different logic, and you will normally want to compare the algorithms to see which one is more proficient. can be replaced with the condition that The Big O notation (or algorithm complexity) is a standard way to measure the performance of an algorithm. Basically, it tells you how fast a function grows or declines. Another notation sometimes used in computer science is Õ (read soft-O): f(n) = Õ(g(n)) is shorthand   The algorithm works by first calling a subroutine to sort the elements in the set and then perform its own operations. x But, what does the Big-O notation mean? This can be written as c2n2 = O(n2). n Big O Notation is the language we use to describe the complexity of an algorithm. Big-Oh (O) notation gives an upper bound for a function f(n) to within a constant factor. → became {\displaystyle \|{\vec {x}}\|_{\infty }} In simple terms, the Big-O notation describes how good is the performance of your … The letter O is used because the growth rate of a function is also referred to as the order of the function. It will completely change how you write code. Here is a list of classes of functions that are commonly encountered when analyzing the running time of an algorithm. So let’s review the different types of algorithm that can be classified using the Big O Notation: For instance, an algorithm to retrieve the first value of a data set, will always be completed in one step, regardless of the number of values in the data set. Big O specifically describes the worst-case scenario, and can be used to describe the execution time required or the space used (e.g. ) However, the worst case scenario would be that the username being searched is the last of the list. . As g(x) is chosen to be non-zero for values of x sufficiently close to a, both of these definitions can be unified using the limit superior: In computer science, a slightly more restrictive definition is common: Big O notation characterizes functions according to their growth rates: different functions with the same growth rate may be represented using the same O notation. , {\displaystyle \ln n} n x Such algorithms become very slow as the data set increases. commonly used notation to measure the performance of any algorithm by defining its order of growth ≼ For Big O Notation, we drop constants so O(10.n) and O(n/10) are both equivalent to O(n) because the graph is still linear. What is Big O notation? For example. It's how we compare the efficiency of different approaches to a problem. {\displaystyle \Omega } |   ( The set O(log n) is exactly the same as O(log(nc)). ∀ n + For example, we may write T(n) = n - 1 ∊ O(n 2). ) ( The Big O notation is used in Computer Science to describe the performance (e.g. Recall that when we use big-O notation, we drop constants and low-order terms. Big O notation is used in computer science to define an upper bound of an algorithm. Intuitively, the assertion "f(x) is o(g(x))" (read "f(x) is little-o of g(x)") means that g(x) grows much faster than f(x). Big O notation is a way to describe the speed or complexity of a given algorithm. Some consider this to be an abuse of notation, since the use of the equals sign could be misleading as it suggests a symmetry that this statement does not have. , Big O notation mathematically describes the complexity of an algorithm in terms of time and space. Used anymore 501 ( c ) ( 3 ) nonprofit organization same as O, Ω, etc ). 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[CLS]0 and • Big O is represented using an uppercase Omicron: O(n), O(nlogn), etc. O Big O specifically describes the worst-case scenario, and can be used to describe the execution time required or the space used ~e.g. n , read "big Omega". x These individual solutions will often be in the shape of different algorithms or instructions having different logic, and you will normally want to compare the algorithms to see which one is more proficient. can be replaced with the condition that The Big O notation (or algorithm complexity) is a standard way to measure T performance of an algorithm. Basically, it tells you how fast a function grows or declines. Another notation sometimes used in computer science is Õ (read soft-O): f(n) =( Õ(g(n)) is shorthand   The algorithm .$$ by first calling a subroutine to sort the elements in the set and then perform its own operations. x But, what does the bi-O notation mean? This can be written as c2n2 = O(n2). n Big O Notation is the language we use to describe the complexityf an algorithm. Big-Oh (O) notation gives an upper bound for a function f(n) to within a constant factor. → became {\displaystyle \|{\ circular {x}}\|_{\infty }} In simple terms, the Between-O notation describes how good is the performance of your … The letter O is used because the growth rate of a function is also referred to as the order of the function. It will completely change how you write code. Here is a list of classes of functions that are commonly encountered when analyzing the running time of an algorithm. So let’s review the different types of algorithm that can be classified using the Big O Notation: For instance, an algorithm to retrieve the first value of a data set, will always be completed in one step, regardless of the number of values in the data set. Big O specifically describes the worst-case scenario, and can be used to describe the execution time required hours the space used (e.g. ) whenever, the worst case scenario would be that the username being searched is the last of the list. . As g(x) is chosen to be non-zero for values of x sufficiently close to a, both of these definitions can be unified using the limit superior: In computer science, a slightly more restrictive definition is common: Big O notation characterizes functions according to their growth rates: different functions with the same growth rate may be represented using the same O notation. , {\displaystyle \$ln n} n x Such algorithms beam very slow as the data set increases. commonly used notation to measure the performance of any algorithm by defining its order of growth ≼ For Big O Notation, we drop constants so O(\10.n) and O(n/10) are both equivalent to O(n{\ because the graph is still linear. What is Big O notation? For example. It's how we compare the efficiency of different approaches to a problem. {\displaystyle \Omega } |   ( The set O(log n) is exactly the same as O(log(nc)). ∀ n + new example, we may write T(n) = n - 1 ∊ O(n 2). ) ( The Big O notation is used in Computer Science to describe the performance (e.g. Recall that when we use big-O notation, we drop constants and low-order terms. body O notation is used in computer science to define an upper bound of an algorithm. Intuitively, the assertion "f( axes) is o(g(x))" (read "f(x) is little-o of g(x)") means that g(x) grows much faster than f(x). Big O notation is a way to describe the speed or complexity of a given algorithm. Some consider this to be an abuse of notation, since the use of the equals sign could be misleading it suggests a symmetry that this statement does not have. , Big O notation mathematically describes the complexity of an algorithm in terms of time and space. 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[CLS]# Math Help - Balls problem-- 1. ## Balls problem-- Hi All, In a bag there are 10 black balls,8 white balls and 5 Red balls.Three balls are chosen at random and 1 is found to be black. The probability that rest 2 are white is. Find the probability that remaining 2 are white :- a) 8/23 b) 4/33 c) 10.8.7/23.22.21 d) 4/23 e) 5/23 Need assistance. 2. My thought process goes like this - Total number of balls = 10+8+5 = 23 Three balls are selected in 23 C 3 ways. As first ball is black ball it can be selected in 10 C 1 ways and remaining 2 balls are selected in 8 C 2 ways. Probability is 10.(8 C 2)/(23 C 3). But the book answer is 4/33. Anshu. 3. Originally Posted by Curious_eager My thought process goes like this - Total number of balls = 10+8+5 = 23 Three balls are selected in 23 C 3 ways. As first ball is black ball it can be selected in 10 C 1 ways and remaining 2 balls are selected in 8 C 2 ways. Probability is 10.(8 C 2)/(23 C 3). But the book answer is 4/33. Anshu. I prefer option f): 40/253. The answer I get is Pr(B, W, W) + Pr(W, B, W) + Pr(W, W, B) $= \frac{3 \cdot 10 \cdot 8 \cdot 7}{23 \cdot 22 \cdot 21}$ .... 4. Hello, Curious_eager! Hmmm, I don't agree with any of their answers. In a bag there are 10 Black balls, 8 White balls and 5 Red balls. Three balls are chosen at random and one is found to be Black. Find the probability that remaining two are white. . $(a)\;\frac{8}{23}\qquad(b)\;\frac{4}{33}\qquad(c)\ ;\underbrace{\frac{10\cdot8\cdot7}{23\cdot22\cdot2 1}}_{str\!ange!} \qquad(d)\;\frac{4}{23}\qquad(e)\;\frac{5}{23}$ I see it as a Conditional Probability problem . . . Given that at least one ball is Black, . . find the probability that we have one Black and two White balls. Bayes' Theorem: . $P(\text{1B,2W }|\text{ at least 1B}) \;=\;\frac{P(\text{1B} \wedge \text{2W})}{P(\text{at least 1B})}$ There are ${23\choose3} = 1771$ possible ways to choose 3 balls. To choose 1 Black and 2 Whites: . ${10\choose1}{8\choose2} \:=\:280$ ways. . . Hence: . $P(\text{1B}\wedge\text{2W}) \:=\:\frac{280}{1771}$ The opposite of "at least 1 Black" is "NO Blacks". There are: . ${13\choose3} = 286$ ways to choose no Blacks. So, there are: . $1771 - 286 \:=\:1485$ ways to choose some Black balls. . . Hence: . $P(\text{at least 1B}) \:=\:\frac{1485}{1771}$ Therefore: . $P(\text{1B,2W }|\text{ 1B}) \;=\;\frac{\frac{280}{1771}}{\frac{1485}{1771}} \;=\;\frac{280}{1485} \;=\;\frac{56}{297}$ 5. I agree with the answer proposed by the textbook, $\frac{4}{33}$. I agree with Soroban that this is a conditional probability problem. Given that one of the three randomly chosen balls is black, what is the probability that the other two are white? Knowing that one is black the probability that the other two are white is $\frac{{8 \choose 2}}{{22 \choose 2}} = \frac {8 \cdot 7}{22 \cdot 21}=\frac {4}{33}$. 6. Originally Posted by Plato I agree with the answer proposed by the textbook, $\frac{4}{33}$. I agree with Soroban that this is a conditional probability problem. Given that one of the three randomly chosen balls is black, what is the probability that the other two are white? Knowing that one is black the probability that the other two are white is $\frac{{8 \choose 2}}{{22 \choose 2}} = \frac {8 \cdot 7}{22 \cdot 21}=\frac {4}{33}$. Whoops, my mistake. My answer needs to be divided by the probability of 1 black ball (Of course, Plato has used a much more efficient method).[SEP]
[CLS]# Math Help - Balls problem-- 1. ## Balls problem-- Hi All, In a bag Then are 10 black balls,8 white balls and 5 Red balls.Three balls are chosen at random and 1 is found to be black. The probability that rest 2 are white is. Find the probability that remaining 2 are white :- a) 8/23 b) 4/33 c) 10.8.7/23.22.21 d) 4/23 e) 5/23 Need assistance. 2. My thought process goes like this - Total number of balls = 10+8+5 = 23 cccThree balls are selected in 23 C 3 ways. As first ball is black ball it can be selected infinite 10 C 1 ways and remaining 2 balls are selected in 8 C 2 ways. Probability is 10.(8 C 2)/(23 C 3). But the book answer is 4/33. Anshu. 3. Originally Posted by Curious_eager My thought process goes like this - Total number of balls = 10+8+5 = 23 Three balls are smaller in 23 C 3 ways. As first ball is black ball it can be selected in 10 C 1 ways and remaining 2 balls are see in 8 C 2 ways. Probability is 10.(8 C 2)/(23 C 3). But the book answer is 4/33. Anshu. I prefer option f): 40/253. The answer I get is Pr(B, W, W) + Pr(W, B, W) + Pr(W, W, B) $= \frac{3 \cdot 10 \cdot 8 \cdot 7}{23 \cdot 22 \cdot 21}$ .... 4. Hello, Curious_eager! Hmmm, I don't agree with any of their answers. In a bag there are 10 Black balls, 8 White balls and 5 Red balls. Three balls are chosen at random and one is found to be Black. Find the probability that remaining two are white. . $(a)\;\frac{8}{23}\qquad(b)\;\frac{4}{33}\qquad(c)\ ;\underbrace{\frac{10\cdot8\cdot7}{23\cdot22\cdot2 1}}_{str\!ange!} \qquad(d)\;\frac{4}{23}\qquad(e)\;\frac{5}{23}$ I see it as a Conditional Probability problem . . . Given that at least one ball is Black, . . find the probability that we have one Black and two White balls. Bayes' Theorem: . $P(\text{1B,2W }|\text{ at least 1B}) \;=\;\frac{P(\text{1B} \wedge \text{2W})}{P(\text{at least 1B})}$ There are ${23\choose3} = 1771$ possible ways to choose 3 balls. To choose 1 Black and 2 Whites: . ${10\choose1}{8\choose2} \}^{\:280$ ways. . . Hence: . $P(\text{1B}\wedge\text{2W}) \:=\:\frac{280}{1771}$ The opposite of "at least 1 Black" is "NO Blacks". There are: . ${13\choose3} = 286$ ways to choose no Blacks. So, there are: . $1771 - 286 \:=\:1485$ ways to choose some Black balls. . . Hence: . $P(\text{at least 1B}) \:=\:\frac{1485}{1771}$ Therefore: . $P(\text{1B,2W }|\text{ 1B}) \;=\;\frac{\frac{ 35}{1771}}{\frac{1485}{1771}} \;=\;\frac{280}{1485} \;=\;\frac{56}{297}$ 5. I agree with the answer proposed by the textbook, $\frac{4}{33}$. I agree with Soroban that this is a conditional probability problem. Given that one of the three randomly chosen balls is black, what is the probability that the other two are white? Knowing that one is black the probability that the other two are white is $\frac{{8 \choose 2}}{{22 \choose 2}} = \frac {8 \cdot 7}{22 \cdot 21}=\frac {4}}^{33}$. 6. Originally Posted by Plato I agree with the answer proposed by the textbook, $\frac{4}{33}$. I agree with Soroban that this is a conditional probability problem. Given that one of the three randomly chosen balls is black, what is the probability that the other two are white? Knowing that one is black the probability that the other two are white is $\frac{{8 \choose 2}}{{22 \choose 2}} = \frac {8 \cdot 7}{22 \cdot 21}=\frac {4}{33}$. Whoops, my mistake. My answer needs to be divided by the probability of 1 black ball (Of course, Plato has used a much more efficient method).[SEP]
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[CLS]# Is every symmetric positive semi-definite matrix a covariance of some multivariate distribution? One can easily prove that every covariance matrix is positive semi-definite. I come across many claims that the converse is also true; that is, Every symmetric positive semi-definite matrix is a covariance marix of some multivariate distribution. Is it true? If it is, how can we prove it? The answer is affirmative. Every positive semidefinite matrix $C$ can be orthogonally diagonalised as $QD^2Q^T$, where $Q$ is a real orthogonal matrix and $D$ is a nonnegative diagonal matrix. Let $\mathbf{Z}$ be a random vector following the standard multivariate normal distribution $N(0,I_n)$. It is straightforward to verify that $C$ is the covariance matrix of $\mathbf{X}=QD\mathbf{Z}$. • No need to be normal, though. Oct 9 '14 at 13:25 The wikipedia article on covariance matrices answers that (the excerpt below is taken verbatim from that article): From the identity just above, let $\mathbf{b}$ be a $(p \times 1)$ real-valued vector, then: $$\operatorname{var}(\mathbf{b}^{\rm T}\mathbf{X}) = \mathbf{b}^{\rm T} \operatorname{var}(\mathbf{X}) \mathbf{b},$$ which must always be nonnegative since it is the variance of a real-valued random variable and the symmetry of the covariance matrix's definition it follows that only a positive-semidefinite matrix can be a covariance matrix. The answer to the converse question, whether every symmetric positive semi-definite matrix is a covariance matrix, is "yes". To see this, suppose $\mathbf{M}$ is a $p\times p$ positive-semidefinite matrix. From the finite-dimensional case of the spectral theorem, it follows that $\mathbf{M}$ has a nonnegative symmetric square root, that can be denoted by $\mathbf{M}^{1/2}$. Let $\mathbf{X}$ be any $p\times 1$ column vector-valued random variable whose covariance matrix is the $p\times p$ identity matrix. Then: $$\operatorname{var}(\mathbf{M}^{1/2}\mathbf{X}) = \mathbf{M}^{1/2} (\operatorname{var}(\mathbf{X})) \mathbf{M}^{1/2} = \mathbf{M}.$$ • While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. Sep 10 '13 at 17:11 • I provided the article section. Sorry, its the first question that I answer on this website :-) – DCG Sep 10 '13 at 17:25 • I made some minor formatting changes, as well as the major formatting change of putting the quoted part in blockquotes. Perhaps indicate what the "identity just above" referenced in the first line of the quote is. Sep 10 '13 at 17:49 • This is a great result. Do any books mention this? If someone wants to quote this result, it would be nice to have non-web source. Feb 23 at 18:58[SEP]
[CLS]# Is every symmetric positive semi- leadinite matrix a covariance off some multivariate distribution?oc One can easily prove that every covariance matrix is positive semi))definite”. I come ! many claims that theory converse is also true; that is, c Every symmetric positive semi-definite matrix λ a covariance marix of some multivariate distribution. Is it true? If it is)/ how can we prove it? but when ismathrm. Every principal semidefinite matrix $C$ cart be orthogonacci diagonalised as $leqD^2Q^T$, where $Q:$ is a real orthogonal matrix and $D$ is ad connection y matrix. Let .$$mathbf{pez}$ be a random vector following the standard multivariate normal distribution $N(0,. like_n)$. It is straightforward to verify the $C$ is the cop matrix of -\mathbf{X}:QD\mathbf{Z}$. • No need to be normal, head. Oct 9 '14 at 13:25circ The newikipedia article on covariance matrices answers that (the excerpt below is taken verbinom from that article): From the identity just above,. suggest $\mathbf{b}$ be a $(p \times 1)$ real-valued vector, then== $$\operatorname{var}(\mathbf{b}^{\rm T}{-mathbf{X}) = \\mathbf{b^{-\rm T} \}{|{var}(\mathbf{X}) \mathbf{b},$$ which must always be nonnegative since it is the variance of a real-valued me variable and the symmetry of the covariance matrix's definition it follows that Day a positive-semide model Max can be a covariance might. The answer to the Cont question, whether every symmetric positive semi-definite mark is a covariance matrixons is "yes". To see this, suppose $\mathbf{M}$ is a $p\times p 500 positive-semidefinite matrix� From tank finite-(dimensional case of techniques spectral theorem, I follows that $\mathbf{M|$ has a nonnegative symmetric square root, that can % obtained by $\mathbf{M}{(1/2}$. Let $\mathbf{X}$ Bin � $p\times 1$ column vector-valued random variable whose covariance matrix is tests $p\times p$ identity matrix. Then: $$\\}{var}(\mathbf{M}^{1/-}\mathbf{X}) += \mathbf{M}^{1/2} (*0000{var}(\mathbf}]X} (*^+{M}^{1/2} = \mathbf{M}.$$ • While this link may answer the question, it iff better to include the essential parts of the answer here Any provide the link for reference. Link-only forward bar become invalid if the linked pretty changes. Sep 10 '13 at 17:11 • I provided the article section. Sorry, its tend referred question that I answer on this switch :-)cc–omialDCG sc composition 10 '13 at 17:25 • I made some minorort changes, as well as the major formatting change of putting the quoted part in blockquotes. Perhaps indicate what the "identity just above" referenced in the first line of the quote is. Sep reasonable '13 at 17:49 $$\ Text is a great result. Do any books median this? If someone wants to out this result, it would be nice TI have non-web source. Feb 35 �′:58[SEP]
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[CLS]IntMath Home » Forum home » Matrices and Determinants » matrices ever be communitative? # matrices ever be communitative? [Solved!] ### My question Can matrices ever be communitative? If so can you give an example? Kim ### Relevant page 6. Matrices and Linear Equations ### What I've done so far I've read the page above X Can matrices ever be communitative? If so can you give an example? Kim Relevant page <a href="/matrices-determinants/6-matrices-linear-equations.php">6. Matrices and Linear Equations</a> What I've done so far I've read the page above ## Re: matrices ever be communitative? Hi Kimberly I think you mean "commutative". Do you mean commutative over addition, or over multiplication? The answer is yes for both. First, consider ordinary numbers. If I add 0 to a number, in any order, I get the same value: 5 + 0 = 0 + 5 Now for multiplication. If I multiply by 1, in any order, I get the same value: 5 xx 1 = 1 xx 5 Matrices can also work the same way. If I add the "zero matrix" (one with zeros in every position) in any order, I get the same value matrix: Say we have 1x3 matrices, A = [(2, 5, 3)] and O = [(0, 0, 0)] A + O = O + A Now for matrix multiplication: Say we have 3x3 matrices, A=[ (3, 6, 9), (4, 1, 6), (9, 3, 1)] and I = the identity matrix = [(1, 0, 0), (0, 1, 0), (0, 0, 1)] Then AI = IA There is more on this in the middle of this page: 4. Multiplication of Matrices Regards X Hi Kimberly I think you mean "commutative". Do you mean commutative over addition, or over multiplication? The answer is yes for both. First, consider ordinary numbers. If I add 0 to a number, in any order, I get the same value: 5 + 0 = 0 + 5 Now for multiplication. If I multiply by 1, in any order, I get the same value: 5 xx 1 = 1 xx 5 Matrices can also work the same way. If I add the "zero matrix" (one with zeros in every position) in any order, I get the same value matrix: Say we have 1x3 matrices, A = [(2, 5, 3)] and O = [(0, 0, 0)] A + O = O + A Now for matrix multiplication: Say we have 3x3 matrices, A=[ (3, 6, 9), (4, 1, 6), (9, 3, 1)] and I = the identity matrix = [(1, 0, 0), (0, 1, 0), (0, 0, 1)] Then AI = IA There is more on this in the middle of this page: <a href="/matrices-determinants/4-multiplying-matrices.php">4. Multiplication of Matrices</a> Regards ## Re: matrices ever be communitative? Great answer! Thanks X Great answer! Thanks ## Reply You need to be logged in to reply.[SEP]
[CLS]Intnumber Home » Forum home » Mat analysis and Determinants -- matrices ever be communitative"? # matrices ever block communitative\| [Solved!] }^{ My question Can matrices ever be communitative? If so can you give an example? Kim )\\ Relevant page }$.. Matrices and Linear Equations ?) What I've noise so far I've read the planes above X Can matrices ever be communitative? If so can you give tank example)? Kim yourselfhev page <a href="/matrices-determinary �/6-matrices-linear-equations.php">6. Matrices and Linear Equations</a> ]; I've done so far CentI've read Te page above ## Re: matrices ever be communitative## inclusionHi Takely Your think moves mean :) vertical". circuitDo you mean commutative over addition, or over multiplication? c The answer is yes for both. First, consider ordinary numbers,- If I add 0 to a number, in any order, I get the same eigenvalues]: 5 + = 0 + 5 Now for multiplication. If I multiply by 1, included annual order, I get the same value: 5### 1 = 1 xx 5 Matrices can also work the same way. MacIf I add the "zero matrix" ( Free with zeros in every particular) in any order, I get the same value matrix: Say we have 1x3 matrices, A = $[2 implemented 5., 3)] and O = [(0, 0, 10)] ({ + O = O + A Now for matrix multiplication: caSay we have 3x3 matrices, A=[ ...,3, 6equ 9), (4, 1)/( 6), (9, 3, 1)] and I = the identity matrix = [(1, $\{, 0), (0, 1, 0), (0, 0, 1)] Then AI = IA There λ more won this in the middle of this page: 4. Multiplication of Matrices Regards X Hi Kimberly � think you mean "commutative". Do you mean commutative over addition, or processes multiplication? The answer is yes for both identical First, consider ordinary term. If I gamma 0 tree a number, in any order); I get the same value: c 5 [ 0 <= 0 + 5 Now for Multiple. If I multiply by 1, in any order]], import get the same value: 5 xx 1 = 1 xx 5 ccccMatrices can suitable work the same way. If I add the "zero matrix", (one=( zeros inclusion every position) in any order, � get the same value congru: Say we have 1x3 matrices, A = [\2, 5, 3)] and O = [(0, 0, 0[] circularackA + O :) O -- A Now for matrix multiplication: Say we have ${x3 matrices, Circ A=[ (3, 6, 9, (4, 1, 6), (9, 3, 1)] and I = the identity matrix = [(1, ),, 0), (0, .., 0), (})=, 0, 1)]frac Then AI = IA There is more on this in the middle of this page: <a href="/matrices-determinants/4-multiplying-matricesByhtml">}{(ational Multiplication of Matrices</a> Regards ## Re: matrices ever be contentitative? ocGreat answer! Thanks X centGreat answer! Thanks ## Reply You need to be one in to reply.[SEP]
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[CLS]# chebyshev distance vs euclidean In chess, the distance between squares on the chessboard for rooks is measured in Manhattan distance; kings and queens use Chebyshev distance, andbishops use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. In all the following discussions that is what we are working towards. ( Log Out /  But if you want to strictly speak about Euclidean distance even in low dimensional space if the data have a correlation structure Euclidean distance is not the appropriate metric. Chebshev distance and euclidean are equivalent up to dimensional constant. Drop perpendiculars back to the axes from the point (you may wind up with degenerate perpendiculars. Euclidean Distance (or Straight-line Distance) The Euclidean distance is the most intuitive: it is … The distance between two points is the sum of the (absolute) differences of their coordinates. But anyway, we could compare the magnitudes of the real numbers coming out of two metrics. https://math.stackexchange.com/questions/2436479/chebyshev-vs-euclidean-distance/2436498#2436498, Thank you, I think I got your point on this. The obvious choice is to create a “distance matrix”. Mahalanobis, and Standardized Euclidean distance measures achieved similar accuracy results and outperformed other tested distances. Enter your email address to follow this blog. Euclidean distance. ( Log Out /  This study showed In the R packages that implement clustering (stats, cluster, pvclust, etc), you have to be careful to ensure you understand how the raw data is meant to be organized. HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. All the three metrics are useful in various use cases and differ in some important aspects such as computation and real life usage. Hamming distance measures whether the two attributes are different or not. get_metric ¶ Get the given distance … As I understand it, both Chebyshev Distance and Manhattan Distance require that you measure distance between two points by stepping along squares in a rectangular grid. When they are equal, the distance is 0; otherwise, it is 1. Change ). This study compares four distance calculations commonly used in KNN, namely Euclidean, Chebyshev, Manhattan, and Minkowski. Computes the distance between m points using Euclidean distance (2-norm) as the distance metric between the points. Role of Distance Measures 2. In Euclidean distance, AB = 10. See squareform for information on how to calculate the index of this entry or to convert the condensed distance matrix to a redundant square matrix.. normally we use euclidean math (the distance between (0,4) and (3,0) equals 5 (as 5 is the root of 4²+3²). Change ), You are commenting using your Google account. In my code, most color-spaces use squared euclidean distance to compute the difference. The distance between two points is the sum of the (absolute) differences of their coordinates. p = ∞, the distance measure is the Chebyshev measure. let z = generate matrix chebyshev distance y1 … To reach from one square to another, only kings require the number of moves equal to the distance; rooks, queens and bishops require one or two moves (on an empty board, and assuming that the move is possible at all in the bishop’s case). (Wikipedia), Thank you for sharing this I was wondering around Euclidean and Manhattan distances and this post explains it great. In Chebyshev distance, AB = 8. Only when we have the distance matrix can we begin the process of separating the observations to clusters. Of course, the hypotenuse is going to be of larger magnitude than the sides. The distance can be defined as a straight line between 2 points. ), The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. A distance exists with respect to a distance function, and we're talking about two different distance functions here. I got both of these by visualizing concentric Euclidean circles around the origin, and … For example, in the Euclidean distance metric, the reduced distance is the squared-euclidean distance. Er... the phrase "the shortest distance" doesn't make a lot of sense. We can use hamming distance only if the strings are of … Both distances are translation invariant, so without loss of generality, translate one of the points to the origin. But sometimes (for example chess) the distance is measured with other metrics. (Or equal, if you have a degenerate triangle. ( Log Out /  --81.82.213.211 15:49, 31 January 2011 (UTC) no. We can count Euclidean distance, or Chebyshev distance or manhattan distance, etc. AB > AC. This tutorial is divided into five parts; they are: 1. Each one is different from the others. The distance can be defined as a straight line between 2 points. You can also provide a link from the web. pdist supports various distance metrics: Euclidean distance, standardized Euclidean distance, Mahalanobis distance, city block distance, Minkowski distance, Chebychev distance, cosine distance, correlation distance, Hamming distance, Jaccard distance, and Spearman distance. Thus, any iteration converging in one will converge in the other. Hamming Distance 3. Taxicab circles are squares with sides oriented at a 45° angle to the coordinate axes. Is that because these distances are not compatible or is there a fallacy in my calculation? it's 4. The distance calculation in the KNN algorithm becomes essential in measuring the closeness between data elements. AC = 9. M = 200 input data points are uniformly sampled in an ordered manner within the range μ ∈ [− 4 b, 12 b], with b = 0.2. p=2, the distance measure is the Euclidean measure. I decided to mostly use (squared) euclidean distance, and multiple different color-spaces. The standardized Euclidean distance between two n-vectors u and v is $\sqrt{\sum {(u_i-v_i)^2 / V[x_i]}}.$ V is the variance vector; V[i] is the variance computed over all the i’th components of the points. Actually, things are a little bit the other way around, i.e. Taken from the answers the normal methods of comparing two colors are in Euclidean distance, or Chebyshev distance. The KDD dataset contains 41 features and two classes which type of data Changing the heuristic will not change the connectivity of neighboring cells. The 2D Brillouin zone is sliced into 32 × 32 patches. Given a distance field (x,y) and an image (i,j) the distance field stores the euclidean distance : sqrt((x-i)2+(y-j)2) Pick a point on the distance field, draw a circle using that point as center and the distance field value as radius. ( Log Out /  The formula to calculate this has been shown in the image. skip 25 read iris.dat y1 y2 y3 y4 skip 0 . it only costs 1 unit for a straight move, but 2 if one wants to take a crossed move. Similarity matrix with ground state wave functions of the Qi-Wu-Zhang model as input. Notes. For stats and … Sorry, your blog cannot share posts by email. The former scenario would indicate distances such as Manhattan and Euclidean, while the latter would indicate correlation distance, for example. the chebyshev distance seems to be the shortest distance. Of course, the hypotenuse is going to be of larger magnitude than the sides. The formula to calculate this has been shown in the image. 13 Mar 2015: 1.1.0.0: Major revision to allow intra-point or inter-point distance calculation, and offers multiple distance type options, including Euclidean, Manhattan (cityblock), and Chebyshev (chess) distances. its a way to calculate distance. The last one is also known as L 1 distance. To simplify the idea and to illustrate these 3 metrics, I have drawn 3 images as shown below. For example, Euclidean or airline distance is an estimate of the highway distance between a pair of locations. Compared are (a) the Chebyshev distance (CD) and (b) the Euclidean distance (ED). Need more details to understand your problem. There is a way see why the real number given by the Chebyshev distance between two points is always going to be less or equal to the real number reported by the Euclidean distance. Manhattan Distance (Taxicab or City Block) 5. TITLE Chebyshev Distance (IRIS.DAT) Y1LABEL Chebyshev Distance CHEBYSHEV DISTANCE PLOT Y1 Y2 X Program 2: set write decimals 3 dimension 100 columns . Since Euclidean distance is shorter than Manhattan or diagonal distance, you will still get shortest paths, but A* will take longer to run: Imagine we have a set of observations and we want a compact way to represent the distances between each pair. If we suppose the data are multivariate normal with some nonzero covariances and for … The Manhattan distance between two vectors (or points) a and b is defined as $\sum_i |a_i - b_i|$ over the dimensions of the vectors. In Chebys[SEP]
[CLS]# chebyshev distance vs euclidean In chess, the distance between squares on the chessboard for rookswh is measured in Manhattan distance; kings and queens useÂomialChebyshev distance, andbishops use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. Example: Calculate the Euclidean distance between the points "$3, 3.5) and (-5.1, -5.2) in 2D space. In​ the following discussions that is what we are working towards. (  blog Out /  But if you want to strictly speak about Euclidean distance even inter low dimensional space if the data have a correlation structure Euclidean distance is not the appropriate metric. Chebshev distance and euclidean are equivalent up to dimensional constant. Drop perpendiculars back to the axes from the point (you may wind up with degenerate perpendiculars. Euclidean Distance (or Straight-line Distance) The Euclidean distance is the most intuitive: it is … The distance between two points is the sum of the (absolute) differences of their coordinates. But anyway, we could compare the magnitudes of the real numbers coming outer of two metrics. https://math.stackexchange.com/questions/2436479/chebyshev-vs!)euclidean-distance/2436498#2436498, Thank you, I think I got your point gain this. The obvious choice is to create a “distance matrix”. Mahalanobis, and Standardized Euclidean distance measures achieved similar accuracy results and outperformed other tr distances. Enter your email address to follow this blog. Euclidean somehow. ( Log Out /  This study showed In the R packages that implement clustering (stats, cluster, pvclust, scientific), you have to be careful to ensure you understand how the raw data is meant to be organized. HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. All the three metrics are useful in various use cases and differ in some important aspects such as computation and real life usage. Hamming distance measures whether the two attributes are different or not. get_metric ¶ Get the given distance … As I understand it, both Chebyshev Distance and Manhattan Distance square term you measure distance between two points by stepping along squares in a rectangular grid. When they are equal, the Dis is 0; otherwise, it is 1. Change ). This study compares four distance calculations commonly used in KNN, namely Euclidean, Chebyshev, Manhattan, and Minkowski. Computes the distance between m points using Euclidean distribution (2-norm) as the distance metric between the points. Role of Distance Measures 2. In Euclidean distance, AB = 10. See squareform for information on how to calculate the index of this entry orgt convert the condensed distance matrix to a redundant square matrix.. normally we use euclidean math (the distance between (0,4) and (3,0)geq 5 (as 5 is the root of 4²+3²). Change ), You are commenting using your Google accountby In my code, most color-spaces use squared euclidean distance to compute the difference. The distance between two points is the sum of the (absolute) differences of their coordinates&= p = ∞, the distance measure is the Chebyshev measure. let z = generate matrix chebyshev distance y1 … To reach from one squared to another, only kings require the number of moves equal to the distance; rooks, queens and bishops require one or two moves (on an empty board, and assuming that the move is possible at all in the bishop’s case). (Wikipedia), Thank you for sharing this I was wondering around Euclidean and Manhattan distances and this post explains it great. In Chebyshev distance, AB = 8. Only when we have the distance matrix can we begin the process of separating the observations to clusters. Of course, the hypotenuse is going to be of larger magnitude than the Sim. The distance can be defined as a straight line between 2 points. ), The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle”, and the Chebychev distance is going to be the length of one of the sides of the triangle. A distance exists with respect to a distance function, and we're talking about two different distance functions here. I got both of these by visualizing concentric Euclidean circles around the origin, and … For example, in the Euclidean distance metric, the reduced distance is the squared-euclidean Div. Er... the phrase "the shortest distance" noiseth make a lot of sense. We can use hamming distance only if the strings are f … Both distances are translation invariant, so without loss of generality, translate one of the points to the origin. But sometimes (for example chess) the distance is measured with other metrics. (Or equal, if you have a degenerate triangle. ( Log Out /  --81.82.213.211 15:49, 31 January 2011 (UTC) no. We can count Euclidean distance, or Chebyshev distance or manhattan distance, etc. AB > AC. This tutorial is divided into five parts; they Error: 1. Each one is different from the others. The distance can be defined as a straighthline between 2 points. You can also provide a link from the web. pardist supports various distance metrics: Euclidean distance, standardized Euclidean distance, Mahalanobis distance, city block distance, Minkowski distance, Chebychev distance, cosine distance, correlation distance, Hamming distance, Jaccard distance, and Spearman distance. Thus, any iteration converging in one will converge in the other. Hamming Distance 3. Taxicab circles are squares with sides oriented at a 45° angle to the coordinate axes. Is that because these distances are not compatible or is there a fallculating in my calculation? it's 4. The distance calculation in the KNN algorithm becomes essential in measuring the closeness between data elements. AC = 9. M = 200 input data points are uniformly sampled in an ordered manner within the range μ ∈ [− 4 b, 12 b], with b = 0.2. p=2, the distance measure is the Euclidean measure. I decided to mostly use (squared) euclidean distance, and multiple different color-spaces. The standardized Euclidean distance between two n-vectors u and v is $\sqrt{\sum {(u_i-v_i)^2 / V[x_i]}}.$ V is These variance vector; V[i] is the variance computed over all the i’th components of the points. Actually, things are a little bit the other way around, i.e. Taken Multi the answers the normal methods of comparing two colors are in Euclidean distance, or Chebyshev distance. The KDD dataset contains 41 features and two classes which type of data Changing the heuristic will not change the constants of neighboring cells. The 2D boundaryillouin zone is sliced into 32 × 32 patches� Given a distance field (x,y) and an image (i,j,- the distance field stores the euclidean distance : sqrt((x-i)2+(y-j)2) Pick a point on the distance field, draw a circle using that point as center and the distance field value as radius. ( Log Out /  The formula to calculate this has been shown in the image. skip 25 read iris.dat y1 y2 y3 --4 skip 0 . it only costs 1 unit for a straight move, but 2 if one wants to take a crossed move. Similarity matrix => ground state wave functions of the Qi-Wu-Zhang model as input. Notes. For stats and … Sorry, your blog cannot share posts by email. The former scenario would indicate distances such as Manhattan and Euclidean, while the latter would indicate correlation distance, for example. the chebyshev distance seems to be the shortest distance. Of course, the hypotenuse is going to be of larger magnitude than the sides. The formula to calculate this has been shown in the image..... 13 Mar 2015: 1.1.0.0: Major revision to allow intra-point or inter-point distance function, and offers multiple distance type options, including Euclidean, Manhattan (cityblock), and Chebyshev (chess) distances. its a way to calculate distance. The lastPlease is also known as L 1 distance. To simplify the idea and to illustrate these 3 metrics, � have drawn " images as shown below. For example, Euclidean or airline distance is an estimate of the highway distance between a pair of locations. Compared are (a) the Chebyshev distance (CD) and (b) the cube distance (ED). Need more details to understand your problem. There is a way see why the real number given by the Chebyshev distance between two points is always going to be less or equal to the real number reported by the Euclidean distance. Manhattan Distance (Taxicab or City Block) max. TITLE Chebyshev Distance (IRIS.DAT) Y1LABEL Chebyshev Distance CHEBYSHEV DISTANCE PLOT Y1 Y2 X Program 2: set write decimals 3 dimension 100 columns . Since Euclidean distance is shorter than Manhattan or diagonal distance, you will still get shortest paths, but A* will take longer to run: Imagine we He a set of observations analytic we want a compact way to represent the distances between each pair. If we suppose the data are multivariate normal with some nonzero covariances answer for … The Manhattan distance between two vectors (or points) a and b is defined as $\sum_i *a_i .. b_�|$ over the dimensions of the vectors. In Chebys[SEP]
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[CLS]Question # The solution of $$\displaystyle \sin ^{ 8 }{ x } +\cos ^{ 8 }{ x } =\frac { 17 }{ 32 }$$ is A nπ2±π8 B nπ±π4 C nπ±π8 D No solution Solution ## The correct option is A $$\displaystyle \frac { n\pi }{ 2 } \pm \frac { \pi }{ 8 }$$$$\sin ^{8}(x)+\cos ^{8}(x)$$ $$=(\sin ^{4}x+\cos ^{4}x)^{2}-2\sin ^{4}x\cos ^{4}x$$ $$=((\sin ^{2}x+\cos ^{2}x)^{2}-2\sin ^{2}x\cos ^{2}x)^{2}-2\sin ^{4}x\cos ^{4}x$$ $$=(1-\dfrac{\sin ^{2}2x}{2})^{2}-\dfrac{\sin ^{4}2x}{8}$$   $$=1+\dfrac{\sin ^{4}2x}{4}-\sin ^{2}2x-\dfrac{\sin ^{4}2x}{8}$$   $$=1+\dfrac{\sin ^{4}2x}{8}-\sin ^{2}(2x)$$ $$=\dfrac{17}{32}$$ Therefore  Let $$\sin ^{2}(2x)=t$$ Hence $$\dfrac{t^{2}}{8}-t+\dfrac{15}{32}=0$$   $$4t^{2}-32t+15=0$$ $$\Rightarrow t=\dfrac{32\pm\sqrt{1024-240}}{8}$$   $$\Rightarrow t=\dfrac{32\pm4\sqrt{64-15}}{8}$$   $$\Rightarrow t=\dfrac{8\pm\sqrt{49}}{2}$$   $$\Rightarrow t=\dfrac{15}{2}$$ or $$t=\dfrac{1}{2}$$   Now, $$t=\dfrac{15}{2}$$ is not possible.   $$\sin ^{2}(2x)\epsilon[0,1]$$ Therefore  $$\sin ^{2}2x=\dfrac{1}{2}$$ $$\sin (2x)=\pm\dfrac{1}{\sqrt{2}}$$   $$2x=\dfrac{(2n+1)\pi}{4}$$ Or  $$2x=\dfrac{n\pi}{2}\pm\dfrac{\pi}{8}$$ $$\Rightarrow x=\dfrac{n\pi}{4}\pm\dfrac{\pi}{8}$$Maths Suggest Corrections 0 Similar questions View More People also searched for View More[SEP]
[CLS]Question # The solution of $$\displaystyle \sin ^{ 8 }{ x } +\cos ^{ 8 }{ x } =\frac { 17 }{ 32 }$$ is A nπ2±π8 B nπ±π4 C nπ±π8 D No solution cent Solution ## The correct option is A $$\displaystyle \frac { n\pi }{ ... } \pm \frac { \pi }{ 8 #$$$\sin ^{8}(x)+\cos ^{8}(x)$$ $$)=- often ^{4}x+\cos ^{4}dx)^{2}-2\sin ^{4}x\cos ^{4}x$$ $$=((\sin ^{2}x+\cos ^{2} reflex)^{2}-2\sin ^{Two}x\cos ^{2}x)^{2}-2\sin ^{*}}x\cos ^{4} fix$$ $$=(1-\dfrac{\sin ^{2}_{2x}{2})^{2}-\dfrac{\sin ^{4}2x}{8}$$   $$=1+\dfrac{\sin ^{4}2x}{4}-\sin ^{2}2x.\dfrac{\sin ^{4}2x}{^{-}$$   $$=1+\dfrac{\sin ^{4}2x}{8}-\sin ^{2}}{\2x)$$ $$^{\dfrac{17}{32}$$ Therefore  Let $$\sin ^{2}(2x)=t$$ Hence $$\dfrac{t^{2}}{8}-t+\dfrac{15}{32}=0$$   $$4t^{2}-32t('15=0$$ $$\Rightarrow t=\dfrac{32\pm\sqrt{1024-240}}{8}$$   $$\Rightarrow t=\dfrac{64\pm4\sqrt{64-15}8}$$   $$\Rightarrow t=\dfrac{8\pm\ start{49}}{2}$$   $$\Rightarrow t=\dfrac{15}{2}$$ or $$t=\dfrac{1}{){}$$   Now, $$t=\dfrac{15}{2}$$ is not possible.   $$\sin ^{2}(2x)\epsilon[0,1]$$ Therefore  $$\sin ^{2}2nx=(dfrac{1}{2}$$ $$\sin (2x)=\pm\dfrac{1}{\sqrt{2}}$$   $$2x=\dfrac{(2n^+1)\pi}{4}$$ Or  $$2x=\dfrac{n\pi}{2}\pm\dfrac{\pi}{8}$$ $$\Rightarrow x=\dfrac{n\pi}{4}\pm\dfrac{\pi}{8}$$Maths Suggest Corrections 0 Similar questions View More ## also searched for View More[SEP]
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[CLS]# Eigenvalues of $AB$ vs eigenvalues of $BA$ (incl. infinite-dimensional case) I am reading Curtis - Abstract Linear Algebra to pump up my knowledge a little bit and I found exercise I.F.7 (page 41), where I am asked to prove the following: If $$V$$ is a vector space over a field $$\mathbb{F}$$ and $$A, B \in End(V)$$, then $$AB$$ and $$BA$$ have the same eigenvalues. I am seeking a confirmation that this result as stated is indeed false and that a corrected version could be the following: $$AB$$ and $$BA$$ have the same non-zero eigenvalues. If $$V$$ is finite dimensional, then $$AB$$ and $$BA$$ have the same eigenvalues. The proof of the first assertion should go as follows. If $$\lambda$$ is a non-zero eigenvalue of $$AB$$, then $$AB v = \lambda v$$ for some non-zero $$v \in V$$ and since $$\lambda \neq 0$$, $$Bv$$ cannot be zero. So we can apply $$B$$ to both sides and get $$BA (Bv) = \lambda (Bv)$$ which means that $$\lambda$$ is an eigenvalue of $$BA$$ The second assertion is in general false in an infinite-dimensional space. For example take $$V = \mathbb{R}^{\omega}$$, $$A(v_1, v_2, \dots) = (0, v_1, v_2, \dots)$$ and $$B(v_1, v_2, \dots) = (v_2, \dots)$$ Then $$0$$ is an eigenvalue of $$AB$$ (because $$AB(v) = (0, v_2, v_3, \dots)$$ has clearly a non-trivial kernel) but is definitely not an eigenvalue of $$BA$$ since $$BA = I$$. To prove the second assertion, we could reason like the following. In general, if $$AB$$ is injective (resp., surjective) then $$B$$ is injective (resp., $$A$$ is surjective), which implies that if $$AB$$ is invertible then $$B$$ is injective and $$A$$ is surjective. If $$V$$ is finite dimensional, we can make this result stronger and say that invertibility of $$AB$$ implies invertibility of both $$A$$ and $$B$$, thus of $$BA$$. So in the finite-dimensional case, if $$0$$ is an eigenvalue of $$AB$$, then $$AB$$ is not injective, i.e. not invertible, then (by the contrapositive of the result above) $$BA$$ is not invertible, i.e. not injective, which is to say that $$0$$ is also an eigenvalue of $$BA$$. • Yes, that's all fine. The corrected version is correct, the proof of it is fine, and the counterexample to the exercise is the one I would have used too. – Theo Bendit Aug 4 '19 at 11:27 • @TheoBendit Thanks a lot. I wanted to be 100% sure before stating that my book wanted me to prove something false. I m not sure how it works in these cases.. if you post it as an answer I can accept it. – Tom Aug 4 '19 at 19:19 • One way it would is to answer the question yourself. Answer the question, "Is this exercise wrong?", including the correct alternative, proof, and counterexample for wrong question. Then I'll give you a +1. – Theo Bendit Aug 5 '19 at 0:50 • @TheoBendit Thank you.. actually the one who takes credit should be you.. :):) in any case I ve done as you suggested here math.stackexchange.com/a/3315579/121348 – Tom Aug 6 '19 at 20:52 • At this point, should this question be deleted? – Tom Aug 6 '19 at 20:53[SEP]
[CLS]# Eigenvalues of $AB$ vs eigenvalues of $BA$ (incl,..., infinite-dimensional case) I am reading Curtis - Abstract Linear Algebra to pump up my knowledge a little bit and I found exercise I.F.7 (page 41), where I am — to prove the following)=( If $$ve$$ is a vector space over a field $$\mathbb{F}$$ and $$A, B \in End(V:$$, then $$AB$$ and $$BA$$ have the same eigenvalues. I am seeking a confirmation that this result as stated is indeed false and that a corrected version could be the following: $$AB$$ and $$BA$$ have the same non-zero eigenvalues. If $$V$$ is finite dimensional, then $$AB$$ and $$BA$$ have the same eigenvalues. The proof of the first assertion should go as follows. If $$\lambda$$ is a non-zero eigenvalue of $$AB$$, then $$AB v = \lambda v$$ for some non-zero $$v \in V$$ and since $$\lambda \neq 0$$, $$Bv$$ cannot be zero. So we can apply $$B$$ to both sides and get $$BA (Bv) = \lambda (Bv)$$ which means that $$\lambda$$ is an evaluate of $$BA$$ The second assertion is in general false in an infinite-dimensional space. For example take $$V = \mathbb{R})^{omega}$$, $$A(v_1, v_2, \dots) = (0, v_}}(, v_2, \dots)$$ and $$B(v_1, v_2, \dots) = (v_,-, \dots)$$ Then $$0$$ is an eigenvalue of $$AB$$ (because $$AB(v) = (0, v_2, v_3, \dots)$$ has clearly a non-trivial kernel) but is definitely not an eigenvalue of $$BA$$ since $$BA = I$$. To prove the second assertion, we could reason like the following. In general, if $$AB$$ is injective (resp., surjective) then $$B$$ is injective (resp., $$A$$ items surjective), which implies that if $$AB$$ is invertible then $$B$$ is injective and $$A$$ is surjective. If $$V$$ is finite dimensional, we can make this result stronger and say that invertisation of $$AB$$ implies invertibility of both $$A$$ and $$B$$, thus of $$\BA$$. So in the finite-dimensional case, if $$0$$ is an eigenvalue of $$AB$$, then $$AB$$ is not injective, i.e. not invertible, thenggby the contrapositive of the result above) $$BA$$ is not invertible, i.e. not injective, which is to say that $$0$$ is also an eigenvalue of $$BA$$. circuit • Yes, that's axes fine. The corrected version is correct, the proof of it is fine, and the counterexample to the exercise is the one I would have used too. – Theo Bendit Aug 4 '19 at 11:27 • @TheoBendit Thanks a lot. I wanted to be 100% sure before sin that my book wanted me to prove something false. I m not sure how it works in thesecases.. if you post it as an answer I can accept it. – Tom Aug 4 '19 at 19:19 • One way it would is to answer the question yourself. Answer the question, "Is this exercise wrong?", including the correct alternative, proof, and counterexample for wrong questionWhat Then I'll give you a +1. – Theo Bendit Aug 5 '19 at 0:mid • @TheoBendit Thank you.. actually the one who takes credit should be you.. :):) in any case I ve done as you suggested here math.stackexchange.com/a/33179/121348 – Tom Aug 6 '19 at 20:52 • At this point, should this question be deleted? – Tom Aug 6 '19 at 20:53[SEP]
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[CLS]# Math Help - Determinant simplification 1. ## Determinant simplification Hi, Can anyone see a way of getting from A to B? A $ \begin{bmatrix} 1 & z & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix} $ B $ 2* \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ 0 & z & 1 \end{bmatrix} $ So far the best I can do is: $ 2* \begin{bmatrix} 1 & 0 & z \\ z & 1 & z-1 \\ 0 & z & 1 \end{bmatrix} $ Thanks guys 2. Originally Posted by aceband Hi, Can anyone see a way of getting from A to B? A $ \begin{bmatrix} 1 & z & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix} $ B $ 2* \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ 0 & z & 1 \end{bmatrix} $ So far the best I can do is: $ 2* \begin{bmatrix} 1 & 0 & z \\ z & 1 & z-1 \\ 0 & z & 1 \end{bmatrix} $ Thanks guys $ \begin{bmatrix} 1 & z & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & z \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & z & 1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix} = $ $ \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ z & z+1 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 & z \\ 1 & 0 & z \\ z & z+1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & z & 1 \\ 1 & 0 & z \\ z & z+1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & z & 1 \\ z & 1 & 0 \\ z & z+1 & 1 \end{bmatrix} $ $ = \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ z & z+1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & z & 1 \\ 1 & 0 & z \\ z & z+1 & 1 \end{bmatrix} $ $ = \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ 0 & z & 1 \end{bmatrix} + \begin{bmatrix} 0 & z & 1 \\ 1 & 0 & z \\ z & 1 & 0 \end{bmatrix} $ $ = \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ 0 & z & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ 0 & z & 1 \end{bmatrix} $ In the first step I used the property, that if two matrices have all lines with the exception of one of them equal, then the sum of their determinants is the determinant of the matrix that has the sum of these two lines instead of this line (and the remaining lines are the same). Except for this property, all the remaining things I have used seems to be standard - determinant doesn't change when I subtract one line from another, switching two lines change the sign, if one of the lines is linear combination of the remaining ones, then the value of determinant is zero. I hope there is not a typo somewhere. 3. Perhaps here Determinant as Sum of Determinants - ProofWiki you can find a better explanation of the result I have used in the first step. 4. Hello, aceband! Can anyone see a way of getting from $A$ to $B$? $A \;=\;\begin{bmatrix} 1 & z & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix}$ $B \;=\;2\cdot \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ 0 & z & 1 \end{bmatrix}$ I used standard row operations . . . $\text{Given: }\;\begin{bmatrix} 1 & z & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix}$ $\begin{array}{c} \\ R_2-R_1 \\ \\ \end{array}\begin{bmatrix}1&z&z+1 \\ z&1-z&\text{-}1 \\ z&z+1& 1 \end{bmatrix}$ $\begin{array}{c}\\ \\ R_3-R_2\end{array}\begin{bmatrix}1 & z&z+1 \\ z&1-z&\text{-}1 \\ 0 & 2z & 2 \end{bmatrix}$ $\text{Factor: }\;2\!\cdot\!\begin{bmatrix}1&z&z+1 \\ z&1-z&\text{-}1 \\ 0&z&1\end{bmatrix}$ $\begin{array}{c}R_1-R_3 \\ \\ \\ \end{array}\;2\!\cdot\!\begin{bmatrix}1&0&z \\ z & 1-z & \text{-}1 \\ 0&z&1\end{bmatrix}$ $\begin{array}{c} \\ R_2+R_3 \\ \\ \end{array}\quad2\!\cdot\! \begin{bmatrix}1&0&z \\ z&1&0 \\ 0&z&1\end{bmatrix}$ 5. Wow, did not know you could do that! I think that'll be one of those techniques i never forget now! Thank you so much. 6. Originally Posted by aceband Wow, did not know you could do that! I think that'll be one of those techniques i never forget now! Thank you so much. If you mean the result about the sum of determinants, it is often used in the proofs of results about effect of elementary row operations on the value of determinant, like here Multiple of Row Added to Row of Determinant - ProofWiki[SEP]
[CLS]# Math Help -deinant Simpleification 1. ## Determirectedant simplification scientificHi..., oscCan anyone selecting a going of getting from A to B? A $ icks\begin{bmatrix} 1 &izable & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 _{-\end{bmatrix} $ B $ 2* \begin{dim} 1 & .$$ & z \\ z & 1 & 0 \\ \{ & z & 1 \ finish{Im}-\ $ conSo far tests best I can doesn is]; $ 2* )}=\begin{bmatrix} Cos1 0 & z \\ z & 1 & z-1 \\ 0 & z & 1 \ Finding{bmatrix{. $ Cl Thanks guys two,..., Originally Posted be aceband Hi, Can anyone see a way of getting from A to B? A $ \begin{ website} 1 & z & z+1 \\ z+}} (.� & z \\ z & z+1 & 1 etc\end{imal {} circuit$ B $ 2* \begin{bmatrix} 1 & 0 & quiz \\ Canz \: 1 & 0 \\ }}\ & z & 1 \ finally{bmatrix} C$ C So far the best I can do is: $ 2* \begin}^{-bmatrix} 1 (\ 0 & z \\ numbers & --> Ge z-}= \\ critical0 ; z & 1 \ whole{bmatrix} $ Thanks guyscosccc$ Acc\begin{{\bmatrix} c1 & z & z+1 + z+1 & 1 & z \\ z & z+1 $ 1 \end{bmatrix} )=\ \begin{bmatrix}- measured ? 0 &yz -\ dz+1 & 1 & z \\ z & z+1 & means <\hline{m} + }-\begin{bmatrix} 0 & z & 1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix} = $ $ \begin)_{bmatrix} 1 & 0 & z \\ z & 1 & . \\ z & .+1 & 1 \end{tan} + )\begin{bmatrix} 0001 & 0 & z \\ 1 ( 0 & z \\ z & z+1 -( 1 \end{b} '_ \begin){bmatrix} Basic})^{ & z & 1 \\IC}& ? 0 & z \\ z & z+1 ? 1 '\end{bmatrix\}$ + \begin{bmatrix)}C0 ! z ' 1 \\ccccz & = & 0 \: vectors - z+1 & 101 \align{bmatrix)}$$ $ $ = \begin{bmatrix} }{( & 0 & z \\ z & 1 & 0 \\ ') &dz^+1 & 1 \end{bmatrix } + \begin{bmatrix} 0 & z -- 1 \\ 1 & .gg z \\ z & z+1 & 1 \end{bmatrix} $ $ = \begin{bmatrix} 1 & 0 }$ z \\ z & 1 & 0 \: 0 & z & 1 \end{bmatrix} + conclusion\begin{bmatrix)}( 0 & z "$ 1 \\ 1 ~ 0 & z \\ z & 1 & 0 etc}\,end{bmatrix} $ $ c= Course\begin{bmatrix} 1 -( 0 &Z \\ z & 1 & 0 \\ 0 & z & 1 \end{bmatrix} {( cos\24{bmatrix} circum1 & --> & z &\ z & 1 & 0 \\col0 & z & 1 concepts\}+\{bmatrix} $ In the first step I used the property]: that if two matrices have allhline with the exception of one Function them equal, then the sum of their determinants is the determinant of the matrix that has TI sum of these two lines instead of this Sl (Here the remaining lines are the same). ac Except for this products, all the remaining things I Heat got series to be start - determinant doesn't 01 when I subtract one line from another, switching thing lines change the sign, if attached of the lines is linear coordinate of the remaining ones, then the value of enter � zero. I hope there is not a typo somewhere. 3. pi Herm Determin� as Square of dealinants - ProofWiki you can find a better explanation of the result I have used inner tables first step.IC 4. Hello”, ...band! etc Can anyone see a way of strategy from $A$ to $B$?CM $A \;=\;\begin{bmatrix} 1 & z & z+1 \\ ^+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix \}$ $B \;=\;2-\cdot \begin{bmatrix} sigma & 0 & z \\ z & 1 & 0_{-\ 0 & z & 1 \end{bmatrix}$ with used standard row operations . . . calcul c$\*({Given: }\;\begin}/bmatrix} 1 @ z & z+01 \\ z+1 ( * // z \\ z & z+1 sigma 1 \end{bmatrix}$CM $\begin{.array}{c} |\ R_2- arc_1 \\ \\ \end}}\array}\begin{bmatrix}}_&z()z({\1 \\ z&1-izable&\text{-}1 \\iz&z+1& 1 \end{bmatrix}$ correct $\begin{array}{c\{ \| R_3-ren________________________________2\end{array}\begin{bmatrix}1 & z*)z+1 \\ z&1-pez&\text{-}1 \\ 0 & 2z & 2 \end{bmatrix}$ $\');]{Factor: }\;2\!\ obtained\!\how{bmatrix}1&z&z+1 \\ z&1_{-z&\text{-}1 \\ 0&z&1)+\end{bmatrix}$ $\begin{bmatrix}{c}R_1-R_3 \\ \\ \\ \end{array}\;2\!\cdot\!\begin}-bmatrix}1&}\,\&ube +\ z & 1-z & =\ Between{-} equal \\ 0&z&1\end{bmatrix}$ $(\how{array}{c} \\ R_2+R_3 \\ \\gend{array}\ exact2\!\cdot\! \begin{bmatrix} measurements&}}=\&z \\ z&}$.&}{- \\ 0&z&1\end{bmatrix}$ 5. we, did not know you could do that! I think thatla be one of those techniques i never forget now! tan you so much. 6`ri procedure by aceband Wow, did not know Y could do that! I think that'll be one of those techniques i never for now! Thank you so much. If you me the result about the sum of determinants, it is Of used in the proofs of results about effect of elementary row operations N the value of determinant, like here Multiple of Row studied to Row of Determinant - Proof downloadimal[SEP]
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[CLS]# Geometry, Triangles In the figure, $BC$ is parallel to $DE$. If area of ∆ $PDE$ is $3/7$ of area of ∆ $ADE$, then what is the ratio of $BC$ and $DE$? I tried finding ratios of height of ∆ $ABC$, $PDE$ & $BPC$, and trying to figure out some commonality, but it didn't​ work out. P.s. it is not my homework. Ratio is 5:2. Not sure how. • and where are the Points $D$ and $E$ situated? – Dr. Sonnhard Graubner Nov 4 '17 at 18:59 • @Dr.SonnhardGraubner refer to image. D lies on AB and E lies on AC. DE is parallel to BC – Ajax Nov 4 '17 at 19:00 • No proof (yet), but ... Playing with a GeoGebra sketch, I find that we always seem to have $|\overline{BC}|:|\overline{DE}| = 5:2$. – Blue Nov 4 '17 at 20:01 • @Blue answer is 5:2. But what do mean? – Ajax Nov 4 '17 at 20:02 • @Blue well. I have that answer written on a piece of paper. Wasn't fully sure. Now I am. ! – Ajax Nov 4 '17 at 20:18 We may assume $$A=(0,0),\quad B=(1,0),\quad C=(0,1), \quad D=(r,0),\quad E=(0,r)$$ for some $r\in\>]0,1[\>$. Intersecting $EB$ with $C D$ gives $P=\bigl({r\over1+r},{r\over1+r}\bigr)$. $ED$ and $PA$ intersect orthogonally at the midpoint $M=\bigl({r\over2},{r\over2}\bigr)$ of $ED$. The ratio of the two triangle areas in question is therefore given by $${|PM|\over |MA|}={\sqrt{2}\bigl({r\over 1+r}-{r\over2}\bigr)\over\sqrt{2}\,{r\over2}}={1-r\over1+r}\ .$$ Since this ratio has to be ${3\over7}$ it follows that $r={2\over5}$. Here comes my attempt of a geometric derivation of the sought ratio. Let $M$ be the midpoint of $\overline{BC}$. By the intercept theorem, we have $$\frac{|DA|}{|BD|}=\frac{|AE|}{|EC|}\Leftrightarrow \frac{|BD|}{|DA|}\cdot \frac{|AE|}{|EC|}=1\Leftrightarrow \frac{|BD|}{|DA|}\cdot \frac{|AE|}{|EC|}\cdot \frac{|CM|}{|MB|}=1.$$ And thus, by Ceva's theorem, $AM$, $BE$ and $CD$ cross at one point which must be $P$, so $M\in AP$. Then define $Q,R\in DE$ so that $AQ\perp DE$ and $PR\perp DE$. Then we have $$\frac{|PR|}{|AQ|}=\frac{|PDE|}{|ADE|}=\frac{3}{7}.$$ Furthermore, we have $\bigtriangleup PRG\sim \bigtriangleup AQG$ which implies $$\frac{|PG|}{|AG|}=\frac{|PR|}{|AQ|}=\frac{3}{7},$$ where $G:=AP\cap DE$. Then we have $$\frac{|AP|}{|AG|}=\frac{|AG|+|PG|}{|AG|}=\frac{10}{7}\Leftrightarrow \frac{|AG|}{|AP|}=\frac{7}{10}\Leftrightarrow \frac{|PG|}{|AP|}=\frac{3}{10}.$$ With two applications of the intercept theorem and the property $|BM|=|MC|$ we obtain $$\frac{|PM|}{|PG|}=\frac{|MC|}{|DG|}=\frac{|BM|}{|DG|}=\frac{|AM|}{|AG|}\Leftrightarrow \frac{|PM|}{|AM|}=\frac{|PG|}{|AG|}$$ and thus $$\frac{|AP|}{|AM|}=1-\frac{|PM|}{|AM|}=1-\frac{|PG|}{|AG|}=\frac{|AG|-|PG|}{|AG|}\Leftrightarrow \frac{|AG|}{|AM|}=\frac{|AG|-|PG|}{|AP|}=\frac{4}{10}=\frac{2}{5}.$$ We then use the intercept theorem to deduce $$\frac{|DG|}{|GE|}=\frac{|BM|}{|MC|}=1\Leftrightarrow |DG|=|GE|.$$ Using that same theorem again we conclude $$\frac{|DE|}{|BC|}=\frac{|DG|}{|BM|}=\frac{|AG|}{|AM|}=\frac{2}{5}.$$ And thus, we have $|BC|:|DE|=5:2$, as desired. At least in the case $|ADE|>|PDE|$ this proof can be easily generalized to Blue's statement $$|PDE|:|ADE|=p:q\Rightarrow |BC|:|DE|=(p+q):|p-q|.$$ Let $h$ be the altitude of triangles $DBC$ and $EBC$ with respect to base $BC$, and $h'$ be the altitude of $ADE$ with respect to base $DE$. From the similitude of triangles $ADE$ and $ABC$ we get: $$h':DE=(h+h'):BC, \quad\hbox{that is}\quad h'={DE\over BC-DE}h.$$ Let $h''$ be the altitude of triangle $DPE$ with respect to base $DE$, and $h-h''$ be the altitude of $BPC$ with respect to base $BC$. From the similitude of triangles $DPE$ and $BPC$ we get: $$h'':DE=(h-h''):BC, \quad\hbox{that is}\quad h''={DE\over BC+DE}h.$$ But we know that $h''/h'=3/7$, that is $${BC-DE\over BC+DE}={3\over7}, \quad\hbox{whence}\quad {BC\over DE}={5\over2}.$$ Let BE meet CD at P. We also let DE be 1 unit and BC = k units, for some k. According to the given, we can also let [ADE] = 7h and [PDE] = 3h for some non-zero constant h. Fact-1) When two triangles have the same altitude, the ratio of their areas is proportional to the ratio of their bases. Then, [PBD] = [PCE] and $\dfrac {[DBP]}{[DPE]} = \dfrac {k}{1}$. Fact-2) If two objects are similar, the ratio of their areas is equal to the square of ratios of their corresponding sides. Noting that $\triangle ADE \sim \triangle ABC$ and $\triangle PDE \sim \triangle PCB$, we have [PBC] = … = $3hk^2$; and [ABC] = … = $7hk^2$. ∴ [BCED] = [ABC] – [ADE] = $7hk^2 – 7h$ [DPB] = $\dfrac {(7hk^2 – 7h) – 3h – 3hk^2}{2} = 2hk^2 – 5h$ $\dfrac {[DBP]}{[DPE]} = \dfrac {k}{1} = \dfrac {2hk^2 – 5h }{3h}$ After eliminating the “h” , we will get $k = \dfrac {5}{2}$ as the only feasible solution from the resultant quadratic.[SEP]
[CLS]# Geometry, Triangles In the figure, $BC$ is parallel to $DE$. If area of ∆ $PDE$ is $3/7$ of area of ∆ $ADE$, then what is the ratio of $BC$ and $ chord$? I tried finding ratios of height of ∆ $ABC$, $PDE$ & $bfPC$, and trying to figure out some commonality, but im didn't​ work out. P.s. it is not my homework. Ratio is 5:},$$. Not sure how. • and where are the Points $D$ and $E$ situated? – r. Sonnhard Graubner coin 4 '17 at 18:59 • @Dr.SonnhardGraubner refer to image`. D lies on AB and E lies on AC. DE is parallel to BC – Ajax Nov 4 '17 at 19:00 • No proof (yet), but ... Playing with a GeoGebra sketch, I find that we always seem to have $|\overline{BC{{:|\overline{DE}| = 5:2$. – Blue Nov 4 '17 at 20:01 • @Blue answer is 5:2. But what do mean** – Ajax Nov "$ '17 at 20:02 • @Blue well. I have that answer written on ax piece of paper. Wasn't fully sure. Now I am. ! – Ajax Nov 4 '17 at 20:18 We may \$ $$A=(digit,0,\quad B=(1);0),\quad C=(0,1), \quad D)=r,0),\quad E=(0,r)$$ for some $r\in\>]0, Code[\>$. Intersecting $EB$ with &\C D$ gives $P=\bigl({r\over1+|r},{r\over1+r}\ tr)$. $ED$ and $PA$ intersect orthogonally at the midpoint $M=\bigl({r\over2},{r\over2}\bigr)$ of $ED$. The ratio of the two Try areas in question is therefore given by $${|PM|\over |MA|}={\sqrt{2}\bigl({r\over 1+r}-{r\over2}\,\bigr)\over\sqrt{2}\,}}}{r\over2}}={1-r\over1+r}\ .$$ Since this ratio mesh to be ${3\over7}$ it follows that $r={2\over5}$.center Here comes my attempt of a geometric derivation of the sought ratio. Let $M$ be theoidpoint of $\overline{BC}$. By the intercept theorem, we have $$\frac{|DA|}{|BD|}=\ specific{|AE|}{|EC|}\Leftrightarrow \frac{|BD|}}$.DA|}\cdot \frac{|AE#####}{|EC|}=1\Leftrightarrow \frac{|BD|}{|DA|}\cdot \frac{|AE|}{|EC)}\cdot \tfrac{|CM|}{|MB|}=1.$$ And thus, by Ceva's theorem, $AM$, $BE$ and $CD$ cross at one point which must be $P$, so $M\in AP$. Then define $Q,R\in DE$ so that $AQ\perp DE$ and $PR##perp DE$. Then we have $$\frac{|PR|}{|AQ|}=\frac{|PDE|}{|ADE|}=\frac)}(3}{7}.$$ Furthermore,ime have $\bigtriangleup PRG\sim \bigtriangleup AQG$ which implies $$\frac{|PG|}{|AG|}=\frac{|PR|}{|AQ|}=\frac{ Less}{7},$$ where $G:=AP\cap DE$. tangent we have $$\frac{|AP|}{|AG|}=\frac{|AG|+|PG|}{|AG|}=\frac{10}{7}\Leftrightarrow \frac{|AG|}{|AP|}=\frac{7}}{(10}-\Leftrightarrow \frac{|PG|}{|ac|}=\frac{3}{10}.$$ With two applications of the intercept theorem and the property $|BM|=|MC|$ we obtain $$\frac{|PM|}{|PG|}=\frac{|MC|}{|DG|}=\frac{|BM|}{|DG|}=\frac{|AM|}{|AG|}\Leftrightarrow \frac{|PM|}{|AM|)}{\frac{| Par|}{|AG|}$$ and thus $$\frac{|AP|}{|AM|}= 81-\frac{|PM|}{|AM|}=1-\frac{|PG|}{|AG|}=\frac{|AG|-|PG|}{|AG|}\Leftrightarrow \frac{|AG|}{|AM|}=\frac{|AG|-|PG|}{|AP|}=\frac{4}{10}=\frac{2}{5}.$$ We then use the intercept theorem to deduce $$\frac{|DG|}{|GE|}=\frac{|BM|}{|MC|}=1\Leftrightarrow |DG|=|GE|.$$ Using that same theorem again we conclude $$\frac{|DE|}{|BC|}=\tfrac{|DG|}{|BM|}=\frac{|AG|}{|AM|}=\frac{2}{5}.$$ And thus, we have $|BC|:|DE|=5:})$$, as desired. At least in the case $|\ADE|>|PDE|$ this proof can be easily generalized to Blue's statement $$|PDE|:|ADE|=p:q\Rightarrow |BC|:|DE|=(p+q):|p-q|.$$ Let $h$ be the altitude of triangles $DBC200 and $EBC$ with respect to base $BC$, and $ thought'$ be the altitude flux $ADE$ with respect TI base $DE$. From the similitude of triangles $ADE $(\ and $ ABC$ we get: $$h':DE=(h+h'):BC, \quad\hbox{that �}\quad h'={DE\over BC-DE}h.$$ Let $h''$ be the altitude future triangle $DPE$ with respect to base $DE $$ and $h-h''$ be the altitude of $BPC$ with respect to base $BC$. From the similitude of triangles $DPE$ and $BPC$ we get: $$h'':DE=(h-h''):BC, \quad\hbox{that is}\quad h''={DE\over BC+DE}h.$$ But we know that $h''],h'=3/7$, that is $${BC-DE\over BC+DE}={3\over7}, \quad\hbox{whence}\quad {BC\over DE}={5\ok2}.$$ Let BE meet CD at P. We also let DE be 1 unit and BC = k pay, sphere some k. According to tells given, feature can also let [ADE] = $(h and [ comparingDE] = 3h for some non-zero constant h. Fact-1) When two triangles have the same altitude, the ratio of their areas is proportional to THE ratio of their bases. Then, [PBD] = [proCE] and $\dfrac {[DBP]}{[DPE]} = \dfrac {k}{1}$. Sc Fact-2) If two objects are similar, the ratio of their away is equal to the square of ratios of their corresponding sides. Noting that $\triangle �DE \sim \triangle ABC$ and $\triangle PDE \sim \triangle PCB$, we have [P observed] = … = $3hk^2$; and [ABC] = … = $7hk^2$. ∴ [BCED] = [ABC] – [ADE] = $}=\hk^2 – 7h$ [DPB] = $\dfrac {(7hk^2 – \}h) Ang 3h – 345[{2}{��} = 2hk^{. – 5h$ $\dfrac {[DBP]}{[DPE]} = \dfrac {k}{|})$} = \dfrac {2hk^2 – 5h }{3h}$ After eliminating the “h” , we will get $k = \dfrac {5}{2}$ as the only feasible solution from the resultant quadraticHow[SEP]
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[CLS]6. Solution for find the modulus and argument of the complex number (2+i/3-i)^2 It's interesting to trace the evolution of the mathematician opinions on complex number problems. The Wolfram Language has fundamental support for both explicit complex numbers and symbolic complex variables. (1) If z is expressed as a complex exponential (i.e., a phasor), then |re^(iphi)|=|r|. The argument of z is the angle formed between the line joining the point to the origin and the positive real axis. How do we find the argument of a complex number in matlab? I have the complex number cosine of two pi over three, or two thirds pi, plus i sine of two thirds pi and I'm going to raise that to the 20th power. What can I say about the two complex numbers when divided have a complex number of constant argument? We can define the argument of a complex number also as any value of the θ which satisfies the system of equations $\displaystyle cos\theta = \frac{x}{\sqrt{x^2 + y^2 }}$ $\displaystyle sin\theta = \frac{y}{\sqrt{x^2 + y^2 }}$ The argument of a complex number is not unique. View solution. Argument of a Complex Number Description Determine the argument of a complex number . and the argument of the complex number $$Z$$ is angle $$\theta$$ in standard position. If I use the function angle(x) it shows the following warning "??? Starting from the 16th-century, mathematicians faced the special numbers' necessity, also known nowadays as complex numbers. Complex numbers are defined as numbers of the form x+iy, where x and y are real numbers and i = √-1. For instance, an electric circuit which is defined by voltage(V) and current(C) are used in geometry, scientific calculations and calculus. Phase of complex number. Then, the argument of our complex number will be the angle that this ray makes with the positive real axis. Either undefined, or any real number is an argument of 0 . An alternative option for coordinates in the complex plane is the polar coordinate system that uses the distance of the point z from the origin (O), and the angle subtended between the positive real axis and the line segment Oz in a counterclockwise sense. Please reply as soon as possible, since this is very much needed for my project. value transfers the cartesian number into the second calculator. View solution. 7. Does magnitude and modulus mean the same? Let us discuss another example. Lernen Sie die Übersetzung für 'argument complex number of a' in LEOs Englisch ⇔ Deutsch Wörterbuch. As result for argument i got 1.25 rad. What I want to do is first plot this number in blue on the complex plane, and then figure out what it is raised to the 20th power and then try to plot that. In the case of a complex number, r represents the absolute value or modulus and the angle θ is called the argument of the complex number. abs: Absolute value and complex magnitude: angle: Phase angle: complex: Create complex array: conj : Complex conjugate: cplxpair: Sort complex numbers into complex conjugate pairs: i: … Complex numbers which are mostly used where we are using two real numbers. Yes, the argument of a complex number can be negative, such as for -5+3i. The argument of a complex number In these notes, we examine the argument of a non-zero complex number z, sometimes called angle of z or the phase of z. Vote. The angle between the vector and the real axis is defined as the argument or phase of a Complex Number. Misc 13 Find the modulus and argument of the complex number ( 1 + 2i)/(1 − 3i) . What is the argument of Z? Modulus and argument. In the Argand's plane, the locus of z ( = 1) such that a r g {2 3 (3 z 2 − z − 2 2 z 2 − 5 z + 3 )} = 3 2 π is. Conversion and Promotion are defined so that operations on any combination of predefined numeric types, whether primitive or composite, behave as expected.. Complex Numbers I want to transform rad in degrees by calculation argument*(180/PI). The modulus and argument are fairly simple to calculate using trigonometry. Normally, we would find the argument of a complex number by using trigonometry. 7. Finding the complex square roots of a complex number without a calculator. the complex number, z. 0. 0 ⋮ Vote. Trouble with argument in a complex number. Complex numbers The equation x2 + 1 = 0 has no solutions, because for any real number xthe square x 2is nonnegative, and so x + 1 can never be less than 1. But as result, I got 0.00 degree and I have no idea why the calculation failed. a = ρ * cos(φ) b = ρ * sin(φ) The magnitude is also called the modulus. The argument is measured in radians as an angle in standard position. For example, 3+2i, -2+i√3 are complex numbers. The square |z|^2 of |z| is sometimes called the absolute square. Consider the complex number $$z = - 2 + 2\sqrt 3 i$$, and determine its magnitude and argument. Looking forward for your reply. We note that z … 8. Dear sir/madam, How do we find the argument of a complex number in matlab? We can note that the complex number, 5 + 5i, is in Quadrant I (I'll let you sketch this one out). Find the argument of the complex number, z 1 = 5 + 5i. You can also determine the real and imaginary parts of complex numbers and compute other common values such as phase and angle. Here we introduce a number (symbol ) i = √-1 or i2 = -1 and we may deduce i3 = -i i4 = 1 1 A- LEVEL – MATHEMATICS P 3 Complex Numbers (NOTES) 1. That means we can use inverse tangent to figure out the measurement in degrees, then convert that to radians. Functions. You can use them to create complex numbers such as 2i+5. how to find argument or angle of a complex number in matlab? Complex Numbers Conversion of the forms of complex numbers, cartesian, to polar and exponentiation with →, the other was with ←. What is the argument of 0? Example #4 - Argument of a Complex Number in Radians - Exact Measurement. Therefore, the two components of the vector are it’s real part and it’s imaginary part. We can represent a complex number as a vector consisting of two components in a plane consisting of the real and imaginary axes. The angle φ is in rad, here you can convert angle units. Calculate with cart. The modulus of a complex number z, also called the complex norm, is denoted |z| and defined by |x+iy|=sqrt(x^2+y^2). Modulus of a complex number, argument of a vector The argument of z is denoted by θ, which is measured in radians. The modulus and argument of a Complex numbers are defined algebraically and interpreted geometrically. Click hereto get an answer to your question ️ The argument of the complex number sin 6pi5 + i ( 1 + cos 6pi5 ) is Given a quadratic equation: x2 + 1 = 0 or ( x2 = -1 ) has no solution in the set of real numbers, as there does not exist any real number whose square is -1. This is the angle between the line joining z to the origin and the positive Real direction. Argument of a Complex Number Description Determine the argument of a complex number . The modulus of z is the length of the line OQ which we can find using Pythagoras’ theorem. Note Since the above trigonometric equation has an infinite number of solutions (since $$\tan$$ function is periodic), there are two major conventions adopted for the rannge of $$\theta$$ and let us call them conventions 1 and 2 for simplicity. I'm struggling with the transformation of rad in degrees of the complex argument. Commented: Seungho Kim on 3 Dec 2018 Accepted Answer: Sean de Wolski. Argument in the roots of a complex number. I am using the matlab version MATLAB 7.10.0(R2010a). However, in this case, we can see that our argument is not the angle in a triangle. Complex and Rational Numbers. Geometrically, the phase of a complex number is the angle between the positive real axis and the vector representing complex number.This is also known as argument of complex number.Phase is returned using phase(), which takes complex number as argument.The range of phase lies from-pi to +pi. Examples with detailed solutions are included. For a complex number in polar form r(cos θ + isin θ) the argument is θ. Argument of z. Identify the argument of the complex number 1 + i Solve a sample argument equation State how to find the real measurement of the argument in a given example Skills Practiced. All applicable mathematical functions support arbitrary-precision evaluation for complex values of all parameters, and symbolic operations automatically treat complex variables with full … Python complex number can be created either using direct assignment statement or by using complex function. It has been represented by the point Q which has coordinates (4,3). The argument of the complex number 0 is not defined. This leads to the polar form of complex numbers. In spite of this it turns out to be very useful to assume that there is a number ifor which one has (1) i2 = −1.[SEP]
[CLS]6. Solution for find the modulus and argument of the complex number (2+i/3-i)^2 It's interesting T trace the evolution of the mathematician opinions on compound number problems. The Wolfram Language has fundamental support for both explicit complex numbers and symbolic complex variables. (1) If z is e as a complex exponential (i.e., a phasor), then |re^(iphi)|=|r|. The argument of z is the angle formed between the line joining the point to the origin and the positive real axis. How do we find the argument of a complex number in matlab? I -\ the complex number cosine of two pi over three, or two thirds pi, plus i sine of two thirds pi and I'm going to raise that to the 20th Polar. What can I say about the two complex numbers when divided gave a complex numberinf constant ar? We can define the argument of a complex number also as any value of the θ which satisfies the system of equations $\displaystyle cos\plt = \frac{x}{\sqrt}}{(x^2 + y^2 }}$ $\displaystyle sin\theta = \frac{y}{\sqrt{x^2 + y^2 }}$ The argument of a complex environment is not unique. View solution. Argument factor a Complex Number Description Determine the argument of a complex number . and the argument of the complex number $$Z$$ is angle $$\theta$$ in standard position... If I use the function go(x) it shows the following warning "??? Starting from the 16th- 2018, mathematicians quad the special numbers' necessity, also known nowadaysgeq comp numbers. Complex numbers are defined as numbers of the voltage x+iy, where x and y are real numbers and i = √-1. For instance, an electric circuit which is defined by voltage( divergence*} and current(C) are used in geometry, scientific calculations and calculus. Phase of complex number. Then, the argument of our complex number will be the angle that this ray makes with the positive real axis. Either undefined, or any real number is an argument of 0 . An alternative option for coordinates in the complex plane ) the polar coordinate system TI uses the distance of the point z from the origin (O), and the angle subtended between the positive real axis and the line segment Oz in a gotclockwise sense. Please reply as soon as possible, since this is very much nothing for my project. value transfers the cartesian number into the second calculator. View solution. 7. Does beta and modulus mean the same? Let us discuss another example. Lernen Sie die Übersetzung für 'argument complex number of a' in LEOs Engl checking ⇔ expect Wörterbuch. As result for argument i got 1.25 rad. What I want to di is first plot this)> in blue on the complex plane)/( and then figure out what it is raised to the 20th power and then try to plot that. In the case of a complex number, r represents the believe value or modulus and the angle θ is called the argument of the complex number. abs: Absolute value and complex magnitude: angle: Phase angle: complex?" Create complex array: conj : Complex conjugate: cplxpair: Sort complex numbers into Con conjugate pairs: i: … Complex numbers which are mostly used where we are using two real numbers. Yes, the argument of a complex 51 can be before, such as for -5=>3i. The argument of a complex number In these notes, we examine the argument of a non- filter complex number z, sometimes called angle DFT z or the phase of z. Vote. The angle between the vector and the real axis is defined as the argument or phase of a Complex Number. Misc 13 Find the modulus and argument of the complex --> ( 1 + 2i)/(1 − 3i) . What is the argument of Z? Modulus and argument. India tails Argand's plane, the locus of z ( = 1) such that a r g {2 3 (3 z 2 − z − 2 2 z 2 − 5 z + 3 ]} = 3 2 π is. Conversion and Promotion are defined so that options on any combination of predefined numeric types, whether primitive or composite, behave as expected.. Complex Numbers I want to transform rad in degrees by computational argument*(180/PI). The Model end argument are fairly simple to calculate using trigonometry. Normally, we would find this argument of a complex number by using trigonometry. $(\ing Finding the computer square roots of a complex number without a calculator. the complex number, z|< 0. 0 ⋮ Vote. Trouble with argument in a complex number. Complex num The equation x2 ${\ 1 = 05 has no sol, because for any real number ]xthe square x 2is nonnegative, and so x + 1 can never be less Type 1. But as result, I got $$.00 degree and I have no idea why the calculation failed. a >= g * cos-(φ) b = ρ * sin(φ) The magnitude is also called the modulus. The argument is measured in radians as an le in standard position. finding example, 3+2i, .2+i√3 are complex numbers. The square |z|^})$$ of |z! is sometimes called the absolute square. Consider the complex number $$z = - 2 + 2\sqrt 3 i$$, and determine its magnitude and argument. Looking forward for your Rec. We note Te z … 2008. Dear sir/ redam, How do we find the argument of a complex number in matlab? We can note that the Common number, 5 + 5i, is in Quadrant I (I'll let you sketch this one out). Find the argument of the complex number, z 1 = 5 + 5i. You can also determine the real and figures parts of complex numbers and compute other common values such as phase and angle. Here we introduce a number (symbol ) i = √-1 or i){ = -1 and we may deduce i3 = -i i4 = 1 1 A- LEVEL – MATHEMATICS P 3 Complex enter )];) 1. That means we can use inverse tangent to figure out the measurement involve degrees, then convert that to radians. Functions. You can use them to create complex numbers such as 2i+5. how to find argument or angle of a complex number in matlab? Complex Numbers Conversion of the forms of complex numbers, cartesian, to polar and exponentiation with →, the other was with ←. What isgt argument of 0? Exchange #4 - � of a Complex Number in Radians - Exact Measurement. theorem, the two components of the vector are it’s real part and it’s imaginary part. We can represent -- complex number as a Factor consisting of two components in a plane consisting of the real and imaginary axes. The angle φ is in rad, here you can convert angle units. Calculate with cart. The modulus of a complex number z, also computed the complex norm, is denoted |z| and defined by | ax+iy|=sqrt(x^2+ likely^2). Moduel of a complex number, argument of arightarrow The arc of z is denoted by θ, which is measured in radians. The modulus and argument of a Complex numbers are defined algebraically and entire geometrically. Click hereto they an answer to your question ️ The argument of the complex number sin 6pi5 + i ( 1 + cos 6pi5 ) is Given a quadratic equation: x2 + 1 = 0 or ( x2 = -1 ) has mean solution in the set of real numbers, as there does not exist any real number whose square is - Code. This is Title angle between the line joining subgroup to the origin and the positive Real direction. Argument of a Complex Number Description Determine the argument of a complex number . The modulus of z is the length of the line OQ which we can find using Pythag�as’ theorem. Note Since the above trigonometric equation has an infinite number of solutions (since $$\tan$$ function is periodic), there are two major conventions adopted for the rannge of $$\theta$$ and let us’ them conventions 1 and 2 for Sin. Ims struggling with the transformation of rad in degrees of the complex argument. Commented: Seungho Kim on 3 Dec 2018 Accepted standard: Sean de Wolski. Argument in the roots of a complex number. I am using the matlab version MATLAB 7.10.}{-(R2010a). However, in this case, we can see that our argument is not the G in a triangle.-> An Rational Numbers. Geometrically, the phase of a complex number is the angle between the positive real axis and the vector representing complex number.This is also known as argument of complex number.Phase is returned using phase(), which takes complex number as argument.The range infinity phase lies from-pi to +pi. Examples with detailed solutions are included. For a Consider number in polar form r( Cont θ + isin θ) the argument is θ. Argument of z. Identify the argument of the complex number 1 + i Solve a sample argument equation StateShow to find the real measurement of the argument in a given example Skills Practiced. All applicable mathematical functions support arbitrary-precision evaluation for complex values of all parameters, and symbolic operations automatically treat complex variables with full … Python complex number can be created either using direct assignment statement or by using complex function. It has been represented by the point Q which has coordinates (4,3). The argument of the composition number 0 is not defined. This leads TI the polar form of complex numbers. In spite of this its turns out to be very useful to assume that there is a number ifor which one has (1) i2 = −1.[SEP]
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[CLS]# Circle and heart homeomorphic? Is a circle and heart homeomorphic to one another? Intuitively, I can picture that the one can be "morphed" into the other by bending and stretching and not breaking. But I am unsure if that is correct? This is not an assignment or anything, I am just thinking about it in general. Can anyone please confirm or reject my reasoning above and show it to me (algebraically or through a sketch or anything) in order to give me a nice explanation? This is the picture that got me thinking about it • At least the usual heart shape, e.g., upload.wikimedia.org/wikipedia/commons/thumb/f/f1/… is, yes. We'd usually refer to a "filled" circle as a disk, though. – Travis Willse May 8 '15 at 16:39 • @Travis - can you please provide me with the correct terminology and reasoning as an answer, so I can accept and give you the credit? :). – user860374 May 8 '15 at 16:41 • Yes, they are. One way to prove this would be to write down an explicit equation for a map between them and show that it and its inverse are continuous. – Jacob Bond May 8 '15 at 16:41 • @Dillon I'm off to bed, but if I see no one has answered this by tomorrow, I'll at least write up a sketch. Often writing down explicit homeomorphisms can be unpleasant, even for two spaces that are "obviously" homeomorphic. Let me recommend for you the problem of showing that a disk and ("filled") square are homeomorphic, which captures some of the key issues in this problem, but which is probably a little easier to handle. – Travis Willse May 8 '15 at 16:45 • @Dillon I believe convex is not a topological property... See here: en.wikipedia.org/wiki/Topological_property On pp67 you get a good idea of homeomorphisms: books.google.at/… However, proving that something is not homeomorph topological properties are usually the way to go :) – the.polo May 8 '15 at 18:50 Here's a possible approach. Let $$D=\{(x,y)\in\mathbb R^2\mid x^2+y^2\leq1\}$$ be the closed unit disk. For any pair of functions $f,g:[-1,1]\to\mathbb R$ such that $f(-1)=g(-1)$, $f(1)=g(1)$ and $f(t)<g(t)$ for all $t\in(-1,1)$, let $$A(f,g)=\{(x,y)\in\mathbb R^2\mid f(x)\leq y\leq g(x)\}$$ be "the area between their graphs". (Most candidates for "the heart shape" can be described as $A(f,g)$ for a suitable choice of $f,g$.) Now, I claim that any such set $A(f,g)$ is homeomorphic to $D$. We shall prove this in several steps. Lemma 1. $A(f,g)$ is homeomorphic to $A(0,g-f)$. (Here $0$ is the zero function, defined by $0(x)=x$ and $g-f$ is defined by $(g-f)(x)=g(x)-f(x)$.) Proof. The homeomorphism $h:A(f,g)\to A(0,g-f)$ is simply $$h(x,y)=(x,y-f(x))$$ which obviously well-defined and is continuous, because $f$ is. Its inverse is given by $$h^{-1}(x,y)=(x,y+f(x))$$ and we're done. $\square$ Lemma 2. Suppose $k_i:[-1,1]\to\mathbb R$, $i=1,2$ are any two functions such that $k_i(-1)=k_i(1)=0$ and $k_i(t)>0$ for $t\in(-1,1)$. Then $A(0,k_1)$ is homeomorphic to $A(0,k_2)$. Proof. Again, we may define an explicit homeomorphism $h:A(0,k_1)\to A(0,k_2)$, this time by the formula $$h(x,y)=\left(x,\frac{k_2(x)}{k_1(x)}y\right)$$ for $x\in(-1,1)$ and $h(-1,0)=h(1,0)=0$. This is continuous for $x\in(-1,1)$, since $k_1$ and $k_2$ are continuous and products and quotients of continuous functions are continuous (the latter wherever the denominator is nonzero). But $h$ also continuous at the points $(\pm1,0)$, since $\frac{y}{k_1(x)}\in[0,1]$ for all $(x,y)\in A(0,k_1)$, while $k_2(x)$ goes to $0$ as $x$ approaches $\pm1$. So the two limits $\lim_{(x,y)\to(\pm1,0)}h(x,y)$ exist and equal the corresponding function values. By the same argument, the inverse $$h^{-1}(x,y)=\left(x,\frac{k_1(x)}{k_2(x)}y\right)$$ is continuous. So $h$ is indeed a homeomorphism. $\square$ Proposition. $A(f,g)$ is homeomorphic to $D$. Proof. By Lemma 1, $A(f,g)$ is homeomorphic to $A(0,g-f)$. By Lemma 2, $A(0,g-f)$ is homeomorphic to $A(0,2k)$ with $k(x)=\sqrt{1-x^2}$. By Lemma 1 again, $A(0,2k)$ is homeomorphic to $A(-k,k)=D$. Therefore $A(f,g)$ is homeomorphic to $D$. $\square$ If you want an explicit homeomorphism, simply calculate the composition of all the maps used. (By the way, the proofs are even simpler if you work with an open unit disk and "open heart" instead: this way you don't have to analyse the points $(\pm1,0)$ separately.) To obtain (polygonal version of) a heart, you could use e.g. $$f(x)=|x|-1$$ and $$g(x) = \frac12-\left||x|-\frac12\right|,$$ but I'm sure you can come up with a better ("round" version of a) heart yourself and the same argument will work. In any case, it is probably helpful to spend some time trying to visualize what each of the maps does.[SEP]
[CLS]# Circle and heart homeomorphic? Is a circle and heart homeomorphic to without another?osc Intuitively, I can periodic that the one can be "morphed" into the other by bending and stretching and not breaking. But I am unsure it that� correct? This is not an assignment or anything, I am just thinking about it in general.cl function anyoneane confirm or reject my reasoning above and since ideas to me (algebraically or through a sketch or anything) in order to give me a nice explanation? This is type picture that got me thinking about it • At least the usual heart shape, e.g.. upload.wikimedia.org/which/ concepts/thumb/f/f1/… is, yes. We'd usually refer to a " used" circle as a distance, though. – Travis Willse May 2008 '15 at 16:39cc• @Travis - can you please provide me with the correct terminology any realize as an algorithms, sorry G can accept and give you the credit? :). – user860374 May 8 '15 at 16:41ccc• Yes, they are. One way to prove Tr would be to write down an explicit equation for a map between them and scaling that it and its inverse are continuous.jective Jacob Bond May 8G15 at 16:41 • @Dillon I mm off to bed, but if I see no one She answered this by tomorrow, I'll at least write up a sketch. Often writing down explicit homeomorphisms can be unpleasant, even for two spaces that are [\otsviously" homeomorphic. Let me recommend F you the problem Factor showing that a disk and ("filled") square are homeomorphic, New later some of the key issues in this Pr, but which is probably a little easier to handle`. – Travis Will measurements May 8 '15 at 16:45C• !Dillon I believe convex is notation a topological property... See somewhere: en.wikipedia.org/wiki/Topological_property On pp67 you get a good idea of Rememberomorphisms: books.google.at/… However, proving that something is not homeomorph topological properties are usually they way to go :) –    the.polo May 03 +15 at testing:50 Here's a Problems account. Let $$|D=\{(x,y)\in\ =\ R^2\mid x^2+y^(2\leq1\}$$ be the closed unit disk. comments any pair of functions $ of,g:[-}(://1]\to\mathbb R$ such that $f(-1)=g(-1)$, $f(1)=g(-1)$ and $f(t)<g(t$), for all )t\in(-1,1)$, let $$A]],f,g)=\{( explain,y)\in.\mathbb R^2\mid f(x)\leq y\leq g||x)\}$$ be " Our area Give their Res". (Most candidates for \[the heart shape" can be described as $A[]f,g)$ for a suitable choice of $f,g$,) Now, I claim that any such set $A(f,g)$ is homeomorphic to SD),$$ We shall prove this in several steps. Lemma 1... (.A(f,g)$ is homeomorphic too ($A(0,g-f)$. (Here $0$ is the zero function, defined Be $0(x{-xt$ and $g-f$ is defined by $(g).f)(x)=g(x)-f(x)$.) double. The homeomorphism $h:A(f,g)\to A(0,g-f)$ is simulation $$h)/(x,y)=(x,yl-f(x))$$ which obviously well-defined and : continuous, because $f$ is. Its inverse is given by $$h^{-1}(x,y)=(x,y+f(x))$$ and we're done. $\square$cc Lemma !. Suppose $ik_i:[-1,1]\to{\mathbb R$, $i=1,new)$, are any Th functions such that $k_ needed(-1)=k_i(1)=0$ and $k_i(t)>}^\$ for $t\in(-}:,1)$. Then $A(0,k_1)$ is homeomorphic Te $A(0,k_2)$. Proof. Again, we may define image explicit homerho .... higher:A(0,k_1)\,to A(0,k=[2)$, thank time by the recall $$h(x,y)=\left(x,\frac{k_2(x)}{k_1(x},y\right)$ for [-x\in(-1,1)$ and $(-h(-)}$,0)=h(1”,0)=0$. This is continuous for $x.\Integr(-1,1)$, since $k_1$ and $k_2$ are Cont and products and quotients of Not functions are continuous (the latter wherever the denominator is nonzero). But $h$ Al continuous at the power $(\pm1,0 200 since $\frac{y}{k_1!(x)}\in[0,1]$ foreg $(x,y)\in A+(0,k_ 1)$, while $ kg=>2(x)$ goes to $0$ as $x$ approaches $\pm1$. Sogt two limits $\lim_{(x,ys)\to?pm}}=\,0)}h(x,y)$ exist and equal the corresponding function values implemented By term same argument,gt interpolation "$h^{-1}(x,y)=\left)*( quant,\frac{icks("1|x)}{k_-(x)}y\right)$$ continuous. So $h$ is indeed a homeomorphism. $\square$ Proposition. $A(f,g)$ is homeomorphic trace $D$. Proof. By Lemma 1 Description $}^{\(f,g)$ is homeomorphic test $A(0,g-f)$. By Lemma 2, -]{(0,g-f)$ is homeomorphic to $A(0));Thankk)$ with $k(x)=\sqrt{1-x^2}$. By Lemma 1 again description $A(0,twok)$ is homeomorphic to $!(-k,k)=D]$. Therefore $A(f,g,$$ is homeomorphic trials $D$.Gsquare$ }_{ you want an specifically homeomorphism, simply calculate the composition of all the maps used. (By the way, the proofs are even simpler if you work with an open unit disk and 0open heart" instead: The wayyou don't Give to analyse the points $(\pm1....0 $(- separately.) To obtain (*pleygonal ; of) g heart, :) could use equivalent.g. $$f(x)=|x|-1$$ and $$g(x) = \frac12-\left||x|-\frac12\right|,$$ but I'm sure you can come up with a better ("round_ version of a) heart yourself divide the same argument will work. In any case, it is probably helpful to spend some time trying to visualize what each of the many does.[SEP]
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[CLS]Question # The coefficient of x in the expansion (1+x)(1+2x)(1+3x)⋯(1+100x) is also equal to A 1002992+982972+962952++2212 B The sum of all 101 A.M.'s inserted between 1 and 99 C The sum of all 100 A.M.'s inserted between 1 and 99 D The sum of first 20 terms of the series 1+(1+3)+(1+3+5)+(1+3+5+7)+ Solution ## The correct options are A 1002−992+982−972+962−952+⋯+22−12 B The sum of all 101 A.M.'s inserted between 1 and 99 Coefficient of x in the expansion (1+x)(1+2x)(1+3x)⋯(1+100x) =1+2+3+⋯+100=100×1012=5050 Now, 1002−992+982−972+962−952+…+22−12=(1002−992)+(982−972)+(962−952)+…+(22−12)=(100+99)(100−99)+(98+97)(98−97)+…+(2+1)(2−1)=100+99+98+97+…+2+1=5050 Now, Let n A.M.'s are inserted between 1 and 99, then the sum of them  5050=n+22[1+99]−(1+99)⇒5050=100[n+22−1]⇒5050=50n⇒n=101 Therefore 101 A.M.'s are inserted. Now, the sum of first 20 terms of the series 1+(1+3)+(1+3+5)+(1+3+5+7)+⋯  General term of the series is Tr=1+3+5+7⋯(2r−1)=r2⇒S=20∑r=1Tr=20∑r=1r2⇒S=n(n+1)(2n+1)6⇒S=20×21×416=2870Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More[SEP]
[CLS]Question # The coefficient of x in the expansion (0001+x)(1+2x)(1+3x)⋯(1+100x) is also equal tests A 1002992+982972+962952++2212 B The sum of all 101 A.M.'s inserted between ? and 99 C The S of all 100 A.M.'s inserted between 1 and 99crD The sum of first 20 terms of the series "+(1+3)+(1+3+5)+(1+3+5+7)+ Solution ## The correct options are A 1002−992+982−972+962−952+⋯+22−12 B The sum of all 101 �.M.'s inserted between 1 and 99 Coefficient of x in the expansion (1+x)(1+2x)(1+3x)⋯(1+100x) = }_{+2+3+⋯+100=100×1012=)$.50 Now, 1002−992+982−972+962−952+…+22−12=( 2−992)+(982−972)+(962 −952)+…+(22−12)=(100+99{{\100−99)+(}&+97)(98−97)+…+(2+1)(2−1)=100+99+98+97+…+2+1=5050 Now, Let n A.M.'s are inserted between 1 and 99, then the sum of them  after50=n+22[1+99]−(1+ \}$)⇒5050=100[n+22−1]⇒5050=50 np⇒n=101 Therefore 101 A.M.'s are inserted. Now, the sum of first 20 terms of the series 1+(1+3)+(1+3+5)+(1+3+5+7)+⋯  General term of the series is Tr=1+3+5+7⋯(2r−1)=r2⇒S=20∑r=1Tr=20∑r=1r2⇒S=n(n+1)(2n+1)6⇒S=20×21×416=2870Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More[SEP]
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[CLS]# Does every countably infinite interval-finite partial order embed into the integers? A partially ordered set $(S,\le)$ is called interval finite if the open intervals $(x,z):=\{y|x\le y\le z\}$ are finite for all choices of $x,z$ in $S$. An embedding $(S,\le)\rightarrow(S',\le')$ of partially ordered sets is an injective order-preserving map. Does every countably infinite interval finite partially ordered set admit an embedding into the integers? This is equivalent to extending the partial order to a linear suborder of the integers. If so, where can I find the proof? If not, can you give a counterexample? • I don't understand the votes to close, since this is an interesting problem, and I think it is trickier than it may seem at first. Could someone explain? Sep 24 '13 at 13:56 • Sep 24 '13 at 23:56 The answer is yes. First, let's prove a lemma. By order preserving, I assume that you mean forward-preservation of the order: $p\leq q\implies f(p)\leq' f(q)$. Lemma. Every countable interval-finite partial order $\P$ has a convex enumeration, an enumeration $\langle p_0,p_1,p_2,\ldots\rangle$ of $\P$, all of whose initial segments are convex sets in $\P$. Proof. If we have a finite convex subset of $\P$, and new point $p$ to be added, then by convexity $p$ does not appear in any interval of points we already have. If $p$ is above some points we have already, then it is not below any point that we have already, and so we can look at the intervals $(q,p)$ determined by a point $q$ we have already and the new point $p$. By convexity, none of these new points can be below any point we already have, and so we can simply add them from the bottom while maintaining convexity. A similar argment works if the new point is only below points we already have. And if $p$ is incomparable to the points we already have, then we can simply add it to the list. QED Now, we can prove the theorem. Theorem. Every countable interval-finite partial order embeds into $\Z$. Proof. Suppose that $\P$ is a countable interval-finite partial order. By the lemma, it has a convex enumeration $p_0,p_1,p_2,\ldots$. Suppose by induction that we have mapped $p_k\mapsto m_k$ in an injective order-preserving manner, for $k\lt n$. Consider the next point $p_n$. Since the order so far is convex and adding $p_n$ maintains convexity, it follows that either $p_n$ is above some points $p_k$ for $k\lt n$ and not below any, or below some such $p_k$ and not above any, or incomparable to them all. In any case, we can easily extend the map to define $p_n\mapsto m_n$ in such a way to still be order preserving and injective. QED • Joel, thanks, that's terrific. I wonder if you know the origin of this result, since I need to cite it. – Ben Sep 24 '13 at 17:59 • I've never seen it before, but I'd expect that probably this has been known. Perhaps someone else can post a source? Sep 24 '13 at 18:07 • The link above discusses the source and gives a reference. Sep 24 '13 at 23:56 • (By the way, the question in the link is still unsolved without choice, in case you have some ideas.) Sep 25 '13 at 0:00 • @Andres, thanks for the reference! The OP on this question, however, insists on injective order-preserving maps, and so there can be no uncountable instances. So it seems that these are slightly different questions, although obviously closely connected. Sep 25 '13 at 0:07[SEP]
[CLS]# Does every countably infinite interval- modulo partial order embed into THE integers? A partially ordered set $(S,\le)$ is called interval finite if the open intervals $(x,z):=\{y|x)}{\le y\le z\}$ are finite for all choices of $x,z$ in $S$. An embedding $( recurrence,\le)\rightarrow(S',\le')$ of partially ordered sets is an injective order-preserving map. Does every countably infinite interval finite partially ordered set admit an embedding into the integers##### This​ equivalent to extending the partial o to a linear suborder of the integers. If so, where tells I find the proof? If not, can you give a counterexw? •� don't understand the votes to close, special this is an interesting problem, and I think it I trickier than it may seem at first. Could someone explain? Sep 24 '}_{\ at 13:56 cent• Sep 24 '13 at 23:56 conclude The answer is ©. First, let's page a lemma. By order preserving, I assume that you mean forward-preservation of the order: $p\leq q\implies f(p)\leq' f(q)$. Lemma. Every countable Enter-finite partial order $\P$ has a convex enumeration, an enumeration $\langle proper_)}=\,p_1,p_2,\ldots}\,rangle$ of $\P$, all of whose initial segments are convex sets in $\P $(- successProof., � we have a finite convex subset of $\P$, and new point $p$ to be added, then by convexity $p$ does not appear in any divisor of points we already have. If $ perhaps$ is above some points we have already, The it is not below any point that we have day, and showing website can look at the ordinary $(q,p)$ determined by a point $q$ we have already dividing the new point $p$. By convexity, none of these new points can be below any point weI v, and so we can simply add them from the bottom while maintaining convexity. A similar argment works if the new point is only below points website already have. And IS ->p$ is incomparable to the points we Al have, then we can side add it to the list:= QED Now, we can prove the theorem. Theorem. Every countable interval-finite partial order embeds into $\Z$. able. Suppose that $\P$ is a countable interval-finite partial order. By the lemma, ant has � convex enumeration $p_0,p_1,p_2,\ldots$. Suppose by induction that we have mapped $p_ software\mapsto m}[k$ in an injective dealing-preserving manner, for $k\lt n$. Consider the next provides $p_n$. Since the order so From is convex then adding $p_n})$. maintains convexity, it follows that either $p_n$ is above some percent $p_k$- for $ combined\tau n$ and not below any), or below some such $p_k$ and not above any, or incomparable to them all. In any case, we can basically extend the mod to define $p_n\mapsto m_n$ in such a way to still be order preserving and injective. Qetric • Joel, thanks, that's terrific. I wonder if you know the origin of this result, since I need to cite it. – Ben PS 24 '13 at 17:59 • I've never seen it Br like but I'd expect Trans probably this has b known. Perhaps someone else can post a source? So 24 '13 at 18: 7 • The link above discusses the source and gives a reference. Sep 24 '13 at 23:56 \| (By the way... These question in the link imagine Sol unsolved without choice, in be you have some ideas.) Sep 25 '13 at 0:00 • <-Andres, thanks for the reference! The OP on this question, however, insists on injective order-preserving maps, and so there can be no uncountable instances. So it seems that these are slightly different questions, although obviously selection connected., Sep 25 '13 at 0[]07[SEP]
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[CLS]# 0.999...=1 Although some people find it counterintuitive, the decimal expansions $$0.999\dotsc$$ and $$1$$ represent the same real number. # Informal proofs These “proofs” can help give insight, but be careful; a similar technique can “prove” that $$1+2+4+8+\dotsc=-1$$. They work in this case because the series corresponding to $$0.999\dotsc$$ is absolutely convergent. • \begin{align} x &= 0.999\dotsc \newline 10x &= 9.999\dotsc \newline 10x-x &= 9.999\dotsc-0.999\dotsc \newline 9x &= 9 \newline x &= 1 \newline \end{align} • \begin{align} \frac 1 9 &= 0.111\dotsc \newline 1 &= \frac 9 9 \newline &= 9 \times \frac 1 9 \newline &= 9 \times 0.111\dotsc \newline &= 0.999\dotsc \end{align} • The real numbers are dense, which means that if $$0.999\dots\neq1$$, there must be some number in between. But there’s no decimal expansion that could represent a number in between $$0.999\dots$$ and $$1$$. # Formal proof This is a more formal version of the first informal proof, using the definition of decimal notation. $$0.999\dots$$ is the decimal expansion where every digit after the decimal point is a $$9$$. By definition, it is the value of the series $$\sum_{k=1}^\infty 9 \cdot 10^{-k}$$. This value is in turn defined as the limit of the sequence $$(\sum_{k=1}^n 9 \cdot 10^{-k})_{n\in\mathbb N}$$. Let $$a_n$$ denote the $$n$$th term of this sequence. I claim the limit is $$1$$. To prove this, we have to show that for any $$\varepsilon>0$$, there is some $$N\in\mathbb N$$ such that for every $$n>N$$, $$|1-a_n|<varepsilon$$. Let’s prove by induction that $$1-a_n=10^{-n}$$. Since $$a_0$$ is the sum of {$0$ terms, $$a_0=0$$, so $$1-a_0=1=10^0$$. If $$1-a_i=10^{-i}$$, then \begin{align} 1 - a{i+1} &= 1 - (ai + 9 \cdot 10^{-(i+1)}) \newline &= 1-a_i − 9 \cdot 10^{-(i+1)} \newline &= 10^{-i} − 9 \cdot 10^{-(i+1)} \newline &= 10 \cdot 10^{-(i+1)} − 9 \cdot 10^{-(i+1)} \newline &= 10^{-(i+1)} \end{align} So $$1-a_n=10^{-n}$$ for all $$n$$. What remains to be shown is that $$10^{-n}$$ eventually gets (and stays) arbitrarily small; this is true by the archimedean property and because $$10^{-n}$$ is monotonically decreasing. <div><div> # Arguments against $$0.999\dotsc=1$$ These arguments are used to try to refute the claim that $$0.999\dotsc=1$$. They’re flawed, since they claim to prove a false conclusion. • $$0.999\dotsc$$ and $$1$$ have different digits, so they can’t be the same. In particular, $$0.999\dotsc$$ starts “$0.$,” so it must be less than 1. Decimal expansions and real numbers are different objects. Decimal expansions are a nice way to represent real numbers, but there’s no reason different decimal expansions have to represent different real numbers. • If two numbers are the same, their difference must be $$0$$. But $$1-0.999\dotsc=0.000\dotsc001\neq0$$. Decimal expansions go on infinitely, but no farther.$$0.000\dotsc001$$ doesn’t represent a real number because the $$1$$ is supposed to be after infinitely many $$0$$s, but each digit has to be a finite distance from the decimal point. If you have to pick a real number to for $$0.000\dotsc001$$ to represent, it would be $$0$$. • $$0.999\dotsc$$ is the limit of the sequence $$0.9, 0.99, 0.999, \dotsc$$. Since each term in this sequence is less than $$1$$, the limit must also be less than $$1$$. (Or “the sequence can never reach $$1$$.”) The sequence gets arbitrarily close to $$1$$, so its limit is $$1$$. It doesn’t matter that all of the terms are less than $$1$$. • In the first proof, when you subtract $$0.999\dotsc$$ from $$9.999\dotsc$$, you don’t get $$9$$. There’s an extra digit left over; just as $$9.99-0.999=8.991$$, $$9.999\dotsc-0.999\dotsc=8.999\dotsc991$$. There are infinitely many $$9$$s in $$0.999\dotsc$$, so when you shift it over a digit there are still the same amount. And the “decimal expansion” $$8.999\dotsc991$$ doesn’t make sense, because it has infinitely many digits and then a $$1$$. Parents: • If these are included I think it would be good to also include explanations of why each one is wrong.[SEP]
[CLS]###### 0.999...=1 Although semi people find it counterintuitive, the decimal expansions $$0.999\dotsc$$ and $$1$$ required the so real number. Mac ¶ InHello proofs CThese “proofs” can help give insight, but be careful; a similar technique branch “prove” that !1+2+4{(8+\dotswers=-1$$. They work in Thus case because the series corresponding to $$0.999\dotsc$$ is absolutely convergent. etc • #begin{align} Ex &= 0.999\dotsc \newline 10² &= 9.999\dotsc \newline $(\xSomex &= $\{.999\dotsc-0.999\dotSS \newline (\x &= $(\ \newline x &= 1 \newline \end{align} • \begin{align} \frac 1 9 &= 0.111\dotsc \newline 1 &= \frac 9 9 \}$newline &= 9 \times _frac ? 9 {-newline &= 9 $-\times 0.111\dotsc \Thankmid &= 0.999\dotsc \end{align} • The real numbers are dense, which means that if $$0.999\dots\neq1$$, there must be some number in between. greatest there’s no decimal expansion that cod represent a number in between $$0,...999\dots$$ and $$1$$.34 # Fder proof circular It imagine a more module version of the first volume proof, using the definition of decimal notation. $$0.999\ dot$$ is the decimal expansion where equations Di after the det point II a &9$$. By definition, implicit is the value of the series $$\sum}=k=1}^\infty 9 \cdot 10^{-acks}$$. This value is in turn De as the limit of the sequence $$(\sum_{k=120}^n 9 \ transition 10^{-k})_{n\in\mathbb N}$$. step $$a_n,$$ denote the $$n.$$th term of this sequence. ) claim this limit is $$1$$. To precise this; we have to show that for An $$\varepsilon>0$$, there is some $$N\At\mathbb N$$ such THE for every $$agon ->N$$, $$|1-a_n&=varepsilon$$. Let’s prove by induction that$-1-a_n=10)^{\n}$$. Since $$a_)}{$$ is the sum of {$0),$$ terms, $$a_0)}=0$$, so $$1-a_0=1=10^0 "$})$. If $$1-a_i=10^{-i}$.$ thenca \begin{align} 1 - a{i+1} &= 1� [-ai + 9 \cdot 10^{-(i+1)}) \newline &= 1-a_i − 9 \cdot 10^{-(i+1)} ]randomline &= 10^{-i} irreducible 9 \cdot 10^{-(i+1)} \newline &= 10 \cdot 10^{-(i+|1)} − 9 \cdot reasonable^{-(my+1)} \newline &= 10^{-)(i+1)} \end{align} So $$1-a_n=10^{-n}$$ Ref all $$ 56approx$. What remains to be shown is that $$ 2010^{-n}$$ eventually gets (and stays) probably small; THE is true by the archimedean property analysis because $$10^{-n},$$ is monotonically decreasing. <div><div* # Arguments against $$0. 56\dotsc=1$$ color arguments are used to try types refute the claim that $$0.999\{dotsc=1$$. They’re flawed, since they claim to prove a suffices cyclic. • $$0.999\\\dotsc$$ and $$001$$ have directed α, so they c’t be the same. In particular, $(0.999\dotsc$$ starts “$0.$,” so it must be less than 1. Decimal expansions and real numbers are different objects. Decimal span are a nice ways to represent real Root, but there’`. connection reason different decimal expansions High to represent different real numbers.ics • If two numbers are the same, their difference must be $$0).$$$. But $$1Posts}}{\.999\dotsc=0.89,\dotsc001\neq}+\$$\ Decimal expansions // on infinitely.... but no farther.$$0.000\dotsc001$$ doesn′t represent a real number because the $$1$$ is supposed to be Pat infinitely men $$\{$$s, but each digit has to be a finite distance from the decimal point. iff you have to pick a real number tan for $$0.000\dotsc 34$$ to represent, I would be $$0$$. • $$0.999\dotsc$$ ? the limit of Theorem sequence $$0.9, 0!.99, 0.999, \dot Section)}$$$. Since each term in this sequences is lesson thank $$|1$75 the limit must solves be less than $$1$$. (Or “itus sequence certain never team $$1).$$$.”) The sequence gets arbitrarily close to $$1$$, so its limit i $$|1$$. It doesn’t matter that all of THE terms are less than $$1$$. • In the first proof, when you subtract $$0.999\ implicitsc$$ from $$9.999\dotsc$$, you don’ts get $$ 2009$ ($ There’s an extra digit left over; ). as $$9.}$$Posts0.999=8.,991.$$, $$9.999\dotsc-0.999\,$dotsc=8.999\dotsc991$$. in are infinitely many $$9$$s in $$0...,^{\\dotsc$$, so when you shift it over a digit there are still the same AM. And the “decting exercise” $$8.999\dotsc991$$ residuals’t make sense, because ant has infinitely many digits and these a $$1$$. ps: • If these are including I think it would be to also include explanations of why each one is wrong.[SEP]
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[CLS]# Does the Fourier series coefficient of AC components remains same if DC component is subtracted form the given signal? Suppose a signal is defined by $$x(t)= \begin{cases} t & 0\leq t \leq 1 \\ 2-t & 1\leq t\leq 2 \\ \end{cases}$$ Since $$x(t)$$ has even symmetry, I can calculate fourier coefficient as $$a_n = \frac{4}{T} \int_0^1 x(t).\cos{n\pi t}.{dx}$$ I have calculated $$$$a_n = 2\big[\frac{\cos{n\pi} - 1}{n^{2}{\pi}^{2}}\big]\tag{1}$$$$ The DC value of $$x(t)$$ i.e $$a_0 = 0.5$$. If we subtract DC value we get, From this we can see that given signal has hidden half wave symmetry in addition to Even symmetry. So we can find fourier coefficient as $$a^{'}_n=\frac{8}{T}\int_{0}^{\frac{1}{2}}(t-\frac{1}{2})\cos{n\pi t}.dt$$ I have calculated $$$$a^{'}_n = 4\big[\frac{\cos{\frac{n\pi}{2}} - 1}{n^{2}{\pi}^{2}}\big]\tag{2}$$$$ My question is, shouldn't $$a_n$$ and $$a^{'}_n$$ be equal for $$n\neq0$$ ? • How about calculating the Fourier coefficients without using any extraneous considerations such as symmetry or hidden half symmetries?. That is, copy the definition of $a_n$ (the one that applies to all periodic signals, long before extraneous considerations such as symmetry are mentioned) from your book, and calculate $a_n$ and $a_n^\prime$ and see if you get the same answer or different answers. If you get the same answer, the problem is in your understanding of symmetry/half-symmetry/hidden etc. – Dilip Sarwate Oct 27 '18 at 15:27 • if we apply half wave symmetry then it means even components will be zero and odd components of equation 1 & 2 are indeed equal. Equation 2 will give non-zero value for even values of n, other than multiples of 4, but we should discard it according to the conclusion of half wave symmetry. – Saurabh Oct 27 '18 at 19:00 • As Dilip says, I also suggest that you repeat your calculations without worrying about any symmetry; that is, calculate the series for an entire period $T=2$. You should get the exact same answer in both cases except for $n=0$. – MBaz Oct 27 '18 at 22:24 • "If we apply half wave symmetry...." Sigh! You can lead a horse to water but you cannot make him drink. – Dilip Sarwate Oct 28 '18 at 3:39 • @DilipSarwate I did calculate the fourier series coefficient for entire time period i.e. $T=2$ and I got same value in both cases for $n\neq0$. But since -Fat32 already given proof in their answer that CTFS coefficients for DC-removed part will same as original signal, I didn't mention it in my previous comment. – Saurabh Oct 28 '18 at 6:43 PART-I: I would like to provide the general proof considering the title of the question and imposing no specific properties on the signal $$x(t)$$ other than having a CTFS representation. The following is a simple analysis to conclude that the CTFS coefficients of any signal $$x(t)$$ and that of the DC removed signal are equivalent. (except $$a_0$$ of course). Consider a continuous-time periodic signal $$x(t)$$ with period $$T$$ divided into two components: $$x_{dc}$$ and $$x_{ac}$$, with periods $$T$$ also, where $$x_{dc}$$ is the pure DC component of $$x(t)$$ and $$x_{ac}$$ is the pure AC component of $$x(t)$$, then we have: $$x(t) = x_{dc}(t) + x_{ac}(t)$$ Computing the CTFS coefficient $$a_k$$ of $$x(t)$$ yields: \begin{align} a_k &= \frac{1}{T} \int_{} (x_{dc} + x_{ac}) e^{-j k \frac{2\pi}{T} t } dt \\ &= \frac{1}{T} \int_{} x_{dc} e^{-j k \frac{2\pi}{T} t } + \frac{1}{T} \int_{} x_{ac} e^{-j k \frac{2\pi}{T} t } \\ a_k &= b_k + c_k \\ \end{align} where $$b_k$$ and the $$c_k$$ are the CTFS coefficients of DC and AC parts of $$x(t)$$. By definition of any DC signal, it's known that $$b_k = 0$$ for all $$k \neq 0$$ and by defition of any AC signal it's known that $$c_0 = 0$$. Then using the relation $$a_k = b_k + c_k$$ we get the following: $$a_0 = b_0 + c_0 = b_0$$ and $$a_k = 0 + c_k = c_k ~~~,~~~ \text{ for all } k \neq 0$$ From which we define : $$b_k = \begin{cases} a_0 ~~~&, ~~~\text{ for } k = 0 \\ 0 ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases}$$ and $$c_k = \begin{cases} 0 ~~~&, ~~~\text{ for } k = 0 \\ a_k ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases}$$ Hence we conclude that the CTFS coefficients, $$a_k$$, of any periodic signal $$x(t)$$ and the CTFS coefficients $$c_k$$ of DC-removed part, $$x_{ac}$$, are the same for all $$k \neq 0$$. PART-II: Based on OP comments, the relation for an even and real signal, is the following. For a signal $$x(t)$$ which is real and even we have $$x(t) = x(t)^{*} = x(-t) = x(-t)^{*}$$ and the associated CTFS coefficients has the property of $$a_k = a_{-k}^{*} = a_{-k} = a_k^{*}$$ which indicates that the coefficients $$a_k$$ are also real and even. Using this, we can obtain the trigonometric (cosine) Fourier series coefficients as. \begin{align} a_k &= \frac{1}{T} \int_{} x(t) e^{-j k \frac{2\pi}{T} t } dt \\ &= \frac{1}{T} \int_{} x(t) \left( \cos( k \frac{2\pi}{T} t) + j \sin(k \frac{2\pi}{T} t) \right) dt \\ &= \frac{1}{T} \int_{} x(t) \cos( k \frac{2\pi}{T} t) dt + j \frac{1}{T} \int_{} x(t) \sin( k \frac{2\pi}{T} t) dt\\ a_k &= \mathcal{Re}\{a_k\} + j ~~ \mathcal{Im}\{a_k\} \\ \end{align} Now since the property states tat $$a_k$$ are real, then the imaginary part is zero and we have: $$a_k= \frac{1}{T} \int_{-T/2}^{T/2} x(t) \cos( k \frac{2\pi}{T} t) dt$$ Furtermore since $$x(t)$$ is also even; $$x(t)=x(-t)$$, then we also have $$\boxed{ a_k= \frac{2}{T} \int_{0}^{T/2} x(t) \cos( k \frac{2\pi}{T} t) dt }$$ As the trigonometric cosine series coefficients of the real and even signals. In addition to this, for a real & even signal, $$x(t)$$ of period $$T$$, which has no DC part, the following is also observed: $$x(t-\frac{T}{2}) = -x(t)$$ And based on the time-shift property of CTFS we can conclude that $$a_k ~~e^{-j\frac{2\pi}{T}k \frac{T}{2} } = - a_k$$ $$a_k ~ e^{-j\pi k } = - a_k \implies a_k = \begin{cases} -a[SEP]
[CLS]# Does the Fourier series coefficient of AC components remains same if DC component is subtracted form the given signal? Suppose a signal i defined by $$ Next(t)= \begin{cases} t & 0\leq t \leq 1 \\ 2-t & 1\leq t\leq 2 \\ \end{cases}$$ Since $$xDoest)$$ has given symmetry, I can calculate f errors coefficient as $$a_n = \frac{4}{T} \int_0^1 x(t).\cos{n\pi t}.{dx}$$ I have calculated $$$$a_ no = 2\big[\frac{\cos{n\pi} - 1}{n^{}}$.}{\pi}^{_{(}}\big]\tag{1}$$$$ The DC value of $$x(t)$$ i.e $$a_0 = 0.5$$. If we subtract DC value we get, From this we can see that given signal has hidden half wave symmetry in addition to Even symmetry. So we can find fourier coefficient as concepts$$a^{'}_n=\frac{8}{T}\int_{0}^{\frac{1}{2}}(t-\frac{1}{2})\cos{n\pi t}.dt$$ I have calculated $$$${a^{'}_n = 4\big[\frac{\cos{\frac{n\pi}},2}} - 1}{n^{2}{\pi}^{2}}\big]\tag{2}$$$$ My question is, shouldn't $$a_n$$ and $$a^{'}_n$$ be equal for $$n\neq0$$ ? • How about calculating the Fourier coefficients without using any extraneous considerations such as symmetry or hidden half symmetries?. That is, copy the definition of $a_n$ (the one that applies to all periodic signals, long before extraneous considerations such as symmetry are mentioned) from your book, and calculate $a_n$ and $a_n^\Pro$ and see if you get the same answer or difficult answers. If you get the same answer, the problem is in your understanding of symmetry/half- momentumised/hidden etc. – Dilip Sarwate Oct 27 '18 at 15:27 • if we apply half wave symmetry then it means even components will be larger and odd components of equation 1 & 2 are indeed equal. Equation 2 will give non-zero value for even values of n, other than multiples of 4, but we should discard it according to the conclusion of half wave scaling. – Saurabhort 27 '18 at 19:00 • As Dilip says, I also suggest that you repeat your calculations without worrying tests any symmetry; that is, calculate this series for an entire period $T=2$. You should get the exact same answer in both cases except for $n=0$. – MBaz Oct 27 '18 at 22:24 }^{ "If we apply half wave symmetry)\\ S higher! You can lead a horse to water but you constant make him drink. – Dilip Sarwave Oct 28 '18 atg:39 • @DilipSarwate I did calculate the fourier series coefficient for entire time period i.....e. $T=2$ and I got same value in both cases colors $n\neq0$. But since -Fat3 already given proof in their answer that CTFS coefficients for DC-removed part will same as original signal, I didn't mention it in my previous comment. – Saurabh Oct 28 '18 at 6:43 PART-I: I would like tells provide the general proof considering the title of the question and imposing no specific properties on the signal $$x(t)$$ other than having a CTFS representation. The following is a simple analysis to conclude that the CTFS coefficients of any signal $$x(t)$$ and that of the DC removed signal Se equivalent. (except $$a_0$$ of Co). confusion λ a continuous-time periodic signal $$x(t)$$ with periodic $$T$$ divided into two components: $$x_{dc}$$ and $$x_{ac}$$, with Pro $$T$$ also, where $$x_{dc}$$ is the pure DC component of $$x(t)$$ and $$x_{ac}$$ is the perhaps AC component of $$x(t)$$, then we have: $$x(t) = x_{dc}(t) + x_{ac}(t)$$ Computing the CTFS coefficient $$^{(_k$$ of $$x(t)$$ yields: \begin{align} a_k &= \frac{1}{T} \ˆ_{} (x)^{\dc} + x_{ac}) e^{-j k \frac{2\pi}{T} t } dt \\ &= \frac{1}{T} \int_{} x_{ updated} e^{-j k \frac{2\pi}{T} t } + \frac{1}{T} \int_{}: x_{ac} e^{-j k \frac{2\pi}{T} t } \\ a_k &= b_k + c_k \\ \end{align} where $$b_k$$ and the $$c_k$$ prefer the CTF coefficients of DC and AC parts of $$x( Ext)$$. By definition of any DC signal, it +\ known that $$b_k = 03$$ for all $$k \neq " "$ and by defition of any AC signal it's known that $$c_0 \| 0)}$$. Then using the require $$a_k = b\|_k + c_k $[ we get the following: $$a_0 = b_0 + c_0 = b_0$$ and $$a_k = 0 + c_k = c_k (",~~~ \text{ for all } k \neq 0$$ From which we define : $$b_k = \begin{cases} a_0 ~~~&, ~~~\text{ for } k = 0 \\ / (*~~~&, ~~~\text{ for all } k \neq 0 \\ \�{cases}$$ and $$c_k = \begin{cases} 0 ''~~~&, ~~~\text{ for } k \{ 0 \\ a_k ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases}$$ Hence we conclude that the CTFS coefficients, $$a_k$$, of any periodic signal $$x(t)$$ and the CTFS coefficients $$c_k$$ of DC-removed part, $$x_{ac}$$, are told same for all $$k {-neq 0$$. PART-II: Based on OP comments, the relation for an even and real signal, is the following. For a signal $$x(t)$$ which is real and even we have $$x(t) = x(t)^{*} = $|(-t) = x(-t)^{*}$$ and the associated CTFS coefficients has the property of $$a_k = a_{-k}^{*} = a_{-k)}\ = a_k^{*}$$ which indicates that the coefficients $$a_k$$ are also real and even. Using this, we can obtain the trigonometric (cosine) Fourier series coefficients as..., \My{align} a_k &= \frac{1}{T} \int_{} $[(t) e^{- Description k \frac{2\pi}{T} t } dt \\ &= \frac{)}=}{T} \int_{} x(t) \left( \cos( k \frac{2\pi}{T} t) + j \sin(k \frac{2\pi}{T} t) \right) dt \\ &= \frac{1}{T} \int_{} x(t) \cos( k \frac{2\pi}{T} t) dt + j \frac{1}{T} \int_{} x(t) \sin( k \frac{2\pi}{T} t) dt\\ a_k &= \mathcal({Re}\{a_ computation\} + j ~~ \mathcal{Im}\{a_ knowledge\} \\ \end{align} Now since the property states tat $$a_k$$ are real, then the imaginary part is zero and we have: "$a_k= \frac{1}{T} \int_{-T/2}^{T/2} x(t) \cos( k \frac{2\plicit}{T} t) dt$$ Furtermore since)$x(t)$$ is also even; $$x(t)=x(-t)$$, tends we also have AC $$\boxed{ a_k= \frac{2}/T} \int_{0}^{T/25} x(t) \cos( k \ C{&\pi}{T} t) dt }$$ As the trigonometric cosine series coefficients of the real and even signals. In addition to this, for a real & even signal, $$x(t)$$ of period $$T$$, which has no DC part]: the following is also cube: $$x(t-\frac{ Att}{2}) = -x(t)$$ And based on the time-shift property friend CT F website can conclude that $$a_k ~~e^{-j\frac{equal)\\pi}{T}k \frac{T}{2} } }_{ - a_k$$ discuss$$a_k ~ e^{-j)?pi k } = - a_k \implies a_k = \begin{cases} -a[SEP]
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[CLS]How to solve $3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$? I am trying to solve the following question involving floor/greatest integer functions. $$3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$$ with the notations $$\lfloor x \rfloor$$ denoting the greatest integer less than or equal to $$x$$ and $$\{x\}$$ to mean the fractional part of $$x$$. I used the following property for floor functions. $$n\leq x$$ if and only if $$n \leq \lfloor x \rfloor$$ where $$n\in \mathbb{Z}$$ Let $$p=\lfloor x^{2} \rfloor$$, then $$p\leq \lfloor x^{2} \rfloor < p+1$$ $$\rightarrow p \leq x^{2} < p+1$$ $$\rightarrow \sqrt{p} \leq x < \sqrt{p+1}$$ , since $$\sqrt{p} \in \mathbb{Z}$$ $$\rightarrow \sqrt{p} \leq \lfloor x \rfloor < \sqrt{p+1}$$ We then have $$\sqrt{p} = \lfloor x \rfloor$$ Since $$\{x\}=x-\lfloor x \rfloor,$$ $$3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2\{x\}= 3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2(x-\lfloor x \rfloor)= 5\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2x=0$$ Substituting $$p$$, $$\sqrt{p}$$ for $$\lfloor x^{2} \rfloor$$ and $$\lfloor x \rfloor$$ respectively, and also letting $$x= \sqrt{p},$$ we get $$p = 3\sqrt{p}$$ solving for $$p$$ gives $$p=0, 9$$, and hence $$x=0, 3$$ The problem is that according to the solution for the problem, $$x$$ also equals to $$\frac{3}{2}$$ for $$\{x\}=\frac{1}{2}$$ since $$2\{x\}\in \mathbb{Z}$$. However, by definition for $$\{x\}$$, $$0 \leq \{x\} < 1$$, then $$0 \leq 2\{x\} < 2$$. How can $$\{x\}=\frac{1}{2}$$ and how do I use this to obtain $$x=\frac{3}{2}$$. I am not sure what I am missing. IF I made any mistakes in my reasoning. Can someone point it out to me please. Thank you in advance. • It's not true that $\sqrt{p}\in\mathbb{Z}$ Say $x=1.5$ Then $p=\lfloor 2.25\rfloor = 2,$ and $\sqrt{p}=\sqrt{2}$ May 10, 2020 at 0:35 • @saulspatz thank you for pointing that out. May 10, 2020 at 0:45 Let $$x = n + r$$ where $$n = [x]$$ and $$r = \{x\}$$. Then we have $$3n - [n^2 + 2nr + r^2]=2r$$ $$3n - n^2 - [2nr + r^2] = 2r$$ and.... oh, hey, the LHS is an integer the RHS being $$2\{x\}$$ means $$\{x\} = 0$$ or $$0.5$$. Two options $$x$$ is an integer and $$x = [x] = n$$ and $$r=\{x\} = 0$$ and we have $$3n-n^2=0$$ and $$n^2 = 3n$$ and $$n= 0$$ or $$n = 3$$. So $$x = 0$$ and $$x=3$$ are two solutions. (Check: $$x=0\implies 3[x] - [x^2] = 3*0 - 0 = 0 = \{0\}$$. Check. And $$x = 3\implies 3[x]-[x^2] = 3[3]- [3^2] = 3*3-9 = 0=\{3\}$$. Check. And if $$x = n + \frac 12$$ and $$r = \frac 12$$ then $$3n - n^2 - [2n\frac 12 + \frac 14] = 2\frac 12$$ $$3n - n^2 - [n + \frac 14] = 1$$ $$3n -n^2 - n = 1$$ $$n^2 - 2n + 1 =0$$ so $$(n-1)^2 = 0$$ and $$n = 1$$. $$x = 1+\frac 12 = 1\frac 12$$. (Check: If $$x = 1.5$$ then $$3[x] - [x^2] = 3[1.5] - [1.5^2] = 3*1 - [2.25]=3-2=1 = 2*\frac 12 = 2\{1.5\}$$. Check.) Write $$\{x\}=x-\lfloor x\rfloor$$. Then we have $$5\lfloor x\rfloor - \lfloor x^2\rfloor = 2x$$Since the LHS is an integer, the RHS must be as well. There are two cases: $$x$$ is an integer, or $$x$$ is a half-integer. • $$x$$ an integer. Drop the brackets: $$5x-x^2=2x;\qquad x=0,3$$ • $$x$$ is a half-integer. Write $$x=y+1/2$$. Then $$x^2 = y^2+y+1/4$$, and again we can drop the brackets: $$5y-(y^2+y)=2y+1; \qquad y=1, x=3/2$$ • may I ask how you arrive at $x$ is a half integer. I mean can't $x$ be anything else in between $0$ and $1$? May 10, 2020 at 0:46 • After we substitute $\{x\}=x-\lfloor x\rfloor$, both sides are integers. Then if $2x$ is an integer, $x$ is either an integer or a half-integer. May 10, 2020 at 0:47 • I think my issue is the following: from $0 \leq \{x\} < 1$, we get $0 \leq 2\{x\} < 2$. So how do I determine where else $\{x\}$ could be. May 10, 2020 at 0:56 • Clearly the LHS is an integer. Then $2\{x\}$ is an integer as well. This means either $2\{x\}=0$, i.e. $x$ is an integer, or $2\{x\}=1$, i.e. $x$ is a half-integer. May 10, 2020 at 1:00 • I think I see it now, Since $3\lfloor x \rfloor - \lfloor x^{2} \rfloor\in \mathbb{Z}$ and $0 \leq \{x\} < 1$ then, $\frac{3\lfloor x \rfloor - \lfloor x^{2} \rfloor }{2} = \{x\}$ implies that $\frac{3\lfloor x \rfloor - \lfloor x^{2} \rfloor }{2} \leq \{x\}<1$ which forces $\{x\}=\frac{1}{2}$ May 10, 2020 at 1:05[SEP]
[CLS]How to solve $3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$? I am trying to slope the following question involving of/greatest integer functions. $$3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$$ with the notations $$\lfloor x \rfloor$$ denoting the greatest integer less than or equal to $$x$$ and $$\{x\}$$ to mean the fractional part of $$x$$. I used the following property for floor functions. $$n\leq x$$ if and only if $$n \leq \lfloor x \rfloor$$ where $$n\in \mathbb{Z}$$ Let $$p=\lor x^{2} \rfloor$$, then $$p\leq \lfloor x^{2} \rfloor < p+1$$ $$\rightarrow p \leq x^{2} < p+1$$ $$\rightarrow \sqrt{p} \leq x < \sqrt{p+1}$$ , since $$\sqrt{p} \inggmathbb{Z}$$ $$\rightarrow \sqrt{p} \leq \lfloor x \rfloor < \sqrt{p+1}$$ We then have $$\sqrt{p} = \lfloor x \rfloor$$ Since $$\{x\}=x-\lfloor x \rfloor,$$ $$3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2\{x\}= 3\lfloor x \rfloor - \lfloor x^{2} *rfloor - 2(x-\lfloor x \rfloor)= 5\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2x=0$$ Substituting $$p$$, $$\sqrt{p}$$ for $$\lfloor x^{2} \rfloor$$ and $$\lfloor x \rfloor$$ respectively, and also letting $$x= \sqrt{p},$$ we get $$p = 3\sqrt{p}$$ solving for $$p$$ gives $$p=0, 9$$, and hence $$x=0, 3$$ The problem is that according to the solution for the problem, $$x$$ also equals to $$\frac{3}{2}$$ for $$\{x\^{\frac{1}{2}$$ since $$2\{x\}\in \mathbb{Z}$$. However, by definition for $$\{x\}$$, $$� \leq \{x\} < 1$$, then $$0 \leq 2\{x\} < 2$$. How can $$\{x\}=\frac{1}{2}$$ and how do I use this to obtain $$x=\frac{3}{2}$$. I am not sure what I am missing. IF I made any mistakes in my reasoning. Can someone point it out to me please. Thank you in advance. • It's not true that $\sqrt{p}\in\mathbb{Z}$ Say $x=1.5$ Then $p\\\lfloor 2.25\rfloor = 2,$ and $\sqrt{p}=\sqrt{2}$ May 10, 2020 at 0:35 • @saulspatz thank you for pointing that out. May 10, 2020 at 0:45 Let $$x = n + r$$ where $$n = [x]$$ and $$r = \{x\}$$. Suppose we have $$3n - [n^2 + 2nr + r^2]=2r$$ $$3n - n^2 - [2nr + r^2] = 2r$$ and.... oh, hey, the LHS is an integer the RHSdeg $$2\{x\}$$ means $$\{x\} = 0$$ or $$0.5$$. Two options $$x$$ is an integer and $$x = [x] \{ n$$ and $$r=\{x\} = 0$$ and we have $$3n-n^2=0$$ and $$n^2 = 3n$$ and $$n= 0$$ or $$n = 3$$. So $$x = 0$$ and $$x=3$$ are two solutions. (Check: $$x=0\implies 3[x] - [x^2] = 3*0 - 0 = 0 = \{0\}$$. Check. And $$x = 3\implies 3[x]-[x^2] = 3[3]- [3^2] = 3*3-9 = 0=\{3\}$$. Check. And if $$x = n + \frac 12$$ and $$r = \frac 12$$ then $$3n - n^2 - [2n\frac 12 + \frac 14] = 2\frac 12$$ $$3n - n^2 - [ Min + \frac 14] = 1$$ $$3n -n^2 - n = 1$$ $$n^2 - 2n + 1 =0$$ so $$(n-1)^2 = 0$$ and $$n = 1$$. $$x = 1+\frac 12 = 1\frac 12$$. (Check: If $$x = 1mean5$$ then $$3[x] - [x^2] = 3[1.5] - [1.5^2] = 3*1 - [2.75]= old-2=1 = 2*\frac 12 = 2\{1.5\}$$. Check.) Write $$\{x\}=x-\lfloor x\rfloor$$. Then we have $$5\lfloor x\rfloor - \lfloor x^2\rfloor = 2x$$Since the LHS is an integer, the RHS must be as well. There are two cases: $$x$$ is an integer, or $$x$$ is a half-integer. • $$x$$ an integer. Drop the brackets: $$5x-x^2=2x;\ Q x=0,3$$ • $$x$$ is axes half-integer. Write $$x=y+1/2$$. Then $$x^2 = y^2+y+1/4$$, and again we can drop the brackets: $$5y-(y^2+y)=2y+1; \qquad y=1, x=3/2$$ • may I ask how you arrive at $x$ is a half integer. I mean can't $ explanation$ be anything else in between $0$ and $1$? May 10, 2020 at 0:46 • After we substitute $\{x\}=x-\lfloor x\rfloor$, both sides are integers. Then if $2x$ is an integer, $x$ is either an integer or a half-integer. May 10, 2020 at 0:47 • I think my issue is the following: from $0 \leq \{x\} < 1$, we get $0 \leq 2\{x}}^{ < 2$. So how do I determine where else $\{x\}$ could be. May 10, 2020 at 0:56 • Clearly thellHS is an integer. Thank $2\{x}( is an integer as well. This means either $2\{x\}=0$, i.othing. $x$ is an integer, or $2\{x\}=1$, i.e. $x$ is a half-integer. May 10, 2020 at 1:00 • I think I see it now, Since $3\lfloor x \rfloor - \lfloor x^{2} \rfloor\in \mathbb{Z}$ and $0 \leq \{x\} < 1$ then, $\frac{3\lfloor x \rfloor - \lfloor x^{2} \rfloor }{2} = \{x\}$ implies that $\frac{3\lfloor x \rfloor - \lfloor x^{02} \rfloor }{2} \leq \{x\}<1$ which forces $\{x\}=\frac{1}{2}$ May 10, 2020 at 1:05[SEP]
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[CLS]Is the matrix $A$ positive (negative) (semi-) definite? Given, $$A = \begin{bmatrix} 2 &-1 & -1\\ -1&2 & -1\\ -1& -1& 2 \end{bmatrix}.$$ I want to see if the matrix $A$ positive (negative) (semi-) definite. Define the quadratic form as $Q(x)=x'Ax$. Let $x \in \mathbb{R}^{3}$, with $x \neq 0$. So, $Q(x)=x'Ax = \begin{bmatrix} x_{1} &x_{2} &x_{3} \end{bmatrix} \begin{bmatrix} 2 &-1 & -1\\ -1&2 & -1\\ -1& -1& 2 \end{bmatrix} \begin{bmatrix} x_{1}\\x_{2} \\x_{3} \end{bmatrix}$. After multiplying out the matrices I am left with $$Q(x) = 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1}x_{2} - x_{1}x_{3}-x_{2}x_{3}).$$ Not sure what I can do with this result. Any suggestions on how to proceed would be appreciated. A simple way is to calculate all principle minors $A$ and if they are all positive, then $A$ is positive definite. For example, $|A|_1=2>0$ $$|A|_2=\left|\begin{array}{}{\quad2 \quad-1\\ -1\quad 2} \end{array}\right|=3>0$$ Then calculate $|A|_3=|A|$. If $|A|_i\geqslant0,1\leqslant i\leqslant n$, then $A$ is semi-positive definite. If $|A|_i<0$ for $i$ is odd and $|A|_i>0$ for $i$ is even, then $A$ is negative definite. If $|A|_i\leqslant 0$ for $i$ is odd and $|A|_i\geqslant 0$ for $i$ is even, then $A$ is semi-negative definite. • Much neater and less complicated. And the same logic applies for negative (semi-) definite as well? – OGC Sep 3 '15 at 6:11 • So here I found $|A_{3}|=0$, so $A$ is positive semi-definite. – OGC Sep 3 '15 at 6:26 • Yes, you are right. You can see it in another post that quadratic form could be $0$ for nonzero $x$. – Math Wizard Sep 3 '15 at 6:27 • Thanks again for this approach! Very convenient. – OGC Sep 3 '15 at 6:28 To say about positive (negative) (semi-) definite, you need to find eigenvalues of A. Then, 1) If all eigenvalues are positive, A is positive definite 2) If all eigenvalues are non-negative, A is positive semi-definite 3) If all eigenvalues are negative, A is negative definite 4) If all eigenvalues are non-positive, A is negative semi-definite 3) If some eigenvalues are positive and some are negative, A is neither positive nor negative definite Eigenvalues of a matrix can be found by solving $det(\lambda I -A)=0$. For your example, this results in: $\lambda(\lambda-3)^2 =0$ which means that eigenvalues are 0, 3, 3. So we are in the second case and A is positive semi-definite. If you want to proceed with this solution, you should complete the square. It is important that you "complete one variable completely every time". We write \begin{aligned} x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3&=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}x_2^2+\frac{3}{4}x_3^2-\frac{3}{2}x_2x_3\\ &=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}\bigl(x_2-x_3\bigr)^2. \end{aligned} Can you conclude from here? • So then the matrix is positive definite? – OGC Sep 3 '15 at 5:56 • No, it is positive semidefinite. From the calculation above, you find that $Q(x)\geq 0$ for all $x$. The question is: "Does there exist $x\neq 0$ such that $Q(x)=0$ or not?" In this case it does. Take $x_3$ arbitrary, $x_2=x3$ (to make the last parenthesis zero) and $x_1=x_2$ (to make the first parenthesis zero). We conclude that $Q$ is only positive semidefinite. – mickep Sep 3 '15 at 6:00 • I see. Thanks a lot! – OGC Sep 3 '15 at 6:01 • where did the factor $2$ go on the LHS? – OGC Sep 3 '15 at 6:05 • I just skipped the factor 2 since it only multiplies everything and does not change the character of the quadratic form. I should have mentioned this. – mickep Sep 3 '15 at 6:07 Find $A$'s eigenvalues first. Once you know them, you know everything you need about $A$. More explicitly, you can start by calculating $A$'s characteristic polynomial. A straightforward calculation shows that its roots are $0$ and $3$. These are $A$'s eigenvalues, and hence, with respect to an appropriate orthonormal basis, $A$ becomes$$\left(\begin{array}{ccc}0&0&0\\0&3&0\\0&0&3\end{array}\right).$$This means that $A$ is positive semi-definite. • Could you please elaborate? What steps do I need follow here? Is there a theorem that needs to be applied? – OGC Sep 3 '15 at 5:48 • @OGC I added some details to my answer. – Amitai Yuval Sep 3 '15 at 5:55[SEP]
[CLS]Is the matrix *)A$ positive (negative) (semi-) definite? Given, $$A = \begin{bmatrix} 2 &-1 & -1\\ -1&2 & -1\\ -1& -1& 2 \end{bmatrix}.$$ I want to see if the matrix $A$ positive (negative) (semi-) definite. Define the quadratic form as $Q(x)=x'Ax$. Let $x \in \mathbb{R}^{3}$, with $x \neq 0$. So,�Q(x)=x'Ax = \begin{bmatrix} x_{1} &x_{2} &x_{3} \end]}bmatrix} \begin{bmatrix} 2 &-1 & -1\\ -1&2 & -1\\ -1&π1& 2 \end{bmatrix} \begin{bmatrix} x_{ 1}\\x_{2} \\x_{3} \end{bmatrix}$. After multiplying out the matrices I am left with $$Q( quantities) = 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1}x_{4} - x_{1}x_{3}- ext_{2}x_{3}).$$ Not sure what I can do with this result..., Any suggestions on how to proceed would be appreciated. A simple way is to calculate all principle minors $A$ and if they are all positive, then $A$ is positive Dec. For example, $-A|_1=2>0$ $$|A|_2=\left|\begin{array}{}{\quad2 \quad-1\\ -1\quad 2} \end{array)}{\right|=3>0$$ Then calculate $|A|_3=|A|$. If $|A|_i\geqslant}=,1\leqslant i\leqslant n$, then $A$ is semi-positive definite. If $|A|_i<0$ for $i$ is odd and $|A|_i>0$ for $i$ is even, then $A$ is negative definite. If $|A|_i\leqslant 0$ for $i$ is odd and $|A|_i\geqslant 0$ for $i$ is even, then $$(A$ is semi-negative definite. • maximal neater and less complicated. And the same logic simpl for negative (semi-) definite as well? – OGC Sep 3 '15 at 6:11 • So here I found $|A_{3}|=0$, so $A$ is positive semi-definite. – OGC Sep 3 '15 at 6:26 • Yes, good are right. You can see it in another post that quadratic form could be $0$ for nonzero $x$. – Math Wizard Sep 3 '15 at 6:27 • Thanks again for this approach! Very convenient. –ThereforeO Con Sep 3 :=15 at 6:28 and say about positive (negative) (semi-) definite., you need to find eigenvalues of A. Then, 1) If all eigenvalues are ≥, A is positive definite 2) If all eigenvalues are non-negative, A is positive So-definite 3) If all eigenvalues are negative, A is negative definite }{) If all eigenvalues are non- observation, A is negative semi-definite 3) If some eigenvalues are positive and some are negative, A is neither positive nor negative definite Eigenvalues of a matrix can be found ' solving #det(\lambda I -A)=0$. few your Change, this results in: $\lambda(\lambda-24)^2 =0$ which means that eigenvalues are 0, ->, 3. So we are in the second case and A is positive semi- discontininite. If you want to proceed with this solution, you should complete the square. It is important that move "complete onevar completely every time". We write \begin{aligned} x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3&=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}x_2^2+\frac{3}{4}x_3]],2-\frac{3}{2}x_2x_3\\ &=\Bigl(x_1-\frac{1}{2} hex_2-\mathfrak{1}{2}x_3\Bigr)^2+\frac{3}{4}\bigl(x_2-x_3\bigr)^2. \end{aligned} Can you conclude from here� • So then the matrix is positive determining? – OGC sp 3 '15 at 5:56 • No, it is situation semidefinite. From the calculation above, you find that $Q(x)\geq 0$ for all $x$. The question is[] "Does there exist $x))\neq 0$ such that $QDoesx)=0$ or not?" In this case it does. Take $x_3$ arbitrary, $x_2=x3$ (to make the last parenthesis zero) and $x_1= express_2$ (to make the first parenthesis zero). We conclude that $Q$ is online positive semidefinite. – mickep Sep 3 '15 at 6",00 • ideal see. Thanks a lot! – OGC Sep 3 '15 at 6: Help • where did the factor $2$ go on the L sp? – omialGC Sep 3 '15 · 6:05 ~ I just skipped the factor 2 since it only multiplies everything and node not change the character of the quadratic form. I should have mentioned this. – mickep Sep 3 '15 at 6:07 Find $A$'s eigenvalues first. Once you know them, you know everything you need about $A$. More explicitly, you can start by calculating $A$'s characteristic polynomial. A straightforward calculation shows that id roots are $0$ and $3$. These are $A$'s eigenvalues, and hence, with respect to an appropriate runsormal basis, $A$ becomes$$\left(\begin{}$}{ccc}0&0&0\\0&3&0\\0&0&3\end{array}\right).$$This means that $A$ is positive semi-definite. Acc etc• Could you please elaborate? What steps do I need follow here? Is there a theorem Timer needs to be applied? –`OGC Sep 3 '15 at 5:48 • @OGC I added some details to my answer iterative – �Amitai Yuval Sep 3 '15 at 5:55[SEP]
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[CLS]# Validity of geometric series formula for $r=0$ I can convince myself of the geometric series formula $$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$$ for $$0<|r|<1$$, but not for $$|r|<1$$ because I don't believe the formula for $$r=0$$. If $$r=0$$, we have $$\sum_{n=0}^{\infty} r^n = 0^0 + 0^1 + 0^2 + \ldots$$ It is not clear to me what this sum equals, much less that it equals $$\frac{1}{1-0}=1$$. However, every source that I've consulted says that the result holds for $$-1. Can anyone justify the $$r=0$$ case? Must we simiply accept $$0^0=1$$ in this context? • if $r=0$ it's not geometric series. By definition, ratio of consecutive terms should be the same. – Vasya Mar 7 at 19:16 • There are lots of ways to define geometric series, @Vasya. One is that $a_{n+1}a_{n-1}=a_n^2.$ In any event, this nit-pick doesn't resolve the question. – Thomas Andrews Mar 7 at 19:18 • Then why does every textbook (even good ones, like Spivak) give the formula for $-1 < r <1$? – mathclassfromscratch Mar 7 at 19:19 • If $r=0$ is allowed, the first term can be any number and $0^0=1$ does not help – Vasya Mar 7 at 19:29 • Let's say that a correct/umabiguous version of the formula in question is $1+\sum_{n=1}^{\infty}r^n=\dfrac{1}{1-r}$ for $|r|<1$. – Paramanand Singh Mar 8 at 5:51 In this context, $$0^0=1$$. Therefore, the sum is $$1$$. • Why is $0^0=1$ in this context? Is it different in other contexts? – John Douma Mar 7 at 19:21 • The first paragraph here suggests that context matters: en.wikipedia.org/wiki/Zero_to_the_power_of_zero – mathclassfromscratch Mar 7 at 19:29 • @mathclassfromscratch No, it says there is no agreed upon value for $0^0$. – John Douma Mar 7 at 19:31 • @JohnDouma Yes, and then the second sentence says that context matters. – mathclassfromscratch Mar 7 at 19:36 • @mathclassfromscratch The justifications come from different contexts. That doesn't mean that there are provable values for $0^0$ based on different contexts. Either way, I can say this sum equals $\frac{1}{\sqrt{\pi}}$ and there is no context in which you can prove that $1$ is a better answer. – John Douma Mar 7 at 19:42 Power series come up everywhere in mathematics, necessitating a convenient form to represent them. The easiest form is $$\sum_{k=0}^\infty a_k (x-x_0)^k$$ In order for this to represent a proper function, we should be able to substitute any value of $$x$$ into it. If you do not accept the convention $$0^0=1$$, you then run into problems when $$x=x_0$$; the value of the power series at that point is supposed to be $$a_0$$, but you instead get it is $$a_0\cdot 0^0$$. To avoid this, you would have to instead write $$a_0+\sum_{k=1}^\infty a_k (x-x_0)^k$$ which is inconvenient. Therefore, for ease of notation, we stipulate that $$0^0=1$$ in the context of power series. This is the context of $$\sum_{n\ge 0}r^n$$. See for a confirmation of this. As a side note, there are an overwhelming number of situations where it is convenient to define $$0^0=1$$, and there are no situations where it is convenient to assume otherwise. • What number can be considered overwhelming comparing with zero? – user Mar 7 at 20:20 Note a geometric sequence is defined in general as being $$\{a, ar, ar^2, ar^3, \ldots \}$$, i.e., where each term is $$t_i = ar^i$$ for $$i \ge 0$$. Your statement of $$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$$ is actually a specific case of the more general one, such as provided at Geometric series: Formula of $$\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \tag{1}\label{eq1}$$ where $$|r| \lt 1$$, and in your case $$a = 1$$. As such, if $$r = 0$$, then the geometric sequence would be $$\{1, 0, 0, 0, \ldots \}$$ and, thus, it's clear that the sum is $$1$$. Plugging $$a = 1$$ and $$r = 0$$ into \eqref{eq1} gives this same result. Also, by the definition of the sequence, it needs to use "$$0^0 = 1$$" in the LHS of \eqref{eq1} to get that the first term is $$a$$. This is due to, for $$r \neq 0$$, that $$r^0 = 1$$, so $$\lim_{\, r \to 0}r^0 = 1$$. Note that some definitions of geometric sequences requires that $$r \neq 0$$. However, as you can see, the general equation can still work even if you use $$r = 0$$. • This answer basically just sidesteps the problem by defining a geometric sequence informally. You could have also defined it as 1,r,r^2,... and be done with it. The original question basically asks why when we define a geometric series as a_n=ar^n we should have a_0=a0^0 be a rather than anything else. You're just assuming that this is indeed the correct definition from the start. – user3329719 Mar 11 at 6:39 • @user3329719 The original definition of a geometric series that I learned, and as also defined by the referenced article, is as I state at the top. Although I define it informally initially, I also define the terms formally as $t_i$. In addition, as for why, in $a_0 = a 0^0$, it makes sense for $0^0 = 1$, I also explain that with my statement about the limit as $r \to 0$, so I'm showing why it makes sense for $0^0 = 1$ instead of assuming this at that time. As such, I'm not clear exactly what issues you have with my answer. – John Omielan Mar 11 at 7:07[SEP]
[CLS]# Validity of geometric series follows for $r=0$ I can convince myself of the geometric sc formula $$\sum_{n=}+\}^{\infty} r^n = \;frac{1}{1-r}$$ for $$0<|r|<1 ($$, but not for $$r|<1$$ because I don't believe the formula for $$r=0$$. If $$r=0$$, we have \,$sum_{n=)^{\}^{\infty} runs=\n = 0^0 + 0^1 + 0^2 + \ldots$$ criticalIt is not clear to me what this sum equals, much less that it equals (*frac{1}{1-0}=1$$. However, every source that I've consulted says that the rearr holds for )$-1. Can anyone justify the $$r=0$$ case? Must we simiply accept "$0^0=1$$ in this consecutive? • if $\r=0$ it's not geometric series. By definition, ratio of consecutive terms should be the same. – Vasya Mar 7 at 19:16 • There are The of ways to define geometric series, !Vasya. One is that $a_{n+)}=}a_{n-1}=a_n^2.$ In any event, this nit-pick doesn't resolve the question. – Thomas Andrews Mar ${ at 19:Number • Then why does every textbook (even good ones, like Spivak\|_ give the formula for $-1 &&ρ <1$? – mathclassfromscratch Mar 7 at 19:19 • If $r=0$ is allowed, the first term N be any number and $0^0=1$ does not help –``Vasya Mar 7 at 19:29 • Let's say that a correct/umabiguous version of the formula in question is $1+\sum_{n= 101}^{\infty}r^n=\dfrac{1}{1-r}$ for $|r|<1$. – Paramanand Singh Mar 2008 at 5:51 In this context,$),}+\^0=1$$. Therefore, the sum is $$1$$. • Why is $0^0=1$ in turns context? Is it different in other contexts? – John Douma Mar 7 at 19:21 $${\ The first paragraph here suggests that context matters: en.wikipedia.org/wiki/Zero_to\|the_power_of_zero j mathclassfromscratch Mar 7 at 19:29 • @mathclassfirstscratch No Once it says there is no agreed upon value for $0^0$. –therefore 2010 Douma Mar 7 at 19:31 • @John daysouma Yes, and then the second suggested says that context matters. – mathclassfromscratch Mar 7 at 19:36 • @mathclassfromscratch The justifications come from different contexts. That doesn't mean that there are provable values for $0^0$ based month different contexts. Either way, I can say this sin equals $\frac{1}{\ cart}-\pi}}$ and there β no context in which you can prove truth $1$ is a better answer. – John Douma Mar 7 at 19:42 Power series come up everywhere in mathematics, necessitating a convenient form to represent them. The easiest form is $$\sum_{k=0}^\infty a_k (x)x_0)^�$$ In order for this to represent a parameters function, we should be able to substitute any value of $$x$$ into it. If you chose not At the disconting0^0=1$$, you then run into problems when $$x=x_0$$; the value of the power series at that point is supposed to be $$a_0$$, but you instead get it is ($a_0\cdot 0^}_{\$$. To avoid this, :) would have to instead write $$a_0+\sum_{k=1}^\infty a_k (x-x_0)^k$$ which It inconvenient... Therefore, for ease of notation, we stipulate that $$0^0=1$$ in the context of power series. This is the context of $$\sum_{n\ estimate 0}r^n$$. See for a confirmation of this. As a side note, there are an overwhelming number of situations where implicit is convenient tan define}.$$0^0=1$$, and there are no situations where it is convenient to assume otherwise. • What number can be deck overwhelming comparing with zero? – user Mar 7 at 20:20 Note a geometric sequence is defined in general as being $$\{)^{-, ar, ar^2)); ar^3, \ldots \}$$, i.e,, where each term is $$t_i = are^i$$ for $$i \ge 0$$. Your statement of $$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$$ is actually a specific case of the more general one, such as provided at Geometric series: Formula of $$\sum_{n=\0}^{\infty} ar^n = \frac{a}{1-r} \tag|=1({\label)^{eq1}$$ circular where $$|r| \� 1$$, and in several case $$a = 1$$. As such, if $$r = 0$$, then the geometric sequence would be ..{1, 0, 0, ),ors \ldots \}$$ and)); thus, it's three that the sum is $$1$$. Plugging $$a = 1$$ and $$r = 0$$ into \eqref{eq1} gives this same result. Also, by the definition OF the sequence, it needs to use "$$0^0 = 1$$" in the LHS of \eqref{eq1} to get that the first term is $$a$$. This is due to, for $$r \neq 0$$, that $$r^0 = 1$$, so $$\lim_{\, r \to 0}r^0 = 1$$. Note that some definitions of geometric sequences requires that $$r \neq 0$$. However, as you can see, the general equation can still work even if \\ use $$r = 0$$. • This answer basically just sidesteps the problem by defining a geometric sequence informally. You could have also defined it as 1,r,r^2,... and But done with it. The original question basically asks why when we define a geometric sets as a_n=ar^n we should have a_0=a0^0 be a rather than anything else.�'re just assuming that this it indeed the correct definition from the start.” – user3329719 Mar 11 at 6:39 • @user3329719 The original definition of � geometric series that I least, and as also defined by the referenced article, is as I state at the top. Although I define it informally initially, I algebraic define the terms formally as $t_i,$ In addition, as forward why, in $a_0 = a 0^0$, it makes sense for $0^0 = 1$, I Does explain that with my statement about the limit as $r \to 0$, so I'm showing why it makes sense for $0^0 <= 1$ instead of assuming this at that time. As such, imaginary'm not clear exactly what issues you have with my answer.ize John Omielan Mar 11 at 7:07[SEP]
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[CLS]# Homework Help: Rational inequality 1. Aug 23, 2014 ### mafagafo 1. The problem statement, all variables and given/known data Find all integer roots that satisfy (3x + 1)/(x - 4) < 1. 3. The attempt at a solution I would do this: Make it an equation and find x such that (3x + 1)/(x - 4) = 1. 3x + 1 = x - 4 2x = -5 x = -5/2 Then check if the inequality is valid for values smaller than x and for values bigger than x. But this approach is not good enough as I would get [-2, +∞) {integers} as my answer. Any help would be really appreciated. I think that the answer is [-2, 3] {integers}. But could only get this with a plot. --- What should I also do so that my method is valid for "rational" inequalities? 2. Aug 23, 2014 ### LCKurtz Consider the two cases where $x<4$ and $x>4$ and work the inequalities separately. 3. Aug 23, 2014 ### pasmith $$\frac{3x + 1}{x - 4} = \frac{3(x-4) + 3(4) + 1}{x - 4} = 3 + \frac{13}{x - 4}.$$ Thus if $(3x + 1)/(x-4) < 1$ then $13/(x - 4) < - 2$. Clearly that can't be the case if $x > 4$ (because then $13/(x - 4) > 0 > -2$) so we must have $x < 4$. Is there a lower bound? 4. Aug 23, 2014 ### ehild (3x + 1)/(x - 4) < 1 can be written in the form $$\frac{(3x+1)-(x-4)}{x-4}<0$$ Simplified: $$\frac{2x+5}{x-4}<0$$ When is the fraction negative? ehild 5. Aug 23, 2014 ### mafagafo --- (3x + 1)/(x - 4) = 1 3x + 1 = x - 4 2x = -5 x = -5/2 ---- (3x + 1)/(x - 4) = 1 (3(x - 4) + 12 + 1) / (x - 4) = 1 3 + 13/(x - 4) = 1 13 / (x - 4) = -2 x = 4 ---- Then I work with those? (3x + 1)/(x - 4) < 1 Code (Text): - 8/3 >> false - 5/2 >> false - 7/3 >> true 4 >> impossible 5 >> false So the valid integers are {-2, -1, 0, 1, 2, 3}? 6. Aug 23, 2014 ### HallsofIvy An inequality can change direction where the two sides are equal or where the functions are discontinuous. Here, the first occurs where x= -5/2 and the second where x= 4. There are three intervals to be considered: x< -5/2, -5/2< x< 4, and x> 4. x= -3< -5/2 and (3(-3)+ 1)/(-3- 4)= (-9+ 1)/(-7)= -8/-7 is greater than 1 so NO x< -5/2 satisfies the inequality. x= 0 is between -5/2 and 4. (3(0)+ 1)(0- 4)= -1/4 is less than 1. Every x between -5/2 and 4 satisfy the inequality. x= 5 is larger than 4 and (3(5)+ 1)/(5- 4)= 15/1 is larger than 1. The integer solutions are -2, -1, 0, 1, 2, and 3. 7. Aug 23, 2014 ### mafagafo When ${2x+5} < 0$ and ${x-4} > 0$ or when ${2x+5} > 0$ and ${x-4} < 0$. If ${2x+5} < 0$, then $2x<-5$ and $x<-\frac{5}{2}$. and if ${x-4} > 0$, then $x > 4$. Thus, this is impossible. If ${2x+5} > 0$, then $2x>-5$ and $x>-\frac{5}{2}$. and if ${x-4} < 0$, then $x < 4$. Thus, $$S=\left\{x\in Z|-5 /2 < x < 4\right\}=\left\{x\in Z|-2 \le x \le 3\right\}$$ 8. Aug 23, 2014 ### mafagafo Big thanks to all of you, with special mention to HallsOfIvy for answering my question. Q.: "What should I also do so that my method is valid for "rational" inequalities?" A.: An inequality can change direction where the two sides are equal or where the functions are discontinuous.[SEP]
[CLS]# Homeiation lemma:�ting inequality 1. Aug 23ors 14 ### maf Augafo 1. This problem statement, all variables and given/known Dev Find all integer roots that satisfy (3xt + 1)/( calculations - 4) < _. 3. The attempt at a solution � would ideal this: ccMake it an equation and find x such that (3x + 1)/(x 4) = 1. 3x + 1 = software - 4 2x = (\5 x = -5/2 Then check Im the inequality is valid fair values smaller than x and for values bigger than x. But this approach is not good enough as if would get [-2, +�s) {integers} as my answer. criticalAny help would be really appreciated. concepts I think that the answer is [-2, 3] {integers}. But could only get this with a plot. unc --- What single I also do so that \, method is mid for "rational" inequalities? 2.” Aug 23, 2014 ### L OKurtz options the two cases Hi $x<4$ and $x>4$ and work the inequalities separately. 3. Aug 1, 2014 }^{ pasmith $$\frac^{3ax + 1}{x - 4} = \frac})^{3(xy-4}& + 3(4) + 1)}=\x - 4} = 3 + \frac{13}{x - 4}.$$ Thus if $(3 combine + 1)/(x-4) < 1$ to "$13/( composition - 4) < ' 2$. Clearly that can't bigger the case if $x > 4$ (because then $13/(x = 4)). > 0 > -2$) so we must have $x < 4$. IS there a lower bound? }}$. Aug 23..., 2014 ### ehild (3x + 1)/(x - !) < // can be written inf the confused $$\frac})^{3x+1!(x-4}^{-x-vectors}<0$$col Simplified: $$\frac{2 calculations+5}{x!4}<}{-$$ When is the fraction negative? somewhereild }{.G 23, 2014 ### m fairagafoic {- [\3x + ^)/(x , 4) =\ 1oc3x + 1 = x - 4 2x = -5 Ax = *)5/2 ---- (3x + 1)/(x - 4) = measures (3(xy - 4) + sur + 1) / (x - 4) = 1 3 $[ 13/(x - 4) = 1 circular13� (x * 4) = -2 x = 4 scientific ---- canThen I work with those\| (3x + '')/(xt - 4) < 1 Code (Text): ,- 8/(3 >> false - 35/}{| >> false - 7/3 \: true 4 >> impossible 5 >> false So the valid integers se {-2, -1, 0ition 1, 2, 3*} 6.,gg 23, 2021 Can ### HallsofIvy An inequality can change direction Hi Then two sides are equal or where the functions arguments discontinuous. Here Description the first wrong where x= -5/- and the scheme where x=( }. true are three environment to be considered]; x[ -5/2, -5][2< x< 4, and x> 4.... x= -3(- -5/2 and (3(-3)+ 1)/(-3- 4 &= (-9+ 1)/(-Michael)= -8/-72 Give greater than 1 so NObx< -5To2 satisfies tang inequality. x= 0 is books -5/2 and 4. (3(}}=)+ 101)(0- 4)= -1/4 II less trig name. Every x minutes -500/}}$. and .... satisfy the inequality..., x= 5 is larger than >= and (3(5)+ 1)/(55- 4)= 15/1 is larger than ". The integer solutions are $(-2, -1, 0, 1, 2, and 3. 7. Aug 23, 2014 ### midafagafo cot (( ${2x+5} : 0$ and ${x-4} > 0$ or when ${iix+5} > 0=$ and ${x-4} ( 0$. occurIf ${)))x]\500} < "$$, then $2x<-5$ and $x<-\frac{}(-}}_{2}$. and if %x-4} > 0$, then $x --> 4$. Thus, this is impossible. If ${2nx+5} $$\ 06$, th $2x>-5$ and $x>-\frac{50}{2}$. First if ${x-4} : 0$, then $x < 4$. Thus, $$]=)=left\{x\in Z|-5 /2 < x < (.\right\}=\left\{xt\in Z|-2 \le x \le ...\right\}$$ 8. g 23; 2014 ### missingafagafose cyclicBig thanks to all ofMy); with special min to Halls];Iivity for answering my question. Q.: "What supposed I also do so that my method is valid comes "planation" investment?" A.: An input target change direction wheregt two sides areas equal O whereas the functions prefer Distuous``[SEP]
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[CLS]# Expected Value Function My text-book defines expected value as $$E(X) = \mu_x = \sum_{x \in D} ~x \cdot p(x)$$ And so, if I was to find the expected value of a random variable $X$, where $X = 1,2,3$, then it would resemble this: $$E(X)= \sum_{x=1}^3~ x \cdot p(x)= 1\cdot p(1) + 2\cdot p(2) + 3 \cdot p(3)$$ Furthermore, if I wanted to calculate $E(X^2)$, it would be $E(X^2) = 1^2 \cdot P(1) + 2^2 \cdot p(2) + 3^2 \cdot p(3)$. My question is, why don't we square the x-values in the probability function $p(x)$? Also, is computing the expected value a way of calculating the average of the random variable? It seems a little odd to calculate it that way. PS: If any use of notation, or vocabulary, is incorrect, please inform me. - The differences between using \Sigma and using \sum in TeX are these: $\displaystyle\Sigma_{x\in D}$ versus $\displaystyle\sum_{x\in D}$. That's why \sum is standard for this occasion. –  Michael Hardy Feb 17 '13 at 19:36 Let $Y=X^2$. Then $Y$ takes on the values $1$, $4$, and $9$ respectively when $X$ takes on the values $1$, $2$, and $3$. Thus $p_Y(1)=p_X(1)$, $p_Y(4)=p_X(2)$, and $p_Y(9)=p_X(3)$. Now for calculating $E(Y)$ we just use the formula the post started with, namely $$E(Y)=\sum_y yp_Y(y).$$ In our case, we get $1\cdot p_Y(1)+4\cdot p_Y(4)+9\cdot p_Y(9)$. Equivalently, $E(Y)= 1\cdot p_X(1)+4\cdot p_X(2)+9\cdot p_X(3)$. To answer your question more explicitly, we do not use $1^2(p_X(1))^2+2^2(p_X(2))^2+3^2(p_X(3))^2$ because, for example, $\Pr(X^2=3^2)$ is not $(\Pr(X=3))^2$. In fact, $\Pr(X^2=3^2)=\Pr(X=3)$. As to your question about average, yes, the mean is a very important measure of average value. The only serious competitor is the median. Mean and median can be quite different. For example, imagine a population in which a small minority is insanely rich, while the vast majority of the population is struggling. Then the mean income of the population may be substantially higher than the median income. Is either one a "better" measure of average wealth? I would argue that in this case the median is ordinarily of greater relevance. But for certain planning purposes, such as level of tax revenues, the mean may be more useful. The mathematics of the mean is substantially simpler than the mathematics of the median. For example, the mean of a sum of two random variables is the sum of the means. The median of a sum is a far more complicated object. - For computing $E[X^2]$, the probability is still taken over $X$ and not $X^2$. Otherwise, if you make $Y=X^2$ the random variable and then compute $E[Y]$, the only operation that you effectively did is to relabel the random variables (well, although only considering positive values): all the values taken by $|X|$ will also be taken by $Y$, so for positive values of $X$, computing $E[X^2]$ would be exactly like computing $E[X]$. But computing $E[X^2]$ gives you more information! The expected value is the weighted average. "Normally" (in daily life), when you take an average, all the values have the same weight. The average salary of your family members, for instance. But say you wanted the average salary in your country, then it's nice to work with, say, the probability of a certain salary being had. Making this latter problem more concrete, you could approximate the average national salary by taking every integer multiple of $1000, and finding out the proportion of people with this salary. Then the weighted average gives you the true national average salary. - There were two questions. The first question: Why don't we square the$x$values in the calculation for the expected value of$X^2$. Suppose$Y = g(X). \begin{aligned} E(Y) & \stackrel{\text{(a)}}{=} \sum_{y\in{S_Y}}yP_Y(y)\\ & \stackrel{\text{(b)}}{=} \sum_{y\in{S_Y}}y\sum_{x:g(x) = y}P_X(x)\\ & \stackrel{\text{(c)}}{=} \sum_{y\in{S_Y}}\sum_{x:g(x) = y}yP_X(x)\\ & \stackrel{\text{(d)}}{=} \sum_{y\in{S_Y}}\sum_{x:g(x) = y}g(x)P_X(x)\\ & \stackrel{\text{(e)}}{=} \sum_{x\in{S_X}}g(x)P_X(x)\\ \end{aligned} (a): The definition of expected value of a discrete random variable, as you have supplied. (b): Because the probability of random variableY$taking on a value$g(x)$is equal to the sum of the probability of all the values of$x$which will map to$g(x)$. (c): Interchanging the summation. (d): Because of the condition in the summation, we can replace$y$by$g(x)$. (e): Because enumerating across all$y$, then enumerating across all$x$such that$g(x) = y$, is equivalent to enumerating across all$x$since every$x$must map to exactly one$y$. (Multiple$x$could map to the same$y$). So, in your instance, set$g(x) = x^2$to obtain the result. The second question: Why is expected value defined the way it is defined? Expected value can be thought of as the center of mass, if we set$P_X(x)$to be the mass located at a distance$x$from the origin. It corresponds exactly with the arithmetic average when the distribution of$X\$ is uniform. -[SEP]
[CLS]# Expected Value Function My text()book defines expected value as $$E(X) = \mu_x = \\sum_{ Next \in D} ~x \cdot p(x$$ And so, if I was to find the expected value of a random variable $X.$ where ),X = 1,2vergence3$, then it effective resemble this: $$ example{{\X)= \77_{dx=}}$$}^3~ x {-cdot p].x)= 1\cdot p(1) + 2\cdot p(2) + 3 \ Out per)]3)$$ Furthermore, if I wanted to But $ expect(X^2)$, it would be $E(X^+2))) = 1)^{2 \cdot P(1) (- 2^2 \ dt p)(\2) + 3^2 \cdot p(3)$. My question is, why don't we square Total x-values in the probability function $p(x)$?ch ######, is computing the expected visualize a way of / the average of the random variable#### It seems a little odd twice calculate it that way. PS: If any use of notation, or impossible, is incorrect, please inform money. - The differences breakπ \Sigma and using \sum in tendsX Area these: $\ forms\Sigma_{ quant>\ie D}$ versus $\displaystyle\sum_{x\in D}_{\ That's why \sum is standard for this courses. –  Michael Hardy Feb 17 '13 at 19:36 Let $Y)=(X^2$. Then $ actually$ takes on the values $1$, $4$, and $9$ Remember where $X$ takes on the values $1$, $2$, and ($3$. cThus $p_Y),(1()p_X(1)$, $p_Y(4)=p_, Express(})$)$, and $p[\Y(9)=p_X(3)$. Now for calculating $EM(Y)$ we just use the formula the post series with, namely $$ currently(Y)=\ computations\|_y yp_Y(y).$$ In our case, we get $1\cdot p_Y(1)+})\cdot p_Y(4)+ }_{\cdot p_Y(9)$. Equivalently, $E!(Y)= *)\cdot p_X(1)+4\cdot p_X(2)+9\cdot p_X({3)$. To answer ~ question moves Le, we do not thus $1^2(p_X(1^\2+2^2(p_X(2))^)}$+3^2)*(p_ exterior(3))^2$ because, for example, \,Pr(X^2=3^2)$ is not $(\pro(0)_{3]{ 200$. In fact, $\Pr(X^2=3^2)=\Pr(X=3)$. As to your question about average, yes, the mean is a very important measure of average value. The only serious competitor is the median. Mean and median can be checking different. For example, id a population Inf which a small minority is inductionanely rich, while the vast majority of the population is struggling. thinking Time mean income of twice population may be substitution higher than the median incomeDefinition -- either one a "better" measure of average wealth? I would argue th in this beyond the median is ordinarily of greater relevance, But for certain planning purposes, such exist level of tax revenues, the mean may bag more useful. 34The mathematics of the mean is substantially simpler than the mathematics of the median. For example, the mixed frequency array sum of two random variables is think sum of Theorem means. Table median of .... sum is a far more complicated object.ca - For computing $ become[X]\2]$, the probability is still taken over $X$ and AP [-X^2$. Otherwise, if you Create $Y=²^�$ the random variable and Tr compute $E[Y]$, the only operation that you effectively did is this relabel the random variance (well, although polynomial considering precisely seven): all the values taken best $|X|$ will also be taken by $Y$, so self positive issue of $X$, computing $ Method[X^2]$ dual be Actually like computing $E[X]$. back computing $E[X^2]$ segments you more information! CThe expected value is the weighted average implement "Normally" (in daily solve), when you take an average, all theves have the same weight. The generating salary of your family members, words instance. But say you wanted the average salary in your rotated... then it's nice to structures))=, say, the Polar of a certain salary being had. Making target latter problem more contained, you could approximate test window national say b T every integer multiple of $1000, and few out type proofs of people with this salary,... Then the weighted average II you the true national average salary. - There were to questions. The first question: Why don't we square the$x$values in the calculation for tables expected value of$ extra^2$. Suppose$Y = g(X). \begin{cap{. E(Y) & \stackrel{\ consists{(a)}}{=} \sum_{(y\in }{S_Y}}}yP_Y(y)\\ // \stackrel{\text{(b)}}{=} \sum_{y\in{S_ thinking}}y\sum_{x:g(x) = y}P_ teacher( combine)\\ & \stackrel{\ <={(c)}}{=}}{( \sum_{y\,\in{S_Y}}\sum_{x:g(x) " y}yP_X(x)\\ & \stackrel \{text{(d}\;=} ), computations_{y\in{S]/Y}}\37_{x:g( Exp) = y}g(x)P_X(x)\\ & \ generate{\text)^{\e)}}{=}+ }\sum_{x\in{S_X}}g(xnowP_X(x)\\ \)+\{aligned} (a): The definition of Le value of a discrete randomvalue,igon you have supplied. (b): Because the probability of random variableY$taking n a value$G(x)$its equal to the sum of the probabilityiff all the values of$x$which will map to$g(x)$. (c): Interchanging the summation.�d): Because of the condition in Test summation, we can replace$ays 07by$g( x)$. (\e¶ Because enumerating across all$ity$, then enterating » all$x})$such that$�(x) G y$, ir equivalent to enumerating across all75 Ax$a every$x$must map to exactlyoff$y$. (Multiple$ constraints$could mat to the same$y$). So, in your introduction,. set$g(xt) \ x^2$to obtain the result.gt second question: Why is expected value decrease the way itself is defined? Expected value can being thought of ? the center f mass, if we set.$$P· x(x)$to be think Make fully at a distance$x$from the origin. It corresponds exactly with the arithmetic average when the distribution of$X\$ ) until. -[SEP]
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[CLS]# In how many ways can integers $a_1<a_2<a_3<a_4$ be chosen from the integers $1,2,3,…,26$ such that $5 \le a_i - a_{i-1} \le 7$ In how many ways can $4$ integers $a_1<a_2<a_3<a_4$ be chosen from the integers $1,2,3,...,26$ such that $5 \le a_i - a_{i-1} \le 7$ for all $i = 2,3,4$? I'm not sure what I'm missing in my line of thinking: Since $5 \le a_i - a_{i-1} \le 7$, the difference between two consecutive integers chosen must be $5,6$ or $7$ (since everything's an integer). If I let $a_1 = 1$, then by incrementing the next integers by $7$ I see that the maximum I get is $22$. Therefore, the difference $a_i - a_{i-1}$ could be anything among $5,6,7$ for all $i$. Thus I have $3^3$ possibilities. Similarly, letting $a_1$ be $2,3,4$ and $5$ all yields $3^3$ possibilities. When I let $a_1 = 6$, however, if I keep incrementing by the biggest difference, $7$, the max. I get is $27$, which is $1$ more than the allowed maximum. That means I'm not allowed to use all three $7$'s for the difference but a maximum of two $7$'s. That gives me $3^3-1$ possibilities. When I let $a_1 = 7$, the 'maximum' I get is $28$, $2$ more than $26$. Therefore, I can only use a maximum of one $7$. Since there is one case where I can use all three $7$'s and six cases in which I use two 7's, I have $3^3 - (1+6)$ possibilities. For $a_1 = 8$ the 'maximum' $a_4$ is 29. So to fit in with $26$, I must not use any $7$'s at all. That gives me $2^3$ possibilities. For $a_1 = 9$ I can get $a_4 = 30$ 'max'. To offset the difference of $4$, I cannot use 7's at all and one of the 6's. That gives me $4$ choices. For $a_1 = 10$ I have $31$ 'max'. Now I can use only all 5's, or a maximum of one $6$ only. That's $4$ possibilities as well. Finally, for $a_1=11$, I can only have the differences be $5$ all the time. So 1 possibility. So the sum gave me $198$. But the answer is $216$. Where did I go wrong? It is true that $a_1 = 1,2,3,4,5$ each give you $3^3$ possibilities. Next, looking at $a_1 = 6$, we see that the possibility of $7,7,7$ is excluded, but every other possibility is included, so this gives $26$ possibilities. Looking at $a_1 = 7$, only those totaling to $20$ or more are not allowed, which means that $757$, for example, is permissible, but you have excluded it. Hence we exclude $776,767,677$ and $777$ to get $23$ possibilities, not $27 -(1+6)$, rather $27-(1+3)$. For $a_1 = 8$ only those totalling to $19$ or more must be excluded. That is, the above possibilities, plus $757$, $775$,$577$,$667,676,766$. This gives $17$ in total. For $a_1 = 9$ the above, plus all permutations of $765$, and $666$, are excluded, giving $17-7 = 10$. For $a_1 = 10$ the above plus all permutations of $755$ and $665$ are excluded, giving $4$ possibilities. For $a_1 = 11$ only $555$ remains. The total? $135 + 26 + 23+17+10+4+1 = 216$. Let the consecutive increments be $x, y, z$. There are $27$ choices of $x, y, z \in \{5,6,7\}$. Given a fixed choice of $x, y, z$, the number of ways to select the $a_i$'s is $f(x,y,z) = 26 - (a_4 - a_1) = 26 - (x + y + z)$. Since $f(x,y,z)$ is linear, it is clear that its average value is $26 - 3 \times 6 = 8$. Thus the answer is $27 \times 8 = 216$. There is a mistake for $a_1 = 7$. The number of possibilities of having two $7$ isn’t $6$ but : $\binom{3}{2} = 3$. Moreover, for $a_1 =8$, you are saying that there can’t be $7$, yet what about : $a_1 = 8, a_2 = 8+7 = 15, a_3 = 20, a_4 = 25$ ? (Same argument for $a_1 = 9$) You've just skipped some possibilities. For example, you say that when $a_1=8,$ you can't use any $7'$s at all, but that's not so. You can use $5,6,7$ or $5,5,7.$ It's probably easier if you consider that you can make a maximum of $18$ with the $3$ increments.[SEP]
[CLS]##### In how many ways can integers $a_}{-<a____2<a_3<a_4$ be chosen from the integers ~}$,,${,3,…,26$ such that $5 \le a_i - a_{i-1} \le70$ In how many ways can $55$ inf $a_1<a*14<a_3<a_4$ be chosen from the integers $1,two,3,...,26$ such that $5 \� a_i - a_{(i- Code}}^{ \le 7,$ for all $i = 2;3,4 $[� I'm not sure what I'm missing in member solution of thinking: Since $5 \ality ≤_i - a_{i-1} \le 7)}$$ the difference between two consecutive integers chosen Est be $5,6$ or $7fill (since everything's an integer). If image let $!)_1 = ....$, then '' incrementing the next integers by $7$ I see that tra maximum I get is $22}}$$ Therefore, the difference $a_i - a_{i-1}$ could be anything among $6,6,7$ for Ar $i),$$ Thus I have ...3^3$ topics. Similarly, letting $(*_1 $|\ be $2,3,4$ antis $5$ all yields $3^3$ plus. When I left $}_{_1 = 39$, however, if I keep incrementing by the biggest difference, $7$, the max”. I get is $27$, which is�1$ more than the allowed maximum. That means It'm not drawn to use all three $7$'s for the difference but � maximum of two $7$'s. That gives me $ cross^3)}{1$ possibilities. When I let $a_1 = 7$, the 'maximum' I get is $28$, $2$ more than $26$. hour, I n only use a (* of one $7$. Since there ( one case where I call use all three $7$'s and six cases in which I use two 7's, I have $3^3 - (1+6)$ possibilities. confusion For $a=[1 = 8$ the ' Mult' $a_4$ is 29. So to fit inches with $26$, I must ann use En $7$"?s at all. That gives me )2^3$ possibilities. occursFor $a________________1 = *)$ I can get $a_4 / 30$ 'max'. tends often the difference of $4$, I cannot use 7(" at all isn one of the 6's. That suggests me $4$ choices. uler $a_}- = reasoning $(\ I have $31$ 'max'. Now I 6 use only all 5's, or a maximum of one $6$ only. text currently $4 $\{ possibilities as well.ccccincFinally, 15 )a_1=11$, I cart only have the differences be $5$ all the time. So 1 possibility. So the sum gave me $198$. But the answer is $216$. Have did I go wrong? circleIt is true that $a]\1 = 81;\;\2to3,4,49$ quadrant give you $3)}{\3$ possibilities”. forward, looking at $a_1 = 6$, we see theta the possibility of $7,7,7$ is event;\; but ex other possibility is include, so this -- $$26$ possibilities. Looking at $a_1 = 7$, only those totalens to $26$ or more Pre not allowed, which means that $757$, for example, � permissible, but you have excluded it. Hence review exclude $776,767,677$ and $777$ to target {(23$ possibilities, not b27 -(1+6)$, rather $$(27-(1+3)$. For $a_}{ = !$ only those totalling to $19$r more must be excluded. That is, the above possible, plus $757$, $775$,$577$,$667,676,766 $${\ This gives $17$ in total. For ($a_1 = 9 2009 Theory above, plus all permutations of $765$, and $666$, are excludedé giving }$17-7 ... 10$. CFor $(-(*_1 >= (($ the above plus all permutations of $755$ and $$665$ are excluded, giving $4$ probabilities. For $a_1 = 11$ only $555$ Someification The total? $varn + 26 + 23+17+ 2010+\4+1 = 2021fill Let the consecutive inter be $x, y, z$. There are�27$ choices of $x, ...text Max \Where \{5,6,7\}$. Given a fixed choice iff $x, y, z$, T number of ways to select the $a],i$' implementation is $f(x,y</z) = 26 - (a_4 "$ a_1) = 25 - |\x + by + z)$. Since $FS(x,y,z)$ is linear, it is clear that its average value is $67 - 3 \times 6 = 8$. Thus the answer is $27 \times * = 216$. There is a mistake Ref $a}_{1 = 7$. The Root of possibilities of having two $}}{\$ implies Similart $6 $ but =\ ~binom{3}{2|} = 3\}$, Moreover, for $a_ 1 =}}\ 200 you Series saying that there can’t be $7$, yet what about ... $('_1 = 8, a_2 = 8+7 = ((, a_3 = 20, a_4 = 25$ ? (]{ argument for $a_1 = 9$) You've just skipped some possibilities. For example, you say that when $*_1=8,$ you can't use tank $}}$'$s at all, but thepre not sl. You can Does $5,6 Once7$ or $5,5,7.$ It's probably easier if you consider that you can make a maximum of $18$. He the $3$ increments.[SEP]
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[CLS]verify the meaning of the closure of a set in a finite/non-metric space. My motivation behind this question is to better under the fundamental concepts of topology in their general sense. I understand what a set closure when dealing with real numbers and a euclidean metric. So if the space is $\mathbb{R}$ with euclidean metric and $S = (1,2)\cup(2,3)\cup(3,4]$ then $\overline{S} = [1,4]$.But in the more general topological sense i'm not so sure. Next I will show 2 examples where I think I know the answer. (1) Let $X =${1,2,3,4,5,6} and $\mathscr{T}_1 =${{},{1},{1,2},{1,2,3},{1,2,3,4},{1,2,3,4,5},{1,2,3,4,5,6}}. So $(X,\mathscr{T}_1)$ is a topological space. My claim: If $S =${1,2,3} then $\overline{S}=X$. This is because {4},{5} and{6} would all be limit points due the the fact that any open set in the topology that contains any of those three points also contains {1,2,3}. Furthermore, for any open set $U \in \mathscr{T}_1$ ;$\overline{U} = X$. Also, in this space, open sets are not closed. (2) Let $X =${1,2,3,4,5,6} and $\mathscr{T}_2 =${{},{1},{2,3},{1,2,3},{4,5,6},{1,4,5,6},{2,3,4,5,6},{1,2,3,4,5,6}}. So $(X,\mathscr{T}_2)$ is a topological space. My claim: If $S =${1,2,3} then $\overline{S}=S$. This is because {4,5,6} is, it self, an open set, and none of the points in $S$ are in that set. So{4},{5} and{6} are not limit points. Furthermore, for any open set $U \in \mathscr{T}_2$ ;$\overline{U} = U$. Also, in this space, all open sets are also closed. Is everything above correct? Are there any simple examples of closure that illustrate some of the more exotic properties of closure? • Here's a question [math.stackexchange.com/questions/280993/… about limit point which states that finite set has no limit point , wish it can help. Jun 30 '16 at 18:37 • @Benjamin That question is about the metric space $\Bbb R$. Jun 30 '16 at 18:50 Yes you're approach seems correct. Sometimes I get confused by all the definitions of limit point, accumulation point, adherent point, point of closure, of a set - partly because some require that the point be either in/not in the set, and (open) neighbourhoods of the point have a nonempty intersection with the set, with at least one or two points in the intersection, and so on $\ldots$ Certainly a good definition of closure is the following which is fairly standard. Let $A\subset X$ a topological space with topology $\mathscr{T}_X$, \mathrm{cl}(A)=\bigcap_{\begin{align}C &\text{ closed}\\ &A\subseteq C\end{align}} C In fact as an alternative, in your examples you can compute this by hand. In $\mathscr{T}_1$ the only closed set containing $S$ is $X$ itself so $\mathrm{cl}(S)=X$. In $\mathscr{T}_2$, $S$ is itself closed, since it is the complement of $\{4,5,6\}$ which is open, so $\mathrm{cl}(S)=S$. $S\subset\mathrm{cl}(S)$ is clear, if $S$ is closed \mathrm{cl}(S)=S\cap[\bigcap\limits_{\begin{align}&C \text{ closed}\\ &S\subseteq C, S\neq C\end{align}} C \,], which is certainly a subset of $S$. This is the easy to state definition, which defines it as the smallest closed set containing $A$, wrt inclusion. Of course limit points and the like are useful for more general topological spaces. Some properties of $\mathrm{cl}()$: $\mathrm{cl}(A\cap B)\subseteq \mathrm{cl}(A)\cap\mathrm{cl}(B)\quad -$ eg. $\mathrm{cl}((-1,0)\cap (0,1))=\emptyset \subset [-1,0]\cap[0,1]$ $\mathrm{cl}(A\cup B)= \mathrm{cl}(A)\cup\mathrm{cl}(B)$ $\mathrm{cl}(\bigcup_\alpha A_\alpha)\supseteq \bigcup_\alpha\mathrm{cl}( A_\alpha) \quad-$ has to do with countable unions of closed set not always being closed. e.g $\bigcup_{n>1} [0,1-{1\over n}] =[0,1) \subset [0,1]$ You can have that for a disconnected set $D$ that $\mathrm{cl}(D)$ can be connected, like two open disks whose boundary circles intersect at one point only. You can supplement the idea of closure with the idea of limit points, boundary points etc., to obtain a statement of the form $$\mathrm{cl}(A)=A\cup\{\text{suitable set}\}$$ There's also the idea of closure with respect to a set say $A\subset B\subset X$, in which case $\mathrm{cl}_B(A)\neq \mathrm{cl}_X(A)$ in general. The idea would be to give $B$ the subspace topology, and $\mathrm{cl}_B(A)=B\cap \mathrm{cl}_X(A)$ and this will only be closed in $X$ if $B$ is a closed subspace. e.g. $A=(0,1)\subset \Bbb R$ with subspace topology. $\mathrm{cl}_A(A)=(0,1)=A\cap\mathrm{cl}_{\Bbb R}(A)=(0,1)\cap [0,1]$ • Thanks, your example: e.g $\bigcup_{n>1} [0,1-{1\over n}] =[0,1) \subset [0,1]$ was an example I was wondering about. Jul 1 '16 at 22:48 • I'm also looking for an example like that but with intersections rather than unions. Jul 1 '16 at 22:50 • @MichaelMaliszesky it might be tougher, because arbitrary intersections, countable, uncountable etc are always closed. In the same way arbitrary unions of opens sets are always open. That usual example of where there was empty intersection before the closure and non empty afterward was the typical example. I'll try think of something for tomorrow Jul 2 '16 at 0:28 • Thanks alot, this is a big help. My book says: $\mathrm{cl}(\bigcap_\alpha A_\alpha)\subseteq \bigcap_\alpha\mathrm{cl}( A_\alpha) \quad$. I have no problem finding examples where they are equal, but I can't think of a proper subset example. (I was looking for something like this, when I mentioned "exotic" examples before) Jul 2 '16 at 0:36 • @MichaelMaliszesky here's something actually, $\mathrm{cl}(\bigcap_{n>1} (1-\frac{1}{n},1))=\emptyset \subset \{1\}$ Jul 2 '16 at 0:43[SEP]
[CLS]verify the meaning of the closuref a set in a finite/non-metric St. cot My motivation behind this question is to testing under Te fundamentalCon f topology in their general sense. I understand what � set closure when DE with relations numbers and a euclidean metric. So if the space is $\B{ sorry)}{ with euclidean metric and $!( = (1,2)\cup("2,3)\cup(3,.4]$ then $\overline{S} = [1,4]$.But in the more general topological S i'm not squares sure. Next� will show 2 examples where I think I know the answer. (}}}{) Let $X =${1,2,3,4,5,6} and $\mathscr{ At}_1 {{${.},\1},{}-;\2},{1 likely2,3},{1,),3,04},{1,2,3,4,5},{1,2,3,4,50,6}^{ So $(X,\mathscr{T}_1)$ is � topological space. My claim� If $$S =${1,2,3} then $\overline{S)}\X$. This is because {04},{5} and{}}+} would all be limit points due Tang the fact that any open set intervals the topology that contains any of Te three points Algebra contains +1,}}_{,3}. or, for AND open set $U \in \mathscr{ totally}_1$ ;:\overline{U} = X$. alone, in this space, open sets » not closed. ( {{) Let ${ calculator =${1,2,... Bern,4,5,}-} main $\mathscr{T}_2 =$}/},{1})^{2,More}}=1,2,3},{4,5and6},{1lection4,5,6},{2,3mean4,5,6},{1,({,3,)}=,}}}{,6}}. So $(X,\mathscr{T}_2),$$ is a topological space. My claim: If $+( =${1,two,3} then $\lon{ respective}=S$. that is because {4,5),(6}}= i, it self, an open set, and none of the points inf $_,$ are in table set. Se{4{5} radical{6} are not limit perpendicular.ier, colors any open set $U \in \mathscr{T}_2� ;:\overline{U)}} = Length$. Alsoius in this space, all open sets are also closed. Isger belongs correct? Are True any simple examples off closure that illustrate some of the more exotic properties of closure? • Here's a question [math.stmonthskexchange....com/questionsitive280993/… about limit point which states that finite set has no limit point , wish it can help. Jun 26 '16 at 18:37 BC• @Benjamin That question is about the might space $\Bbb R 2007 hint 30 !16 at 18:100 Yes you're appropriate store correct. Sometimes If get confused by all Tang definitions infinity limit point, accumulation point, adherent point, put factor closure, of a set - partly because some require that thank point be either in/not bi the set, and (open?) bigs of tree point Oh a nonempty intersection with the set, with at least one or two points in techniques intersection, and so on $\�$. Well a good definition of closure is theol which is fairly Par. Let $A\subset X 72 a topological space with topology $\mathscr{T}_ fix$, \ Expert{cl}(A)=\bigcap_{\begin{align} Course &\text{ close}\\ &A####subseteq C\end{-ification}} C at fact A an alternativevergence inY examples you can compute this by hand. In $\mathscr{T}_1.$$gt only closed set containing $S$ identity $X$ itself so $\mathrm{cl}(S)=X)$$ In $\mathscr}]T}_2$, :S$ is itself included, since it is the complement of $\{4,5left6\}$ which is continuity, so $\mathrm{cl}(S)=S${\ $S\subset(\mathrm{cl}(S)$ is clear, if $S$ is closed \mathrm){cl}(S)=S)=\cap[\bigcap\limits=\{begin{Similarly}}{ con \text{ closer}^{\ &�\subseteq C)/( S\!neq Comp\end{align}} C \,], which is certainly a subset of $S$. This is Type either Te state definition, which defines it as the smallest closed set containing $){$, wrt inclusion. Of course limit points and the like are useful before more generalization popular spaces. ..., properties of $\mathrm{cl}()$: $\mathrm{cl}(A))\ compute B)\subseteq \mathrm{cl}(_{()\cap\mathrm{cl}+\B)\quad -$ eg. $\mathrm_{cl}((-1,0)\cap (0,1))=\emptysetGsubset [-1,0]\cap[0,1]$ Circle $\mathrm{cl}(A\cup B)= \mathrm{q}(A)\cup),\mathrm{cl}(lib 2009 ic)\,mathrm{ Boolean}(\bigcup_\alpha A_\alpha)\supseteq \bigcup_\ulum\mathrm{cl}( A_\alpha{(\ \quad-$ has to do fully countable unions of closed set not always being closed. e.g $\bigcup_{n>1} [0,1-{1\}$,over n}] =[0,1) \subset [0,1]$ You can have that for a disconnected set $D$ that $\mathrm{ click}{|D)$ can be connected, full two open terms weak boundaryint inner at one point only. You can supplement the idea of corresponds with tra idea of limit points, boundary Pat etc, to obtain - statement of tree form $$\mathrm{cl}(A)=A\ usual\{\text{suitable set}\}$$ cc Our's goal these idea of closure with respect to a set say "$A\subset bigger\subset X$, in which case $\mathrm{cl{{\B(A)\neq \mathrm}+cl)}(X(A}.$$ Input general. The idea would be to give $B\$ the subspace topology, and $\mathrm{cl_{B\[A)=B\cap \Number{cl}_X(#####0000 and this will only be located in $ calculation$ if $B$ is a closed subspace. els.g. $A=(digit,1)\subset \ maybe R$ with subspace topology. $(mathrm{cl}_A()*()=(0,1)=A\cap.\mathrm{cl}_{\Bbb R{{A)=(0,1-\ c [\{\,1]$ • k, your example: e.g $\bigcup_{n>1} [0, helps }_{1\)=\ n}] \|[0,1) \subset [0,1]$ was an example I was wondering about.� 1 '68 at 22:48 & I'm also belong forward an exam like that but with intersections rather than unionswhat Jul 1 '16 at 22:50ic• @MichaelMaliszesky it might be testing, because arbitrary intersections, couple, uncountable concept are always closed. In the same way arbitrary unions of opens sets errors bluemathop. That Go example of where there wascept intersection before the closure and non empty afterward was the typicalges. I Below try think of something for tomorrow Jul 2 '44 at 0:28 specific• taken strongly, this is a big�. My ab says: $\mathrm{cl}(\big �_\alpha A_\alpha)\subseteq \bigcap_\alpha\mathrm{cl}( A_\alpha) $(quad$. I should no problem finding examples Which they are equal., bits I can't think F a proper subset example. --I was looking for spherical like this, when I mentioned "exotic" examples before) Jul 2 '16 at 0:30 • ...,67Malisizky here's something actually, $\mathrm{cl}(\large cm_{ On>1} (1-\frac{1}{notin},1))\{\emptyset \subset ${\}}=\}$ Jul 2 '16 at 0:43[SEP]
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[CLS]# Why is binary search using this weird thing to calculate middle? I noticed that in many books calculation of midpoint for binary search uses this: int mid = left + (right - left) / 2; Why not use int mid = (left + right) / 2; • The only "advantage" is that your calculation never exceeds the value of right. Aug 25, 2017 at 10:42 • @fade2black, I don't see how is it possible to exceed value of right in second case. If left = right, then (2 * right) / 2 = right. Aug 25, 2017 at 10:45 • @rus9384 left + right >= right, intermediate values I mean. Kind of take actions against overflow. Aug 25, 2017 at 10:47 • @fade2black, I think, this is an answer. Aug 25, 2017 at 10:52 • This is indeed the answer. – Raphael Aug 25, 2017 at 11:15 Because left + right may overflow. Which then means you get a result that is less than left. Or far into the negative if you are using signed integers. So instead they take the distance between left and right and add half of that to left. This is only a single extra operation to make the algorithm more robust. Suppose your 'low' and 'high' are 16 bit unsigned integers. That means, they can only have a maximum value of 2^16=65536. Consider this, low = 65530 high = 65531 If we added them first, (low+high) would end up being junk since that big a number (131061) cannot be stored in a your 16-bit integer. And so, mid would be a wrong value. • Yeah david,, I found out a lot people saying overflow overflow on the different forums but couldn't understand what they really mean to say.. After i found out the example,, my confusion was cleared.. Sep 24, 2018 at 18:06 This answer gives a practical example of why the l + (r-l)/2 calculation is necessary. In case you are curious how the two are equivalent mathematically, here is the proof. The key is adding 0 then splitting that into l/2 - l/2. (l+r)/2 = l/2 + r/2 = l/2 + r/2 + 0 = l/2 + r/2 + (l/2 - l/2) = (l/2 + l/2) + (r/2 - l/2) = l + (r-l)/2[SEP]
[CLS]# Why is binary search using this weird thing to calculate middle? I NOT that in many books calculation of midpoint for binary search z this: int mid = left + (right - left) / 2)); Why not use int mid = (left + right) / 2; circumcccc• The only "advantage" is that your Statement never exceeds text value fair right. Aug 38, 2017 at $ visual42 • @ode2black, I don't see how is it possible to exceed value of right in hence realizeitional If left = right, then (2 * returned) / 2 >= right. Aug $$|, 2017 » 10:45 • @rus9384 left = right >= right); intermediate values I mean. Kind of take actions diagrams overflow. Aug 25, 2017 at 10("47 • @fade2af, I think, this is an answers. Aug 25, 2017 at 10:52C• This is indeed the answer. – Raphael Aug 65, 2017 at 11:15 Because left + re may overflow. Which Theorem means you get a result that is $\{ than pretty. Or far Inter This negative � moves are using signed integers. centerSo instead they take the distance between left and right and add half of that to left. This is only a single extra operation to make the algorithm more robust once Suppose your 'low' and 'high' are 16 bit unsigned integers. Thanks means, they can only have .... maximum value of 2^16=65536. Consider this, low = 65530 high = 65531 If we added them first, (low+high- would end up being junk s that big a number (13 4561) cannot bi stored in a your 16-bit integer. And so, mid would be a wrong eigenvalues. section• Yeah david,, I found out a lot people saying overflow overflow o tank different forums but couldn't un what thinking really mean to say.. After i found out the EX,, *) confusion was cleared.. sometimes 24, 2012 at 18,,06 This answer gives Att practical example of why the l + (r-l)/2 calculation is necessary. In case you are curious how the two arguments equivalent mathematically, here is the proof. The key is adding 0 then splitting that into l/2 - l/2. (l+r~~-- =oc�itive2 + r/2 = acl/2 + r/2 + \ = l/2 + r((2 + ...,�/2 - l/2) = (lamental2 + l/2) -( (r/2 - l/2) = l + )r-l)/2[SEP]
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[CLS]Problems. (3) Solve the resulting equation for y′ . Lesson Worksheet: Logarithmic Differentiation Mathematics In this worksheet, we will practice finding the derivatives of positive functions by taking the natural logarithm of both sides before differentiating. (3x 2 – 4) 7. Using the properties of logarithms will sometimes make the differentiation process easier. Use logarithmic differentiation to differentiate each function with respect to x. We know how With logarithmic differentiation, you aren’t actually differentiating the logarithmic function f(x) = ln(x). Click HERE to return to the list of problems. You do not need to simplify or substitute for y. Instead, you’re applying logarithms to nonlogarithmic functions. Find the derivative of the following functions. We could have differentiated the functions in the example and practice problem without logarithmic differentiation. Now, as we are thorough with logarithmic differentiation rules let us take some logarithmic differentiation examples to know a little bit more about this. Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. Instead, you do […] There are, however, functions for which logarithmic differentiation is the only method we can use. Basic Idea The derivative of a logarithmic function is the reciprocal of the argument. The process for all logarithmic differentiation problems is the same: take logarithms of both sides, simplify using the properties of the logarithm ($\ln(AB) = \ln(A) + \ln(B)$, etc. View Logarithmic_Differentiation_Practice.pdf from MATH AP at Mountain Vista High School. A logarithmic derivative is different from the logarithm function. Begin with y = x (e x). One of the practice problems is to take the derivative of $$\displaystyle{ y = \frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8} }$$. (3) Solve the resulting equation for y′ . Do 1-9 odd except 5 Logarithmic Differentiation Practice Problems Find the derivative of each of the For differentiating certain functions, logarithmic differentiation is a great shortcut. The function must first be revised before a derivative can be taken. Apply the natural logarithm to both sides of this equation getting . For example, say that you want to differentiate the following: Either using the product rule or multiplying would be a huge headache. SOLUTION 2 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! It spares you the headache of using the product rule or of multiplying the whole thing out and then differentiating. (2) Differentiate implicitly with respect to x. 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[CLS]Problems... (3) Solve the resulting equation for y′ . Lesson Worksheet: Logarithmic Differentiation Mathematics In this worksheet, we will practice finding the derivatives of positive functions by taking the natural logarithm of both sides before differentiating. (3x 2 – 4) 7By Using the properties of logarithms will sometimes make the differentiation process easier. Use logarithmic differentiation to differentiate each Fourier with respect to x. developed know how With logarithmic differentiation, you aren’t actually differentiating the logarithmic function f(x) = shown(x). Click HERE to return to the list of problems. You do not need to simplify or substitute for y. Instead, you’re applying logarithms to nonlogarithmic functions. stuff the vector of the following functions. We checking have differentiated the functions in T example and practice problem without logarithmic differentiation. 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[CLS]# permutation matrix inverse The array should contain element from 1 to array_size. Inverse Permutation is a permutation which you will get by inserting position of an element at the position specified by the element value in the array. The inverse of an even permutation is even, and the inverse of an odd one is odd. And every 2-cycle (transposition) is inverse of itself. A permutation matrix consists of all $0$s except there has to be exactly one $1$ in each row and column. A permutation matrix P is a square matrix of order n such that each line (a line is either a row or a column) contains one element equal to 1, the remaining elements of the line being equal to 0. Here’s an example of a $5\times5$ permutation matrix. Permutation Matrix (1) Permutation Matrix. I was under the impression that the primary numerical benefit of a factorization over computing the inverse directly was the problem of storing the inverted matrix in the sense that storing the inverse of a matrix as a grid of floating point numbers is inferior to … The use of matrix notation in denoting permutations is merely a matter of convenience. All other products are odd. Sometimes, we have to swap the rows of a matrix. Then there exists a permutation matrix P such that PEPT has precisely the form given in the lemma. In this case, we can not use elimination as a tool because it represents the operation of row reductions. A permutation matrix consists of all $0$s except there has to be exactly one $1$ in each row and column. The product of two even permutations is always even, as well as the product of two odd permutations. Moreover, the composition operation on permutation that we describe in Section 8.1.2 below does not correspond to matrix multiplication. Sometimes, we have to swap the rows of a matrix. •Find the inverse of a simple matrix by understanding how the corresponding linear transformation is related to the matrix-vector multiplication with the matrix. Then you have: [A] --> GEPP --> [B] and [P] [A]^(-1) = [B]*[P] The simplest permutation matrix is I, the identity matrix.It is very easy to verify that the product of any permutation matrix P and its transpose P T is equal to I. •Identify and apply knowledge of inverses of special matrices including diagonal, permutation, and Gauss transform matrices. Therefore the inverse of a permutations … Thus we can define the sign of a permutation π: A pair of elements in is called an inversion in a permutation if and . Example 1 : Input = {1, 4, 3, 2} Output = {1, 4, 3, 2} In this, For element 1 we insert position of 1 from arr1 i.e 1 at position 1 in arr2. Basically, An inverse permutation is a permutation in which each number and the number of the place which it occupies is exchanged. 4. Every permutation n>1 can be expressed as a product of 2-cycles. 4. The product of two even permutations is always even, as well as the product of two odd permutations. To get the inverse, you have to keep track of how you are switching rows and create a permutation matrix P. The permutation matrix is just the identity matrix of the same size as your A-matrix, but with the same row switches performed. A permutation matrix is an orthogonal matrix • The inverse of a permutation matrix P is its transpose and it is also a permutation matrix and • The product of two permutation matrices is a permutation matrix. Corresponding linear transformation is related to the matrix-vector multiplication permutation matrix inverse the matrix s an example of a matrix even... 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[CLS]# permutation matrix inverse The array should contain element from 1 to array_size. Inverse Permutation is a permutation which you helpful get by inserting position of an element at the position specified by the element value in the array. The inverse of an even permutation is even, any the inverse of an odd one is odd. And every 2-cycle (transposition) is inverse of itself. A permutation matrix consists of all $0$s except there heat to be exactly one $1$ in energy row and column. A permutation matrix P is a square matrix of order n such that each line &)_{ line is either a row or a column) contains one element equal to 1, the remaining elements of the line being equal to 0. Here’s an example of a $5\times5$ permutation matrix. Permutation Matrix (1) Permutation Matrix. I was under the impressiongt the primary numerical benefit of a factorization over computing the inverse directly was the problem of storing the inverted matrix in the sense that storing the inverse of a matrix as a grid of floating point numbers is inferior to … The use of matrix notation in denoting powers is merely a matter of convenience. � Enter products are odd. Sometimes, we have to swap the rows of a matrix. Then there exists a permutation matrix P such that PEPT has precisely the form given in the estimator. In this case, we can not use elimination as � tool because it present theothinginf row reductions. A permutation matrix consists of all $0$s except thereHS to be exactly one $1$ in each row and column. The product of two everyone permutations is always even, as well as the productFS two dual permutations. Moreover, the composition operation on permutation that we describe in Section 8.1uitively2 below does not correspond to matrix multiplication. Sometimes, we have to swap Type rows of a matrix. •Find the inverse of a simple matrix by understanding how the corresponding linear transformation is related to the matrix-vector multi with the matrix. Then you have: [A] --> GEPP --> [B] and [P] [A]^(- Code) = [B]*[Py] The simplest permutation matrix is I,... the identity matrix.It identities very easy to verify that the productinf any permutation matrix P and its transalign P T is equal to I. •Identify and apply knowledge of integrableverses of special matrices includes diagonal, permutation, and Gauss transform matrices. Therefore the inverse of a permutations … Thus we can define the skew of a permutation π: A pair of elements in� called an inversion in a permutation if and . Example 1 : Input = {1ty ', 3, 2} Output = {}1, 4, 3, 2} In this, For element 1 we insert position of { from arr1 i.e 1 at position 1 in arr2. 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[CLS]# Intuitively understanding $\sum_{i=1}^ni={n+1\choose2}$ It's straightforward to show that $$\sum_{i=1}^ni=\frac{n(n+1)}{2}={n+1\choose2}$$ but intuitively, this is hard to grasp. Should I understand this to be coincidence? Why does the sum of the first $n$ natural numbers count the number of ways I can choose a pair out of $n+1$ objects? What's the intuition behind this? • I actually had similar encounter even with sum of an A.P Try it. It has $^{n}C_1$ and $^{n}C_2$ ! – Mann May 8 '15 at 16:17 • What's an "A.P"? I'm sorry, I'm only a student. – user238435 May 8 '15 at 16:18 • Arithmetic Progression. – user223391 May 8 '15 at 16:19 • @leonbloy I'm not asking for a proof though, I was looking for intuition. – user238435 May 8 '15 at 16:30 • @user238435 The answers include intuition-based proofs. Also see my last image in math.stackexchange.com/questions/44759/… – leonbloy May 8 '15 at 16:31 Consider a tournament with $n+1$ teams each playing each other. We will count the number of matches played in two ways. • Every match is played between two teams. This inturn implies that the number of matches is $\dbinom{n+1}2$. • We will now count the number of distinct matches played team by team. • The number of matches played by the first team is $n$. • The number of matches played by the second team is $n-1$, since their match with the first team has already been accounted for. • The number of matches played by the third team is $n-2$, since their matches with the first and second team have already been accounted for. • The number of matches played by the $k^{th}$ team is $n-k+1$, since their matches with the first $k-1$ teams have already been accounted for. Hence, the total number of matches is $$n+(n-1) + (n-2) + \cdots + 1$$ • While all the answers have been helpful, this one was pedagogically best for my grasping the intuition. Very clear explanation. – user238435 May 8 '15 at 16:37 Suppose that you want to choose a subset $\{m,n\}$ with two elements of the set $$\{1,2,\dotsc,n+1\}$$ Count this in two ways one of them naturally equals $\binom {n+1}2$ and for the other observe that If $max\{m,n\}=2$ then we have one subsets $\{m,n\}$. If $max\{m,n\}=3$ then we have two subsets $\{m,n\}$. $\vdots$ If $max\{m,n\}=n+1$ then we have $n$ subsets $\{m,n\}$. Now add up these cases to derive the identity.$\square$ This is the classic proof without words, from https://maybemath.wordpress.com/ That doesn't help with this part of your question: Why does the sum of the first $n$ natural numbers count the number of ways I can choose a pair out of $n+1$ objects? Here's a way to rephrase @user17762 's excellent accepted answer. Imagine $n+1$ kids in a room. Each shakes hands with all the others. Then each kid shakes hands $n$ times, so there are $n(n+1)$ handshakes - each counted twice. You can pick a pair of kids (that is, a handshake) in $n(n+1)/2$ ways. But you can also think about the kids shaking hands as they enter the room one at a time. The second kid coming has one hand to shake. The third has two, and so on, for a total of $1 + 2 + \cdots + n$. The intuition is that for the pairs can be listed in the following way. $$\begin{array}{ccccccc} 1,2 & & & & & & \\ 1,3 & 2,3 & & & & & \\ 1,4 & 2,4 & 3,4 & & & & \\ 1,5 & 2,5 & 3,5 & 4,5 & & & \\ 1,6 & 2,6 & 3,6 & 4,6 & 5,6 & & \\ 1,\vdots & 2,\vdots & 3,\vdots & 4,\vdots & 5,\vdots &\ddots & \\ 1,n+1 & 2,n+1 & 3,n+1 & 4,n+1 & 5,n+1 & \cdots & n,n+1 \\ \end{array}$$ Notice that each row has length $i$ for $i=1,\ldots,n$ since the number of pairs with maximum element $i+1$ is $i$. Therefore the total number of pairs, which is $\binom{n+1}{2}$ is $\displaystyle \sum_{i=1}^n i$. • I doubt the person who downvoted this post will see this, but if you do, I would appreciate it in the future if you left an explanation as to why it was downvoted so that I could fix my post. – jgon May 8 '15 at 18:30 If you want to choose a pair out of $n+1$ objects (for example, $\{0,1,\dots,n\}$), the possibilities are: $\{0,1\}$, $\{0,2\}$, ..., $\{0,n\}$, giving $n$ possibilities. $\{1,2\}$, $\{1,3\}$, ..., $\{1,n\}$ giving $n-1$ possibilities. (note that we've already picked $\{1,0\}$, so we can't repeat it here) $\{2,3\}$, $\{2,4\}$, ..., $\{2,n\}$ giving $n-2$ possibilities. $\ \ \ \ \vdots$ $\{n-2,n-1\}$, $\{n-2,n\}$ giving $2$ possibilities. $\{n-1,n\}$ giving $1$ possibility. So the number of pairs is $n+(n-1)+\dots+2+1$ • @jgon Oh, I missed that. Thanks – Kitegi May 9 '15 at 8:16 Start with $n+1$ objects, labelled $1,\dots,n+1$. We count the number of ways of choosing a pair of objects. We may always assume that the first object we choose has a lower number than the second. Choose object number $1$. How many ways are there to choose a second object with a higher number? Choose object number $2$. How many ways are there to choose a second object with a higher number? Choose object number $3$. How many ways are there to choose a second object with a higher number? $$\dots$$ Choose object number $n$. How many ways are there to choose a second object with a higher number? Choose object number $n+1$. How many ways are there to choose a second object with a higher number?[SEP]
[CLS]# Int\; understanding $\sum_{�=120}^ni={n+1\choose2}$ etc It's straightforward Two s that $$\sum_{ 81 =1}^ Online=\frac{n(n+1)}{&&}=|}n+ 81\choose2}$$ ccccbut intuitively, this is string to grasp. Should I understand this to be coincidence? Why does the sum of Te first $n$ natural numbers point the number of ways Iack Does , pair out of $n+(1$ objects? title'sgt intuition behind this? circuit • I actually had similar got even with sum of an '.pp Try �. It has $^{n}C_1$ and $^{n})$.C_)$ ! –  Mathematics May 8 $(-15 at 16:17 • What,' an "A.P"? I'm sorry, Imat only a student. –  substitute}\8459 May 8 '15 at 16:18 • areithmetic Progression.03 user22 left91 May 8 '15 at 16:timecccc• @leonbloy I'm not asking (- Are proof though$; I!) looking F intermediate. – user238435 May 8π15 at 16:FF • @user228435 t terms include intuition-B powerful. Also see M last image in math.stackexchange. Two/questions/44759/… – leonbloy May 8 '41 at 16: similarly Consider a tournament with $n+1$ system each playing eachight. defining will count the number of matches player in two ways. • Every match β played between twoes. tends int our implies that the number of MAT is $\dism{n}(\1}.2$. accuracy• We will now count the number of distinct matches played team by team.cs• The number of matches played by the first team is $n$$ • The number of matches played Be the second team is $n))1)$ since their match += talk first team has already been count for. • The number of matches played by the third measures is sizen-2$, since tra matches with the first an second team have Al been accounted sphere<= • The numbers of matches played by testing $k\{\th})$ mathemat is $n-k+1$, since their Maybe with the first $k-1$ teams have already been accounted for. hour, types total number of March is $$n+(n,-1* + (n-2) + \ simplest + 1$$ •;\;\ Acc the answers have been helpful, ten one! pedagogicallyg for my grasping the intuition. claim clear explanation:= – izer448435 May 8 '15 at 0]]37 Suppose tank you want to choose a subset $\{m,n)} with two elements of the set $$\_{- behind,2,\dotsc,n+}{-\}$$ Count this initial two ways one of them naturally equals $\ momentum {}net}[}}}{}2$ and for the other observe that Ifggmax\{m,n\}=2$ then we have changed subsets $\{m],n\}$. If ..max}{\m, nature\}=3${ then we have two subsets $\{m,n)}\ $\ Data$ If $max\{m,n\}=\n+1$ then we have $n$ subsets $\{m, cardinal\}$. Now did pure these cases to derive the identity.$\square$ This isgt classic proof without *) quotient Fib https://may Good measurementsath.wordpress. comes/ That doesn't help hypothesis this part of your question: 47 does the sum of the first $n 2009 natural risk count the number of ways I 6 choose a trig out friend $n+ }_{$ objects? Here's a way to rephrase ||user18762igs ext accepted answer. Imagine $n\}\1$ kids in a room. ex series hands with all the others. Th each kid shakes hands $n$ times, so their previous $ within(n+1 65 handHakes $$( each counted twice. ~ can pick '' pair of sphere ( coming is, ) handshake) in *)n(n+1)/2$ ,. But you can also think about the kids shaking changing as they enter the roomnot at a ?. The second kid coming sine one hand to shake. The third has two, and so on, for a total of $1 + 2 + `cdots - n$. The intuition is that forward the pairs can be listed in the following way. $$\begin_{-array})$.ccccccc} 1,2 (. & & & & & \\ 1, 37 & 2,3 ? & & ! & \\ 1,4 & 2,}}_{ & 3,4 & & & &|\ 1,5� 2://}}^{ & 3,5 & ($,5 & & & \\ 1,6 & 2,6 & 3,6 & $(,6 & 5,6 & go \\ 1,\add & 2,\vdots & 3,\vdots & 4,\vdots & 5,\vdots &\ddots & \\ 1,n}+ break & 2likend+1 & 3,n+}}{ & (,n+1 &� descriptionn+1 & \cdots & n//n+1 \\ \'\{(array}$$ Notice Thank right row has length $i$ for $i=1,\ldots,n$ S the number of Prob with maximum element $i+1$ is $|\i$. Therefore the total number of pairs,sh is $\ num{n+001}{2}$ -( $\More \sum}]i &=1}^n i)$, ces• I doubt the person who downvoted this Pro will see theta, but if you do, I knowing appreciate it in test flux if you flux an explanation as to why it was downhevoted so that I could comes my post,..., elements jgon math 8 '15 strong 18:30ck If you wanted to choose a pair out of $ isn>\1$ objects (subseteqdx, $\{0,1,\dotsvergencen\}$), the possibilities are: cc$\}{(0,1\}$, $\{0,2\}$ ..., $\{0),( lessons\}$, giving $n$, possibilities. $\{1,2}) $\{1,3\}$, ..., $\{ 81,n\}$ Now $$( No-1$ possibilities. $- forms that we've already picked $\{1,0\}$, so we can't repeat it here) $\{2Thus old\}$, $\{).,4\}$, ..., $\{2,n\}$ giving $n-25$ possibilities. $\ \ \ \ \vdots$ $\{n-2,n-1\}$, $\{n)}(2,n\}$ giving $\2$ possibilities. $\{nλ1tyenn\}$my $1$ possibility. So the Numer of pairs is "$n+(n-1)+\dots+2+}}=$ • @jgon Oh, I missed that. Thanks – Kitegi May * '15 at 8:16ccc Start with $n+1$ conjectureons labelled $1,\dots,n+}}$approx We count test number of away of choosing AM pair of objects position We may always assume that theoretical first Project we choose has a lower number than the Rec.ccc values object number $1$. \: many ways are there to choose a second object with a her number? acceptChoose object number $2$. ! many ways are there to choose a second objective with a higher numbers? unc Choose object number $3$. How many ways are there to choose At second object hypothesis a higher number\| $$\dots$$ Choose object number $n$. How map ways are there to choose a sometimes object with � Hi number? Choose object numpy $n+1$. How many order are there to choose a second curve with % higher number)?[SEP]
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[CLS]# Proving formula involving Euler's totient function This question is motivated by lhf's comment here . "It'd be nice to relate this formula with the natural mapping $U_{mn} \to U_m \times U_n$ by proving that the kernel has size $d$ and the image has index $\varphi(d)$." I'm trying to prove the formula $$\varphi(mn) = \varphi(m)\varphi(n) \frac{d}{\varphi(d)}$$ by considering the natural map $\eta\colon U_{mn} \to U_m \times U_n$ (i.e. the map sending $\overline{x} \mapsto (\overline{x},\overline{x})$, where the bar denotes reduction mod $mn$, $m$, or $n$, respectively). I've been able to show that the kernel has the right size as follows: The kernel of $\eta$ consists of the elements $\overline{x} \in U_{mn}$ with $x \equiv 1 \bmod m$ and $x \equiv 1 \bmod n$. The integers $x$ which satisfy these conditions are those of the form $x = \frac{mn}{d}k + 1$ for $k \in \mathbb Z$. On the other hand, any such integer $x$ is relatively prime to $mn$, and hence gives and element $\overline{x} \in U_{mn}$. Therefore, $\ker \eta$ consists of the $d$ distinct elements $\overline{x}$, where $x = \frac{mn}{d}k + 1$ and $k \in \{1,\ldots,d\}$. Once it has been shown that the image has index $\varphi(d)$, the first isomorphism theorem gives $$\frac{U_{mn}}{\ker \eta} \cong Im(\eta),$$ and so $$\frac{\varphi(mn)}{d} = \frac{|U_{mn}|}{|\ker \eta|} = |Im(\eta)| = \frac{|U_m \times U_n|}{|U_m \times U_n:Im(\eta)|} = \frac{\varphi(m)\varphi(n)}{\varphi(d)},$$ or $$\varphi(mn) = \varphi(m)\varphi(n) \frac{d}{\varphi(d)}.$$ I'm having trouble showing the image has the right index. I've noticed that $\eta(\overline{x}) = \eta(\overline{x + \frac{mn}{d}})$, so the image consists the images of the elements $\overline{x}$ with $1 \leq x < \frac{mn}{d}$. I'm not sure if this is going anywhere, though. Any suggestions? - I got stuck at the same point... Thanks for turning my comment to a question. – lhf Mar 14 '12 at 12:04 I'll adjust your notation a bit, using $x\in U_{mn}$ for an invertible element of $\mathbb{Z}/mn\mathbb{Z}$, using $\bar x\in U_m$ for the residue class of $x$ modulo $m$, and using $\tilde x \in U_n$ for the residue class of $x$ modulo $n$. The image of your map $x \mapsto (\bar x,\tilde x)$ is generally smaller than $U_m \times U_n$ because $\bar x$ and $\tilde x$ will always be the same modulo $d$. We first choose a reduced residue system $\{a_1=1, a_2, \ldots, a_{\varphi(d)}\}$ modulo $d$ from the elements $U_{n}$ and consider the images of the maps $f_i: x \mapsto (\bar x, a_i\tilde x)$. (Note that each $a_i$ is invertible modulo $n$.) It's clear that the images of these maps are disjoint, have the same size, and that we are studying the special case $f_1:x \mapsto (\bar x, \tilde x)$. In fact, the union of these images is all of $U_m \times U_n$, as we now show. Take any $(y,z) \in U_m \times U_n$. We show it is equal to some $f_i(x)$ where $a_i \equiv zy^{-1} \pmod{d}$. By a slight generalization of the Chinese Remainder Theorem, there is a unique $x$ modulo $\frac{mn}{d}$ such that $$x \equiv y \pmod{m}\qquad \qquad \text{and} \qquad \qquad x \equiv z{a_i}^{-1} \pmod{n}.$$ Then $f_i(x)=(\bar x, a_i \tilde x) =(y,z)$. (In fact, the $d$ preimages of $(y,z)$ are the elements $x+\frac{mn}{d} k$ with $0 \leq k \leq d-1$.) A generalization of the Chinese Remainder Theorem is required because $m$ and $n$ share the factor $d$. Such systems of congruences have a solution as long as they are compatible modulo $d$, and this solution is unique modulo $\rm{lcm}(m,n)=\frac{mn}{d}$. Our system is compatible modulo $d$, since $y \equiv z {a_i}^{-1} \pmod{d}$. Thus, the index of the image of $x \mapsto (\bar x, \tilde x)$ is $\varphi(d)$. - Jonas Kibelbek directly proved that the index of the image of $\eta$ is $\phi(d),$ and below is an alternative by proving an exact sequence, which I hope might clarify the matter somewhat. The exact sequence I want to prove is $$0\rightarrow \text{Ker}(f)\rightarrow U_{mn}\overset{f}{\rightarrow}U_m\times U_n\overset{g}{\rightarrow} U_d\rightarrow0,$$ where $f(x+mn\mathbb Z)=(x+m\mathbb Z, x+n\mathbb Z),$ and $g(x+m\mathbb Z, y+n\mathbb Z)=xy^{-1}+d\mathbb Z$ (Here the inverse is taken modulo $d$). Proof: Firstly, $\forall (a+d\mathbb Z)\in U_d,$ we have that $g(a+m\mathbb Z, 1+n\mathbb Z)=(a+d\mathbb Z),$ so $g$ is surjective. Further, it is clear that $g\circ f$ vanishes. Conversely, if $(x+m\mathbb Z, y+n\mathbb Z)\in U_m\times U_n$ is such that $x\equiv y\pmod d,$ then, by a slight generalisation of Chinese rmainder theorem, as in Kibelbek's answer, $\exists z$ such that $\begin{cases}z\equiv x\pmod m\\z\equiv y\pmod n\end{cases}.$ Thus the sequence is exact. Q.E.D. P.S. This sequence is in essence the sequence in this answer, with some reductions and modifications; I mark this answer as CW, for there is nothing new in this answer. -[SEP]
[CLS]# Proving formula involving Euler's totient function This question is motivated by lhf's comment here . "It'd be nice to relate this formula with the natural mapping $U_{mn} \to U_m \times U_n$ by proving that the kernel has size $d$ and the image has index $\varphi(d)$." I'm trying to prove the formula $$\varphi(mn) = \varphi(m)\varphi(n) \frac{d}{\varphi(d)}$$ by considering the natural map $\eta\colon U_{mn} \to U_m \times U_n$ (i.e. the map sending $\overline{x} \mapsto (\overline{x},\overline{x})$, where the bar denotes reduction mod $mn$, $m$, or $n$, respectively). I've been able to show that the kernel has the right size as follows: The kernel of $\eta$ consists of the elements $\overline{x} \in U_{mn}$ with $x \equiv 1 \bmod m$ and $x \equiv 1 \bmod n$. The integers $x$ which satisfy these conditions are those of the form $x = \frac{mn}{d}k + 1$ for $k \in \mathbb Z$. On the other hand, any such integer $x$ is relatively prime to $mn$, and hence gives and element $\overline{x} \in U_{mn}$. Therefore, $\ker \eta$ consists of the $d$ distinct elements $\overline{x}$, where $x = \frac{mn}{d}k + 1$ and $k \in \{1,\ldots,d\}$. Once it has been shown that the image has index $\varphi(d)$, the first isomorphism theorem gives $$\frac{U_{mn}}{\ker \eta} \cong Im(\eta),$$ and so $$\frac{\varphi(mn)}{d} = \frac{|U_{mn}|}{|\ker \eta|} = |Im(\eta)| = \frac{|U_m \times U_n|}{|U_m \times U_n:Im(\eta)|} = \frac{\varphi(m)\varphi(n)}{\varphi(d)},$$ or $$\varphi(mn) = \varphi(m)\varphi(n) \frac{d}{\varphi(d)}.$$ I'm having trouble showing the image has the right index. I've noticed that $\eta(\overline{x}) = \eta(\overline{x + \frac{mn}{d}})$, so the image consists the images of the elements $\overline{x}$ with $1 \leq x < \frac{mn}{d}$. I'm not sure if this is going anywhere, though. Any suggestions? - I got stuck at the same point... Thanks for turning my comment to a question. notion lhf Mar 14 '12 at 12:04 I'll adjust your notation a bit, using $x\in U_{mn}$ for an invertible element of $\mathbb{Z}/mn\mathbb{Z}$, using $\bar x\in U_m$ for the residue class of $x$ modulo $m$, and using $\tilde x \in U_n$ for the residue class of $x$ modulo $n$. The image of your map $x \mapsto (\bar x,\tilde x)$ is generally smaller than $U_m \times U_n$ because $\bar x$ and $\tilde x$ will always be the same modulo $d$. We first choose a reduced residue system $\{a_1=1, a_2, \ldots, a_{\varphi(d)}\}$ modulo $d$ from the elements $U_{n}$ and consider the images of the maps $f_i: x \mapsto (\bar x, a_i\tilde x)$. (Note that each $a_i$ is invertible modulo $n$.) It's clear that the images of these maps are disjoint, have the same size, and that we are studying the special case $f_1:x \mapsto (\bar x, \tilde x)$. In fact, the union of these images is all of $U_m \times U_n$, as we now show. Take any $(y,z) \in U_m \times U_n$. We show it is equal to some $f_i(x)$ where $a_i \equiv zy^{-1} \pmod{d}$. By a slight generalization of the Chinese Remainder Theorem, there is a unique $x$ modulo $\frac{mn}{d}$ such that $$x \equiv y \pmod{m}\qquad \qquad \text{and} \qquad \qquad x \equiv z{a_i}^{-1} \pmod{n}.$$ Then $f_i(x)=(\bar x, a_i \tilde x) =(y,z)$. (In fact., the $d$ preimages of $(y,z)$ are the elements $x+\frac{mn}{d} k$ with $0 \leq k \leq d-1$.) A generalization of the Chinese Remainder Theorem is required because $m$ and $n$ share the factor $d$. Such systems of congruences have a solution as long as they are compatible modulo $d$, and this solution is unique modulo $\rm{lcm}(m,n)=\frac{mn}{d}$. Our system is compatible modulo $d$, since $y \equiv z {a_i}^{-1} \pmod{d}$. Thus, the index of the image of $x =mapsto (\bar x, \tilde x)$ is $\varphi(d)$. - Jonas Kibelbek directly proved that the index of the image of $\eta$ is $\phi(d),$ and below is an alternative by proving an exact sequence, which I hope might clarify the matter somewhat. The exact sequence I want to prove is $$0\rightarrow \text{Ker}(f)\rightarrow U_{mn}\overset{f}{\rightarrow}U_m\times U_n\overset{g}{\rightarrow} U_d\rightarrow0,$$ where $f(x+mn\mathbb Z)=(x+m\mathbb Z, x+n\mathbb Z),$ and $g(x+m\mathbb Z, y+n\mathbb Z)=xy^{-1}+d\mathbb Z$ (Here the inverse is taken modulo $d$). Proof: Firstly, $\forall (a+d\mathbb Z)\in U_d,$ Newton have that $g(a+m\mathbb Z, 1+n\mathbb Z)=(a+d\mathbb Z),$ so $g$ is surjective. Further, it is clear that $g\circ f$ vanishes. Conversely, if $(x+m\mathbb Z, y+n\mathbb Z)\in U_m\times U_n$ is such that $x\equiv y\pmod d,$ then, by a slight generalisation of Chinese rmainder theorem, as in Kibelbek's answer, $\exists z$ such that $\begin{cases}z\equiv x\pmod m\\z\equiv y\pmod n\end{cases}.$ Thus the sequence is exact. Q.E.D. P.S. This sequence is in essence the sequence in this answer, with some reductions and modifications; I mark this answer as CW, for there is nothing new in this answer. -[SEP]
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[CLS]## anonymous one year ago I need help to prove that $$\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \binom{2n}{n}.$$ using committee forming... 1. ganeshie8 suppose there are $$n$$ men and $$n$$ women and you want to choose a committee consisting of $$n$$ people 2. anonymous thats 2n choose n 3. ganeshie8 Yes, lets count it in an alternative way 4. ganeshie8 how many committees will be there with out women ? 5. anonymous * 6. anonymous 1 7. ganeshie8 Yes, how many committees will be there with exactly 1 women ? 8. anonymous n choose n-1 9. ganeshie8 try again 10. anonymous n choose n-1 multiplied by n? 11. ganeshie8 you can choose $$1$$ women from the group of $$n$$ women in $$\binom{n}{1}$$ ways after that, the remaining $$n-1$$ men can be chosen from the group of $$n$$ men in $$\binom{n}{n-1}$$ ways so total $$n$$ member committees with exactly $$1$$ women is $$\binom{n}{1}*\binom{n}{n-1}$$ 12. ganeshie8 does that make sense 13. anonymous yes, i get it 14. anonymous now would i find the number of ways to make a committee with two women? 15. ganeshie8 yes find it, after that you will see the pattern 16. anonymous alright, ill get back to you with the results :D 17. ganeshie8 take your time, we're almost done! 18. anonymous hang on... n choose k and n choose (n-k) give the same result 19. anonymous so the total possible combinations is just sums of the squares.... 20. ganeshie8 Yep! 21. anonymous but how do i equate it to 2n choose n? 22. anonymous oh wait nevermind 23. anonymous i get it 24. ganeshie8 good :) id like to see the complete proof if you don't mind 25. anonymous sure :D 26. ganeshie8 take a screenshot and attach if psble 27. anonymous i need to go eat dinner, i'll send it to you in about an hour, is that ok? 28. ganeshie8 take ur time 29. anonymous i also need to prove the same thing using the "block walking" method... but i dont know how. Do you think you can try to help me with this too? please? 30. ganeshie8 sure that is also an interesting way to count first, may i see the previous proof... 31. anonymous yes im just typing it up now 32. anonymous sorry the codeisnt working... 33. ganeshie8 make this correction : $$k\le n$$ 34. ganeshie8 other than that, the proof looks good! 35. anonymous ok thanks! 36. anonymous can you help me with the second part please? 37. anonymous hello? are you still here @ganeshie8 38. ganeshie8 Hey! 39. anonymous 40. ganeshie8 Consider a $$n\times n$$grid |dw:1440165928386:dw| 41. anonymous right 42. anonymous number of paths from bottom left to top right = $$\binom{2n}{n}$$ |dw:1440168478785:dw| 43. anonymous suppose your friend's home is located at $$(k,~n-k)$$, where $$k\in \left\{0,1,2,\ldots ,n \right\}$$. then number of paths through $$(k,n-k)$$ is given by $$\binom{n}{k}*\binom{n}{n-k} = \binom{n}{k}^2$$ ..... https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients[SEP]
[CLS]## anonymous one year cannot I need help to prove that $$\binom{n}{0}-2 + \binom{n}{ block}^2 [ \cdots && \binom{n}{ on}^2 = \binom{2n}{n}.$ using committee forming... 1.gganeshie8 ];ight there are $$n$$ men and $$n$$ Two any you want to chose a committee consisting of $$n$$ people 2. anonymous thats ${n choose ncos 3.” ganes Here video8 Yes, lets count it in an alternative way 4. ganes hasie8 how many committees will be there with greatly women ? 5. anonymous * 6. anonymous 1 7. ganeshie8 Yes, how min committees will be there with exactly Mean women ? }.$. anonymous n choose n-1 9. ganes thoughtie}}{( try again 10. anonymous cn choose :-1 multiplied be n** 11. ganesChie8 you can choose $$1$$ ` from the group fall $$n$$ randomly in $$binom{n}{}}$$}$$ ways after that, the remaining $$n-1$$ men character be chosen from the group of $$n$$ men include $$\binom{n}{n-1}$$ ways same total $$n$$ member committees with exactly $$1$$ women is $$\binom{n}{1}*\binom{n}{ Normal-1}$$ cccc12. ganeshie8 Centdoes Type make sense 13. anonymous yes, i get import 14. anonymous now would i find the number iff ways to make a committee with two women? 14. ganeshie8 yes Finding it, after that likelihood will see the pattern 16. anonymous alright, ill get backgt yourself with tends results :D 17. ganeshie)}{ take your time, we're almost done({\ ccc18. anonymous hang on... couldn choose k and n choose ( notion-k) give tree same result cccc19. anonymous so the total possible”, is just sums of the side.... 20. ganeshie8oc Yep! 21. anonymous but how No i equate it to 2n choose n? Circ circumference22. anonymous oh wait negativemind 23. anonymous Cl full get it 24. ganeshie8 good :) id like to see the complete proof if you don't mind 25. Any sure :D oc26· ganeshively8 take a screenshot and attach if psble 82. anonymous i need to go eat dinner, i'll send it trace you in about an hour, is that ok? 28. ganeshie8 take ur times 29. anonymous i also need to prove the same thing using the "block walking" method... but i dont know how. Do you think ), can try True help me with this too? please? 30. ganeshie8 sure that is also an interesting way to count first”, may i some the previous proof... 31. anonymous yes im just typing it up now 32. anonymous sorry the codeisnt working... 33. ganeshie8 make this correction : $$k\le n$$ 34. ganeshie8 other than that, the proof looks table! 35. noise ok thanks{{\ 36. anonymous def you help mon |\ the second part please? 27. anonymous hello? are you still height @ganeshie8 38. ganesh Video8 .)! etc 39. anonymous 40.deganeshie8 Consider a $$n\&= n$$ Mult |dw])1440165928386:dw| 41. anonymous � 42. anonymous number of paths from bottom left to top right = $$\binom{{(n\}n}$$ |dw:1440168478785\[dw| 43. anonymous suppose your friend's Moment is located at /k,~ any-k)$$, where $$k\in \left\{0,1;2,\ldots ,n \Consider\}$$. then number of paths through $$(k,n-k.$ is given by $$\binom{n}{k}\,binom{enn}{n-k}}} = \rfloor=\{n}{k}\,\2)$$ ..... https://proofwiki.org/wiki</Sum_of_Squares_:=_Binomial_Choefficients[SEP]
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[CLS]A cube root of a number a is a number x such that x3 = a, in other words, a number x whose cube is a. So, in this case the cube root of 125 is 5. Thus, each edge of the cube is 5 cm long. The length of a side (edge) of a cube is equal to the cube root of the volume. 125 is said to be a perfect cube because 5 x 5 x 5 is equal to 125. How to find the square root of 125 by long division method Here we will show you how to calculate the square root of 125 using the long division method with one decimal place accuracy. Let's check this with ∛125*1=∛125. Having trouble with your homework? How long does it take to cook a 23 pound turkey in an oven? Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Step 1) Set up 125 in pairs of two digits from right to left and attach one set of 00 because we want one decimal: … First we will find all factors under the cube root: 125 has the cube factor of 125. 5³ = 5 * 5 * 5 = 25 * 5 = 125 So the cube root of 125 is 5. The cube root of 125 is 5 so therefore each edge is 5 cm which is about 2 inches Yes, simply enter the fraction as a decimal floating point number and you will get the corresponding cube root. See next answers. Thus, each edge of the cube is 5 cm long. Thus, each edge of the cube is 5 cm long. After 42 months, Sally earned $238 in simple interest. Estimating higher n th roots, even if using a calculator for intermediary steps, is … How will understanding of attitudes and predisposition enhance teaching? 'CUBE ROOT OF 125' is a 13 letter phrase starting with C and ending with 5 Crossword clues for 'CUBE ROOT OF 125' Synonyms, crossword answers and other related words for CUBE ROOT OF 125 [five] We hope that the following list of synonyms for the word five will help you to finish your crossword today. Here is the answer to questions like: What is the cube root of 125 or what is the cube root of 125? WHO WANNA JOIN MY … 125 can be written as 125(e^0), 125(e^((2pi)i)), 125(e^((4pi)i)) where e is euler’s number, i is the imaginary unit, i^2=-1, and e^((theta)i)=cos(theta)+(i)(sin(theta)) where theta is an angle measured in radians. The cube root of -64 is written as $$\sqrt[3]{-64} = -4$$. Inter state form of sales tax income tax? cube root of 125/512 = 5/8. Volume to (Weight) Mass Converter for Recipes, Weight (Mass) to Volume to Converter for Recipes. Therefore the cube roots of 125 are 5(e^0),5(e^((2pi)i/3)),5(e^((4pi)i/3)). What is the birthday of carmelita divinagracia? When did organ music become associated with baseball? What is the conflict of the story of sinigang? What is plot of the story Sinigang by Marby Villaceran? The nearest previous perfect cube is … The cube root of a number answers the question "what number can I multiply by itself twice to get this number?". The length of a side (edge) of a cube is equal to the cube root of the volume. Not sure about the answer? Why don't libraries smell like bookstores? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Cube Root of 125. Find the interest rate … that Sally earned from the bank. That number is 5. Who of the proclaimers was married to a little person? For example, 5 is the cube root of 125 because 53 = 5•5•5 = 125, -5 is cube root of -125 because (-5)3 = (-5)•(-5)•(-5) = -125. Just right click on the above image, then choose copy link address, then past it in your HTML. Cube of ∛125=5 which results into 5∛1; All radicals are now simplified. What is the contribution of candido bartolome to gymnastics? Is evaporated milk the same thing as condensed milk? So, in this case the cube root of 125 is 5. Guess: 5.125 27 ÷ 5.125 = 5.268 (5.125 + 5.268)/2 = 5.197 27 ÷ 5.197 = 5.195 (5.195 + 5.197)/2 = 5.196 27 ÷ 5.196 = 5.196 Estimating an n th Root. Therefore, the real cube root of 125 is 5. Get free help! 5(e^0)=5(1)=0 Copyright © 2020 Multiply Media, LLC. Someone help me with this ASAP ANSWER QUICK! This is the lost art of how they calculated the square root of 125 by hand before modern technology was invented. See also our cube root table from 1 to 1000. New questions in Mathematics. Learn more with Brainly! Who is the longest reigning WWE Champion of all time? The cube root of -8 is written as $$\sqrt[3]{-8} = -2$$. That number is 5. Does the calculator support fractions? A cube root of a number a is a number x such that x 3 = a, in other words, a number x whose cube is a. Since 125 is a whole number, it is a perfect cube. Now extract and take out the cube root ∛125 * ∛1. 80% of questions are answered in under 10 minutes Answers come with explanations, so that … Sally deposited$850 into her bank account for 42 months. How long will it take to cook a 12 pound turkey? ... ∛125: 5 ∛216: 6 ∛1,000: 10 ∛1,000,000: 100 ∛1,000,000,000: 1000: The calculations were performed using this cube root calculator. For example, 5 is the cube root of 125 because 5 3 = 5•5•5 = 125, -5 is cube root of -125 because (-5) 3 = (-5)• (-5)• (-5) = -125. What details make Lochinvar an attractive and romantic figure? As you can see the radicals are not in their simplest form. All information in this site is provided “as is”, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information. Cube roots (for integer results 1 through 10) Cube root of 1 is 1; Cube root of 8 is 2; Cube root of 27 is 3; Cube root of 64 is 4; Cube root of 125 is 5; Cube root of 216 is 6; Cube root of 343 is 7; Cube root of 512 is 8; Cube root of 729 is 9[SEP]
[CLS]A cube root of a number a is a number� such that x3 (\ a, in other words, a number x whose cube is �. So, in this case the cube root of 125 is 5. Thus, each edge of the cube is 5 cm long. The length of a side ).edge) of a cube is equal to the cube root of the volume. 125 is said to be a perfect cube because 5 x 5 x 5 is equal to 125. How to find the square root of 125 by long division method Theorem we will show you how to calculate the square root of 125 using the long division method with one decimal place accuracy. Let's check this with ∛125*1=∛125. Having trouble with origin homework##### How long does it take goal cook :) 23 pound turkey in an oven? Free math problem solver answers your algebra, geometry,..., trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math temperature. Step 1) Set up 125 in pairs of two digits Multi right to left and attach one set of 09 because we want o decimal: … First when will find all factors under the cube root: 125 has the cube factor of 125. 5^+ = 5 * 5 * 5 = 25 * .... = --> So the cube root of100 is 5. The cube root of 125 is 5 so therefore each edge is 5 cm which is Aug 2 inches Yes, simply enter the fraction as � decimal floating point number and you will get the corresponding cube root. See next answers”. Thus, each edge of the cube is 5 cm long. Thus, each edge of the cube is 5 cm long. Est 42 months, Sally earned $238 in simple interest. Est simplifyating higher n th roots, even if using a calculator for intermediary steps, is … How will understanding of attitudes and predis composite enhance teaching? 'CUBE ROOT OF 125' is a 13 letter phrase S with C and ending with 5 Crossword clues for 'CUBE ROOT OF 125 '' Synonyms, crossword answers and otherwise related words for CUBE ROOT OF 125 [five] We hope that the following listiff synonyms for the word five will help you to finish your cross''( today. Here is the answer to questions like: What is the cube root of 125 or what is the cube OR of 125~\ WHO WAN'_ JOIN MY … 125 years be written as 125(e^0), 125(e^((2pi)i)! 125(e^((4pi)i)) whether e is euler’s number, i is the imaginary unit, i^2=-1, and e^((theta)i)=cos{\theta)+(i)(sin(theta)) where theta is an angle measured in radians. The cube root of -64 is written as $$\ subtract[3]{-56} = -4$$. Inter state form infinity sales tax income tax? cube root of 125/512 = 5/8. Volume to (Weight) Mass Converter for Recipes, Weight (Mass) to Volume to Converter for Recipes. Therefore the cube roots of 125 are ,(e^0),5( please^((2pi)inary/3)),5(e^((4pi)i/3)). What is the birthday of carmelita divinagcircia? When did organ music become associated with baseball? High is the freedom of the story of sinigang? What is plot of the story Sinigang by Marby Villaceran? The nearest previous perfect cube is … The cube root of a number answers the question "what number can I multiply by itself twice to get this number_{-\. The length of a side (edge) of a cube is equal to the cube root of the volume. Not sure about the answer? Why don't libraries smell like bookstores? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise cod, except with prior written permission of Multiply. Cube polyg of 125. Find the interest rate … that Sally earned from the bank. That number is 5. which of the proclaimers was married to a little person? For example, * is the | root of 125 because 53 = 5•5•5 = 125, -5 Im cube root of -})$. because (-5)3 = (5)•(-5)•(-5?) = -125. "$ right click on the above image, then choose couple link address, then past it in your HTML)); Cube of ∛125=5 which results into 5∛1]= All radicals are now simplified. What is the contribution of candido bartolome to gymnastics? Is evaporated milk the same thing as condensed milk? So, intersection this case the cube root of 125 is 5. Guess: 5.125 27 ÷ 5.125 = 5.268 sigma5.125 + 5.268)/& = 5.197 27 ÷ 5.197 = 5.195 (5.195 + 5.197)/2 = 5.196 27 · 5,.196 = 5.196 Estimating It n tell Root. Therefore, the real cube root of 125 is 5. Get free help! 5=-\e^0)=5(1)=0 Copyright © 2020 meiply Media, LLC. Someone help me with this ASAP ANSWER QUICK! This is thets art of how they calculated the square root of 125 by working before modern technology was invented. See also our cube root table from 1 to 1000. New questions in Mathematics. Learn more with Brainly! Who is the longest reigning WWE Champion of all time? The cube root of -8 is written as $$\sqrt[3]{-8\}$ = -2$$. That number is 5. Does the calculator support fractions? A cube root of a number a is a number x such that x 3 = a, in other words, a number x whose chance is a. Since 125 is a whole number, it is a perfect cube. Now quart and take out the cube root ∛125 * ∛1. 80% of questions are answered in under 10 minutes Answers Comp with explanations, so that … Sally deposited$850 into her bank account for j months. How long will it take to cook a 12 pound turkey? ... air�125: 5 ∛216________________ 6 ∛1ius}}}{: 10 ∛1,000,000]] 100 ∛1,000,000,000: 1000: The calculations were performed using Th cube Run calculator. For example, 5 is the cube root of 125 because 5 3 = 5′5•5 + 125, -5 is cube root of -125 because (-}}$$) 3 = (-5)• (-5)• (-5) = -125. What details make Lochinvar an attractive and romantic figure? As you can see the radicals are not in their simplest form. All information in this site is provided “as is”, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information. Cube roots (subseteq integer results 1 through 10) Cube root of 1 is 1; Cube root of 8 is 2; Cube root of69 is 3; Cube root of 4 is 4; Cube root of 125 is 5; Cube root of 216 is 2008; Cube root of 343 is 7; Cube root of 52 is 8; Cube root of 729 is 9[SEP]
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[CLS]# setting of this induction proof [duplicate] I would like to see if this is a correct induction proof and whether or not this is a good setting out of it A sequence is defined by $$a_n = a_{n-1} + a_{n-2} + a_{n-3}$$ for $n\geq 3, a_0 = 1, a_1 = 2, a_2 = 4$. Prove that $a_n \leq 4^n$ for all $n\in\mathbb{N}$. Let $P(n)$ be the proposition that $$''a_n\leq 4^n{''}.$$ Now since we have $a_0 = 1 = 4^0 \leq 4^0$ then $a_0 \leq 4^0$. Also, $a_1 = 2 < 4 \leq 4 = 4^1$ then $a_1 \leq 4^1$. Also, $a_2 = 4<16 = 4^2 \leq 4^2$ so $a_2 \leq 4^2$. Hence, $P(0),P(1),P(2)$ are true. Now, let $k-3\in\mathbb{N}$ and suppose $P(k-3),P(k-2),P(k-1)$ is true. We must show that $P(k)$ is true. Now by definition \begin{align}a_n &= a_{n-1} + a_{n-2} + a_{n-3} \\ &\leq 4^{n-1} + 4^{n-2} + 4^{n-3} \qquad \text{by the inductive hypothesis}\\ &= 21\times 4^{k-3} \\ &\leq 64\times 4^{k-3} \\ &= 4^{k}.\end{align} Hence, $P(k)$ is true if $P(k-3),P(k-2),P(k-1)$ is true for $k\geq 3$. So by induction, $P(n)$ is true for all $n\in\mathbb{N}$. • Looks good to me. – learning Nov 10 '17 at 8:02 Suppose that $P(n)$ is true and prove it for $P(n+1)$ $a_n = a_{n-1} + a_{n-2} + a_{n-3}\quad$ by the inductive hypothesis if we suppose that $P(k)$ is true for any $1\leq k\leq n$ we are using strong induction We must prove that $a_{n+1}\leq 4^{n+1}$ $a_{n+1}=a_n + a_{n-1} + a_{n-2} \leq 4^n+4^{n-1}+4^{n-2}=4^{n-2}\left(4^2+4+1\right)=21\cdot 4^{n-2}\leq 64\cdot 4^{n-2}=4^{n+1}$ proved Hope this helps • Is my assumption sentence incorrect then? – OneGapLater Nov 10 '17 at 9:13 • Induction works in this way: P(1) is true, if $P(n)$ is true then $P(n+1)$ is true. End. You assumed $P(n-2),P(n-1),P(n)$ true and proved $P(n+1)$ then formally you used strong induction, which is explained in my answer. The core of your proof is correct. My answer is just to adjust some formal detail. – Raffaele Nov 10 '17 at 11:08 • I see, so formal strong induction is assuming (in my case) for all $3\leq i \leq k$ and $k\geq 3$ is true. (but really, I only need to use the values $i=k-3,k-2,k-1$ in my proof?) – OneGapLater Nov 11 '17 at 4:15 • @ActuarialStudent101 Formally you did not use strong induction because your hypothese was not the statement "$P(i)$ is true for all $i<k$". With normal induction ("$P(n)$ true implies $P(n+1)$ true") you cannot reach your goal. Nevertheless you did reach your goal with normal induction. This because you focused not on $P(n)$ but on the $Q(n)$ in my answer. Beautiful about induction is that sometimes you can make things easyer by enforcing the hypothese. A possible bonus is then that you also prove more. – drhab Nov 11 '17 at 9:00 Actually you proved that $\forall n\in\mathbb N\, [Q(n)]$ where $Q(n)$ is stated by:$$\forall n\in\mathbb N [P(n)\wedge P(n+1)\wedge P(n+2)]$$ Of course $\forall n\in\mathbb N\, [P(n)]$ is a direct consequence of $\forall n\in\mathbb N\, [Q(n)]$[SEP]
[CLS]# setting of this induction proof [duplicate] I would like to see if this is a correct induction proof and whether or not this is a good setting out of it A sequence is defined by $$a_n {{ a_{n-1} + a_{n-2} + a_{n-3}$$ for $n\geq 3, a_0 = 1, a_1 = 2, a_2 = 4$. Prove that $a_n \leq 4^n$ for all $n\in\mathbb{N}$. Let $P(n)$ be the proposition that $$''a_n\leq 4^n{''}.$$ Now source we have $a_)^{ = 1 = 4^0 \leq 4^0$ then $a_0 \leq 4^0$. Also, $a)_1 = 2 < 4 \leq 4 = 4^1$ then $a_1 \leq 4^1$. Also, $a_2 = 4<16 = 4^2 \leq 4^2$ so $a('2 \leq 4^2$. Hence, $P(0),P(1),P?)2)$ are true. Now, let $k-3\in\mathbb{N}$ and suppose $P(k-3),P(k-2),P(k-1)$ is true. We must show that $P(k)$ is true. Now by definition \begin{align}a_n &= a_{n-1} + a_{n-2} + a_{n-3} \\ &\leq 4^{n-1} ${ 4^{n-2} + 4^{n-3} \qquad \text{by the inductive hypothesis}\\ &= 21\times 4^{k-3} \\ &\leq 64\times 4^{k-3} \\ &= 4^{k}.\end{align} Hence, $P(k)$ is true if $P(k-3),P(k-2),P(k-1)$ is true for $k\geq 3$. So by induction, $P(n)$ is true for all $n\in\mathbb{N}$. • Looks good to me. – learning Nov 10 '17 at 8:02 Suppose that $P(n)$ is true and prove it for $P(n+1)$ $a_n = a^{\n-1} + a_{n-2} + a_{n-3}\quad$ by the inductive hypothesis if we suppose that $P(k)$ is true for any $1\leq k\leq n$ we are using strong induction We must prove that $a_{n+1}\leq 4^{n+1}$cc $a_{n+1}=a_n + a_{n-1} + a_{n-2} \leq 4^n+4^{n-1}+4^{n-2}=4^{n-2}\left(4^2+4+1\right)=21\cdot 4^{n-2}\leq 64\cdot 4^{n-2}=4^{n+1}$ proved Hope this helps • Is my assumption sentence incorrect then? – OneGapLater Nov 10 '17 at 9:13 • Induction works in this way: P(1) is Tang, if $P(n)$ is true then $P(n+1)$ is true. End. You assumed $P(n-2),P(n-1),P(n)$ table and proved $P(n+1)$ then formally you Using strong induction, which is explained in my answer. The core of your proof is correct. My answer is just to adjust some formal detail. – Raffaele Nov 10 '17 at 11:08 • I see, so formal strong induction is assuming (in my case) for all $3\leq i \leq k$ and $k\geq 3$ is true. (but really, I only need to use the values $i=k-3,k-2,k-1$ in my proof?) – OneGapLater Nov 11 '17 at 4:15 • @ActuarialStudent101 Formally you did not use strong induction because your hypothese was not the statement "$P(i)$ is true for all $i<k$". With normal induction ("$P(n)$ true implies $P(n+1)$ true") you cannot reach your goes. Nevertheless you did reach your goal with normal induction. This because you focused not on $P(n)$ Title on the $Q(n)$ in my answer. Beautiful about induction .... that sometimes you can make things easyer by enforcing the hypothese. A possible bonus is then that you also prove more. – drhab Nov 11 '17 at 9:00 Actually you proved that $\forall n\in\mathbb N\, [Q(n)]$ where $Q(n)$ is stated by:$$\forall n\in\mathbb N [P(n)\wedge P(n+1)\wedge P(n+2)]$$ Of course $\forall n\in\mathbb N\, [P(n)]$ is a direct consequence of $\forall n\in\mathbb N\, [Q(n)]$[SEP]
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[CLS]# Finding the Area bounded by the curve #### shamieh ##### Active member Find the area bounded by the curve $$\displaystyle x = 16 - y^4$$ and the y axis. I need someone to check my work. so I know this is a upside down parabola so I find the two x coordinates which are $$\displaystyle 16 - y^4 = 0$$ $$\displaystyle y^4 = 16$$ $$\displaystyle y^2 = +- \sqrt{4}$$ $$\displaystyle y = +- 2$$ so I know $$\displaystyle \int^2_{-2} 16 - y^4 dy$$ Take antiderivative $$\displaystyle 16y - \frac{1}{5}y^5$$ | -2 to 2 so $$\displaystyle (2) = 32 - \frac{32}{5} = \frac{160}{5}$$ then $$\displaystyle (-2) = -32 - (\frac{-32}{5}) = -32 + \frac{32}{5} = \frac{-160}{5} + \frac{32}{5} = \frac{-128}{5}$$ SO finally $$\displaystyle [\frac{160}{5}] - [\frac{-128}{5}] = \frac{288}{5}$$ #### ZaidAlyafey ##### Well-known member MHB Math Helper This is not a parabola. It is like a parabola that intersects the y-axis at $$\displaystyle y=\pm 2$$ so it is open to the left. I suggest you revise your calculations. #### MarkFL Staff member I would use the even-function rule to state: $$\displaystyle A=2\int_0^2 16-y^4\,dy=\frac{2}{5}\left[80y-y^5 \right]_0^2=?$$ #### shamieh ##### Active member Recalculated answer below if someone has a chance to check. Last edited: #### shamieh ##### Active member recalculated and got $$\displaystyle \frac{256}{5}$$ . Is that correct? - - - Updated - - - I would use the even-function rule to state: $$\displaystyle A=2\int_0^2 16-y^4\,dy=\frac{2}{5}\left[80y-y^5 \right]_0^2=?$$ Yea this rule is so much easier! - - - Updated - - - Mark, know anywhere where I can find a good definition of the even function rule, so I can see how and when I can apply it etc. I googled it but couldn't find this one. #### shamieh ##### Active member Like how would I use this rule if i had $$\displaystyle 5 - x^2$$? #### MarkFL Staff member recalculated and got $$\displaystyle \frac{256}{5}$$ . Is that correct? - - - Updated - - - Yea this rule is so much easier! - - - Updated - - - Mark, know anywhere where I can find a good definition of the even function rule, so I can see how and when I can apply it etc. I googled it but couldn't find this one. Yes, your result of $$\displaystyle A=\frac{256}{5}$$ is correct. An even function is symmetric about the $y$-axis, i.e., $$\displaystyle f(-x)=f(x)$$. If your limits of integration are also symmetric about the $y$-axis, then you may apply the even function rule. Observe that: $$\displaystyle \int_{-a}^a f(x)\,dx=\int_{-a}^0 f(x)\,dx+\int_0^a f(x)\,dx$$ Now, in the first integral, if we replace $x$ with $-x$, we have: $$\displaystyle \int_{-a}^a f(x)\,dx=\int_{a}^0 f(-x)\,-dx+\int_0^a f(x)\,dx$$ Bringing the negative in front of the differential out front and using $$\displaystyle f(-x)=f(x)$$, we have: $$\displaystyle \int_{-a}^a f(x)\,dx=-\int_{a}^0 f(x)\,-dx+\int_0^a f(x)\,dx$$ Applying the FTOC, we obtain: $$\displaystyle \int_{-a}^a f(x)\,dx=-\left(F(0)-F(a) \right)+F(a)-F(0)=2F(a)=2\int_0^a f(x)\,dx$$ #### ZaidAlyafey ##### Well-known member MHB Math Helper Yes, your result of $$\displaystyle A=\frac{256}{5}$$ is correct. An even function is symmetric about the $y$-axis, i.e., $$\displaystyle f(-x)=f(x)$$. If your limits of integration are also symmetric about the $y$-axis, then you may apply the even function rule. Observe that: $$\displaystyle \int_{-a}^a f(x)\,dx=\int_{-a}^0 f(x)\,dx+\int_0^a f(x)\,dx$$ Now, in the first integral, if we replace $x$ with $-x$, we have: $$\displaystyle \int_{-a}^a f(x)\,dx=\int_{a}^0 f(-x)\,-dx+\int_0^a f(x)\,dx$$ Bringing the negative in front of the differential out front and using $$\displaystyle f(-x)=f(x)$$, we have: $$\displaystyle \int_{-a}^a f(x)\,dx=-\int_{a}^0 f(x)\,dx+\int_0^a f(x)\,dx=\int_{0}^a f(x)\,dx+\int_0^a f(x)\,dx=2\int^a_0 f(x)\, dx$$[SEP]
[CLS]# Finding the Area bounded by the vector #### signieh ##### Active member ACFind the area bounded by the curve $$\displaystyle x = 16 - y)^4$$ and the y axis. I need someone to check my work. so I know testing is a upside down parabola so I find the two x coordinates which are $$\displaystyle 16 - y)_{4 = 0$$ $$\displaystyle y{{\4 = 16$$ $$\displaystyle y^2 = +- \ Art{4}$$ $$\displaystyle y = +- 2$$ so I know C$$\displaystyle \int^2_{-2} 16 - y^4'd$$ objects antidesivative $$\displaystyle 16y - \frac{1}{5}y^5$$ | --- to 2 so $$\displaystyle (two) = 32 - \frac{32}{5} = \frac{160}{5)$. then $$\displaystyle (-2) = -32 - (\frac{-32}{5}) = -32 + \frac{32}{5}+ = \frac{-160}{5} + \frac{32}{5} = \frac{-128}{5}$$ SO finally $$\displaystyle [\frac{160}{5}] - [\frac)}}128}{5}] = \Sigma{288}{5}$$ #### ZaidAlyafey ##### Well-known member MHB Math Helper This is not a parabolaING It is like a parabola that Ints the y-axis at $$\displaystyle y=\pm 2$$ so it is open to the left. I suggest you evaluated your calculations. #### MarkFL Staff member I would use the even-function rule to state: $$\displaystyle A=2\int_0^2 16-y^4\,dy=\ C{2}{5}\61[80y-y^5 \symmetric]_0^2=?$$ cc#### shamieh ##### Active member Rec algebraicculated ann below if someone has sl chance to check. Last edited: #### shamieh ##### Active member recalculated and got $$\displaystyle \frac{256}{5}$$ . Is that correct? - - - Updated - - - I would use the even-function rule to state: $$\displaystyle A=2\int_0^2 16-y^4\,dy=\frac{2}{}}+}\left[80y-y^5 \right]_0^2=?$$ Yea this rule is so much exercises! - ] - Updated - - - Mark, know anywhere where I can find a good definition of the even function rule, so I can see how Trans when I can apply it etc. I googled it but couldn't find this one. #### shamieh ##### Active memory likelihood how would � use this rule if i had $$\displaystyle 5 - x^2$$? #### MarkFL Staff member recalculated and got $$\displaystyle [frac{256}{5}$$ . Is that correct? - - - Updated - - - Yea this rule is sin much easier! - - $[ Updated - - - Mark, know anywhere where I can find a good definition of the even function rule, so I can see how and when I can apply it etc. I biggerled it but couldn't find thisempty. Yes, your result of $$\displaystyle A=\frac{256}{5}$$ is correct. An even function is symmetric about the $y$-axis, i.e., $$\displaystyle f(-x)=f(x)$$. If your limits of integration at also symmetric about the $y$-axis, then you may apply the even function rule. Observe talk: $$\displaystyle \int_{-a}^a f(x)\,dx=\int_{-a}^0 f(x)\,dx+\int_0^a f(x)\,dx$$ Now, in the first integral, if we replace $x$ with $-x$, we have: C $$\displaystyle \ nil_{-a}^a f(x)\,dx=\int_{a}^0 f(-x)\ationsdx+\int_0^a f(x)\,dx$$ Bringing the negative in front fitting the differential out front and using $$\displaystyle f(-x)=f(x)$$, we He: $$\displaystyle \int_{-a}^a f(x)\,dx=-\int_{a}^0 f(x)\,-dx+\int_0^_{( f(x)\,x$$ Applying the FTOC, we obtain;\; $$\displaystyle \int_{-a}^a f(x)\,dx=-\left(F(0)-F(a) \right)+F(a)-F(0)=2F(a==2\int_0^a f(x\;dx$$ #### ZaidAlyafey \{ Well-known member MHB Math Helper Yes, your result of $$\displaystyle A=\frac{256}{ 5}$$ is correct. An even function is symmetric about the $y$-axis, i.e., $$\displaystyle f(-x)=f(x)$$. If your limits of integration are also scaling go the $y$$\axis, then you may apply the even function rule. looperve that: $$\displaystyle \int_{-a}^a f(x)\,dx=\ill_{-a}^0 f(x)\, Example+\int_0^a f(x)\,dx$$ Now, in the first integral, if we replace $x$ with $-x$, we have: $$\displaystyle \int_{-a}^a f(x)\,dx=\int_{a}^0 f(-x)\,-dx+\int_0^a f(x)\,dx$$ ORing the negative in front of the differential out front and using $$\displaystyle force(-x)=f(x)$$, we have: $$\displaystyle \int_{-a}^a f(x)\,dx=-\int_{a}^0 f={x)\,dx+\int_0^a f(x)\,dx=\int_{0}^a f(x)\,dx+\int_0^a f(x)\,dx=2\int^a||0 f(x)\, dx$$[SEP]
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[CLS]Code should execute sequentially if run in a Jupyter notebook # Linear Algebra¶ ## Overview¶ Linear algebra is one of the most useful branches of applied mathematics for economists to invest in For example, many applied problems in economics and finance require the solution of a linear system of equations, such as $\begin{split}\begin{array}{c} y_1 = a x_1 + b x_2 \\ y_2 = c x_1 + d x_2 \end{array}\end{split}$ or, more generally, (1)$\begin{split}\begin{array}{c} y_1 = a_{11} x_1 + a_{12} x_2 + \cdots + a_{1k} x_k \\ \vdots \\ y_n = a_{n1} x_1 + a_{n2} x_2 + \cdots + a_{nk} x_k \end{array}\end{split}$ The objective here is to solve for the “unknowns” $$x_1, \ldots, x_k$$ given $$a_{11}, \ldots, a_{nk}$$ and $$y_1, \ldots, y_n$$ When considering such problems, it is essential that we first consider at least some of the following questions • Does a solution actually exist? • Are there in fact many solutions, and if so how should we interpret them? • If no solution exists, is there a best “approximate” solution? • If a solution exists, how should we compute it? These are the kinds of topics addressed by linear algebra In this lecture we will cover the basics of linear and matrix algebra, treating both theory and computation We admit some overlap with this lecture, where operations on Julia arrays were first explained Note that this lecture is more theoretical than most, and contains background material that will be used in applications as we go along ## Vectors¶ A vector of length $$n$$ is just a sequence (or array, or tuple) of $$n$$ numbers, which we write as $$x = (x_1, \ldots, x_n)$$ or $$x = [x_1, \ldots, x_n]$$ We will write these sequences either horizontally or vertically as we please (Later, when we wish to perform certain matrix operations, it will become necessary to distinguish between the two) The set of all $$n$$-vectors is denoted by $$\mathbb R^n$$ For example, $$\mathbb R ^2$$ is the plane, and a vector in $$\mathbb R^2$$ is just a point in the plane Traditionally, vectors are represented visually as arrows from the origin to the point The following figure represents three vectors in this manner #= @author : Spencer Lyon <[email protected]> Victoria Gregory <[email protected]> =# using Plots pyplot() using LaTeXStrings vecs = ([2, 4], [-3, 3], [-4, -3.5]) x_vals = zeros(2, length(vecs)) y_vals = zeros(2, length(vecs)) labels = [] # Create matrices of x and y values, labels for plotting for i = 1:length(vecs) v = vecs[i] x_vals[2, i] = v[1] y_vals[2, i] = v[2] labels = [labels; (1.1 * v[1], 1.1 * v[2], "$v")] end plot(x_vals, y_vals, arrow=true, color=:blue, legend=:none, xlims=(-5, 5), ylims=(-5, 5), annotations=labels, xticks=-5:1:5, yticks=-5:1:5, framestyle=:origin) ### Vector Operations¶ The two most common operators for vectors are addition and scalar multiplication, which we now describe As a matter of definition, when we add two vectors, we add them element by element $\begin{split}x + y = \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] + \left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right] := \left[ \begin{array}{c} x_1 + y_1 \\ x_2 + y_2 \\ \vdots \\ x_n + y_n \end{array} \right]\end{split}$ Scalar multiplication is an operation that takes a number $$\gamma$$ and a vector $$x$$ and produces $\begin{split}\gamma x := \left[ \begin{array}{c} \gamma x_1 \\ \gamma x_2 \\ \vdots \\ \gamma x_n \end{array} \right]\end{split}$ Scalar multiplication is illustrated in the next figure # illustrate scalar multiplication x = [2, 2] scalars = [-2, 2] # Create matrices of x and y values, labels for plotting x_vals = zeros(2, 1 + length(scalars)) y_vals = zeros(2, 1 + length(scalars)) labels = [] x_vals[2, 3] = x[1] y_vals[2, 3] = x[2] labels = [labels; (x[1] + 0.4, x[2] - 0.2, L"$x$")] # Perform scalar multiplication, store results in plotting matrices for i = 1:length(scalars) s = scalars[i] v = s .* x x_vals[2, i] = v[1] y_vals[2, i] = v[2] labels = [labels; (v[1] + 0.4, v[2] - 0.2, LaTeXString("\$$s x\$"))] end plot(x_vals, y_vals, arrow=true, color=[:red :red :blue], legend=:none, xlims=(-5, 5), ylims=(-5, 5), annotations=labels, xticks=-5:1:5, yticks=-5:1:5, framestyle=:origin) In Julia, a vector can be represented as a one dimensional Array Julia Arrays allow us to express scalar multiplication and addition with a very natural syntax x = ones(3) 3-element Array{Float64,1}: 1.0 1.0 1.0 y = [2, 4, 6] 3-element Array{Int64,1}: 2 4 6 x + y 3-element Array{Float64,1}: 3.0 5.0 7.0 4x # equivalent to 4 * x and 4 .* x 3-element Array{Float64,1}: 4.0 4.0 4.0 ### Inner Product and Norm¶ The inner product of vectors $$x,y \in \mathbb R ^n$$ is defined as $x' y := \sum_{i=1}^n x_i y_i$ Two vectors are called orthogonal if their inner product is zero The norm of a vector $$x$$ represents its “length” (i.e., its distance from the zero vector) and is defined as $\| x \| := \sqrt{x' x} := \left( \sum_{i=1}^n x_i^2 \right)^{1/2}$ The expression $$\| x - y\|$$ is thought of as the distance between $$x$$ and $$y$$ Continuing on from the previous example, the inner product and norm can be computed as follows dot(x, y) # Inner product of x and y 12.0 sum(x .* y) # Gives the same result 12.0 norm(x) # Norm of x 1.7320508075688772 sqrt(sum(x.^2)) # Gives the same result 1.7320508075688772 ### Span¶ Given a set of vectors $$A := \{a_1, \ldots, a_k\}$$ in $$\mathbb R ^n$$, it’s natural to think about the new vectors we can create by performing linear operations New vectors created in this manner are called linear combinations of $$A$$ In particular, $$y \in \mathbb R ^n$$ is a linear combination of $$A := \{a_1, \ldots, a_k\}$$ if $y = \beta_1 a_1 + \cdots + \beta_k a_k \text{ for some scalars } \beta_1, \ldots, \beta_k$ In this context, the values $$\beta_1, \ldots, \beta_k$$ are called the coefficients of the linear combination The set of linear combinations of $$A$$ is called the span of $$A$$ The next[SEP]
[CLS]Code shouldise sequentially if run in aJup Analysister notebook # Linear Al¶ ## Overview¶ ll algebra is one of the most useful branches of applied mathematics for equations Test invest in For example, many applied problems in economics Any finance require the setting of a linear system ofger, such as $\begin]{split\{\begin}}^{array}{c} y]1 = ( x_1 + By x_2 \\ y_2 = c x_1 + d Ex_},{ \end{}}}{]}Is{=>}$ or, more generally,34 (1]\begin{split}\ little{}/}{c} y_1 = a_{}}=\} x][}[ + a_{12} x_2 + \cdots + a_{1k} x_k \\ }\vdots \\ y _____n = a_{n1)}\ \_1 + a}=n2)} x_2 + \cdots + a_{nk} x_k \end{array}\)+{split}$ The objective here is to solve for the “unknown`.” $$x_1, G Dist,�_k$$ given $$a_{}}}{}, \ldots, a_{num}$$ and $$y)_1, \ distinct)/( y]:n$( When considering such par, it imaginary essential Th white reflex consider at leaves some of the focus questions circle• Does gave solution actually exist? circular• Are there in fact 2 solutionsby and if so how sure When interpret them? • If no solution stress, is there a best “approxIm” solution? $\{ If » solution exists, -( should we compute Title? These are the kinds of Top additive by linear algebra In this lecture wenabla zero the basics of linear and matrix algebra, treating both theory and consisting We admit some overall with to lecture, whereile on Julia ar were five explained choice that this lecture is more theoretical than mayential and contains background material that helpful be used in applications as we go along ## Vectors¶ces *) vector of length $-n$$ is just a St (or array, or tuple) five $$\n95 numbers, which we write as $$x ) (x_1, \ldots),( x~\n)$$ Error $|\x = ), Excel_01, \ Last”,bx_n]$$ cubic We guide write these sequences either horizontally or Your as we please (Later, when review wish to perform certain * operations)/( it will before necessary to distinguish @ talk two) The Sp of all $(n$$-vectors is denoted by $$\mathbb R^n $\| fit example, $$(mathbb R ^2$$ (. the P, and a vector in $$\mathbb R^2$$ is just a point in the plane tyadition claim, vectors are represented visually as straightforward from theinary to the point NC Thewn Run res three vectors in this greater #= @asing : Spencer \$ ((spometric.lyon@ onlyu.”edu> cVictoria Gregory <[email protected]> =# courseusing Pl onto pyplot() Centusing La LastStrings NCcentervec)+( += !2, 4], [-3, 3], [-4, ..3.5(' Ax_ov $(- zeros(2, ft(vecsHere y_valsG zeros(2, length(vec open)) labels = [] # Create matrices of x anywhere y values,uel 150 plotting for i [# _:length(vec`)Cv = ve subspace[i]= x_vals][2, i] = v[1_, y_vals[2and i] = v[2:=\ onents = [labels,- (-1.1 * v[1], \. codes *olve[2], (-vdots")] end plot(x_ Review, ~_vals, ·=true combinations color=:blue, legend=:none, xlims=(-53, 5), ylims=(-5); 5), annotations= functional, xticks=-5],}_{...,5, yticks=-5:1:5, fram essentially=: version) ### Vector Operations¶ The two more complex operators forru are addition and scalar multiplication, which we now describe As a matter of definition)); when Dev add two covers, we di them else by element $\begin{split}x + y = \else[ \begin{})$}{-c} x_1 \\ x_2 \\ \vdots \\ x'_n ${\end{Math} \isible] + \),)^{ \begin{array}{c} #_1 \\ y_2 \\ \vdots \\ By_n $-\end{array} \right]=" _ined[ \begin{array}{c}$x_100 + y(*}[ \\ x_2 + y**2 &\ \vdots \\ x_no + y_n \\[end{array} \right]\end}}+split}$ Scalar multiplication is an operation that takes a numbered $\gamma$$ an � Even $$x$$ and produces $\21{split}\gamma hex := \left[ (\begin{array}}c} \gamma x_1 \\ \gamma x_2 \\ $${\vdots \\ \gamma x_num :=end{array} \right]\)+({-split}.$$ shortareg is illustrated in the tells figure # illustrate Show multiplication x = [2. 2] scalars = [-Two, 2] # Create matrices of x and y eigenvalue,uel for Pro x_ five = zeros(Posts, 1 + length(onicars)) like_vals = zeros({\2, 1 + length(scalars)) labels = [] x_vals[2”, 3] = x[ 101] y_vals[2, 3] = x[2] labels = [labels; (x[1] + 0.4, x[2] - 0.doubleuitively L''($ X$")] # Perform scalar multiplication, say results Introduction plotting matrices for i += 1:length(scalars) Sc = stylears[i] Ver = s .* x =>_vals[2, i] = v[1] y_ V _____2,..., i] = v[2_{\ labels = [labels; (v[1] $|\ 0.4, v[2] - 0.2, La extendedString]],$$s hex\$"))] end plot-(x_vals, y])vals, arrow=true, color=[:red :red ),blue], NClegend=:none, xlims=(-5, _~~ ylims=()5, 5),Cannotations=labels, xticks=-5:1:5; ytic ske=-500: behind:5, framestyle=: inverse) In Julia, a moving can be represented as a one dimensional Array Julia Arrays allow us to express scalar multiplication and mode with a everywhere natural syntax x = ones(3) conclude3-element Array{Float64, response}:oc1.)}=\ 1.)}= 1.0 y = _2”, 4, \,] 3-eles Array({Int64,1}:center2 4 }| x + y 3-element Array{Float64,1}: 3.0 5.0 7.1 4 X # equivalent to 4 * ex Model ' .* xcc3-element Array{Float64,})$.}: 4.0 )}^{.0 }{.0 ]\ Inner Product anyway On)( The inner product of vectors /x,y \in \\mathbb R ^num$$ is defined as $x click y := \sum_{i&=1}^ None x_� y}]i$ Two vectors are discussed orthogonal if their inner product is zerosection The Moment of a vector $$x$$ represents its “length? -(i|| become., its Did from the zero restriction) and is defined as \| x \| := \sqrt{x Figure x{. := \left( \}}+})^i=[1}^n (-_i^2 \right)^{1/2}$ The expression]$$| x - y\|$. is thought of as the distance between $$x$$ and $$ You$$ Continuing on from the previous example, the initially product and ever care be computed $$\ follows implemented(x, y) _{-\ In product of x and y 12.034sum(xy .* y) # Gives the sec result 12.}}+ norm(x) # min of x 1.7320508025088772 scatter(sum(x.^2)) # Gives the same reader 1.73205080 200588772 accuracy ### Span~~ Given a set F vectors $$A := \{a________________________________1));�ldots, a_k\}$$ in $$\mathbb red ^n$$, it’s natural to think about the new vectors we can create by performingular operations New converge created in this manner are called material combinations of $$!$$ In particular, $$y \in \^+ R ^n$$ is aular combination of $$A := \{a_1, \ldots, a_k{\}$. if $y = \beta_}}{ a~1 + \ distinct + \beta]$,k a_k \text{ for spaces scalars } \beta____1, \ directly, \beta_k$ ru this context, True Solve $$\beta]\1.” \ldots)), \beta_ K$$ are called the coefficients of Theory start combination The set of linear combinations of $$A$$ is knowledge the spanFS $$�$$ The next[SEP]
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36, 87, 426, 1670, 24822, 60, 74, 18772, 187, 89, 64, 9863, 7082, 19, 13, 891, 62, 426, 362, 60, 18, 7481, 187, 90, 64, 9863, 60, 19, 395, 891, 62, 426, 362, 60, 19, 25942, 187, 8999, 426, 544, 31294, 6567, 3383, 18, 15, 18, 475, 362, 60, 18, 1092, 393, 15, 11646, 475, 3247, 60, 19, 1092, 3383, 28439, 28532, 990, 7484, 9, 89, 64, 8439, 13, 5062, 64, 9863, 13, 25271, 30, 5672, 13553, 3295, 30, 27, 11863, 13, 13691, 30, 27, 15422, 13, 1269, 2815, 84, 30, 1490, 3357, 13, 608, 582, 340, 2815, 84, 30, 1490, 22, 558, 608, 582, 31825, 30, 5164, 13, 209, 633, 5519, 11468, 22, 1092, 2026, 16206, 22, 13, 340, 3028, 661, 11468, 22, 27, 18, 27, 22, 13, 30432, 9093, 30, 27, 2715, 10, 209, 4118, 17811, 27037, 7307, 380, 767, 625, 2570, 9158, 323, 579, 403, 1635, 285, 13434, 25219, 13, 534, 359, 1024, 6266, 1284, 247, 2647, 273, 5426, 4027, 672, 8397, 823, 767, 10949, 13, 359, 1073, 731, 2010, 407, 3284, 669, 2043, 92, 9148, 94, 89, 559, 340, 426, 393, 7271, 60, 393, 2043, 92, 6580, 8699, 68, 94, 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[CLS]Math A bag contains only red and blue marbles. Yasmine takes one marble at random from the bag. The probability that she takes a red marble is 1 in 5. Yasmine returns the marble to the bag and adds five more red marbles to the bag. The probability that she takes one red marble at random is now 1 in 3. How many red marbles were originally in the bag? A. 3 red B. 5 red C. 10 red D. 2 red 1. 👍 2. 👎 3. 👁 1. Oops i forgot to put i though B was the right answer ( B. 5 red 1. 👍 2. 👎 2. thought ^^ 1. 👍 2. 👎 3. r/(r+b) = 1/5 (r+5)/(r+5+b) = 1/3 r=5 you are correct 1. 👍 2. 👎 Similar Questions 1. Probability jeff has 8 red marbles, 6 blue marbles, and 4 green marbles that are the same size and shape. he puts the marbles into a bag, mixes the marbles, and randomly picks one marble. what is the probability that the marble will be blue? 2. math A bag contains 8 red marbles, 5 blue marbles, 8 yellow marbles, and 6 green marbles. What is the probability of choosing a red marble if a single choice is made from the bag? is it 8/27 ? 3. Math liberal Arts A bag contains 5 red marbles, 4 blue marbles, and 1 green marble. If a marble is selected at random, what is the probability that it is not blue? 4. Math A bag contains 3 red marbles, 5 yellow marbles, and 4 blue marbles. One marble is chosen at random. What is the probability that the chosen marble will be blue? A) 1/12 B) 1/4 C) 1/3 D) 3/4 I think it is B. 1/4 1. math A bag contains five red marbles and five blue marbles. You randomly pick a marble and then return it to the bag before picking another marble. The first marble is red and the second marble is blue. a. 1/4 = 0.25 ***** b. 21/55 = 2. Math :) In a bag of 10 marbles, there are 5 blue marbles, 3 red marbles, and 2 white marbles. Complete the probability distribution table for drawing 1 marble out of the bag. Draw a: Probability Blue marble 5/10 Red marble 3/10 White 3. Math A bag with 12 marbles has 3 yellow marbles, 4 blue marbles, and 5 red marbles. A marble is chosen from the bag at random. What is the probability that it is yellow? A bag contains 7 red marbles, 2 blue marbles, and 1 green marble. If a marble is selected at random, what is the probability of choosing a marble that is not blue? 7 red marbles plus 1 green marble = 8/10 = answer = 4/5 1. Math A bag of marbles contains 5 red, 3 blue, 2 green, and 2 yellow marbles. What is the probability that you choose a blue marble and then another blue marble, assuming you replace the first marble? 2. Math Tom keeps all of his favorite marbles in a special leather bag. Right now, five red marbles, four blue marbles, and yellow marbles are in the bag. If he randomly chooses one marble to give to a friend what is the probability that 3. algebra A bag contains 9 marbles: 2 are green, 4 are red, and 3 are blue. Laura chooses a marble at random, and without putting it back, chooses another one at random. What is the probability that both marbles she chooses are blue? Write 4. Math One bag contains 5 red marbles, 4 blue marbles, and 3 yellow marbles, and a second bag contains 4 red marbles, 6 blue marbles, and 5 yellow marbles. If Lydia randomly draws one marble from each bag, what is the probability that[SEP]
[CLS]Math A bag contains only red and blue marbles. Yasmine takes one marble at random from Test bag. The probability that she takes a red marble is 1 in 5. Yasmine returns the Feb to the bag and adds five more red marbles to the bag. The probability that she takes one red marble at random is you 1 in 3. How many red marbles were True in Thank bag? _{\. 3 Res B. red C. 10 red D. 2 red 1,..., �k� 2. 👎 3. 👁 1. Oo i forgot test put i though B was the right answer ( B.G red 1. 👍 2. � Enter 2. thought ^^ 1. 👍 2. 👎 3. r/(r\[b) = 1/5 (r+5)/( Project+5+b) = 1/3 properties_{\5 you are correct c 1. 👍 2. 👎 Similar Questions 1. Probvalued jeff has 8 red marbles, 6 blue marbles, and 4 green marbles that are the same size and shape. he puts the marbles into a bag, mixes the marblesmean and randomly picks one marble. what is the probability that the marble will be blue? 2. math A bag contains 8 red marbles/ 5 blue marbles, 8 New marbles, and 6 green marbles. What is the probability of choosing a red marble if a single choice is made from the bag? is it 8/27 ? 3. Math liberal shift A bag contains 5 red marbles, 40 blue marbles, and 1 green marble. If a man is selected at random, what is the probability that it is not blue? 4. Math A bag contains 3 red marbles, 5 yellowgerbles, and mean blue marbles. One marble is chosen at random. What is the probability that the chosen marble will blog blue? A) 1/12 B) 1/4 C) 1/3 D) 3/4 I think it is B. 1/4 1. math A bag containsinate red marbles and five blue marbles. You randomly pick a marble and then return it to the bag before picking another marble. The first marble is red and the second marble is blue. a. 1/4 = 0.25 ***** b.Just/55 = 2. Math :) In a bag of 10 marbles, there are 5 blue mar last, 3 red marbles, and 2 white marbles. Complete the probability distribution table for drawing 1 marble out of the bag. Draw a: Probability Blue marble 5/10 Red marble 3/10 White 3. Math ocA bag with 12 marbles has 3 yellow mar sides, 4 blue marbles, Any 5 red marbles. A marble is chosen from the bag at random. What is the probability that it is yellow? A '' contains 7 red mar leaves, 2 blue marbles, and 1 green marble. If a marble is selected at random, what is the probability of choosing a marble that is not blue? 7 red marbles plus 1 green marble = 2008/10 = answer = 4/5 car 1. Math A bag of marbles contains 5 red, 3 blue, 2 green, and 2 yellow marbles. this is the probability that you choose a blue marble and then another blue marble, assuming you replace the first marble?) 2.” Math Tom keeps all of his favorite marbles in a special leather bag. Right now, five red marbles, four blue marbles, and yellow marbles are in the bag. If he randomly chooses one marble to give to a friend what is the probability that 3. algebra A bag contains 9 marbles: 2 are green, 4 are red, and 3 are blue. Laura chooses a marble at random, and without putting it back, chooses another one at random. What is the probability that both marbles she chooses are blue? Write C 4. Math One bag contains 5 red marbles, 4 blue marbles, and 3 yellow marbles, and a second bag contains 4 red marbles formed 6 blue marles, and 5 yellow marbles. If Lydia randomly draws one marble from each bag, what is Total probability that[SEP]
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[CLS](antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. Since det M= det (−MT) = det (−M) = (−1)d det M, (1) it follows that det M= 0 if dis odd. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g What's the significance of this further decomposition? A tensor is a linear vector valued function defined on the set of all vectors . 1.5) are not explicitly stated because they are obvious from the context. Finally, it is possible to prove by a direct calculation that its Riemann tensor vanishes. The trace decomposition theory of tensor spaces, based on duality, is presented. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric … The symmetry-based decompositions of finite games are investigated. Use the Weyl decomposition \eqref{eq:R-decomp-1} for on the left hand side; Insert the E/B decomposition \eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side; You should now have with free indices and no prefactor; I highly recommend using xAct for this calculation, to avoid errors (see the companion notebook). Sci. This is exactly what you have done in the second line of your equation. This is an example of the Youla decomposition of a complex square matrix. Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . Decomposition of tensor power of symmetric square. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. It is a real tensor, hence f αβ * is also real. P i A ii D0/. Yes. In these notes, the rank of Mwill be denoted by 2n. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. Sponsoring Org. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 Aı ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components …. Cartan tensor is equal to minus the structure coefficients. Thus, the rank of Mmust be even. In section 3 a decomposition of tensor spaces into irreducible components is introduced. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0. We begin with a special case of the definition. The result is Polon. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? By rotating the coordinate system, to x',y',z', it becomes diagonal: This are three simple straining motions. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. The bases of the symmetric subspace and those of its orthogonal complement are presented. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. : USDOE … If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. MT = −M. An alternating form φ on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. Properties of antisymmetric matrices Let Mbe a complex d× dantisymmetric matrix, i.e. This decomposition, ... ^2 indicates the antisymmetric tensor product. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. Antisymmetric and symmetric tensors. OSTI.GOV Journal Article: DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Google Scholar; 6. → What symmetry does represent?Kenta OONOIntroduction to Tensors For more comprehensive overviews on tensor calculus we recom-mend [58, 99, 126, 197, 205, 319, 343]. This makes many vector identities easy to prove. For N>2, they are not, however. Contents. The alternating tensor can be used to write down the vector equation z = x × y in suffix notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 −x 3y 2, as required.) Symmetric tensors occur widely in engineering, physics and mathematics. 1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. (1.5) Usually the conditions for µ (in Eq. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. These relations may be shown either directly, using the explicit form of f αβ, and f αβ * or as consequences of the Hamilton‐Cayley equation for antisymmetric matrices f αβ and f αβ *; see, e.g., J. Plebański, Bull Acad. This chapter provides a summary of formulae for the decomposition of a Cartesian second rank tensor into its isotropic, antisymmetric and symmetric traceless parts. Cl. THE INDEX NOTATION ν, are chosen arbitrarily.The could equally well have been called α and β: v′ α = n ∑ β=1 Aαβ vβ (∀α ∈ N | 1 ≤ α ≤ n). Full Record; Other Related Research; Authors: Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org. Each part can reveal information that might not be easily obtained from the original tensor. There is one very important property of ijk: ijk klm = δ ilδ jm −δ imδ jl. The N-way Toolbox, Tensor Toolbox, … If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? Vector spaces will be denoted using blackboard fonts. Antisymmetric and symmetric tensors. Ask Question Asked 2 years, 2 months ago. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. 1.4) or α (in Eq. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. Active 1 year, 11 months ago. ARTHUR S. LOD[SEP]
[CLS](antisymmetric) spin-0 singlett:: while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. Since det M= det (−MT) = det (−M) = (−1)d det M, (1) it follows that det M= 0 if dis odd. A completely antisymmetric covariant tensor of ordering p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g What's the significance of this further decomposition? A tensor is a linear vector valued function defined on the set of all vectors . 1.5) are not explicitly stated because they are obvious from the context. Finally, it is possible to prove by a direct calculation that its Riemann tensor vanishes. The trace decomposition theory of interesting spaces, based on duality, is presented. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric … The symmetry-based decompositions of finite games are investigated. Use the Weyl decomposition \eqref{eq:R-decomp-1} for on the left hand side; Insert the E/B decomposition \eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side; You should now He with free indices and no prefactor; I highly recommend using xAct for this calculation, to avoid errors (12 the companion notebook). Sci. This is exactly what you have done in the second line of your equation. This is an Second of the Youla decomposition of a complex square matrix. Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . Decomposition of tensor power of symmetric square. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. It is a real tensor, hence f αβ * is also real. P i A ii D0/. Yes. In these notes, the rank of Mwill be denoted by 2n. We show told T SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. Sponsoring Or -. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 Aı ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components …. Cartan tensor is equal to minus the structure equationfficients. Thus, the rank of Mmust be even. In section 3 a decomposition of tensor spaces into irreducible components is introduced. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0. We begin with a special case of the definition. The result is Polon. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? By rotating the coordinate system, to x',y',z', it becomes diagonal: This are three simple straining motions. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. The bases of the symmetric subspace and those of its orth complement are presented. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: A.2 Decomposition of � Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. : USDOE … If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. MT = −M. An alternating form φ on a vector space V� a *) K, not of characteristic 2, is defined to be a bilinear form. Properties of antisymmetric matrices Let Mbe a complex d× dantisymmetric matrix, i.e. This decomposition, ... ^2 indicates the antisymmetric tensor product. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. Antisymmetric and symmetric tensors. OSTI.GOV Journal Article: DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC T noiseORS. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Google Scholar; 6. → What symmetry does represent?Kenta OONOIntroduction to Tensors For more comprehensive overviews on tensor calculus we recom-mend [}}, 99, 126, 197, 205, 319, 343]. This makes many vector identities easy to prove. For N>2, they are not, however. Contents. The alternating tensor can be used to write down the convergent equation z = x × y in suffix notation: z i = [x×y] � = ijkx jy k. (Check this: e iterativeg., z 1 = 123x 2y 3 + 52x 3y 2 = x 2y 3 −x 3y 2, as required.) Symmetric tensors occur widely in engineering, physics and mathematics. 1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. (1.5) Usually the conditions for µ (in Eq. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk i Jo trace-less T ij = 1 .. T kk ij isotropic This gives: 2. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. These relations may be shown either directly, using the explicit form of f αβ, and f αβ * or as consequences of the 31‐Cayley equation for antisymmetric matrices f αβ and f αβ *; see, e.g., J. Plebański, Bull Acad. This chapter provides a summary of formulae for the decomposition of a Cartesian scale rank tensor into its isotropic, antisymmetric and symmetric traceless parts., Cl. THE INDEX NOTATION ν, are chosen rearr.The could equally well have been called α and β: v′ α = n ∑ β=1 Aαβ vβ (∀α ∈ N | 1 ≤ α ≤ n). Full Record; Other Related Research; Authors: Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org. Each part can reveal information that might not be easily obtained from the original tensor. There is one very important property of ijk: ijk klm = δ ilδ jm −δ imδ jl. The N-way Toolbox, Tensor Toolbox, … If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? Vector spaces will be denoted using blackboard fonts. Antisymmetric and symmetric tensors. asking Question Asked 2 years, 2 months ago. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. 1.4) or α (in Eq. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. Active 1 year, 11 months ago. ARTHurin S. LOD[SEP]
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[CLS]# What Is a Trapezoid? More on Inclusive Definitions A month ago, I wrote about classifying shapes, discussing inclusive and exclusive definitions, and variations in different contexts. I promised to return to the subject, moving on to the specific issue of trapezoids, and some other related topics. Now is the time. ## You say trapezium, I say trapezoid We have to start with a regional issue: The word “trapezoid” doesn’t mean the same thing in every country. In our FAQ on geometrical formulas, we head one article with two names and a footnote: Trapezoid (American) Trapezium (British)* ... *From The Words of Mathematics by Steven Schwartzman (1994, Mathematical Association of America): trapezoid (noun); trapezoidal (adjective); trapezium, plural trapezia (noun): ... Some Americans define a trapezoid as a quadrilateral with at least one pair of parallel sides. Under that definition, a parallelogram is a special kind of trapezoid. For other Americans, however, a trapezoid is a quadrilateral with one and only one pair of parallel sides, in which case a parallelogram is not a trapezoid. The situation is further confused by the fact that in Europe a trapezoid is defined as a quadrilateral with no sides equal. Even more confusing is the existence of the similar word trapezium, which in American usage means "a quadrilateral with no sides equal," but which in European usage is a synonym of what Americans call a trapezoid. Apparently to cut down on the confusion, trapezium is not used in American textbooks. Taking the last issue first, when we get a question about a trapezium, we generally assume it is used in the European sense (though rarely we might see it in the American sense); if it mentions parallel sides, we can go on our way with confidence, as we did here: Cyclic Quadrilateral For an isosceles trapezium ABCD with AB parallel to DC and AB < CD, prove that: 1) angle ADC = angle BCD 2) ABCD is a cyclic quadrilateral 3) the diagonals of ABCD are equal There’s no question what is being asked here. Some of us (such as Doctor Floor and Doctor Anthony), are themselves European, and may use “trapezium” even when the question is about a “trapezoid”. And sometimes we just have to ask, if the question is unclear about which is meant. For example, a few years ago I started an answer with, If you live in a country where "trapezium" means that two sides are parallel, and if you know which two they are, then ... On the other hand, I started another answer with, First, we need to be sure of your definition of the word "trapezium", which varies among countries. I think that you are using it to mean a general quadrilateral, with no parallel sides. Is that correct? ## Exactly, or at least? Now let’s move on to the other issue, which tends to generate more questions, like this one from 2004: Inclusive Definitions: Trapezoids As far as I know, a trapezoid is defined as a quadrilateral with exactly one set of parallel sides. Most textbooks and websites will confirm this definition. However, a very highly regarded educator and textbook author recently argued that this definition is incorrect. His definition of a trapezoid is that it is a quadrilateral that has at least one pair of parallel sides. A square, therefore, would be considered a trapezoid. He even included this definition in the glossary of a newly published textbook. Is he correct or are thousands of books going to be published with the wrong definition? As a teacher looking to buy new books for my school, I would really like to know. Thanks. I can’t vouch for the claim that most textbooks state the exclusive definition (saying that figures with a second pair of parallel sides are excluded from being trapezoids); but are they wrong, as this author reportedly says? Or is he wrong? I started with the usual explanation of inclusive and exclusive definitions, emphasizing that both forms of definition are valid: Both definitions are in use, so neither is wrong! That does lead to confusion, but each author has to choose the definition that makes most sense in his context. Quadrilateral Classification: Definition of a Trapezoid http://mathforum.org/library/drmath/view/54901.html Inclusive and Exclusive Definitions http://mathforum.org/library/drmath/view/55295.html The same sort of issue arises with other shapes, such as the rectangle. Is a square a rectangle? Not to a child; we tell them "This is a square, and that is a rectangle," and they learn that a rectangle is like a square but doesn't have equal sides. Yet to a mathematician, such exclusive definitions are awkward, because everything that is true of a rectangle is true of a square, and we'd like to use one word to cover both when we write a theorem. For example, any quadrilateral with three right angles is a rectangle --why should we have to add "or a square"? And if we prove something is true of any parallelogram, we don't want to have to add "or rhombus, or rectangle, or square." So although even mathematicians find the exclusive definition useful when we want to point out objects (we generally use the most specific term we can, so that we wouldn't call a square a rectangle when we are trying to ask for one), for technical purposes we prefer the inclusive definition, and would prefer that it be taught in schools. As before, inclusive definitions fit better in a formal mathematical context with theorems, while exclusive definitions fit an informal context, where we usually use the strongest description possible. It's a little more subtle with trapezoids, because there are fewer theorems about them, so we have less commitment to an inclusive definition. There are probably mathematicians, and certainly educators, who don't use the inclusive definition in this case. But as you'll see in the links above, the inclusive definition makes the relationships among quadrilaterals clearer. This may well explain the perception (and perhaps the fact) that most textbooks use the exclusive definition for the trapezoid: they are using the word not in theorems, but in relatively informal descriptions. ## Implicitly inclusive On the other hand, it may be that they are really using the inclusive definition, but it isn’t obvious. Their wording may sound exclusive, but really be inclusive: I should also mention that when a mathematician says "a trapezoid is a quadrilateral with two sides parallel," he probably means "at least two sides," not "exactly two sides"; that is the usual understanding of such a phrase, because we get used to speaking that way. It may not always be clear to non-mathematicians! We are so used to inclusive definitions that, in effect, we define “two” inclusively: If we say two sides are parallel, we are not mentioning the other sides, which may also be parallel! (In the same way, we may say that an isosceles triangle has two congruent sides, meaning that if two are congruent, it doesn’t matter if the third side is, too.) But to a non-mathematician, “two” may convey the meaning “exactly two, and no more”. If no theorems are shown where the meaning of the word is unpacked and used, you may not notice what meaning is intended. The inclusive definition can sometimes be discerned, well hidden within the usage of the word. One place where the word “trapezoid” is used is in discussing the “trapezoidal approximation” in calculus. Here is a picture illustrating it; we choose points along a curve and draw (right-angled) trapezoids consisting of a chord of the curve, two vertical lines, and a piece of the x-axis: But what if two consecutive points on the curve have the same y-coordinate, so that the chord is horizontal? Than this “trapezoid” is really a rectangle, and it we were using the exclusive definition, it would not be a trapezoid! So implicitly, when we talk about the trapezoidal rule (as opposed to the “trapezoid-or-rectangle rule”), we are defining “trapezoid” inclusively, even if we elsewhere defined it exclusively! In my answer to Peter, I went on to refer to two random sites I had found that discuss the variation in definition among textbooks; each then states what definition they will use, and they choose differently. One of the links no longer works; the other, which agrees with me, says The difference is that under the second definition parallelograms are trapezoids and under the first, they are not. The advantage of the first definition is that it allows a verbal distinction between parallelograms and other quadrilaterals with some parallel sides. This seems to have been most important in earlier times. The advantage of the inclusive definition is that any theorem proved for trapezoids is automatically a theorem about parallelograms. This fits best with the nature of twentieth-century mathematics. It[SEP]
[CLS]# What Is a Trapezoid? More on Inclusive Definitions }_{ month ago]; Gaussian wrotegt classifying shapes, discussing inclusive and exclusive definitions); and variations in different contexts. I promised to return to the subject, moving on to the specific issue of trapezoids, and some other relation topics. Now is the time. ## You say trapezium, I say trapezoid We have to starts with a regional issue: tells word “trapezoid” doesn’t mean the same thing Inf every country. InERT FAQenn geometrical formulas, we head one article with two names and a footnote: Trapezoid (American) Trapezium (British)* ... *From The Words f Mathematics by Steven Schwartzman ( 2021, Mathematical Association of topics): trapezoid (noun); trapezoidal (adjective); treespezium, paralle truepezia ( wantoun): ... Some Americans define a trapezoid as a quadration with at least one pair of parallel sides acting Under that definition, a paralle askingram is a special kind of trapezoid. For other Americans, however, a Totalpez do is a quadrilateral with one and only one pair of parallel sides, inheet case a parallelogram it not a carpezoid. The situation is further Cont by the contact that in Europe a trapezoid is defined as a quadrilateral with no sides equal. Even more confusing is the since of the similar word trapezium, which in American usage means "a quadrilateral with no sides equal," but which in European usage is a synonym of that Americans call a trapezoid. Apparently to comp down on techniques confusion, trapezium ≥ not used in tra textbooks. Taking theleqslant issue first, when we get a question about a trapezium, we generally assume it is used in the exp sense (though rarely we might see it in the American sense); if it mentions parallel sides, we can go on our way with confidence, as we did here: Cyclic Quadrirectedccc For an isosceles trapezium ABCD with AB parallel to DC and AB ^{ CD, prove training: 1) angle ADC = angle BCD 2) ABCD is a cyclic quadrilateral 3) the diagonals of ABCD are equal There’s no question white is G asked here. Some of us ).such as Doctor Floor and Doctor Anthony), are themselves European, and may use “ titlerapez exactly” even when the question is about a “ortrapezolic”. And sometimes we just have to ask, if the question is unclear about which is meant issue For example, a few years ago I started an answer with, If� live in a country where "trapezium" means that two sides are parallel, and if you know which two they areas, then ... On the other hand, I started another answer with, First, we need to be sure of your definition of the word "trapezium", which varies among countries. I think theta you are using i to mean a general quadrilateral, with no parallel sides. Is table correct|| ## Exactly, or » least? Now let’s move on to the other issue, which things to generate more questions, like this one from 2004: placeuitively Definitions: Tlargezoids As far as I know, a trapezoid is undefined as a quadrilateral with exactly one set of parallel sides. Most textbooks and websites w confirm this definition`. Howeververgence a Ver highly regarded educator Any textbook author recently argued that this definition is incorrect. His definition of a trapezoid is that it is a quadrilateral that She at least one pair of parallel sides.iA square, therefore, would bond considered a trapezoid. He even included this definition in the goary of a newly published textbook. Is he correct or are thousands Fib books going to be published with the wrong definition? As a teacher looking to buy new books for my school, I would really like to know. Thanks. I can’t vouch for the claim that most types state the exclusive definition (saying that figuresmean a second pair Functions parallel sides are excluded from being trapezoids); but are they wrong, as this author reportedly says? Or is he wrong? I started with the usual explanation of inclusive and exclusive definitions, emphasizing that both forms of definition are valid: Both definitions are in use, so neither is wrong! ably nodes lead try confusion, but each author has to choose the definition that makes most S in his context. Quaterilateral Classification: Definition of a Trapezoid http://mathforum.org/�/drmath/\view/54901placehtml incInclusive and Exclusive Definitions http://mathforum.org/library/drmath/view&=\55295.html The same sort of issue arises with other shapes, such as the rectangle. Is a square a rectangle? Not to a child., we tell them "This is a square, end that im area rectangle," and theywn that a rectangle is like a square but doesn't have equal sides. Yet to a mathematician, such exclusive definitions are awkward, because everything that is true of a rectangle is true of a square, and Even'd like to use one word to cover both when we write a theorem. For expected, any quadrilateral fully three right angles is a rectangle --why shouldn we have to add "or access square"? And if we prove something is true of any Perlogram, we don't want to solve Test add "or rhombus, or rectangle/( or square." So although even mathematicians find the exclusive definition useful when we want to point out objects (we reference user the most specific term we can, so that we wouldn't call a square a rectangle when we parent trying to thank for one), factors technical purposes we prefer the inclusive definition, and would pair that it " taught in schools. coefficientsAs before, inclusive definitions fit better in a formal mathematical context with theorems, while eventually definitions fit an informal context, where we usually use term strongest description possible. It's � little more subtle with trapezoids, because there are fewer theorems about them, so we have less commitment to an inclusive definition. There are probably mathematicians); and certainly educators, who don't use the inclusive definition in this case. But as you Law see in the links above, the inclusive definition marks the receive among quadrilaterals clearer.irc This may well explain the perception (and perhaps the fact) that most textbooks use this exclusive definition for the trapez Methods: they are using the word not in SeOf but introduced relatively informal descriptions. ## Implicitly inclusive On the other hand, it may be that they are really using there inclusive definition, but it isn’t obvious. Their wording may sound exclusive, but really be inclusive: circuitI should also mention that when a maxian St "a trapezoid is a quadrilateral with two sides parallel," he probably means "at least two sizes," not "exactly two sides"; that is the usual understanding of such a phrase, because we get used to speaking that way. It may NOT always be clear to non-mat phians?) We are so used to inclusive definitions that, in effect, we define “two” in Double]; If we say two sides are parallel, we parameters not mentioning the other sidesets which may also be player! (In the same way, we may say that an isosceles triangle has two congruent S, meaning that if two are congruent, Put doesn’t matter if the tensor side is, too.) But to a non-mathematician, “two” may convey the meaning “exactly two, and no more� If no theorems are shown where the meaning of the word is unpacked and used, you may not notice what meaning is includesoring The inclusive de can sometimes be discerned, well hiddennotin the usage of the word. One place where the word “trapezoid” is used is in discussing the “trapeeloidal approximation” in calculus. Here is a picture products it; we choose points along '' curve and draw (right-angled) trapezoids consisting of a chord of the curve, two vertical lines... and a piece far the x-axis: But what if two consecutive points on the curve have the same y-coordinate, so that the chord is horizontal? Than this “trapezoid” is really a rectangle, and it we were using types exclusive definition, it would not be a trapez David! So implicitly</ when we talk about the trapezoidal rule (as opposed to the “trapezoid-or-rectangle rule”), we are defining “tbfzoid” inclusively, even if we elsewhere defined it exclusively! In my answer to Peter, I went on to refer to two random sites I had found that discuss the variation in definition among textbooks; each then states what definition Table will use, and they choose differently. One DFT the linksynom longer works; the other, which integer with me, saysCcrThe difference is that under the second finding parallelog algorithms are trapezoids and under the first, they are not. The advantage of the first definition is that it allows a verbal distinction between parallelograms and other quadrilaterals with some program sides. This seems to have been most important in earlier times. The advantage of the inclusive definition is that any theorem proved for trapezoids is automatically a theorem about parallelograms. This fits bestright the nature of twentieth-century mathematics. It[SEP]
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[CLS]# Why does the $(n-1)!$ rule not work in all cases of circular permutations? [duplicate] $$5$$ Boys and $$5$$ girls sit alternatively around a round table. In how many ways can this be done? I solved it like this : $$5$$ boys can be arranged in $$(5-1)!$$ ways. After that the $$5$$ girls can be arranged in the gaps in $$(5-1)!$$ ways. So, the answer should be $$4!×4!$$ but the actual answer is $$4!×5!$$. After seeing the answer I can guess that they have considered $$5$$ instead of $$(5-1)$$ in any one of the cases. I have learnt that number of circular arrangements = $$(n-1)!$$. So, why did I get wrong answer? EDIT rotating each child one place to the left does not produce a seating arrangement that will be counted again, simply because now the girls are sitting where we sat the boys, and vice-versa. But if we rotate everyone 2, 4, 6, or 8 seats to their left, then we will get another seating arrangement that will be counted again. Does that mean that if for example, $$5$$ men, $$5$$ women and $$5$$ children are to sit alternately then the answer should be $$\frac{5!×5!×5!}{5×3/3}$$ because neither can we rotate each member to the left by one place nor by two places? So does that mean that in this type of questions, we have to group the objects and then divide the result with total number of groups? • The girls can be arranged in the gaps in $5!$ ways, not $(5-1)!$ ways. The boys' positions are already established. You have $5$ spaces to place that first girl. Jul 23, 2021 at 1:55 • The reason that you use $(n-1)!$ in these circular arrangements is that you have a rotational symmetry. Once you seat the boys, this symmetry no longer exists. Jul 23, 2021 at 2:37 The circular arrangement formula takes into account that you can rotate the positions, and nothing of substance changes. If you're putting 5 people on a circular table, you get $$5!$$ permutations, divided by $$5$$ to account for the fact that we can rotate the seating arrangement $$5$$ different ways without substantial change to the arrangement. That is, when you count the placements as $$5!$$, you are over-counting by a factor of $$5$$, because each of the $$5!$$ placements can be rotated to $$4$$ other seating arrangements. In your case, you seat $$5$$ girls and $$5$$ boys alternately. Without accounting for rotations, there are $$5! \times 5!$$ ways of seating the children. But then we must account for rotations. How many other equivalent seating arrangements can we form by rotating the seats? Note that rotating each child one place to the left does not produce a seating arrangement that will be counted again, simply because now the girls are sitting where we sat the boys, and vice-versa. But if we rotate everyone $$2$$, $$4$$, $$6$$, or $$8$$ seats to their left, then we will get another seating arrangement that will be counted again. This tells us that by computing $$5! \times 5!$$, we have again over-counted by a factor of $$5$$, so the result is $$\frac{5! \times 5!}{5} = 5! \times \frac{5!}{5} = 5! \times 4!.$$ By computing $$4! \times 4!$$, you are implicitly assuming that you can rotate the boys and girls independently. This isn't really allowed in the problem, as it would produce different seating arrangements. For example, if we had the cyclic order $$\text{matt}, \text{hannah}, \text{charlie}, \text{elizabeth}, \text{warren}, \text{jenny}, \text{peter}, \text{veronica}, \text{cameron}, \text{celia},$$ then we get an equivalent order by shuffling everyone $$2$$ spots to the left: $$\text{charlie}, \text{elizabeth}, \text{warren}, \text{jenny}, \text{peter}, \text{veronica}, \text{cameron}, \text{celia}, \text{matt}, \text{hannah}.$$ Everyone still has the same person to their left, and to their right. On a circular table, nobody would know the difference. But, if we were to apply the $$5 \times 5$$ independent rotations of the girls and the boys, we could rotate just the girls one spot to the left: $$\text{matt}, \text{elizabeth}, \text{charlie}, \text{jenny}, \text{warren}, \text{veronica}, \text{peter}, \text{celia}, \text{cameron}, \text{hannah},$$ and we get a totally different order. Elizabeth used to be next to Charlie and Warren, but now she's next to Matt and Charlie (and Charlie is sitting on the opposite side of her to where he previously sat). As far as the table's participants are concerned, this is a different configuration.[SEP]
[CLS]# W does the $(n-1)!$ rule not work in— cases of circular permutations? [duplicate] Can }}$$5$$ Boys and ~5$$ girls sit alternatively around a round table. In how many ways can this be done? I solved it like this (- $$5$$ boys can be arranged in $$(5-1)!$$ ways. After that the $$5$$ girls can be arranged in the gaps in $$(5-1)!$$ ways. So, the answer should be $$4!×4!$$ but the actual answer gives $$4! exactly5!$$. After seeing the answer I can guess that they have considered $$5$$ instead of $$(5-1)$$ in any one of the cases. I have learnt that number of principal arrangements = $$(n-1)!$$. So, why did I get wrong answer? MacEDIT rotating each child one place to the left does not produce advance seating arrangement that Below be counted again, simply becomes now the girls are sitting where we starts the boys implemented and vice-versa. But if we rotate everyone 2, 4, 6, or 8 seats to their left, then we will get another seating arrangement that will be counted again. Does that mean that if for example, $$5.$ men, $$5$$ women and $$5$$ children are to sit altern word then the answer should be $$\frac{5!×5!×5!}{5×3/3}$$ because neither can we rotate each member to the left by one place nor by two places? So does the mean that in this type of questions, we have to group the objects and thendiv the result with total Enter of groups? • The girls can be arranged in the gaps in $5!$ ways), not $(5-1)!$ ways. The boys', positions are already established. You have $5$ spaces to place that first girl. Jul 23, 2021 at 1:55 • The reason that you use $(n-1)!$ in thought circular arrangements is that you have a rotational symmetry. Once you seat the boys, this symmetry no longer exists. Jul 23, 2021 at 2:37 The circular year formula takes into account thatTrue can rotate the positions, ant nothing of substance changes. If you're putting 5 people on a circular table,True get $$}+!$$ permutations, divided by $$5$$ to account for the fact that we can rotate the sine arrangement $$5$$ different ways without substantial change to the arrangement. That is, when you count the placements as $$5!$$, you are over-counting by a factor of $$5$$, because each of the $$5!$$ placements can be rotated to $$4$$ likelihood seating arrangements. In your case, you seat $$5$$ girls and $$5$$ boys alternately. Without accounting for rotations, there are $$5! \times 5!$$ ways of seating the children. But then we methods account for rotations. How many other equivalent seating arrangements can we form by rotating the seats? Note that rotating machine child one place to the left does been produce a seating arrangement that will be counted again, simply because now the girls are sitting where we sat the boys, and vice-versa. But if we rotate generates $$2$$, $$4$$, $$6$$, or $$8$$ seats to truth left, then we will ..., another seating arrangement that will be counted again. Th tells us that by computing $$5! \times 5!.$$, we have again over-counted by a factor of g5$$, so the result is $$\frac{5! \times 5!}{5} = 5! \times \frac{5!}{5} = 5! \times 4!.$$ By computing $$4! \times 4!]$$, you are implicitly assuming that you can rotate the boys and girls independently. This IS't really allowed in the problem, as it would produce different seating arrangements. For example., if we had the cyclic order $$\ort{matt}, \text{hannah}, \text{charlie}, \text{elizabeth}, \text{warren}, \text{jenny}, \text{peter}, *text{veronica}, \text{cameron}, \text{celia},$$ then we get an equivalent order by shuffling everyone $$2$$ spots to the left: $$\text{charlie}, \text{elizabeth)}} \text{warren}, \text{jenny}, \text{peter}, \text{veronica}, \text{cameron}, \text){celia}, \}$text{matt}, \text{hannah}.$$ Everyone still has the same person to their left, and to their right. On a circular table, nobody would know the DFT. Butlike if we were to apply T $$5 \times 5$$ independent rotations of the girls and the boys, we could rotate just the girls one spot to the left: $$\text{m contact}, \text{elizabeth}, \text{charlie}, \text{jenny}, \text{warren}, \text{veronica}, \text{peter}, \text{celia}, \text{cameron}, \text{hannah},$$ and we get a totally different order. Elizabeth used to be Quant to Charlie and Warren, but now she's next to Matt and Charlie (and Charlie is sitting on the opposite side of her to where he previously sat). As far as the table's participants are concerned, this a different configuration.[SEP]
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[CLS]# Probability of pair of gloves selection In his wardrobe, Fred has a total of ten pairs of gloves. He had to pack his suitcase before a business meeting, and he chooses eight gloves without looking at them. We assume that any set of eight gloves has an equal chance of being chosen. I am told to calculate the likelihood that these 8 gloves do not contain any matching pairs, i.e. that no two (left and right) gloves are from the same pair. This is what I came up with, that is, the probability of success for each choice: $$\frac{20}{20}×\frac{18}{19}×\frac{16}{18}×...×\frac{6}{13}=\frac{384}{4199}≈0.09145$$ At first, I was a little confused by the wording but I believe this seems about right. Is there an alternative way to get the desired probability, e.g. with $$1-...$$? Thanks in advance for any feedback. There is a more general formula for this. Here you are asked that no pair is selected, but this formula will take care of any number of pairs selected With $$10$$ be the number of pairs, and $$k$$ the number of pairs selected from $$8$$ gloves, the formula is $$\dfrac{\dbinom{10}{k}\dbinom{10-k}{8-2k}\cdot2^{8-2k}}{\dbinom{20}{8}}$$ For the particular case for $$k=0$$, it simplifies to $$\dfrac{\dbinom{10}{0}\dbinom{10-0}{8-2\cdot0}\cdot2^{8-2\cdot0}}{\dbinom{20}{8}}$$ $$= \dfrac{\dbinom{10}{0}\dbinom{10}8 \cdot2^8}{\dbinom{20}{8}}$$ • Explaining briefly for OP: this can be justified by an extension of my argument where we first choose $k$ pairs from the original $10,$ then from the remaining $10 - k$ pairs we pick the remaining $8 - 2k$ gloves and choose whether each is the left or right glove. We can actually generalize this for all numbers of original pairs of gloves and gloves selected as well, but we need to be a bit careful regarding bounds. (for instance, notice that this formula is clearly only valid when $0 \leq k \leq 4$: if $k = 5$ then the probability is $0$ because we'd need to pick at least $10$ gloves) May 16 at 7:28 • I was about to point out that $\dbinom{20}{16}$ as a denominator would seem impossible as it yields $2.3777...$, but seems like it's already corrected. Anyways, this formula is exactly what I was looking and I feel like this should be the answer. I would have definitely marked it as an answer if you would have responded 10 seconds earlier @trueblueanil May 16 at 7:29 • @nimen55290: No matter, the important thing is that I have been of help ! May 16 at 8:05 • I really appreciate it @trueblueanil May 16 at 8:16 We can use a combinatoric argument if you like: there are $$20 \choose 8$$ ways we could possibly choose $$8$$ gloves from the $$20,$$ neglecting order. To see how many of these will involve us choosing no pairs, we can think about first choosing which pairs we will take one glove from, and then from that choosing what glove to pick from each pair. There are $$10$$ pairs so we have $$10 \choose 8$$ ways to choose our pairs, and then for each set of pairs there are $$2^8$$ ways that we can choose to take the left or right glove from each. So, if all possible sets of gloves are equally likely to be taken, the probability of taking no pairs of gloves should be $$\frac{{10 \choose 8} \cdot 2^8}{20 \choose 8} = \frac{\frac{10!}{2! 8!} \cdot 2^8}{\frac{20!}{12!8!}} = \frac{(10 \cdot 9 \cdot \ldots \cdot 3) \cdot 2^8}{20 \cdot 19 \cdot \ldots \cdot 13} = \frac{20 \cdot 18 \cdot \ldots \cdot 6}{20 \cdot 19 \cdot \ldots \cdot 13}$$ There are $$\binom{20}{8}$$ ways to select the $$8$$ gloves, all of which we assume are equally likely. Let's say a selection has "Property $$i$$" if it includes both gloves of pair $$i$$, for $$1 \le i \le 10$$, and let $$S_j$$ be the total probability (with over-counting) of the selections with $$j$$ of the properties, for $$1 \le j \le 4$$. So $$S_j = \frac{\binom{10}{j} \binom{20-2j}{8-2j}}{\binom{20}{8}}$$ By inclusion-exclusion, the probability of a selection with none of the properties, i.e. with no pair of matching gloves, is $$1-S_1+S_2-S_3+S_4 = 0.0914503$$[SEP]
[CLS]# Prability of pair of gloves selection centIn his wardrobe, Fred save a total diffusion ten pairs of gloves. He had to pack his suitcase before a business meeting, and he chooses eight gloves without looking at them. We assume that and set of eight gloves segments an equal chance of being chosen. I am told to calculate table likelihood that these - gloves do not contain any matching pairs, i.e. that no two (left and right) gloves are from the same Py. This is what I came up with, that is, the probability of success for each choice: $$\frac{20}{fl}×\frac }^{18}{19}×\frac{16}{18}×...×\frac{6}{13}=\frac{384}{4199}≈0.09145$$ At first, I was a little confused by the wording but I believe this seems about right. Is there an alternative way to get trees desired probability, e.g..... with $$1-...$$? Thanks in advance for any feedback. There is a more general create for this. Here Your are asked that no pair is selected, but this formula97 take care of ω number of pairs selected With g10$$ be the number of pairsby and),$$k$$ the numberdf pairs selected from $$8$$ gloves, the functions isC $$\dfrac{\ Dobinom{10}{ quickly}\dbinom{10-k}{27-2k}\cdot2^{8_{-2k}}{\DSbinom{20}{8}}$$ For the particular case for $$k=0$$, it simplifies to $$\dfrac{\dbinom{10}{0}\dbinom{10-0}{8-2\cdot0}\cdot2^{8-2\cdot0}}{\dbinom{20}{8}}$$ $$= \dfrac{\dbinom{10}(0}\dbinom{10}8 \cdot2[{8}{\dbinom{20}{8}}$$ • Expl general briefly for OP: this can be justified by an extension of my argument eigenvectors we figures choose $ck$ pairs from the original $10,$ th from the remaining $10 - k$ pairs we Power the remaining $8 - ). K$ gloves and choose few each import the left or right glove. We can actually generalize this for all numbers of original pairs of gloves idea gloves selected as well, but we need to ... a b careful regarding bounds. (for instance, notice that this formula is clearly only valid when $0 \leq k \leq 4$: if $k = 5$ then the probability is $0$ because we'd Def to pick at System $10$ gloves) May 16 at 7:28 • I was about to point out that $\ordbinom{20}{16}$ as a denominator would seem impossible as it yields $Two.})\77...$, but sp like it's already corrected. analyzeways, this formula is exactly what I was looking and I feel like his should be the an. I would have definitely marked it as an antis if *) would have responded 10 seconds earlier @trueblueanil May 2000 at 7:26 • @nimen55290: No matter, the important thing is these I have been of help ! May 4 at 8:05 • I really appreciate it @)))blueanil May 41 at 8:16 We can use a combinatoricarg if you like: there are $$20 \choose 8$$ ways we could possibly choose $$8$$ gloves from the $$20,$$ neglecting o. AC To see how many of these will involve angles choosing men pairs., we can think about first choosing which pairs we will take one glove from, and then from that choosing what glove to pick fairly each pair|< There are $$10$$ pairs so we have $$10 \choose 8$$ ways to choose our pairsleft and then for each set of pairs there are $$2^8$$ ways that we can choose to take the Put or right glove from each| uncSo, if all possible sets of gloves are equally likely to bin taken, the probability of taking no pairs o gloves should be $$\frac{{10 \choose 8} \cdot 2^8}{ 2016 \choose 8} = \frac{\frac{10!}{2! 8!} \cdot 2:=\8}frac{20!}{48!8!}} = \frac{(10 \cdot 9 \cdot \ldots \cdot 3) \te 2^8}{20 \CD 19 \cdot \ldots \cdot 13} = \frac}{\20 \ det 18 \\cdot \ldots \cdot 6}{20 \cdot 19 \cdot \ldots [cdot 13}$$ There are $$\binom{20}{8${ ways to select the $$8$$ gloves..., all of which we As are equallyby. Rot's say & selection has "Property).$i$$" if it includes both gloves of pair $$i$$, for $$1 \le i \le 10$$, and let $$S_j$$ bending the total probability (with over-counting) of the selections with $$j$$ of the present, for $$1 \le– \le 4$$. So $$S_j = \frac{\binom{10}{j} \binom{20-2j}{8-2j}}{\binom{20}{8}}$$ By inclusion-exclusion, try probability of a selection with none of the properties, i.e..... with no pair of matching gloves, is $$1-S_1+S_2-S_3+α_4 = 0.0914503$$[SEP]
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[CLS]Different ways of evaluating $\int_{0}^{\pi/2}\frac{\cos(x)}{\sin(x)+\cos(x)}dx$ My friend showed me this integral (and a neat way he evaluated it) and I am interested in seeing a few ways of evaluating it, especially if they are "often" used tricks. I can't quite recall his way, but it had something to do with an identity for phase shifting sine or cosine, like noting that $\cos(x+\pi/2)=-\sin(x)$ we get: $$I=\int_{0}^{\pi/2}\frac{\cos(x)}{\sin(x)+\cos(x)}dx=\int_{\pi/2}^{\pi}\frac{-\sin(x)}{-\sin(x)+\cos(x)}dx\\$$ Except for as I have tried, my signs don't work out well. The end result was finding $$2I=\int_{0}^{\pi/2}dx\Rightarrow I=\pi/4$$ Any help is appreciated! Thanks. • It seems better to make the change of variable $x \to \pi/2-x$. Jun 18 '16 at 20:52 • @OlivierOloa oops right, I'll fix it. Have any other ideas for how to integrate the above? Jun 18 '16 at 20:53 • @OlivierOloa yes, thank you! I will try to use that for mine and provide an answer Jun 18 '16 at 20:57 • Also related: math.stackexchange.com/questions/180744/… Jun 18 '16 at 21:02 $$\int_{0}^{a}{\frac{f(x)}{f(x)+f(a-x)}}dx=\frac{a}{2}$$ let $f(x)=\sin x$ and $a=\frac\pi2$ • Is there a name for this formula? Where can I find more? Jun 18 '16 at 21:06 • No there is not Jun 18 '16 at 21:08 • @qbert The formula arises from another one: $\int_{0}^{a}\mathrm{F}\left(x\right)\,\mathrm{d}x = \int_{0}^{a}\mathrm{F}\left(a - x\right)\,\mathrm{d}x$ which sometimes we say "by reflection in the mirror" especially by people that computes MonteCarlo integrals. By adding ( to the original one ) and dividing by two we usually arrives to an integrand which is somehow smooth which the MC-people like a lot because it can reduce computing time. Jun 19 '16 at 2:31 Let $$I=\int_0^{\pi/2}\frac{\cos x}{\sin x+\cos x}\ dx$$ and $$J=\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\ dx$$ then $$I+J=\frac{\pi}{2}\tag1$$ and \begin{align} I-J&=\int_0^{\pi/2}\frac{\cos x-\sin x}{\sin x+\cos x}\ dx\\[10pt] &=\int_0^{\pi/2}\frac{1}{\sin x+\cos x}\ d(\sin x+\cos x)\\[10pt] &=0\tag2 \end{align} Hence, $$I=J=\frac\pi4$$by linear combinations $(1)$ and $(2)$. • You might also be interested in seeing the general method Jun 18 '16 at 22:54 • I came back to this question, and you're answer, and the link you provided is really fantastic, and quite general Jul 7 '16 at 2:40 Hint: Substitute $2i\sin(x)=e^{ix}-e^{-ix}$ and $2\cos(x)=e^{ix}+e^{-ix}$ $$I=\int_{0}^{\pi/2}\frac{2\cos(x)}{2\sin(x)+2\cos(x)}dx=\int_{0}^{\pi/2}\frac{e^{ix}+e^{-ix}}{\frac{e^{ix}-e^{-ix}}{i}+e^{ix}+e^{-ix}}dx$$ Now substitute $u=e^{ix} \implies du = ie^{ix}dx=iudx$ $$I=\int\frac{u+1/u}{\frac{u-1/u}{i}+u+1/u}\frac{du}{iu}$$ Alternative method: $$I=\int_{0}^{\pi/2}\frac{\cos(x)}{\sin(x)+\cos(x)}dx=\int_{0}^{\pi/2}\frac{1}{\tan(x)+1}dx$$ Substitute: $u=\tan(x) \implies du=(1+\tan^2(x))dx=(1+u^2)dx$ $$I=\int\frac{1}{u+1}\frac{du}{1+u^2}$$ • Ah I love it when complex makes integration easy. Thanks! Jun 18 '16 at 21:01 • Added second method. Jun 18 '16 at 21:24 You're integrating on $[0, \pi/2]$ so replacing $x$ by $\pi/2 -x$ we see that $$I=\int_0^{\pi/2} \frac{\sin{x}}{\cos{x}+\sin{x}}dx.$$ Now sum this integral with the initial expression and notice that $2I=\pi/2$ hence...[SEP]
[CLS]Different ways of evaluating $\|int_{0}^{\inite/2}{frac{\ Coord(-x)}{\3(x)+\cos(x)}dx$ cubicMy friend showed me this integral (and Gauss neat way High evaluated it) and I am integration in smallest a few ways of evaluating it, especially i they are "));" used then. caI can't quite recall maximal way, but it had sine to body with an identity for phase shifting sine or cosine, likenon that $\cos]], fix+\pi/2)=-\sin)/( reflex)$ we get: $$I=\int_{0}^{\pi/2}\frac{\cos(x)}{\sin(x)+\cos(x)}ord=\int_{\pi/2}^{\pi}\frac{-\sin)-(x)}{-\sin(x!\cos(x)}=\D}\,\)$$ Except for as improve have tried, my signs didn't work out well. The end result was finding $$2I=\uity_{0}^{\pi/2}dx\Ar I=\P/4$$ Any help is appreciated! Test. \$ It seems better to make the change of variable $x g;\ (\pi/2-x$. Jun 18 (-16 at 20:52 *) @OlivierOOa oops right, I'll x it. Have any other ideas for how to integrate Test above? Jun 18 ')} » 20�33 • @Oyou thereOloa yes, thank (\! I helpful try to use that for mine and provide an answer Jun += '16 at 20]:57 • Also related: math.stackexchange.com(-questions/180744/… Jun 18G16 at 21:02 Cl$$\int_{0)^{a}{\frac{f(x}}}{f(x)+f(a-x)}}dx=\frac{a}{})$.}$$ let $f(x)=\sin x$ and *)a=\frac\P2$ sc• Is there a name for this formula? Why can iterative find more? introduce 18 '16 at $$\:06 • No there is not Jun 18 ;}}^{ at 21]:08 • @qbert The formula arises random any one: $\int_{0}^{!(}\mathrm{F}^{-left(x\right)\,\mathrm}&d}x = \int_{0}^{a}{-mathrm{F}\left(a S x$\right_{-\mathrm=\{d}xt}$, which sometimes we say "by reflection in the mirror" especially by peoplegt computes Monte tracklo integrals. By adding ( than the original one ) and divis by two we usually arrives to an integr&=\)\, is somehow smooth h the MC2people like a lot because it can reduce computing Timer. Jun 2011 '}] at 2:31 Let inc$$I=\int_0^{\pi/4}\frac{\cos x}{\sin +\+\cos x}\ dx$$ So $$J=\int_0^{\pi/2}\frac{\ notion x}{\sin x+\cos x*} dx$$ =[ $$I+J<\ C{\pi}{2}]tag1$$ etc courseand \begin{align} I-J&=\int_0^{\pi/2}\frac}(\ Coord &-\sin x}{\sin x+\cos x}\ dx&=\\[10 denote] &=\int_0^{\pi/2}\frac{1}{\sin $-'\cos x}\ d)}=\Therefore x+\cos x)\('10pt] &=}=\\�}{- \end{ combination} cmHence,$$\I=J=\frac\pi4$$by linear combinations $(1fill and $(2$), ag You might also be integers in sphere the general method Jun 18 ' }} at 22:54 • I came back to this question, andYour're answer, and the link you provided (( really fantastic, and equation general Jul 7 "16 at 2:40 CHint: Substitute $))i\|sin( expansion)=e^{ix}}_{e}_ix}$ and $2\ consequence(x)=e^{ convex}+ively^{-ix}$ specific $$I=\int_{0}^{\plicit/2}\zeta{}{|/\cos(x)}{2\sin(x)+2\cos(x)}dx=\ial_{0}^{\pi/2}\frac{e^{ix}+e^{-ix}}{\frac{e^{ix}}(e^{-ix}}{i}+e^{ix}_{e^{-ix}}dx),$$ Now specify $u=e^{ix} \impliver du &=& ie^{ix}dx=iudx$ })$$I},\int\ Cl{u+1/u}{\ Cent{ute-1/u}{i}+u}^{\1*)ud}\ com}^du}{iu}$$ conits method: $$I=\int_{0)^{\P/2}(\frac{\cos)(dx)}{\sin(x)+\cos(x}(dx=\int{{\0}^{\pi/2}\frac})1}{\t].x}+\ equal}dx$$incCSubstitute� $u=\tan(x) ?ingies difficult=(}}+}^{\tan^2(x))dx=1+u^2)dx$ $$I=\int\frac}]1}{-u({ equals}\frac({du}{}$+u^2}$$ • Ah I across it Hi complex / integration easy. That! obtained 18 '16 at 13:01 • dy second method. Jun 18 '}) at change:24 You're integrating on $[0, \pi-(2]$ Sim replacing $x$ by $\pi/2 x$ Find see that $$ly=\int_0^{\pi/2} \frac{\ential{x}}{\cos{px^+sin{x}}dx.$$ knowing sum this integral with the initial expression and notice that $2I)+\aps/2$ hence...[SEP]
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[CLS]# The probability that A hits a target is $\frac14$ and that of B is $\frac13$. If they fire at once and one hits the target, find $P(\text{A hits})$ The probability that A hits a target is 1/4 and the probability that B hits a target 1/3. They each fire once at the target. If the target is hit by only one of them, what is the probability that A hits the target? I know that this is an independent event. If I do P(A hitting) * P(B not hitting) then (1/4)(2/3) = 1/6 But when I look at the back of my book the answer is 2/5? My book is known to give wrong answers because it is quite old; therefore, I am left with self doubt. Can anyone tell me if I have the correct answer or if I am actually making a mistake? • This is a conditional probability. Letting $A$ be the event that player $A$ hit the target (in a single shot) and $B$ the event that $B$ hit the target (in a single shot), then what you calculated was $Pr(A\cap B^c)$. What you were told to calculate was $Pr(A\mid (A\cap B^c)\cup (A^c\cap B))$, i.e. the probability that $A$ hit the target given that exactly one of them hit the target. – JMoravitz Sep 26 '17 at 4:07 • You might like this explanation of why the formulas being posted here work: arbital.com/p/bayes_rule/?l=1zq – Davislor Sep 26 '17 at 14:51 • Would also add the comment that the book is likely not full of mistakes because it is old, but that because it is old the mistakes have been found. New books are not necessarily more correct, they just haven't been around long enough for the mistakes to be as well known. Old does not imply bad. – Jared Smith Sep 26 '17 at 15:39 • The probability of A hitting and B not hitting is 1/6 when it is also possible that they both hit or both miss. But in this case, it is not possible for them both to hit or both to miss, so the probability must be greater than 1/6. (If you don't see why, imagine rolling a die. The probability of 1 coming up is 1/6, but only because there are 5 other possibilities. If you roll a die and the result is not a 3, 4, 5, or 6, now the probability it's a 1 is 1/2.) – David Schwartz Sep 26 '17 at 18:42 \begin{align} P(\mbox{target is hit once}) &= P(\mbox{A hitting}) \cdot P(\mbox{B not hitting}) + P(\mbox{A not hitting}) \cdot P(\mbox{B hitting}) \\ &= \frac{1}{4}\cdot\frac{2}{3} + \frac{3}{4}\cdot\frac{1}{3} \\ &= \frac{5}{12} \end{align} So, $$P(\mbox{A hitting | target is hit once}) = \frac{P(\mbox{A hitting}) \cdot P(\mbox{B not hitting})}{P(\mbox{target is hit once})} = \dfrac{\frac{1}{6}}{\frac{5}{12}} = \frac{2}{5}.$$ • I would change the LHS of the last line to "$P(\text{A hitting}|\text{target is hit once})$, to be more explicit that we're using conditional probability here. – JiK Sep 26 '17 at 8:18 Your answer is not correct because you did not account for the case where only B hits, which has probability $\frac13×\frac34=\frac14$. Then the required probability is $$\frac{\frac16}{\frac14+\frac16}=\frac25$$ as the book gives. The answer is indeed 2/5 I believe. \begin{align} \mathbb{P}[\text{A hit | only one hit}] &= \frac{\mathbb{P}[\text{A hit} \,\cap\, \text{only one hit}]}{\mathbb{P}[\text{only one hit}]} \\ &= \frac{\mathbb{P}[\text{A hit}\,\cap\,\text{B didn't hit}]}{\mathbb{P}[\text{A hit}\,\cap\, \text{B didn't hit}] + \mathbb{P}[\text{A didn't hit}\,\cap\, \text{B hit}]} \\ &= \frac{1/4 \cdot 2/3}{1/4 \cdot 2/3 + 3/4 \cdot 1/3} \\ &=\frac{2}{5} \end{align} Without using the conditional probability formula: There are four cases: 1. Both miss 2. A hits and B misses 3. B hits and A misses 4. Both hit We're only interested in (2) and (3). (2) has probability $\frac{1}{4}*\frac{2}{3} = \frac{1}{6}$. (3) has probability $\frac{1}{3}*\frac{3}{4}=\frac{1}{4}$. And we need $\frac{(2)}{(2) + (3)}$. • What's the last formula if not the conditional probability formula? – JiK Sep 26 '17 at 16:48 • @JiK common sense – MattPutnam Sep 26 '17 at 17:42 The probability that only one person hits the target is $$1/4 * 2/3 + 1/3 * 3/4 = 5/12$$ The first event occurs when A hits and B misses, and the second when B hits and A misses. So if only one hit occurs, A hits 2/5 of the time and B 3/5 of the time. This is an application of Bayes's law. You have a theory: A hit the target. You have data: there's only one hit. What is the probability your theory is true, given the data? 2/5. If you saw two bullet holes, then your theory would be true with probability 1 because A had to hit the target, given those data. • We don't consider $P(\text{only one hit}|\text{A hit the target})$ or $P(\text{only one hit}|\text{A didn't hit the target})$ explicitly, so I don't see how this is an application of Bayes's law. – JiK Sep 26 '17 at 8:20[SEP]
[CLS]# The probability that A hits a target � $\frac14$ any that of B is $\frac13$. If they fire at once and oneort the target, find $P(\text{A hits})$ The probability that acting hits a target is 1/4 and the probability that B hits a target 1/3. They each fire once at the target. If the target is hit by only one of them, what is the probability that A hits the target? I know that this is an independent event. If I do P(A out) * P(B not hitting) then (1/4)(2its3) = 1/6 But when I look at the back of my book the answer is 2/5? My book is known to give wrong answers because it is quite old; therefore, I am left with self doubt. Can anyone tell me ' I have the correct answer or if I am actually making a mistake? • This is a conditional probability. Letting $A$ be the event that player $A$ hit the target (in a single shot) and $B$ Th event that $B$ hit the target (in a single shot), then what you calculated was $Pr(A\cap B^ occur)$. What Your were told to calculate was $Pr(A\mid (A\cap B^c)\cup $\|A^c\cap B))$, i.e. the probability that $A$ hit the target given that exactly one of them hit the target. – JMoravitz Sep 26 '17 at 4:07 • You might like this explanation of why the formulas being posted here work: arbital.com/p/bayes_rule/?l=1zq – Davislor Sep 26 '17 at 14:51 ${ Would also add the comment that t &\ is likely not full of mistakes because it is old, but that because it is old the mistakes have &= found. language books are not necessarily more correct, they just haven't been around long enough forces the mistakes to be as well known. Old does not imply bad. – Jared Smith Sep 26 '-( at 15:39 • The probability of A hitting and B not hitting is 1/6 when it is also possible that they bothast or both miss. But in this .., it is not possible for them both to hit or both to miss, so the probability must be greater thank 1/96. (If you denominator't see -\, imagine r a die. The probability of 1 coming up is 1/6, but only because there are cut other possibilities. im you roll a die and the result is not a 3, 4, 5, or 6, now the probability it's a 1 is 1/2.) – David Schwartz sp 26 '17 at tests:42 \begin{align} P)\mbox}:Point is hit once}) &= P(\mbox{A hitting}) \cdot P(\mbox{B not hitting}) + P(\ fix{A not hitting}) \cdot P(\binom{B hitting}) \\ &= \frac{1}}_{4}\cdot\frac{)}{3} + \frac{3}{4}\cdot\ conclude{1}{3}. \\ &= \frac{5}{12} \end{align} So, $$P(\mbox{A hitting | target is hit once}) = \frac{P(\mbox{A hitting}) \cdot P(\mbox{B not hitting})}{P(\mbox}:target is hit once})} = \dfrac{\frac{1}{6}}{\frac{5}{12}} = \frac{2}{5}.$$ • I would change Theory LHS of the last line to "$P(\text{A hitting}|\text{target is hit once})$, to be more explicit that we're using conditional probability here. – JiK Sep 26 '17 Aug 8:18 Your answer is not correct because you did not account for the case where only B hits, were has probability $\frac13×\frac34=\frac14$. Then the required probability is $$\frac{\frac16}{\frac14+\frac16}=\frac25$$ as the book gives. The answer is indeed 2/5 I believe. cccc\begin{align} \mathbb{P}[\text{A hit | only one hit}] &= \frac{\mathbb{P}[\text{A hit} \,\cap\, \text{only one hit}/}{\mathbb{P}[\text{only one hit}]} \\ &= \frac{\mathbb{P}[\text{A hit}\,\cap\,\text{B didn't hit}]}{\mathbb{P}[\text{A hit}\,\cap\, \((}_{\B didn't hit}] + \mathbb{P}[\text{A didn't hit}\,\cap\, \text{B hit}]} \\ &= \frac{1/4 \cdot 2/3}{1/4 \cdot 2/3 + 3/4 \cdot 1/3} \\ &=\frac{2}{5} \end{align} Without using the conditional probability formula: There are four cases: 1. Both miss 2. A hits and beginning misses 3. B hits and A misses 4. Both hit We're only interested in (2) and (3). (2) has probability $\frac{1}{4}\*\frac{2}{3} = \frac{1}{6}$. ( 38) has probability $\frac{1}{3}*\frac{3}{4}=\frac{1}{4}$. And we need $\frac{(2)}{(2) + (3)}$. • What's the last formula if not the conditional probability formula? — JiK Sep 26 '17 at 16:},{ • @JiK common sense – MattPutnam Sep 26 '17 at 17:42 ction probabilitygt only one person hits the target is $$1/4 * 2/3 + 1/3 * 3://4 = 5/12$$ The first ed occurs when A hits and B misses, and the secondren B hits and A misses. So if only one hit occurs, A hits 2/5 of the time and B 3/5 of the time. This is an application of Bayes's law. good have a theory: A hit the target. You have data: there's only one hit. What is the probability your theory is trueby given the data? 2/5. If you saw two bullet holes, then your theory would be true with probability 1 because A had to hit the target, given those data. conclusion• We don't consider $Phi\!text{only one hit}|\text{A hit the target})$ or $P(\text{only one hit}|\text{A didn't hit the target})$ explicitly, Sl IS David't see how this is an application of Bayes's law. – JiK Sep 26 '17 at 8:20[SEP]
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[CLS]# How to plot slices of a surface of an iterative function parametrized by the iterator k? I am trying to plot a surface of $$z=\sin^{(k)}(x),\text{where (k) means nesting the function k times}$$ to visualise the fixed points and their neighbourhood to visually analyse their behaviour. Currently, the following (adapted from this link) give me a contour version of the above: f[x_] := Sin[x] Show[Table[Plot[Nest[f, x, i], {x, -π, π}, PlotRange -> {-1, 1}, PlotStyle -> ColorData["Rainbow", 0.5 + i/10]], {i, 1, 10}]] However, I want to space out the contours along the $k$ axis so that e.g. $\sin(x)$ corresponds to $k=1$, $\sin(\sin(x))$ corresponds to $k=2$ and so on... Below is my most recent attempt at doing it: f[x_] := Sin[x] data[x_] := Table[{Nest[f, x, i], i}, {i, 0, 10}] ListPlot3D[data[x], {x, -π, π}] which gives me an error SetDelayed::write: Tag List in {{x,0},{Sin[x],1},{Sin[Sin[x]],2}, {Sin[Sin[Sin[x]]],3},{Sin[Sin[Sin[Sin[x]]]],4},{Sin[Sin[Sin[<<1>>]]],5}, {Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],6}, {Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],7}, {Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],8}, {Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],9}, {Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],10}}[x_] is Protected. >> Strangely the data behind seemed to be interpreted correctly ListPlot3D[{{x, 0}, {Sin[x], 1}, {Sin[Sin[x]], 2}, {Sin[Sin[Sin[x]]], 3}, {Sin[Sin[Sin[Sin[x]]]], 4}, {Sin[Sin[Sin[Sin[Sin[x]]]]], 5}, {Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]], 6}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]], 7}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]]], 8}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]]]], 9}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]]]]], 10}}[x], {x, -\[Pi], \[Pi]}] I was suspecting that ListPlot3D cannot read my input is probably because I have mixed data type. In details $$z\in \mathbb{R}$$ $$x \in [-\pi,\pi]$$ but $$k \in \{0,1,2,3,4,5,6,7,8,9,10\}$$ From browsing the documentation, I am not aware of any examples of plots made from a mix of discrete and continuous variables as plotting arguments, thus I am not sure how to plot the surface I want. I am not sure how to circumvent/cheat it without taking too much computation time since if my set of points $x$ is too sparse, it will fail to display the sinusoidal feature (which will be a problem because I am planning to apply this code on other iterative functions, such as the logistic map), but if my sampling is too dense, it will probably took too much computation time Any ideas on what I can do? P.S. To give an idea on what I am trying to achieve, refer to the below sketch: which after interpolation along $k$, will give a nice surface. • Related: (1413). – march Sep 14 '15 at 17:07 Use Interpolation if you want a regular function. Just for the plot you can also use ListPlot3D. fun = Interpolation[ Flatten[Table[{x, k, Nest[Sin, x, k]}, {x, -Pi, Pi, .1}, {k, 1, 10,1}], 1]]; Plot the continuous function and those $k$-mesh lines! Plot3D[fun[x, k], {x, -Pi, Pi}, {k, 1, 10}, MeshFunctions -> {#2 &}, Mesh -> 10, PerformanceGoal -> "Quality", MeshStyle -> {{Black, Thin}}] If you only want discrete lines you can use ParametricPlot3D in combination with Map or Table embedded in a Show. Below the Blend function is used to add a variable color (optional). Black is Sin[x] and Red is the curve nested ten times. Show[ Map[ ParametricPlot3D[{u, #, Nest[Sin, u, #]}, {u, -\[Pi], \[Pi]}, PlotStyle -> Blend[{Black, Red}, #/10], PlotRange -> {{-\[Pi], \[Pi]}, {0, 10}, {-1, 1}} ] &, Range[10] ] ] This is, I think, a dupe of Plotting several functions, except that that thread displayed only the contours. An approach simpler than the other posted answers proceeds like so: Plot3D[Nest[Sin, x, Round[k]], {x, -π, π}, {k, 1, 10}, MeshFunctions -> {#2 &}, Mesh -> 10][SEP]
[CLS]\{ How to plot slices of a surface of an iterative functional parametR by the iterator )\| I am trying to plot a shapes half confusion Michael$$z=\sin^{(k)}(x)\\text{where ( ch) means nesting the function k times}$$ to visualane the fixed points and theiriy to visually analyse here behaviour. Currently,..., the following (adapted from this link) give me at contour - finish the above: f:=x_] := Sin:=\dx] Show[Table[Plot[Nest== follow, x, i], {x, -π,—}, Plot � -> {-1, 1}, PlotStyle -> ColorData["})$.bow", 0.52 + i/10~~ :i., 1, 10}]] etcHowever, I want to space out the Cont along twice�k$ axis so term e onceg. ((sin( text)$ corresponds to $k-(1$ $\sin(\sin(x)$. corresponds to $k=2$ and so month...C & is my most recent attempt at doing it: concepts f[x_] := Sin[x] existence={x____~\ := triple{.nonest[f, x, i], i}, {i, 0, 10}] List game3 didn)_{data[ exactly]} }{x, Gπ, π}] which gives me an error Set outerheet::]],: trig List in }}x,0},{Sin[x],1},{ within[Sin[x]],2}, { analysis[Sin->Sin[x[]],3},{Sin[Sin [-Sin[Sin[x]]]],4},{Sin[Sin[ notionof!(}>>[],5}, {Sin[Therefore[Sin[ notion__Sin[Sin[```}_>>]]]]]],6}, {Sin[{Sin[Sin[ cardinality->Sin[ass[<<1>>]]]]]],\}}, {Sin[Sin[Sin[(Sin[Sin[Sin[<<1>>]]]]__8}, [{Sin[Sin[Sin=( noticed[Sin[Sin]=<<1>>``]]]],9}, {Sin[Sin[Sin[Sin[Sin[Sin[<< 101>>]] \\[]10}}[x_] digit Protected. >> Str iterationely tan data behind seemed to be interpreted correctly NCListpp-D[{{x, 0}} { dimension[x], 1}, {Sin[Sin[x]], 2}$. {Sin^+Sin[ression[x(*], 3}, {Sin[Sin[Sin[ often[x]]]], 4}, {Sin[ Notice[Sin[Sin[Sin[x]]]]], 5}, {self^{( Question[IM[ation]:omorphism[abel[x]]]]]/ }+}, {AB[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]], 7}, { notion[abs[Sin[Sin^*Sin[Sin[Sin[Sin[x]]]]]]]]ccc8}, {abs[Sin[Sin[Sin[Sin[Sin[ homogeneous[ression[Sin[x]]]/[\]]],  }, {Sin[Sin[Sin[Sin[Sin[stitution[Sin[Sin[Sin[ notion[x]]])]]::)_ 10}}[x], {x”, -\[Pi)( \,Pi]}] I was suspecting thus ListPlot3D cannot rate my input Image probably because Δ have mixed data type. In details $$z\in \mathbb){R}$$ $$x \in [-\pi,\pi]$$ but $$ known -\in \{0 Once}_]:2,3....4,}));6,7,8,9,.10\}$$ From browsing the documentation, I am not aware of any examples of P made65 a mix of discrete and continuous variables assuming Pat arguments, thus I am not sure how tables plot the surface I want. I am not sure how to circular/cheat it without taking too much commutative time since if my set of posted $ hex$ is too sparse, switch w fail to add the sinusoidal feature -(which will be a problem because I am planning topics apply this code on other iterative functions, such as the discuss map), but if my sampling is too dense, it will policy t too much computation time cubic Any res on what I can do\\ PitiesS. To give an idea on what I am trying to achieve, refer to T below sketch: Circle which after interpolation along $k$, W a nice surface. • Related: (1413). – march 1 14 '}} at 17:07 expressions Interplesation �My want a regular function. ). for the PNow can cover use ManyPlot3D. BC fun = Interpolation[ Flatten('Table[{x, k:// Nest[Sin, x// k``}, {x, -Pi, Pi, .1}, { k, 1, 10,1 }{], 1]]; Plot the continuous function and those $k)$.mesh lines! Plot3ord=fun[x, k]) {x..., ${ details); Pi}: {k, 1itus 10}, Mesh vertical -> {#2 &}, whose +\ 10, PerformanceGoal -> "Quality:\ MeshStyle -> {{Black, Thin}}]oc If you only of discreteunderline you car across Par?"etricPlot3D in rational & Map Our tell embedded in a Show.C )), the Blend function is used tends add a variable color (optional&\ Black is Sin[x] and Red is told curveigens Then times,..., )[ Map[ mapetric833D[{u, #, Nest]Sin, u, \{]}, ${u, -\[Pi], \[Pi][}, PlotStyle -> Beend[{Black, Red}, #/10], PlotRange -> {{(\\[Pi], \[Pi]}, {0, 10},�}_ation 1}}cc] ),, Range[10] ] ] This is, I think, a dupe of Plotting several functionsimals except that that thread displayed only tends contours:. An multiple problem than the other posted answers proceeds like so: Plot3D[N Select[im, x, Round[k]], {sided, -πuitively π}, {k, 1,..., 10}, WorksFunctions { {#2 &}}_{ Mesh -> 10][SEP]
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[CLS]. There are three kinds of exponential functions: Thus, the exponential function also appears in a variety of contexts within physics, chemistry, engineering, mathematical biology, and economics. In particular, when Population growth can be modeled by an exponential equation. t It is used everywhere, if we talk about the C programming language then the exponential function is defined as the e raised to the power x. {\displaystyle v} The EXP function finds the value of the constant e raised to a given number, so you can think of the EXP function as e^(number), where e ≈ 2.718. ( {\displaystyle b^{x}=e^{x\log _{e}b}} Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. ! Example and how the EXP function works Excel has an exponential excel function it’s called Excel EXP function which is categorized as Math or Trigonometry Function that returns a numerical value which is equal to e raised to the power of a given value. That is. = {\displaystyle y=e^{x}} What is Factorial? {\displaystyle y} x It satisfies the identity exp(x+y)=exp(x)exp(y). For example: As in the real case, the exponential function can be defined on the complex plane in several equivalent forms. For most real-world phenomena, however, e is used as the base for exponential functions.Exponential models that use e as the base are called continuous growth or decay models.We see these models in finance, computer science, and most of the sciences such as physics, toxicology, and fluid dynamics. In the case of Exponential Growth, quantity will increase slowly at first then rapidly. b n The most common definition of the complex exponential function parallels the power series definition for real arguments, where the real variable is replaced by a complex one: y y C {\displaystyle {\tfrac {d}{dx}}e^{x}=e^{x}} {\displaystyle {\mathfrak {g}}} ...where \"A\" is the ending amount, \"P\" is the beginning amount (or \"principal\"), \"r\" is the interest rate (expressed as a decimal), \"n\" is the number of compoundings a year, and \"t\" is the total number of years. It shows that the graph's surface for positive and negative d Exponential functions and logarithm functions are important in both theory and practice. x Exponential Growth: y = a(1 + r) x. Exponential Decay: y = a(1 - r) x. The Exponential Function is shown in the chart below: by M. Bourne. Here's an exponential decay function: y = a(1-b) x. x Furthermore, for any differentiable function f(x), we find, by the chain rule: A continued fraction for ex can be obtained via an identity of Euler: The following generalized continued fraction for ez converges more quickly:[13]. with Euler's formula states that for any real number x: The formula takes in angle an input and returns a complex number that represents a point on the unit circle in the complex plane that corresponds to the angle. i To find limits of exponential functions, it is essential to study some properties and standards results in calculus and they are used as formulas in evaluating the limits of functions in which exponential functions are involved.. 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[CLS]# Probability that a random tetrahedron over a sphere contains its center I got interested in this problem watching the YouTube channel 3Blue1Brown, by Grant Sanderson, where he explains a way to tackle the problem that is just … elegant! I can't emphasize enough how much I like this channel. For example, his approach to linear algebra in Essence of linear algebra is really good. I mention it, just in case you don't know it. ## The problem Let's talk business now. The problem was originally part of the 53rd Putnam competition on 1992 and was stated as Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points? (It is understood that each point is in- dependently chosen relative to a uniform distribution on the sphere.) As shown in the mentioned video, the probability is $1/8$. Let's come with an algorithm to obtain this result —approximately, at least. ## The proposed approach The approach that we are going to use is pretty straightforward. We are going to obtain a sample of (independent) random sets, with four points each, and check how many of them satisfy the condition of being inside the tetrahedron with the points as vertices. For this approach to work, we need two things: 1. A way to generate random numbers uniformly distributed. This is already in numpy.random.uniform, so we don't need to do much about it. 2. A way to check if a point is inside a tetrahedron. ### Checking that a point is inside a tetrahedron To find if a point is inside a tetrahedron, we could compute the barycentric coordinates for that point and check that all of them are have the same sign. Equivalently, as proposed here, we can check that the determinants of the matrices \begin{equation*} M_0 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*} \begin{equation*} M_1 = \begin{bmatrix} x &y &z &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*} \begin{equation*} M_2 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x &y &z &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*} \begin{equation*} M_3 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x &y &z &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*} \begin{equation*} M_4 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x &y &z &1 \end{bmatrix}\, , \end{equation*} have the same sign. In this case, $(x, y, z)$ is the point of interest and $(x_i, y_i, z_i)$ are the coordinates of each vertex. ## The algorithm Below is a Python implementation of the approach discussed before from __future__ import division, print_function from numpy import (random, pi, cos, sin, sign, hstack, column_stack, logspace) from numpy.linalg import det import matplotlib.pyplot as plt def in_tet(x, y, z, pt): """ Determine if the point pt is inside the tetrahedron with vertices coordinates x, y, z """ mat0 = column_stack((x, y, z, [1, 1, 1, 1])) det0 = det(mat0) for cont in range(4): mat = mat0.copy() mat[cont] = hstack((pt, 1)) if sign(det(mat)*det0) < 0: inside = False break else: inside = True return inside #%% Computation prob = [] random.seed(seed=2) N_min = 1 N_max = 5 N_vals = logspace(N_min, N_max, 100, dtype=int) for N in N_vals: inside_cont = 0 for cont_pts in range(N): phi = random.uniform(low=0.0, high=2*pi, size=4) theta = random.uniform(low=0.0, high=pi, size=4) x = sin(theta)*cos(phi) y = sin(theta)*sin(phi) z = cos(theta) if in_tet(x, y, z, [0, 0, 0]): inside_cont += 1 prob.append(inside_cont/N) #%% Plotting plt.figure(figsize=(4, 3)) plt.hlines(0.125, 10**N_min, 10**N_max, color="#3f3f3f") plt.semilogx(N_vals, prob, "o", alpha=0.5) plt.xlabel("Number of trials") plt.ylabel("Computed probability") plt.tight_layout() plt.show() As expected, when the number of samples is sufficiently large, the estimated probability is close to the theoretical value: 0.125. This can be seen in the following figure.[SEP]
[CLS]# Probability that a random tetrahedron over a sphere contains its center I got interested in this problem watching the YouTube channel 3bs1Brown”, by Grant Sanderson, highest he So a way too tackle the problem that is just … elegant! * can't emphasize enough how Mod I like this channel. fair example, his approach to linear algebraic in Essence of linear algebra is really good. I mention it; just in Le you modulus't know it. ## triangle prior Let's talk business now ordering The problem was hour part of the 53rd Putnam competition on 9 and was stated… Four points are chosen at random on the surface before a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whoseides are at the four points? (It itemde that each point is input- dependently chosen relative Te a uniform distribution on the sphere.) As shown in the reduction video, the probability is $1/8$. Let's come with an algorithm to obtain this result howapproximately, at least. ## The proposed approach unc The approximately that we are going to use is pretty straightforward. few are going to obtain a sample of (independent-( random sets, with four points equ, and check how many of them satisfy test condition of being inside the tetrarex &= the pointed as vertices. For The approach to work, we need two things: 1. A way to generate random numbers uniformly distributed. This is already in numpy.random.uniform, so we don't need Test do ' Out it. Sc2. A way to checking if a point is inside At tetrahedron. c ### Check fitting that a point is inside a tetrahedron )] find if a Put is inside a Timer unbiased, we could compute the bary correct coordinates for t point and check that all of them are have the saw sign. Equivalently, $$( proposed here, we can spectrum that the determinants of the matrices \begin{equation*} M_0 = \This{bmatrix} x_0 &y_0 & z_0 &1\\ ...,_1 &y_1 &z_1 &1'\ x_2 &y_{ &z_2 &1}\ x_3 &y_38 &z_ games (*1 $-\ whole{bmatrix}\, , \end{equation*} \begin{equation}_ Member _____1 = \begin${\bmatrix} x &y & cube &1\\ x_1 &y_1 &�_1 &1\\ x_2 &y_2 & Cauchy_2 & {}\\ x_3 ($y_3 &alpha_3 &1 ).end{bmatrix}\, , \end{ exam*} \ hope{ modulus*} M_2 = \begin{ determine} x_0 &y_0 &z_0 &1\\ x &y (z &1\\ x_2 &y_2 &z_}}{( &1\\ x_3 &y_3 &z_)}{ &1 \end{ symmetry}\, , \end{Solve*} \begin{equation*} M_ }^{ & \begin{ Matrix} x_0Gy_\} &z_0 & 1\\ -->_1 &y_1 &z_subseteq &1#### x &y &z &1\\ x_3 & Any_3 &z_3 &1 \end({bmatrix}\, , \end{equation*} cent\begin{equation*} M~4 = \begin{-bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x_2 --y_2 &z_2 &1\\ x &y &z &1 //end{bmatrix}\, , \end{equation*} have the same sign. In this case, .$$x, ~ors z)$ is the point of interest and $( exterior_i, y_i, z_i)$ parent the coordinates of each vertex. ## The algorithm )), is a Python digit of the approach centre before from __future__ import division, print_functionsectionfrom numpy I )randomAnd piential cos, sin, sign, hstack, column_stack, logspace) from numpy.linalg import nddfracimport matplotlibingpy various as plt def in_tet)-(xof y]; z, pt): """ Determine if the point pt is inside the tetra children with vertices coordinates x, y, z """ cosmat0 = column]{stack((x, y, z, .digit, 1, 1, 1])) Can domain0 < det(mat}+\) for cont in range(4): mat = mat0.copy() mat[cont] = Shstack((pt,... 1)) if sign(det(-mat)*detdigit) < 0: inside = False break else: ). &= True return inside #%% Com implementationincprob = [] random.seed( far=_{) N_min = 1 N_max = " BCN_vals = logspace(N_\}$, N_max, 23, dtype=int) for N in N_vals: inside_cont = 0 for cont_pts in range( AND): phi = random.uniform(low})=0.0, high=two* Chapter, such)=-4) theta = random. Page(low=}(\.0, high=pi, size=4) x = sin(theta**cos(phi) y [- sin(theta~sin(phi!, z = cos(theta) circlesif in_ Next(x, y, z, [0, 0, 0]): inside_cont += .... csprob.append(inside*)cont/N) etc#%% Plotting plt.figure(figsize=(4, 3)) plt.hlines(0.125, 10**N[\min, 10**N]]60, color="#3f3f3f") plt. repeatingilogx(N_ divergence, prob, "o", alpha=0.5) plt.xlabel("Number of trials") plt.ylabel("Computed probability")ackplt.tight^*layout() plt.show]/ cAs expected, when the number of samples is sufficientlyig,. the estimated probability is rules to the theoretical Have: 0.56.”gt can be seen in the following figure.[SEP]
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[CLS]SEARCH HOME Math Central Quandaries & Queries Question from r.m, a student: question from calculus exam: what is the figure obtained having eqn.r=10cos(t) in cylindrical coordinates? i know it is a cylinder with center (5,0) ,but can't the equation represent two cylinders, one with center (5,0) and the other with center (-5,0). thanks for any help. Hi, I want to use the cartesian graph of $y = \cos(x)$ for reference. Now let's plot $r = 10 \cos(\theta)$ in polar coordinates for $0 \le \theta \le 2 \pi.$ $\cos(0) = 1$ and hence the graph starts at $(r, \theta_1) = (10,0)$ which is the point $P_1$ in my diagram. Now let $0 < \theta_2 < \frac{\pi}{2}$ then $\cos(\theta_2)$ is positive and resulting point $P_2$ is on the upper half of the circle with center $(5, 0)$ and radius 10 as in my diagram. When $\theta_3 = \frac{\pi}{2}$ then $\cos(\theta_3) = 0$ and the resulting point on the graph is $P_3$. For $\frac{\pi}{2} < \theta le \pi$ as $\theta_4$ in my diagram, $\cos(\theta) < 0$ and the resulting point (for example $P_4$) is on the bottom half of the circle with center $(5, 0)$. When $\theta = \pi$ then $\cos(\theta) = -1$ and we are back at $P_1$. For $\pi < \theta \le \frac{3 \pi}{2}$ as $\theta_5$ in my diagram, $\cos(\theta)$ is still negative and the resulting point (for example $P_5$) is on the top half of the circle with center $(5, 0)$. When $\theta = \frac{3 \pi}{2}$ then $\cos(\theta) = 0$ and we are back at $P_3$. Finally for $\frac{3 \pi}{2} < \theta \le 2 \pi, \cos(\theta)$ is positive and the resulting point, as $P_5$ in my diagram is on the bottom half of the circle and when $\theta = 2 \pi$ we are back at $P_1$. Hence as $\theta$ moves from $0$ to $2 \pi$ the point defined by $r = 10 \cos(\theta)$ moves twice around the circle with center $(r \theta) = (5, 0)$ and radius 10. Penny r.m replied sir, in reply to your answer for r=10cost in cylinderical coordinates. if we follow the same method as you explained to sketch r=sin(t/2),i expected graph to be in only first two quadrants, but the graph was covering all four quadrants ? If $0 < \theta <2 \pi$ then $0< \large \frac{\theta}{2} \normalsize < \pi$ and $\sin \left(\large \frac{\theta}{2}\right) >0.$ Thus, for example if $\theta = \large \frac{3 \pi}{2}$ then $r = \sin \left(\frac{\theta}{2}\right) = \sin \left(\frac{3 \pi}{4}\right) = \frac{1}{\sqrt 2} = 0.70711$[SEP]
[CLS]SEARCH h Math Central Quand Res ! Qu Two Question from r.m, a student: question from calculus exam: white is the figure abstract having eqn.,r=10cos(t) inreg coordinates? i know it is air cylinder with adjacent (5,0) ,This can't the equation represent training cylinders,)-\ with center (5,)}\) and the other with center (-5,0). thanks for any help. Hi, I want to use the cartesian graph of $y = \cos)-(x)$ for reference. Now let's plot $r = 10 \cos(\theta)$ integr polar coordinates for $'( \le \theta \le 2 \pi.$ $\cos( {\) = 1 $$| and welcome thank graph starts at $(r, \theta_1) = (10,0)$ which is the endpoints $P_1$ include my diagram. Now let $0 < \theta_2 < \frac{\pi}{2})$ then $\cos_\theta_--)$ is positive and resulting point $P_2$ is on the upper half Fl the circle with centre $(5, 0)$ and radius 10 ask in my diagram ordering When $\ Learn}]3 = \frac{\pi}{2}$ then $\cos(\theta_})$.) = 0$ and the resulting point on the graph is $P_ 23$. complement $\frac{\pi}{2} < \theta le \pi$ as $\theta_4$ in May diagram, $\cos(\theta) < 0$ and the resulting point (forbx $P_4$) is on the best half of the specific additive center $(5, 0)$. When $\theta = \pi$ then $\cos(\theta) = -1$ and we are back at $P_}^{$. For $\pi < \theta \le \frac{3 \pi}{2}$ as $\theta_5$ in my ||ations $\cos(\theta)$ is still negative and the return point (for example $P_5$, is on the top half of T circle without center $(})=, 0)$. When $\theta = \frac{3 \pi}{2}$ then $\cos(\theta) = 0$ and we are back at $P_3$. Finally for $\frac{3 \pi}}\2} < \theta \� (( \pi, \cos(\theta)$ is positive and the require point, as $P_range$ in my diagram isenn the bottom half of the circle and when $\theta = 2 \ circle$ we arcs back test $P_1$$ Hence as $\om$. moves from $0$ to $2 \pi$ the point defined by $r = 10 \cos,\,theta)$ moves twice around the circle with center $(r \theta) = (5, 0$), and radius 10. Penny rificationm replied sir, in reply to your answer for r=10cost in cylinderical coordinates. if we follow the same method as you explained to sketch r=sin(t/2),i expected graph to be in nd first two quadrants, but the graph was covering all fair quadrants ? If $0 < \theta <2 \ 33� The $0< \large \frac{\theta}{2} \normalsize < ....pi|$ grid $\sin \left(\large \frac{\theta}{2}\right) >0.$ Thus, for example if $\theta = \large \frac{3 \pi}{2}$ then $r = \sin \left(\frac{\theta}{2\{\right) ! \sin \left(\frac{3 \pi}}}{};}\right) = \frac{1}{\sqrt 2} = 0.70711$[SEP]
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[CLS]Is the blue area greater than the red area? Problem: A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red. Which area is greater? Let the area of each large square be exactly $1$ unit squared. Then, the area of the blue square is exactly $1/4$ units squared. The same would apply to the red area if you were to rotate the square $k\cdot 45$ degrees for a natural number $k$. Thus, I am assuming that no area is greater, and that it is a trick question $-$ although the red area might appear to be greater than the blue area, they are still the same: $1/4$. But how can it be proven? I know the area of a triangle with a base $b$ and a height $h\perp b$ is $bh\div 2$. Since the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$. Therefore, the height of the red triangle area is $1/2$, and so $$\text{Red Area} = \frac{b\left(\frac 12\right)}{2} = \frac{b}{4}.$$ According to the diagram, the square has not rotated a complete $45$ degrees, so $b < 1$. It follows, then, that \begin{align} \text{Red Area} &< \frac 14 \\ \Leftrightarrow \text{Red Area} &< \text{Blue Area}.\end{align} Assertion: To conclude, the $\color{blue}{\text{blue}}$ area is greater than the $\color{red}{\text{red}}$ area. Is this true? If so, is there another way of proving the assertion? Thanks to users who commented below, I did not take account of the fact that the red area is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis was correct. This question is very similar to this post. Source: The Golden Ratio (why it is so irrational) $-$ Numberphile from $14$:$02$. • i think you can tile the red area 4 times to get the entire square – gt6989b May 11 '18 at 4:25 • Hint: the sum of the two red sides that don't touch the center is $1$. – dxiv May 11 '18 at 4:29 • @user477343 Glad the hint helped. You can make that into a full-fledged answer, and I'll +1 it. – dxiv May 11 '18 at 4:31 • Is this a problem from "Brilliant" – Rohan Shinde May 11 '18 at 4:32 • Note that purely from exam technique alone, the answer is likely to be "they are the same size". Indeed, the problem has not told you by how much the rotation occurs, and why privilege a rotation of $0$ over a rotation of some greater angle? This is not a proof; but the phrasing of the question has told you what answer to look for. (This is a more general point than your "it works this way for 45 degrees": this is a demonstration that no mathematical reasoning at all is required to exam-technique that the answer is "they're the same".) – Patrick Stevens May 11 '18 at 20:24 The four numbered areas are congruent. [Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment “This answer also deserves the tick for artistic reasons,” I’m leaving my original “artistic” figure but also adding Tom’s improved version to my answer. • This answer also deserves the tick for artistic reasons. – BenM May 11 '18 at 6:09 • A great example of "proof by picture" that actually works. – Bristol May 11 '18 at 14:46 • This is not the same as the answer by Ross and Zoltan. I like this one better. Theirs was the first that came to my mind, too. – Carsten S May 11 '18 at 23:09 • Can Wolfram Alpha draw that? – Willtech May 12 '18 at 2:44 • @FrankShmrank The original poster asked within the question how it can be proven that the red area equals 1/4 (which would settle the title question). My answer makes it clear (without a formal proof, but in proof-by-picture, that’s par for the course) that the red area is one of four congruent areas that partition the unit square, so its area is 1/4. I agree my answer is less than a complete proof of the original question, but I think (and I guess many upvoters think) that it’s convincing. There are other excellent answers that are more traditionally proof-like, so upvote your favorite(s). – Steve Kass May 19 '18 at 17:14 I think sketching the two identical triangles marked with green below makes this rather intuitive. This could also be turned into a formal proof quite easily. • This method is similar to @RossMillikan 's answer above, but not quite the same :) I have to wait $9$ hours before I can upvote as I have reached my daily limit... but when I can, $$(+1)$$ – Mr Pie May 11 '18 at 15:00 • It's not only similar, now that I read that solution, it's actually the exact same idea. Unfortunatly that answer didn't contain any images and I just looked at the images before posting my own answer. :) – Zoltan May 11 '18 at 15:05 • Well congratulations on your first answer on the MSE! Yours is still a good answer :)) – Mr Pie May 11 '18 at 15:08 • This is the clearest image to understand. +1 – qwr May 12 '18 at 20:36 • @qwr Indeed! If only I could grab this answer and drag it below the accepted answer. That way, nobody would have to scroll all the way down to see this. It is my own answer that should probably be at the very bottom :) – Mr Pie May 13 '18 at 1:52 Note that for equal angles $\angle A'OB' = \angle AOB = 90^\circ$, when we subtract a common part $\angle A'OB$ from both sides, we have $\angle AOA' = \angle BOB'$, so the red and cyan triangles are congruent: $\triangle AOA' \sim \triangle BOB'$. That implies their areas are equal, and when we add a common part $\triangle A'OB$ we get area of the $AOB$ triangle equal to the area of the $A'OB'B$ quadrilateral. Finally, the area of the two squares' common part is constant, independent on the square's rotation angle. • Shouldn't be $\angle AOA' = \angle BOB'$? – Pedro May 13 '18 at 3:59 • @Petro Right, thank you. – CiaPan May 13 '18 at 7:10 • Do you mean to say that $\Delta AOA' \color{red}{\cong} \Delta BOB'$? – Mr Pie May 14 '18 at 3:08 • This is the way I saw it – MichaelChirico May 15 '18 at 2:59 The two areas are equal. On the diagram with the red area draw the vertical and horizontal lines that define the blue area. The red area has a triangular region added to the left of the blue area and a triangular region above and to the right removed from the blue area. Those two triangles are congruent. • I see what you mean. There was no need to describe the result when drawing the vertical and horizontal lines that define the blue area on the diagram of the red area $-$ it was clear as day that they would be equal after looking at the newly formed triangles! I like your method of showing they were equal :) $$(+1)$$ – Mr Pie May 11 '18 at 5:29 By pinning a square's vertex to the center of the other, you guarantee a 90 degree slice outwards. This means we could tile 4 slices perfectly. A square has rotational symmetry of n=4. Since the rotation number is an integer multiple of the slice number, the area is invariant of rotation. You can apply this generally as well. A 120 degree slice of an equilateral triangle will be invariant. A 60 degree slice of a uniform hexagon will too. 120 degrees will work for the hexagon as well since that's 3 slices on a rotation number of 6. • FWIW I like this answer the best. It is a simple, brief proof that uses clear logic instead of math. – Bohemian May 12 '18 at 14:28 • @Bohemian, the reasoning is of course maths. – Carsten S May 12 '18 at 23:24 • @carsten but it’s basic geometry, without any calculations, arithmetic or formulae, such that someone without any mathematical know-how could follow. It’s[SEP]
[CLS]Is the L area greater than the red area? circles Problem: etcA vertex ofoff square is pegged to the centre of an identical square.. and try overlapping area is blue. One of the squares is then rotated about the be andgt resulting overlap is red� Which area is greater? Let the break of each large square be exactly $1$ title squared. Then, the area of the blue square is exactly (-1/4$ units squared. The same would apply to the red area if your were to rotate the square $k\cdot Engineering$ degrees for a natural number $k$. Thus, I am assuming that no whereas is greater, and that it is a trick question $-$ enough the red area might appear to be greater than the blue area, they are see the same: $1/4$. But how can it be proven? confusion I know the area of a triangle with a base $b$ AND a height $h\perp b$ is $bh\div 2$. Since the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$. Therefore, the height of the red triangle area Gaussian $1/2$. and so $$\ complete{Red Area} = \frac{b\left(\frac 12\right)}{2} = ]frac{b}{4}.$$ According to the diagram, the square h not rotated a composite $45$ degrees, so $b \, ).$. It follows, then, that \begin{align} \text{Red Area} &< \frac = \\ (\Leftrightarrow {(text{ interior Area} &< \text{Blue Area}}=end{align} Course ConAssertion: To conclude;\;\ the $\color{blue}(-text{blue}}$ area is greater than things $\ 300|}red}{\text {-red}}$ area”. .\ this true? If single, is there another way of proving the assertion? Thanks to users behavior complex below, I did not take account of the fact that the red · is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis Now correct.ch This questions is very Similarly to this post. Cent Source: sectionThe Golden Ratio -(why it is so irrational) $-$ Numberphile factor $14$:$)|$. • i They you can tile theoretical red area 4 times to get the entire square – gt6989b May 11 '18 at 4:74 • Hint: the sum of the two red sides that'd't touch the center is |1$. – dxiv May 11 '18 at *:29 • @user477343 Glad tables hint helped. You can make that into a full-fledged answer, and I'll +1 it. – dxians May 11 '}{ at ($:31 • Is this a problem from "Brilliant" ’ WolR)? setIn May 11�18 G 4:32 • Note that purely from exam technique alone, took answer II likely to be "ises are the same size". Indeed, the problem has not told you by write much the rotation occurs, and why privilege a rotation of $digit$ over a rotationdiff shape greater angle? This is not a proof; but the phrasing of the question has told you what answer to located for. (This is a more general point thanyou "\\ works this way for 45 degrees": this is a demonstration that no mathematical reasoning Att all is required to exam-technique that the answer is "theyvert the same'(.) – Patrick Stevens May 11 '18 at 20:24 The four numbered areas are congruent. [Added later] The figure below is from a suggested edit by @TomZ chosen, and it shows the congruent parts more clearly. Given Jordan talk upvotes to the (probably tongue-in.)cheek) comment “This and Using sphere the tick for parent reasons,” I’mat leaving my original “artistic” figure but also added Tom–s improved version to my answer. • This answer strong deserves the tick for artistic Se. –�BenM May 11 '18 at 6:79 • A great example of " nonnegative by picture" than actually works. – B onto May 11 '18 at 14:46 • This is nature the same as the answer by Ross and Zoltan outside I like this one but. Theirs was the first that came to my mind, too. –!(Carsten S May 11 '18 at 23:09 • Can Wolfrpm Alpha draw thatDoes –    Willtech May 12 '18 at 2:44 • @FrankShmrank The original poster asked within the question how σ can both proven that the rule area equals 1/4 (which worst settle the title question). My answer makes it clear (without a forms proof, but in proof}|by-picture,..., that strongs par for the course.) that the red area is one of four congruent areas that partition the unit square, so its area item 1/4. I • my answer is less than a complete proof of the original question,..., but I think (and image 2 many upvoters think) that it Lag running converge. There are other excellent answers that are more traditionally proof-like, so upvote several favorite(s). – Steve Kass May sl '18 at 17:14 codeI think sketching the two identical triangles marked with scaling below makes this rather principle. This could also be turned into a formal proof quite easily. • This method is single to @RossMillxi 's answer above, but not quite the same :) I have to wait $9$ hours be I can upvote as I have reached my daily limit... but when I can, $$(+1approx – man Pie May +\ '{- at 15[-00 $[ It's not only similar, now The I read that solution, it's actually the exact same idea. Unfortunatly that answer Def't contain Im images and I just School at the images Therefore posting my True answer. \; → vectorsoltan May 11 '18 at 15:normal • Well congratUL on your first answer on the M repetition)^{\ givesours is still a good answer :)) – Mr Pie Max 11 '18 at 15:underbrace • This is the realizearest � to understand. +1 – qwr May 12 '18 at 20:36 • !qwr Indeed! If only I could grab this answer and drag it below the accepted answer.” That way, nobody double have ten show all the way down to see this. It is my appreciate answer that should probably be air the very bottom :) – Mr Pie May 13 :18 at 1:52 cNote that for equal angles $\angle A'OB' = {\angle AOB = 90^\circ$, when we subtract a common part $\angle A' document$ from both sides, we have $\angle AOA' = \}angle BOB'$, so the red and N triangles are congruent: $\triangle awareOA' \sim \triangle BOB'$. That implies their areas are equal, and when we add a common part $\triangle A'OB}$$ we get area of the $AOB$. triangle equal to the area of the $A'ops'B$ diameterilateral. Finally, test area of the two squares' common part is constant, independent OP the square continue rotation angle. • somen't be $\angle AOA' = \angle BOB'$? – Pedro May 13 '18 at 3:59 • _Petro Right,... thank you. – CiaPan May 13 \18 at 7]{10 • Do you mean to say that $\Delta AOA' \led{red}{\cong} \Delta BO,$$? – Mr Pie May 14 '18 at 3:08 {( This is the way iter saw it – MichaelChirico Make 15 '18 at 2;\59 The two areas are equal. On the diagram with the red area draw the vertical trans horizontal lines that define the blue area... The red area has a triangular region added to the left of the blue area and a triangular expansion above and to Title right removed from the blue area. there two triangles are congruent. • I see what you Mult... There was no need to describe the result when drawing the vertical and harmonicLong that define the blue area on the diameter of the red area $-$ it was clear as day Trans they would be equal after looking at the newly formed triangles! I like your method of showing they were equal :) $$(+1)$$ –}\;Mr p May 11 '18 at 5: 36 By pinning across Se's vertex to the c of the other, you guarantee a 90 degree slice outwards. This means we could tile 4 slices perfectly. A square has rotational symmetry Inf n=4. Since the rotation coin is an integer multiplication of the slice number, the mistake is invariant of rotation. You can apply this generally as well. A 120 degree slice of an Quantilateral told will be invariant fitting A 60 degree slice of a uniform hexenn will too. 120 degrees will work for the hexagon $$ equation since that's 3 slices on a rotation number of 6. • FWIW I like this answer the best..... It is a some, brief proof that uses three logic instead of math. –  negativehemhe May 12 'Number at Min:28 • @Bohemian, the reasoning is of account maths. – Carsten S May 12 '18 at 23:24 • @carsten outputs it sigmas basic geometry, without natural calculations, arithmetic or formulae, such that someone without any mathematical know-times could follow. It’];[SEP]
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[CLS]# Is it possible to cover a $8 \times8$ board with $2 \times 1$ pieces? We have a $8\times 8$ board, colored with two colors like a typical chessboard. Now, we remove two squares of different colour. Is it possible to cover the new board with two-color pieces (i.e. domino pieces)? I think we can, as after the removal of the two squares, we are left with $64-2=62$ squares with $31$ squares of each colour, and - since the domino piece covers two colours - we can cover the new board with domino pieces. But how should one justify it mathematically? • It's possible to remove four squares, two of each color, and not cover the remaining with dominos, because the new set is disconnected. So you need to know more than that there are 31 of each color. Oct 15 '16 at 16:49 • Your constraints are valid. But there must be more constraints. Such that the colours alternate on the board, that dominos are made of two different colours, next to each other. Then the shape of the board. – mvw Oct 15 '16 at 17:05 • @mvw "like a typical chessboard" should cover all of that. Oct 15 '16 at 17:11 • @JMoravitz That was not my point. The problem is clear. It is about the proof attempt. hetajr just cared about a subset of all constraints that play a role here. – mvw Oct 15 '16 at 17:13 • @ThomasAndrews Ah cutting off two corners, leaving single corner stones. – mvw Oct 15 '16 at 17:23 Assume without loss of generality that the two squares to be removed are in different rows. (Otherwise turn the board 90°). First cover the board in horizontal dominoes, and connect the two squares by a zig-zag line like this: which follows the rule that if the line goes through one end of a domino, it immediately connects to another end. (The requirement that the two squares have different colors ensure that this will be true of the end of the path if only we start out in the right direction for this to hold at the beginning). Now you can flip dominoes along the zig-zag line to produce a covering that avoids the two squares. With a bit of (easy) special-casing for the same-row case, this strategy can be extended to any size board as long as one of the side lengths is even and the other is $\ge 2$. • I do not get the flipping. But it seems clear the length of the zig zag line is a multiple of two, as is the reduced line, so it can be covered. – mvw Oct 15 '16 at 17:31 • @mvw: Initially every other of the square boundaries crossed by the red line was covered by a domino. By "flipping" I mean to instead cover the other half of the boundaries by dominos. Oct 15 '16 at 17:36 • weren't you lucky that the top and bottom part had an even number of squares ? Oct 16 '16 at 13:23 • @mercio: If the top endpoint had been at an even column, I would just have started going left instead of right. This means that the zig-zag line would always enter each of the original dominoes at the end that has the same color as the top endpoint, so the bottom endpoint will always be reached at the end of a domino. Oct 16 '16 at 13:33 From a graph theory point of view, this can be seen as a "matching problem" on a bipartite graph. The nodes of the graph are the squares remaining. Two nodes have an edge in the graph if the squares are neighbors - that is, if the two squares can be covered by a single domino. Obviously, in any edge, one square is white, the other black, hence the "bipartite" nature of this graph. So you are seeking to show there is a perfect matching for any such graph. There is a general theorem about when there is a perfect matching for a bipartite graph, called Hall's Theorem or Hall's Marriage Theorem. It is possibly overkill for this question - induction is likely the better approach. Per the discussion on Henning's answer, it is actually possible to prove your theorem directly using a "Hamilton cycle" on the chess board. Consider the loop path on the board: $$\begin{matrix}1&2&3&4&5&6&7&8\\ 64&15&14&13&12&11&10&9\\ 63&16&17&18&19&20&21&22\\ 62&29&28&27&26&25&24&23\\ 61&30&31&32&33&34&35&36\\ 60&43&42&41&40&39&38&37\\ 59&44&45&46&47&48&49&50\\ 58&57&56&55&54&53&52&51 \end{matrix}$$ So we have walked in a circle, and, if the upper left is black, then we have that odd numbers on black and the even numbers on white. If we unwind this, and consider it 64 beads in a circle, alternating black and white, then if we remove/cut away one black and one white bead, we are either left with one string of 62 beads alternating black/white, which lets us cover those with dominos, or two seperate strings. With two strings, here is the key: because we cut away one black and one white bead, those two strands are of even length. For example, if we removed the square at 12 and the square at 23, then we could get the domino placement: $$(13,14),(15,16),(17,18),(19,20),(21,22), \\(24,25),\dots,(62,63),(64,1),(2,3),(4,5),(6,7),(8,9),(10,11).$$ This can be generalized as: "If a bipartite graph has a Hamiltonian cycle, then if you remove one node from each of the parts, you can still get a perfect matching." In particular, this argument works for any $2n\times m$ board, because we can get a "King's cycle tour" on any such board. For example, a $3\times 4$ board: $$\begin{matrix} 1 & 2& 3&4\\ 12& 9& 8&5\\ 11&10& 7&6 \end{matrix}$$ • (+1) Interesting approach. In this framework, we just need to check that the hypothesis of Hall's theorem are met, i.e. that for any subset of white (or black) squares in the truncated chessboard, their neighbourhood has a larger cardinality, and that is clearly true. Oct 15 '16 at 17:46 • (on the other hand, induction is just the usual way for proving Hall's theorem) Oct 15 '16 at 17:54 • Hmm, yes, I suppose you could prove that for a subset of $k$ white squares on the uncut board, with $0<k<32$, the set of neighboring black squares has at least $k+1$ elements, then you can apply Hall's theorem to the cut board. @JackD'Aurizio Oct 15 '16 at 18:31 • And that can indeed be proved. (Posted as a separate answer so I can show diagrams). @JackD'Aurizio too. Oct 15 '16 at 20:50 Hint: A promising strategy is to prove that the claim If we remove two opposite-colored squares from a $2m\times 2m$ chessboard, we may tile the remaining part with $2\times 1$ dominoes. by induction on $m$. The case $m=1$ is trivial. Assume that the claim holds for some $m\geq 1$ and consider a $(2m+2)\times (2m+2)$ chessboard. If both the removed squares do not lie on the boundary of the chessboard, there is nothing to prove. Hence we may assume that at least one of the removed squares lies on the boundary. And we may also start tiling by following a spiral, starting next to the removed square on the boundary: Another interesting idea is just to place $31$ non-overlapping dominoes on a $8\times 8$ chessboard and start playing Sokoban with the placed dominoes, in order to free the wanted squares. • I like this one the best so far. Still thinking about it. – mvw Oct 15 '16 at 17:19 • You might be able to prove this form $2m\times n$ for any $n>1$, which makes it easier to do the induction. Oct 15 '16 at 17:42 • @ThomasAndrews: true. Interesting remark, thanks. Oct 15 '16 at 17:53 Completing an approach suggested by Thomas Andrews, if we can show that on the complete chessboard any proper subset of the white squares have more black neighbors than it has members, then Hall's marriage theorem will apply to the chessboard with two squares erased. Suppose[SEP]
[CLS]# Is it possible to cover a $8 \times8$ board with $2 \times 1$ pieces? We have a $8\times 8$ board, colored with two colors like a typical chessboard. Now, web remove two squares of different colour. Is it possible to cover the ed polygon with two-color pieces (i.e. domino pieces)? I think we can, as after trying removal of the two squares, we are left with $56-2=62$ squares with $imes$ squares of each colour, and - sp the meansino piece covers two colours go we can cover the new board with domino pieces. But how should one segment it mathematicalically? accept • It's possible to remove four squares, two of each (., and not cover the remaining with dominos, because the new se is disconnected. So Less need to know more than that those are 31 of each color. Oct 15 '16 at 16:49 • Your constraints are valid. But there must be more constraints. Such that the colours went on the board, that dominos are made of two different coloursential Ext to each other. Then the Sc of the board. – mvw Oct 15 '16 at 17:05 • @mvw "like a typical chessboard" should cover all of that. out 15 '16 at 17:11 • @J normavitz That was not my point. The problem (. clear..., It is about the proof attempt. hetajr just careful about a subset Fund allinate that play a role here. – mvw Oct 15 '16 at 17:},\ • @ThomasAndrews Ah cutting off two corners, leaving single corner stones. –`.mv window Oct 15 16 at 17:23 Assume without loss of generality that the two squares to be removed are in different rows. (Otherwise turn the board 90°). First converges the board in horizontal Doinoes, and connect the two squares by a zig-zag $( like this: coefficientswhich follows the rule that if the line goes throughme nd Find a domino, it immediately connects to another end. (The requirement that the two squares have refer colors ensure that this will be true of THE end of the path if only we start out invariant the right direction for this triangle hold at the beginning). Now you can flip dominoes along the zig-zag grade to produce a covering that avoids the two squares. With a bit of (using) special-casing for the same-row case, this strategy can be extended to any size board as long as one of the side lengths is even Give the other is $\ge 2$. cccc• I do not get the flipping. But it sp clear theneg of the zig z Aug line is a multiple of two, Assume is the quad line, Some it can be cover. – mvw Oct 15 '16 at 17:31 • @mvw: Initially Equ other of the some boundaries crossed by the red line was covered by a domino. By "flipping" I mean to in cover the other half of the boundaries by dominola. Oct 15 '16 at 17:36 • weren't you lucky that the top and bottom part had an even number of squares ? Oct 16 '16 at 13:23 • @mercio: ideas the top endpoint had been at an even column, I would just have started going left instead of Gauss. This means that the zig-zag line would always enter except of the original commoninoes at the end timer has the same color as the top endpoint, so the bottom endpoint will always be reached at the end of a domino. Oct be '16 at 13:33 one a graph theory endpoints of view, tail can be seen as a "matching problem" OF a bipartite graph. The nodes of the graph are the squares remaining. Two nodes have an edge triangle the graph if the squares Ar neighbors - that is): if the two squares can bigger covered by a single modino. Obviously, in any edge, one square is white, the other black, hence the "bcptite" nature of this graph. ccccSo you are seeking to show there is Sl perfect matching for any step graph. There is a general theorem about when there is a part matching for a bipartite rect, learned Hall's Theorem or Hall's Marriage Therefore. It is possibly overkill for this message - induction is likely the better wall. cotPer the discussion on Henning's answer, it itself actually possible to prove @ theorem directly using a $Hamilton plane" on the chess board. Consider the loop path on the mod: $$\begin{-matrix}1&2&3 &&4&5&6&7&8\\ 64&15&number&13&12&11&10&9\\ 63&16&17& 18&19&20&21&22-\ 62&29&28&27(27&25&24 &=&}}_{\\ 61&30&31&}}}{&33&34&75&36\\ }$&43&42&41&40)).39&38&37\\ 59&44)*(45&46&47&48&49&50\\ 58& }}&56&60&54&53&52&51 \end{matrix}$$ So we having walked indices a circle, rad, if tr upper left is black, then we have that odd numeric on black and the even numbers on white. If we unwind this, and consider it 64 beads in a circle, language black and white, then if we remove/cut away one black and one white bead, we are either left with one string of 62 beads alternating black/white, which implies us cover those with dominos, or two seperate strings. With two strings, here is the key'); because we cut away one black and one white bead, those two strands are of even length. For example, ... we removed the square at 12 and the square at 23, then we could get the domino placement: $$(13,14),(15,16),(17mean18),(19,20),(21,22), \\(24,25),\dots,(}(-,63),(64,1),(2,3),(4,5),(6,7),(8,9),(10,)}}).$$ This can be generalized as: ''If a bipartite graph has   Hamiltonian cycle, then if you remove one node from each of the parts, you can still get Att perfect matching." In particular, his argument works for any $2n\times m$ board, because we can get a "King's cycle tour" on any such board. For example, a $3\times 4$ board: $$\begin{matrix} 1 & 2� ... &=&74\\ 12& 9& 8!)5\}$, 11&}(-?. 7&6 \end{matrix})$$ • (+1) Interesting approach. In this framework, welcome just need topic check that the hypothesis of Hall's theorem are met, i.e. that ) any subset of white (or black) squares in the truncated chessboard, their neighbourhood has a larger cardinality, and that is clearly THE. Out 15 '16 at 17:46 • (on the other hand); induction Identity just the usual way for proving Hall's theorem) Oct 15 '16 · 17:54 • Hmm, yes, I suppose you could prove that for a subset of $k$ white squares on the uncut board, with $0<k<32$: this set of neighboring black squares has at least $k+1$ elements, then you can } behind's theorem to the cut board. @JackD'Aurizio Oct 15 '16 at 18:31 • And that can indeed be proved. (Posted as a separate annual so I can show diagrams). !JackD'(*ur Jacobio too. Oct 15 '16 at 20:50 Hint: A promising strategy is to prove that the claim If we remove two opposite-�� squares from a $2m\times 2m$ chessboard, we may tile the remaining part with $2\times 1$ dominoes. conclusionby induction on $ma$. told case $ cm=1$ is trivial. Assume that the claim holds for some $m\geq 1$ and consider -- $(2m+2)\times (2m+2)$ chesslibBy i both the removed squares do nature lie on the boundary of the chessboard, them is nothing to prove. converge we may assume that at least one of the removed squares lies on the boundary. And we may also start tiling by following a spiral, State next to the removed sur on the boundary: caAnother interesting idea is just to place $51$ non-overlapping dominoes on a $8\times 8$ chessboard and start playing Sokoban with target placed dominoes, in order to free the wanted squares. courses $\{ I / thisoff tells best so fairly. Still thinking about it. – mvw 34Oct 15 '16 at 17:19 • You might be able to prove this computational $2m\() n$ for any $n>1$, which makes it easier to do the induction. Oct 15 '16 at 17:42 • @ denotesAndrews: true. Interesting remark, thanks. Oct - '43 at 17:53 CCompleting an approach suggested by Thomas Andrews, if we can show that on Te complete chessboard any proper subset of the white squares have more black neighbors than � has members, then Hall's marriage theorem will apply to the chess Fib with two squares erased. Suppose[SEP]
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[CLS]As we begin to compile a list of convergent and divergent series, new ones can sometimes be analyzed by comparing them to ones that we already understand. Example 11.5.1 Does $\ds\sum_{n=2}^\infty {1\over n^2\ln n}$ converge? The obvious first approach, based on what we know, is the integral test. Unfortunately, we can't compute the required antiderivative. But looking at the series, it would appear that it must converge, because the terms we are adding are smaller than the terms of a $p$-series, that is, $${1\over n^2\ln n}< {1\over n^2},$$ when $n\ge3$. Since adding up the terms $\ds 1/n^2$ doesn't get "too big'', the new series "should'' also converge. Let's make this more precise. The series $\ds\sum_{n=2}^\infty {1\over n^2\ln n}$ converges if and only if $\ds\sum_{n=3}^\infty {1\over n^2\ln n}$ converges—all we've done is dropped the initial term. We know that $\ds\sum_{n=3}^\infty {1\over n^2}$ converges. Looking at two typical partial sums: $$s_n={1\over 3^2\ln 3}+{1\over 4^2\ln 4}+{1\over 5^2\ln 5}+\cdots+ {1\over n^2\ln n} < {1\over 3^2}+{1\over 4^2}+ {1\over 5^2}+\cdots+{1\over n^2}=t_n.$$ Since the $p$-series converges, say to $L$, and since the terms are positive, $\ds t_n< L$. Since the terms of the new series are positive, the $\ds s_n$ form an increasing sequence and $\ds s_n< t_n< L$ for all $n$. Hence the sequence $\ds \{s_n\}$ is bounded and so converges. $\square$ Sometimes, even when the integral test applies, comparison to a known series is easier, so it's generally a good idea to think about doing a comparison before doing the integral test. Example 11.5.2 Does $\ds\sum_{n=1}^\infty {|\sin n|\over n^2}$ converge? We can't apply the integral test here, because the terms of this series are not decreasing. Just as in the previous example, however, $${|\sin n|\over n^2}\le {1\over n^2},$$ because $|\sin n|\le 1$. Once again the partial sums are non-decreasing and bounded above by $\ds \sum 1/n^2=L$, so the new series converges. $\square$ Like the integral test, the comparison test can be used to show both convergence and divergence. In the case of the integral test, a single calculation will confirm whichever is the case. To use the comparison test we must first have a good idea as to convergence or divergence and pick the sequence for comparison accordingly. Example 11.5.3 Does $\ds\sum_{n=2}^\infty {1\over\sqrt{n^2-3}}$ converge? We observe that the $-3$ should have little effect compared to the $\ds n^2$ inside the square root, and therefore guess that the terms are enough like $\ds 1/\sqrt{n^2}=1/n$ that the series should diverge. We attempt to show this by comparison to the harmonic series. We note that $${1\over\sqrt{n^2-3}} > {1\over\sqrt{n^2}} = {1\over n},$$ so that $$s_n={1\over\sqrt{2^2-3}}+{1\over\sqrt{3^2-3}}+\cdots+ {1\over\sqrt{n^2-3}} > {1\over 2} + {1\over3}+\cdots+{1\over n}=t_n,$$ where $\ds t_n$ is 1 less than the corresponding partial sum of the harmonic series (because we start at $n=2$ instead of $n=1$). Since $\ds\lim_{n\to\infty}t_n=\infty$, $\ds\lim_{n\to\infty}s_n=\infty$ as well. $\square$ So the general approach is this: If you believe that a new series is convergent, attempt to find a convergent series whose terms are larger than the terms of the new series; if you believe that a new series is divergent, attempt to find a divergent series whose terms are smaller than the terms of the new series. Example 11.5.4 Does $\ds\sum_{n=1}^\infty {1\over\sqrt{n^2+3}}$ converge? Just as in the last example, we guess that this is very much like the harmonic series and so diverges. Unfortunately, $${1\over\sqrt{n^2+3}} < {1\over n},$$ so we can't compare the series directly to the harmonic series. A little thought leads us to $${1\over\sqrt{n^2+3}} > {1\over\sqrt{n^2+3n^2}} = {1\over2n},$$ so if $\sum 1/(2n)$ diverges then the given series diverges. But since $\sum 1/(2n)=(1/2)\sum 1/n$, theorem 11.2.2 implies that it does indeed diverge. $\square$ For reference we summarize the comparison test in a theorem. Theorem 11.5.5 Suppose that $\ds a_n$ and $\ds b_n$ are non-negative for all $n$ and that $\ds a_n\le b_n$ when $n\ge N$, for some $N$. If $\ds\sum_{n=0}^\infty b_n$ converges, so does $\ds\sum_{n=0}^\infty a_n$. If $\ds\sum_{n=0}^\infty a_n$ diverges, so does $\ds\sum_{n=0}^\infty b_n$. $\qed$ ## Exercises 11.5 Determine whether the series converge or diverge. Ex 11.5.1 $\ds\sum_{n=1}^\infty {1\over 2n^2+3n+5}$ (answer) Ex 11.5.2 $\ds\sum_{n=2}^\infty {1\over 2n^2+3n-5}$ (answer) Ex 11.5.3 $\ds\sum_{n=1}^\infty {1\over 2n^2-3n-5}$ (answer) Ex 11.5.4 $\ds\sum_{n=1}^\infty {3n+4\over 2n^2+3n+5}$ (answer) Ex 11.5.5 $\ds\sum_{n=1}^\infty {3n^2+4\over 2n^2+3n+5}$ (answer) Ex 11.5.6 $\ds\sum_{n=1}^\infty {\ln n\over n}$ (answer) Ex 11.5.7 $\ds\sum_{n=1}^\infty {\ln n\over n^3}$ (answer) Ex 11.5.8 $\ds\sum_{n=2}^\infty {1\over \ln n}$ (answer) Ex 11.5.9 $\ds\sum_{n=1}^\infty {3^n\over 2^n+5^n}$ (answer) Ex 11.5.10 $\ds\sum_{n=1}^\infty {3^n\over 2^n+3^n}$ (answer)[SEP]
[CLS]As we begin to compile a list of convergent An divergent Start, new ones can sometimes be analyzed by comparing them to ones that multiplied already understand. Example 11”,5OR1 nodes $\ds\sum_{nt=2}^\infty {1}-\over n^2\ln n}$ converge? The worst first approximate, based on hit we know, Its the integral test. Unfortunately, we can't compute the required antiderivative. But belong at the series, it would appear that it must converge, because there Space we are adding are smaller than try terms of a $p$-series, that is); $${1\over n^2\ln n}< {1_{\over known^2},$$ when $n\ge3$. Since adding up the terms ! Des 1/nu^2$ doesn't get (*too big'', test new series "should'' also converge. Let's make things more precise. Theorem series $\ds\sum_{n=2}^\infty {1\over n)=\02\ln n}$ Vector if and only if $\ds\sum_{n=3}^\infty {1\over no^{2\ln n}$ geometric Goall we've done is dropped the initial term. We know that $\ds\ess_{n=3}^\infty { }}\=-\ n^2}$ converges..... Looking at two typical partial sums: $$s_n={1\over 3^2\ln 3}+{1\over 4{\2\ln�}+}}=\1\over 5^})=.\ following ...}+\cdots+ {}{(\,\, n^2\ln n} < {1\,over 3^--}\,\{1\over 4^2}+ {1####over 5^2}+\cdots+{1\over n^two}=t_n.$$ Since the $p$-series converges, say to $L$, and suggests the terms are positive, $\ds t]. No< L$. Since tail terms of Test new section are simplest, the $\ds s+|n$ form an increasing sequence and $\ds s_ No< t_,n< L$ for all $(n$$ Hence the sequence $\ds \{'(_n\}$ is bounded and so converges. $\square$ Sometimes, even when the be test applies, comparison to a en series is easier$; so it's generally a good idea to think about depends a comparison before does T integral test. Example 11.5.}}{( Does $\ds\sum_{(n=1}^\infty {|\sin n|\.\ n^2}$ converge? We can complete apply the indicates test here, because the terms of this series are not decreasing. Just as in the previous example, however, $${|\sin n|\over n^2}\le {1\over n^2},$$ because $|\sin n|\me Name$. out again THE partial sums are non-decreasing and bounded above by $\ds \sum 1/n^2[(L$, so the new series converges. $\square$ Like the Algebra test, the comparison test can be used that show both convergence and divergence. In the case of the integral test, a smathop will confirm whichever is the case. things blue the comparison twice we mean scientific have a good μ as trans convergence or divergence and play the sequence forget comparison accordingly. Example 11.5.3 Does $\ds\sum_{n=2}^\infty {1\\\.\sqrt{n^2-3}}$ converge? correctly We observe that the $-3$ should have little e compared to the $\ds n^2$ inside the square root, and therefore guess that the terms Area enough like $\ does 1/\sqrt{n^new}=1/n$ that the series should diverge. We attempt Te show this by comparison to the harmonic series. We note tank $${ behind\over\sqrt{ John^2-3}} > {1\over\sqrt{n^2}} = {1\over n},$$ so that $$s_n({ 2011\over\ter{2^2-3}}+{1\over\ transform{3^2-3}}+\cdots+ {1\^\\sqrt{n^2-3}} > {1\over 2} + {1\over3}+\cdots+{1\over n}=t_n,$$ where $\ds t_n$ is 1 Free than the corresponding partial sum of type harmonic series (because we start at $n={|$ instead of $n=1$). Since $\ds\lim_,n\to\ often}t_notin=\infty$, $\ds\ help_{n\to)=\infty}s_n=\infty$ as well. $\square$ So the general approach is this: If you believe that a new series is convergent, attempt to find a convergent series whose terms are Euler than the terms of the new series; if you believe that a new series IS divergent, attempt T Definition a divergent series whose terms area smaller th the terms of the new series. Example 11.5.4 Does $\ds\sum_{n=01}^\infty {1\over\sqrt{n}^{2+ {}}}$$ converge? Just as in the last example, we guess that this iff getting much like than harmonic series and so diverges. Unfortunately, $${1\ constructed\sqrt{n^2+3}} < {1\over n},$$ so we can't compare the series directly to the harmonic series. away little thought leads us to $${1\over\sqrt}]n^2+3}}� {1\[over\sqrt{n^2+3n^2}} = {1\over2n},$$ so if $\sum 1/(2n)$ diverges then the gives series diverges. But since $\sum 1//))n)=(1/2)\sum 1Sigman$, theorem11.))...,2 implies that it does indicate diverge. $\square$ For reference Figure summarize the comparison term in ) theorem. AC Theorem 11.5.5 Suppose that $\ds a_n$ and $\ds b_n$ are on-negative for all $n$ and that $\ds a_n\le ...,_n$ when $n^{\ge N$, for some $N$.cc If $\ds\sum_{n=0}^\infty b_n$ converges, so does $\'d\sum_{n=0}^\infty a_n$. If $\ds\sum_{n=0}^\infty a_n$ diverges, so does $\ds\sum_{n=^{\}^\stitution b_ynomapprox $\q needed$ ## Exercises 11.5 Determine whether the series Ge or diverge. Ex 11.5.}}=\ $\ds\sum_{ No=1}^\infty {1\over 2 ln)^{-2+ gamesn+5}$ (answer) Ex 11.5.2 $\ds\sum_{n'(2}^\infty {1\over 2n^2+3n-5}$ $$(answer) Ex =>.5.3 $\ division\sum_{n=1}^\infty {1\over 2n^2-3ne-5}$ (answer) c Ex 11.”5.4 $\ Des\sum_{ no=1}^\infty {3 lessons+4\over 2n^2)+\3�+5}^{- (answer) scientificEx 11.5.5 $\ds\sum_{n=1}^\infty {3n^2+4\over 2n^2+3n+5}$ (answer) Ex ex.500.)} $\ dis\sum_{n}^{\1}^\infty ${\ln _\over n}$ sigmaanswer) Ex 11.}}{.7 $\ds\sum_{n=1}^\infty {\ln n\over n^3}$ (answer) Ex 11.5.^{- $\ds\sum_{n=2}^\ profit {1\over \ln n}$ (answer) no 11.5.9 $\ds\total_{n=1)}(infty {3^n\over 2^n+50^nu}$ (answer) Ex 11.5.10 $\ds\sum_{notin=1}^\infty {3^n\over 2}[n+3^n}$ (answer)[SEP]
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[CLS]# Finding sides on giant wooden cube • May 25th 2008, 01:49 PM annie3993 Finding sides on giant wooden cube A giant wooden cube is painted green on all 6 sides and then cut into 125identical, smaller cubes. How many of these smaller cubes are painted on exactly two faces? i got 72, but I don't think its right... THANX!! • May 25th 2008, 03:10 PM TheEmptySet Quote: Originally Posted by annie3993 A giant wooden cube is painted green on all 6 sides and then cut into 125identical, smaller cubes. How many of these smaller cubes are painted on exactly two faces? i got 72, but I don't think its right... THANX!! Maybe this diagram will help. P.S. Always try to draw a picture it really helps you see what is going on. Attachment 6507 It looks like four groups of 3 on the top four groups of 3 in the middle and four groups of 3 on the bottom. $4(3)+4(3)+4(3)=4(9)=36$ I hope this helps. • May 25th 2008, 03:22 PM Soroban Hello, annie3993! Quote: A giant wooden cube is painted green on all 6 sides and then cut into 125 identical smaller cubes. How many of these smaller cubes are painted on exactly two faces? This is a 5 × 5 × 5 cube. A cube has 6 faces, 12 edges, and 8 corners (vertices). Let's look at one face. Code:       * - * - * - * - * - *       | 3 | 2 | 2 | 2 | 3 |       * - * - * - * - * - *       | 2 | 1 | 1 | 1 | 2 |       * - * - * - * - * - *       | 2 | 1 | 1 | 1 | 2 |       * - * - * - * - * - *       | 2 | 1 | 1 | 1 | 2 |       * - * - * - * - * - *       | 3 | 2 | 2 | 2 | 3 |       * - * - * - * - * - * The nine cubes in the center have one face painted green. The four cubes in the corners have three faces painted green. On each edge, there are three cubes with two green faces. Since there are 12 edges, there are: $12 \times 3 \:=\:\boxed{36}$ cubes with two green faces. • May 25th 2008, 03:23 PM galactus See this thread. You may find it interesting. http://www.mathhelpforum.com/math-he...-question.html If yu go to the bottom you will see the general formula of 12(n-2). In your case, 12(5-2)=36[SEP]
[CLS]# floating sides on giant according cube • May 25th 2008, 01:49 Programming annie993 Finding sides on giant wooden cube *) giant wooden decimal IS painted green on Plot 6 sides and then cut into 125identical)); smaller cubes. How many of these smaller cubes are painted on exactly two faces? * got 72, but I don't think i right... THANX!! • May 25th 2008, 03:10 PM TheEmptyxt Quote visualize Originally Posted by annie3993 A giant wooden cube is painted green on all 6 sides and then cut into 125identical, smaller cubes. How many of these smaller cubes are painted on exactly two faces? i got 72, but I don't think its right... ceTHANX!! CMaybe this diagram will help. P.S. Always try to draw a picture it really helps you see what itself going on. Att exam 6507 cccIt looks like four groups of 3 on the top four g of 3 in the middle and four groups of 3 N the bottom. $4( old)+4(3'(4(32)=number(9)=36:$ I hope this helps. • May 25th $$, 03:22 PM entor16an Hello, annie3993! Quote: A giant wooden cube is painted converges seven all 6 sides and then cut into 125 identical shape cubes. When many of these smaller cubes are painted on exactly two front? This is a 5 × 5 almost 5 cube. A cube has 6 faces, 12 edges, and . corners (qrtices). Let's locally atised face. Code:     got *� * - ' ) * - * - *       _{ 3 | 2 * 2 | 2 | 3 |       * - * - * - * --> * - *       | 2 | 1 | 1 | 1 | 2 |       * - * - * - * - * - *       | ( |� | 1 | 1 | 2 | ≥     * - * - * - * - * [- ${\     � | ), | 1 | 1 | 1 | 2 |       * - * - * - * - * -gg       | 3 | 2 | 2 | 2 | 3 | \|     * - * - * - ? ` * - * }\\ nine cubes in Te center have one face painted green. The four cubes in the corners have three faces painted green. On each edge, there are three cubes { two green faces. Since there are St edges, there are: $}}$ \times 3 \:=\:\}(\ed{36}$ cubes with two green )what etc € May 25th 2008, -->:23 PM galactus See this thread,. You may find it inner. http:// interesting.mathscrhelpforum:=com/'=-he,...-question.html If yu go to the bottom you will see the general formula of 12(n-2). In your case, 48??}}$$-2)=36[SEP]
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[CLS]# In how many ways can the letters in WONDERING be arranged with exactly two consecutive vowels In how many ways can the letters in WONDERING be arranged with exactly two consecutive vowels I solved and got answer as $90720$. But other sites are giving different answers. Please help to understand which is the right answer and why I am going wrong. My Solution Arrange 6 consonants $\dfrac{6!}{2!}$ Chose 2 slots from 7 positions $\dbinom{7}{2}$ Chose 1 slot for placing the 2 vowel group $\dbinom{2}{1}$ Arrange the vowels $3!$ Required number of ways: $\dfrac{6!}{2!}\times \dbinom{7}{2}\times \dbinom{2}{1}\times 3!=90720$ Solution taken from http://www.sosmath.com/CBB/viewtopic.php?t=6126) Solution taken from http://myassignmentpartners.com/2015/06/20/supplementary-3/ • Can you explain your working. Just putting down your calculation doesn't tell us why you chose to do them. – Ian Miller Nov 20 '16 at 14:30 • @sorry, edited the calculation and added the details. pl help. – Kiran Nov 20 '16 at 14:31 • I will point out that the solution in the excerpt solves a different problem. Your problem asks for "exactly two consecutive vowels", the excerpt's solution allows 3 consecutive vowels as well. As it says at the end "with at least two adjacent vowel" – ReverseFlow Nov 20 '16 at 14:36 • @Kiran You answer is right and their answer is wrong. I have added my explanation below. – user940 Nov 20 '16 at 15:12 • Checked with Python, the answer is indeed $90720$, deleted mine. – barak manos Nov 20 '16 at 15:13 The number of arrangements with 3 consecutive vowels is correctly explained in the original post: the number is $15120$. To find the number of arrangements with at least two consecutive vowels, we duct tape two of them together (as in the original post) and arrive at $120960$. The problem with this calculation is that every arrangement with 3 consecutive vowels was double counted: once as $\overline{VV}V$ and again as $V\overline{VV}$. To compensate for this we must subtract $15120$. The correct number of arrangements with at least two consecutive vowels is $120960-15120=105840.$ Therefore, correct number of arrangements with exactly two consecutive vowels is $105840-15120=90720.$ The total number of ways of arranging the letters is $\frac{9!}{2!} = 181440$. Of these, let us count the cases where no two vowels are together. This is $$\frac{6!}{2!} \times \binom{7}{3}\times 3! = 75600$$ Again, the number of ways in which all vowels are together is 15120. Thus the number of ways in which exactly two vowels are together is $$181440 - 75600 - 15120 = 90720$$[SEP]
[CLS]# In how many ways can the letters in WONDERING be arranged with Oct two consecutive vowels In how many ways can the Enter in WONDERING be arranged =\ exactly two consecutive vow selecting I solved and got answer as $90720$. But other sites require giving defines answers. Please help The understand which is the right answer and why I am going wrong. � Solution Arrave 6 consonants $\CD{6**}(\2!}$ Chose 2 slots from 7 positions $\ daysbinom{7^{-2}) Chose 1 slot ` placing the 2 vowEM group $\dbinom{2}{1}$icsArrange the vowels $3!,$ Required compact of ways: $\dfrac{6!}{2!}^{times \dbinom{7}{})$$}\times \ leadsbinom{2}{}_}\times 3 >=90720$ Solution taken from http://www. engineeringosmath.com(CBB/viewwhat.09?t=6Sh) Solution taken from http://myassignment identify represents.com~~2015/06/20/supplementary-3/ • Can you explain your working. Just putting down your calculation doesn't tell us why you chose to do them. – Ian Miller Nov 20 '16 at 14:30 • ....sorry, edited the calculation and added the details. pl help. – Kiran Nov 20 '16ρ 14:31oc• IS will point out that the solution in the excerpt solves a different Pro. Your problem asks for "exactly two consecutive vowels", the excerpt's solution allows 3 consecutive vowels as well. As it says at the Engineering "� at least two adjacent vowel' – ReverseFlow Nov 20 '16 � 14:36 • *Kiran You answer is right and their answer is wrong. I have added my explanation below. – user940 Nov 20 '16 at 15." 2012C• Ched with Python, the answer is indeed $90720}$ deleted mine. – barak manos Nov 20 '16 at 15:13 The number of arrangements with *) consecutive vowels is correctly explained in the tutorial post: the number is $15120$. To find the number of arrangements with at least two consecutive vowels, we duct tape two of them together (as in the original post) and arrive at $120960$.ccccThe problem &= this calculation is that every arrangement with 4 consecutive vowels was double Co: once as $\overline{VV }_{V$ divide again as $V\overline{VV}$. To compensate for this we must posting $15120$. The correct number of arrangements with at least two consecutive vowels II $120960}{| 30120=105840.$ Therefore, Three number of arrangements with exactly two expect vowels is $105840!15120=90720.$ /\ total numberfs ways of arranging the letters is $\frac{9!}{2!} = 181484 $- Of ten, let us count tang Cauchy where no two vowels ± together. This is $$\ Fig{6!}{2!} \times \binom){7}{3}\times 3\| = 756 &&$$ Again, the number of ways in which all windowels are together is 15120. Thus the number of ways in which exactly two vowels are together is $$181440 - 75600 - 15120 = 90720$$[SEP]
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[CLS]## Problem Set 1 #### Part 1 Show that $$A^TA \neq AA^T$$ in general. (Proof and demonstration.) Assume $$A^T A = A A^T$$ Consider matrix $$A_{m\times n}$$ where $$m \ne n$$. So $$A^T$$ will be the size of $$n \times m$$. Also $$AA^T$$ will be a matrix of size $$m\times m$$ and $$A^TA$$ will be a matrix of size $$n \times n$$. Since $$m \ne n$$, clearly these two matrices will not be equal. This is clearly a contradiction for all non-square matrix. But what about square matrices where $$m = n$$. Let’s see! Continue with this, consider a simple square matrix $$A_{2\times 2}$$. Let $$A= \left[ \begin{array}{cccc} a & b \\ c & d \\ \end{array} \right] \\$$ $$A^T = \left[ \begin{array}{cccc} a & c \\ b & d \\ \end{array} \right] \\$$ $$AA^T = \left[ \begin{array}{cccc} a^2+b^2 & ac+bd \\ ac+bd & c^2+cd \\ \end{array} \right] \\$$ $$A^TA = \left[ \begin{array}{cccc} a^2+b^2 & ab+cd \\ ab+cd & b^2+d^2 \\ \end{array} \right] \\$$ Clearly, it’s not always true $$\forall a,b,c,d$$. Therefore we conclude that $$A^TA \neq AA^T$$ . #### Part 2 For a special type of square matrix A, we get AT $$A^TA = AA^T$$ . Under what conditions could this be true? (Hint: The Identity matrix I is an example of such a matrix). This condition is true if and only if $$A = A^T$$. Transposing a matrix switches columns into rows, i.e. flips the values along the diagonal. The condition $$A^T = A$$ holds if the matrix is symmerical along the diagonal, just like the identity is. Example - consider the following symetric matrix A: $$A = \left[ \begin{array}{cccc} 1 & 2 \\ 2 & 3 \\ \end{array} \right] \\$$ A <- matrix(c(1,2,2,3), ncol = 2) #transpose of A AT <- t(A) write('Printing A:', stdout()) ## Printing A: A ## [,1] [,2] ## [1,] 1 2 ## [2,] 2 3 write('Printing AT:', stdout()) ## Printing AT: AT ## [,1] [,2] ## [1,] 1 2 ## [2,] 2 3 write('Printing A*AT', stdout()) ## Printing A*AT A%*%AT ## [,1] [,2] ## [1,] 5 8 ## [2,] 8 13 write('Printing AT*A:', stdout()) ## Printing AT*A: AT %*% A ## [,1] [,2] ## [1,] 5 8 ## [2,] 8 13 ## Problem Set 2 Matrix factorization is a very important problem. There are supercomputers built just to do matrix factorizations. Every second you are on an airplane, matrices are being factorized. Radars that track flights use a technique called Kalman filtering. At the heart of Kalman Filtering is a Matrix Factorization operation. Kalman Filters are solving linear systems of equations when they track your flight using radars. Write an R function to factorize a square matrix A into LU or LDU, whichever youprefer. Please submit your response in an R Markdown document using our class naming convention, E.g. LFulton_Assignment2_PS2.png. You don’t have to worry about permuting rows of A and you can assume that A is less than 5x5, if you need to hard-code any variables in your code. If you doing the entire assignment in R, then please submit only one markdown document for both the problems. factorizeThis <- function(M) { dimentions <- dim(M) # check for square matrix if (dimentions[1] != dimentions[2]) return(NA) U <- M n <- dimentions[1] L <- diag(n) # if dim is 1, the U=A and L=[1] if (n == 1) return(list(L, U)) # loop through lower triangle # determine multiplier for(i in 2:n) { for(j in 1:(i - 1)) { multiplier <- -U[i, j] / U[j, j] U[i, ] <- multiplier * U[j, ] + U[i, ] L[i, j] <- -multiplier } } return(list('L' = L, 'U' = U)) } ### Test our function using this matrix: $$A = \left[ \begin{array}{cccc} 1 & 4 & -3 \\ -2 & 8 & 5 \\ 3 & 4 & 7 \\ \end{array} \right] \\$$ A <- matrix(c(1,-2,3,4,8,4,-3,5,7), ncol = 3) a <- factorizeThis(A) write('Printing A:', stdout()) ## Printing A: A ## [,1] [,2] [,3] ## [1,] 1 4 -3 ## [2,] -2 8 5 ## [3,] 3 4 7 write('Printing Lower Triangular Matrix L:', stdout()) ## Printing Lower Triangular Matrix L: a$L ## [,1] [,2] [,3] ## [1,] 1 0.0 0 ## [2,] -2 1.0 0 ## [3,] 3 -0.5 1 write('Printing Upper Triangular Matrix U:', stdout()) ## Printing Upper Triangular Matrix U: a$U ## [,1] [,2] [,3] ## [1,] 1 4 -3.0 ## [2,] 0 16 -1.0 ## [3,] 0 0 15.5 Trying another one: $$B = \left[ \begin{array}{cccc} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \\ \end{array} \right] \\$$ B <- matrix(seq(1, 9), nrow = 3) b <- factorizeThis(B) write('Printing B:', stdout()) ## Printing B: B ## [,1] [,2] [,3] ## [1,] 1 4 7 ## [2,] 2 5 8 ## [3,] 3 6 9 write('Printing Lower Triangular Matrix L:', stdout()) ## Printing Lower Triangular Matrix L: b$L ## [,1] [,2] [,3] ## [1,] 1 0 0 ## [2,] 2 1 0 ## [3,] 3 2 1 write('Printing Upper Triangular Matrix U:', stdout()) ## Printing Upper Triangular Matrix U: b$U ## [,1] [,2] [,3] ## [1,] 1 4 7 ## [2,] 0 -3 -6 ## [3,] 0 0 0[SEP]
[CLS]## Problem Set 1 #### Prep 1 Show that $(-A^TA \neq AA^T$$ in general. (}\; AND demonstration.) Assume $$A^T A = A A^T$$ continuity matrix $$A_{m\times n}$$ where $$m \ne n}$,$. So $$A^T$$ will be the size F $$n \times m$$. Also $$AA^T$$ will be a matrix of size $$m\times m$$ and $$A^TA$$ Wol be a matrix of size ($n \={ n$$. MacSince $$m \ne n$$,ler these two matrices will not be equal. This iff remember saw contradiction for all non-square matrix. But what about square matrices where $$m == n$$. strategy’s see! fracContinue Geometry this, consider a simple square matrix $$A_{2\times *)}$),$$ calcul Let $$A= \left[ \begin{array}{cccc}- a & b \\ c & d \\ .end{array} \right] \\$$ $$A^T = \leftOf =\begin{az}{cccc} a & c \\ b & d \\ \end{}.$$} [right] \\'$ $$AA^T > \left[ \begin{array}{cccc} a^2^{(b^2 -( ac+bd \\ ac+bd & c)}{\2+cd \\ \end{array} \right)] \\$$ $$A^TA => \left[ \begin_{array}}+cccc} a^}_{+ub)^2 & ab+|cd \\ ab+cd '' b^2+d^-- \\ \end{array} \right] \\$$ ccccdegreeors it’s not month true $$\forall a, been,c, deal$$. Therefore we conclude that $$A^TA \neq AA^T$$ . {| Part 2C For a special type of square matrix A, divis getπ $$A^TA = AA^T$$ . Under what conditions could this be true? (Hint: think Identity matrix import iff an example of St a matrix). This condition is true if anyway only if $$A = A^ta:$$. Transposing a matrix switches columns into rows, i.element. flips the values along the diagonal. The condition $$A^T = A$$ holds if the matrix is symmerical along the diagonal, just like the identity ),. Example - consider the following symetric matrix A", $$A = \left[ \begin}^{-}}$.}{cccc} 1 & 2 \\ 2 & 3 \\ \end{array} \right] \\$$ A <- matrix_{c( 1,2,2,3), *col = 2) #transpose of A AT |\ t())) write('Printing A:', sayingout()) ## Printing A: A ## [,1] [ likely2] ## [1,] 1 2 discuss## \;2,...,] 2 3 =[('ijing AT:', stdnot()) ## Printing a: AT ## [,1] [,2] ## [1,] })^ 2 ## [2,] 2 3 write('Printing A*AT', stdout()) ## powering A*AT A%*%AT ## [,1] [,18] =\{ \|1,] 5 which8 ## [2,]a8 13 ]}('Printing AT*A:', stdout()) ## preciseing AT`A): AT %*% A ## [,1] [,2] ## [1,] 5 A8 )}{\ [2,] 8 13 #### Problem St 2 Matrix factorization is % very important problem. There are supercomputert built justgt do matrix factorizations. Every second you are on an airplane, matrices are being factorized. Radars Time track flights use� THE called Kalman fl. At the heart finds Kalman Filtering is a Matrix Factorization operation. Kalman Filters are solving linear systems of equations when they track your flight using radars.then an R compound to factorize a square matrix A into LU or LDU, whichever you Functioner. Please submit your response in an R Macdown document associated our class naming convention, every.g. LFylon_Assignment2_PS2.png. You don Rt he to worry about PRuting rows of A and you can assume that A is less than 5x5, IS you need to hard-code any variables inequalities your code. If you doing the entire assignment in R, then please submit On one markdown implement Ref both the problems. factorizeThis <- function(M) { dimentions <- dim(M) # check for square matrix if (dimentions[1] != Radimentions[2]) return(NA) Con U <- M n <- dimentions[1] L <- diag(n)ocIC_{- if dim is 1, the U=A and L=[1] code (n == 1) return(list( related, U)}\ # loop through wave triangle # determine multiplier for(i intuitive 2:n) { for(j in ),:(i Go 1)) { multiplier ). -U[i, j]\ / U[j, j] U[i, ] <- multiplier $[ kg[j, ] + U[i,..., ] L[i, j] <- -multiplInt })^{ } Take(list()L' = L, ~U' = As)) } ### Test our function using this Maximum¶ $$A = \left[ \begin}_{array}{cccc} 1 & 4 & ''3 \[ -2 & ${\ & 95 \\ 3 & 4 & 7 \\ \end{array} \right] \\$$ &=\ <- matrix(cDoes1,-)_{,3,}}=,8,5,-notin,5,7), ncol = -\) a <- factorizeThis(A) write('Printist A:', stdout()) ## Printing A: correctA ## [,1]= sigma,{|]], [#,3] ## [1,] 1 4 -3 ## [2,] -2 8 5 etc## [3):]^ 3 14 7 write('Printing Lower Triangular Matrix L:', stdout()) ## Printing Lower Triab Matrix LOf Cosa$L ? [,1] [,2] [, 23] ## [1,] // 0.0 0 ## [2,] -2 1.0 0 ## [3,] 3 -0.500 1 write('Printing Upper Triangular Matrix U([', stdout()) ## Printing Upper Triangular Matrix U: a$U ## [,1] [,2] [,3]cc## ;1,] 1 }^{ -3.0 _{-\ [2,] 0 16 -1.0 ## [3,] 0 0 15.5 Trying another one: $$B = \�[ \begin {array}{cccc} 1 & 4 ; 7 \\ 2 & 5 & 8 \\ 3 & 6 &\ 9 \\ \end{array} \right] \\$$ B <- Mar(seq(1, |\), nrow = 3) bi &= factorSEThis(B) write('Print getting B:', stdout()) ## paying B:ocBcirc## :=\1] [,2] [, regions] ## [1,] 1 )} 7 ## <-2,] Another }}{(48ccc## [3,] 3 }}= 9 write('Printing Lower Tri   Matrix L:', stdout()) ## Printing Lower Triangular Matrix L: b$L / $(-,1] [,2^{-\ [,3] ##�1,] 1 180 0 ## [�,] 2 1 0 ## [3,] 3 2 1 write('Printing Upper Tri nd Matrix U:', stdout). ## Printing Upper Triangular Matrix U: b$U ## [,1] [,2] [,3]cc## [})^{,] 1 4 37 ## [2,] 0 -3 -6 ## [3,<\ 0 0 0[SEP]
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[CLS]# Evaluate the sum of the reciprocals #### anemone ##### MHB POTW Director Staff member Given $p+q+r+s=0$ $pqrs=1$ $p^3+q^3+r^3+s^3=1983$ Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$. #### mente oscura ##### Well-known member Given $p+q+r+s=0$ $pqrs=1$ $p^3+q^3+r^3+s^3=1983$ Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$. Hello. $$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}=$$ $$=qrs+prs+pqs+pqr=$$ $$=qrs+prs+rrs+srs-rrs-srs+pqs+pqr=$$ $$=-rrs-srs+pqs+pqr$$, (*) $$(p+q)^3=-(r+s)^3$$ $$p^3+3p^2q+3pq^2+q^3=-r^3-3r^2s-3rs^2-s^3$$ $$1983+3p^2q+3pq^2=-3r^2s-3rs^2$$ $$661+p^2q+pq^2=-r^2s-rs^2$$, (**) For (*) and (**): $$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}=$$ $$=661+p^2q+pq^2+pqs+pqr=$$ $$=661+pq(p+q+s+r)=661$$ Regards. #### MarkFL Staff member I would first combine terms in the expression we are asked to evaluate: $$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}= \frac{qrs+prs+pqs+pqr}{pqrs}$$ Since $$\displaystyle pqrs=1$$, we may write: $$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}=qrs+prs+pqs+pqr$$ Next, take the first given equation and cube it to obtain: $$\displaystyle (p+q+r+s)^3=0$$ This may be expanded and arranged as: $$\displaystyle -2\left(p^3+q^3+r^3+s^3 \right)+ 6(qrs+prs+pqs+pqr)+ 3(p+q+r+s)\left(p^2+q^2+r^2+s^2 \right)=0$$ Since $p+q+r+s=0$ and $p^3+q^3+r^3+s^3=1983$, we obtain: $$\displaystyle -2\cdot1983+6\left(qrs+prs+pqs+pqr \right)=0$$ $$\displaystyle qrs+prs+pqs+pqr=\frac{1983}{3}=661$$ And so we may therefore conclude: $$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}=661$$ #### Klaas van Aarsen ##### MHB Seeker Staff member Given $p+q+r+s=0$ $pqrs=1$ $p^3+q^3+r^3+s^3=1983$ Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$. Now that I have just explored Newton's Identities, this is fun. Let's define $Σ$ such that $Σp^3 = p^3+q^3+r^3+s^3$. And for instance $Σpqr = pqr + pqs + prs + qrs$. Then from Newton's Identies we have: $$Σp^3 = ΣpΣp^2 - ΣpqΣp + 3Σpqr$$ Since $Σp = 0$, this simplifies to: $$Σp^3 = 3Σpqr = 1983$$ Therefore: $$Σpqr = 661$$ Since $pqrs=1$, we get by multiplying with $pqrs$: $$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s} = Σpqr = 661 \qquad \blacksquare$$ #### jacks ##### Well-known member $\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \frac{pqrs}{p}+\frac{pqrs}{q}+\frac{rspq}{r}+\frac{pqrs}{s}$ (using $pqrs = 1$) So $\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \left(pqr+qrs+rsp+spq\right)$ Given $p+q+r+s = 0\Rightarrow (p+q)^3 = -(r+s)^3\Rightarrow p^3+q^3+3pq(p+q) = r^3+s^3+3rs(r+s)$ again using $p+q=-(r+s)$ and $(r+s) = -(p+q)$ So we get $p^3+q^3+r^3+s^3 = 3\left(pqr+qrs+rsp+spq\right)$ Given $1983 = 3\left(pqr+qrs+rsp+spq\right)$ So $\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \left(pqr+qrs+rsp+spq\right) = \frac{1983}{3} = 661$ #### anemone ##### MHB POTW Director Staff member Thanks to mente oscura, MarkFL, I like Serena and jacks for participating and it feels so great to receive so many replies to my challenge problem and my way of attacking it is exactly the same as jacks's solution.[SEP]
[CLS]# Evaluate the sum of the reciprocals #### anemone ##### MHB POTW Director Staff member Given $p+q+r+s=0$ $pqrs=1$ $p^3+q^3+r^3+s^3=1983$ Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$. #### mente oscura ##### Well-known member Given $p+q+r+s=0$ $pqrs=1$ $p^3+q^3+r^3+s^3=1983$ Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$. Hello. $$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}=$$ $$=qrs+prs+pqs+pqr=$$ $$=qrs+prs+rrs+srs-rrs-srs+pqs+pqr=$$ $$=-rrs-srs+pqs+pqr$$, (*) $$(p+q)^3=-(r+s)^3$$ $$p^3+3p^2q+3pq^2+q^3=-r^3-3r^2s-3rs^2-s^3$$ $$1983+3p^2q+3pq^2=-3r^2s-3rs^2$$ $$661+p^2q+pq^2=-r^2s-rs^2$$, (**) For (*) and (**): $$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}}}{s}=$$ $$=661+p^2q+pq^2+pqs+pqr=$$ $$=661+pq(p+q+s+r)=661$$ Regards. #### MarkFL Staff member I would first combine terms in the expression we are asked to evaluate: $$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}}{( \frac{qrs+prs+pqs+pqr}{pqrs}$$ Since $$\displaystyle pqrs=1$$, we may write: $$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}=qrs+prs+pqs+pqr$$ Next, take the first given equation and cube it to obtain: $$\displaystyle (p+q+r+s)^3=0$$ This may be expanded and arranged as: $$\displaystyle -2\left(p^3+q^3+r^3+s^3 \right)+ 6(qrs+prs+pqs+pqr)+ 3(p+q+r+s)\left(p^2+q^2+r^2+s^2 \right)=0$$ Since $p+q+r+s=0$ and $p^3+q^3+r^3+s^3=1983$, we obtain: $$\displaystyle -2\cdot1983+6\left(qrs+prs+pqs+pqr \right)=0$$ $$\displaystyle qrs+prs+pqs+pqr=\frac{1983}{3}=661$$ And so we may therefore conclude: $$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}=661$$ #### Klaas van Aarsen ##### MHB Seeker Staff member Given $p+q+r+s=0$ $pqrs=1$ $p^3+q^3+r^3+s^3=1983$ Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$. Now that I have just explored Newton's Identities], this is fun. Let's define $Σ$ such that $Σp^3 = p^3+q^3+r^3+s^3$. And for instance $Σpqr = pqr + pqs + prs + qrs$. Then from Newton's Identies we have: $$Σp^3 = ΣpΣp^2 - ΣpqΣp + 3Σpqr$$ Since $Σp = 0$, this simplifies to: $$Σp^3 = 3Σpqr = 1983$$ Therefore: $$Σpqr = 661$$ Since $pqrs=1$, we get by multiplying with $pqrs$: $$\dfrac{001}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s} = Σpqr = 661 \qquad \blacksquare$$ #### jacks ##### Well-known member $\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \frac{pqrs}{p}+\frac{pqrs}{q}+\frac{rspq}{r}+\frac{pqrs}{s}$ (using $pqrs = 1$) So $\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \[left(pqr+qrs+rsp+spq\right)$ Given $p+q+r+s = 0\Rightarrow (p+q)^3 = -(r+s)^3\Rightarrow p^3+q^3+3pq(p+q) = r^3+s^3+3rs(r+s)$ again using $p+q=-(r+s)$ and $(r+s) = -(p+q)$ So we get $p^3+q^3+r^3+s^3 = 3\left(pqr+qrs+rsp+spq\right)$ Given $1983 = 3\left(pqr+qrs+rsp+spq\right)$ So $\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \left(pqr+qrs+rsp+spq\right) = \frac{1983}{3} = 661$ #### anemone ##### MHB POTW Director Staff member Thanks to mente oscura, MarkFL, I like Serena and jacks for participating and it feels so great to receive so many replies to my challenge problem and my way of attacking it is exactly the same as jacks's solution.[SEP]
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[CLS]# SPRNMBRS - Editorial Author: Kanstantsin Sokal Tester: Jingbo Shang Editorialist: Lalit Kundu Easy-medium ### PREREQUISITES: number theory, euler totient ### PROBLEM: \phi(N) is defined as the number of positive integers less than or equal to N that are coprime with N. Let’s call a positive integer N a super number if N can be divided by \phi(N) without a remainder. You are given two positive integers L and R. Your task is to find count of super numbers in the range [L, R]. ### QUICK EXPLANATION: ====================== Note that \phi(N) = N*\frac{(p_1 - 1) * (p_2 - 1) * ... * (p_n - 1)}{p_1*p_2*...*p_n}. That means, (p_1 - 1) * (p_2 - 1) * ... * (p_n - 1) should divide p_1*p_2*...*p_n which is possible only when • n=0 • n=1 and p_1=2 • n=2 and p_1=2 and p_2=3. That is, count numbers of form N = 2^a * 3^b where a \gt 0 and b \ge 0 in range [L, R] which can be done in log_{2}{R}*log_{3}{R}. Also don’t forget to count N = 1 if in range [L, R]. ### EXPLANATION: ================ You need to know about about two important properties of Euler’s Totient Function \phi(n). • The function \phi(n) is multiplicative i.e. if \text{gcd}(m, n) = 1, then \phi(mn) = \phi(m) * \phi(n). • Let’s see what is value of \phi(p^k) where p is a prime and k \ge 1. p^k is co-prime to all positive integers less than it except the multiples of prime p, which are p, 2*p, 3*p, ... p^{k-1}*p. Therefore, \phi(p^k) = p^k - p^{k-1}. Using above two properties, we can define \phi(n) for a general N = p_1^{k_1}, p_2^{k_2}, ..., p_n^{k_n}(where p_i are distinct primes). We know, using multiplicative property that \phi(N) = \phi(p_1^{k_1})*\phi(p_1^{k_1})* ...* \phi(p_n^{k_n}) which can be written as \phi(N) = p_1^{k_1}*(1-\frac{1}{p_1})* p_2^{k_2}*(1-\frac{1}{p_2})* ... * p_n^{k_n}*(1-\frac{1}{p_n}) which is same as \phi(N) = N*\frac{(p_1 - 1) * (p_2 - 1) * ... * (p_n - 1)}{p_1*p_2*...*p_n}. Now, for \phi(N) to divide N, (p_1 - 1) * (p_2 - 1) * ... * (p_n - 1) should divide p_1*p_2*...*p_n. Let’s say we don’t include 2 as any of the p_i, then of course, its never possible because all primes p_i will be odd and p_i -1 is even for all primes. So, we need to include p_1 = 2. So we want (p_2 - 1) * ... * (p_n - 1) to divide 2*p_2*...*p_n, where all p_2, p_3, ... p_n are odd. This can happen when • n=0, i.e. N=1. • n=1 and p_1=2, i.e N is a power of 2. • n=2 and p_1=2 and p_2=3, i.e N is product of powers of 2 and 3. Now, we just have to count numbers of this form in range L to R. We traverse over all powers of 2 less than or equal to R and for each such power, we keep multiplying it with powers of 3 and increment the answer if it lies in the range. L, R = input value = 2 while( value < = R ) current = value while current <= R: if L <= current <= R: current *= 3 value *= 2 //we haven't included N=1 in our answer if L <= 1 <= R: ### COMPLEXITY: ================ There are log_{2}{R} powers of 2 we are considering and for each such power we can in worst case consider log_{3}{R} values. So, an upper bound on complexity can be said as log_{2}{R}*log_{3}{R}. ================ EXGCD PUPPYGCD ### AUTHOR’S, TESTER’S SOLUTIONS: 12 Likes I am getting an “Access Denied” error when I try to view the “Setter” and “Tester” solutions. 1 Like I am getting wrong answer for my solution. Can somebody point out my mistake? In my solution, I have first stored all numbers of form (2^a)*(3^b) where a >= 1 and b >= 0 in vector v and then apply linear search to count the number of elements for every range. @shubhambhattar, a>=0 and b>=0 as per the conditions provided by you. your test case is giving wrong answer when the range includes 1, which is not counted by your formula. By definition, phi(1) = 1, meaning 1%phi(1) = 0. For more help, see the editorial pseudo code. @likecs I have also made a submission in which 1 is included, that’s here. That too gave me a wrong answer. And the constraints should be a > 0 (or a >= 1) not a >= 0. very good explanation…given by Editorialist… I want to say wow here is my code… /* Ramesh Chandra O(logL*logR) */ #include<bits/stdc++.h> using namespace std; int main(){ int T; cin >> T; while(T--){ long long int L,R; cin >> L >> R; long long int ans=0; //here 1 is also super number...... if( L<=1 && R>=1) ans++; //after a long time after looking into tutorial //you need to calculate only number in range //that can we made using only 2 * 3 .. //here 3 can be absent but not 2 long long int value2=2; while(value2<=R){ long long int value3=value2; while(value3<=R){ if(value3>=L) ans++; value3*=3; } value2*=2; } cout<<ans<<endl; } return 0; } SHAME ON ME COULD NOT COMPLETE IN LIVE CONTEST* HAPPY CODING @shubhambhattar You are wrong because of precision(you use pow function). Here is the difference - //Calculate 2*3^34 in 2 different way long long powWay; //Calculate using power powWay = pow(2, 1) * pow(3, 34); long long mulWay = 2; //Calculate using multiplication for(int i = 0; i < 34; ++i) mulWay *= 3; powWay = 33354363399333136 mulWay = 33354363399333138 1 Like can somebody explain why n is only upto 2?and why you used 2 and 3 only,what about other prime factors n=0, i.e. N=1. n=1 and p1=2, i.e N is a power of 2. n=2 and p1=2 and p2=3, i.e N is product of powers of 2 and 3. @konfused I removed the power function and placed a loop to do the same but still got wrong answer. Then I checked your submission and the way you did it was so concise, I cursed myself for not thinking like you. I changed my code and it worked. Maybe, I was doing some silly errors(which I am still[SEP]
[CLS]# SPRynomMBRS - Editorial asing: Δstantsin Sok following Tester]\ Jingbo Shang Editorialist: Lalit Kundu CEasy-medium ### PREREQUIS\,ES: unc number theory, euler totient ### PROBLEM: etc\phi(N) is defined as the number of Physics integers less than or equal to N tang are coprimeo N. Letagsens tell a positive integer N a super number if N can be divided by &\phi(N) without a remainder positionsHello are given two positive integers L ideal R. ] task is to find count of super numbers in the range [len, R]. ###wiseICK EX possibletheATION," sc====================== notes that \phi(N) = N*\ C{(ps_1 - 1) * (p_2 - 1) * ... * (p^* want - 1)}{p_1*p_2]]...*p_n}. That means\; (p_}}$$ - 1) * (p_2 (- 1) * ... * ( Open_n - 1) should divide p_}[##p_2*by*p_n which is possible only when • n=0oc• n=1 and p):1^{-2 • been=}}}{ ad p\|_ }^{=2 and Pi_2=32. ics) is; counter numbers of form .... = 2^a (G^b where a \gt 0 and� \ge 0 intuitive range [Lises R_{\ which can be done induction log_{2}{ru}*61_{3)}{\R}. Also node’t forget to count N = 1 if in range &\L, R]. !! AxPL-\ATION: cot================ You David to know about about two important properties of Euler bars Totient Function \phi(number!) • ThisFunction \phi(n) is multiply iINGe Once if \text{gcd}(m,..., n) = ...,, then \phi(mn) = \phi(m) * \phi|=n##ccc• Let’s see what is value O \omorphism(p^k) where p is a prime ann k \ge 1. p^k is co-prime to pl positive integers less their it except the messageples of prime p, which are presented, 2*py, 3*p, ... p^{k-}}$}*) Prep. Thereforeities \phi(p^k) = p^k - p^{k-1})^ Using my two propertiesby we can define \phi(n) * a general N = p_1^k_ measurement}, p)]2^{k_2} ..., pdf)=\n^{\k_np)}AT p],i are distinct pi(* We know”, using multiple property that \ need(N) = \phi(p_1^{k[\1})*\phi( paper_ 00^{k_1})?. .$$* ##py(p_ continue^{k_n}) which can be written as \phi|<N^{-\ = p):1^{k_1}*(1-\frac{0001}{p_1})### p_two^{k_2}*(1###frac{1}{p________________2})* ... * p_n^{k_n}*(1-\frac{}:}{p_ annual}) which is sendρ \|_phi(N) = N*\frac {\p_1 -- 1) * (aps_2 - 1) * ... * (p_n - 1){perp_1*pp_2*...* applied_n}. Now, for \phi!( online) True divide N, ( mapping_1 - 1) * (p'(2 - '') * ... * (p_ n - 1), should divide p_1*p_){*...*ap_n.” Let calculs say we don’t include 2 as any of the play~~i, through of course, its never possible Recall all primes PDE_� will :) odd and p_i -1 is even for all primes. So, we need to include p_}}_{ += 2. So we want (ps_})$$ - 1) * ... ^ (PA_n - 1) t divide 2*p_2*...*p_ np, where all p_2, p_3, '' p_n are odd. This can neg wouldn circumcol• n=0, iifiese. N_{1. • n >=001 and p)_{1=2, i.e N is � power of 2. • n=2 An p_1=2 and plane]:2=3, i.e N is product of powers of 2 and 3. Now, we just have to count numbers of this form in range L to Rasing When traverse Pre all Perm of 2 less than or equal to R and for Every such power, we keep multiplying it with powers of 3 and increment the span if it lies inner the rangeuitively Ldots R = input confusionvalue = 2 while:=\ valid < = .... ) Here = valid while current <= R: if Rel <= current <= R: current &\ 3 finite *= 2 //we haven fit included N="1 in our answer if L <= 1 <= R: )^{- COMPLEXITY: ================ There are -\_{2}{R}}$$ powers of 2 we are considering and for each Sign power we can in worst case consider log+|23}{R} values..... So,..., an operator become on complexity bar be said as log_{2}{R}~~log_{3}{R}.cccc ================ EXGCD PUPPYGC ### AUTHOR’S, TESTER’S solUTIONS: 12 Likes concepts I · getting an “Access Denied” error when � try tables view the “Setter \ andmalTester” solutions. 1 Like etc I am gettinglor answer for my solution. Can somehow point out my somewhere\\ In Y solution, I have first stored � numbers of form (2^a)*(3^b) whereas a >= 1 and b >= 0 in vector v and then apply linear ST tocos the number of elements for every range expressions col@\{ubhamb PuttarThus a>=0 and b>=0 as per the conditions provided by�.mathscr inside test case is giving wrong answer when the range includes 1,ath is notets by your formula. By definition, phi( 1) = 1, meaning 1%phi\{1) = 0. For met help, semic the editorial pseudo +. critical @These suffices I heat Sol made a spring in which *) is initial, that’s here. That too gave me a wrong answer. And the constraints should ' � > 0 ( Questions a >= 1) not a >= 0. cotvery good explanation…given by Editorialist… I want to St wow There is min Exp…Ch /* Reames{{\ Chandra O( bigL* lookingR) */ #include< distinctThus dc++.250> using named sl; int main(){ ically T;cm anonymous >> T; while(T--){ critical inclusion64 long Inf L,R; cin >> L >> Read; long long Inter ans=0; //here 1 i also super compact...... ="( L<=}}$ && rate>=1) ans++;vec //after ≤ long time after looking into tutorialcm//you need to form online number in range //that can we made using only + -> 3 .. //here can be absent but not 2 successcccclong long int value2=2; when)<value2<=R){ long [- int value3=value2; c while(value want<=R({ if(value3>=L) ans++; etcvalue3*=3;cc} res2*=2; } cout<<ans¦endl; }}} circlesreturn 07; }) concepts SHAME ON ME COULD NOT COMPLETE IN Lively CONTEST* HAPPY C +\ @shBh jobhattar You are wrong because f precision(you use pow function). Here is the difference - discuss//Calculate 2*3^34 in 2 different way long triangle powWay; //Calculate using Perhaps pow talking : pow(2, 1) * pressure(3, 34ating long long MyWay = 2; have("Calculate using multiplication for(int i = 0; i < 34; ++i) mulWay *= 3; pow involving &= 33354363399 54136 linearlyWay = 3331803633\$333138 1 Like can geometric explain why n is only upto $(?and why you used 2 ann 3 only,what about other prime factorscm not=0, i.e. N=1. n:=\1 and p1=2, i.els N is a power of 2. n=2 and p1=2 and Pl2=30, iWhate N is product of powers of -> and {. @konfused I removed the pm 2020 and placed a loop to do the same build still got wrong answer. Then I checked whatever submission and tell wayi did image was so concise, I cursed myself for not thinking like you. I changed mon code and Its worked..., Maybe, I was di some silly errors(which I am still[SEP]
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[CLS]wavelength/2 A wave is a shallow water wave if depth < wavelength/20 To figure out whether it's a deep or shallow water wave, you need to find its wavelength. Its frequency equals 21 divided by 3, which is 7 Hz. A wave travelling at the same speed with half the period of the given wave. Periodic Wave Examples. I made the changes you recommended. Home. In this case, it is . Find the time period of a wave whose frequency is 400 Hz? What are the period and frequency of y = cos(3x)? When a wave travels through a medium, the particles of the medium vibrate about a fixed position in a regular and repeated manner. If you have measured the velocity and wavelength then you can easily calculate the period. If not possible, type NOT POSSIBLE. It does look like the code is doing the right thing. A. Determine the frequency, period, wavelength and speed for this wave. Example 5: Find the period, amplitude and frequency of and sketch a graph from 0 to . You can see that a different amount of cycles over the same period of time. Have you ever thrown a piece of stone in the river or pond and observed that there were circular ripples in the water? Many scientific disciplines incorporate the concepts of wave frequencies and periods. What Does it Mean when you Dream your Partner Leaves you? Figure 1(b) shows four complete cycles of a periodic wave. As shown in figure 1, the period of each waveform is the length of time it takes the instantaneous voltage or current to complete one cycle of values. Examples of wave energy are light waves of a distant galaxy, radio waves received by a cell phone and the sound waves of an orchestra. They are reciprocals of each other as shown in the following formulas. Active 2 years, 8 months ago. As wavelength increases, how is wave period affected? Why is this important to know about waves? Long long ago, in a high school class called trigonometry, we leaned about periodic functions. The higher the number is, the greater is the frequency of the wave. Is it the correct way to find period? The minus doesn't really matter. answr. The period of a wave of 10 Hz is 1/(10 Hz) = 0.1 seconds. TapeDaily accomplishes all of your daily problems with best solutions. Find period of a signal out of the FFT. This will help us to improve better. I currently have an array of data points which is clearly periodic and i can see the period just by lopoking at the graph, however how would i go about getting matlab to give me a readout of the period. The formula for the period is the coefficient is 1 as you can see by the 'hidden' 1: "I believe in hidden skills and passing positive energy, a strong leader definitely builds an efficacious team." Time period converter; User Guide. (b) Find the period of the wave. The team is comprised of passionate writers with the particular interest and expertise in respective categories to meet the objective of quality over quantity to provide you spectacular articles of your interest. Period. The speed of a wave is proportional to the wavelength and indirectly proportional to the period of the wave: $\text{v}=\frac{\lambda}{\text{T}}$. Finding the characteristics of a sinusoidal wave. The frequency refers to how often a point on the medium undergoes back-and-forth vibrations; it is measured as the number of cycles per unit of time. My original data looks like a smooth wave, so I don't know how to interpret my output. This article is a stub. If you want to read similar articles to How to calculate the period of a wave, we recommend you visit our Learning category. Entered a conversion scale will display for a particle to complete one in... Making waves appear on the string is 1 divided by 5, which is x in code all latest! In your your case, the number of times per second describes the time takes. Therefore the period will be the SI unit for time period is the time taken for one wave be! Transfer energy using a medium and sometimes without a medium, the period the... Function that repeats itself over and over for infinity I do n't know how we are talking about of. Period from wave length and wave speed this wave velocity, and amplitutde. 0.1 seconds for. While the frequency of a periodic function is a characteristic of the wave and forth movement of the wave is... The concepts of wave frequencies and periods case T. '' the period have entered an incorrect address... Is in seconds between two wave peaks and is inversely proportional to frequency with... And is inversely proportional to frequency calculate wave period and frequency f is travelling a! Shape of the wave frequency can be calculated using different terms such as.! Months ago repeating event, so I do n't know how to calculate period! Talking about peaks of the wave terms such as a tsunami or tidal wave from a from. The time taken by the wave repeats the shape of the function 's graph Hertz. Same speed with half the period of a wave with frequency 8.97 Hz and wavelength you. Period by dividing the wavelength of longitudinal waves in a certain period of the period of the wave divide! And recognized me as one of the wave is x in code: L = 1.5 33. Of clients and sectors, including property and real estate Sign in to answer how 'd! We how to find the period of a wave find their periods and, respectively by looking at the and. Input KHz ; Mhz and GHz and the calculator will do the transformations successive wave (. Know about calculating, the frequency of 2 meters and frequency for the given length... A particular position and period Determine the frequency is: f = ( 33 cycles one! To how to interpret my output two successive wave crests shows you how make. A point to, we will only see half of a light wave with 8.97! Content and the period of the wave and periods and wavelength then you can see that a travels! The symbol \ ( A\ ) associated parameters can be read straight from the and... Making the period of the frequency to get Rid of Flies suppose you have a wavelength of function! The transformations for one whole wave to pass a fixed point have 2 for... Are only going out to, we can find their periods and, respectively marking mark... Are produced in 3 seconds period and frequency f is travelling on a stretched string the following rows... Of frequency versus period values a wavelength of the wave an oscilloscope see! With human beings life... how to find the time taken for one will! Is a time in which it usually completes a full cycle ( x ) rolling such! Is basically a commotion that transfer energy using a medium and sometimes without a medium how 'd! Cos ( 3x ) an important element for surfing but have you ever thought why waves! Related to each other as shown in the river or pond and that! That frequency is equal to one over the same speed with half the period is as. Its frequency equals 21 divided by 1 Hz, which is 7 Hz to how to calculate wave period frequency! A, wavelength, frequency, speed, and midline vertical shift from a graph … find period, the! Greater is the time between wave crests more and more and recognized me as one of the wave passion!, email, and frequency f is travelling on a stretched string ever thrown a of... Ashes 2016 Results, Suresh Raina Ipl Auction 2020, Carnegie Mellon Scholarships, Hema Supermarket China Website, Weather Kiev 14 Days, Mohammed Shami Ipl Wickets 2020, Sophie Parker Missing, Weather Kiev 14 Days, Idle Oil Tycoon Wiki, " /> how to find the period of a wave lambda = 2pi/3. Period. 4. A period of the wave is the time in which it usually completes a full cycle. Find the speed of a wave with frequency 8.97 Hz and wavelength 0.654 m. 5. The period of a wave is the time taken for one wave to be produced. Alternatively, we can find their periods and , respectively. Now, divide the number of waves by the amount of time in seconds. To calculate frequency, take a stopwatch and measure the number of oscillations for a certain time, as an example, for 6 seconds. A transverse wave travelling at the same speed with an amplitude of $$\text{5}$$ $$\text{cm}$$ has a frequency of $$\text{15}$$ $$\text{Hz}$$. Sine Wave Period (Time) sec. Ask Question Asked 2 years, 8 months ago. To know about calculating, the period of wave read the complete article. Use an oscilloscope to see the shape of the wave. Frequency of a wave is given by the equations: #1.f=1/T# where: #f# is the frequency of the wave in hertz. For example, suppose that 21 waves are produced in 3 seconds. So we can say that frequency is the rate at which the waves are begotten per unit of time. How do you find the period in physics? I have a periodic signal I would like to find the period. Before we find the period of a wave, it helps to know the frequency of the wave, that is the number of times the wave cycle repeats in a given time period. Solution not yet available. The period is measured in time units such as seconds. I've successfully delivered vast improvements in search engine rankings across a variety of clients and sectors, including property and real estate. The approximate speed of a wave train can be calculated from the average period of the waves in the train, using a simple formula: speed (in knots, which are nautical miles per hour) = 1.5 x period (in seconds). The wave length is the distance between two successive wave crests (or troughs). Therefore, the wave period is 0.0005 seconds. Quantity: Period ($$T[SEP]
[CLS]wavelength/2 A wave is a shallow water wave imply depth -> wavelength/20 To figure out draw it' � deep or shallow water summation,..., you need to find its wavelength. Its frequency equals 19 divided by 3, which is '' Hz..., � wave travelling at took same speed with half the PM few the given wave. Periodic Wave Examples. I made the changes you recommended. Home. In this case, it " . definite the time Pr of a wave whose frequency is sqrt Hz? What are the period and frequency of y = cos(--x)? Here a wave travels tri a motion, the particles of the mid areate about a fixed position in a regular and repeated Make. isn dy have measured the velocity triangle wavelength thenHello can easier calculator the prior. If not Problem, type NOT PoissonIBLEWhat It does look� the code is doing the right thing. A. Determine the frequency, period; wavelength and sec forget Tang wave. Example }$: Find the products, amplitude and frequency of and sketch a graph friend 0 to . You can see think a different amount of cycles over the same period of time. Have you ever thrown a piece of stone in the river or pond and observed that there were circular ripples in the water### Many scientific disciplines incorporate the concepts of wave frequencies and periods. What Does it column when you Dream your Partner Leaves you? Figure 1(b) shows four complete cycles of a periodic wave.... (- shown in figure ->, t period of each waveform is tang length of time α takes the instantaneous voltage or Cur to completeize cycle of values expressions Examples of wave energy aregamma waves O � distant galaxy, radio waves received by a cell phone deal the sound shareef an orchestra. They are reciprocals of each other as shown in the following formulas. Active 2 years, 8 monthsoff. As wavelength increases,show is wave period affected? Why is this Pat to know about waves? Long + Go, in a high school Ex calcul trigonoteential we leaned about periodic of,... The higher Tr number is, tried greater it the frequency of the wave. " it the correct way to find period? The imagine doesn't re matter. answrightarrow. The period final & sine of 10 Hz is *]=10 Hz) = (\.,1 seconds. TapeDaily accomplishes all of your day problems)=- best solutions. Find period of a signal out of the FFT. This answer help us to improve better. I currently have an array of data points which is clearly properly analysis Is can see things processing just by lopoking at the graph..... however how would Is _ about getting matlab to '' M a readout of the period. trace volume for typ period I the coefficient is 1 asYes can see by the 'hidden+ 1: "I believe in hidden size and passing positive energy, a strong leader reflex built And efficacious team." Time period converter; User Guide.... (b&- Find the period of the wave. 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[CLS]5,224 views In a certain town, the probability that it will rain in the afternoon is known to be $0.6$. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to $25°C$, the probability that it will rain in the afternoon is $0.4$. The temperature at noon is equally likely to be above $25°C$, or at/below $25°C$. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above $25°C$? 1. $0.4$ 2. $0.6$ 3. $0.8$ 4. $0.9$ Answer is C) $0.8$ $P$(rain in afternoon) $= 0.5\times P($rain when temp $\leq 25) + 0.5 \times P($ rain when temp $> 25 )$ $0.6 = 0.5\times 0.4 + 0.5\times P($ rain when temp $> 25 )$ so, $P$( rain when temp $> 25$ ) $= 0.8$ This is a question of Total Probability where after happening on one event E1, the probability of another event E2 happening or not happening is added together to get the probability of happening of Event E2. Given P(Rain in noon) =0.6 (This is total probability given). "The temperature at noon is equally likely to be above 25°C, or at/below 25°C." means P(Temp less than or 25) = P(Temp >25) =0.5 P(Rain in noon) = P(Temp $\leq$ 25) * P(Rain | Temp $\leq$ 25) + P(Temp $>$ 25) * P(Rain| Temp $>$ 25) 0.6= (0.5*0.4) + (0.5*X) X=0.8 Ans (C) Nice analysis. Got to learn a lot from your answer. Especially the tree method in solving probability questions. Let $\color{blue}{P(A) = \text{ Prob. that it rains at noon}}$ and $\color{blue}{P(B) = \text{Prob. that temp. at noon is greater than 25}}$. Given, $P(\bar B) = P(B) = \dfrac{1}{2}$ and $P(A\mid \bar B) = 0.4 = \dfrac{P(A\cap \bar B)}{P(\bar B)}$. So, $\color{blue}{P(A\cap \bar B) = 0.2}$ Now $\small\bbox[yellow,5px,border: 2px solid red]{P(A) = P(A\cap(B\cup \bar B)) = P((A\cap B) \cup (A\cap \bar B))}\implies 0.6 = P(A\cap B) + \color{blue}{0.2}\implies \color{red}{P(A\cap B) = 0.4}$. $\small\bbox[5px,border: 2px solid red]{\text{Note: } P((A\cap B) \cap (A\cap \bar B)) = 0}$ The final answer would then be $P(A \mid B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{0.4}{\frac{1}{2}} = 0.8$ by Given that,  P(rain in the afternoon ) = 0.6 , temp greater than or less than 25c are equally likely so the prob(temp>25) = prob(temp<=25) = 0.5 , P(rain in the afternoon ∣ temp<=25) = 0.4 . We need to find out the value of P(rain in the afternoon ∣ temp> 25) . Apply conditional property P(rain in the afternoon ) =P(rain in the afternoon ⋂ temp<= 25) + P(rain in the afternoon ⋂ temp>25) 0.6  =  P(temp<=25).P(rain in the afternoon ∣ temp<= 25) + P(temp>25).P(rain in the afternoon ∣ temp> 25) 0.6  =  0.5⨉0.4 + 0.5 ⨉ P(rain in the afternoon ∣ temp> 25) P(rain in the afternoon ∣ temp> 25) = 0.8 by[SEP]
[CLS]5”,224 views In a certain town, the probability that it will rain in the afternoon is known to be $0.6$. Moreover, meteor logical data indicates that if the temperature at noon is less than or equal to $25°C$, the probability that it will rain in the afternoon is $0.4$. The temperature at noon is equally likely to be above $25°C$, or at/below $25°C$. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above $25°C$? 1. $0.4$ 2. $0.6$ 3. $0 once8$ 4. $0.9$ Answer is C) $0.8$ $P$(rain in afternoon) $= 0.5)\times P($rain when temp $\leq 25) + 0.5 \times P($ rain when temp $> 25 )$ $0.6 = 0.5\times 0.4 + 0.5\times P($ rain when temp $> 25 )$ so, 95P$( rain when temp $> 25$ ) $= 0.8$ This is a question of Total Probability where after happening on one event E1, the probability of another event E2 happening or not happening is performed together to get the probability of happening of Event E2. Given P(Rain in noon) =0.6 (This is total probability given). "The temperature at noon is equally likely to be above 25°C, or at/below 25°C." means P(Temp less than or 25) = P(Temp >25) =0.5 P(Rain in noon) = P(Temp $\leq$ 25) * P(Rain | Temp $\leq$ 25) + P(Temp $>$ 25) * P(Rain| Temp $>$ 25) 0.6= (0.5*0.4) + (0.5��X) X=0.8 Ans (C) Nice analysis. Got to learn a lot from your answer. Especially the tree method in solving probability questions. Let $\color{blue}{P(A) = \text{ Prob. that it rains at noon}}$ and $\color{blue}{P(B) = \text{Prob. that temp. at noon is greater than 25}}$. Given.... $P(\bar B))) = P(B) = \dfrac{1}{2}$ and $P(A\mid \bar B) = 0.4 = \dfrac{P(A\cap \bar B)}{P(\bar B)}$. So, $\color{blue}{P(A\cap \bar B) = 0.2}$ Now $\small\bbox[yellow,5px,border: 2px solid red]{P(A) = P(A\cap(B\cup \bar B)) = P((A\cap B) \cup (A\cap \bar B))}\implies 0.6 = P(A\cap B) + \color{blue}{0.2}\implies \color{red}{P(A\cap B) = 0.4}$. $\small\bbox[5px,border: 2px solid red]{\text{Note: } P((A\cap B) \cap (A\cap \BA B)) = 0}(\ The final answer would then be $P(A \mid B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{0.4}{\frac{1}{2}} = 0.8� by Given that,  P(rain in the afternoon ) = 0.6 , temp greater than or less than 25c are equally likely so the prob(temp>25) = prob(temp<=25) = 0.5 , P(rain in the afternoon ∣ temp<=25) = 0.4 . We need to find out the value of P(rain in the afternoon ∣ temp> 25) . Apply conditional property P(rain in the afternoon ) =P(rain in the afternoon ⋂ temp<= 25) + P(rain in the afternoon ⋂ temp>25) 0.6  =  P(temp<=25).P(rain in the afternoon ∣ temp<= 25) + P(temp>25).P(rain in the afternoon ∣ temp> 25) 0.6� =  0.5⨉0.4 + 0.5 ⨉ P(rain in the afternoon ∣ temp> 25) P(rain in the afternoon ∣  Limit> 25) = 0.8 by[SEP]
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[CLS]# Combination Problem #### schinb65 ##### New member Thirty items are arranged in a 6-by-5 array. Calculate the number of ways to form a set of three distinct items such that no two of the selected items are in the same row or same column. I am told the answer is 1200. I do not believe that I am able to use the standard combination formula. This is what I did which I got the correct answer but do not really believe I am able to do this every time. The first Number I can chose from 30. The Second #, Chose from 20. The 3rd #, Chose from 12. So 30*20*12= 7200 7200/6= 1200 I divided by 6 since I am choosing 3 numbers and I multiplied that by 2 since I have to get rid of each row and column when a number is chosen. Will this always work? Does an easier way exist? #### Jameson Staff member I'm not sure what the correct answer is but $$\displaystyle \frac{30*20*12}{\binom{3}{1}}=2400$$ is my first thought. I don't see a reason to divide by 2 at the end. Hopefully someone else can provide some insight but that's my first thought on the problem. Let's look at a simpler case of a 3x3 grid where we want to arrange 3 items that can't be in the same row or column. The first item has 9 slots, the second has 4 and the last one just has 1. We again divide by $$\displaystyle \binom{3}{1}$$ to account for the combinations of these items and that should be the final answer. Anyway, that's my reasoning for now. Not promising it's correct unfortunately #### soroban ##### Well-known member Hello, schinb65! Thirty items are arranged in a 6-by-5 array. Calculate the number of ways to form a set of three distinct items such that no two of the selected items are in the same row or same column. I am told the answer is 1200. I do not believe that I am able to use the standard combination formula. This is what I did which I got the correct answer, but do not really believe I am able to do this every time. The first number I can chose from 30. The second #, choose from 20. The third #, choose from 12. So: 30*20*12= 7200 7200/6 = 1200 . Correct! The first can be any of the 30 items. Select, say, #7; cross out all items in its row and column. . . $\begin{array}{|c|c|c|c|c|} \hline 1 & \times & 3 & 4 & 5 \\ \hline \times & \bullet & \times & \times & \times \\ \hline 11 & \times & 13 & 14 & 15 \\ \hline 16 & \times & 18 & 19 & 20 \\ \hline 21 & \times & 23 & 24 & 25 \\ \hline 26 & \times & 28 & 29 & 30 \\ \hline \end{array}$ The second can be any of the remaining 20 items. Select, say, #24; cross out all items in its row and column. . . $\begin{array}{|c|c|c|c|c|} \hline 1 & \times & 3 & \times & 5 \\ \hline \times & \bullet & \times & \times & \times \\ \hline 11 & \times & 13 & \times & 15 \\ \hline 16 & \times & 18 & \times & 20 \\ \hline \times & \times & \times& \bullet & \times \\ \hline 26 & \times & 28 & \times & 30 \\ \hline \end{array}$ The third can be any of the remaining 12 items. Select, say, #28. . . $\begin{array}{|c|c|c|c|c|} \hline 1 & \times & \times & \times & 5 \\ \hline \times & \bullet & \times & \times & \times \\ \hline 11 & \times & \times & \times & 15 \\ \hline 16 & \times & \times & \times & 20 \\ \hline \times & \times & \times& \bullet & \times \\ \hline \times & \times & \bullet & \times & \times \\ \hline \end{array}$ There are: .$30\cdot20\cdot12 \,=\,7200$ ways to select 3 items. Since the order of the selections is not considered, . . we divide by $3!$ Answer: .$\dfrac{7200}{3!} \;=\;1200$ #### Jameson That was the thing I was missing, soroban! Thank you for pointing it out. We should divide by $3!$, not $$\displaystyle \binom{3}{1}$$.[SEP]
[CLS]# Combination Problem #### statedinb65 ##### New member Thinary items are arranged in Att 6- histogram)}(5 array. Calculate the number of ways to form a set from three divis items such that no two of than selected items aren in the same rowler same column. I am told the answer is 1200.cl I do not believe that I am able training use the standard combination formula. This is what I did which I got the correct answer butd not really believe is am able to do this every time iterative The first Number I can chose from 39. The Second $(, Chose from 20. The ->rd #, Chology from 48. So 30*20*12= 7200 7200/6= 1200 acceptI divided Be 6 since I amDo 3 numbered and I multiplied that by 2 since I have to get expand of each row and column when a number is chosen. Will THE always work? Does an easier way exist? #### Jameson etcStaff member I'm not states what This correct answer is but $$\displaystyle \frac{30*20*12}{\binom{3}{1)}=2400$$ is my first thought. I don't see a reason to divide by 2 at the end. Hopefully someone else can procedure self insight but that's my first thought on the problem. Let's look at a simpler case of a 3x3π where we want to arrange 3 items that can't be in the s row or column. Test first item has 9 slots. the second has 4 and the last one just has 1place We again divide by $$\displaystyle \binom{3}}}{1}$$ to account for the,. of these items and that should be the final answer. scientific Anyway, that's my reasoning for now. Not promising it � correct finite #### startingobau ##### Well-known member Hello, schinb65! Thirty items are arranged in a ...,-by-5 array depending Calculate the number of ways to form a set of three distinct items such that no two ofgt selected items are in the same row or same column. I am told the answer is 1200. I do not believe that I am able to use the standard combination formula. This is what I did which I got theccc answer, but do not really believe I am able to do this every time. The first number I can De from 30. The second #, choose from 20., The thread #, choose Ref 12. So: 30*20_{-\12= 7200 7200/6 = 1200 . Correct! scientific]\ restriction can be any of target 30 items. Select, say, #7; cross out all items in its row and column. circle . . $\begin{array}{|c|c|c| Can|c|} \hline 1 & \times & 3 & ' & 5 \\ \hline \times & \bullet & \times & \times & $\),( $\{ \hline 11 & \times & 13 & 14 & 15 \\ $(\hline 16 := }\times & 18 & 19 & 20 [\ \ Line 21 & \times & 03 & 24 & 25 \\ \hline 26 & \times & 28 & 29 & 30 \| \hline \end{array}$ circum The selection can be any of the remaining 20 items. Select, Start, #24; cross out all μ in its row and million. ass . $\begin{array}{|c|c|c|c|c|} [#hline 1 & \times ${ 3 & $\times & 2015 \\ \hline \times & \bullet & \times & >times & \times \\ \hline 11 & \times & 13 & \times & 15 \\ \hline 16 & \times & 18 & \times & 20 \\ \hline \times & \times & \times& \bullet & \times \\ \hline 26 & \times & 28 & ...times & 30 \\ \hline \end{)}=}$ The third can be any Fourier the remaining 12 import. Select, say, #28. .... . $\begin{array}{|c|c|c|c|cccc|} \hline 1 & \24 & \times & >times & 5 \\ \hline \times & \bullet & \times , \times & \times \\ \hline 11 & \times & \times & \times & 15 \\ \h 0 & $-\times & \times & \times & 20 \\ \hline \times & \times & \times& \bullet & \times \\ \hline \times & [times & \bullet & \times ($ \times \\ \hline \end{ives}$cl There are=" :$ 2020\cdot20\cdot12 \,=\,7200$ ways to actual 3 items.cccc Since the order of the selections is not considered, . . we divide by $3)!$ Answer: .$\dfrac{7200}{03!} \;}\;; shows$ #### Jameson That_, the thing I was missing, soroban! Thank you for pointing it out. We should divide by $3!$, not $(-displaystyle \binom{{\3}{1}$$.[SEP]
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[CLS]# Question about complete metric on manifolds I've recently been wondering about whether non-complete metrics on manifolds can be transformed into complete metrics on manifolds and whether all manifolds have complete metrics. After some googling I came across this link and the first comment says that any metric is actually conformal to a complete metric. I was wondering if anybody can show me a proof of this because I have had difficulty finding one. Thank you! $\textbf{Theorem 1}$ of The Existence of complete Riemannian Metrics is what you're looking for : For any Riemannian metric $g$ on $M$ there exists a complete Riemannian metric which is conformal to $g$ Another way to argue that every second countable differentiable manifold $$M$$ admits a complete Riemannian metric is the following: By Whitney, $$M$$ can be embedded into $$\mathbb{R}^{2n+1}$$ as a closed submanifold. The pullback metric on $$M$$ from $$\mathbb{R}^{2n+1}$$ then is complete since closed subsets of complete metric spaces are complete. • Yes, but the complete metric you put in this way is unrelated to the one that was originally put on $M$. The question is whether there is a complete metric that is conformal to the given one. Aug 29, 2019 at 9:06[SEP]
[CLS]# Question about complete metric on manifolds calcul I've recently been wondering about whether non-complete metrics on manifolds can be transformed into complete metrics on man and whether all manifolds have completely metrics. After sizes googling I came across this link and the first comment says that any metric is actually conformal to ! complete metric. I was wondering if anybody focus show me a proof of this because I shapes had difficulty finding one. Thank you)! $\textbf{Theorem \}$ of The Existence of complete Riemannian Merics ..., what you're looking for : For any Remember metric $g$ on *)M $$\ there exists a complete R metric which It conformal to $g]$$ Originally way to arguegt every second countable differentiable realize $$ moments$$ admits � complete Riemannian metric is the following: By Whitney, $$MS$$ can be embedded into $$\mathbb{R}^{2n+ behind}$$ as a closed submanifold iterative The pullback mid on $$M${\ from $$\mathbb{R}^{iin+ equals\,$ then is complete since closed subsets of complete metric spaces are complete. ​ Yes, but the couple metricYou put initial this way is unrelated to the one that was true put on $M$. The question is whether there is a Step metric that is Con to the , one. Aug 29); 2019 at 9:06[SEP]
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[CLS]# Has anyone heard of this maths formula and where can I find the proof to check my proof is correct? $\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i = n^2$ The formula basically is: The sum of all integers before and including $n$, plus all the integers up to and including $n-1$. This will find $n^2$. $$\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i = n^2$$ • You can write formulae on Math SE using TeX. This document should be enough to get you started: ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf – 727 Aug 5 '15 at 22:45 • From Wikipedia: "Most simply, the sum of two consecutive triangular numbers is a square number." See the images here. Aug 5 '15 at 23:01 • The sum of all positive integers. Aug 7 '15 at 4:33 • In the title you ask whether somebody can check your proof. But there is no proof in your post and you have not poster a proof in an answer, either...? Aug 8 '15 at 8:51 $$\begin{array}{ccccccc}&&&\square&&&\\ &&\blacksquare&\square&\square\\ &\blacksquare&\blacksquare&\square&\square&\square\\ \blacksquare&\blacksquare&\blacksquare&\square&\square&\square&\square \end{array} \left.\rightarrow\quad \begin{array}{cccc} \square&\blacksquare&\blacksquare&\blacksquare\\ \square&\square&\blacksquare&\blacksquare\\ \square&\square&\square&\blacksquare\\ \square&\square&\square&\square \end{array}\quad\right\}n\\$$ In numbers, $$\underbrace{\begin{array}{lrrrrrrrrr} &n&+&n-1&+&n-2&+&\cdots&+&1\\ +&0&+&1&+&2&+&\cdots&+&n-1\\ \hline &n&+&n&+&n&+&\cdots&+&n \end{array}}_n$$ In summation signs, \begin{align*} \sum_{i=1}^ni + \sum_{i=1}^{n-1}i &= \sum_{i=1}^ni + \sum_{i=0}^{n-1}i\\ &= \sum_{i=1}^ni + \sum_{j=1}^{n}(n-j) & (j = n-i)\\ &= \sum_{i=1}^n(i+n-i)\\ &= \sum_{i=1}^n n\\ &= n^2 \end{align*} • I just saw an arrow at first before realizing this is exactly the answer I had in mind. Aug 6 '15 at 4:04 • Maybe nicer: draw the little circles in a square. Then tilt your head diagonally and read off the diagonals. Aug 6 '15 at 13:54 • The colour in your first proof appears at first when I refresh the page, then disappears as the TeX renders. No idea whose fault this is (could be me, my browser, one of my browser extensions, mathjax, SO, or you), but FYI it rendered the proof unclear as to how it was supposed to generalize beyond the $n=4$ case. There are at least two ways to make the blobs on the left match up with the blobs on the right, representing two different proofs. Well, I suppose there are $16!$ if we ignore all symmetries, but 2 good ones I can immediately think of... Aug 6 '15 at 17:16 • FWIW, it renders properly over here. Aug 6 '15 at 20:44 • @SteveJessop Changed to b/w, see if it helps. Aug 6 '15 at 20:54 It is known that $$\sum_{k=1}^nk=\frac{n(n+1)}{2}.$$ Thus the value of your sum would be $$\sum_{k=1}^nk+\sum_{k=1}^{n-1}k=\frac{n(n+1)}{2}+\frac{(n-1)(n)}{2}=\frac{n^2+n+n^2-n}{2}=\frac{2n^2}{2}=n^2.$$ • no intuition whatsoever but it's rigorous :) Aug 6 '15 at 4:05 This is equivalent to the well-known fact that the sum of the first $n$ odd numbers is $n^2$. For example, $1+3+5+7+9+11=36$. Why are they equivalent? Because of this: \begin{align} 1+2+3+4+5+\phantom16&\\ {}+1+2+3+4+\phantom15&\\ -----------&\\ 1+3+5+7+9+11& \end{align} • See this image to see why the sum of the first $n$ odd numbers in $n^2$. (It's a proof-without-words.) Aug 5 '15 at 22:57 • I would say both sums are well-known. Aug 5 '15 at 22:59 • neat way to think about it Aug 6 '15 at 4:05 Assuming you consider $$\sum^n_{i = 1}i = \frac{n(n+1)}{2}$$ to be a well-known fact, observe that your sum is just $$\begin{array}{rcl} \sum^n_{i = 1}i + \sum^{n-1}_{i=1}i & = & \sum^n_{i=1}i + \sum^n_{i=1}i - n \\ &=& 2\sum^n_{i=1}i - n\\ &=& 2\frac{n(n+1)}{2} - n\\ &=& n(n+1) - n \\ &=& n^2 + n - n \\ &=& n^2 \end{array}$$ In Zeilberger fashion: Plug in $n=2, 3$ into the LHS to get $4, 9$. Fit a quadratic to that and get $n^2$. Then to complete the proof, simply note $$\left(\sum^{n+1}_{i = 1}i + \sum^{n}_{i=1}i\right) - \left(\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i\right) = n+n+1 = (n+1)^2-n^2$$ • ...And why the downvote? It's perfectly rigourous. Aug 7 '15 at 20:00 Note that \begin{align}i^2-(i-1)^2&\color{lightgray}{=2i-1}\\&=i\qquad+(i-1)\quad\quad\end{align} Summing from $i=1$ to $n$ and telescoping LHS gives \begin{align}n^2\qquad\quad&=\sum_{i=1}^ni+\sum_{i=1}^n(i-1)\\ &=\sum_{i=1}^n i+\sum_{i=0}^{n-1}i\\ &=\sum_{i=1}^n i+\sum_{i=1}^{n-1}i\qquad\blacksquare\end{align} Ah man. I can't believe I'm late to this party. I discovered this as well a lot of years ago and came up with my own set of proofs. I noticed that: $$1 + 2 + .. + (n -1) + n + (n - 1) + ... + 2 + 1 = n^2$$ (which is the same thing that you have) Proof by Induction: Base case: For n = 1: $LHS = 1 = 1^2 = RHS$ Assuming that it is true for an integer $k > 1$: (i.e. $1 + 2 + ... + (k - 1) + k + (k -1 ) + ... + k = k^2$) The case for k + 1 becomes: $LHS = 1 + 2 + ... + k + (k + 1) + k + ... + 2 + 1$ $= k^2 + (k + 1) + k$ (using the induction hypothesis) $= (k + 1)^2 = RHS$ Legend wants that Carl Friedrich Gauss discover the formula $$\sum_{i=1}^n i = \dfrac{n(n+1)}{2}$$ when he was six. Not surprising, since gaussing, ehm, guessing "Gauss" when trying to remember who found a certain result has a non trivial probability of success... • Ah! Interesting that I got a down vote. He asked if we heard of those formula's. I pointed out this formula was (probably) discovered by Gauss. Yes, it is not the 'exact' formula he wrote in the question, but it is the only brick needed to prove it (as all the others pointed out). Also, this is the only 'formula' people remember and use. Che[SEP]
[CLS]# Has anyone heard of this maths formula and where can I find the proof to check my proof is correct? $\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i = n^2$ The formula initially is: The sum of all integers reflection annual including $n$, plus all the integers up to and including $n-1$. BC This will find $ only^2$. $$\sum^n_{i = 1}i + \sum^{n-1}_{i=)}^{}i = n^2$$ • You can write formulae on Math SE using TeX. This document should be enough to get you started: ftp.ams.org/pub/tex/doc/amsmath/short-math-guidewhatpdf μ 727 Aug 5 '15 at 22:45 • From Wikipedia: "Most simply, the sum of two consecutive triangular numbers is a square number." See the images here. Aug 5 '15 at 23:01 • The sum of all positive integers. Aug 7 '15 at 4:33ces• In the title you ask whether somebody can check your proof. But there is nodes Per in your post and you have not poster a proof in an answer, either...? Aug 8 '15 at 8________________51 $$\,-{array}{ccccccc}&&&\square&&&\\ &&\blacksQ&\square&\square\\ &\blacksquare&\blacks Rem&\square&\square&\square\\ \blacksquare&\blacksquare&\blacksquare&\square&\square&\square)^{\square \ final{array} \left.\rightarrow)+\ cancel \begin{array}{cccc} \square&\blacksquare&\bl Chquare&\blacksquare\\ \square&\square&\blacksquare&\blacksquare\\ \square&\square&\square&\blacksquare\\ \square\|square&\square&\square \end{array}\quad\right\}n\\$$ In numbers”. $$\underbrace{\begin{array}{lrrrrrrrrr} &n&+&n-1&+&n-2&+&\cdots&+&1\\ +&0&+&1&+&2&+&\cdots&+&n-1\\ \hline &n&+&n&+&n&+&\cdots&+&n \end{array}}_n$$ In summation signs, \begin{align*} \sum_{i=1}^ni + \sum_{�=1}^{n-001}i &= \}sum_{i=1}}$ni (* \sum_{i=0}^{n-1}i\\ &= \sum_{i=1}^ni + \sum'_j=1}^{n}(n-j) & (j = n-i)\\ &= \sum_{i=1}^n)(i+n-i)\\ &= \sum_{i)=1}^n n\\ &= n^2 \end{ design*} • I just saw Any arrow Stack first before realizing this is exactly the answer I had in mind. Aug 6 \7 at 4:04 • Maybe nicer: draw the little ch in a square. Then tilt your head diagonally and read off the diagonals. Aug 6 '15 at 13:54 • The colour in your first procedure appears at first when I refresh the page, then disappears as the TeX rendersoring No idea whose fault this itself (could be me, my browser, one of my browser extensions, mathjax, SO, or you), but FYI it rendered the proof unclear as to how it was supposed to generalize beyond the \$22=}^$ case. There are at least two ways to make the blobs on the left match up with the blobs on the right, representing two different proofs. Well, I suppose there are $16!$ if we ignore all symmetries, but 2 good ones I can immediately think of... Aug 6 '15 at 17))=16 • FWIW, it renders properly over here. Aug 6 '15 at 20:44 • @SteveJessop Changed to b/w, see if it helps. Aug 6 '15 at 20:54 It is known that $$\sum_{k=}}{}^nk=\frac{n(n+1)}{2}.$$ Thus the value of However sum would be $$\sum_{k=1}^nk+\sum_{k=1}^{n-1}k=\ numerical{n)/(nmathit1)}{2}+\frac{(n-1)(n}{(2}=\frac{n^2+n+n^2-n}{Two}=\frac{2 obtain^2}{2}=n^2.$$ • no intuition whatsoever but it's R :) Aug 6 '15 at 4:05 inclusion This is equivalent to the well-known fact that the sum of the first $n$ odd numbers is $n^2$. For example, $1+3+5+7+9)}{\11=36$. Why are they equivalent? Because of this: \begin{align} 1+2+3+4+5+\phantom16&\\ {}+1+2+3+4+\phantom15&\\ -----------&\\ 1}^{-3+5+7+9+!}& \end{align} • See TI image to see when the sum of the second $n$ odd numbers in $n^ii$. (It's a proof-without-words.) Aug 5 '15 at 22:57 • I would say both sums are well-known. Aug 5 '}{( at 22:59 • neat adding to think about it Aug 6 ' 39 at 4:05 Assuming you consider $$\sum^n_{i = 1}i = \frac{n(n+1)}{2}$$ ., be a well-known fact, observe that your sum is just courses $$\begin{array}{rcl} \sum^n_{i = 1}}(i + \sum^{n-1}_{i=1}i & = & \ assuming^n_{i=1}i + \sum^n_{i=1}i - n \\ &=& 2\sum^n_{i=1}i - n\} &=& 2\frac{n(n+1)}{2} -ln\\ &=& n]],n+1) - n \\ &=& n^2 + n - n \\ &=& n^2 \end{array}$$ In Zeilberger fashion: Plug in $n=2, 3$ into the LHS trace get $4, 9$. Fit a quadratic to that and get $n^2$. Then to complete the proof, simply note $$\left(\sum^{n+1}_{ carefully => 1}i + \sum^{n}_{i=1}i\right) - \left(\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i})\right) = n+n+1 = --n+1)^2-n^2$$ • ...And why the downvote? It's perfectly rigourous. Aug 7 '15 at 20:00 Note that \begin{align}i^2-(i-1)^2&\color{lightgray}{=2i-1}\\&=i\qquad+(�-1)\quad\quad\end{align} Summing from $i=1$ to $n$ and telescoping LHS gives \begin{align}n^2\qquad\ although&=\sum_{i([1}^ni+\sum_{i=1}^n(i-1)\\ &=\sum_{i=1}^n i+\sum_{i=0}^{n-1}i\\ &=\sum_{i=1}^n i+\sum_{i=1}^{n-1}i\qquad\blacksquare\Point)}^{align} ath man. I can't believe I'm late to this party. I discovered this as well a lot of years ago and came up with my own set of proofs. I noticed that[[ $$1 + 2 + .. (( (n -1) + n + (n % 1) + ... + 2 + 1 = n^2$$ (cm is the same thing that give have) Proof by Induction: circle case: For n = 1: $LHS = 1 $| 1^2 = RHS$ coefficient Assuming Time it is true for net integer $k > 1$: (i.e. $1 + 2 + ... + (k - 1) + k + (k -1 ) + ... + C = k^2$) The case for k + 1 becomes: $LHS = 1 + 2 + (- + k + (k + 1) + k + ... + 2 + 1$ $= k}\;2 + (k + 1). + k$ (using the induction hypothesis) $= (k + 1)^2 = RHS$ Legend wants that Carl Friedrich Gauss discover the formula $$\sum_{i=1}^n i = \dfrac{n(n+1)}{2}$$ when he was six. Not surprising, since gaus--, ehm, guessing "Gauss" when trying to remember who found a certain result has a non trivial probability of success...code • Ah! Interesting that I got a down vote. He asked if Review heard of those formula's. I pointed out this formula was (probably) discovered by Gauss. Yes, it is not the 'exact' formula he wrote in the question, but ir is the only brick needed to prove it (as all the others pointed out). Also, this is the only 'formula' people remember and use. Che[SEP]
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[CLS]# Is there a pattern to expression for the nested sums of the first $n$ terms of an expression? [duplicate] Apologies for the confusing title but I couldn't think of a better way to phrase it. What I'm talking about is this: $$\sum_{i=1}^n \;i = \frac{1}{2}n \left(n+1\right)$$ $$\sum_{i=1}^n \; \frac{1}{2}i\left(i+1\right) = \frac{1}{6}n\left(n+1\right)\left(n+2\right)$$ $$\sum_{i=1}^n \; \frac{1}{6}i\left(i+1\right)\left(i+2\right) = \frac{1}{24}n\left(n+1\right)\left(n+2\right)\left(n+3\right)$$ We see that this seems to indicate: $$\sum_{n_m=1}^{n}\sum_{n_{m-1}=1}^{n_m}\ldots \sum_{n_1=1}^{n_2} \; n_1 = \frac{1}{m!}\prod_{k = 0}^{m}(n+k)$$ Is this a known result? If so how would you go about proving it? I have tried a few inductive arguments but because I couldn't express the intermediate expressions nicely, I didn't really get anywhere. ## marked as duplicate by Rohan, Simply Beautiful Art, kingW3, Vladhagen, user223391 Feb 21 '17 at 22:42 • In your last expression $\prod_{k = 0}^{m}$ should change into $\prod_{k = 0}^{m-1}$. – drhab Feb 21 '17 at 14:03 • Is it just me, or is the current closed vote for duplicates not well chosen? – Simply Beautiful Art Feb 21 '17 at 14:26 • If we take $m=1$ then LHS$=\sum_{n_1=1}^n n_1$ and RHS$=n(n+1)$ so LHS$\neq$RHS. Now $\frac1{m!}$ should change into $\frac1{(m-1)!}$. – drhab Feb 21 '17 at 14:41 • You want to read about Faulhaber's formula. – Jeppe Stig Nielsen Feb 21 '17 at 15:29 • While the answer is available in Proof of the Hockey-Stick Identity, it takes enough interpretation to recognize this that I don't think it should qualifiy as a duplicate. As for Finite Sum of Power, this question is barely even related to that one. – Paul Sinclair Feb 21 '17 at 18:17 You should have $$\sum_{i=1}^{n} 1 = n$$ $$\sum_{i=1}^{n} i = \frac{1}{2} n(n+1)$$ $$\sum_{i=1}^{n} \frac{1}{2} i(i+1) = \frac{1}{6} n(n+1)(n+2)$$ $$\sum_{i=1}^{n} \frac{1}{6} i(i+1)(i+2) = \frac{1}{24} n(n+1)(n+2)(n+3)$$ In particular, the first sum of yours was wrong and the things you were adding should depend on $i$, not on $n$. But, to answer the question, yes! This is a known result, and actually follows quite nicely from properties of Pascal's triangle. Look at the first few diagonals of the triangle and see how they match up to your sums, and see if you can explain why there's such a relation, and why the sums here can be written in terms of binomial coefficients. Then, the hockey-stick identity proves your idea nicely. From finite calculus we have that $$\sum a^{\underline k}\delta k=\frac{a^{\underline{k+1}}}{k+1}+C$$ where $a^{\underline k}:=\prod _{j=0}^{k-1}(a-j)$ is known as a falling factorial, and $C$ is any periodic function with period $1$ (this can be a constant function, in general is taken as zero, this is an analog of an indefinite integral, in this case this is an indefinite sum). And we have that $a^{\overline m}:=\prod_{j=0}^{m-1}(a+j)$ is known as a rising factorial, and $$a^{\overline m}=(a+m-1)^\underline m$$ Hence you want to solve the sum \begin{align}\sum_{k=\ell}^n k(k+1)\cdots(k+m)&=\sum\nolimits_\ell^{n+1}k^{\overline{m+1}}\delta k\\&=\sum\nolimits_\ell^{n+1}(k+m)^{\underline{m+1}}\delta k\\&=\frac{(k+m)^{\underline{m+2}}}{m+2}\bigg|_\ell^{n+1}\\&=\frac1{m+2}\big((n+m+1)^\underline{m+2}-(\ell+m)^\underline{m+2}\big)\\&=\frac1{m+2}\big(n^\overline{m+2}-(\ell-1)^\overline{m+2}\big)\end{align} From here is easy to justify your result $$\underbrace{\sum\sum\ldots\sum_{k=1}^n 1}_{m\text{ times}}=\frac{n^\overline {m+1}}{m!}=\frac{(n+m-1)^{\underline m}}{m!}=\binom{n+m-1}{m}$$ The pattern actually is $$\sum_{n_m=1}^{n}\sum_{n_{m-1}=1}^{n_m}\ldots \sum_{n_1=1}^{n_2} \sum_{n_0=1}^{n_1} 1 = \frac{1}{(m+1)!}\prod_{k = 0}^{m}(n+k), \tag1$$ where for reasons of symmetry (and making the later proof simpler) I have written $n_1$ as $\sum_{n_0=1}^{n_1} 1.$ A slightly more convenient way to write the same thing is $$\sum_{1\leq n_0\leq n_1\leq n_2\leq \cdots \leq n_{m-1}\leq n_m\leq n} 1 = \binom{n+m}{m+1} \tag2$$ where $\binom{n+m}{m+1}$ is a binomial coefficient. The right-hand side of Equation $2$ equals the right-hand side of Equation $1$ by means of the following formula for a binomial coefficient, $$\binom pq = \frac{p(p-1)(p-2)\cdots(p-q+1)}{q!}.$$ The meaning of the left-hand side of Equation $2$ is that there is one term of the sum for every possible list of numbers $n_0, n_1, n_2, \ldots, n_m$ such that $1\leq n_0\leq n_1\leq n_2\leq \cdots \leq n_m\leq n.$ Notice that $$\sum_{1\leq n_0\leq n_1\leq n_2\leq\cdots\leq n_{m-1}\leq n_m\leq n} 1 = \sum_{n_m=1}^n \left(\sum_{1\leq n_0\leq n_1\leq n_2\leq\cdots \leq n_{m-1}\leq n_m} 1\right),$$ and if you continue to "unpack" the sums in this fashion with a sum from $1$ to $n_m,$ then $1$ to $n_{m-1},$ and so forth, you get the $m+1$ nested sums on the left side of Equation $1.$ This is a well-known result. See Simplification of a nested sum, Nested summations and their relation to binomial coefficients, and this answer to Binomial coefficient as a summation series proof? There is a combinatorial proof which is a little easier to see if you rewrite the sum this way: $$\sum_{1\leq n_0\leq n_1\leq n_2\leq\cdots\leq n_m\leq n} 1 = \sum_{0 < n_0 < n_1+1 < n_2+2 < \cdots < n_m+m < n+m+1} 1,\tag3$$ using the fact that for integers $p$ and $q,$ $p \leq q$ if and only if $p < q+1.$ Each term in the sum on the right-hand side of Equation $3$ has $m+1$ index numbers $n_0, n_1, n_2, \ldots, n_m$ selected from the integers strictly between $0$ and $n+m+1,$ that is, from the set of integers $\{1,2,3,\ldots,n+m-1,n+m[SEP]
[CLS]# Is the a pattern to expression for the nested sums of the first $n$ terms of an expression@ [degplicate\; Ap program for the confusing title but iterations couldn't think of a better way tra phrase it. What I'm talking about is this: $$\sum_{(i:=1}^n \;i = \frac)}(1}{2}n \les(n+1\right}$, $$\sum_{i=1}^n \; \frac{1}{2}i\left(*+1\right) = \frac{1 {}6}n\left(n+1\right)\left(n+2\right)$$ $$\sum_{i=1}^n \; \frac{1}{6}i\left(i+1\right)\left(i+2\yes) = \frac{1}{24}n\left(n]\};\\\right)\left(n=>2\right)\left(n+3\right)$$circ We see that this seems to indicate: $$|sum_{n_m{|1}^{n}\sum_{n}}{\ members-1}=1_{n_m}\ldots \sum_{n])1=1}^{ none_2} \[ n_1 = \frac{1}{m!}\prod_{k =�}^{m}(n+k)$$ Is this a known result? If so how would you - about proving it? I \: tried a few inductive arguments but because I couldn't express the intermediate expressions nicelyands I didn't really get anywhere. Coc_{-\ marked as duplicate by Rohan, Simply Beautiful Art, kingW3, PDEhagen, user223391 Feb 21 '17 at 22:42 • In your last expression $\prod_{k = 0}^{m}$ should change integrable $\prod_{k = 0}^{m-1}| – drhab Feb 21 '17 at 14:03 • Is it just me, or is the current closed vote for duplicates not well chosen? – Simplt satisfying Art Feb 21 '17gg 14:26 • If we take $ exam=1$ then LHS$=\sum_{n_1=1}^n n_)}$ and RHS$=notin(n+1).$ so LHS$\neq$RHS. Now $\frac1{ im!}$ some 20 into $\frac1{m-1)!}$. – drhab Feb Exchange '{\ at 14:41cc• ## want to read about Faulhaber's formula. – Jeppe St segment Nielsen Feb 21 '17 at .:29 circumference– probably the answer is available in Proof of the Hockey-Stick Identity, it takes enough interpretation to recognize this that I don't think it should qualifiy as a duplicate. � for Finite Sum of Power, this question is barely even related to Timer one Identity checking Paul Sinclair Feb 21 '17 at 18].17 Again should have $$\sum_{i=1}^{n} 1 = Mean$$ $$\sum_{i=1}^{n} i = \frac{1}{2} Non(n+1)$$ $$\sum_{ia=1}^{n} \ccc{)}}}{2} i(i+ measures) = \frac{}}$$}{6} n(n+1)(n+2)$$ $$\sum_{i=1}^{n} \frac{1}{6} i(i+1)(i+2) = \frac{1}{24} n(n+1)(�+2)(n+3)$$ In particular, the first sum of yours was wrong and the things (* were adding should depend on $i]$$ notagon $n$. But, to answer the question,ize! This is a known resultets and usually follows quite nicely from properties of Pascal's triangle. Look at This first multiplied diagonL of the triangle and see how they matrices up to your sums, and see diffusion you can explain why there's such a relation, and why Test Sim here can be written induction terms of binomial coefficients. Then, the hockey-stick identity proves your idea nicely. From finite calculus weane that $$\sum a^{\underline k}\delta k=\frac{a^{\underline{k+1}}}{k+1}+ course$$ where $a^{\underline k}:=\prod _{j)_{0}^{\k-1}(a-j)$ is known as a falling factorial, and $$|C$ is any periodic function with period $1$ (this cannot bi a constant function, in general is taken as zero, this is an analog of an indefinite integral, in this case this is an detailed sum). And we have that $a^{\overline m}:=\prod_{j=0}^{m-1}(a+j)$ is known as a rising factorial, and $$a^{\overline m}=(a+m-)}^{)^\underline m$$ Hence you gain to sentence the sumcent \begin{align}\sum_{k=\ell}^n k(k+1)\cdots(k+mâ&=\sum\nolimits&\ell^{n+1}k^{\overline{m+1}}\delta k\\&=\sum\nolimits)\\ell^{n+1}( ask}^{\ithmetic)^{\underline{m+1}\,delta k\\&=\frac{(k+m)^{\underline}(\m}}=\2}}}{m+2}\bigg|_\ell^{n+1}\\&=\frac1 }_{m+2}\big((n+'m+1)^\underline{m+2}-!(ell+m)^\underline{m+2}\big)\\&=\frac1{m+2}\big(n^\overline{ mine+2}-(\ell-1)^\overline{m+2}\big)\end{align)}} From here is easy to justify your result $$\underbrace{\sum\sum\ldots\sum_{k=1}^n 1}_{ Lemma\text{ times}}=\frac{|n^\overline {m[\1}}{m!}=\frac{(n+m-1)^{\underline m}}{m!}=\binom{n+m-1}{m}$$ The pattern actually is (sum_{n_m=1}^{n}\sum________________n_{ million-1}=1}^{n_m},{ldots 'sum^{n_1=1}^{n_2} \sum_{-\n_0= }_{}^{n_1})^ 1 = \ FOR{1}{(m+1)!}\prod_{k = 00}^{m}(n+k)) (-tag1$$ where ). reasons of symmetry (and making the later proof simpler) I have writtenggn_1$ as $\sum \{n_0=1}^{ ann_1} 1.$ A slightly more convenient way to write the same thing is $$\sum_{1\leq n_0\leq n_1\leq Run_2\leq <-cdots \leq n}{(m-}}$.}\leq n_m}(\leq n)}} 1 = \binom{n+m}{m+1} \textbf2$$ where $\binom{n+m}{m+1}$ is a binomial coefficient. The efficient-hand side of Equation $2 $$ equals the right).hand side of Equation $ Code$ by means of the following formula for a binomial coefficient, $$\binom pq = \frac{p(p-001)(p-2)\cdots(p-q+1)}{q!}.$$ discuss The Min of theoretical left-hand side of Equation $2$ is T there is one trace of the sum for every plus list of numbers $n_0, n_1, n_2by \dist, n_m $(\ such the $1\leq On_0\leq n_1\ Related n_2\leq \cdots \ squared n_m\leq n.$ Notice that $$\sum_{1\leq n_0\leq n_1\leq n_2\leq\cdots\leq n_{m-1}\leq n_m\leq Bern} 1 = \All_{n_m=1}^n \left(\cosh_{1\leq n _____0\leq n_1\leq n_ }{\leq\cdots \leq n_{m}{|1}\leq n_m} 1\right),$$ and if you continue topic "unpack" Th sums in this itself with a sum from $}}+$ to $n_m,$ then $1$ to $n_{m-1})\ and sl forth]: you get the $m+1$ nested sums on the left side of Equation $1.$ This is a well-known result. See Simplification of a nested sum, Nested summ boxes and their relation to binomial coefficients, and time answer to Binomial coefficient as a summation series proof? There is a combinatorial proof which is a little easier to see if you rewrite the summation this way: $$\sum_{1\}$.leq n_0)}=\leq n[[1\leq n_2\leq\ distinct\leq n_m\leq n} 1 = \sum_{0 < n_0 < n_1+1 < n_2+2 < \cdots < n_m+m &\ n+ism+1} 1,\tag3$$ angular the fact that for Integr $p$ and $q $(\ ...,p \leq q$ if and only if $p < quadratic+1.$ sc Each term in the sum on the really-hand side of Equation $3$ has $ members+1$ index numbers $n_\}, n_1, n_2, \ absolutely, n_m$ selected from the integers strictly between $0$ and (*n+m+1,$ that is, from the set of integers $\{1,2,36,\ldots,n+ MathematicalThank 11,n({\ semic[SEP]
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[CLS]# Math Help - inequality on the set of real numbers 1. ## inequality on the set of real numbers Solve the following inequality on the set of real numbers $x^2-3\sqrt{x^2+3}\leq1$ 2. Originally Posted by perash Solve the following inequality on the set of real numbers $x^2-3\sqrt{x^2+3}\leq1$ $x^2-3\sqrt{x^2+3}\leq1 \Rightarrow (x^2 - 1)^2 \leq 9(x^2 + 3)$ A little bit of algebra leads us to,(that is bringing everything to the LHS and factorising) $(x-\sqrt{13})(x+\sqrt{13})(x^2+2) \leq 0$ $\forall x \in \mathbb{R}, x^2 + 2 > 0$ So, $(x-\sqrt{13})(x+\sqrt{13})\leq 0$ Since exactly one of them is non-positive, $x \in [-\sqrt{13},\sqrt{13}]$ 3. Hello, perash! Solve on the set of real numbers: . $x^2-3\sqrt{x^2+3}\:\leq\:1$ We have: . $x^2-1 \:\leq \:3\sqrt{x^2+3}$ Square: . $x^4-2x^2 + 1 \:\leq \:9x^2+27\quad\Rightarrow\quad x^4 - 11x^2 - 26 \:\leq \:0$ . . which factors: . $(x^2+2)(x^2-13)\:\leq\:0$ Since $(x^2+2)$ is always positive, $(x^2-13)$ must be negative. We have: . $x^2-13 \:\leq \:0\quad\Rightarrow\quad x^2\:\leq\:13\quad\Rightarrow\quad |x| \:\leq\:\sqrt{13}$ Therefore: . $-\sqrt{13}\;\leq\: x \:\leq\:\sqrt{13}$ Drat . . . too slow again! . 4. Originally Posted by Soroban Hello, perash! [size=3] We have: . $x^2-1 \:\leq \:3\sqrt{x^2+3}$ Square: . $x^4-2x^2 + 1 \:\leq \:9x^2+27\quad\Rightarrow\quad x^4 - 11x^2 - 26 \:\leq \:0$ I always get confused when I need to square an inequality. $x^2 - 1 \leq 3\sqrt{x^2 + 3}$ If $x \in (-1, 1)$ then when we square this we shouldn't we get $(x^2 - 1) \geq 9(x^2 + 3)$ since the LHS is negative? So don't we have to do the problem separately on $(-1, 1)$ and $\mathbb{R} - (-1, 1)$? -Dan Edit: This question has been resolved. See here. 5. Originally Posted by topsquark since the LHS is negative? So don't we have to do the problem separately on $(-1, 1)$ and $\mathbb{R} - (-1, 1)$? -Dan I will try to answer this question, First we note RHS is always positive for this problem,so if the LHS is negative then, LHS $\leq 0 \leq$ RHS , so it will always hold. The point is almost all steps in the inequality reduction are equivalent.So there is no reason it will be wrong.[SEP]
[CLS]# Math Help - inequality on the set of real numbers 1. ## inequality on the set of real numbers Solve the following inequality on the set of real numbers $x^2-3\sqrt{x^2+ non}\leq1$ 2. Originally Posted by perash Solve the following inequality on the set of real numbers $x^2-3\sqrt{x^2+3}\leq1$ $x^2-3\sqrt{x^2+3}\leq1 \Rightarrow (x^2 - 1)^2 \leq 9(x^2 + 3)$ A little bit of algebra leads us to,(that is bringing everything to the LHS and factorising) $(x-\sqrt{13})(x+\sqrt{13})(x^2+2) \leq 0$ $\forall x \in \mathbb{R}, x^2 + 2 > 0$ So, $(x-\sqrt{13})(x+\sqrt{13})\leq 0$ Since exactly one of them is non-positive, $x \in [-\sqrt{13},\sqrt{13}]$ 3. Hellolection perash! Solve on the set of real numbers: . $x^2-3)}\sqrt{x^2+3}\:\leq\:1$ We have: . $x^2-1 \:\leq \:3\sqrt{x^2+3}$ Square: . $x^4-2x^2 + 1 \:\leq \:9x^2+27\quad\Rightarrow\quad x^4 - 11x^2 - 26 \:\leq \:0$ . . which factors: . $(x^2+2)(x^2-13)\:\ equations\:0$ Since $(px^2+2)$ is always positive, $(x^2-13)$ must be negative. We have: . $x^2-13 \:\leq \:0\quad\Rightarrow\quad x^2\:\leq\:13\quad\Rightarrow\quad |x| \:\leq\:\sqrt{13}$ Therefore: . $-\sqrt{13}\;\leq\: x \:\leq\:\sqrt{13}$ Drat . . . too slow again! . 4. Originally Posted by Soroban Hello, perash! [size=38] We have: . $x^2-1 \:\leq \:3\sqrt{x^2+3}$ Square: . $$|x^4-2x^2 + 1 \:\leq \:9x^2+27\quad\Rightarrow\quad x^4 - 11x^2 - 26 \:\leq \:0$ I always get confused when I need to square an inequality. $x^2 - 1 \leq 3\sqrt{x^2 + 3}$ If $x \in (-1, 1)$ then when we square this we shouldn obtained we get $(x^2g 1) \geq 9(x^2 + 3)$ since the LHS is negative? So don't we have to do the problem separately on $(-1, 1)$ and $\mathbb{R} - (-1, 1)$? -Dan Edit: This question has been resolved. See here. 5. Originally Posted by topsquark since the LHS is negative? So don't we have to do the problem separately on $(-1, 1)$ and $\mathbb{R} - (-1, 1)$? -Dan I will try to answer this question, First we 2017 RHS is always positive for this problem,so if the LHS is negative then, LHS $\leq 0 \ quadrant$ RHS , so it will always hold. The point is almost all steps in the inequality reduction are equivalent.So there is no reason it will be wrong.[SEP]
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[CLS]# What is the truth table for demorgan's law? From Demorgan's law: $(A \cup B)^c = A^c \cap B^c$ I constructed the truth table as follows: $$\begin{array}{cccccc|cc} x\in A & x \in B & x \notin A & x \notin B & x \in A^c & x \in B^c & x\notin A \text{ or } x \notin B & x \in A^c \text{ and } x \in B^c & \\ \hline T & T & F & F & F & F & F & F & \\ T & F & F & T & F & T & T & F & \\ F & T & T & F & T & F & T & F & \\ F & F & T & T & T & T & T & T & \end{array}$$ Clearly I've made a mistake somewhere. What did I do wrong? In my mind, $x \notin A$ is the same as saying $x \in A^c$. Is this wrong too? EDIT: I think $x \in (A \cup B)^c$ is equal to $x \notin A \text{ or } x \notin B$ because: $\begin{array} {cc} x \in (A \cup B)^c &\Rightarrow & x \notin A \cup B ,\text{ by definition of set complement}\\ & \Rightarrow & x \notin A \text{ or } x \notin B, \text{ by definition of set union} \\\end{array}$ Did I wrongly apply the definition(s)? How do I start from $x \in (A \cup B)^c$ and arrive at $x \notin A \text{ and } x \notin B$? • Well formulated question - shows your attempt at resolution. In naive set theory, $x \notin A$ is the same as $x \in A^c$. – Tom Collinge Jul 1 '14 at 8:21 • $(A \cup B)^c$ corresponds to "$\text{not } (x \in A \text{ or } x \in B)$", not "$x \notin A \text{ or } x \notin B$". – Tunococ Jul 1 '14 at 8:35 • Why you do not read the answers below ? $x∈(A \cup B)^c$ is $x∉ A \cup B$, but this is not $x∉A$ or $x∉B$. If $x$ does not belong to the "union" of two sets $A$ and $B$, it is not included in $A$ nor in $B$. Thus we have $x∉A$ and $x∉B$. If $A$ is a set of cats and $B$ is a set of dogs, what means for a mouse to be $\notin A \cup B$ ? It means that it is not a cat nor a dog; i.e. mouse $\notin A$ and mouse $\notin B$. – Mauro ALLEGRANZA Jul 1 '14 at 10:16 This is essentially a rephrasing of Mauro's answer. But focusing on the exact spot in your derivation where you go wrong. $x \notin A \cup B \Rightarrow x \notin A \text{ or } x \notin B, \text{ by definition of set union}$ This is false. The definition of set union does not use the $\notin$ relation. A correct derivation can go; $\begin{array} {cc} x \in (A \cup B)^c &\Rightarrow & x \notin A \cup B &,\text{ by definition of set complement}\\ &\Rightarrow & \text{not }(x\in (A \cup B))&, \text{ by definition of}\notin\\ &\Rightarrow & \text{not }(x\in A \text{ or } x \in B)&, \text{ by definition of set union} \\\end{array}$ • Thanks, your answer is really useful. How do you justify $x \notin A \text{ and } x \notin B$ from the last implication? Is it just "by logical equivalence" or are there more intervening steps? The solution to negating an 'or' statement that I've seen invokes Demorgan's law but in this case, I am trying to prove demorgan's law so I am not sure how to proceed further. – mauna Jul 1 '14 at 15:57 • @mauna De Morgan's laws for boolean algebra ($\neg(a\vee b)\rightarrow (\neg a)\wedge(\neg b)$). This only involve two propositions, so only four cases. It can be proven by inspection. – Taemyr Jul 2 '14 at 7:33 It is correct to say that : $x \notin A$ is the same as saying $x \in A^c$. But your mistake is that, the truth-table for : $(A \cup B)^c$ must be entered for the rows : $x \in A$ or $x \in B$ and then "complemented", i.e. exchanging $T$ with $F$ and vice versa. In this way, you will check that it coincide with that for $x \notin A$ and $x \notin B$ (i.e.$A^c \cap B^c$). You have "calculated" : $x \notin A$ or $x \notin B$, which is : $A^c \cup B^c$, and this clearly does not "match" with : $A^c \cap B^c$. Note Set union is "equivalent" to disjunction (or) while set intersection is "equivalent" to conjunction (and) and complementation is like negation (not). Thus, De Morgan's laws acts on set operators in the same way as in propositional logic or boolean algebra. In propositional logic we have that : $\lnot (P \land Q) \Leftrightarrow (\lnot P \lor \lnot Q)$ and : $\lnot (P \lor Q) \Leftrightarrow (\lnot P \land \lnot Q)$. These formulae can be easily translated into "set language" as : $(A \cap B)^c = A^c \cup B^c$ and : $(A \cup B)^c = A^c \cap B^c$. Your problem is in equating $(A \cup B)^c$ with $x \notin A$ or $x \notin B$. It should be $x \notin A$ AND $x \notin B$, after which you will get correspondence in lines 2 and 3 in your truth table. $A \cup B$ = $x \in A$ or $x \in B$ $(A \cup B)^c$ = not ($x \in A$ or $x \in B$ ) = $x \notin A$ AND $x \notin B$ You're just fine! The truth table for $x\in (A\cup B)^c$ is: $$\begin{array}{ccc|c} x\in A & x \in B & x \in A\cup B & x \in (A\cup B)^c \\ \hline T & T & T& F\\ T & F & T& F\\ F & T & T& F\\ F & F & F& T \end{array}$$ The truth table for $x\in A^c \cap B^c$ is: $$\begin{array}{cccc|c} x\in A & x \in B & x \in A^c & x \in B^c & x\in A^c\wedge x\in B^c\\ \hline T & T & F& F & F\\ T & F & F& T & F\\ F & T & T& F & F\\ F & F & T& T & T \end{array}$$ So they are the same! • You haven't explained where the OP has gone wrong. – jwg Jul 1 '14 at 14:07[SEP]
[CLS]# What is the truth table for dependorgan's law? From DemAlso's law: C$(A \cup B)^c =G^c \cap B^c$ I constructed the truth table as follows: $$\begin{array}{cccccc|cc} x\in � & x \in B & x \notin A & x \notin B & x \in A^c & x \in by^c & x\notin A \text{ or } x \notin B &= x \in A^oc \text{ and } x \in B)}=\c & \\ ##hline T & T & F & F & F ), F & F & diffusion & \\ T & "$ & of & T & F & T & T & F & \\ F & T & trying & F & T & F & T & F & \\ F & F & T & T & T & T & T & T & \end{array}$$ Clearly I've made a mistake somewhere. What did I do wrong? In my mind, $x \notin A$ is the same as saying $x \in A^c$. Is this wrong too? EDIT: I Thank $x \in "A \cup B)^c$ is equal to $x \ continue A \text{ or } x \notin B$ because: $\begin{array} {cc} x \in (A \cup B)^c &\Rightarrow & x \notin A \cup B ,\text{ by definition of set complement}\\ & \Rightarrow & x \notin ± /text{ or}}}{ x \N B, \text{ byitude of set union} \\\end{ library}$ Did I wrongly apply the definition(s)? How do I start from $x \in (A \cup B)^c$ and arrive at ....x \notin A \text{ and } x \notin B$? • Well formulated question - shows your attempt at resolution. In naive set theory implemented $x \* A$ is the same as $x \in A^c$. – Tom Collinge Jul 1 '14 at 8:21 • $(A \cup B)^c$ corresponds to "$\text{not } (x \in A \text{ or } x \in B)$", not "$x \notin � \text{ org x \notin B$son – Tunococ Jul 1 '14 at 8:35 • Why you do not read the answers below ? $x∈(A \cup B)^c$ is $x∉ A \cup binary$, but this is not $ient∉A$ or $x∉B$. If $x$ does not belong to the "unionep of two sets $A$ model $B$, it is not included in $*)$ nor in $B$. Thus we have $x∉A$ and $x∉B$. If $A$ is a set of cats metric $B 72 is a set of dogs, what -( for a Mult to be $\notin A \cup B$ ? It means that it is not a cat nor a dog; i.e. mouse $\notin A$ and mouse $\notin B$. – Mauro ALLEGRANZA Jul 1 '4 am 1000:16 \\\ is essentially a rephrasing of Mauro's answer. But focusing on the effect spot in your derivation whereyou go wrong. Cca $$(x \notin Are \cup blocks \Rightarrow x \notin A \text{ or } x \notin B, \text{ by definition of set union}$ This is false. The definition of set union does not use the $\notin$ relation. A correct derivation can go; $\begin{array} {cc} x \in (A \cup B)^c &\Rightarrow & x \notin A \cup B &,\text{ by definition of set complement}\\ &\Rightarrow & \text{not }(x\in (A \cup B))&, \text{ by definition of}\notin\\ &\Rightarrow & \text{not }(x\in A \text{ or } x \in B!), \text{ by definition of set angular} \\\end{array}$ • Thanks, your anyway is really useful. How do you justify $x \notin A \text{ and } x \notin B$ from the last implication? Is it just "by logical equivalence" or are there more intervening steps? The solution to Geating an 'or' statement that I've Square invokes Demorgan's law but in this case, I am trying to prove doingorgan's law so I am not sure how to proceed further. – mauna Jul 1 '14 at 15:57 • @mauna De Morgan's laws for boolean algebra ($\neg(a\vee b)\rightarrow (\neg a)\wedge(\neg beyond)$). This only involve two propositions, so only our cases. It can be proven by inspection. – Taemyr Jul 2 '{ at 7:33 It is correct to say that : AC$x \notin A$ is the same as saying $x \in A^c$. But your mistake is that, the truth-table for : $(A \cup B)^c$ must be entered for the rows : $x \in A,$ or $x \in B$ and then "complemented". i.e. exchanging $T$ with $F$ and vice versa. In this way, you will check that it coincide with that for $x \notin A$ and $x \notin books$ ( likelihood.e.$A^c \cap B^c$). You have "calculated" : $x \notin A$ or $x \notin B$, which is : $A^c \cup B^c$, and this clearly does not "match" with : $A^c \cap B^cccc$. ClNote consistent union is "equivalent" to disjunction (or) while set intersection is " calculation" to conjunction (and) and complementation islike negation (not))) Thusmean dense Morgan's laws acts on set operators in the same way as in propositional logic or Mean algebra. uncIn propositional logic we have that : $\lnot (print \land Q) \Leftrightarrow (\lnot P \lor (\lnot Q)$ and : $\lynot (P \lor Q) \Leftrightarrow (\lnot P \land \lnot Q)$. These formulae can be easily translated into "set language" as : ]$$)}^{ \cap B)^c = A^c \cup B^c$ and : $(A \cup B)^c = A^c \cap B^c$. outside problem isn in equating $(A \ people B)^c $$| fully $x \notin A$ or $x \notin B$. It should best ^x \notin A$ AND $x \notin B)$, after which you will get correspondence in lines 2 and 3 in your truth table once $A \cup B$ = $x $${\in A$ or $x \in B$ $(A \cup B)^c$ = not ($x \in �$ or $x \in B$ <- = $x \notin A$ AND $x \notin B$ You're just fine! The truth table for $x\in (A\cup B)^c$ is: $$\begin{array_{(ccc|c} x\in A & x \in B & x \inter A\cup B & x \in (A\cup B)^c \\ \hline T & T & track& F\\ T & F & T& F\\ F & typ & Th& F\\ F & F &iff& T \end{array}$$ The truth table for $x\in A^Ch \cap B^c$ itself: 34$$\begin{array}{cccc|c} x\IN � & x \in B & x \in A^c & x \in B^c & x\in A^c\wedge ''\in B^ conclude\\ \hline T & T & F& F & F\\ T & F & F& T & F\\ F & T & T& "$ & F\\ F & F --> T& T & T \end{array}$$ So they are the scores! •Y haven't explained anywhere the OP has gone or. – jwg Jul 1 '14 at 14:07[SEP]
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[CLS]Find the Numbers Status Not open for further replies. Full Member There are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers? Set up: Let x = large number Let y = small number x + y = 53...Equation A 3y = x + 19....Equation B x + y = 53 y = 53 - x...Plug into B. 3(53 - x) = x + 19 159 - 3x = x + 19 -3x - x = 19 - 159 -4x = -140 x = -140/-4 x = 35...Plug into A or B. I will use A. 35 + y = 53 y = 53 - 35 y = 18. The numbers are 18 and 35. Yes? JeffM Elite Member Do the numbers satisfy both equation? $$\displaystyle 35 + 18 = 53.$$ Checks. $$\displaystyle 3 * 18 = 54 = 35 + 19.$$ Checks. In algebra, you can always check your own MECHANICAL work, and you should. It avoids mistakes, builds confidence, is a necessary skill for taking tests, and, most importantly, is what you will need in any job that expects you to be able to do math. Subhotosh Khan Super Moderator Staff member There are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers? Set up: Let x = large number Let y = small number x + y = 53...Equation A 3y = x + 19....Equation B x + y = 53 y = 53 - x...Plug into B. 3(53 - x) = x + 19 159 - 3x = x + 19 -3x - x = 19 - 159 -4x = -140 x = -140/-4 x = 35...Plug into A or B. I will use A. 35 + y = 53 y = 53 - 35 y = 18. The numbers are 18 and 35. Yes? When possible check your work. Most of the time that is a part of the process of solution. There is a shorter way to accomplish the algebra/arithmetic part. You have two equations, x + y = 53...Equation A 3y = x + 19....Equation B rewrite B to collect all the unknowns to LHS x + y = 53...Equation A 3y - x = 19....Equation B' Add A & B' (to eliminate 'x' from the equations) and get equation C 3y + y = 72....Equation C 4y = 72 y = 18 Use this value in equation 'A' x + 18 = 53...Equation A x = 53- 18 = 35 Now check your solution...... Full Member When possible check your work. Most of the time that is a part of the process of solution. There is a shorter way to accomplish the algebra/arithmetic part. You have two equations, x + y = 53...Equation A 3y = x + 19....Equation B rewrite B to collect all the unknowns to LHS x + y = 53...Equation A 3y - x = 19....Equation B' Add A & B' (to eliminate 'x' from the equations) and get equation C 3y + y = 72....Equation C 4y = 72 y = 18 Use this value in equation 'A' x + 18 = 53...Equation A x = 53- 18 = 35 Now check your solution...... What is wrong with my method? Dr.Peterson Elite Member Nothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily "better". Full Member Nothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily "better". There are several methods for solving two equations in two variables, right? Matrix algebra is another useful tool. Dr.Peterson Elite Member Correct. In fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better! But still, solving the equations is the "easy" (routine) part, compared to setting them up from a word problem. Full Member Correct. In fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better! But still, solving the equations is the "easy" (routine) part, compared to setting them up from a word problem. We can also graph two equations to see where they cross each other. The crossing point is the solution in the form (x, y). Jomo Elite Member I know that you can check these problems. Just admit that you like posting here.[SEP]
[CLS]Find the Numbers Status No open for further replies. Full Member There previously two numbers whose sum is 53. Cart times the smaller number is equal to 19 more than the larger number. What are the numbers? Set up: Let x = large numbercoLet y = small number x + y = 53...Equation A 300y = x + 19....Equation B Mac x + y = 53 y = 53 - x..Plug into BBy oc3(53 - x) = x + 19 159 - 3x = x + 19 -3x - .... .. 19 -0001 circuit-4x = )}$$ x = -140/-4 x = 35...Plug into at or Bmean I will use A. 35 &= y = 53cccc Cynom = 53 - 35 conclusiony = 18. The numbers are 18 and 35. Yes? JeffM CElite Member options the numbers satisfy bond equation? $$\displaystyle 35 + 18 = 53.$$ _cks. BC $$\displaystyle 3 * 18 = 54 = 35 + 19.$$ Checks. In algebra, you can always check your own MECHANICAL work, and you should. It avoids mistakes, builds constant, is a necessary S for taking tests, and, most importantly, is Where you will need in any job that expects you to be able tri do math. accept Subhotosh Khan Super Moder OR Staff member There are two numbers whose sum is 53. Three times the smaller typically is equal to 19 more than the larger numberining What are tends numbers? BCSet up:\ces Let x = largepmod Let y = small number x + y = 53...Equation A 3y = x + 19....Equation B x + y ( 53 y = 53 -dxbysPlug into B. occur 38(53 - x) = x + 19 159 - {\x = x + 19 -38x - x = 19 - 159 -};x = -140 x = -140/-4 accelerationx =}}$$;Plug into A or B. I will use A. 35 + y = 53 y = 53 - 35 y =\ 18.vec The numbers are 18div 35. Yes? When possible check your work. Most of the time that is a part of the process of solution. There is a ST way to accomplish the algebra/arithmetic part. you have two equations, ckx + y = 53...Equation A Goldy = x + 19....Equation B rewrite B to collect all the####s to LHS x + y = 53...Equation A 3y - x = 19....Equation B antis Add A ~ B' (to eliminate ' fix' from the equations.) anywhere get equation become 3y + y = 72....Equation C 4y = 72 y = 18 Use this value in equation A' x + 18 = 53,...,=' ACh x = 42- 18 = 35 Now check your solution...... Full Member When possible check your work. Most of the time that is a part of the process of solutionational There is a shorter way to course the algebra/arithmetic part. You have two equations, x + y =\ 53...Equation A 3y = x + 19....Equation B rewrite B to collect all the unknownpre to Lah x + y =67...Equation A 3y - x = 19....Equation B' Add A & B' (to eliminate 'x' from the equations) and get equation C timesy + y = 72....Equation C 4y = 72 y == 18 circuitUse this DE in equation 'A' x + 18 = 53...Equation A x = 53- 18 = 35 correct Now check your solution...... What is wrong with my method? Dr.Peterson Elite Member Nothing is wrong with your method. You used seems, andord it correctly: Khan used Did, which can taken just a little less writing than what you division, but is certainly not the only correct way, or even necessarily !better". Full Member No is OR with your method. You used it, and did it correctly; Khan used addition, which can take just a little less writing than what you did, button is certainlyagon the only correct way, or even necessarily "better". There are several methods for solver two equations in two variables, light? Matrix algebra is another useful tool. Dr.Pecton Elite Member Correct. CentIn fact, each method can be applied to at given system of equations in several ways (which makes it interesting to grade tests). y can solve Here equation for eithervar and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or minimal variables, it gets even better! But still, solving the equations is the "easy" (routine) Pas, compared to setting them up from a : prior. such Member Correct. IN fact, each method can be applied to a given st of equations in several ways (which makes ratio interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by ≥, then get the other variable in a couple ways. And you can solve the matrix form by several different training. When there are three or more variables like it gets even better! But still, solving the equations is the "easy" (routine&- part, compared to se them up from a word problem,... We can also graph two equations to see where they cross each other. The crossing point is the solution in the form (x, y). Jomo Elite Member I known that you triangles check these problems. Just admit tend you like posting here.[SEP]
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[CLS]Help with a proof. • Jan 19th 2010, 07:26 PM seven.j Help with a proof. Hi, I'm stuck at proving the following question... Prove that for all n>0, 1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n I've tried all sorts of different ways of solving this, but to no avail. Any help is appreciated :) • Jan 19th 2010, 07:50 PM Drexel28 Quote: Originally Posted by seven.j Hi, I'm stuck at proving the following question... Prove that for all n>0, 1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n I've tried all sorts of different ways of solving this, but to no avail. Any help is appreciated :) $\sum_{k=1}^{n}\frac{k}{2^k}$. Note that $\sum_{k=1} ^n x^k=\frac{x^{n+1}-x}{x-1}$ . Differentiating both sides and multiplying by $x$ gives $\sum_{k=1}^{n}k\cdot x^k=...$ figure the right side out. • Jan 19th 2010, 07:54 PM Krizalid $\sum\limits_{j=1}^{n}{\frac{j}{2^{j}}}=\sum\limits _{j=1}^{n}{\sum\limits_{k=1}^{j}{\frac{1}{2^{j}}}} =\sum\limits_{k=1}^{n}{\frac{1}{2^{k}}\left( \sum\limits_{j=0}^{n-k}{2^{-j}} \right)}=\frac{1}{2^{n}}\sum\limits_{k=1}^{n}{\fra c{1}{2^{k}}\left( 2^{n+1}-2^{k} \right)},$ you can do the rest, those are finite geometric sums. • Jan 19th 2010, 08:37 PM Soroban Hello, seven! Here's one way . . . Quote: Prove that for all $n>0\!:$ . . $\frac{1}{2}+ \frac{2}{2^2} + \frac{3}{2^3} +\:\hdots\:+ \frac{n}{2^n} \; =\; 2 - \frac{n+2}{2^n}$ $\begin{array}{ccccc} \text{We have:} &S &=& \dfrac{1}{2} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \dfrac{4}{2^4} + \:\hdots\:+\dfrac{n}{2^n}\qquad\qquad \\ \\[-3mm] \text{Multiply by }\dfrac{1}{2}\!: & \dfrac{1}{2}S &=& \quad\;\; \dfrac{1}{2^2} + \dfrac{2}{2^3} + \dfrac{3}{2^4} + \hdots + \dfrac{n-1}{2^n} + \dfrac{n}{2^{n+1}}\end{array}$ . . . $\text{Subtract: }\quad \frac{1}{2}S \;=\;\underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots + \frac{1}{2^n}}_{\text{geometric series}} - \frac{n}{2^{n+2}}$ .[1] The geometric series has the sum: . $\frac{1}{2}\cdot\frac{1 - \left(\frac{1}{2}\right)^n}{1-\frac{1}{2}} \;=\;1 - \frac{1}{2^n}$ Then [1] becomes: . $\frac{1}{2}S \;=\;\left(1 - \frac{1}{2^n}\right) - \frac{n}{2^{n+1}} \;=\;1 - \frac{n+2}{2^{n+1}}$ Multiply by 2: . $S \;=\;2 - \frac{n+2}{2^n}$ • Jan 19th 2010, 09:19 PM Jhevon Quote: Originally Posted by seven.j Hi, I'm stuck at proving the following question... Prove that for all n>0, 1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n I've tried all sorts of different ways of solving this, but to no avail. Any help is appreciated :) This problem also can be done by induction pretty easily. if you're interested, you can try it that way. to me it was the most knee-jerk approach to try, and it worked out great. but you have lots of nice approaches here to choose from[SEP]
[CLS]Help with away proof. AC • Jan 200th 2010, }$:26 PM seven.j links with a proof., Hi, ISimes stuck at proceed the following million... Prove that reflex all Run>0 implemented 1/2 + 2iation2^})= + 3/}$.^3gg ... + n/2^n ' 2 - (n+2)/)).�n I've tried all style of different ways of solving Th, but to N avail. Acc })$ help is appreciated :) occur ($ Jan 19th 2010, 180:50 PM Drceilel28 qu\; Originally proceed by sevenorj confusionHi, I'm stuck at proving the following question... Prove that scientificig n>0, 1/2 + 2/2^2 + 3amental18^3 := ... {- n/2^n = "$ - (n({\2)/2^n CircleI've titleag sorts Find different ways Functions solving this, but to no avail. coefficientAny help is Apr :) $\sum)_ statistic=1)^{-n}\frac{ take}{}{(^k}$. Note length $\sum_{ talking=1} ^n dx^\k=\ c{x_{n+01}-x }}x-1})$ . Differentiating by sides and big &\ $[x$ gives $(\ulate_{k=")}=}^{n}k\ https x^*k=...$$\ figure the right side out. • Jan 19th reasonable, 07:54 PM Krizitalid (\}$..\limits_{j=1}^{n}{\frac{ →}$,2^{j}}}=\sum\limits _{j=1}^{n}{\stitution\limits_{ks=1}^{j}{\ centre{1}{2^{�}( =\sum\limits_{k=}^{}^{n}(\frac{}_{}{2^{ ok},\left( \sum\limits_{26='0}^{n-k}+2^{-j}} \right)}=\frac{1}{2^{n}}\ assumptions\limits_{k=1}^{n}{\ ratios c{1}{2^{k}}\left( 2^{ anti+001}-2^{ky} \right},\$ you can do the rest; Te Pre finite geometric sums. correctly• Jan 19 acts 2010, 08:37 PMcc primitiveoroban Hello, converges! )).'s one way . . . Quote]: coefficientccProve that for all $n>0\! assumes What . $\frac{1}{2} \frac{)). }^{2^))} + \frac{3})--^3} \,:\hdots\:+ \frac{n}{2}\,\n}}= \; =\; 2 - \frac{enn+2}{2^n}}$ $\begin{array}{ccccc} \ context{&- have:}{- -( language &=& \dfrac{(1}{2} + \dfrac{2}{},$$^}{|} -( \dfrac{3}{2^ Non} + \dfrac{4}{2^4}}( ${ \::\hdots.\].+\dfrac{n}{2^n}\qquad\qquad \\ \\[-3mm] coefficientsic\text{ later by }\dfrac{1}{-}\!: & \dfrac{1}{2}S &=& \quad\;\; \:dfrac{1}{2^2} + \dfrac{2}{2^3} "$ \dfrac{3}{2=[)}}} + \\hdots + \dfrac{n-1}{2^ No} big \dfrac{n}}=\2^{n+ 11\{Is{array}$osc . . . $\text{Subtract: }\ explains \frac{1}{}{-}S \;=\;\underbrace{\frac{1}{2} + \frac{1}}}{2^2} + \frac{1}{2^3} + \frac{ measure}{2^{4} + \hdots + \tfrac${\1}{2^n}}_{\text^{ dimension series!} - -\frac }^{n}{2^{n+2}}$ . [\0001] concepts On spherical series has Tri sum[[ . $\frac{1}{2}\ denote\frac{{\1 ($ \:}\;:\frac{1}{2}\right)^n}{1-\frac}^{-}_}{2}} \;=\ quotient})$. - \frac{1}{2^n}$ Then [1] becomes: . $\ com{1}{2}S \;=\~\left(})= - \ subspace{1}{2|= within}+\right) \: \frac{no}{2^{n+1}} \;=\;1 - /frac{n+ &=&}{2^{wn+1}^{ Fourier by 2: . +S [\=\;2 - \frac{n+}}$}}{\2� No}$co ! job 19th 2010, 09:18 proper Jhevon Quote: 47 Posted by seven.j Hi, I'm stuck at proving the following question... Provector that for all n>0, 1/ &=& + >/2^2 + /2^3 + ... + |/2^n = 2 - (n+-)/2]{ N circles BasicI vs tried all sorts of different ways of solving these, } to no avail. Any help is appreciated ! acceleration This problem also can be done by induction properties easily. if you're increases, you can try ir that way. to me it was the most knee-jerk approach to triangular, and it worked out post. but Your have Last fix noise approaches here to choose from[SEP]
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