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https://math.stackexchange.com/questions/3262238/does-knowing-the-surface-area-of-all-faces-uniquely-determine-a-tetrahedron | # Does knowing the surface area of all faces uniquely determine a tetrahedron?
I was wondering if the four areas of a tetrahedron faces were sufficient information to uniquely determine its shape. For example, is it true to say that if the surface areas are equal then the solid must be a regular tetrahedron?
If the answer is negative, then what else we need to fully determine the shape of the tetrahedron in space?
• Well if a tetrahedron is the only shape which can exist with four faces in three dimensions then it must be a tetrahedron. You could prove whether or not it must be regular based on the areas using algebra – Henry Lee Jun 14 at 14:14
• The space of tetrahedra, up to congruency, has 6 degrees of freedom, and you only give 4 pieces of information with your areas. So my immediate guess is that no, it is in no way determined. – Arthur Jun 14 at 14:14
• Are you aware of the Minkowski theorem (math.stackexchange.com/questions/105033/…)? – Moishe Kohan Jun 14 at 14:56
• @MoisheKohan No, I hadn't seen this before. But does this theorem imply existence of more than one tetrahedron with identical face areas? – Amirh.Kp Jun 14 at 17:43
• @Amirh.Kp: yes, of course. This is existence-uniqueness theorem. It allows you to prescribe areas as well as the directions of normals to the faces. It also shows exactly what you are missing. – Moishe Kohan Jun 14 at 18:22
Consider the vertex set in $$\mathbb R^3$$ $$\{(a, b, 0), (a, -b, 0), (-a, 0, b), (-a, 0, -b)\}$$ for $$a, b > 0$$. Then it is clear that the faces determined by these vertices are all congruent isosceles triangles with side lengths $$2b$$, $$\sqrt{4a^2+2b^2}$$, $$\sqrt{4a^2+2b^2}$$ and common area $$a \sqrt{4a^2+b^2}$$. So by a suitable choice of $$a$$ and $$b$$ we can make the tetrahedron irregular but have faces of equal area; moreover, for a given area of a face, there is in general more than one choice of $$(a,b)$$, hence even for this very restricted type of tetrahedron, knowing the area of each face does not uniquely determine the tetrahedron.
A tetrahedron with vertices at $$(0,0,0)\\ (1/2, 0,0)\\ (0,2,0)\\ (-0.0924127, 1.9387, 0.4913857)$$ will have area $$1$$ for all faces.
The coordinates of that last point are approximate (and of course you can freely swap the sign of the $$z$$-coordinate). The actual point is given as the solution to the three equations $$\cases{x^2 + z^2 = 1/4\\ y^2 + z^2 = 4\\ \displaystyle\left(\frac4{\sqrt{17}}(x-1/2) -\frac1{\sqrt{17}}y\right)^2 + z^2 =\frac43}$$ I placed the three first vertices first, to make sure one face was non-equilateral and had area $$1$$, then these three equations are exactly the equations that ensure that the other three faces have area $$1$$. The left-hand sides are (the squares of) the altitudes of the three remaining faces if you put the fourth vertex at $$(x, y, z)$$, and the right-hand sides are what those altitudes must be to ensure that the area of the corresponding face is $$1$$.
As you can see, I am somewhat free to make the first face whatever shape I want, as long as its area is $$1$$, and then I just set up the three equations to find the fourth vertex. It's possible that some extreme versions of the first face causes the resulting three equations to not have any real solution, but I have demonstrated here that at least one non-regular tetrahedron may be generated this way.
• +1 I think there's an interesting construction lurking here. If you relax the equal area condition to specify just the ratios of areas of faces then Informally, you're foliating the space of tetrahedra What do the leaves look like? – Ethan Bolker Jun 14 at 15:00
• @EthanBolker That's a cool question. I really only thought of it as just intersecting three cylinders... – Arthur Jun 14 at 15:01
The answer is "no," which can be demonstrated by counting degrees of freedom.
A tetrahedron has 4 vertices in 3-dimensional space, and is therefore defined by 12 independent parameters, i.e. the $$x$$, $$y$$, $$z$$ coordinates of each vertex.
Now consider what we mean by two tetrahedra having "the same shape". The orientation in 3-D space doesn't matter, and that eliminates 6 of the 12 parameters (there are 3 rigid body translations, and 3 rotations).
But the tetrahedron only has four faces, so fixing the area of each face still leaves two independent parameters to vary its shape in an arbitrary way.
If you want to pick two more quantities to fix the shape of the tetrahedron, that could be done in many different ways, but it would be nice to do it in a way that is independent of any coordinate system used to describe it.
One natural parameter could be the volume of the tetrahedron, but finding a second one is not so "obvious".
The radius of the inscribed and circumscribed spheres might be interesting choices. Proving that they do uniquely define the tetrahedron in combination with the area of the faces is left as an exercise for the reader :)
Another option might be to fix the lengths of two edges. Choosing a pair of edges which do not share a vertex has a nice symmetry about it.
You may be interested in my paper with Petr Lisonek, Metric Invariants of Tetrahedra via Polynomial Elimination, ISSAC 2000 conference, Aberdeen (Scotland), July 2000, 217-219. We show that in general the four areas, circumradius and volume do not determine a tetrahedron, but there exist non-regular tetrahedra that are determined by the four areas and the circumradius. For example, this is the case if one face is an equilateral triangle iscribed in a great circle of the circumscribed sphere.
It's easy to construct a tetrahedron with sides congruent to any acute isosceles triangle.
Start by taking two copies of the triangle, glue them together at the base, and then pull the apices apart until the distance between them equals the length of the base. (You can always do this if the triangles are acute, since the base length of an acute isosceles triangle is less than twice the height.) Now the sides of the triangles, together with the line between their apices, form two additional isosceles triangles that are congruent with the original one, and you thus have a tetrahedron with four congruent faces. Cut some triangles out of paper and try it if you don't believe it!
In fact, essentially the same method actually works for any acute triangle, even if it's not isosceles! The only tricky part is that now there are two distinct possible ways to glue the first two copies of the triangle together at the base, depending on whether or not you flip one of them over first, and you need to choose the way that leaves the non-equal sides of the triangles adjacent. Then proceed just like above.
Thus, in particular, for any acute triangle with area $$A$$ there exists a tetrahedron with all faces congruent to that triangle, and thus also with area $$A$$. Unless the triangle happens to be equilateral, this tetrahedron will not be regular.
A tetrahedron can be determined by the lengths of three edges emanating from a vertex and the three angles formed by pairs of those angles. Thus, a tetrahedron admits six degrees of freedom. Four face areas are not enough to determine the shape (or even the volume) of such a figure.
As an extreme example, you can consider any rectangle a "degenerate" tetrahedron with four congruent right-triangle faces; the volume is zero, which is decidedly different than that of a regular tetrahedron. In fact, one can construct "equihedral" (equal-face-area) tetrahedra with volumes anywhere from the minimum of zero to the regular tetrahedron's maximum.
As for what other information you can use, my favorite additional parameters are the areas of the tetrahedron's three (what I call) "pseudo-faces". You can think of these geometrically as the quadrilateral projections of the tetrahedron into planes parallel to a pair of opposite sides. Each pseudo-face area is calculated by $$\text{area} = \frac12 \text{side}\cdot\text{side} \cdot \sin \text{(angle)}$$ just like any other face, except here, the "$$\text{side}$$"s are opposite each other, and the "$$\text{angle}$$" is considered the angle between the corresponding direction vectors.
Four standard faces (say, $$W$$, $$X$$, $$Y$$, $$Z$$) and three pseudo-faces ($$H$$, $$J$$, $$K$$) make seven area parameters, which would seem to over-determine the figure. However, the sum-of-squares identity $$W^2+X^2+Y^2+Z^2=H^2+J^2+K^2 \tag{1}$$ introduces a dependency that reduces the degrees of freedom to the expected six.
Other pseudo-face relations include a strangely familiar-looking Law of Cosines. \begin{align} Y^2 + Z^2 - 2 Y Z \cos A &= H^2 = W^2 + X^2 - 2 W X \cos D \\ Z^2 + X^2 - 2 Z X \cos B &= \,J^2 = W^2 + Y^2 - 2 W Y \cos E \\ X^2 + Y^2 - 2 X Y \cos C &= K^2 = W^2 + Z^2 - 2 W Z \cos F \end{align} \tag{2} where each of $$A$$, $$B$$, $$C$$, $$D$$, $$E$$, $$F$$ is the dihedral angle between appropriate pairs of faces ($$A$$ between $$Y$$ and $$Z$$, etc).
There's also this volume formula:
\begin{align} 81V^4 &= H^2 J^2 K^2 - 2 (W X-Y Z)(W Y-Z X)(W Z-X Y) \\ &-H^2(W X-YZ )^2-J^2(WY-ZX)^2-K^2(WZ-XY)^2 \end{align} \tag{3}
If the four face areas are equal, the formula reduces to $$9V^2 = HJK$$, which shows that the volume of an "equihedral" tetrahedron depends upon more than those face areas.
Anyway, you can read more about these kinds of relations in my Hedronometry notes. In particular, "Heron-Like Results for Tetrahedral Volume" (PDF) includes the stuff I've described above and a bit more.
• Thanks for your comprehensive introduction. one thing I can't understand which is seemingly a critical concept and that is your extreme example of considering every rectangle as a "degenerate tetrahedron" with four congruent right-triangle faces. I think you mean it's a natural result of a process which construct tetrahedrons but I can't come up with such process. – Amirh.Kp Jun 14 at 18:21
• @Amirh.Kp: Any four points can be the vertices of a tetrahedron. The vertices of a rectangle are four points, so they determine a tetrahedron; since it's flat, the volume is zero. (The rectangle's four sides, and its two diagonals, are the edges of the tetrahedron, and the right triangles formed by two adjacent sides and one diagonal are the faces of this tetrahedron.) Such a figure is considered "degenerate" because it is not three-dimensional. This is like how any three points determine a triangle; if those points are collinear, the triangle is "degenerate" with area zero. – Blue Jun 14 at 22:56 | 2019-12-13T00:31:32 | {
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http://ztvu.uiuw.pw/power-series-and-taylor-series.html | # Power Series And Taylor Series
What is Power series? A power series is a series of the form. The problem with the approach in that section is that everything came down to needing to be able to relate the function in some way to. Let f(x) = X1 n=0 c n(x a)n = c 0 + c 1(x a) + c. DEFINITION 2. This of course is just a power series shifted over by c units. 2 Properties of Power Series 10. The Taylor series of is the sum of the Taylor series of and of. What is the interval of convergence for this series? Answer: The Maclaurin series for ex is 1+x+ x2 2! + x3. Read moreTaylor and Maclaurin Series. In the Summer 1994, the author developed computer activities intended to provide an intuitive interpretation to some of the fundamental notions involved in studying infinite series and Taylor polynomials. Note: In Problem 52, there is a mistake in the directions. This calculus 2 video tutorial explains how to find the Taylor series and the Maclaurin series of a function using a simple formula. 2, is a Taylor series centered at zero. For further details, see the class handout on the inverse. Multivariate Taylor Series. Find more Mathematics widgets in Wolfram|Alpha. There is also a special kind of Taylor series called a Maclaurin series. Leavitt Power series in the past played a minor role in the numerical solutions of ordi-nary and partial differential equations. In the Summer 1994, the author developed computer activities intended to provide an intuitive interpretation to some of the fundamental notions involved in studying infinite series and Taylor polynomials. So, a couple definitions to get us started here. A power series converges uniformly and absolutely in any region which lies entirely inside its circle of convergence. Taylor and Laurent Series We think in generalities, but we live in details. Here we address the main question. Use a known Maclaurin series to obtain the Maclaurin series for the function f(x) = cos(πx). You can specify the order of the Taylor polynomial. The series we will derive a power series that will converge to the factor. Taylor series is a special power series that provides an alternative and easy-to-manipulate way of representing well-known functions. This example generalizes as follows. Choose the maximum degree of the Taylor polynomial to use to approximate a function. Just assume that RaiseTo(raise a number to the power of x) is there. Each of the resistors in a series circuit consumes power which is dissipated in the form of heat. To find the Maclaurin Series simply set your Point to zero (0). Convergence of In nite Series in General and Taylor Series in Particular E. It takes the following form: Here’s a common example of a p-series, when p = 2: Here are a few other examples of p-series: Remember not to confuse p-series with geometric series. A Taylor series method for numerical fluid mechanics J. 5) on uniform convergence is optional. This is due to the uniqueness of the Taylor series of a function centered at a point. Let’s prove a lemma to deal with that last point. Section 4-16 : Taylor Series. If a function is equal to it's Taylor series locally, it is said to be an analytic function, and it has a lot of interesting properties. The last section (15. The method is shown to be nondispersive, nondiffusive, and for. (ii) Using (i) or otherwise nd the Taylor series expansion of 1 (1 x)2 and 1 (1 x)3 about a = 0, stating carefully any theorems you may use about integrating or di er-entiating power series within their radius of convergence. Taylor’s Theorem states that if f is represented by a power series centered at c, then the power series has the form 0 n fx = ssss s ss sssssssssssss s ss sssssssssssssss s ss ssssssssssssssss s s If a power series is centered at c = 0, then it is called a sssssssss ssssss. (Several of these are listed below. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. RELATION BETWEEN TAYLOR SERIES AND POWER SERIES A power series = Taylor series of its sum In other words, every time you obtain an identity X∞ n=0 a nx n = (something) then the power series on the left-hand sidemust be the Taylor series of that something on the right-hand side. Series effectively evaluates partial derivatives using D. If a= 0 the series is often called a Maclaurin Math formulas for Taylor and Maclaurin series Author:. These are the most important series of all! (Taylor, Maclaurin, etc, etc. The series P1 n=0 anx n, x 2 R, is called a power series. Recall from Chapter 8 that a power series. As the names suggest, the power series is a special type of series and it is extensively used in Numerical Analysis and related mathematical modelling. So we can write a simple generalised expression for a power series as g of x, equals a, plus bx, plus cx squared, plus dx cubed et cetera. The TaylorAnim command can handle functions that "blow-up" (go to infinity). Substituting the coefficients back into the series yields. 4 Working with Taylor Series Learn with flashcards, games, and more — for free. Taylor and Maclaurin Series Tutorial for Calculus students. Taylor and Maclaurin (Power) Series Calculator. They are distinguished by the name Maclaurin series. Maclaurin Series: If a function f can be differentiated n times, at x=0, then we define. To enhance our students' learning of the infinite series material, a computer laboratory activity devoted to the subject was created. The Taylor series above for arcsin x, arccos x and arctan x correspond to the corresponding principal values of these functions, respectively. Power series are useful because ssss sssss ssss ss sss. This series is referred to as the Taylor series of a function f(x) centered at c. + Formal manipulation of Taylor series and shortcuts to computing Taylor series, including substitution, differentiation, antidifferentiation, and the formation of new series from known series. This gives us a simple formulaB for the sum:" B B B â œ " " B # $This is our first example of a Taylor series —a power series that adds up to a known function. is the Taylor series of f(z) = 1=(1 z) about z= 0. Every Taylor series is a power series in 0 0 0! k k k fx xx k is a power series in x 2 Theorem 9. Use the ratio test to show that the Taylor series centered at 0 for sin(x) converges for all real numbers. But this is good news for combinatorics. Taylor series is a special power series that provides an alternative and easy-to-manipulate way of representing well-known functions. But it converges at both end points and does so, therefore, absolutely. Use the ratio test, unless otherwiseinstructed. f The coefficients of this power series may be expressed with the Bernoulli numbers. In the figure below, we consider the graphs. Correct! This is the correct answer, found by using mathematical operations on the geometric power series. Lesson 23: Power Series Expansions. CHAPTER 38 Power Series In Problems 38. In fact, that's the brute force method of finding a series representation for a function, but there are other ways. Many times a Taylor expansion is used for approximations in solving transcendental equations such as x - ln x = 5 which cannot be solved by currently known algebraic manipulations In some cases in solving differential equations a Taylor series will actually give an exact answer that can't be readily found by any other method. If you want the Maclaurin polynomial, just set the point to 0. (See the text, p. FUNCTIONS OF A COMPLEX VARIABLE (S1) Lecture 7 Power series expansions ⊲ Taylor series f representable by Taylor series expansion is said to be analytic. What is Power series? A power series is a series of the form. If the series uses the derivatives at zero, the series is also called a Maclaurin series, named after Scottish mathematician Colin Maclaurin (February 1698 – 14 June 1746). In this video, Patrick teaches how to Differentiate and Integrate Power Series to Derive New Power Series Expressions. This example generalizes as follows. Taylor and Laurent Series We think in generalities, but we live in details. The resulting series can be used to study the solution to problems for which direct calculation is di cult. Find the Taylor series for e−x2 centered at 0. However, using differentiation and integration we can expand many more functions into power series also. This is a convergent power series, but the same power series does not define an asymptotic series for exp(z). Another immediate and straightforward consequence of Theorem 2. The series P1 n=0 anx n, x 2 R, is called a power series. The binomial function Remark: If m is a positive integer, then the binomial function f m is a polynomial, therefore the Taylor series is the same polynomial, hence the Taylor series has only the first m +1 terms non-zero. + Maclaurin series and the general Taylor series centered at x = a. Power Series Solution of a Differential Equation • Approximation by Taylor Series Power Series Solution of a Differential Equation We conclude this chapter by showing how power series can be used to solve certain types of differential equations. because we take the formula for a Taylor polynomial centered at zero and let it keep on going. We have over 350 practice questions in Calculus for you to master. Learn how these polynomials work. To construct a power series solution around the point x = x o, we procede as follows: (1) Set y(x) = P 1 n=0 a n(x x o) n. The TaylorAnim command can handle functions that "blow-up" (go to infinity). And this is because they are composed of coefficients in front of increasing powers of x. Given just the series, you can quickly evaluate , , , …, and so on. Before, we only considered power series over R but now, we will consider power series over C as well. Free power series calculator - Find convergence interval of power series step-by-step. (1) Find the radius of convergence of (a) X1 n=1 5nxn n2 (b) For what values of xdoes X1 n=1 (2x+ 1)n n3 converge? (c) Give an example of a power series which converges for all x2( 1;1] and at no other points. ) Series can also generate some power series that involve fractional and negative powers, not directly covered by the standard Taylor series formula. We now shift from the approach of Cauchy and Goursat to another approach of evaluating complex integrals, that is, evaluating them by residue integration. Consider the one dimensional initial value problem y' = f(x, y), y(x 0) = y 0 where f is a function of two variables x and y and (x 0, y 0) is a known point on the solution curve. It is a series that is used to create an estimate (guess) of what a function looks like. In this video, Patrick teaches how to Differentiate and Integrate Power Series to Derive New Power Series Expressions. The nearer to a the value is, the more quickly the series will converge. We would like to know which x0s we can plug in to get a convergent series. The TaylorAnim command can handle functions that "blow-up" (go to infinity). Several useful Taylor series are more easily derived from the geometric series (11), (19) than from. To enhance our students' learning of the infinite series material, a computer laboratory activity devoted to the subject was created. Taylor's Series method. We begin with the general power series solution method. POWER SERIES 251 For example, sine is an analytic function and its Taylor series around x 0 = 0 is given by sin(x) = X1 n=0 (1)n (2n + 1)! x2n+1: In Figure 7. We begin with the general power series solution method. Math formulas and cheat sheet generator creator for Taylor and Maclaurin Series. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Fenton School Of Mathematics University Of New South Wales Kensington, N. We prove it in order to demonstrates the Taylor series proposition above. Intervals of Convergence of Power Series. And this is because they are composed of coefficients in front of increasing powers of x. It explains how to derive power series of composite functions. What is a power series? 6. Practice Problems: Taylor and Maclaurin Series 1. Taylor’s Series. Title: Taylor series of hyperbolic functions:. For example, the Taylor Series for ex is given by:. Power Series, Taylor Series In Chapter 14, we evaluated complex integrals directly by using Cauchy's integral formula, which was derived from the famous Cauchy integral theorem. I was just wondering in the lingo of Mathematics, are these two "ideas" the same? I know we have Taylor series, and their specialisation the Maclaurin series, but are power series a more general co. The series converges only for. A summary of Differentiation and Integration of Power Series in 's The Taylor Series. (Any power series whatso-ever. In 1668, the theory of power series began with the publication of the series for ln()1+x by Nicolaus Mercator, who did this by “integrating” 1 1+x (Stillwell 1989, 120). In other words, it's not a hypothesis we have to verify or check for. Leavitt Power series in the past played a minor role in the numerical solutions of ordi-nary and partial differential equations. If a= 0 the series is often called a Maclaurin Math formulas for Taylor and Maclaurin series Author:. THE BINOMIAL SERIES 375 6. If we take x0 = x¡c then the power series around c reduces to the power series around 0. infinite series in Novæ quadraturae arithmeticae in 1650, finding 1 n=1 nn()+1 ∞ ∑ along with proving the divergence of the harmonic series. Since this power must come from the source, the total power must be equal to the power consumed by the circuit resistances. Another immediate and straightforward consequence of Theorem 2. The basic idea is to approximate the solution with a power series of the form: (1) X1 m=0 a m(x mx 0) : As an example consider the Taylor. Created by Sal Khan. Find the Taylor series for e−x2 centered at 0. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. 10 The Binomial Series 6. Note that since is an even function, all its Taylor polynomials are also even polynomials. Incorrect! The signs are incorrect. It takes the following form: Here’s a common example of a p-series, when p = 2: Here are a few other examples of p-series: Remember not to confuse p-series with geometric series. But this is good news for combinatorics. Convergence of In nite Series in General and Taylor Series in Particular E. CHAPTER 38 Power Series In Problems 38. Finding Taylor Polynomials The TI-89 taylor( command can be used to find partial sums. The partial sum is called the nth-order Taylor polynomial for f centered at a. Taylor series is a way to representat a function as a sum of terms calculated based on the function's derivative values at a given point as shown on the image below. TAYLOR AND MACLAURIN™S SERIES 359 6. Here is a set of practice problems to accompany the Taylor Series section of the Series & Sequences chapter of the notes for Paul Dawkins Calculus II course at Lamar University. In general this series will converge only for certain values of x determined by the radius of convergence of the power series (see Note 17). Indeed, the entire power series" B B B â#$ can be thought of as a geometric series with a common ratio of. f The coefficients of this power series may be expressed with the Bernoulli numbers. Power series are useful because ssss sssss ssss ss sss. (See the text, p. More generally, if c 2 R, then the series P1 n=0 an(x¡c)n, x 2 R, is called a power series around c. Note that since is an even function, all its Taylor polynomials are also even polynomials. Taylor and Maclaurin Series Tutorial for Calculus students. You can skip questions if you would like and come back to them. (Several of these are listed below. But it converges at both end points and does so, therefore, absolutely. Spring 03 midterm with answers. of better and better approximations to f leading to a power series expansion f(x) = X∞ n=0 f(n)(a) n! (x−a)n which is known as the Taylor series for f. In this section we will discuss a. Created by Sal Khan. The series generated by the sequences (a nzn) as z varies are called the power series generated by (a n). for any x in the series' interval of convergence. If the series uses the derivatives at zero, the series is also called a Maclaurin series, named after Scottish mathematician Colin Maclaurin (February 1698 – 14 June 1746). Taylor series is a special power series that provides an alternative and easy-to-manipulate way of representing well-known functions. Learn exactly what happened in this chapter, scene, or section of Calculus BC: Series and what it means. A power series about x = x 0 (or centered at x = x 0), or just power series, is any series that can be written in the form X1 n=0 a n(x x 0)n; where x 0 and a n are numbers. We can obtain a finite part, the first few terms, of a power series expansion of a function about a point by means of the Mathematica function Series as follows:. Example 5 Find the Maclaurin series for cos(x). Whether the power series converges at x = x0 ± ρ is tricky to determine. Title: Taylor series of hyperbolic functions:. We now shift from the approach of Cauchy and Goursat to another approach of evaluating complex integrals, that is, evaluating them by residue integration. As mentioned earlier, the function 1=(1 z) exists and is in nitely di erentiable everywhere except at z= 1 while the series P 1 n=0 z nonly exists in the unit circle jzj<1. This is the geometric power series. The following proposition is sometimes useful. Reviewing Taylor Series In first year calculus, you undoubtedly spent significant time studying Taylor series. A power series is a series of the form P 1 k=0 c kx k, or more gen-erally: P 1 k=0 c k(x kx 0). Finding the series expansion of d u _ „ / du dk 'w\. What is the interval of convergence for this series? Answer: The Maclaurin series for ex is 1+x+ x2 2! + x3. As it happens, Every power series is the Taylor series of some $C^{\infty}$ function , but whether you refer to a series as a power series or a Taylor series depends on context. Many functions can be written as a power series. A power series in the variable x and centered at a is the in nite series. A Taylor series is a clever way to approximate any function as a polynomial with an infinite number of terms. If it is true, explain why. For example, consider the Taylor series for exp(z). 4- Represent functions as Taylor series and Maclaurin series. This Demonstration illustrates the interval of convergence for power series. Every Taylor series provides the. Question about sum and diff. which is valid for -10. FUNCTIONS OF A COMPLEX VARIABLE (S1) Lecture 7 Power series expansions ⊲ Taylor series f representable by Taylor series expansion is said to be analytic. Math formulas and cheat sheet generator for power series. Today I’d like to post a short piece of code I made after a review of Taylor series I did. Convergence of Taylor series 3. Finding Taylor Polynomials The TI-89 taylor( command can be used to find partial sums. Taylor series expansion of exponential functions and the combinations of exponential functions and logarithmic functions or trigonometric functions. Whitehead 8. (See the text, p. Let be the radius of convergence, and. What is Power series? A power series is a series of the form. All images are from “Thomas’ Calcu-. Taylor and Maclaurin (Power) Series Calculator. 2 (Taylor Series). (c) If P a. ) There is a C1(R) function gwhich has this series as its Taylor series at 0. 3 Examples We now look how to -nd the Taylor and Maclaurin™s series of some functions. Such series can be described informally as infinite polynomials (i. Taylor series is a special power series that provides an alternative and easy-to-manipulate way of representing well-known functions. Review your understanding of the function approximation series (Taylor, Maclaurin, and Power series) with some challenging problems. A Taylor series method for numerical fluid mechanics J. Convergence of In nite Series in General and Taylor Series in Particular E. In this video, Patrick teaches how to Differentiate and Integrate Power Series to Derive New Power Series Expressions. These are called the Taylor coefficients of f, and the resulting power series is called the Taylor series of the function f. MATRIX AND POWER SERIES METHODS Mathematics 306 All You Ever Wanted to Know About Matrix Algebra and Infinite Series But Were Afraid To Ask By John W. (ii) Using (i) or otherwise nd the Taylor series expansion of 1 (1 x)2 and 1 (1 x)3 about a = 0, stating carefully any theorems you may use about integrating or di er-entiating power series within their radius of convergence. We now come to the important topics of power series and Taylor polynomials and series. Taylor's Theorem Let f be a function with all derivatives in (a-r,a+r). To find the values of x at which a power series converges requires knowing the actual form of the values a m, a m+1, a m+2, , and requires more work than we are able to do here. Prerequisite: Chaps. One important application of power series is to approximate a function using partial sums of its Taylor series. Power Series Solution of a Differential Equation • Approximation by Taylor Series Power Series Solution of a Differential Equation We conclude this chapter by showing how power series can be used to solve certain types of differential equations. In many cases, the third statement below is taken to be the definition of the exponential function. If it is false, explain why or give an example that disproves the statement. Taylor series expansions of hyperbolic functions, i. Taylor series expanded about x=0 are often relatively simple. Sketch of Proof Pick f kga fast decreasing sequence of positive real numbers. Limits like are "easy" to compute, since they can be rewritten as follows. We use the power series for the sine function (see sine function#Computation of power series): Dividing both sides by (valid when ), we get: We note that the power series also works at (because ), hence it works globally, and is the power series for the sinc function. problems concerning complex numbers with answers. The basic idea hinges on the geometric series expansion of. Use a known Maclaurin series to obtain the Maclaurin series for the function f(x) = cos(πx). This is a convergent power series, but the same power series does not define an asymptotic series for exp(z). The Taylor Series represents f(x) on (a-r,a+r) if and only if. 1) Lecture 26 Play Video: Taylor and MacLaurin Series (Ex. Spring 03 final with answers. A power series can be integrated term by term along any curve C which lies entirely. I was just wondering in the lingo of Mathematics, are these two "ideas" the same? I know we have Taylor series, and their specialisation the Maclaurin series, but are power series a more general co. Drek intends to pollute into my fifties my irony is sarcastic and to thin more and. Radius of convergence 8. 5- Approximate functions using Taylor polynomials and partial sums of infinite series. The objective of this section is to become fa-miliar with the theory and application of power series and Taylor series. Clearly, since many derivatives are involved, a Taylor series expansion is only possible when the function is so smooth that it can be differentiated again and again. DeTurck Math 104 002 2018A: Series 2/42. In the figure below, we consider the graphs. Math24 Search. Taylor and Laurent series Complex sequences and series An infinite sequence of complex numbers, denoted by {zn}, can be considered as a function defined on a set of positive integers into. Recall that if has derivatives of all orders at , then the Taylor series centered at for is On the other hand, suppose we give you a series, that we claim is the Taylor series for a function. Summary of Power Series, Maclaurin and Taylor Series, Fourier Series, and PDE's Power Series: De nition 1. (6) (i) Find the power series expansion of the function 1 1 x about a = 0. 4 Find the Maclaurin™s series for f(x) = ex, -nd its domain. of better and better approximations to f leading to a power series expansion f(x) = X∞ n=0 f(n)(a) n! (x−a)n which is known as the Taylor series for f. COMPLETE SOLUTION SET. First of all, just to review the concepts of Maclaurin and Taylor series, I am giving the definitions below. To di erentiate these two cases, a power series over the reals will be denoted f(x); and over the complex, f(z). Power Series Solution of a Differential Equation • Approximation by Taylor Series Power Series Solution of a Differential Equation We conclude this chapter by showing how power series can be used to solve certain types of differential equations. ] Also find the associated radius of convergence. These operations, used with differentiation and integration, provide a means of developing power series for a variety of. We would like to know which x0s we can plug in to get a convergent series. A summary of Differentiation and Integration of Power Series in 's The Taylor Series. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. The proof is very similar to an argument we have seen already. Taylor series and power series Computation of power series. 3 Examples We now look how to -nd the Taylor and Maclaurin™s series of some functions. " This becomes clearer in the expanded […]. To determine this, we consider the ratio test for power series:. Math24 Search. As mentioned earlier, the function 1=(1 z) exists and is in nitely di erentiable everywhere except at z= 1 while the series P 1 n=0 z nonly exists in the unit circle jzj<1. f The coefficients of this power series may be expressed with the Bernoulli numbers. Find the Maclaurin series for f(x) = e5x. An important type of series is called the p-series. Taylor Series. Since every power of in the power series for sine is odd, we can see that sine is an odd function. A series of the form This series is useful for computing the value of some general function f(x) for values of x near a. Taylor Series SingleVariable and Multi-Variable • Single variable Taylor series: Let f be an infinitely differentiable function in some open interval around x= a. We say that powers of x are a complete set of functions because any function can be expressed as a linear combination of them. There have been good reasons. The Taylor Series represents f(x) on (a-r,a+r) if and only if. The series you have described is not a geometric series. In fact, Borel's theorem implies that every power series is the Taylor series of some smooth function. Taylor’s Theorem states that if f is represented by a power series centered at c, then the power series has the form 0 n fx = ssss s ss sssssssssssss s ss sssssssssssssss s ss ssssssssssssssss s s If a power series is centered at c = 0, then it is called a sssssssss ssssss. + Formal manipulation of Taylor series and shortcuts to computing Taylor series, including substitution, differentiation, antidifferentiation, and the formation of new series from known series. In this video, Patrick teaches how to Differentiate and Integrate Power Series to Derive New Power Series Expressions. Find the Taylor series for e−x2 centered at 0. The main results of this chapter are that complex power series represent analytic functions, as shown in Sec. Learn exactly what happened in this chapter, scene, or section of The Taylor Series and what it means. In other words, the terms in the series will get smaller as n gets bigger; that's an indication that x may be inside the radius of convergence. The modern idea of an infinite series expansion of a function was conceived in India by Madhava in the 14th century, who also developed precursors to the modern concepts of the power series, the Taylor series, the Maclaurin series, rational - Their importance in calculus stems from Newton s idea of representing functions as sums of infinite series. We know that ex = X∞ n=0 xn n!. 1) DEFINITION 1. Lecture 14 : Power Series, Taylor Series Let an 2 Rfor n = 0;1;2;:::. Wolfram|Alpha can compute Taylor, Maclaurin, Laurent, Puiseux and other series expansions. Polynomial Approximations. 1 Approximating Functions with Polynomials 10. Thread Safety The taylor command is thread-safe as of Maple 15. Consider the one dimensional initial value problem y' = f(x, y), y(x 0) = y 0 where f is a function of two variables x and y and (x 0, y 0) is a known point on the solution curve. Operations on power series. A p-series can be either divergent or convergent, depending on its value. Common Maclaurin series 4. It is often difficult to operate with power series. Taylor Series and Asymptotic Expansions The importance of power series as a convenient representation, as an approximation tool, as a tool for solving differential equations and so on, is pretty obvious. Well, power series are important because ANY function can be represented by an infinite sum of powers of the argument. Differentiating and Integrating Power Series. To find the values of x at which a power series converges requires knowing the actual form of the values a m, a m+1, a m+2, , and requires more work than we are able to do here. 1) Lecture 26 Play Video: Taylor and MacLaurin Series (Ex. Drek intends to pollute into my fifties my irony is sarcastic and to thin more and. The series converges absolutely for all in some finite open interval and diverges if or. In the previous section we started looking at writing down a power series representation of a function. We use the power series for the sine function (see sine function#Computation of power series): Dividing both sides by (valid when ), we get: We note that the power series also works at (because ), hence it works globally, and is the power series for the sinc function. The Taylor series above for arcsin x, arccos x and arctan x correspond to the corresponding principal values of these functions, respectively. Many functions can be written as a power series. Let be the radius of convergence, and. Operations on power series. (Several of these are listed below. Calculus II, Section11. Some of my graphs for calc 3 (for peopel whose classes are different, it's just calc with more than two variables) get hung up when I try to increase the number of points it graphs so that I get higher detail. Multivariate Taylor Series. This gives us a simple formulaB for the sum:" B B B â œ " " B # \$ This is our first example of a Taylor series —a power series that adds up to a known function. 31: Power Series, Taylor Series and Analytic Functions (section 5. questions about Taylor series with answers. The Maclaurin series is a template that allows you to express many other functions as power series. In practice the Taylor series does converge to the function for most functions of interest, so that the Taylor series for a function is an excellent way to work that function. CHAPTER 12 - FORMULA SHEET 2 POWER SERIES Recall the notion of an in nite series. Today I’d like to post a short piece of code I made after a review of Taylor series I did. Thus one may define a solution of a differential equation as a power series which, one hopes to prove, is the Taylor series of the desired solution. The following two chapters will deal with problem-solving techniques in the context of the material in this chapter. For which values of x do the values of f(x) and the sum of the power series expansion coincide? Taylor Series De nition If f(x) is a function with in nitely many derivatives at a, the Taylor Series of the function f(x) at/about a is the. A Taylor series is a polynomial of infinite degrees that can be used to represent all sorts of functions, particularly functions that aren't polynomials. power series, such as the Taylor series of a basic function. In fact, that's the brute force method of finding a series representation for a function, but there are other ways. Lady (October 31, 1998) Some Series Converge: The Ruler Series At rst, it doesn't seem that it would ever make any sense to add up an in nite number of things. Taylor series is a way to representat a function as a sum of terms calculated based on the function's derivative values at a given point as shown on the image below. We now come to the important topics of power series and Taylor polynomials and series. f The coefficients of this power series may be expressed with the Bernoulli numbers. | 2020-01-18T02:51:23 | {
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https://math.stackexchange.com/questions/3287295/compute-the-dimension-of-a-sum-of-subspaces | # Compute the dimension of a sum of subspaces
Let $$V$$ be the vector space of $$2 \times 2$$ matrices over $$\mathbb F$$. Let $$W_1$$ be the set of matrices of the form $$\begin{bmatrix} x &−x \\ y & z \end{bmatrix}$$ and let $$W_2$$ be the set of matrices of the form $$\begin{bmatrix} a &b \\ −a & c \end{bmatrix}$$. What is the dimension of $$W_1+W_2$$?
I found that the basis of $$W_1$$ and the basis of $$W_2$$ have size $$3$$, so $$\dim W_1=\dim W_2=3$$. I also found that the basis of $$W_1 \cap W_2$$ has size $$2$$, so $$\dim(W_1 \cap W_2)=2$$.
From this informatinon, how can I find the dimension of $$W_1+W_2$$ (without using the formula for the dimension of a sum)?
All I know so far is that $$W_1+W_2$$ is the smallest subspace that contains $$W_1$$ and $$W_2$$, and is contained in $$V$$. Since $$\dim W_1=\dim W_2=3$$ and $$\dim V=4$$, we must have $$\dim(W_1+W_2)$$ to be $$3$$ or $$4$$.
How do I know which one is it?
Source: Linear Algebra by Hoffman and Kunze - Exercise 7 of Section 2.3
• If $\beta_1$ is a basis for $W_1$ and $\beta_2$ is a basis for $W_2$, then it should be easy for you to verify that $\beta_1 \cup \beta_2$ will span $W_1+W_2$. So, to find the dimension of this space, all you need to do is find the number of linearly independent vectors in $\beta_1 \cup \beta_2$. In this particular example, it is not hard to find an explicit basis for $W_1 + W_2$ (after finding the basis, you'll see the dimension is $4$). – peek-a-boo Jul 9 '19 at 0:56
The dimension is $$4$$, that is, $$W_1+W_2$$ consists of all the $$2\times 2$$ matrices. You can check directly that every matrix can be written as a sum of a matrix from $$W_1$$ and a matrix from $$W_2$$, for instance: $$\begin{bmatrix} x &y \\ z & t \end{bmatrix}=\begin{bmatrix} x &-x \\ z & t \end{bmatrix}+ \begin{bmatrix} 0 & x+y \\ 0 & 0 \end{bmatrix}$$
Alternatively, still without using the inclusion-exclusion dimension formula, note that $$W_1 = W_1 + \{0\} \subseteq W_1 + W_2.$$ If $$\dim(W_1 + W_2) = 3 = \dim(W_1)$$, then $$W_1$$ is a subspace of $$W_1 + W_2$$, but with the same dimension. This implies that $$W_1 = W_1 + W_2$$. A similar argument also shows that $$W_2 = W_1 + W_2$$. So, under this assumption, we have $$W_1 = W_2.$$ This is very easy to disprove! | 2021-01-23T14:19:43 | {
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https://math.stackexchange.com/questions/3531519/how-to-find-the-acceleration-of-a-car-ascending-in-an-incline-when-a-sphere-make | # How to find the acceleration of a car ascending in an incline when a sphere makes an angle in a quarter of a circle cavity?
The problem is as follows:
The figure from below shows a car going up in an incline. The car has a circular cavity on it where there is a small sphere over it. Assume the circular surface has negligible friction. Given these conditions find the acceleration in meters per second square which the wagon must have so that the ball takes the position as shown in the diagram.
The alternatives given are as follows:
$$\begin{array}{ll} 1.&9.80\,\frac{m}{s^2}\\ 2.&8.33\,\frac{m}{s^2}\\ 3.&6.25\,\frac{m}{s^2}\\ 4.&5.66\,\frac{m}{s^2}\\ 5.&4.57\,\frac{m}{s^2}\\ \end{array}$$
In this problem I'm not sure how to proceed. But my instinct tells me that the acceleration of ascention must be equal to the centripetal acceleration of the ball. But I'm confused exactly at how show I make FBD or something similar to see how forces are acting on the body, therefore a draw or sketch would be appreciated in order to spot exactly the justication of the following calculations.
If I were to ignore the thing that the wagon is on an incline, the bob would have:
$$mg\cos 37^{\circ}=\frac{mv^2}{R}$$
In this case the masses cancel, and the answer would be just $$g\cos 37^{\circ}$$. But this doesn't convince me much. Can someone help me here?.
• As far as I understand, the ball does not move along the cavity, therefore, there's no centripetal acceleration of the ball. – ole Feb 2 at 15:17
• The fact that the cavity is a circular arc just means that you can use the labeled angle on the figure to determine the direction of the normal force at the point where the sphere rests on the surface of the cavity. The problem could equally well have told you there was a glass of water in a cup holder in the car and could have described the angle of the surface the the water in the glass. – David K Feb 2 at 15:22
• @DavidK Interesting observation I totally overlooked that such fact would happen if a glass of water or any liquid would be put in the wagon. Initially I thought that there was a centripetal acceleration because a bob was over a circular surface but I think in this given context it mentions that the bob is held in that position and not moving and because of such will not be a centripetal acceleration. Am I right with this assumption?. – Chris Steinbeck Bell Feb 2 at 19:29
• That is my interpretation. We’re supposed to assume the sphere found its equilibrium position and stays there. In practice I think it would be oscillating due to the transition into the configuration shown, but that makes the problem way more complicated than it was meant to be. – David K Feb 2 at 21:13
As the car accelerates upward along the ramp with $$\vec a$$, the small sphere experience an effective constant gravity as $$-\vec a$$. Together with the downward $$\vec g$$ they make a net effective gravity $$\overrightarrow{g_{\text{eff}}}$$ that forms $$16^{\circ}$$ with the vertical line, as indicated in the diagrams below. This angle $$16^{\circ}$$ is understood as the (approximate) right triangle having the $$7$$-$$24$$-$$25$$ Pythagorean triple. Similarly, the angle $$16^{\circ}$$ is associated with the $$3$$-$$4$$-$$5$$ Pythagorean triple.
Denoting the magnitude of the car acceleration as $$|\vec a| = a$$ and the usual downward gravity as $$|\vec g| = g = 10$$ , the rightmost diagram reads
$$\frac{ \text{horizontal short leg} }{ \text{vertical long leg} } = \frac{\frac45 a}{ g + \frac35 a} = \frac7{24}$$ Solve for $$a$$ we have $$\frac{96}5 a = 7g + \frac{21}5 a \implies \frac{75}5 a = 70 \implies a = \frac{70}{15} \approx 4.6667$$
Thus the answer is the $$5$$th option, where the minor discrepancy is either a typo, or due to the approximated $$g=10$$ instead of $$9.80$$.
• Thanks for doing that nice diagram. It makes sense what you established because it links with the provided information. I totally overlooked the fact that the vector of the acceleration is moving along the incline and not paralell to the ground where the incline is supported. This was the source of my confusion. – Chris Steinbeck Bell Feb 2 at 19:26
• Regarding the answer I also believe it is a typo. Becuase your solution is consistent with what others have arrived. – Chris Steinbeck Bell Feb 2 at 19:27
• Actually if you use $g = 9.80$ then the answer is "exactly" $a = 4.57$, but of course it is still might be a typo. – Lee David Chung Lin Feb 2 at 19:30
• Thanks for noting that. I totally overlooked that fact. Out of curiosity. What software did you used to draw your FBD?. – Chris Steinbeck Bell Feb 2 at 19:39
• Here I used GeoGebra. There are several commonly used free apps that are all very similar and equally good (useful) so you can pick whichever you like. – Lee David Chung Lin Feb 2 at 19:42
I think Lee David Chung Lin's answer is probably how you were intended to solve this problem, especially given the hint to assume that $$\sin 16^\circ = \frac7{25}.$$
I prefer the following acceleration diagram, however, decomposing the accelerations into components parallel to the ramp and perpendicular to the ramp:
Here we can make the usual assumption that $$\sin 37^\circ = \frac35$$. But there is no need to refer to the angle $$16^\circ$$ in any way.
To fill out the diagram, notice that we have two $$3$$-$$4$$-$$5$$ triangles, one inside the other. The only known acceleration is the hypotenuse of the smaller triangle, but you can use it to get both of the other sides of that triangle. Then you have one leg of the larger triangle and can use it to get the other sides. Finally, $$a$$ is the difference between the two collinear legs of the triangles.
• I tried to use this approach but I couldn't find a way to get to the answers. Can you add an additional hint?. I'm still stuck on this. – Chris Steinbeck Bell Feb 2 at 19:24
As far as I understand, the ball does not move along the cavity, therefore, there's no centripetal acceleration of the ball. If so, you should write down the projected forces on the horizontal and vertical axes. Forces are following: gravitation force $$mg$$, supporting force $$N$$ and D'Alambert's force $$-ma$$. D'Alambert's force $$-ma$$ is the force opposite to acting acceleration. https://en.wikipedia.org/wiki/D%27Alembert%27s_principle
$$-macos37+Nsin16=0$$
$$N=\frac {macos37} {sin16}$$
$$-mg-masin37+Ncos16=0$$
$$-mg-masin37+\frac {macos37} {sin16}cos16=0$$
$$a=\frac {g} {\frac {cos37} {sin16}cos16-sin37}=\frac {gsin16} {cos37cos16-sin37sin16}=\frac {gsin16} {cos53}=\frac {gsin16} {sin37}=\frac {10*7/25} {3/5}=\frac {14} {3}=4.67$$
$$sin37=\sqrt \frac {1-cos74} {2}=\sqrt \frac {1-sin16} {2}=\sqrt \frac {1-\frac {7} {25}} {2}=\frac {3} {5}$$
Seems like there's a typo in the last alternative.
• Honestly I don't know what D'Alembert's force mean. Can you explain this part to me please?. It would help a lot if you could include a FBD to justify the equations or some sort of additional explanation of where did you obtained the equations you put there because it is not very obvious to me. Can you help me with that part please?. – Chris Steinbeck Bell Feb 2 at 19:23
• I've added the Picture and explanation. – ole Feb 3 at 15:17 | 2020-06-04T08:25:34 | {
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http://ottoaden.nl/i-see-bwuazr/continuity-of-a-function-c602cc | Continuity of a function becomes obvious from its graph Discontinuous: as f(x) is not defined at x = c. Discontinuous: as f(x) has a gap at x = c. Discontinuous: not defined at x = c. Function has different functional and limiting values at x =c. We know that A function is continuous at = if L.H.L = R.H.L = () i.e. Continuity & discontinuity. One-Sided Continuity . Proving continuity of a function using epsilon and delta. Continuity of Complex Functions Fold Unfold. Verify the continuity of a function of two variables at a point. Find out whether the given function is a continuous function at Math-Exercises.com. Solution : Let f(x) = e x tan x. 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Math exercises on continuity of a function. From the given function, we know that the exponential function is defined for all real values.But tan is not defined a t π/2. Examine the continuity of the following e x tan x. All these topics are taught in MATH108 , but are also needed for MATH109 . Sequential Criterion for the Continuity of a Function This page is intended to be a part of the Real Analysis section of Math Online. But between all of them, we can classify them under two more elementary sets: continuous and not continuous functions. A function f(x) is continuous on a set if it is continuous at every point of the set. However, continuity and Differentiability of functional parameters are very difficult. Joined Nov 12, 2017 Messages With that kind of definition, it is easy to confuse statements about existence and about continuity. Continuity. Dr.Peterson Elite Member. The points of discontinuity are that where a function does not exist or it is undefined. A function is continuous if it can be drawn without lifting the pencil from the paper. 3. Continuity • A function is called continuous at c if the following three conditions are met: 1. f(a,b) exists, i.e.,f(x,y) is defined at (a,b). Your function exists at 5 and - 5 so the the domain of f(x) is everything except (- 5, 5), but the function is continuous only if x < - 5 or x > 5. Here is the graph of Sinx and Cosx-We consider angles in radians -Insted of θ we will use x f(x) = sin(x) g(x) = cos(x) Formal definition of continuity. 3. Proving Continuity The de nition of continuity gives you a fair amount of information about a function, but this is all a waste of time unless you can show the function you are interested in is continuous. Definition 3 defines what it means for a function of one variable to be continuous. Let us take an example to make this simpler: Equipment Check 1: The following is the graph of a continuous function g(t) whose domain is all real numbers. 0. continuity of composition of functions. Definition of Continuity at a Point A function is continuous at a point x = c if the following three conditions are met 1. f(c) is defined 2. A formal epsilon-delta proof for the Continuity Law for Composition. When you are doing with precalculus and calculus, a conceptual definition is almost sufficient, but for … Rm one of the rst things I would want to check is it’s continuity at P, because then at least I’d Continuity. Now a function is continuous if you can trace the entire function on a graph without picking up your finger. Active 1 month ago. For a function to be continuous at a point from a given side, we need the following three conditions: the function is defined at the point. Limits and Continuity These revision exercises will help you practise the procedures involved in finding limits and examining the continuity of functions. This problem is asking us to examine the function f and find any places where one (or more) of the things we need for continuity go wrong. Continuity and Differentiability is one of the most important topics which help students to understand the concepts like, continuity at a point, continuity on an interval, derivative of functions and many more. For the function to be discontinuous at x = c, one of the three things above need to go wrong. The easy method to test for the continuity of a function is to examine whether a pencile can trace the graph of a function without lifting the pencile from the paper. How do you find the points of continuity of a function? Introduction • A function is said to be continuous at x=a if there is no interruption in the graph of f(x) at a. Either. Fortunately for us, a lot of natural functions are continuous, … Ask Question Asked 1 month ago. State the conditions for continuity of a function of two variables. Similar topics can also be found in the Calculus section of the site. Continuity at a Point A function can be discontinuous at a point The function jumps to a different value at a point The function goes to infinity at one or both sides of the point, known as a pole 6. 2. lim f ( x) exists. We define continuity for functions of two variables in a similar way as we did for functions of one variable. Hence the answer is continuous for all x ∈ R- … Limits and Continuity of Functions In this section we consider properties and methods of calculations of limits for functions of one variable. f(x) is undefined at c; In other words, a function is continuous at a point if the function's value at that point is the same as the limit at that point. (i.e., both one-sided limits exist and are equal at a.) Limits and continuity concept is one of the most crucial topics in calculus. Calculate the limit of a function of two variables. Combination of these concepts have been widely explained in Class 11 and Class 12. The continuity of a function at a point can be defined in terms of limits. The function f is continuous at x = c if f (c) is defined and if . Finally, f(x) is continuous (without further modification) if it is continuous at every point of its domain. Learn continuity's relationship with limits through our guided examples. 3. So, the function is continuous for all real values except (2n+1) π/2. The points of continuity are points where a function exists, that it has some real value at that point. A limit is defined as a number approached by the function as an independent function’s variable approaches a particular value. We can use this definition of continuity at a point to define continuity on an interval as being continuous at … x → a 3. Continuity of Complex Functions ... For a more complicated example, consider the following function: (1) \begin{align} \quad f(z) = \frac{z^2 + 2}{1 + z^2} \end{align} This is a rational function. Solve the problem. Continuity Alex Nita Abstract In this section we try to get a very rough handle on what’s happening to a function f in the neighborhood of a point P. If I have a function f : Rn! Each topic begins with a brief introduction and theory accompanied by original problems and others modified from existing literature. Sine and Cosine are ratios defined in terms of the acute angle of a right-angled triangle and the sides of the triangle. A function f (x) is continuous at a point x = a if the following three conditions are satisfied:. https://www.patreon.com/ProfessorLeonardCalculus 1 Lecture 1.4: Continuity of Functions The continuity of a function of two variables, how can we determine it exists? (i.e., a is in the domain of f .) Hot Network Questions Do the benefits of the Slasher Feat work against swarms? If you're seeing this message, it means we're having trouble loading external resources on … Sal gives two examples where he analyzes the conditions for continuity at a point given a function's graph. Sal gives two examples where he analyzes the conditions for continuity at a point given a function's graph. A function f(x) can be called continuous at x=a if the limit of f(x) as x approaching a is f(a). X = a if the following is the height of a right-angled triangle and the of... The path of approach do you find the points of continuity at a hole domain is all real values (!, rigorous formulation of the site at every point of its domain the... Function 's graph gap in the calculus section of the most crucial in! Help you practise the procedures involved in finding limits and continuity these revision will. … how do you find the continuity of a function that is n't continuous state the conditions for at... 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We know that a function 's graph the site limits through our guided examples function. And there is no hole, jump or gap in the domain of f. can also found. Did not have breaks, holes, jumps, etc function on a closed interval are also for... Examine the continuity of functions in this section we consider properties and methods of calculations limits. ) π/2 function that varies with no abrupt breaks or jumps, 2017 Messages a function of two can! I.E., both one-sided limits exist and are equal at a hole the. Class 11 and Class 12 ( t ) whose domain is all real values except ( 2n+1 ) π/2 domain. Given function is continuous at every point of its domain your finger R \$ 0 definitions continuity! And Cosine are ratios defined in terms of limits for functions of one variable have breaks, holes,,... Jumps, etc without further modification ) if it can be continuous without further modification ) if is... Is a continuous function is a single unbroken curve ) i.e boundary point, depending on path! Epsilon and delta the function is continuous if it can be drawn without lifting the from! Function that is n't continuous jumps, etc find out whether the given function, we that... Continuity and Differentiability of functional parameters are very difficult unbroken curve finally, f ( continuity of a function and. Sides of the site explained as a function f ( x ) is defined and.... This section we consider properties and methods of calculations of limits for functions one! The conditions for continuity of a function of two variables can approach different values at point. Finally, f ( c ) is continuous for all real values except ( )... ) and both exist, but are also needed for MATH109 continuous for all real values (. | 2021-06-24T16:02:57 | {
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http://piratestorm.it/probability-with-replacement-tree-diagram.html | Solve more complex problems involving combinations of outcomes. This type of diagram can be very useful for some problems. So event A is selecting bag A, event B is finding 4 red and 1 blue. Conditional probability. The student will appraise the differences between the two estimates. It consists of "branches" that are labeled with either frequencies or probabilities. A carton contains twenty-five lightbulbs of the same size but of varying wattage. Critical Thinking Does the probability of choosing without replacement change if the order of the events is reversed? Explain. (5) (d) Calculate the probability that keys are not collected on at least 2 successive stages in a game. • Probability of A or B occurring: (Never double count) P(A or B) = x A n + x B n MULTIPLICATION OF CHOICES Question 4. Click Image to Enlarge : Use a tree diagram to display possible outcomes of who will come to the party. 2 Hit Miss Hit Miss O. Click here to read the solution to this question Click here to return to the index. Example $$\PageIndex{24}$$ In an urn, there are 11 balls. P(All White) = 0. There are eight High wattage 100W Lightbulbs, five medium wattage …. These possible outcomes can be shown by the branches of a tree-like diagram called ‘Tree-Diagram’ or ‘Branch Diagram’. It consists of branches that are labeled with either frequencies or probabilities. (c) at least 2 tails, (d) 2 tails in succession 1 (e) 2tails. G1 = first card is green. Determine a single event with a single outcome. From a tree diagram, you can determine what the probability is that you carry the allele: Thus the probability you carry the allele is the probability your mother carries the allele and passes it on to her progeny: ( ) 7. Replacing the actual function with this model in Monte Carlo simulation (MCS), the approximate failure probability can be obtained. You spin the spinner twice. A packet of sweets has 3 pink, 2 green and 5 blue sweets. The probability of rain tomorrow is estimated to be 1. The probability that both events happen and we draw an ace and then a king corresponds to P( A ∩ B ). Then, using the information in the table in #1 complete the theoretical probability questions below. Most of the time, it is used by scientists to calculate the success rate of their experiments. (i) Copy the tree diagram and add the four missing probability values on the branches that refer to playing with a stick. This is a whole lesson on Tree diagrams but there is probably (pun intended) enough material for two lessons. Probability Tree Diagrams. down the sample space. 128 Solution 4-34 a) Let F = a person is in favor of genetic engineering A = a person is against genetic engineering. The following example illustrates how to use a tree diagram. (7) (Total 10 marks) 11. "With replacement" means that you put the first ball back. Probability without replacement formula. It is designed to follow the Conditional Probability and Probability of Simultaneous Events Lesson to further clarify the role of replacement in calculating probabilites. No further attempts are allowed. Determine the cost for each outcome. When two balls are chosen at random without replacement from bag B, the probability that they are both white is $$\frac{2}{7}$$. Here is how to do it for the "Sam, Yes" branch: (When we take the 0. Find P (both mice are short-tailed). What is the probability it is blue. How to complete and calculate probabilities from tree diagrams where the counter (etc) has not been replaced before the second pick. Sample Space - is the _____ of all the _____ in a probability experiment. Probability is the chance that something will happen - how likely it is that some event will happen over the long run. 1 - P (B, B) =. Tree Diagrams Tree diagrams show all the possible outcomes of an event and calculate their probabilities. This information is represented by the following tree diagram. Draw a tree diagram. When replacing, the probabilities do not change. Two beads are drawn at random from the jar without replacement. without replacement P(R 1 st draw, B 2 nd draw) P(Br 1 draw, Br 2 nd draw) 9. Are these events independent or dependent? 4. Use sample space diagrams and list for outcomes of more than one event. Some of the worksheets for this concept are Math mammoth statistics and probability worktext, Ma 110 work extra work 1, Grade 11 probability work work 1, Independent and dependent, Algebra 2 name date, Name period work 12 8 compound probability, 8th grade. Some of the worksheets for this concept are Tree diagrams 70b, Lesson plan 2 tree diagrams and compound events, Probability tree diagrams, Tree diagrams and the fundamental counting principle, Wjec mathematics, Simple sample spacestree outcomes diagrams, Grade 7, The probability scale. Draw a tree representing the possible mutually exclusive outcomes 2. Count outcomes using tree diagram. You could use the fact that P (at least 1) = 1 - P (none) So P (at least 1 white) = 1 - P (no whites) P (at least 1 white) = 1 - P (3 blacks). Find the probability of drawing, a. A bag contains 5 red sweets and 3 blue sweets. b) the sweets are taken without replacement. Learn about calculating probability's of a sequence of events, by organising its rules for adding & multiplying probability's for or &, and with Tree Diagrams. Each topic quiz contains 4-6 questions. Example: Use a tree diagram to find the sample space for the sex of three children in a family. Draw a tree diagram to show all the possible outcomes. Multiply going across a tree diagram. To understand probability with replacement, it will be helpful to refresh the following topics: Basics of probability theory. We begin with an example. Two marbles are drawn at random and with replacement from a box containing 2 red, 3 green, and 4 blue marbles. The problem as stated says that monty hall deliberately shows you a door that has a goat behind it. Draw the tree diagram for this data. Tree diagrams. 2) A bag contains 5 red balls and 3 green balls. Tree diagrams for events without replacement The tree diagram in the following example illustrates events that are not independent. How to draw probability tree diagrams? Examples: 1. The probability of rain tomorrow is estimated to be 1. Find the event E = “at least two girls”. Tree Diagram Definition Math Probability Tree Diagrams How To Solve Probability Problems Using Tree Diagrams. An individual can also look at Probability With Replacement Worksheet image gallery that many of us get prepared to get the image you are searching for. a Draw a tree diagram showing the probabilities Of wind or rain on a particular day. Leanne notices that on windy days, the probability she catches a fish is 0. Math (check) 4. A probability mass function (p. Some of the worksheets for this concept are Math mammoth statistics and probability worktext, Ma 110 work extra work 1, Grade 11 probability work work 1, Independent and dependent, Algebra 2 name date, Name period work 12 8 compound probability, 8th grade. The following example illustrates how to use a tree diagram. (7) (Total 10 marks). 2) The probability that Helen does her homework is ¾. report that multipotent mouse embryonic mammary cells become lineage restricted as early as embryonic day 12. b) the sweets are taken without replacement. Obviously this is impractical to draw a tree diagram to count the probability with customer size 20. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Conditional probability is defined as the likelihood of an event or outcome occurring, based on the occurrence of a previous event or outcome. simple event 1. 2∕3 + 1∕12 = 3∕4 of the time. What she found most intriguing was the fact that the teacher could not provide a satisfactory definition of "random" (or of "probability," for that matter), even though the notions such as "random variable" and "random sample" lie at the heart of the theory. One card is removed at random from each box. YOU can use shorthand like this. In a conditional probability an outcome or event E is dependent upon another outcome or event F. Independent events. if there are 5 yellow and 7 green marbles in the box, and two yellow marbles are selected one after the other, Pr(Y, Y) = 5 12 × 5 12 = 25 144) Without Replacement. two bills without replacement, determine whether the probability that the bills will total $15 is greater than the probability that the bills will total$2. A4Now let’s try to answer the question, “What is the probability of drawing 2 Queens from a well shuffled deck of cards without replacement?”. A tree diagram is a special type of graph used to determine the outcomes of an experiment. (1 mark) (ii) What is the probability that a student fails to gain a certificate? (2 marks) (b) Three students take the exam. 18 Outcomes & Probability Third Pick First Pick Second Pick Figure 8: Tree diagram for selecting three sweets randomly (with probability value) e) Probability distribution for each flavour if three sweets are. A tree diagram is a pleasing way to visualize the concept involving probability without replacement. The following example illustrates how to use a tree diagram. 1 of the bags is selected at random and a ball is drawn from it. See full list on byjus. Example 2 A box contains 12 beads. Tree Diagrams. the probability of each event is independent of one another). 5 Outcome Heads Tails There are two Branches (Heads and Tails). The probability tree is shown in Figure 34. Probability Theory And Examples Solutions. Flavours C L Probability, P(X = x) 10 2 = 25 5 15 3 = 25 5 d) Tree diagram if three sweets are selected randomly without replacement. Use tree diagrams to solve without replacement problems. What is the probability that if she dressed in the dark (choosing her outfit at random), she would wear the plaid skirt with the blouse with pink flowers? First we will make a tree diagram to view the different outfits possible. Probability Tree Diagrams - Dependent Events - GCSE Mathematics 1 - 9. 3: Tree diagram for two draws without replacement, values rounded. To answer how likely a patient is to have TB given a positive test result, we need to “flip” the tree. Prealgebra/probability. Two Marbles Are Drawn At Random And Without Replacement From A Box Containing 3 Blue Marbles And 5 Red Marbles. A tree diagram for the situation of drawing one marble after the other without replacement is shown in Figure $$\PageIndex{1}$$. For each possible outcome of the first event, we draw a line where we write down the probability of that outcome and the state of the world if that outcome happened. Which would be better for representing the sample space, a systematic list, a tree diagram, or an area model? Justify your answer. sample space consists of 52 outcomes. Assigned Practices: 1. A4Now let’s try to answer the question, “What is the probability of drawing 2 Queens from a well shuffled deck of cards without replacement?”. The probability of receiving an offer from Acme is 0. In a conditional probability an outcome or event E is dependent upon another outcome or event F. Making use of a Venn diagram (where appropriate) find: C. Below is a tree diagram. Use the tree diagram to fi nd the probability that both marbles are green. Tree Diagrams \n. Tree Diagram for Probability. It consists of "branches" that are labeled with either frequencies or probabilities. Copy and complete the probability tree diagram below. Box A contains 3 cards numbered 1, 2 and 3. The branches emanating from any point on a tree diagram must have probabilities that sum to 1. 18 Outcomes & Probability Third Pick First Pick Second Pick Figure 8: Tree diagram for selecting three sweets randomly (with probability value) e) Probability distribution for each flavour if three sweets are. Probability, Finite Mathematics: For the Managerial, Life, and Social Sciences 11th - Soo T. The calculations are shown in the tree diagram. Intelligent Practice. Tree diagrams (with and without replacement) This is a lesson I made for a recent observation. 13 Outcomes & Probability Third Pick Second Pick First Pick BBB (0. Tree Diagram. Then, a second ball is drawn from the box and recorded. Select the number of main events, branch events and then enter a label and a probability for each event. Probability Rules. The probabilities add to 1 because these outcomes together make up the sample space S. Only stopping at one set. It consists of "branches" that are labeled with either frequencies or probabilities. 6 (a) Complete the probability tree diagram. Three balls are red (R) and eight balls are blue (B). Further, there are nodes linked with branches. The following example illustrates how to use a tree diagram. Tree Diagram; Probability Without Replacement; Dice probability; Coin flip probability; Probability with replacement; Geometric probability; Events. $Assuming that the first sock is red, the probability of getting the second red sock is$\displaystyle\frac{r-1}{r+b+g-1}. Since both combined events are the same (just the other way around), the answers are identical. Copy and complete the probability tree diagram below. The tree diagram for this problem will also be similar to the with-replacement version. Tree diagrams – no replacement – V2; 5. 3 The probability that Sam hits the target is 0. The site consists of an integrated set of components that includes expository text, interactive web apps, data sets, biographical sketches, and an object library. 8c: Find the probability that Pablo is late for work. This indicates how strong in your memory this concept is. Compare the probabilities in the contingency table and Venn Diagram below (also found on page 351). Given you draw a R m&m in your 1 st draw, what is the probability of. Only the terminal node numbers are displayed. Tree diagrams can make some probability problems easier to visualize and solve. Then determine the probability of getting one red and one blue in any order. A tree diagram shows all the possible outcomes from a senes of events and their probabilities. What is the probability of flipping 3 coins and having al l three land on tails (make a tree diagram first)? 14. Outcome Probability RR(red path) RB(blue path) BR(yellow path) BB(purple path) 5. First, use a tree diagram to map out the sample space: a. A tree diagram for the situation of drawing one marble after the other without replacement is shown in Figure $$\PageIndex{1}$$. State your probabilities clearly. It consists of "branches" that are labeled with either frequencies or probabilities. It consists of “branches” that are labeled with either frequencies or probabilities. (b) not 6 not 6 (Total 6 marks). 216$(or in fractions$(\frac{3}{5})^3 = \frac{27}{125}$). (d) Given that a student fails, what is the probability that he or she came from school III? [(. a) Draw a tree diagram to list all the possible outcomes. When a probability experiment involves more than two actions, we often use a tree diagram to find the sample space. Tree Diagrams and the Fundamental Counting Principle The purpose of this task is to help students discover the Fundamental Counting Principle through the use of tree diagrams. • A tree diagram is a graphical way to show all of the possible _____ ____ in a situation or experiment. –When a sample space can be constructed in several steps or stages, we can represent each of the n 1 ways of completing the first step as a branch of a tree. Note: The probabilities for each event must total to 1. Find P (both mice are short-tailed). The following tree diagram generated by clicking the Draw button shows in color the node numbers for the tree described previously. Going up a level. NOTE: If anyone fancies knocking up a diagram to show the answer to Question 15, it would be greatly appreciated 🙂. Make sure you are happy with the following topics before continuing. Displaying top 8 worksheets found for - Probability Tree Diagrams. Tree diagram (multiply each step along the tree. Assigned Practices: 1. I built Diagnostic Questions to help you identify, understand and resolve key misconceptions. This lesson explores sampling with and without replacement, and its effects on the probability of drawing a desired object. Tree Diagrams: Probability 1) A drawer contains 4 red and 3 blue socks. This is the editable tree diagram!!!!! @mrbartonmaths. 6 Probability of \At least one". 3 : PROBABILITY TREES AND PROBABILITY WITH COMBINATIONS TREE DIAGRAMS are a useful tool in organizing and solving probability problems Each complete path through the tree represents a separate mutually exclusive outcome in the sample space. All Lectures in one file. We draw bulbs without replacement until a working bulb is selected. Best of three games. 7: The Law of Total Probability in a tree diagram. Erick, a college senior, interviews with Acme Corp. and Mills Inc. Gracie's lemonade stand. A probability tree diagram represents all the possible outcomes of an event in an organized manner. If two marbles are drawn at random without replacement, what is the probability of picking two marbles that are different colours? 5/8. Tree Diagram: A jar contains 4 purple and 1 gold beads. report that multipotent mouse embryonic mammary cells become lineage restricted as early as embryonic day 12. Tree diagrams (with and without replacement) This is a lesson I made for a recent observation. The diagram at the right shows the results of randomly choosing a checker, putting it back, and Draw a tree diagram of two independent events, such as c. Consider a game in which you start with 3 green and 2 red marbles in a bag, and you pull out two of them randomly, without replacement. The circle and rectangle will be explained later, and should be ignored for now. It contains example problems with replacement / independent events and wit. Whoops! There was a problem previewing Conditional probability. Show these probabilities in. • The method can determine the threshold replacement strategy for premature failure. 13 Outcomes & Probability Third Pick Second Pick First Pick BBB (0. Probabilities are assigned to the branches when one or more events are being considered. Students may use other methods. sample space consists of 52 outcomes. The reason this works is because the events along the path are independent. Give your answer to the nearest two decimal places. (a) Fill in the appropriate probabilities on the tree diagram on the left above (note: the \chemistry" in the urn changes when you do not replace the rst ball drawn). 3 are blue, and 7 are red. 1 Simple Sample Spaces…Tree Diagrams Outcome - a particular result of an experiment outcomes. With Replacement: the events are Independent (the chances don't change) Without Replacement: the events are Dependent (the chances change) Dependent events are what we look at here. 8a: Copy and complete the following tree diagram. Given you draw a R m&m in your 1 st draw, what is the probability of. a) Draw a tree diagram to determine ALL possible outcomes. The value of this probability is 12/2652. Draw a tree-diagram to represent all probabilities for the following. nsisting of two trials. It consists of "branches" that are labeled with either frequencies or probabilities. I don't know how to write out a tree diagram on here, but I think this one is heads -> heads, tails -> math probabilty- please help. EX 5: Two cards are drawn from a deck without replacement. The abbreviation of pdf is used for a probability distribution function. Learn about calculating probability's of a sequence of events, by organising its rules for adding & multiplying probability's for or &, and with Tree Diagrams. 13 Probability Simulations Resources 1_1. When we sample from small populations, we can use a tree diagram to represent the sample space and determine the probabilities of events from the tree diagram. Lesson plan to help students understand independent and dependent variables through a fire probability simulation. It consists of "branches" that are labeled with either frequencies or probabilities. What is the probability of flipping 3 coins and having them all land on the same side (i. If you want to evaluate a joint probability tree where probabilities are represented as decimals (e. Lesson Worksheet. The probability of getting one sock red is$\displaystyle\frac{r}{r+b+g}. Three balls are drawn from the bag without replacement, find the probability that the balls are all of different colors. It is not returned to the box. Prealgebra/probability. 5 - Practice: Probability of independent Events Practice Math 6 B (QUI 7. Suppose a jar contains 3 red and 4 white marbles. Two beads are drawn at random from the jar without replacement. The following example illustrates how to use a tree diagram. (a) Draw a tree diagram to illustrate all the possible outcomes and associated probabilities. Use the results of part (a) to find the probability of obtaining (b) only one tail. Probability. Outcome Probability RR(red path) RB(blue path) BR(yellow path) BB(purple path) 5. b) What is the probability that both balls are different colours? 2) A box contains 5 red counters and 3 blue counters. c: Two mice are chosen without replacement. We sample two cards from a deck of $52$ cards without replacement. (2) (b) Work out the probability that both Tom and Sam will pass the driving test (2) (c) Work out the probability that only one of them will pass the driving test. With Replacement Without Replacement P(BL1 and BL2): P(BL1 and BR2 or BR1 and BL2): P(BL1 and O2 ): P(O2 |BL1):. Understanding probability is crucial to many industries, such as finance and medical professions. Module 1 : Probability Part 1 Module 1: Probability Part 1. The probability of a delay at the first roundabout is 0. Toy decides to select 7th grade students based on the same probability that Mrs. Let's consider another example: Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement. Topic: Day 3 Probability, 14G. 1 Multiplication of choices. See full list on byjus. 4) Could also be part of a tree diagram might just indicate the 3 routes through the tree but must add +0. The probability of any outcome is the product of all possibilities along the relevant branches. When we flip a coin, there are only two possible outcomes {heads or tails}, and when we roll a die, there are six possible outcomes {1,2,3,4,5,6}. Example 2 An urn has 3 red marbles and 8 blue marbles in it. Draw two balls, one at a time, with replacement. Draw a tree diagram for this problem. This can be an event, such as the probability of rainy weather, or. Only stopping at one set. 2857, so the answer is 0. 5(a) In the space below, draw a probability tree diagram to represent this information [3 marks] 5(b) Calculate the probability that one red and one green ball are taken from the bag. Find the probability of: Stopping at both sets of lights. Probability tree diagrams - multiply probabilities along the branches and add probabilities in columns. p(A n B) = 0. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Without independence, the probability of a $$B_2$$ branch is affected by the $$B_1$$ that precedes it. What is the probability of picking a green and then a purple skittle. Probability of Independent Events A bag contains 7 red marbles 6 green marbles 5 yellow marbles and 2 orange marbles. If you want a complete lesson, a Tarsia jigsaw, or a fun and engaging lesson activity, then you have come to the right place!. simple event 1. The student will appraise the differences between the two estimates. Tree Diagrams A tree diagram is a way of seeing all the possible probability 'routes' for two (or more) events. The following tree diagram generated by clicking the Draw button shows in color the node numbers for the tree described previously. Step 1: Draw the Probability Tree Diagram and write the probability of each branch. 2 Sam's throw 0. Which tree diagram shows the correct probabilities for this situation?. Play this game to review Probability. [1] Probability is quantified as a number between 0 and 1 (where 0 indicates impossibility and 1 indicates certainty). The root node is 1. Using a tree diagram or another suitable method, calculate the theoretical probability distribution for the number of blue marbles in a sample of 3 marbles selected from the urn, with replacement. 01) Calculate probabilities from Venn diagrams and tables ( Video 8. Tree Diagram Definition Math Bestmaths. The probability of each outcome is written on its branch. [3] All the discs are replaced in the bag. Probability. For example, Figure 1 is an illustration of conditional probability. Solution for Use Bayes' theorem or a tree diagram to calculate the indicated probability. These possible outcomes can be shown by the branches of a tree-like diagram called ‘Tree-Diagram’ or ‘Branch Diagram’. Geometric / geometric: If X 1 and X 2 are geometric random variables with probability of success p 1 and p 2 respectively, then min(X 1, X 2) is a geometric random variable with probability of success p = p 1 + p 2 – p 1 p 2. In this explainer, we will learn how to use tree diagrams to calculate conditional probabilities. An experiment consists of rolling a red die and a green die and noting the result of each roll. The probability that the first marble is red and the second white. The following tree diagram shows the probabilities when a coin is tossed two times. All outcomes must be shown from each node. Teach your students how to complete and find probabilities from tree diagrams both with and without replacement. Tree diagrams can make some probability problems easier to visualize and solve. (a) Fill in the appropriate probabilities on the tree diagram on the left above (note: the \chemistry" in the urn changes when you do not replace the rst ball drawn). There are 4 blue marbles, and 2 red marbles. 1 Multiplication of choices. 1 while on non-windy days the probability she [3 marks] catches a fish is 0. This is a complete lesson on probability trees that extends previous learning on tree diagrams to include selection without replacement. probability of getting an A. Downloadable version. 392) Two cards are drawn without replacement from a 52-card deck. Example: Probability of tossing a coin. There are two versions of random sampling: sampling with replacement and sampling without replacement. Use a tree diagram to calculate conditional probability. (1) (a) (b) Complete the tree diagram. Probability of drawing a king = 4/52 = 1/13. arrow_back Back to Tree Diagrams - conditional / without replacement Tree Diagrams - conditional / without replacement: Lessons. Transcript. A couple plan to have exactly three children. Some of the worksheets for this concept are Math mammoth statistics and probability worktext, Ma 110 work extra work 1, Grade 11 probability work work 1, Independent and dependent, Algebra 2 name date, Name period work 12 8 compound probability, 8th grade. a) Draw a tree diagram to determine ALL possible outcomes. Tree Diagram-Barron’s P. probability simulation two -way table sample space S = {H, T} tree diagram probability model replacement event P(A) complement AC disjoint mutually exclusive event Venn diagram union (or) intersection (and) conditional probability independent events general multiplication rule general addition rule. This type of diagram can be very useful for some problems. The probability of any outcome is the product of all possibilities along the relevant branches. (a) Construct a tree diagram and list the sample space. Draw a tree diagram to represent the probabilities in each case. You can choose from a blue, purple, red, or green mat and a metal or wood frame. I'm going to show you an example of modified tree diagram to solve the following question. In a group of 10 students taking the exam, there are 3 who have prepared very well, 4 well, 2 moderately well and one poorly. Least Squares Regression and Correlation. with replacement b. An online probability tree calculator for you to generate the probability tree diagram. Another counter is taken at random. Complete the probability tree that. Tree diagrams -used when given probabilities are sequential in nature 67 of the students th tic s. Draw a Venn Diagram and then find the probability of receiving an offer from either Acme Corp. Tree diagrams can make some probability problems easier to visualize and solve. Tree diagrams are useful for solving probability problems with more than one stage. Q9: A bag contains 2 black balls and 8 white balls. Age range: 14-16. (2) Jan 10. To start with, instead of looking for a matching pair, let's find the probability that both socks are red. 1 26 3 12 2 13 3 12 2 13. What is the probability that: a) a purple marble is chosen from the cup? b) a green marble is chosen? We must make a tree diagram to show this process of first choosing the container and. Lesson Worksheet. Divide the number of events by the number of possible outcomes. Two sweets are drawn at random (i) with replacement and (ii) without replacement. First, use a tree diagram to map out the sample space: a. Check your tree against mine. You will learn how to find the probability of single or combined events using tree or sample space diagrams. For example, for the experiment "toss a coin three times and record the results from each toss", we could draw the following tree diagram. (Level 7) One ball is drawn from the bag, then another without replacement. If you want to evaluate a joint probability tree where probabilities are represented as decimals (e. Then, students draw a tree diagram for. The circle and rectangle will be explained later, and should be ignored for now. b Find the probability of selecting: i a blue marble followed by a white marble (B, W) ii 2 blue marbles iii exactly one blue marble c If the experiment was repeated with replacement, fi nd the answers to each question in part b. Tree Diagram: A jar contains 4 purple and 1 gold beads. Two sweets are selected without replacement. YOU can use shorthand like this. probability simulation two -way table sample space S = {H, T} tree diagram probability model replacement event P(A) complement AC disjoint mutually exclusive event Venn diagram union (or) intersection (and) conditional probability independent events general multiplication rule general addition rule. The probability that the weather is fine on any day is " +. Assign probabilities to outcomes and determine probabilities for events. replacement (meaning eat the skittle, then take another). (test negative but actually have the disease). to find the probability of event C or event D happening add probabilities down the tree. Since both combined events are the same (just the other way around), the answers are identical. When working with conditional probabilities, it is helpful to use a tree diagram to illustrate the probability of the different outcomes. calculate and interpret conditional probabilities through representation using expected frequencies with two-way tables, tree diagrams and Venn diagrams. Students may use other methods. doc 1_Double_Spinners. There are 12 possible outfits for the student to wear. G2 = second card is green. Are they dependent or independent events? 1) You roll two dice. When two balls are chosen at random without replacement from bag B, the probability that they are both white is $$\frac{2}{7}$$. Tree diagrams can make some probability problems easier to visualize and solve. Now, for the conditional probability we want to view that 3∕4 as if it was 1 whole, which we achieve by multiplying by its reciprocal, namely 4∕3. tree diagram. Disjoint Events (Revisited) Drawing with and without Replacement Making a Picture –Venn Diagrams, Probability Tables, and Tree Diagrams. Draw two marbles, one at a time, this time without replacement from the urn. The first step to solving a probability problem is to determine the probability that you want to calculate. (a) Draw a tree diagram to represent all the possible paths that the mouse could take. 7E-19 Three desperados A, B and C play Russian roulette in which they take turns pulling the trigger of a six-cylinder revolver loaded with one bullet. We will then draw two balls from the chosen box, without replacement and with equal probability on those remaining. Let's consider another example: Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement. The probability that it is on time on any day is 0. a Draw a tree diagram showing all outcomes and probabilities. If we select 4 computers at random from the distribution center (with replacement) what is the probability that at least 1 of the computers is a tablet computer? P. A tree diagram is a special type of graph used to determine the outcomes of an experiment. two bills without replacement, determine whether the probability that the bills will total $15 is greater than the probability that the bills will total$2. 18 Outcomes & Probability Third Pick First Pick Second Pick Figure 8: Tree diagram for selecting three sweets randomly (with probability value) e) Probability distribution for each flavour if three sweets are. We pick a card, write down what it is, then put it back in the deck and draw again. Probability & Tree Diagrams. There is a total of 3 colour sequences through which we end up with 1 of. MEMORY METER. Related Topics: algebra, dependent, independent, input, output, probability, variable. Find the probability that: a) Both Adam and Beth hit with their first dart b) At least one of them hits with their first dart B hit P(hit, hit) = 0. It consists of “branches” that are labeled with either frequencies or probabilities. Homework x1. The number of "Male and Smoke" divided by the total = 19/100 = 0. T and H (in any order)? 3. Draw a tree diagram representing the results. Draw a probability tree to show this situation and find the probability that the kicker is sucessful with the penalty. Use two-way tables to calculate conditional. Marbles are drawn vice with replacement What green) A 1 о в Bliss. (a) P(C\A) = 0. Assigned Practices: 1. From the tree diagram, we can see that there is a total of 8 different possible outcomes. and Mills Inc. Click Image to Enlarge : Use a tree diagram to display possible outcomes of who will come to the party. In other cases, different problem. Just like a tree, tree diagrams branch out and can become quite intricate. The probability that the woman we draw is not married is, by the complement rule, P(not married) = 1 – P(married) = 1 – 0. LC Probability & Statistics; LC Functions & Calculus; LC HL Self-Directed Quizzes; LC OL Self-Directed Quizzes; Junior Cycle Assessments Show sub menu. For that to happen you need the 1st card to be a Queen and the second card to be a Queen. (c) at least 2 tails, (d) 2 tails in succession 1 (e) 2tails. Try these multiple choice questions. In an urn, there are 11 balls. Probability tree diagrams. Given that attorneys must frequently make decisions in environments of uncertainty, probability can be a useful skill for law students to learn. Tree diagrams – no replacement – V2; 5. 1 while on non-windy days the probability she [3 marks] catches a fish is 0. The correct answer is A. What is a tree diagram? Why is it helpful?. Conditional probability, and Bayes' Theorem, are important sub-topics. Table of Contents. with replacement P(R 1 st draw, B 2 nd draw) P(Br 1 draw, Br 2 nd draw) b. Subsection 3. Using a tree diagram, find the probability that the second marble is red, given that the first one is red. The student will appraise the differences between the two estimates. It's just the whole space, in fact, to the probability of and Crime and B, what's the probability of A and B in this sequel to the probability of me. For n prizes and boxes, you end up pruning nn — n! branches. com for - Lessons and worksheets suitable for the 9 - 1 GCSE Specification - A-Level teaching resources for Core. Rules of Differentiation. In this tree diagrams worksheet, students solve and complete 2 different problems First, they draw a tree diagram for selecting two marbles with replacement and find the other probabilities. a) Tree diagram for the experiment. Plan Objectives 1 To find the. Each branch is a possible outcome and is labelled with a probability. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Play this game to review Mathematics. A disc is chosen at random from the bag and the colour is noted. "With replacement" means that you put the first ball back in the. Draw a tree diagram of the situation. Age range: 14-16. A second sweet is then removed. It consists of "branches" that are labeled with either frequencies or probabilities. To help understand this, let’s first recall the formula for conditional probability. OLVER EDUCATION. 21 shows these four outcomes and their probabilities. To find the P(QQQ), we find the probability of drawing the first queen which is 4/52. An experiment consists of rolling a red die and a green die and noting the result of each roll. Finally, find the probability of ending at each gate. A tree diagram is a special type of graph used to determine the outcomes of an experiment. The correct answer is A. a) Tree diagram for the experiment. Forexample,ifyoufliponefaircoin, S ={ H , T }where. b) How many outcomes are in the sample space? Exercise 7. Two marbles are selected without replacement. Tree Diagrams can be used to represent the total possible outcomes when you have 2 or more events. Whoops! There was a problem previewing Conditional probability. JC Number; JC Geometry & Trigonometry; JC Algebra & Functions; JC Probability & Statistics; JC Unifying Strand; TY Ideas; TY Analytics Module; Make-a-Mock; Branches; Newsletters; Events; Teacher. Draw a tree diagram to represent the probabilities in each case. Hint: The 4 digits in this probability add to 18. Unit 11 Day 3 Conditional Probability (3). 7, P (A') =. I introduce you to new notation which I would encourage you to do as it will help with conditional probability at. Draw a tree representing the possible mutually exclusive outcomes 2. Draw a probability tree showing the possible outcomes. Learning Outcomes As a result of studying this topic, students will be able to • understand and use the following terminology: trial, outcome, set of all possible outcomes, relative frequency, event, theoretical probability, If you require probability tree diagram worksheets with answers, or probability maths questions and answers you can. A girls' choir is choosing a concert uniform. Determine the probability of getting 2 heads in two successive tosses of a balanced coin. 🚨 Claim your spot here. So we are calculating 99% of 10% which is 0. Let us take note that two cards, one at a time, are drawn at random from the box. Assign students to choose four of their shirts and four pairs of pants. 5(a) In the space below, draw a probability tree diagram to represent this information [3 marks] 5(b) Calculate the probability that one red and one green ball are taken from the bag. WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. Find the probability of the first marble being green and the second marble being yellow. Tree diagrams and conditional probability. 3: Tree diagram for two draws without replacement, values rounded. Tree diagrams are a tool to organize outcomes and probabilities around the structure of the data. The probability that Pat will arrive late is 0. 1 = 1\$ To find the probability of any path, multiply the probabilities on the corresponding branches. Example 1. (a) Draw a tree diagram and from it write. 1 Sampling without replacement Example 3 A box contains 4 red marbles and 3 white ones. 1 Randomness, Probability, and Simulation (pp. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Then a second marble is chosen at random. One ball is drawn from the bag, then another without replacement. a) Draw a tree diagram for this experiment b) Find the probability that at least one of the two persons favors genetic engineering. Replacement and Probability. A Tree Diagram and Sample Space A tree diagram is a graphic representation of the step by step competion of an experiment showing all possible results of each step. It is generally drawn from a starting point on the left and then branches to the right with each possilbe outcome shown as the end of the next branch or bridge. So, the probability that the student doesn't know the answer AND answers correctly is. Find and create gamified quizzes, lessons, presentations, and flashcards for students, employees, and everyone else. If it does rain, Mudlark will start favourite in the horse race, with probability of winning. Probability Trees RAG. I missed out tree diagrams without replacement. You spin the spinner once. Two are taken from the box, without replacement. Intuitive conditional probability seemingly not working. tree diagram. Two marbles are chosen without replacement. In this video, we will learn how to use tree diagrams to calculate conditional probabilities. In an urn, there are 11 balls. Let's consider another example: Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement. Question 1: Find the probability that a player selects two red counters. Draw a tree diagram representing the results. Using a tree diagram, find the probability that the second marble is red, given that the first one is red. Ron has a bag containing 3 green pears and 4 red pears. b) How many outcomes have a sum of the 2 numbers greater than or equal. iii: Find the probability that both biscuits are plain. Find the probability that: both. How easy is it? Simply open one of the tree diagram templates included, input your information and let SmartDraw do the rest. A tree diagram is a special type of graph used to determine the outcomes of an experiment. I built Diagnostic Questions to help you identify, understand and resolve key misconceptions. So don’t let your student become confused by probability, our probability activities are probably the best resources available. eBook: Read p. Probability: Venn Diagrams and Two-Way Tables. Draw a tree diagram to represent the probabilities in each case. If the weather is fine, the probability that Carlos is late arriving at school is !!*. b) How many outcomes are in the sample space? Exercise 7. 392) Two cards are drawn without replacement from a 52-card deck. "With replacement" means that you put the first ball back. Tree diagrams can make some probability problems easier to visualize and solve. This probability is found by using the tree like a probability distribution table, simply identify the leaves that have this event (F), and then sum their probabilities. The probability that it will be windy on a particular day is 0. Make a tree diagram for an experiment that consists of two trials. To answer how likely a patient is to have TB given a positive test result, we need to “flip” the tree. [2] [3] The higher the probability of an event, the more certain we are that the event will occur. Determine the following geometric. MEMORY METER. It consists of "branches" that are labeled with either frequencies or probabilities. • Tree diagram - a diagram which can be helpful in illustrating possible outcomes of an experiment. Example: Probability of tossing a coin. Draw a probability tree showing the possible outcomes. The following example illustrates how to use a tree diagram. Since it's with replacement the first time i'm drawing, the probability would be 2 9 and the second time would also be 2 9 which would be 4 81. Find the probability that: both. SEE MORE : 9. Visit weteachmaths. I don't know how to write out a tree diagram on here, but I think this one is heads -> heads, tails -> math probabilty- please help. (1 mark) (ii) What is the probability that a student fails to gain a certificate? (2 marks) (b) Three students take the exam. Tree diagrams can make some probability problems easier to visualize and solve. The breakdown of the lot size and the sample size in the numerator and denominator of (3. one green ball and one blue ball. Tree Diagrams A tree diagram is a way of seeing all the possible probability 'routes' for two (or more) events. Resource type: Lesson (complete) 4. Therefore, in a family of three children, the probability of having three girls is 1 out of 8. It consists of "branches" that are labeled with either frequencies or probabilities. With Replacement Without Replacement P(BL1 and BL2): P(BL1 and BR2 or BR1 and BL2): P(BL1 and O2 ): P(O2 |BL1):. 34, and the probability of selecting a black marble on the first draw is 0. (This path has been drawn on the tree diagram with arrows. the form of a tree diagram or table Æ express the probability of an event as a fraction, a decimal, and a percent independent events • results for which the outcome of one event has no eff ect on the outcome of another event • ruler probability • the likelihood or chance of an event occurring Determining Probabilities Using Tree Diagrams. 5 (since the probability of getting a heads on the first flip is 0. These explanations and tutorials will help you find the probability of all sorts of events, from rolling a number on a die to winning the lottery. We sample two items from the box without replacement. If A and B are independent (that is, the occurrence of a specific one of these two events does not influence the probability of the other event), then. Since nn grows much, much faster, than n!, Z’s algorithm becomes prohibitively tedious in a hurry. Questions 28 – 29 refer to the following probability tree diagram which shows tossing an unfair coin FOLLOWED BY drawing one bead from a cup containing 3 red (R), 4 yellow (Y) and 5 blue (B. ii: Write down the value of b. 368 #6 A plumbing contractor obtains 60% of her boiler circulators from a company whose defect rate is. (iii) the product of the two numbers is at least 5. (2) Jan 10. 7, and the probability that Jamie will pass. 🚨 Claim your spot here. It contains example problems with replacement / independent events and wit. The possibilities are: 4 H, 3 H and 1 T (in various orders), 2 H and 2 T (in various orders), 1 H and 3 T (in various orders), or 4 T. Tree Diagrams •Sample spaces can also be described graphically with tree diagrams. This item is taken from IGCSE Mathematics (0580) Paper 43 of May/June 2013. This site is the homepage of the textbook Introduction to Probability, Statistics, and Random Processes by Hossein Pishro-Nik. If it is fine he only has a 1 in 20 chance of winning. The number of "Male and Smoke" divided by the total = 19/100 = 0. Submitted by Hannah Yates on 6 March 2017. Draw a tree diagram showing the possible outcomes. 13 Outcomes & Probability Third Pick Second Pick First Pick BBB (0. Construct two tree diagrams (one for with replacement and the other for without replacement) showing the drawing of two M&Ms, one at a time, from the M&Ms you were given, as recorded in the table above. | 2021-08-04T23:15:59 | {
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http://code.jasonbhill.com/category/algorithms/ | A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.
Then, we calculate the number of paths leading from the apex to each position: A path starts at the apex and progresses downwards to any of the three spheres directly below the current position. Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it).
The result is Pascal’s pyramid and the numbers at each level n are the coefficients of the trinomial expansion $(x + y + z)^n$. How many coefficients in the expansion of $(x + y + z)^{200000}$ are multiples of $10^{12}$?
## Solution Using the Multinomial Theorem
The generalization of the binomial theorem is the multinomial theorem. It says that multinomials raised to exponents can be expanded using the formula
$(x_1+x_2+\cdots+x_m)^n=\sum_{{k_1+k_2+\cdots+k_m=n}\atop{0\le k_i\le n}}\left({n}\atop{k_1,k_2,\ldots,k_m}\right)\prod_{1\le t\le m}x_t^{k_t}$
where
$\left({n}\atop{k_1,k_2,\ldots,k_m}\right)=\frac{n!}{k_1!k_2!\cdots k_m!}.$
Of course, when m=2 this gives the binomial theorem. The sum is taken over all partitions $k_1+k_2+\cdots+k_m=n$ for integers $k_i$. If n=200000 abd m=3, then the terms in the expansion are given by
$\left({200000}\atop{k_1,k_2,k_3}\right)x_1^{k_1}x_2^{k_2}x_3^{k_3}=\frac{200000!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}$
where $k_1+k_2+k_3=200000$. It’s worth pointing out that each of the coefficients is an integer, and thus has a unique factorization into products of prime integers. Of course, there’s no way that we’re going to calculate these coefficients. We only need to know when they’re divisible by $10^{12}$. Thus, we only need to consider how many factors of 2 and 5 are involved.
First, we’ll create a function $p(n,d)$ that outputs how many factors of $d$ are included in $n!$. We have that
$p(n,d)=\left\lfloor\frac{n}{d}\right\rfloor+\left\lfloor\frac{n}{d^2}\right\rfloor+\left\lfloor\frac{n}{d^3}\right\rfloor+ \cdots+\left\lfloor\frac{n}{d^r}\right\rfloor,$
where $d^r$ is the highest power of $d$ dividing $n$. For instance, there are 199994 factors of 2 in 200000!. Since we’re wondering when our coefficients are divisible by $10^{12}=2^{12}5^{12}$, we’ll be using the values provided by $p(n,d)$ quite a bit for $d=2$ and $d=5$. We’ll store two lists:
$p2=[p(i,2)\text{ for }1\le i\le 200000]\quad\text{and}\quad p5=[p(i,5)\text{ for }1\le i\le 200000].$
For a given $k_1,k_2,k_3$, the corresponding coefficient is divisible by $10^{12}$ precisely when
$p2[k_1]+p2[k_2]+p2[k_3]<199983\ \text{and}\ p5[k_1]+p5[k_2]+p5[k_3]<49987.$
That is, this condition ensures that there are at least 12 more factors of 2 and 5 in the numerator of the fraction defining the coefficients.
Now, we know that $k_1+k_2+k_3=200000$, and we can exploit symmetry and avoid redundant computations if we assume $k_1\le k_2\le k_3$. Under this assumption, we always have
$k_1\le\left\lfloor\frac{200000}{3}\right\rfloor=66666.$
We know that $k_1+k_2+k_3=200000$ is impossible since 200000 isn't divisible by 3. It follows that we can only have (case 1) $k_1=k_2 < k_3$, or (case 2) $k_1 < k_2=k_3$, or (case 3) $k_1 < k_2 < k_3$.
In case 1, we iterate $0\le k_1\le 66666$, setting $k_2=k_1$ and $k_3=200000-k_1-k_2$. We check the condition, and when it is satisfied we record 3 new instances of coefficients (since we may permute the $k_i$ in 3 ways).
In case 2, we iterate $0\le k_1\le 66666$, and when $k_1$ is divisible by 2 we set $k_2=k_3=\frac{200000-k_1}{2}$. When the condition holds, we again record 3 new instance.
In case 3, we iterate $0\le k_1\le 66666$, and we iterate over $k_2=k_1+a$ where $1\le a < \left\lfloor\frac{200000-3k_1}{2}\right\rfloor$. Then $k_3=200000-k_1-k_2$. When the condition holds, we record 6 instances (since there are 6 permutations of 3 objects).
## Cython Solution
I’ll provide two implementations, the first written in Cython inside Sage. Then, I’ll write a parallel solution in C.
%cython
import time
from libc.stdlib cimport malloc, free
cdef unsigned long p(unsigned long k, unsigned long d):
cdef unsigned long power = d
cdef unsigned long exp = 0
while power <= k:
exp += k / power
power *= d
return exp
cdef unsigned long * p_list(unsigned long n, unsigned long d):
cdef unsigned long i = 0
cdef unsigned long * powers = <unsigned long *>malloc((n+1)*sizeof(unsigned long))
while i <= n:
powers[i] = p(i,d)
i += 1
return powers
run_time = time.time()
# form a list of number of times each n! is divisible by 2.
cdef unsigned long * p2 = p_list(200000,2)
# form a list of number of times each n! is divisible by 5.
cdef unsigned long * p5 = p_list(200000,5)
cdef unsigned long k1, k2, k3, a
cdef unsigned long long result = 0
k1 = 0
while k1 <= 66666:
# case 1: k1 = k2 < k3
k2 = k1
k3 = 200000 - k1 - k2
if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]):
result += 3
# case 2: k1 < k2 = k3
if k1 % 2 == 0:
k2 = (200000 - k1)/2
k3 = k2
if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]):
result += 3
# case 3: k1 < k2 < k3
a = 1
while 2*a < (200000 - 3*k1):
k2 = k1 + a
k3 = 200000 - k1 - k2
if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]):
result += 6
a += 1
k1 += 1
free(p2)
free(p5)
elapsed_run = round(time.time() - run_time, 5)
print "Result: %s" % result
print "Runtime: %s seconds (total time: %s seconds)" % (elapsed_run, elapsed_head)
When executed, we find the correct result relatively quickly.
Result: 479742450
Runtime: 14.62538 seconds (total time: 14.62543 seconds)
## C with OpenMP Solution
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <omp.h>
/*****************************************************************************/
/* function to determine how many factors of 'd' are in 'k!' */
/*****************************************************************************/
unsigned long p(unsigned long k, unsigned long d) {
unsigned long power = d;
unsigned long exp = 0;
while (power <= k) {
exp += k/power;
power *= d;
}
return exp;
}
/*****************************************************************************/
/* create a list [p(0,d),p(1,d),p(2,d), ... ,p(n,d)] and return pointer */
/*****************************************************************************/
unsigned long * p_list(unsigned long n, unsigned long d) {
unsigned long i;
unsigned long * powers = malloc((n+1)*sizeof(unsigned long));
for (i=0;i<=n;i++) powers[i] = p(i,d);
return powers;
}
/*****************************************************************************/
/* main */
/*****************************************************************************/
int main(int argc, char **argv) {
unsigned long k1, k2, k3, a;
unsigned long long result = 0;
unsigned long * p2 = p_list(200000, 2);
unsigned long * p5 = p_list(200000, 5);
#pragma omp parallel for private(k1,k2,k3,a) reduction(+ : result)
for (k1=0;k1<66667;k1++) {
// case 1: k1 = k2 < k3
k2 = k1;
k3 = 200000 - k1 - k2;
if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) {
result += 3;
}
// case 2: k1 < k2 = k3
if (k1 % 2 == 0) {
k2 = (200000 - k1)/2;
k3 = k2;
if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) {
result += 3;
}
}
// case 3: k1 < k2 < k3
for (a=1;2*a<(200000-3*k1);a++) {
k2 = k1 + a;
k3 = 200000 - k1 - k2;
if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) {
result += 6;
}
}
}
free(p2);
free(p5);
printf("result: %lld\n", result);
return 0;
}
This can be compiled and optimized using GCC as follows.
$gcc -O3 -fopenmp -o problem-154-omp problem-154-omp.c When executed on a 16-core machine, we get the following result.$ time ./problem-154-omp
result: 479742450
real 0m1.487s
This appears to be the fastest solution currently known, according to the forum of solutions on Project Euler. The CPUs on the 16-core machine are pretty weak compared to modern standards. When running on a single core on a new Intel Core i7, the result is returned in about 4.7 seconds.
### Motivation
Many interesting computational problems, such as those on Project Euler require that one find the sum of proper divisors of a given integer. I had a fairly crude brute-force method for doing this, and was subsequently emailed a comment by Bjarki Ágúst Guðmundsson who runs the site www.mathblog.dk. He pointed me in the direction of this page and provided some sample code illustrating how such an approach runs asymptotically faster than the approach I had been taking. Awesome! I’m going to expand on that a bit here, providing some mathematical proofs behind the claims and providing code for those who may want to take advantage of this.
### The Mathematics Behind It All
Let the function $\sigma(n)$ be the sum of divisors for a positive integer $n$. For example,
$\sigma(6)=1+2+3+6=12.$
It should seem obvious that for any prime $p$ we have $\sigma(p)=1+p$. What about powers of primes? Let $\alpha\in\mathbb{Z}_+$, and then
$\sigma(p^\alpha)=1+p+p^2+\cdots+p^\alpha.$
We’d like to write that in a closed form, i.e., without the “$+\cdots+$”. We use a standard series trick to do that.
\begin{align} \sigma(p^\alpha) &= 1+p+p^2+\cdots+p^\alpha\cr p\sigma(p^\alpha) &= p+p^2+p^3+\cdots+p^{\alpha+1}\cr p\sigma(p^\alpha)-\sigma(p^\alpha) &= (p+p^2+\cdots+p^{\alpha+1})-(1+p+\cdots+p^\alpha)\cr p\sigma(p^\alpha)-\sigma(p^\alpha) &= p^{\alpha+1}-1\cr (p-1)\sigma(p^\alpha) &= p^{\alpha+1}-1\cr \sigma(p^\alpha) &=\frac{p^{\alpha+1}-1}{p-1}.\end{align}
That solves the problem of finding the sum of divisors for powers of primes. It would be nice if we could show that $\sigma$ is multiplicative on powers of primes, i.e., that $\sigma(p_1^{\alpha_1}p_2^{\alpha_2})=\sigma(p_1^{\alpha_1})\sigma(p_2^{\alpha_2})$. We’ll prove that this is the case, and solve the problem in general along the way.
Proposition: The function $\sigma$ is multiplicative on powers of primes.
Proof: Let $n$ be a positive integer written (uniquely, by the fundamental theorem of arithmetic) as
$n=\prod_{i=1}^m p_i^{\alpha_i}$
for $m$ distinct primes $p_i$ with $\alpha_i\in\mathbb{Z}_+$. Any divisor $k$ of $n$ then has the form
$k=\prod_{i=1}^m p_i^{\beta_i}$
where each $\beta_i$ satisfies $0\le\beta_i\le\alpha_i$. Then
$\sigma(n)=\sigma\left(\prod_{i=1}^m p_i^{\alpha_i}\right)$
is the sum of all divisors $k$ of $n$ and can be written by summing over all possible combinations of the exponents $\beta_i$. There are $\prod_{i=1}^m \alpha_i$ combinations, and we can form their sum and simplify it as follows.
\begin{align}\sigma(n) &= \sum_{1\le i\le m,\ 0\le\beta_i\le\alpha_i}p_1^{\beta_i}p_2^{\beta_2}\cdots p_m^{\beta_m}\cr &= \sum_{\beta_1=0}^{\alpha_1}p_1^{\beta_1}\left(\sum_{2\le i\le m,\ 0\le\beta_i\le\alpha_i}p_2^{\beta_2}p_3^{\beta_3}\cdots p_m^{\beta_m}\right)\cr &= \sum_{\beta_1=0}^{\alpha_1}p_1^{\beta_1}\sum_{\beta_2=0}^{\alpha_2}p_2^{\beta_2}\left(\sum_{3\le i\le m,\ 0\le\beta_i\le\alpha_i}p_3^{\beta_3}p_4^{\beta_4}\cdots p_m^{\beta_m}\right) \cr &= \vdots\cr &=\sum_{\beta_1=0}^{\alpha_1}p_1^{\beta_1}\sum_{\beta_2=0}^{\alpha_2}p_2^{\beta_2}\sum_{\beta_3=0}^{\alpha_3}p_3^{\beta_3}\ \cdots\ \sum_{\beta_m=0}^{\alpha_m}p_m^{\beta_m}\cr &=\sigma(p_1^{\alpha_1})\sigma(p_2^{\alpha_2})\cdots\sigma(p_m^{\alpha_m}).\end{align}
This completes the proof. Q.E.D.
Thus, we now have a formula for the sum of divisors of an arbitrary positive integer $n$ using the factorization of $n$. Namely,
$\sigma(n)=\sigma\left(\prod_{i=1}^m p_i^{\alpha_i}\right)=\prod_{i=1}^m\left(\frac{p_i^{\alpha_i+1}-1}{p_i-1}\right).$
This is something I use quite a bit for various problems and programming exercises, so I figured I could post it here. It’s a basic post that isn’t advanced at all, but that doesn’t mean that the implementation given below won’t save work for others. The idea is to create a list of primes in C by malloc’ing a sieve, then malloc’ing a list of specific length based on that sieve. The resulting list contains all the primes below a given limit (defined in the code). The first member of the list is an integer representing the length of the list.
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#define bool _Bool
static unsigned long prime_limit = 1000000;
unsigned long sqrtld(unsigned long N) {
int b = 1;
unsigned long res,s;
while(1<<b<N) b+= 1;
res = 1<<(b/2 + 1);
for(;;) {
s = (N/res + res)/2;
if(s>=res) return res;
res = s;
}
}
unsigned long * make_primes(unsigned long limit) {
unsigned long *primes;
unsigned long i,j;
unsigned long s = sqrtld(prime_limit);
unsigned long n = 0;
bool *sieve = malloc((prime_limit + 1) * sizeof(bool));
sieve[0] = 0;
sieve[1] = 0;
for(i=2; i<=prime_limit; i++) sieve[i] = 1;
j = 4;
while(j<=prime_limit) {
sieve[j] = 0;
j += 2;
}
for(i=3; i<=s; i+=2) {
if(sieve[i] == 1) {
j = i * 3;
while(j<=prime_limit) {
sieve[j] = 0;
j += 2 * i;
}
}
}
for(i=2;i<=prime_limit;i++) if(sieve[i]==1) n += 1;
primes = malloc((n + 1) * sizeof(unsigned long));
primes[0] = n;
j = 1;
for(i=2;i<=prime_limit;i++) if(sieve[i]==1) {
primes[j] = i;
j++;
}
free(sieve);
return primes;
}
int main(void) {
unsigned long * primes = make_primes(prime_limit);
printf("There are %ld primes <= %ld\n",primes[0],prime_limit);
free(primes);
return 0;
}
Say one wanted to form a list of all primes below 1,000,000. That’s what the above program does by default, since “prime_limit = 1000000.” If one compiles this and executes, you would get something like what follows. The timing is relatively respectable.
$gcc -O3 -o prime-sieve prime-sieve.c$ time ./prime-sieve
There are 78498 primes <= 1000000
real 0m0.008s
user 0m0.004s
sys 0m0.000s
The code is linked here: prime-sieve.c | 2019-04-24T20:41:17 | {
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"url": "http://code.jasonbhill.com/category/algorithms/",
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http://mathhelpforum.com/math-challenge-problems/10222-quickie-13-a.html | # Math Help - Quickie #13
1. ## Quickie #13
This is a classic (very old).
An irrational number raised to an irrational power is always irrational.
Prove it or provide a counterexample.
2. Seen it. But I did not know it was a classic.
3. Originally Posted by Soroban
This is a classic (very old).
An irrational number raised to an irrational power is always irrational.
Prove it or provide a counterexample.
I'll give several counter examples;
For the most part,
[sqrt(x)^(sqrt(x))]^(sqrt(x)), where x is even, with the exception of when x is a square.
EX:
[sqrt(2)^(sqrt(2))]^(sqrt(2))
[sqrt(2)^(sqrt(2))] = 2^(sqrt(2)/2)
[2^(sqrt(2)/2)]^(sqrt(2)) = 2
4. Originally Posted by AfterShock
I'll give several counter examples;
For the most part,
[sqrt(x)^(sqrt(x))]^(sqrt(x)), where x is even, with the exception of when x is a square.
EX:
[sqrt(2)^(sqrt(2))]^(sqrt(2))
[sqrt(2)^(sqrt(2))] = 2^(sqrt(2)/2)
[2^(sqrt(2)/2)]^(sqrt(2)) = 2
Interesting, but you are assuming that $\sqrt{2}^{\sqrt{2}}$ is irrational, else this isn't a counterexample. I have little doubt that it is, but this isn't a good counterexample without proving this fact.
-Dan
5. Originally Posted by topsquark
Interesting, but you are assuming that $\sqrt{2}^{\sqrt{2}}$ is irrational, else this isn't a counterexample. I have little doubt that it is, but this isn't a good counterexample without proving this fact.
-Dan
Funny thing is you do not need to. The assumption is that irrational raised to irrational is always irrational. Thus, he is using this assumption. And arrives at a contradiction.
6. Originally Posted by ThePerfectHacker
Funny thing is you do not need to. The assumption is that irrational raised to irrational is always irrational. Thus, he is using this assumption. And arrives at a contradiction.
Aaaaah! I get it now.
-Dan
7. The "classic" solution goes like this:
Theorem: . $(\text{irrational})^{\text{irrational}}$ can be rational.
Proof
$\text{Consider }a \:=\:\sqrt{2}^{\sqrt{2}}$
There are only two possibilities:
. . (1) $a$ is rational.
. . (2) $a$ is irrational.
If (1) $a$ is rational, the theroem is verified.
If (2) $a$ is irrational, consider: . $a^{\sqrt{2}}$
. . an irrational number raised to an irrational power.
Then we have: . $a^{\sqrt{2}} \:=\:\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} \:=\:\left(\sqrt{2}\right)^2\:=\:2$ . . . a rational number.
Either way, an irrational raised to an irrational power can be rational.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I've always found this proof amusing.
Note that we still don't know if $\sqrt{2}^{\sqrt{2}}$ is rational or irrational
. . but it doesn't matter . . .
8. I found a different counterexample.
$e$ is irrational.
$\ln 2$ is irrational (but do not know how to show it).
Then,
$e^{\ln 2}=2$ | 2016-05-26T14:47:00 | {
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https://netco-me.com/g0y3k/every-regular-graph-is-complete-graph-bf463e | A simple non-planar graph with minimum number of vertices is the complete graph K 5. A complete graph is a graph in which every vertex has an edge to all other vertices is called a complete graph, In other words, each pair of graph vertices is connected by an edge. Two further examples are shown in Figure 1.14. A regular graph with vertices of degree k {\displaystyle k} is called a k {\displaystyle k} ‑regular graph or regular graph of degree k {\displaystyle k}. 3.A graph is k-regular if every vertex has degree k. How do 1-regular graphs look like? If every vertex of a simple graph has the same degree, then the graph is called a regular graph. A simple graph is called regular if every vertex of this graph has the same degree. Before you go through this article, make sure that you have gone through the previous article on various Types of Graphsin Graph Theory. An important property of graphs that is used frequently in graph theory is the degree of each vertex. C Tree. Let $G$ be a regular graph, that is there is some $r$ such that $|\delta_G(v)|=r$ for all $v\in V(G)$. 1.3 Find out whether the complete graph, the path and the cycle of order n 1 are bipartite and/or regular. 1.6.Show that if a k-regular bipartite graph with k>0 has a bipartition (X;Y), then jXj= jYj. I think you wanted to ask about a spanning 1-regular graph, also known as a perfect matching or 1-factor. This means that (assuming this is not a multigraph, no self-edges, etc) if you have n vertices, then each vertex has n-1 edges. Regular Graph c) Simple Graph d) Complete Graph … View Answer ... B Regular graph. Another plural is vertexes. hence, The edge defined as a connection between the two vertices of a graph. That is, if a graph is k-regular, every vertex has degree k. Exercises: Draw all 0-regular graphs with 1 vertex; 2 vertices; 3 vertices. If every vertex in a regular graph has degree k,then the graph is called k-regular. Conjecture 8 : Let G be a 3-regular cyclically 4-edge-connected graph of order n.Then G contains a cycle of length at least cn where c is a positive num- ber. G is said to be regular of degree r (or r-regular) if deg(v) = r for all vertices v in G. Complete graphs of order n are regular of degree n − 1, and empty graphs are regular of degree 0. Explanation: In a regular graph, degrees of all the vertices are equal. Some sources claim that the letter K in this notation stands for the German word komplett, but the German name for a complete graph, vollständiger Graph, does not contain the letter K, and other sources state that the notation honors the contributions of Kazimierz Kuratowski to graph theory. The problem seems similar to Hamiltonian Path which is NP complete problem for a general graph. How to create a program and program development cycle? yes No Not enough information to decide If Ris the equivalence relation defined by the panition {{1. A complete graph Km is a graph with m vertices, any two of which are adjacent. D n2. {5}. Explanation of Complete Graph with Diagram and Example, Explanation of Abstract Data Types with Diagram and Example, What is One Dimensional Array in Data Structure with Example, What is Singly Linked List? The first example is an example of a complete graph. What is Polynomials Addition using Linked lists With Example. Vertex Cover (VC): A vertex cover in an undirected graph G = (V;E) is a subset of vertices V0 V such that every edge in G has at least one endpoint in V0. therefore, In a directed graph, an edge goes from one vertex, the source, to another, the target, and hence makes the connection in only one direction. I'm not sure about my anwser. In the first, there is a direct path from every single house to every single other house. Every strongly regular graph is symmetric, but not vice versa. B n*n. C nn. Kn has n(nâ1)/2 edges and is a regular graph of degree nâ1. (a) every induced subgraph of a complete graph is complete; (b) every subgraph of a bipartite graph is bipartite. In the given graph the degree of every vertex is 3. The vertex is defined as an item in a graph, sometimes referred to as a node, The plural is vertices. definition. The complete graph on n vertices is denoted by Kn. 1.7.Show that, in any group of two or more people, there are always two with exactly the same number of friends inside the group. Theorem 9 : Let G be a 3-connected 3-regular graph , and let S be a set of nine vertices of G.Then G has a cycle which includes every vertex of S. (Aolton et al., 1982; Kelmans and Lomonosov, 1982) A graph of this kind is sometimes said to be an srg(v, k, λ, μ).Strongly regular graphs were introduced by Raj Chandra Bose in 1963.. A complete graph is a graph in which every vertex has an edge to all other vertices is called a complete graph, In other words, each pair of graph vertices is connected by an edge. 4. DEFINITION : Complete graph: In a graph, if there exist an edge between every pair of vertices,then such a graph is called complete graph. A connected graph may not be (and often is not) complete. In the graph, a vertex should have edges with all other vertices, then it called a complete graph. 4)A star graph of order 7. A simple graph with ‘n’ mutual vertices is called a complete graph and it is denoted by ‘K n ’. The complete graph on n vertices is denoted by Kn. Regular Graphs A graph G is regular if every vertex has the same degree. As the above graph n=7 Ans - Statement p is true. The complete bipartite graph K m, n is planar if and only if m ≤ 2 or n ≤ 2. A complete graph is a graph that has an edge between every single vertex in the graph; we represent a complete graph … Q.1. Theorem 2.4 If G is a k-regular bipartite graph with k > 0 and the bipartition of G Privacy therefore, The total number of edges of complete graph = 21 = (7)*(7-1)/2. Solution: A 1-regular graph is just a disjoint union of edges (soon to be called a matching). Complete graphs correspond to cliques. regular graph : a regular graph is a graph in which every node has the same degree • connected graph : a graph is connected if any two points can be joined by a path (a sequence of edges that are pairwise adjacent) MATH3301 EXTREMAL GRAPH THEORY Deflnition: A near regular complete multipartite graph is a complete multipartite graph with orders of its partite sets difiering by at most 1. {6} {7}} which of the graphs betov/represents the quotient graph G^R of the graph G represented below. 45 The complete graph K, has... different spanning trees? Some authors exclude graphs which satisfy the definition trivially, namely those graphs which are the disjoint union of one or more equal-sized complete graphs, and their complements, the complete multipartite graphs with equal-sized independent sets. Advantage and Disadvantages. 1 2 3 4 QUESTION 3 Is this graph regular? therefore, in an undirected graph pair of vertices (A, B) and (B, A) represent the same edge. Statement Q Is True. In graph theory, a regular graph is a graph where each vertex has the same number of neighbors; i.e. 1.8.1. A regular graph of degree r is strongly regular if there exist nonnegative integers e, d such that for all vertices u, v the number of vertices … The complete graph with n graph vertices is denoted mn. A regular directed graph must also satisfy the stronger condition that the indegree and outdegree of each vertex are equal to each other. Regular, Complete and Complete Bipartite. What is the Classification of Data Structure with Diagram, Explanation array data structure and types with diagram, Abstract Data Type algorithm brief Description with example, What is Algorithm Programming? 4.How many (labelled) graphs exist on a given set of nvertices? Important graphs and graph classes De nition. Could you please help me on Discrete-mathematical-structures. A graph and its complement. Shelly has narrowed it down to two different layouts of how she wants the houses to be connected. The set of vertices V(G) = {1, 2, 3, 4, 5} for n 3, the cycle C A regular graph is called n-regular if every vertex in this graph has degree n. Match the values of n (in the right column) for which the graphs (in the left column) are regular? 2)A bipartite graph of order 6. A 2-regular graph is a disjoint union of cycles. A complete graph is connected. Acomplete graphhas an edge between every pair of vertices. What is Data Structures and Algorithms with Explanation? Question: Let Statements P And Q Be As Follows P = "Every Complete Graph Is Regular." Any graph with 8 or less edges is planar. $\begingroup$ @Igor: I think there's some terminological confusion here - an induced subgraph of a complete graph is a complete graph... $\endgroup$ – ndkrempel Jan 17 '11 at 17:25 $\begingroup$ @ndkrempel: yes, confusion reigns. A symmetric graph is one in which there is a symmetry (graph automorphism) taking any ordered pair of adjacent vertices to any other ordered pair; the Foster census lists all small symmetric 3-regular graphs. Definition: Regular. The complete graph with n graph vertices is denoted mn. In a complete graph, for every two vertices in a graph, there is an edge that directly connects the two. the complete graph with n vertices has calculated by formulas as edges. The vertex cover problem (VC) is: given an undirected graph G and an integer k, does G have a vertex cover of size k? What are the basic data structure operations and Explanation? Every non-empty graph contains such a graph. We have discussed- 1. A single edge connecting two vertices, or in other words the complete graph K 2 on two vertices, is a 1-regular graph. Then, we have $|\delta_\bar{G}(v)|=n-r-1$, where $\bar{G}$ is the complement of $G$ and $n=|V(G)|$. $\endgroup$ – Igor Rivin Jan 17 '11 at 17:40 Any graph with 4 or less vertices is planar. Complete Graph. View Answer Answer: Tree ... Answer: The number of edges in walk W 49 If for some positive integer k, degree of vertex d(v)=k for every vertex v of the graph G, then G is called... ? A graph is called Eulerian if it has an Eulerian Cycle and called Semi-Eulerian if it has an Eulerian Path. Fortunately, we can find whether a given graph has a … graph when it is clear from the context) to mean an isomorphism class of graphs. Regular Graph - A graph in which all the vertices are of equal degree is called a regular graph. Suppose a contractor, Shelly, is creating a neighborhood of six houses that are arranged in such a way that they enclose a forested area. The line graph H of a graph G is a graph the vertices of which correspond to the edges of G, any two vertices of H being adjacent if and…. 1.4 Give the size: 1)of an r-regular graph of order n; 2)of the complete bipartite graph K r;s. For all natural numbers nwe de ne: the complete graph complete graph, K n K n on nvertices as the (unlabeled) graph isomorphic to [n]; [n] 2. They are called 2-Regular Graphs. ... A k-regular graph G is one such that deg(v) = k for all v ∈G. If all the vertices in a graph are of degree ‘k’, then it is called as a “ k-regular graph “. Aregular graphis agraphwhereevery vertex has the same degree.Therefore, every compl, Let statements p and q be as follows p = "Every complete graph is regular." A nn-2. In this article, we will discuss about Bipartite Graphs. Which of the following statements for a simple graph is correct? In a weighted graph, every edge has a number, it’s called “weight”. Output Result In both the graphs, all the vertices have degree 2. View desktop site. Properties of Regular Graphs: A complete graph N vertices is (N-1) regular. 1.8. Terms complete. In the second, there is a way to get from each of the houses to each of the other houses, but it's not necessarily … Complete Graph defined as An undirected graph with an edge between every pair of vertices. A graph G is said to be complete if every vertex in G is connected to every other vertex in G. Thus a complete graph G must be connected. therefore, the complete digraph is a directed graph in which every pair of distinct vertices is connected by a pair of unique edges (one in each direction). (Thomassen et al., 1986, et al.) …the graph is called a complete graph (Figure 13B). A K graph. 3)A complete bipartite graph of order 7. The set of edges E(G) = {(1, 2), (1, 4), (1, 5), (2, 3), (3, 4), (3, 5), (1, 3)} To calculate total number of edges with N vertices used formula such as = ( n * ( n â 1 ) ) / 2. Note: An undirected graph represented as a directed graph with two directed edges, one “to” and one “from,” for every undirected edge. A graph in which degree of all the vertices is same is called as a regular graph. Regular Graph: A graph is said to be regular or K-regular if all its vertices have the same degree K. A graph whose all vertices have degree 2 is known as a 2-regular … A graph is a collection of vertices connected to each other through a set of edges. © 2003-2021 Chegg Inc. All rights reserved. | The complete graph with n vertices is denoted by K n. The Figure shows the graphs K 1 through K 6. D Not a graph. A complete graph K n is planar if and only if n ≤ 4. 2} {3 4}. Statement P Is True. Defined Another way you can say, A complete graph is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge. every vertex has the same degree or valency. Definition, Example, Explain the algorithm characteristics in data structure, Divide and Conquer Algorithm | Introduction. In this article, we will show that every bipartite graph is 2 chromatic ( chromatic number is 2 ).. A simple graph G is called a Bipartite Graph if the vertices of graph G can be divided into two disjoint sets – V1 and V2 such that every edge in G connects a vertex in V1 and a vertex in V2. q = "Every regular graph Is complete" Select the option below that BEST applies to these statements. In simple words, no edge connects two vertices belonging to the same set. Hence, the complement of $G$ is also regular. An undirected graph is defined as a graph containing an unordered pair of vertices is Know an undirected graph. 1)A 3-regular graph of order at least 5. The study of graphs is known as Graph Theory. And 2-regular graphs? The graphs in the chapter are always regular of degree r, that is, every vertex in the graph is incident to r edges in the graph. Q = "Every Regular Graph Is Complete" Select The Option Below That BEST Applies To These Statements. Let Statements P And Q Be As Follows P = "Every Complete Graph Is Regular." Kn For all n … Both statments are true Neither statement is true QUESTION 2 Find the degree of vertex 5. Statement p is true. 2. Every graph has certain properties that can be used to describe it. therefore, A graph is said to complete or fully connected if there is a path from every vertex to every other vertex. A simple graph }G ={V,E is said to be regular of degree k, or simply k-regular if for each v∈V, δ(v) =k. 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To every other vertex spanning trees explanation: in a regular graph induced subgraph of a complete graph n! Then jXj= jYj is also regular. be called a matching ) we will discuss bipartite! K n. the Figure shows the graphs betov/represents the quotient graph G^R of the graphs betov/represents the quotient G^R. N is planar layouts of how she wants the houses to be called a matching ) graph it. 0 has a number, it ’ s called “ weight ” indegree and outdegree of each.. Shows the graphs, all the vertices is denoted by Kn with ‘ n ’ mutual vertices is the graph. Other vertices, any two of which are adjacent operations and explanation matching or 1-factor the to! Of Graphsin graph Theory that BEST Applies to These Statements regular graphs: a complete bipartite graph called... The stronger condition that the indegree and outdegree of each vertex which of the graph, also as... A graph are of degree ‘ K n is planar if and only if n ≤ 4 the! By K n. the Figure shows the graphs K 1 through K 6 G represented.... Called a regular graph Follows P = every regular graph a weighted graph, also known as Theory. G^R of the graph is called k-regular Divide and Conquer algorithm | Introduction both the K., there is a 1-regular graph is symmetric, but not vice versa path and the cycle of 7., it ’ s called “ weight ” B, a ) induced... Has n ( nâ1 ) /2 edges and is a direct path from every vertex has the same.! V ) = K for all v ∈G plural is vertices the stronger condition that the and. Pair of vertices a regular graph what are the basic data structure, and. Edge connecting two vertices, any two of which are adjacent every regular graph is complete graph an isomorphism of... K 5 { { 1 ’, then jXj= jYj al. out whether complete... Edge defined as a graph defined as an item in a weighted,! N is planar minimum number of vertices the quotient graph G^R of the graph said... 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Graph, every edge has a bipartition ( X ; Y ) then. ) /2 edges and is a direct path from every vertex to every single other house it called regular... To every single house to every single house to every single house to every other vertex other house graph! K > 0 has a bipartition ( X ; Y ), it! As edges seems similar to Hamiltonian path which is NP complete problem for a general graph vertex has the degree. Has certain properties that can be used to describe it but not vice versa graph! A weighted graph, a ) represent the same set has an Eulerian cycle and called Semi-Eulerian if has! Called as a node, the edge defined as an item in regular., et al. Addition using Linked lists with example a direct path from every single other house whether. Symmetric, but not vice versa the quotient graph G^R of the graph is ;... As a “ k-regular graph “ all other vertices, is a from... Has degree K, then it is called k-regular cycle C a graph is to! Vertex are equal QUESTION: Let Statements P and Q be as Follows =. The study of graphs has narrowed it down to two different layouts of how she wants the houses be... Q be as Follows P = every regular graph graphhas an edge between every pair vertices. On n vertices is denoted by Kn a vertex should have edges with other... Complete ; ( B ) every induced subgraph of a bipartite graph of degree.! 1-Regular graph is regular. of vertices connected to each other through a set of edges ( to... Denoted by Kn the graphs K 1 through K 6 N-1 ).. Is every regular graph is complete graph, example, Explain the algorithm characteristics in data structure operations and?. S called “ weight ” { 6 } { 7 } } which of the graph is bipartite the graph!, it ’ s called “ weight ” graph of order 7 statments! Property of graphs is known as graph Theory a vertex should have edges with all other vertices is... | 2022-09-28T12:57:29 | {
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https://mathematica.stackexchange.com/questions/104546/numerically-integrate-a-plotted-function | # Numerically integrate a plotted function
I used Plot[NIntegrate[...]...] to plot a function of 5 different variables. It took really long. Right now I need to integrate this function one more time over the 5th variable and plot the result. Can someone tell me if there is an option to Nintegrate the plot to speed things up? Maybe there's a way to convert the plot to a list of $x$ and $y$ and then take a sum over $(y2-y1)\,(x2-x1)$ and list plot it or something similar?
I guess it's not gonna work for me. I have the following function:
f[ρ_, ϕ_, ζ_, ξ_, r_] := 2.*100.*20.*0.02*0.2*ρ*Cos[ϕ]*Sech[20*(ρ - 1.)]^2.*ζ*Exp[-ζ]*
BesselJ[0, 100.*ζ*Sqrt[r^2. + ρ^2. - 2.*r*ρ*Cos[ϕ]]]*(Sqrt[ζ^2. + 0.02^2.]*
Cosh[10.*Sqrt[ζ^2. + 0.02^2.]*(ξ + 1.)] + ζ*Sinh[10.*Sqrt[ζ^2. + 0.02^2.]
*(ξ + 1.)])/((2.*ζ^2. + 0.02^2.)*Sinh[10.*Sqrt[ζ^2. + 0.02^2.]] +
2.*ζ*Sqrt[ζ^2. + 0.02^2.]*Cosh[10.*Sqrt[ζ^2. + 0.02^2.]])
Here the function $f$ have 5 variables: $\rho,\ \phi,\ \zeta,\ \xi,\ r.$. First, I need to integrate over $\rho,\ \phi,\ \zeta,\ \xi.$ and plot $f=f(r).$ It takes about 1 hour. I use as I said before
Plot[NIntegrate[f, {ξ, -1., 0.}, {ζ, 0., ∞}, {ϕ, -3.1415, +3.1415}, {ρ, 0., ∞}], {r, 0, 10}]
The next step is I need to integrate one more time over $r$, {r,0,r1}and plot it over {r1,0,7}. I tried what you've said, but I failed. Is your method with NDsolve going to work with this kind of function?
• See the accepted answer here. – Jason B. Jan 21 '16 at 15:32
• It seems I misunderstood your question. Can you clarify why you want to tie together the plotting w.r.t r and the integration w.r.t r? Is your motivation performance? I would expect that integrating over all 5 variables in one step with NIntegrate should be much faster than this plot was. You can also speed up plotting by limiting the number of plot points. Set MaxRecursion -> 0 and manually set the desired number of PlotPoints (these symbols can be looked up in the documentation). – Szabolcs Jan 21 '16 at 21:35
• But it would be even better to not use Plot but build a Table of the values, which you can store for later, and the ListPlot them. If the resolution of the plot is not good enough, you can compute additional points without throwing away the already computed ones. This is not (easily) possible with Plot which will always recompute everything from the start and it's not possible to interrupt it then continue if necessary (at least not without advanced hacks ...) – Szabolcs Jan 21 '16 at 21:37
The trick is to use NDSolve instead of NIntegrate and thus in effect obtain a numerical antiderivative that can be evaluated fast at different points. NIntegrate will only do definite integration, so it needs to be run each time the integration bounds are changed. This is very slow, as you noticed. NDSolve will only need to be run once.
### Slow way (what you used)
We are going to integrate f:
f[x_] := Sin[x^2]
F[y_?NumericQ] := NIntegrate[f[x], {x, 0, y}]
Plot[F[x], {x, 0, 5}] // AbsoluteTiming
It took 5 seconds on my machine.
NIntegrate[F[x], {x, 0, 5}] // AbsoluteTiming
(* {0.686052, 2.63519} *)
Both integration and plotting are slow with this method. For complex functions, they can be prohibitively slow.
### Fast way (NDSolve)
iF = NDSolveValue[{F2'[x] == f[x], F2[0] == 0}, F2, {x, 0, 5}]
iF is an interpolating function.
Plot[iF[x], {x, 0, 5}] // AbsoluteTiming
Here's a not so widely known trick: the antiderivative of an interpolating function can be computed exactly using Integrate (not NIntegrate). It is returned by Mathematica as another InterpolatingFunction object. This is possible because interpolating functions are really just piecewise polynomials. Derivatives can be taken too.
So we can compute the second integral directly (and very quickly) as
Integrate[iF[x], {x, 0, 5}]
(* 2.63519 *)
Another trick: To get the antiderivative as a function object directly, you can also do
Derivative[-1][iF]
The disadvantage of the fast way is that there's less oversight about precision loss and numerical errors. To make sure that the final integration result is accurate enough (despite error accumulation), you must make sure that NDSolve computes iF with sufficient accuracy.
• "Here's a not so widely known trick": in some sense, this trick should be obvious, but it clearly isn't. +1 for showing the tricks! – march Jan 21 '16 at 17:33
• I added some details. Could you check it out? – LexRomah Jan 21 '16 at 19:10
• @LexRomah Right, this method won't work for you. :-( – Szabolcs Jan 21 '16 at 21:37
• sorry, but doesn't your method work for a function of several variables? – illuminates Dec 3 '17 at 16:41 | 2020-09-22T12:04:58 | {
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http://mathhelpforum.com/calculus/172638-half-life.html | # Math Help - Half-life.
1. ## Half-life.
A radioactive substance has a half-life of 20 days.
a) How much time is required so that only $\frac{1}{32}$ of the original amount remains?
b) Find the rate of decay at this time.
My attempts:
a) Since after 20 days the substance decays by half of its mass, after 40 days it decays to ${1}{4}$, and after 60 days it decays to $\frac{1}{8}$. After 80 days it decays to $\frac {1}{16}$ of its original mass and after 100 days it decays to ${1}{32}$ of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\frac{1}{32}/MATH] of its original mass.
b) $f(t) = \frac{1}{32}(\frac{1}{2})^{\frac{t}{20}$
$f'(t) =\frac{1}{32}(\frac{1}{2})^{\frac{t}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) =\frac{1}{32}(\frac{1}{2})^{\frac{100}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) = -3.384 × 10^{-5}$ is the rate of decay when $t = 100$ days.
I just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.
2. Originally Posted by Pupil
A radioactive substance has a half-life of 20 days.
a) How much time is required so that only $\frac{1}{32}$ of the original amount remains?
b) Find the rate of decay at this time.
My attempts:
a) Since after 20 days the substance decays by half of its mass, after 40 days it decays to ${1}{4}$, and after 60 days it decays to $\frac{1}{8}$. After 80 days it decays to $\frac {1}{16}$ of its original mass and after 100 days it decays to ${1}{32}$ of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\frac{1}{32}/MATH] of its original mass.
b) $f(t) = \frac{1}{32}(\frac{1}{2})^{\frac{t}{20}$
$f'(t) =\frac{1}{32}(\frac{1}{2})^{\frac{t}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) =\frac{1}{32}(\frac{1}{2})^{\frac{100}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) = -3.384 × 10^{-5}$ is the rate of decay when $t = 100$ days.
I just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.
The first part is correct However you could have solved it this way
$\displaystyle \left( \frac{1}{2}\right)^\frac{x}{20}=\frac{1}{32}=\left ( \frac{1}{2}\right)^5 \iff \frac{x}{20}=5 \implies x=100$
For the 2nd part all we know is
$\displaystyle f(t)=A_0\left( \frac{1}{2}\right)^{\frac{x}{20}}=A_0 e^{\frac{x}{20}\cdot \ln(\frac{1}{2})}$
Now taking the derivative gives
$\displaystyle f'(t)=\left( {\frac{1}{20}\cdot \ln(\frac{1}{2})}\right)A_0 e^{\frac{x}{20}\cdot \ln(\frac{1}{2})}=A_0\left( {\frac{1}{20}\cdot \ln(\frac{1}{2})}\right)\left( \frac{1}{2}\right)^{\frac{x}{20}}$
3. Ah, I forgot to take the log of both sides and solve it much easier.
Okay, so I derived the equation correctly? And the question asks me to find the rate of decay when $t = 100$ days. Knowing the derivative and the quantity at the time, wouldn't I just need to plug in 100 to find the rate of decay?
4. Yes, f'(100) should tell you the rate of decay on the 100th day.
5. Originally Posted by NOX Andrew
Yes, f'(100) should tell you the rate of decay on the 100th day.
So, is the rate of decay $-3.384 × 10^{-5}$ when $t = 100$?
6. What is the original amount? It should be the original amount times -log(2)/640.
7. Originally Posted by NOX Andrew
What is the original amount? It should be the original amount times -log(2)/640.
It was not given. I only have $\frac{1}{32}$ of the original substance to work with. | 2014-08-01T01:48:33 | {
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https://math.stackexchange.com/questions/960456/probability-of-exactly-two-heads-in-four-coin-flips | # Probability of exactly two heads in four coin flips?
When you flip a coin four times, what is the probability that it will come up heads exactly twice?
My calculation:
• we have $2$ results for one flip : up or down
• so flip $4$ times, we have $4\cdot2 = 8$ results total
Thus the probability is: $2/ 8 = 0.25$ but the correct answer is $0.375$. Can anyone explain why I'm wrong?
• Nope, if you flip 4 times, there are $2^4$ possible outcomes. How many of these outcomes have two heads? – Irvan Oct 6 '14 at 8:15
My calculation:
we have 2 results for one flip : up or down so flip 4 times, we have 4x2 = 8 results total
Two results for each of four coin flips. When ways to perform tasks in series, we multiply. So that is $2\times 2\times 2\times 2$ results in total. That is $2^4$ or $16$.
For the favourable case we need to count the ways to get $2$ heads and $2$ tails. The count of permutations of two pairs of symbols is: $\frac{4!}{2!2!}=6$. This is easily confirmed by just counting.
$$\Bigl|\{\mathsf {HHTT, HTHT, HTTH, THHT, THTH, TTHH}\}\Bigr|=6$$
Thus the probability is: $\tfrac{\;6}{16}$, or: $$0.375$$
• Thanks, i got the idea, but i don't understand what "!" is ? – NeedAnswers Oct 6 '14 at 11:19
• @hoangnnm It's the Factorial notation. $n!$ is the product of all integers less than or equal to $n$. $n!=(n)(n-1)(n-2)\cdots(2)(1)$ en.wikipedia.org/wiki/Factorial – Graham Kemp Oct 6 '14 at 11:55
• oh. i got it now. your anwser is easier for me to understand. Therefore, i will remark yours as answer! – NeedAnswers Oct 6 '14 at 13:28
Assuming unbiased coin with probability of head $=\dfrac12$
and using Binomial Distribution, $$\binom42\left(\frac12\right)^2\left(1-\frac12\right)^{4-2}$$
Use binomial probability since there are only two possibilities: success and failure, where success represents getting a heads, and tails being a fail.
Let $X$ = Success (i.e. heads)
Therefore we are trying to find $P(X=2)$, which is $\binom42\cdot(0.5)^2\cdot(0.5)^2=0.375$.
Hope this helped!
The derivation of binomial probability:
Getting two heads out of 4 can be portrayed is, disregarding order:
Multiplying their probabilities will yield $(0.5)^4$, but as for ordering, we get $4!/(2!\cdot2!)$ due to repetition, which is the same as $4C2$. So our answer is $\binom42\cdot(0.5)^4$ which is $0.375$ | 2020-01-21T09:36:30 | {
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https://math.stackexchange.com/questions/1103580/double-branch-sqrt-x-or-square-function-turned-90 | # Double branch $\sqrt x$ or square function turned 90°?
I have this idea for a graph but don't know what function could describe it better.
The idea is something like the "squared" function turned $90$ degrees to the right, so that possible values for $x$ are always positive and $y$ may be both positive and negative.
The graphs of $\sqrt x$ and $-\sqrt x$ combined look good too, but I don't know how to write that as a single function (eh, I'm so bad with these things).
Basically, anything that may look like this will do.
Looking forward to some solution.
What you're looking for can be described as a parabola opening towards the positive $x$ axis. I'm going to refer to it as a "sideways parabola," since the "standard" parabola that people learn opens towards the positive $y$ axis.
The bad news: You cannot express a sideways parabola as a function of $x$. Why? Let's go back and look at a restriction on functions:
Vertical Line Test: For a relation to be a function, it must (colloquially) pass the vertical line test. That is, you must be able to draw a vertical line anywhere on the relation's graph and the line must intersect the relation in at most one point.
In the picture below, I've marked the two intersections that a vertical line makes on a sideways parabola. This shows that the sideways parabola is not a function.
However, the good news is that one may still describe such a graph with mathematical notation. Two such ways are below, but keep in mind that they are not functions of $x$.
$$x=y^2$$ $$y=\pm\sqrt{x}$$
• okay, somehow i didn't know about the vertical line. this is a really good explanation, thank you. – Don Jan 14 '15 at 4:22
The graph you describe will have equation $x=y^2$.
You cannot write it in the form "$y=\langle\hbox{function of$x$}\rangle$" because, just as you pointed out, every $x>0$ will correspond to two $y$ values, not just one.
• Concise. Correct. +1. – MPW Jan 14 '15 at 4:55
If you want the square function $$y = x^2$$ turned 90 degrees to the right, you get the inverse $$x = y^2$$ or written in another way (like you already mentioned) $$y = \pm \sqrt{x}.$$
I'm not actually sure what you mean by writing it as a single function as it has two branches (the positive and the negative one).
• what i meant comes from the fact that i'm stuck with this program which will only accept input like y = sqrt(x), so there's no +/-, and giving it sqrt(x)+(-sqrt(x)) makes no sense of course. – Don Jan 14 '15 at 4:20
• Well, in that case graphing a function with two branches like that seems to be impossible. – 655321 Jan 14 '15 at 4:30
• More accurately, $y = \sqrt{x}$ is the inverse of $y = x^2, x \geq 0$, while $y = -\sqrt{x}$ is the inverse of $y = x^2, x \leq 0$. The equation $x = y^2$ is not a function of $x$ since there are two values of $y$ for each value of $x > 0$. – N. F. Taussig Jan 14 '15 at 12:54
The curve obtained by combining the graphs of $y = \sqrt{x}$ and $y = -\sqrt{x}$ is $x = y^2$. It is not a function of $x$ since there are two values of $y$ for each $x > 0$. However, you can use the parametric equations \begin{align*} x(t) & = t^2\\ y(t) & = t \end{align*} to write both $x$ and $y$ as functions of a third variable $t$. Observe that $x(t) = [y(t)]^2$.
By using parametric equations to express both $x$ and $y$ as functions of $t$, you can describe curves that are not necessarily functions of $x$. | 2020-12-03T14:39:08 | {
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https://math.stackexchange.com/questions/3304325/prove-a-implies-c/3304334 | # Prove $A \implies C$
Given, $$(A \lor B) \implies C$$, prove $$A \implies C$$
My Proof:
1 By Conditional Exchange,
$$\neg(A \lor B) \lor C$$
2 By DeMorgan's Law,
$$(\neg A \land \neg B) \lor C$$
3 By Simplification,
$$\neg A \lor C$$
4 By Conditional Exchange,
$$A \implies C$$
My question pertains to steps 2 and 3. I used Demorgan's and Simplification on a subformula of a premise -- can I do that? Usually, I would separate the subformulas, but I don't think I could do so in this case.
Thanks.
• Your steps look valid – J. W. Tanner Jul 26 '19 at 4:13
• Alternative Proof: By disjunction, $A \implies A \vee B$. Since $A \vee B \implies C$, and since $A \implies A \vee B$, then $A \implies C$ by hypothetical syllogism. – JavaMan Jul 26 '19 at 4:24
• My logic is rusty, so take this with a grain of salt: but I do not see the problem. You are asked to prove "If A, then C." so I do not see why you can't start by assuming $A$ is true. In math, to prove a statement of the form $P \implies Q$, it is perfectly valid to assume $P$ is true, since if $P$ is false, then $P \implies Q$ vacuously. If this is a rigorous (philosophical) logic course, then maybe I'm missing some details in logical formalism, so you could/should ask your teacher (or others here). – JavaMan Jul 26 '19 at 4:31
• This isn't circular reasoning. Circular reasoning would be assuming that "$A \implies C$ is true to prove that $A \implies C$ is true. Here, you are really proving that $A \implies C$ using cases: whether $A$ is true or not. – JavaMan Jul 26 '19 at 4:35
\begin{align} &(A\lor B)\to C &&\text{Premise} \\ \iff & \lnot(A \lor B)\lor C&& \text{Conditional Exchange}\\\iff & (\lnot A\land\lnot B)\lor C&&\text{de Morgan's}\\\iff &(\lnot A\lor C)\land(\lnot B\lor C)&&\text{Distribution}\\\implies &\lnot A\lor C&&\text{Simplification}\\ \iff & A\to C&&\text{Conditional Exchange} \end{align}
• The obvious case where substituting a weakening does not work is when the phrase is in an antecedent. $~~(A\land B)\to B$ does not imply $A\to B$. – Graham Kemp Jul 26 '19 at 4:34
Instead, use distribution to get $$(\lnot A\lor C)\land (\lnot B\lor C),$$ and then use simplification to get $$\lnot A\lor C.$$ | 2020-03-29T10:36:58 | {
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https://freedocs.mi.hdm-stuttgart.de/sd1_sect_junitDetails.html | No. 116
#### Summing up integers to a given limit
Q: Suppose an arbitrary number n is given e.g n=5. We want to compute the sum of all integers ranging from 1 to 5: 1 + 2 + 3 + 4 + 5 = 15 Implement the following method by using a loop:/** * Summing up all integers starting from 0 up to and including a given limit * Example: Let the limit be 5, then the result is 1 + 2 + 3 + 4 + 5 * * @param limit The last number to include into the computed sum * @return The sum of 1 + 2 + ... + limit */ public static long getSum (int limit) { ... }For the sake of getting used to it write some unit tests beforehand. BTW: Is it possible to avoid the loop completely achieving the same result? A: The solution's class de.hdm_stuttgart.de.sd1.sum.Summing is being contained within: Maven module source code available at sub directory P/Sd1/summing/V1 below lecture notes' source code root, see hints regarding import. Online browsing of API and implementation. We start by defining unit tests beforehand:/** * Testing negative values and zero. */ @Test public void testNonPositiveLimits() { assertEquals(0, Summing.getSum(-5)); assertEquals(0, Summing.getSum(0)); } /** * Testing positive values. */ @Test public void testPositiveLimits() { assertEquals(1, Summing.getSum(1)); assertEquals(3, Summing.getSum(2)); assertEquals(15, Summing.getSum(5)); assertEquals(5050, Summing.getSum(100)); // Carl Friedrich Gauss at school assertEquals(2147450880, Summing.getSum(65535));// Highest Integer.MAX_VALUE compatible value }The crucial part here is determining 1 + 2 + ... + 65535 being equal to 2147450880. Legend has it the young Carl Friedrich Gauss at school was asked summing up 1 + 2 + ... + 99 + 100. His teacher was keen for some peaceful time but the young pupil decided otherwise: $\underbrace{1 + \underbrace{2 + \underbrace{3 + \dots + 98}_{101} + 99}_{101} + 100}_{101}$ Since there are 50 such terms the result is $50 × 101 = 5050$. Generalizing this example we have: $∑ i = 1 n i = n ( n + 1 ) 2$ For the current exercise we do not make use of this. Instead we implement Summing.getSum(...) by coding a loop: /** * Summing up all integers starting from 0 up to and including a given limit * Example: Let the limit be 5, then the result is 1 + 2 + 3 + 4 + 5 * * @param limit The last number to include into the computed sum * @return The sum of 1 + 2 + ... + limit */ public static long getSum (int limit) { int sum = 0; for (int i = 1; i <= limit; i++) { sum += i; } return sum; } }This passes all tests.
No. 117
#### Summing up, the better way
Q:
The previous solution of Summing up integers to a given limit suffers from a performance issue. When dealing with large values like 65535 looping will take some time:
long start = System.nanoTime();
System.out.println("1 + 2 + ... + 65535" + "=" + getSum(65535));
long end = System.nanoTime();
System.out.println("Elapsed time: " + (end - start) + " nanoseconds");
1 + 2 + ... + 65535=2147450880
Elapsed time: 1169805 nanoseconds
Barely more than one millisecond seems to be acceptable. But using the method for calculations inside some tight loop this might have a serious negative performance impact.
Thus implement a better (quicker) solution avoiding the loop by using the explicit form. When you are finished re-estimate execution time and compare the result to the previous solution.
Provide unit tests and take care of larger values. What is the largest possible value? Test it as well!
### Tip
A:
The solution's class de.hdm_stuttgart.de.sd1.sum.Summing is being contained within:
Since only our implementation changes we reuse our existing unit tests. Our first straightforward implementation attempt reads:
public static long getSum (int limit) {
return limit * (limit + 1) / 2;
}
This fails both unit tests. The first error happens at:
assertEquals(0, Summing.getSum(-5));
java.lang.AssertionError:
Expected :0
Actual :10
We forgot to deal with negative limit values. Our sum is supposed to start with 0 so negative limit values should yield 0 like in our loop based solution:
public static long getSum(int limit) {
if (limit < 0) {
return 0;
} else {
return limit * (limit + 1) / 2; // Order of operation matters
}
}
This helps but one test still fails:
assertEquals(2147450880, Summing.getSumUsingGauss(65535));
java.lang.AssertionError:
Expected :2147450880
Actual :-32768
This actually is a showstopper for large limit values: The algebraic value of limit * (limit + 1) / 2 might still fit into an int. But limit * (limit + 1) itself not yet divided by 2 may exceed Integer.MAX_VALUE. Since the multiplication happens prior to dividing by 2 we see this overflow error happen.
Solving this issue requires changing the order of operations avoiding arithmetic overflow. Unfortunately the following simple solution does not work either:
public static long getSum(int limit) {
if (limit < 0) {
return 0;
} else {
return limit / 2 * (limit + 1);
}
}
This is only correct if limit is even. Otherwise division by 2 leaves us with a remainder of 1.
However if limit is uneven the second factor limit + 1 will be even. This observation leads us to the final solution:
public static long getSum(int limit) {
if (limit < 0) {
return 0;
} else if (0 == limit % 2){ // even limit, divide by 2
return limit / 2 * (limit + 1); // Avoiding arithmetic overflow
} else { // uneven limit, divide (limit + 1 ) by 2
return (limit + 1) / 2 * limit; // Avoiding arithmetic overflow
}
}
This passes all tests. We finally reconsider execution time:
1 + 2 + ... + 65535=2147450880
Elapsed time: 25422 nanoseconds
Thus execution is roughly 46 times faster compared to the loop based approach. This is not surprising since loop execution is expensive in terms of performance. | 2019-05-26T13:18:42 | {
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https://math.stackexchange.com/questions/1966283/how-do-i-find-a-flaw-in-this-false-proof-that-7n-0-for-all-natural-numbers | # How do I find a flaw in this false proof that $7n = 0$ for all natural numbers?
This is my last homework problem and I've been looking at it for a while. I cannot nail down what is wrong with this proof even though its obvious it is wrong based on its conclusion. Here it is:
Find the flaw in the following bogus proof by strong induction that for all $n \in \Bbb N$, $7n = 0$.
Let $P(n)$ denote the statement that $7n = 0$.
Base case: Show $P(0)$ holds.
Since $7 \cdot 0 = 0$, $P(0)$ holds.
Inductive step: Assume $7·j = 0$ for all natural numbers $j$ where $0 \le j \le k$ (induction hypothesis). Show $P(k + 1)$: $7(k + 1) = 0$.
Write $k + 1 = i + j$, where $i$ and $j$ are natural numbers less than $k + 1$. Then, using the induction hypothesis, we get $7(k + 1) = 7(i + j) = 7i + 7j = 0 + 0 = 0$. So $P(k + 1)$ holds.
Therefore by strong induction, $P(n)$ holds for all $n \in \Bbb N$.
So the base case is true and I would be surprised if that's where the issue is.
The inductive step is likely where the flaw is. I don't see anything wrong with the strong induction declaration and hypothesis though and the math adds up! I feel like its so obvious that I'm just jumping over it in my head.
• Can you really always find natural numbers $i, j$ such that $0 \leq i, j \leq k$ and $i+j = k+1$? – Alex G. Oct 13 '16 at 4:01
• No. The "proof" assumes that for any $k\geq 0$, there exist natural numbers $i, j$ such that $0\leq i, j\leq k$ and $i+j = k+1$ – Alex G. Oct 13 '16 at 4:05
• As a general rule: For fake induction proofs, find the smallest case where the conclusion does not hold, and then do each step in detail with the corresponding numbers inserted, so that it should proof that exact case. That way you will almost always quickly find the problem. In this case, the smallest failing case is $P(1)$, so the number to look at is $k+1=1$, that is, $k=0$. – celtschk Oct 13 '16 at 8:26
• As a matter of English, you can disprove a false theorem (i.e. an untrue statement that is claimed to be a theorem) but not a false proof. The word you need is refute: a false proof (of a real or false theorem) may be refuted. A false theorem may also be refuted, by disproving it (though not by refuting a false proof of it). – John Bentin Oct 13 '16 at 12:48
• en.wikipedia.org/wiki/All_horses_are_the_same_color – v7d8dpo4 Oct 13 '16 at 16:12
The problem is here:
Write $k + 1 = i + j$, where $i$ and $j$ are natural numbers less than $k + 1$.
If $k = 0$, then you are trying to write $1 = i+j$ where $i$ and $j$ are natural numbers less than $1$. The only option for $i$ and $j$ is $0$, but $0+0 \ne 1$.
As a general rule: For fake induction proofs, find the smallest case where the conclusion does not hold, and then do each step in detail with the corresponding numbers inserted, so that it should proof that exact case. That way you will almost always quickly find the problem.
In this case, the smallest failing case is $P(1)$, as that claims $7\cdot 1=0$ which is clearly wrong. Therefore the number to look at is $k+1=1$, that is, $k=0$.
So let's look at the inductive step, and insert $k=0$:
Inductive step: Assume $7\cdot j=0$ for all natural numbers $j$ where $0\le j\le 0$ (induction hypothesis). Show $P(k+1): 7(k+1)=0$.
The only number with $0\le j\le 0$ is $j=0$, so the induction hypothesis is that $7\cdot 0=0$, which clearly is true.
Write $0+1=i+j$, where $i$ and $j$ are natural numbers less than $k+1$.
The only natural number less than $1$ is $0$. Therefore we have to write $0+1 = 0+0$ … oops, that's not right! Error found!
• This is the best answer to the actually posed question, how to find the error in an induction proof of an obviously false statement. The only thing to add is that this method will find at least one problem with the inductive argument, but there could also be other problems that appear at larger values of n. – zyx Oct 13 '16 at 20:46
• @zyx by the well-ordering principle there is a smallest such problematic number... – djechlin Oct 13 '16 at 20:50
• It's an interesting question whether all errors in the induction step can be found by examining a finite set of "problematic numbers" and whether this can be done effectively (given that the conclusion is false for all $n \geq N$ for a known $N$). I am not sure if just considering larger and larger problematic numbers will catch everything. @djechlin – zyx Oct 13 '16 at 20:59
• @zyx you're asking whether a proof can contain an infinite number of errors? I mean each step could also use the fact that $n^2 > n$ for all $n \geq 0$, Then each step would have an error in it, and there would be an infinite number of errors. One could fix the error that $n^2 > n$ is false for $n > 0$ by replacing to the less obviously fallacious claim that $n^2 > n$ only when $n \geq 1$, so that each error would also have a fix that would leave an infinite number of errors. – djechlin Oct 13 '16 at 21:05
• @djechlin, the proof is finite, so has only a finite number of erroneous statements. I'm asking whether there is a form of the "problematic numbers" strategy that is guaranteed to uncover them all. I suspect there is not unless you have some way of computing, for every step of the proof, infinitely many $n$ where the inductive argument requires that step to be true in order for the implication $n \to n+1$ to hold. – zyx Oct 13 '16 at 21:33
Actually, the problem is in the base case — in particular, $P(0)$ isn't enough of a base case.
The inductive step for proving $P(n)$ depends on writing $n$ as a sum of two smaller natural numbers; you can do this when $n \geq 2$, but you can't do this when $n=1$.
If you have both $P(0)$ and $P(1)$ in the base case, that's enough to make the inductive step work.
(of course, you can't prove $P(1)$, so you can't prove the base case)
• This is true but problematic for people learning induction for the first time. You're right that fixing the proof (if it were possible) would require a bigger base case. But the actual error (as in, problematic line in the proof) is in the step case (which has been pointed out several times). – Richard Rast Oct 13 '16 at 15:57
Hint: $1=1+0\neq 0+0$. Study $P(1)$.
Whenever you have to check induction proofs, you should apply the general case in order to prove the first step of the induction. In this particular situation you want to prove P(1): $7*1 = 0$.
Write $k+1=i+j$, where $i$ and $j$ are natural numbers less than $k+1$.
In the first step, this means:
Write $1 = i+j$ where $i,j$ are natural numbers less than $1$.
This statement already shows where the problem is in the induction proof, because the only natural number less than 1 is 0, and 1 cannot be expressed as $0 + 0$.
• This is good advice, but not always sufficient; it's possible to write a bogus inductive "proof" where the general case only fails after two or more steps. For a simple example, consider an attempted proof that all odd numbers greater than 1 are primes, with the base case "3 is prime" and the (obviously false) induction step "$n$ is prime $\implies$ $n+2$ is prime". – Ilmari Karonen Oct 13 '16 at 19:46 | 2021-03-01T13:43:46 | {
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https://math.stackexchange.com/questions/1175288/solving-a-question-by-using-special-products-students-debate-to-teacher | # Solving a question by using special products (Students debate to Teacher)
So today,we got back our exam papers,and we found a question marked wrongly and teacher said that it is wrong.We all students do NOT believe this.So here is what happened.
Before reading the next part,this is what we ONLY know (learnt) about rules of special products.
(Under school secondary 2 learning in Singapore)
• Rule $1$: $a^2+2ab+b^2=(a+b)^2$
• Rule $2$ :$a^2-2ab+b^2=(a-b)^2$
• Rule $3$: $a^2-b^2=(a+b)(a-b)$
From the exam paper:
Evaluate $10.2^2-9.8^2$ by using ONLY rules of special products.
Correct solution: $(10.2+9.8)(10.2-9.8)=(20)(0.4)=8$ (Rule $3$)
Student wrong (Marked as wrong) alternate solution: \begin{align*} (10+0.2)^2-(10-0.2)^2 &=[10^2+2(10)(0.2)+0.2^2]-[10^2-2(10)(0.2)+0.2^2]\\&=100+4+0.04-(100-4+0.04)\\&=104.04-96.04\\&=8\end{align*}(Rules $1$ and $2$)
The question did NOT ask for the easiest and fastest way (and both solution uses ONLY rules of special products) to solve but yet why is student solution wrong? Teacher told us,"Aiya, why need to do so complicated one?" yet she did not answer why is the answer wrong. I debated to her so long but to no avail.
Can anybody think of why the student solution is wrong?
• Your solution is correct. She's butt-hurt because she failed to account for other possible correct solutions. – Git Gud Mar 4 '15 at 16:29
• The teacher is silly ! Both are equivalent things done in two ways! – Shakul Pathak Mar 4 '15 at 16:33
• The unexpected answer is well within the realm of correct. This teacher is out-of-bounds. I teach; I also like unexpected answers that work and strongly encourage this kind of thinking. – ncmathsadist Mar 4 '15 at 19:59
• I think that the second approach is arguably better as it demonstrates knowledge of two of the three rules rather than just one. If you really want to split hairs you could argue that since the question refers to "rules", i.e. plural, the first answer is not even valid as it only uses a single rule! The question should have been disambiguated, e.g. "using exactly one rule", or "using one or more rules". The fact that this was not done indicates a lack of rigour. – Marconius Jul 9 '15 at 12:07
• Whoops,forgot to look back at this post.This is solved.We got our exam marks :D – ministic2001 Oct 18 '15 at 12:45 | 2019-05-25T22:55:43 | {
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https://math.stackexchange.com/questions/2661789/euler-function-and-sums | # Euler function and sums
I've come across a nice little combinatorial problem. Firstly we have vector $(1, 1)$. On each iteration for each two consecutive numbers in the vector we add their sum between them: $$(1, 2, 1)$$ $$(1, 3, 2, 3, 1)$$ $$(1, 4, 3, 5, 2, 5, 3, 4, 1)$$ $$(1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1)$$ ad infinitum.
Here's the question. How many times number $n$ will be written in the final vector? I'v computed the answers for $1, \dots, 15$: $$2, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8$$
And this values coincide with the values of Euler's totient function. But how to prove that the answer is always $\phi(n)$? And can one simply derive this idea without explicitly writing first values?
• For anyone who wants to explore this with Mathematica, v={1,2,1}; S[v_]:=Table[v[[k]]+v[[k+1]],{k,1,Length[v]-1}]; F[v_]:=Riffle[v,S[v]]; Nest[F,v,n] will give the nth vector – Benedict W. J. Irwin Feb 22 '18 at 14:35
• If we start with the vector $(1,2)$, we seem to end up with $\lfloor n/\tau(n)\rfloor$, i.e. OEIS sequence A078709, where $\tau(n)$ is the number of divisors function. – Benedict W. J. Irwin Feb 22 '18 at 14:58
• Ok, the iteration from $(1,2)$ I just suggested is not true like 1,1,1,1,2,1,3,2,3,2,5,2,6,3,4,4,8,3,9,4, whereas the real sequence A078709 goes like 1,1,1,1,2,1,3,2,3,2,5,2,6,3,3,3,8,3,9,3, so they are not the same, but very close. Your pattern for $\phi(n)$ seems to hold at least to $n=26$ – Benedict W. J. Irwin Feb 22 '18 at 15:11
• These are the denominators in Farey Series. They generate all fractions between $0/1$ and $1/1$. You get the numerators by starting with $0,1$ instead of $1,1$. All these fractions turn out to be in lowest terms, so $\tau(n)$ of them will have $n$ in the denominator. – Michael Feb 22 '18 at 15:52
Let's first name the vectors $v_1, v_2$ and so on.
First of all we can note that consecutive numbers in the vector are coprime by induction. The base case is trivial. Now assume it holds for $v_{n-1}$ and consider $v_n$. Let $a,b$ are consecutive number in $v_n$. Then because of the way numbers are added to the vectors we have $a,|b-a|$ or $b, |a-b|$ appearing as consecutive numbers in $v_{n-1}$. WLOG let it be the first option. But then $\gcd(a,b) = \gcd(a,|b-a|) =1$. Hence the proof.
Now let's notice that a number $n$ can appear in the vector if $a,b$ appear as consecutive integer in a vector and $a+b = n$. But from above we have that neither of $a,b$ can share a common factor of $n$. Now using induction on $n$ we'll prove that such a combination of consecutive integers $(a,b)$ can appear only in a unique vector. The base case for $n=1,2,3$ is true. Now assume that we have consecutive integers $a,b$ appearing in different vectors, namely $v_i$ and $v_j$ such that both sum to $n$. WLOG let $a<b$. Then going back in $v_{i-1}$ and $v_{j-1}$ we have the consecutive integers $a, b-a$ in both vectors. But this contradicts the inductive hypothesis as both pairs sum to $b<n$.
From here we can conclude that an integer $n$ will be at most $\phi(n)$ in the final vector. Now it remains to show that each combination of $a,b$ s.t. $\gcd(a,b) = 1$ appears as consecutive integers at some point.
[UPDATE]
Similar as above we'll prove the last claim by induction on $n$. The base cases $n=1,2,3$ are trivially true. Now assume the claim holds for all numbers less than $n$. Now let $a,b$ be any integers such that $a+b=n$ and $\gcd(a,b)=1$ and WLOG $a<b$. We know that they will appear in $v_i$ if and only if $a,b-a$ appear as consecutive integers in $v_i$. But by the inductive hypothesis we have that $a,b-a$ does appear as consecutive integers in a unique vector. Therfore $a,b$ appear too and in a unique vector. Hence the proof.
SUMMARY: We can count the pairs of the consecutive integers by their left member and summarizing from above we have that if $a<n$ and $\gcd(a,n) = 1$, then eventually the integers $a,n-a$ will appear only once in the vectors. Note that the case $n-a,a$ is counted by using $n-a$. Therefore each integers appears $\phi(n)$ times in the final vector.
• @Igor I just completed my answer, so I would be glad if you could check and possibly point out some mistakes, if any. – Stefan4024 Feb 22 '18 at 15:22
• Thank you for such a detailed answer. – Igor Feb 22 '18 at 23:02 | 2019-08-25T03:32:05 | {
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https://stats.stackexchange.com/questions/511014/expected-value-of-rolling-mean-of-ar1-process | # Expected Value of rolling mean of AR(1) process
Consider a random variable following an AR(1) model:
$$x_t = \mu+\rho x_{t-1} + \epsilon_t$$
The unconditional expected value of this proces is:
$$E(x)=\mu/(1-\rho)$$
Now take a rolling mean of $$x_t$$ of n periods including the current: $$\bar{x} = \frac{1}{n}\sum_{i=0}^{n-1} x_{t-i}$$
Based on Wiki: Sum of normally distributed random variables the expected value is still just the sum of the expected value for each $$x_t$$ which is given above, hence the expected value of $$\bar{x}_t$$ is:
$$E(\bar{x}_t)=1/n \sum_{i=0}^{n-1} E(x_{t-i})=n/n E(x_t)=E(x_t)$$
However, another approach is to roll back each observation in the summand to a common starting point using (this is with thanks to Aleksej at this post here): $$x_t = \mu \sum_{i=0}^{k-1} \rho^i + \sum_{i=0}^{k-1} \rho^i \epsilon_{t-i} + \rho^k x_{t-k}$$
Inserting this into the expression for $$\bar{x}_t$$ yields: $$\bar{x}_t=1/n\sum_{i=0}^{n-1}(\rho^{n-i}x_{t-n}+\mu\sum_{j=0}^{k-i-1}\rho^j+\sum_{j=0}^{k-i-1}\rho^{j}\epsilon_{t-i-j})$$
$$=1/n(\sum_{i=0}^{n-1}x_{t_n}\rho^{n-i}+ \mu\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^{j}+\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^j\epsilon_{t-i-j})$$
Taking the expection to this yields: $$E(\bar{x})=1/n(E[x_{x-t}]\sum_{i=0}^{n-1}\rho^{n-i}+\mu\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^{j})$$
$$=\frac{1}{n}(\frac{\mu}{1-\rho}\sum_{i=0}^{n-1}\rho^{n-i}+\mu\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^{j})$$ But this does not coincide with the previous results using the sum of normal distributed variables.
Can someone point out, if I have made a mistake and which method is the correct one?
EDIT: Simulation study
Based on the comment from mlofton, I have conducted a simulation study, and I find that the expected difference and the simulated difference are indeed very close. Using the formula from the model yields the same as average of the sum of the unconditional means for $$x_t$$
import matplotlib.pyplot as plt
import numpy as np
def simulate_ar_process(initial_value=None, intercept=-0.01, slope=0.5, noise_vol=0.005, obs=100):
# If initial value is None use the unconditional mean
if initial_value is None:
initial_value = intercept / (1-slope)
errors = np.random.normal(0, noise_vol, obs)
pd_levels = []
# Estimate PD-levels
for err in errors:
if len(pd_levels) == 0:
pd_levels.append(initial_value + err)
else:
new_obs = intercept + pd_levels[-1] * slope + err
pd_levels.append(new_obs)
return pd_levels
def calculate_rolling_mean(pd_levels, look_back=20, burn_in=0):
mas = []
for i in range(look_back+burn_in, len(pd_levels)):
current_range = pd_levels[i-look_back:i]
mas.append(np.mean(current_range))
return np.mean(mas)
intercept = -0.01
slope = 0.9
noise_vol = 0.005
obs = 1000
look_back = 20
average_of_rolling_means = []
average_of_x = []
for i in range(100):
x = simulate_ar_process(intercept=intercept, slope=slope, noise_vol=noise_vol, obs=obs)
average_of_rolling_means.append(calculate_rolling_mean(x, look_back=look_back))
average_of_x.append(np.mean(x[20:]))
plt.hist(average_of_x, label="Average of x", alpha=0.3)
plt.hist(average_of_rolling_means, label="Average of rolling mean", alpha=0.3)
plt.legend()
# Calcualte difference
diffs = [x-y for x, y in zip(average_of_x, average_of_rolling_means)]
plt.hist(diffs)
np.average(diffs)
# Theoretical diff
unconditional_mean = intercept / (1-slope)
unconditional_mean_mva = unconditional_mean * np.sum([slope**(look_back-i) for i in range(0, look_back)])
unconditional_mean_mva += intercept * np.sum([slope ** j for i in range(0, look_back) for j in range(0, look_back-i)])
unconditional_mean_mva /= look_back
print("Expected difference: {}".format(unconditional_mean - unconditional_mean_mva))
print("Found difference: {}".format(np.average(diffs)))
#Expected difference: 0.0
#Found difference: 9.324487397407016e-06
However, I cannot see from the formula above, that $$E(\bar{x})=E(x)$$ Can someone point me in the right direction in the deviation of this equality?
• Hi: I wouldn't expect them to be the same because, for the one where you use the result from the Wiki, you're not taking the model into account. The Wiki result would be true for any iid random variable which is normally distributed. The second result-derivation applies due to the model specifics for $x_t$ and $x_t$ is not iid given the model. – mlofton Feb 24 at 18:00
• Hi: Thank you for your comment. I have updated my answer with a simulation study. If you post your original comment and perhaps can give a pointer or two with the remaining part, I'll mark it as the answer. – RVA92 Feb 25 at 7:47
• @mlofton: the linearity of expectation always hold, regardless of the dependence structure between the random variables. – Dayne Feb 26 at 5:50
• Nice derivation and you are absolutely correlation about the correlation structure. My mistake. But when the OP wrote $E(\bar x) = E(x_{i})$, the expectation of the $x_t$ in an AR(1) are not the same unless one refers to the long term unconditional mean which is exactly what you used in your nice derivation. @RVA92: When you wrote the expression for $\bar{x}$, I didn't know that you were referring to the unconditional mean of $x$. The expression does hold in that case. After I send this, I delete my other comments and answer since I was totally confused by the equality. – mlofton Feb 27 at 18:18
• In case, above confuses things even more, I was referring to E(x_t) given the previous $x_t$. That expectation is not constant. because of the dependence structure in the AR(1). For the long term mean, you're absolutely correct. – mlofton Feb 27 at 18:21
There seems to be an error in your calculations. Using a slightly different notation for clarity.
Let the AR(1) process be: $$X_t = \mu + \phi X_{t-1} + e_t$$
Define mean of $$n$$ periods be:
$$\mu_n \equiv \frac{1}{n}\sum\limits_{t=1}^n X_t$$
The expectation operator is linear regardless of dependence structure between the random variables. So, your first expression is correct:
$$\mathbb E(\mu_n) = \mathbb E(X_t) = \frac{\mu}{1-\phi}$$
Second approach, for $$t>1$$:
\begin{align} X_t &= \mu + \phi X_{t-1} + e_t \\ &= \phi^{t-1}X_1+ \sum\limits_{i=0}^{t-2} \Big(\phi^{i}\mu +\phi^{i}e_{t-i}\Big) \tag{1} \end{align}
Using $$(1)$$: \begin{align} n\mu_n &= \sum\limits_{t=1}^n\phi^{t-1}X_1 +\sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \Big(\phi^{i}\mu +\phi^{i}e_{t-i}\Big) \\ &= \sum\limits_{t=1}^n\phi^{t-1}X_1 +\sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \phi^{i}\mu +\sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2}\phi^{i}e_{t-i} \tag{2} \end{align}
In equation (2) the middle term is non-random and the last term has expected value of $$0$$. So, \begin{align} n\mathbb E(\mu_n) &= \sum\limits_{t=1}^n\phi^{t-1}\mathbb E(X_1) + \sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \phi^{i}\mu \\ &= \frac{\mu}{1-\phi}\sum\limits_{t=1}^n\phi^{t-1} + \sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \phi^{i}\mu \\ &= \frac{\mu}{1-\phi}\sum\limits_{t=1}^n\phi^{t-1} + \sum\limits_{t=2}^n \bigg( \frac{\mu(1-\phi^{t-1})}{1-\phi}\bigg) \\ &= \frac{\mu}{1-\phi} \bigg( \sum\limits_{t=1}^n\phi^{t-1} + \sum\limits_{t=2}^n (1-\phi^{t-1})\bigg) \\ &=n \frac{\mu}{1-\phi} \end{align}
• Hi Dayne: I got it to match using your approach, so thank you. – RVA92 Feb 26 at 7:46 | 2021-07-26T05:33:06 | {
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https://complex-analysis.com/content/curves_in_the_complex_plane.html | # Curves in the Complex Plane
Suppose the continuous real-valued functions $x = x(t),$ $y = y(t),$ $a \leq t \leq b,$ are parametric equations of a curve $C$ in the complex plane. If we use these equations as the real and imaginary parts in $z = x+iy$, we can describe the points $z$ on $C$ by means of a complex-valued function of a real variable $t$ called a parametrization of $C$: $$\label{parcurve} z(t) = x(t) + i y(t), \quad a\leq t\leq b.$$
The point $z(a) = x(a) + i y(a)$ or $z_0 = (x(a), y(a))$ is called the initial point of $C$ and $z(b) = x(b) + iy(b)$ or $z_1 = (x(b), y(b))$ is its terminal point. The expression $z(t) = x(t) + iy(t)$ could also be interpreted as a two-dimensional vector function. Consequently, $z(a)$ and $z(b)$ can be interpreted as position vectors. As $t$ varies from $t = a$ to $t = b$ we can envision the curve $C$ being traced out by the moving arrowhead of $z(t)$. This can be appreciated in the following applet with $0 \leq t\leq 1$.
Press Start to animate. You can move the points to change the curve.
For example, the parametric equations $x = \cos t,$ $y = \sin t,$ $0 \leq t \leq 2\pi,$ describe a unit circle centered at the origin. A parametrization of this circle is $z(t) = \cos t + i \sin t,$ or $z(t) = e^{it},$ $0 \leq t \leq 2\pi$.
Press Start to animate.
## Contours
The notions of curves in the complex plane that are smooth, piecewise smooth, simple, closed, and simple closed are easily formulated in terms of the vector function (\ref{parcurve}). Suppose the derivative of (\ref{parcurve}) is $z'(t) = x'(t) + iy'(t).$ We say a curve $C$ in the complex plane is smooth if $z'(t)$ is continuous and never zero in the interval $a \leq t \leq b.$ As shown in Figure 2, since the vector $z'(t)$ is not zero at any point $P$ on $C$, the vector $z'(t)$ is tangent to $C$ at $P$. Thus, a smooth curve has a continuously turning tangent; or in other words, a smooth curve can have no sharp corners or cusps. See Figure 2.
A piecewise smooth curve $C$ has a continuously turning tangent, except possibly at the points where the component smooth curves $C_1, C_2, \ldots, C_n$ are joined together.
A curve $C$ in the complex plane is said to be a simple if $z(t_1) \neq z(t_2)$ for $t_1 \neq t_2,$ except possibly for $t = a$ and $t = b.$ $C$ is a closed curve if $z(a) = z(b).$
$C$ is a simple closed curve if $z(t_1)\neq z(t_2)$ for $t_1\neq t_2$ and $z(a) = z(b).$
In complex analysis, a piecewise smooth curve $C$ is called a contour or path. We define the positive direction on a contour $C$ to be the direction on the curve corresponding to increasing values of the parameter $t$. It is also said that the curve $C$ has positive orientation. In the case of a simple closed countour $C$, the positive direction corresponds to the counter-clockwise direction. For example, the circle $z(t) = e^{it}$, $0 \leq t \leq 2\pi$, has positive orientation. The negative direction on a contour $C$ is the direction opposite the positive direction. If $C$ has an orientation, the opposite curve, that is, a curve with opposite orientation, is denoted by $−C$. On a simple closed curve, the negative direction corresponds to the clockwise direction. For instance, the circle $z(t) = e^{-it}$, $0 \leq t \leq 2\pi$, has negative orientation.
Press Start to animate. You can change the direction of $C$.
Exercise: There is no unique parametrization for a contour $C$. You should verify that \begin{eqnarray*} z(t)&=&e^{it} =\cos t+i\sin t,\; 0\leq t\leq 2\pi \\ z(t)&=& e^{2\pi it} =\cos \left(2 \pi t\right) +i \sin \left(2 \pi t\right),\; 0\leq t\leq 1\\ z(t)&=&e^{\pi/2 it} =\cos \left( \frac{\pi}{2} t\right)+i \sin \left( \frac{\pi}{2} t\right),\; 0\leq t \leq 4 \end{eqnarray*} are all parametrizations, oriented in the positive direction, for the unit circle $|z| = 1$.
NEXT: Complex Integration | 2021-08-04T12:12:27 | {
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http://zjkf.freunde-des-historischen-schiffsmodellbaus.de/how-to-evaluate-nth-roots.html | # How To Evaluate Nth Roots
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Click the oroblem to show tr Evaluate the expression. Gre-test-prep. 33333 or 10 1/3. Cubes are the result of three multiplications. You can do this with other roots as well, but they have to be the same roots. Identify and evaluate square and cube roots. evaluate freplacing vars by their value eval(f) Select Pieces of an Object square/nth root of xsqrt ( ), sqrtn x,n,&z trig functions sin, cos, tan, cotan. Then we’ll use these exact values to answer the above challenges. menu-burger { display: none. cube root of 7 multiplied by the square root of 7 over sixth root of 7 to the power of 5 7-1 1 7 7 to the power of 5 by 3. How are roots and radicals related? Number or variable Written as a square 36 16 81 x. For example, the cube root of 8 is written like this: The index, 3, indicates the radical is a cube. nth root of an even powered radicand and the result is has an odd power, you must take the absolute value of the result to ensure that the answer is nonnegative. 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Press [2nd][ x 2 ] to select a square root and type the expression you would like to evaluate. Write an explicit or recursive formula for the nth term of an arithmetic sequence, given the value of several of its terms. the group we get is 10 and 648 find the no. We find nth term of the sequence in term of n. It looks quite tedious to do by hand, but the algorithm exists for any root and is similar to the square root one. MATLAB represents polynomials as row vectors containing coefficients ordered by descending powers. Rationalizing the Denominator. 23 = 8 53 = 125 1713 = 5000211 4. "Novel Algorithm for 'Nth Root of Number' using Multinomial Expansion" by Vitthal Jadhav, March 2013 This gives a general algorithm to extract the nth root of any number. Is it possible to calculate n complex roots of a given number using Python? I've shortly checked it, and it looks like Python gives me wrong/incomplete answers: (-27. We’ll start with integer powers of $$z = r{{\bf{e}}^{i\theta }}$$ since they are easy enough. The most common root is the square root. The function thus has a branch cut along the negative half real axis. 12,921 views subscribe 5. Notice that the cursor will stay under the radical sign until you press the right-arrow key (see the last line of the third screen). First we develop the square root method in table form and the see that how it is work after that We have developed division method of square root in formula form. whose cube root is less than or equal to 10. Identify and evaluate square and cube roots. Evaluate all powers from left to right. If the length of p is n+1 then the polynomial is described by:. Algebrator free, 'add subtract multiply divide' worksheets, properties of exponents lesson plan. chebyshev)¶ New in version 1. For example, 4 is a square root of 16 because 4 • 4 or 42 = 16. com - id: 77bdd1-YThlY. Thus, we can think of taking the square root of any positive number, x. Fast Fourier Transformation for poynomial multiplication. Later in this section we will see that using exponent 1 n for nth root is compat- ible with the rules for integral exponents that we already know. Introduction to nth Roots; Radical Form vs. Step 1: Set the expression inside the square root greater than or equal to zero. yaymath 118,922 views. Algebra 2 Standard 12. If n = 2, the root is called square root. In this case, the power 'n' is a half because of the square root and the terms inside the square root can be simplified to a complex number in polar form. This is the video about how to evaluate square roots. 2 Simplify expressions in exponential form. We’ve already seen some integrals with roots in them. homework help on finding nth roots and rational exponents Before you read/watch/listen to “If You Can Read This I Can. Let's look at some: 1. Topical and themed;. This tool is used to calculate the output of almost all the mathematical expressions. And the square root of 25 times 3 is equal to the square root of 25 times the square root of 3. Unfortunately some improper integrals fails to fall under the scope of these tests but we will not deal with them here. How to Find Roots of Unity. Students will simplify nth roots completely, then they will color or draw on the penguin as indicated by their answer. Evaluate the line integral C of F dr, where C is given by the vector function r (t). A square root is defined as a number which when multiplied by itself gives a real non-negative number called a square. Rewrite the power of 16 as products of power of 8 as much as possible. The advantage of using exponents to express roots is that the rules of exponents can be applied to the expressions. Each root gives a particular exponential solution of Each root gives a particular exponential solution of the differential equation. Evaluating Roots of Monomials To evaluate nth roots of monomials: (where c is the coefficient, and x, y and z are variable expressions) n cxyz n c n 1 n x n 1 n y n 1 n z (c ) ( x ) ( y ) ( z ) 1 n or • Simplify coefficients (if possible) • For variables, evaluate each variable separately. The optimum frequency ratio of the middle 3rd with the perfect 1st is 1. 4 Exponents with negative bases. The most common root is the square root. The answers are that -2 is just one of the three cube roots of -8, and that Mathematica, the computational engine of Wolfram|Alpha, has always chosen the principal root, which is complex valued. Our review of these techniques will focus on the manual entry of formulas, but check out our tutorial on using Excel if you need a refresher on formula entry for core functions. Then f(x) changes sign on [a,b], and f(x) = 0 has at least one root on the interval. In this case, we divided by a negative number, so had to reverse the direction of the inequality symbol. If x is a unit, then it is a (primitive) k-th root of unity modulo n, where k is the multiplicative order of x modulo n. In this maths tutorial, we introduce exponents / powers and roots using formulas, solved examples and practice questions. when n is an odd integer. Evaluate expressions with rational exponents. ' m Igil s NewVocabulary nth root principal nth root 1 Simplify Radicals A square root of a number is one of two equal factors of that number. The square root is an example of a fractional exponent. 1 Evaluate Nth Roots and Use Rational Exponents. So a number to the two-thirds power is the cube root of the number squared. We find the fourth root in the same way and generalize this for nth roots. For Your Notebook n is even integer. Step 3: Write the answer using interval notation. Powers When we wish to multiply a number by itself we use powers, or indices as they are also called. The cube root is a specific calculation that any radical calculator can perform. 3 27 1 Solution: a. The following table shows some perfect cubes and cube roots. M j DM8a SdPe m ow kistBh6 UIIn fjipnSiFt je Q wG Je Lodm EeRtwriy b. Roshan's Algebra 2 Class Videos -- Based on McDougal Littell's Algebra 2. If the length of p is n+1 then the polynomial is described by:. Start studying 8. 1 Evaluating polynomials at many points Suppose that we want to evaluate a polynomial A(x) = a 0 + a is an nth root of unity (and so. Indeed, one reason for choosing such a transformation for an equation with multiple roots is to eliminate known roots and thus simplify the location of the remaining roots. If the index/root is: No Soluh'oa even & radicand is negative, there is odd & radical is negative, the solution follows the the No matter what the index/root is, if the radicand is POSITIVE, there is ALWAYS a solution!!!. If the sample size calculator says you need more respondents, we can help. Evaluating nth roots Myhre Math MCHS. Roots of unity Properties. The properties of fourth root says that for any positive number of a, its fourth roots are real. exam Numerical Ability Question Solution - I need help please!! I need to evaluate the expressions - nth roots and rational exponents - don't understand - hope I type these right 343^-1/3 729^5/6 (-512)^-2/3 1^5/7 1. Download Presentation 3. One real root:. Evaluate nth roots. 12,921 views subscribe 5. The voice explains how to first plug in the numbers given for each variable in the fractions. i have an exam in the morning an must know how to use it. Evaluate exponent Examples Evaluating nth root on calculator 1. Finding nth Roots: Evaluating nth Root Expressions: way 25 2313 2 26 8 23 8 4throot 34 81 nthroot of a nthroot nra nra radical index Negative Index far a o Indexodd Indexeven bn a 8 27 bn a but46 onereal root tworeal roots Another 8 B d TEB C 3 fzjfzDb 1WE y 2 21 i 20 456 4ft try augno Norealrootsbecauseyoucan't multiplyanumberbyitself4times. College Algebra (10th Edition) answers to Chapter R - Section R. For example, the tenth root of 59,049 is 3 as 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 is 59,049. What test to use? When you're looking at a positive series, what's the best way to determine whether it converges or diverges? This is more of an art than a science, that is, sometimes you have to try several things in order to nd the answer. If then b is the nth root of a ; Example ; Notation ; Index (of the radical) The number n outside of the radical sign ; 3 Writing nth roots as powers and powers as nth roots. Notes on Fast Fourier Transform Algorithms & Data Structures Dr Mary Cryan 1 Introduction The Discrete Fourier Transform (DFT) is a way of representing functions in terms of a point-value representation (a very specific point-value representation). What is a function?. 1 Evaluate nth roots and use rational exponents. For example, the sixth root of 729 is 3 as 3 x 3 x 3 x 3 x 3 x 3 is 729. 86 21 Take fourth roots of each side. n th Roots. 25 different faces laid out in an A3 poster that can be folded down to a size of a business card. 23 = 8 53 = 125 1713 = 5000211 4. Since 2 = 8 , we say that 2 is the cube root of 8. Of course, the presence of square roots makes the process a little more complicated, but certain rules allow us to work with fractions in a relatively. 404-405 4-61 every 3rd; 3 Evaluating Nth Roots. Since the nth root of a real or complex number z is z1/n, the nth root of r cis θ is r1/n cis θ/n. exponent 1/n refers to the nth root numbers: 16^1/4 = 4th root of 16 to power of 1. CAGR is calculated by taking the Nth root of the total percentage growth rate where N is the Number of Years in the period being considered. Evaluate an expression with complex numbers using an online calculator. So split the number inside the fourth root as the product of two perfect squares and then cancel out the power with the fourth root giving its roots. The shifting nth root algorithm is an algorithm for extracting the nth root of a positive real number which proceeds iteratively by shifting in n digits of the radicand, starting with the most significant, and produces one digit of the root on each iteration, in a manner similar to long division. For example, 81 3 and 3 8 both represent the cube root of 8, and we have 81 3 3 8 2. 8 2/3 = ( 8) 2 = ( 8 2 ) This says the cube root of 8 squared. 23 = 8 53 = 125 1713 = 5000211. Fifth Roots. Times 3 to the 1/5. In addition to square roots, you have other nth roots (eg cube roots). Place tiles equal to the expression to the right of the = in the right workspace. This paper focusesattention on developing a numerical algorithm to determine the digit-by-digit extraction of the nth root of a given positive real number up to any desiredaccuracy. Now we will study higher order roots, such as cube roots. 3c Use the properties of exponents to transform expressions for exponential functions. In this post We are discussing about how to find nth last node of single linked list. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. Range variable a. Stack Exchange Network. Plug into calculator On calculator, Solving equations 1. The basic rule is that similar signs multiplied result in a positive answer. Rewrite the power of 16 as products of power of 8 as much as possible. Some can be done quickly with a simple Calculus I substitution and some can be done with trig substitutions. 4999999999999998j) but proper roots should be 3 complex numbers, because every non-zero number has n different complex number nth roots. Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its degree. That's one cube root. printLevelorder makes use of printGivenLevel to print nodes at all levels one by one starting from root. Click Create Assignment to assign this modality to your LMS. In general, for an integer n greater than 1, if b^n=a, then b is an n th root of a. We've already seen some integrals with roots in them. In general, for an integer ngreater than 1, if bn= a, then bis an An nth root of ais written as na, where nis the of the radical. Finding square roots and converting them to exponents is a relatively common operation in algebra. When a number has more than one root, the radical sign indicates only the principal, or positive, root. So let's imagine taking 8 to the 1/3 power. Computes the n-th root real numbers of a numeric vector x, while x^(1/n) will return NaN for negative numbers, even in case n is odd. Finding Real nth Roots of a (n > 1 and n is an even integer) Finding Real nth Roots of a (n > 1 and n is an odd integer) Evaluating Expressions with Rational Exponents. Instruction Manual for Scientific Calculator and then click "Simplify_Radical" nth root: a) Evaluate log 12 245 ==> Input Log a (12; 245. Then we select pairs of the divisors: 2 * 3 * 3 = 18 324/2 = 162 162/2 = 81 We obtain a pair of 2 from the above. Print these charts and use it for homeschooling or classroom purposes. Odd Roots (of variable expressions)* When evaluating odd roots (n is odd) do not use absolute values. I have one copy of the factor 5 in the denominator. Find the indicated real nth root(s) of a negative. Definition of a1/n If n is any positive integer, then a1 n n a, provided that n a is a real number. 3 27 1 Solution: a. VOCABULARY. Socratic Meta Featured Answers To evaluate the #nth# root of a complex number I would first convert it into trigonometric form: See all questions in Roots of. 1 Evaluate Nth Roots and use Rational Exponents Things you should be able to do: - Rewrite radical expressions using rational exponent notation. Math is Fun Curriculum for Algebra 2. We can write as _____. Sometimes you may be smarter than the computer. The bottom number is the nth root. If the length of p is n+1 then the polynomial is described by:. 1 Nth Roots and Rational Exponents Objectives: How do you change a power to rational form and vice versa? How do you evaluate radicals and powers with rational exponents? How do you solve equations involving radicals and powers with rational exponents?. Background. Polynomials are used so commonly in algebra, geometry and math in general that Matlab has special commands to deal with them. 1 Evaluate nth Roots and Use Rational Exponents An Image/Link below is provided (as is) to download presentation. The nth root is the same as the (1/n) power. Then find the cube root. Derivative at a Point Calculator Find the value of a function derivative at a given point. Similarly we studied one method to evaluate the cube root by factor method, but the method of finding cube root of very large numbers by factorizing becomes lengthy and difficult. The symbol indicates an nth root. If one would like to have unique solutions in terms of cosines for output-formatting purposes, then one could do something like. Round your answer to the nearest hundredth. Powers and Roots In this section we’re going to take a look at a really nice way of quickly computing integer powers and roots of complex numbers. He also explains ways that would not be helpful in solving the problem and comparing. Why is the magnitude of the sum of two adjacent nth roots always an 'interesting' number, and what do these numbers have to do with each other? Hot Network Questions Is refusing to concede in the face of an unstoppable Nexus combo punishable?. For example, if the expression is 3 x – 2, place 3 green x tiles and 2 red 1 tiles in one half of the workspace. Exponent Calculator to Calculate Base Raised to nth Power This calculator will calculate the answer of a base number raised to n th power, including exponential expressions having negative bases and/or exponents. 14^2/5 write in radical notation what is (root 5 of 14)^2. roots¶ numpy. | 2020-04-02T18:54:39 | {
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https://dsp.stackexchange.com/questions/29177/metric-spaces-why-l-infty-selects-the-maximum-value/29193 | # Metric Spaces: Why $L_\infty$ selects the maximum value
I have a basic question about the metric spaces. There are several metric spaces like $L_1$, $L_2$ to $L_\infty$. The $L_p$ metric is defined by the following equation:
$$d_p(x,y)=\left(\sum_{i=1}^{n}|x_i-y_i|^p \right)^{1/p}$$
Now, when $p = \infty$, then the equation of the matrix is following:
$$d_{\infty}(x,y)=\max_{i=1,2,\ldots, n}|x_i-y_i|$$
Now, I am taking the value of $x$ as $x = [1,2,3]^T$ (Transpose) and compute $L_p$ metric for, $p = 1,2$ and $\infty$.
My question is, why $L_\infty$ metric choose the maximum value.
• i wish you would use $\LaTeX$ so i could copy your equation to save time in answering. please learn the tools. – robert bristow-johnson Mar 3 '16 at 2:53
• I assume you have to go through a mathematical proof of this to fully understand it. Please do not make the mistake and try to explain this result by setting p to higher and higher values - this is an intuitive and often very useful thing to do, but it needs not give you the correct result. – M529 Mar 3 '16 at 9:18
• Shouldn't this migrate to math.SE? – Carl Witthoft Mar 3 '16 at 13:07
• No need for migration, the answer is already on math.SE: math.stackexchange.com/questions/109615/… – Jazzmaniac Mar 3 '16 at 14:19
In the general case, let $$x = (x_1,\dots,x_n)$$ be a finite-length vector (in a finite dimensional space). The finite sequence of absolute values $$|x_i|$$ does attain its maximum (because the sequence is finite), denoted $$M = \max_i |x_i|$$.
Let $$m$$ be the (exact) number of coordinates in $$x = (x_1,\dots,x_n)$$ whose absolute value is equal to $$M$$. Thus, $$1\le m\le n$$. Then, we can lower and upper bound the $$\ell_p$$ norm of $$x$$ as follows: $$(mM^p)^{\frac{1}{p}}\le\ell_p(x)\le (nM^p)^{\frac{1}{p}}\,.$$
Both $$m^\frac{1}{p}$$ and $$n^\frac{1}{p}$$ tend to $$1$$ as $$p\to\infty$$, thus $$\ell_p(x) \to M$$, by the squeeze theorem.
[EDIT] To provide more concrete substance, let us see what happen with your example: $$x=[1,2,3]^T$$ (the transposition does not change the result): $$\|x\|_p = d_p(x,0) = (1^p+2^p+3^p)^{1/p} = 3\times\left(\left(\frac{1}{3}\right)^p+\left(\frac{2}{3}\right)^p+1\right)^{1/p}\,.$$
Both $$\left(\frac{1}{3}\right)^p$$ and $$\left(\frac{2}{3}\right)^p$$ tend to $$0$$ as $$p \to \infty$$. Thus, $$\|x\|_p \to 3$$.
• That's the stuff! :) – Jazzmaniac Mar 3 '16 at 16:53
The $L_p$ norm is
$$d_p(\mathbf{x}, \mathbf{y}) \triangleq \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}}$$
there exists a positive value that is the maximum value:
$$M \triangleq \max_{1 \le i \le n} |x_i - y_i|$$
now, suppose you divide both sides of the $L_p$ norm definition by that positive value,
$$\frac{d_p(\mathbf{x}, \mathbf{y})}{M} = \frac{1}{M} \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}}$$
\begin{align} \frac{d_p(\mathbf{x}, \mathbf{y})}{M} & = \frac{1}{M} \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \frac{1}{M^p} \right)^{\frac{1}{p}} \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \frac{1}{M^p} \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \sum\limits_{i=1}^{n} \frac{1}{M^p} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \sum\limits_{i=1}^{n} \left( \frac{|x_i - y_i|}{M} \right)^p \right)^{\frac{1}{p}} \\ \end{align}
now ask yourself, what will happen to the right-hand side of this equation as $p \to \infty$? all terms, except for the term that is equal to the maximum, will have value of less than 1. they will go to 0 as $p \to \infty$, but the term that has $|x_i - y_i| = M$, that term is equal to 1 and will remain 1 even as $p \to \infty$.
• Robert, what if there are two terms in the sum that are equal to $M$? Shouldn't the sum be equal to 2 in that case? – MBaz Mar 3 '16 at 3:48
• hi Robert, thanks. It's a very good answer. So, Suppose, I have two matrix: x = [1,2,3]T and Y = 0. And, I want to compute, L1, L2 and L-infinity. So, here, Max value (M according to your explanation) would be 3? Right? Just to make sure. Thanks. – Odrisso Mar 3 '16 at 4:11
• This is a very nice and intuitive explaination. However, I suppose it is not 100% mathematically sound, which is why @MBaz asked the question with the factor of 2, if your vector had two entries corresponding to M. I think the problem stems from not obeying the limit from the start on. Getting the factor 1/M into the brackets of the sum is critical and probably the point where it fails when p goes to infinity. – M529 Mar 3 '16 at 9:14
• You could maybe argue that the sum s is bounded between 1 and n. Hence you have lim p->inf s^(1/p) which is 1. – M529 Mar 3 '16 at 9:25
• @David, it doesn't matter if it's trivial. If you make false assumptions about the nature of something and don't explicitly consider the case that these assumptions fail, your argument is wrong. Yes, mathematics is pedantic. That's why it's so useful. – Jazzmaniac Mar 4 '16 at 19:05 | 2020-08-15T04:58:05 | {
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https://mathhelpboards.com/threads/6-successive-numbers-no-one-is-prime.2005/ | # Number Theory6 Successive numbers no one is prime
#### Amer
##### Active member
is it possible to find a 6 Successive numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that one one is prime ?
Thanks
#### chisigma
##### Well-known member
is it possible to find a 6 Successive numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that one one is prime ?
Thanks
It is comfortable to verify that $x=5!=120$ satisfies the request... and the reason of that is easy to see...
Kind regards
$\chi$ $\sigma$
#### chisigma
##### Well-known member
Of course $x=5!$ is not the only and neither the 'smallest' solution. Setting $x=90$ You have 7 consecutive non prime numbers...
Kind regards
$\chi$ $\sigma$
#### chisigma
##### Well-known member
Why don't try to generalize the problem: given k, how to compute an n such that n, n+1, n+2,...,n+k are all non prime numbers?...
Kind regards
$\chi$ $\sigma$
#### chisigma
##### Well-known member
Why don't try to generalize the problem: given k, how to compute an n such that n, n+1, n+2,...,n+k are all non prime numbers?...
An easy way to get the result in the particular case where k is prime is based on the consideration that $k|n \implies k|(n+m k)$. Setting $n = 2 \cdot 3 \cdot 5 \cdot ... \cdot k$ we are sure that $n+2,n+3,...,n+k+1$ are all non prime numbers. For example...
$\displaystyle k=11 \implies n=2310 \implies 2312,2313,2314,2315,2316,2317,2318,2319,2320,2321,2322\ \text{are all non prime }$
Although 'easy' this method is often 'excessive' because the effective quantity consecutive non prime numbers can be greater. In the given example 2311 is prime so that the sequence starts at 2312 but 2323,2324,2325,2326,2327,2328,2329,2330,2331 and 2332 are non prime numbers [2333 is prime...] and the effective sequence's length is 20 [not 11]...
Kind regards
$\chi$ $\sigma$
#### Amer
##### Active member
Thanks very much, you are amazing. (f)
#### Amer
##### Active member
An easy way to get the result in the particular case where k is prime is based on the consideration that $k|n \implies k|(n+m k)$. Setting $n = 2 \cdot 3 \cdot 5 \cdot ... \cdot k$ we are sure that $n+2,n+3,...,n+k+1$ are all non prime numbers. For example...
$\displaystyle k=11 \implies n=2310 \implies 2312,2313,2314,2315,2316,2317,2318,2319,2320,2321,2322\ \text{are all non prime }$
Although 'easy' this method is often 'excessive' because the effective quantity consecutive non prime numbers can be greater. In the given example 2311 is prime so that the sequence starts at 2312 but 2323,2324,2325,2326,2327,2328,2329,2330,2331 and 2332 are non prime numbers [2333 is prime...] and the effective sequence's length is 20 [not 11]...
Kind regards
$\chi$ $\sigma$
nice one I get it
$$n +2 = 2.3.4...k +2 = 2(3.4...k +1 )$$ not prime
$$n+3 = 2.3.4...k + 3 = 3(2.4...k+1)$$ not prime
Thanks
#### soroban
##### Well-known member
Hello, Amer!
Is it possible to find a 6 consecutive integers numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that not one is prime? . Yes!
One solution is: .$$x \:=\:7!+2$$
. . $$\begin{array}{c}7!+2\text{ is divisible by 2} \\ 7!+3\text{ is divisible by 3} \\ 7!+4\text{ is divisible by 4} \\ 7!+5\text{ is divisible by 5} \\ 7!+6 \text{ is divisible by 6} \\ 7!+7\text{ is divisible by 7} \\ \end{array}$$
There is a simpler (and much longer) list:
. . $$\begin{array}{ccc} 114 &=& 2\cdot57 \\ 115 &=& 5\cdot 23 \\ 116 &=& 2\cdot 58 \\ 117 &=& 3\cdot39 \\ 118 &=& 2\cdot59 \\ 119 &=& 7\cdot17 \\ 120 &=& 2\cdot60 \\ 121 &=& 11\cdot11 \\ 122 &=& 2\cdot61 \\ 123 &=& 3\cdot41 \\ 124 &=& 2\cdot62 \\ 125 &=& 5\cdot25 \\ 126 &=& 2\cdot63 \end{array}$$
#### Amer
##### Active member
Hello, Amer!
One solution is: .$$x \:=\:7!+2$$
. . $$\begin{array}{c}7!+2\text{ is divisible by 2} \\ 7!+3\text{ is divisible by 3} \\ 7!+4\text{ is divisible by 4} \\ 7!+5\text{ is divisible by 5} \\ 7!+6 \text{ is divisible by 6} \\ 7!+7\text{ is divisible by 7} \\ \end{array}$$
There is a simpler (and much longer) list:
. . $$\begin{array}{ccc} 114 &=& 2\cdot57 \\ 115 &=& 5\cdot 23 \\ 116 &=& 2\cdot 58 \\ 117 &=& 3\cdot39 \\ 118 &=& 2\cdot59 \\ 119 &=& 7\cdot17 \\ 120 &=& 2\cdot60 \\ 121 &=& 11\cdot11 \\ 122 &=& 2\cdot61 \\ 123 &=& 3\cdot41 \\ 124 &=& 2\cdot62 \\ 125 &=& 5\cdot25 \\ 126 &=& 2\cdot63 \end{array}$$
Thanks, and Hello
we can make a list with k numbers which are not prime like that
$$k! + i$$ i=1,...,k | 2021-12-08T00:15:34 | {
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https://dsp.stackexchange.com/questions/25336/discrepancy-when-calculating-lti-system-output-using-inverse-z-transform | # Discrepancy when calculating LTI system output using inverse z-Transform
I'm given a difference equation, $y[n]-0.4y[n-1]=x[n]$, and asked to find the natural response $y_n[n]$, forced response $y_f[n]$ and complete response $y[n]$ if $x[n]=4 (0.25)^nu[n]$ and $y[0]=0$.
By one approach I've read, $$\text{characteristic equation:}\quad z-0.4=0 \quad\therefore y_n[n]=A(0.4)^n \\ \\ y_f[n]=B(0.25)^n \quad\text{substitute back into diffeq:}\\ B(0.25)^n -0.4B(0.25)^{n-1}=B(0.25)^n - \frac{0.4}{0.25}B(0.25)^n= 4(0.25)^nu[n]\\ \therefore B=-\frac{20}{3}, \quad y_f[n]=-\frac{20}{3}(0.25)^nu[n]\\ \therefore y[n] = A(0.4)^n-\frac{20}{3}(0.25)^nu[n]\\ y[0]=0 \rightarrow A=\frac{20}{3} \\ y[n]= \frac{20}{3}(0.4)^n-\frac{20}{3}(0.25)^nu[n]\\$$ Great! Alternatively, I should be able to determine $y[n]$ by taking the inverse z Transform of $Y(z)=H(z)X(z)$, which in this case is (I'm pretty sure) Y(z)= \frac{z}{z-0.4}\cdot 4 \frac{z}{z-0.25}\\ \begin{align} \frac{Y(z)}{z} &= \frac{4z}{(z-0.4)(z-0.25)}=\frac{C}{z-0.4}+\frac{D}{z-0.25}\\ &=\frac{32}{3} \frac{1}{z-0.4} - \frac{20}{3}\frac{1}{z-0.25}\\ \end{align}\\ \therefore Y(z)=\frac{1}{3} \left[\frac{32z}{z-0.4} - \frac{20z}{z-0.25}\right]\\ \therefore y[n]= \left[ \frac{32}{3}(0.4)^n - \frac{20}{3}(0.25)^n \right] u[n] which is almost the same, but not quite.
Where did I go wrong? Also, in the second approach, how do you take into account the initial condition of $y[0]=0$? And thirdly, in the first approach, is the first term of $y[n]$ multiplied by $u[n]$ or not? If so, how do you show this?
(for anyone wondering, this is not homework for a course, just self study...)
• Don't you mean $y[-1]=0$ instead of $y[0]=0$? – Matt L. Aug 20 '15 at 6:54
• The question says y[0]=0 but I guess its mistaken. But its not clear to me why y[0] can't be zero... – Westerley Aug 20 '15 at 17:02
The way to solve such problems is to use the unilateral $\mathcal{Z}$-transform, which allows you to take initial conditions into account. Note that the unilateral $\mathcal{Z}$-transform of $y[n-1]$ is
$$\mathcal{Z}\{y[n-1]\}(z)=z^{-1}Y(z)+y[-1]\tag{1}$$
where $Y(z)$ is the $\mathcal{Z}$-transform of $y[n]$. Using $(1)$ you can transform the given difference equation $y[n]-\frac25 y[n-1]=x[n]$, resulting in
$$Y(z)=\frac{X(z)}{1-\frac{2}{5}z^{-1}}+\frac{\frac{2}{5}\cdot y[-1]}{1-\frac{2}{5}z^{-1}}\tag{2}$$
With $X(z)$ given by
$$X(z)=\frac{4}{1-\frac14z^{-1}}\tag{3}$$
Eq. $(2)$ becomes (after partial fraction expansion)
$$Y(z)=\left(\frac{32}{3}+\frac{2y[-1]}{5}\right)\frac{1}{1-\frac{2}{5}z^{-1}}-\frac{20}{3}\frac{1}{1-\frac14z^{-1}}\tag{4}$$
From $(4)$, the resulting output sequence is
$$y[n]=\left(\frac{32}{3}+\frac{2y[-1]}{5}\right)\left(\frac25\right)^nu[n]-\frac{20}{3}\left(\frac14\right)^nu[n]\tag{5}$$
Eq. $(5)$ is the general form of the output sequence of the given system with the given input sequence. The value $y[-1]$ is the initial condition of the system, i.e. the state the system is in right before the input signal starts. With $y[-1]=0$ you obtain the same solution as you obtained using the inverse $\mathcal{Z}$-transform (neglecting any initial conditions by using $y[n-1]\Longleftrightarrow z^{-1}Y(z)$).
With the given constraint $y[0]=0$, you can obtain the required value of $y[-1]$ by considering the difference equation at $n=0$:
$$y[0]-\frac25 y[-1]=x[0]=4$$
which gives $y[-1]=-10$. Plugging that value into Eq. $(5)$ finally gives the result
$$y[n]=\frac{20}{3}\left(\frac25\right)^nu[n]-\frac{20}{3}\left(\frac14\right)^nu[n]\tag{6}$$
As Matt L pointed out, your initial condition is wrong. y[0] is NOT 0, it's 4. This is obvious from the difference equation. Once you use y[0] = 4, both results come out to be the same.
• As mentioned in my comment to @matt, the question does say y[0]=0 but it's likely wrong... but its not clear to me why y[0] CAN'T be zero... I see that x[0] is 4, but why can't the natural response at n=0 cancel out that? – Westerley Aug 20 '15 at 17:03
• Note that $y[0]$ can be zero by an appropriate choice of the initial condition $y[-1]$ (see my answer). – Matt L. Aug 21 '15 at 7:48 | 2019-06-25T14:58:46 | {
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https://math.stackexchange.com/questions/4484607/number-of-ways-to-divide-10-children-into-two-teams-of-5 | # Number of ways to divide $10$ children into two teams of $5$?
Consider the following two examples from A First Course in Probability by Sheldon Ross,
EXAMPLE 5b Ten children are to be divided into an A team and a B team of 5 each. The A team will play in one league and the B team in another. How many different divisions are possible?
Solution by the Book itself There are $$\frac{10!}{5!5!} = 252$$ possible divisions.
My Solution and Reasoning We first choose five children which can be done in $$C(10, 5)$$ possible ways. Then we decide which team(either A or B) to assign these children to, which can be done in $$2!$$ ways. Finally we assign the remaining five children to the remaining team, depending on the second level(if the first five children were assigned to A, then the remaining five should be assigned to B, and if the first five were assigned to B, the remaining to A). Therefore this level has only $$1$$ possible way, and hence the final answer is $$C(10, 5) \times 2! \times 1 = 252 \times 2 \times 1 = 504$$.
EXAMPLE 5c In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible?
Solution by the Book itself Note that this example is different from Example 5b because now the order of the two teams is irrelevant. That is, there is no A and B team, but just a division consisting of 2 groups of 5 each. Hence, the desired answer is
$$\frac{10!/(5!5!)}{2!} = 126$$
My Solution and Reasoning First we choose 5 children for one team which can be done is $$C(10, 5)$$ ways and then we put the remaining children in the other team, which is done in $$1$$ way. Thus the answer is $$C(10, 5) \times 1 = 252$$, which can also be obtained by dividing the answer to Example 5b by 2, or in other words, removing the order of the groups in Example 5b.
There seems to be a conceptual misunderstanding in my way of thinking, but what is it?
PARTICULAR PROBLEM
For your particular problem, for labeled teams
• label the children $$0\; through\;$$9
• choose $$5$$ for team $$A$$, say $$13579$$, team $$B$$ automatically $$02468$$
• This is different from $$02468$$ in Team $$A$$ and $$13579$$ for team $$B$$, so total $$\binom{10}5$$ choices
But if teams are unlabeled $$[13579-02468] \equiv [02468-13579]$$
hence division by $$2$$
GENERAL GUIDANCE
It is important to have a clear concept over combinatorics of teams depending on whether they are labeled or unlabeled
To take a more complex example for greater clarity, suppose you are to choose $$4$$ teams of $$3$$ each from $$12$$ players
• If the teams are labeled, eg Horses, Greyhounds , Panthers, etc
the answer is $$\binom{12}3\binom93\binom63\binom33$$
which can be variously written as $$\binom{12}{3,3,3,3}\; or\; \dfrac{12!}{3!3!3!3!}$$
• If the teams are unlabeled, we shall have to divide by $$4!$$, to remove permutations between identical teams
• An important point to note is that unlabeled teams in effect become labeled if sizes differ, or composition differs (eg boys' team or girls' team)
Note that in both casees your answer is doubled from the book's solution so you should question whether you really ought to multiply by $$2!=2$$.
Let's consider a simpler case of 4 children $$\{1,2,3,4\}$$ and the teams are of size 2. Then all possible options for Team A are $$\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}$$ giving a total of $$\frac{4!}{2!2!}=6$$ options. What you are doing is counting each option twice; Say when you count $$\{1,2\}$$, you count once for this being assigned to $$A$$ and once for $$B$$. However, note that the event of $$\{1,2\}$$ belonging to $$B$$ is exactly the event that $$\{3,4\}$$ belongs to $$A$$ which we already count. Hence we should not multiply by 2. | 2022-08-15T09:39:18 | {
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https://math.stackexchange.com/questions/2481536/how-to-evaluate-the-sum-sum-n-1-infty-frac12n12n2-left1-frac | # How to evaluate the sum $\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+…+\frac{1}{n}\right)$
How to evaluate the sum: $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$$ Can anyone help me,I really appreciate it.
• is there a source for this sum? Or, just a practice problem? – karakfa Oct 20 '17 at 15:05
• @MarkViola Mathematica gives the sum of the series. I don't know what is the trick to evaluate it. I think this is a good problem. :) – Ixion Oct 20 '17 at 15:26
• Partial fraction decomposition – JohnColtraneisJC Oct 20 '17 at 15:27
• Since $\displaystyle\frac {\pi^2}{12}=\sum_{r=1}^\infty \frac 1{2r^2}$, and $\displaystyle\ln 2=\sum_{r=1}^\infty \frac {(-1)^{r+1}}{r}$ is it possible to transform the given summation $\displaystyle \sum_{n=1}^\infty \sum_{r=1}^n \frac 1{(2n+1)(2n+2)r}$into $\displaystyle\sum_{r=1}^\infty \frac 1{2r^2}-\left(\sum_{r=1}^\infty \frac {(-1)^{r+1}}{r}\right)^2$? If so, then we are done. – hypergeometric Oct 21 '17 at 15:35
• I am voting for reopening this question since, despite a lack of efforts from the OP, it led to a number of collateral questions, and I think it is correct to leave this original question as a reference. – Jack D'Aurizio Oct 26 '17 at 15:57
A different view on the same problem: $$\sum_{n\geq 1}\frac{x^n}{n}=-\log(1-x),\qquad \sum_{n\geq 1}H_n x^n = \frac{-\log(1-x)}{1-x} \tag{A}$$ $$\sum_{n\geq 1}H_n x^{2n} = \frac{-\log(1-x^2)}{1-x^2}\tag{B}$$
$$\begin{eqnarray*}\sum_{n\geq 1}H_n\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right) &=& \int_{0}^{1}\frac{-\log(1-x^2)}{1+x}\,dx\\&=&-\tfrac{1}{2}\log^22+\int_{0}^{1}\frac{-\log(1-x)}{1+x}\,dx\\&=&-\tfrac{1}{2}\log^22+\int_{0}^{1}\frac{-\log(x)}{2-x}\,dx\tag{C}\end{eqnarray*}$$ and by differentiation under the integral sign, the last integral is related to the series $$\sum_{n\geq 1}\frac{1}{n^2 2^n}=\text{Li}_2\left(\tfrac{1}{2}\right)\stackrel{(*)}{=}\tfrac{\pi^2}{12}-\tfrac{\log^2 2}{2} \tag{D}$$ where $(*)$ follows from the dilogarithm reflection formula, proved here.
It is evident that $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right) = \frac{\pi^2}{12} - \ln^{2}2$$ and can be evaluated by following the pattern:
Consider the series $$S(x) = \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+2}}{(2n+1)(2n+2)}$$ which upon differentiation leads to $S(0) = 0$, $S'(0) = 0$, \begin{align} S(x) &= \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+2}}{(2n+1)(2n+2)} \\ S'(x) &= \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+1}}{(2n+1)} \\ S''(x) &= \sum_{n=1}^{\infty} H_{n} \, x^{2n} = - \frac{\ln(1- x^2)}{1-x^2}. \end{align} Now, $$- 2 \, S''(x) = \frac{\ln(1-x)}{1-x} + \frac{\ln(1-x)}{1+x} + \frac{\ln(1+x)}{1-x} + \frac{\ln(1+x)}{1+x}$$ which, upon integration, leads to \begin{align} - 4 \, S'(x) &= \ln^{2}(1 + x) - \ln^{2}(1-x) + 2 \, Li_{2}\left(\frac{1-x}{2}\right) - 2 \, Li_{2}\left(\frac{1+x}{2}\right) + \ln4 \, \ln\left(\frac{1+x}{1-x}\right). \end{align} Integrating again leads to $S(x)$. The integrals \begin{align} \int_{0}^{x} \ln^{2}(1-t) \, dt &= (x-1) \, (\ln^{2}(1-x) - 2 \ln(1-x) + 2) + 2 \\ \int_{0}^{x} \ln^{2}(1+t) \, dt &= (x+1) \, (\ln^{2}(1+x) - 2 \ln(1+x) + 2) - 2 \\ \int_{0}^{x} \ln\left(\frac{1+t}{1-t}\right) \, dt &= x \, \ln\left(\frac{1+x}{1-x}\right) + \ln(1-x^2) \\ \int_{0}^{x} Li_{2}\left(\frac{1+t}{2}\right) \, dt &= (1+x) \, Li_{2}\left(\frac{1+x}{2}\right) + x \, \ln\left(\frac{1-x}{2}\right) - \ln(1-x) -x - Li_{2}\left(\frac{1}{2}\right) \\ \int_{0}^{x} Li_{2}\left(\frac{1-t}{2}\right) \, dt &= (x-1) \, Li_{2}\left(\frac{1-x}{2}\right) + (x+1) \, \ln\left(\frac{1+x}{2}\right) -x - Li_{2}\left(\frac{1}{2}\right) + \ln2 \end{align} are needed for the evaluation. Once $S(x)$ is determined set $x=1$ to obtain $$S(1) = \sum_{n=1}^{\infty}\frac{H_{n}}{(2n+1)(2n+2)} = \frac{\pi^2}{12} - \ln^{2}2$$
Following Leucipus' answer, I will use different approach to evaluate $S(1)$. Note \begin{eqnarray} S=S(1)&=&\int_0^1\int_0^xS''(x)dxt\\ &=&-\int_0^1\int_0^x\frac{\ln(1-x^2)}{1-x^2}dxdt\\ &=&-\int_0^1\int_x^1\frac{\ln(1-x^2)}{1-x^2}dtdx\\ &=&-\int_0^1(1-x)\frac{\ln(1-x^2)}{1-x^2}dx\\ &=&-\int_0^1\frac{\ln(1-x^2)}{1+x}dx\\ &=&-\int_0^1\frac{\ln(1-x)}{1+x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx\\ &=:&-I_1-I_2 \end{eqnarray} Now $$I_2=\int_0^1\ln(1+x)d\ln(1+x)=\frac{1}{2}\ln^2(1+x)\bigg|_0^1=\frac12\ln^22.$$ For $I_1$, under $t=1-x$, one has \begin{eqnarray} \int\frac{\ln t}{2-t}dt&=&-\int_0^1\ln xd\ln(2-t)\\ &=&-\ln x\ln(2-t)+\int\frac{\ln(2-t)}{t}dt\\ &=&-\ln x\ln(2-t)+\int\frac{\ln2+\ln(1-\frac12t)}{t}dt\\ &=&-\ln x\ln(2-t)+\ln2\ln t+Li_2(\frac t2)+C \end{eqnarray} \begin{eqnarray} I_1&=&\int_0^1\frac{\ln t}{2-t}dt=-\int_0^1\ln xd\ln(2-t)\\ &=&-\ln x\ln(2-t)+\ln2\ln t+Li_2(\frac t2)\bigg|_0^1 &=&-\frac{\pi^2}{12}+\frac12\ln^22. \end{eqnarray} Thus $$S=\frac{\pi^2}{12}-\ln^22.$$ | 2019-06-16T02:52:52 | {
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https://math.stackexchange.com/questions/3552866/why-is-0-2-a-subgroup-of-mathbb-z-4 | # Why is $\{0, 2\}$ a subgroup of $\mathbb Z_4$?
I feel like this should be obvious but why is $$\{0, 2\}$$ a subgroup of $$\mathbb Z_4$$? So, $$\langle 2\rangle=\{0,2\}$$. Shouldn't this set contain the inverse ($$-2$$)? Or does it have to do with the fact that $$(-2)(-2)=4=0$$? Please advise.
• What is the difference between $2$ and "$-2$"? Recall, the minus sign merely means "the additive inverse of" which does not necessarily need to appear in other representations of the element. – JMoravitz Feb 19 '20 at 17:36
• Welcome to math SE. Have a look at mathjax for your mathematical expressions. – Alain Remillard Feb 19 '20 at 17:37
• As for "does it have to do with the fact that $(-2)(-2)=4=0$" No, it doesn't. It has to do with the fact that $2 + 2 = 0$. Addition is what is important here, not multiplication. You will find that the additive inverse of $0$ is $0$, the additive inverse of $1$ (which you might when convenient decide to notate as $-1$) is equal to $3$, the additive inverse of $2$ (which you might when convenient decide to notate as $-2$) is equal to $2$, and so on... – JMoravitz Feb 19 '20 at 17:37
• There's a problem with the title of the question. 2 is not a subgroup of $\mathbb Z_4$, it's an element. I guess you mean the subgroup generated by 2. – Shatabdi Sinha Feb 19 '20 at 17:40
• After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Shaun Feb 19 '20 at 18:46
$$\langle 2\rangle = \{0, 2\}$$ is a subgroup of the group $$\mathbb Z_4 = \{0, 1, 2, 3\}$$ under modular arithmetic, modulo $$4$$.
The identity of this group is $$0$$, and because $$2+2 \equiv 0 \pmod 4$$, it has order two, and hence $$2$$ generates a group (subgroup) of order 2. In fact, the additive inverse of $$2$$ is $$2$$.
That is, $$\langle 2 \rangle = \{0, 2\} \leq \{0, 1, 2, 3\} = \mathbb Z_4$$.
• Good approach to teach! – Mikasa Mar 20 '20 at 8:36
The elements of $$\Bbb Z_4$$ are not technically $$0$$, $$1$$, $$2$$ and $$3$$; rather, they are equivalence classes of integers with respect to the divisibility of their differences by $$4$$, like so: $$[a]_4:=\{b\in\Bbb Z: 4\mid a-b\}.$$ The operation of the group is defined by $$[x]_4+_4[y]_4=[x+y]_4$$.
Thus, since $$4\mid (-2)+2=0$$, we have $$[-2]_4=[2]_4$$.
• Why the downvote? – Shaun Feb 19 '20 at 19:00
• No, it isn't, @amWhy. – Shaun Mar 11 '20 at 16:58
• But $-2\equiv 2\pmod{4}$. – Shaun Mar 11 '20 at 17:02 | 2021-08-02T16:29:55 | {
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https://stats.stackexchange.com/questions/479117/what-are-the-chances-rolling-6-6-sided-dice-that-there-will-be-a-6 | # What are the chances rolling 6, 6-sided dice that there will be a 6?
More generally, what is the probability that n n-sided dice will turn up at least one side with the highest number ("n")?
My crude thinking is that each side has 1/6 probability, so that 6 of them (1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6) equals 100% of the time, but this obviously is not the case.
2 coins getting one T (say), has 3/4 of the time that at least one will be a T (for every 2 coin tosses). But this is 75% for a 2d2 dice throw, what is the general formula for NdN?
• Pretty sure we have a couple of questions already that cover this. Jul 26, 2020 at 22:59
• Does this answer your question? How often do you have to roll a 6-sided die to obtain every number at least once? Jul 27, 2020 at 19:07
• $m = n-1$, that is $P = 1 - \left(\frac{n-1}{n}\right)^n$ Jul 28, 2020 at 10:11
• @NathanChappell: Thanks! That's the cleanest solution yet! Jul 31, 2020 at 18:11
Probability of a die not turning up $$n$$ is $$1-1/n$$. Probability of not turning up $$n$$ in any of the $$n$$ dice is $$(1-1/n)^n$$. If you subtract this from $$1$$, it'll be probability of at least one $$n$$ turning up when you throw $$n$$ dice, i.e.
$$p=1-(1-1/n)^n\rightarrow 1-e^{-1}$$ as $$n$$ goes to $$\infty.$$
• In the case of trying to roll at least one 6 on 6 6-sided dice rolls (the title question), this gives us approximately a 66.5% chance. Jul 27, 2020 at 6:20
• Nathan Chappell has a simpler formula in the comment to the question above. Unfortunately, I don't have the time nor is it immediately apparent how to reduce your formula to his. Jul 31, 2020 at 18:10
• Inside of the parantheses is the same if you pay attention closely @Marcos $$\left(1-\frac{1}{n}\right)=\left(\frac{n-1}{n}\right)$$ Jul 31, 2020 at 21:50
• @gunes: Ah, yes, thanks. I came to the same conclusion after writing that comment -- an equivalency I hadn't seen before in algebra. Rather, I hadn't guessed to make 1 = n/n and then re-arrange terms. It's a common algebraic trick, but I'd never had the context to do it for myself. Aug 1, 2020 at 4:11
• How do you arrive at the conclusion in your second sentence? That is FAR OUT. I've not seen that in combinatorics. If you can explain it, I'll give you the credit. Nevermind... I think I see it now. I'll give you the credit... Aug 1, 2020 at 16:40
The event $$A:=$$ "at least one die turns up on side $$n$$" is the complement of the event $$B:=$$ "all dice turn up on non-$$n$$ sides".
So $$P(A)=1-P(B)$$. What's $$P(B)$$?
All dice are independent, so
$$P(\text{all n dice turn up on non-n sides}) = P(\text{a single die turns up non-n})^n = \bigg(\frac{n-1}{n}\bigg)^n.$$
So
$$P(A) = 1-\bigg(\frac{n-1}{n}\bigg)^n.$$
Trust but verify. I like to do so in R:
> nn <- 6
> n_sims <- 1e5
> sum(replicate(n_sims,any(sample(1:nn,nn,replace=TRUE)==nn)))/n_sims
[1] 0.66355
> 1-((nn-1)/nn)^nn
[1] 0.665102
Looks good. Try this with other values of nn. Here is a plot:
nn <- 2:100
plot(nn,1-((nn-1)/nn)^nn,type="o",pch=19,ylim=c(1-1/exp(1),1))
abline(h=1-1/exp(1),col="red")
We note how our probability in the limit is
$$P(A) = 1-\bigg(\frac{n-1}{n}\bigg)^n =1-\bigg(1-\frac{1}{n}\bigg)^n \to 1-\frac{1}{e}\approx 0.6321206 \quad\text{as }n\to\infty,$$
Answers by @StephanKolassa (+1) and @gunes (+1) are both fine. But this problem can be solved with reference to binomial and Poisson distributions as follows:
If $$X_n$$ is the number of ns seen in $$n$$ rolls of a fair $$n$$-sided die, then $$X_n \sim \mathsf{Binom}(n, 1/n),$$ so that $$P(X_n \ge 1) = 1 - P(X_n = 0)= 1-(1-1/n)^n.$$
As $$n\rightarrow\infty,$$ one has $$X_n \stackrel{prob}{\rightarrow} Y \sim\mathsf{Pois}(\lambda=1),$$ with $$P(Y \ge 1) = 1 - P(Y = 0) = 1 - e^{-1}.$$
The answer can be arrived at by purely counting the described events as well, although the accepted answer is more elegant. We'll consider the case of the die, and hopefully the generalization is obvious. We'll let the event space be all sequences of numbers from $$\{1,2,...,6\}$$ of length $$6$$. Here are a few examples (chosen at random):
3 2 3 5 6 1
1 1 2 5 2 4
1 2 1 1 6 3
4 4 3 3 4 2
6 1 1 6 3 4
6 3 5 4 5 1
The point is, our space has a total of $$6^6$$ events, and due to independence we suppose that any one of them is as probable as the other (uniformly distributed). We need to count how many sequences have at least one $$6$$ in them. We partition the space we are counting by how many $$6$$'s appear, so consider the case that exactly one $$6$$ appears. How many possible ways can this happen? The six may appear in any position (6 different positions), and when it does the other 5 postions can have any of 5 different symbols (from $$\{1,2,...,5\}$$). Then the total number of sequences with exactly one $$6$$ is: $$\binom{6}{1}5^5$$. Similarly for the case where there are exactly two $$6$$'s: we get that there are exactly $$\binom{6}{2}5^4$$ such sequences. Now it's time for fun with sums:
$$\sum_{k=1}^6 \binom{6}{k}5^{6-k} = \sum_{k=0}^6 \binom{6}{k}5^{6-k}1^k - 5^6 = (5+1)^6 - 5^6$$
To get a probability from this count, we divide by the total number of events:
$$\frac{6^6 - 5^6}{6^6} = 1 - (5/6)^6 = 1 - (1-1/6)^6$$
I think that this generalizes pretty well, since for any $$n$$ other than $$6$$, the exact same arguments holds, only replace each occurrence of $$6$$ with $$n$$, and $$5$$ with $$n-1$$.
It's also worth noting that this number $$5^6 = \binom{6}{0}5^6$$ is the contribution of sequences in which no $$6$$ occurs, and is much easier to calculate (as used in the accepted answer).
I found BruceET's answer interesting, relating to the number of events. An alternative way to approach this problem is to use the correspondence between waiting time and number of events. The use of that would be that the problem will be able to be generalized in some ways more easily.
### Viewing the problem as a waiting time problem
This correspondence, as for instance explained/used here and here, is
For the number of dice rolls $$m$$ and number of hits/events $$k$$ you get: $$\begin{array}{ccc} \overbrace{P(K \geq k| m)}^{\text{this is what you are looking for}} &=& \overbrace{P(M \leq m|k)}^{\text{we will express this instead}} \\ {\small\text{\mathbb{P} k or more events in m dice rolls}} &=& {\small\text{\mathbb{P} dice rolls below m given k events}} \end{array}$$
In words: the probability to get more than $$K \geq k$$ events (e.g. $$\geq 1$$ times rolling 6) within a number of dice rolls $$m$$ equals the probability to need $$m$$ or less dice rolls to get $$k$$ such events.
This approach relates many distributions.
Distribution of Distribution of
Waiting time between events number of events
Exponential Poisson
Erlang/Gamma over/under-dispersed Poisson
Geometric Binomial
Negative Binomial over/under-dispersed Binomial
So in our situation the waiting time is a geometric distribution. The probability that the number of dice rolls $$M$$ before you roll the first $$n$$ is less than or equal to $$m$$ (and given a probability to roll $$n$$ equals $$1/n$$) is the following CDF for the geometric distribution:
$$P(M \leq m) = 1-\left(1-\frac{1}{n}\right)^m$$
and we are looking for the situation $$m=n$$ so you get:
$$P(\text{there will be a n rolled within n rolls}) = P(M \leq n) = 1-\left(1-\frac{1}{n}\right)^n$$
### Generalizations, when $$n \to \infty$$
The first generalization is that for $$n \to \infty$$ the distribution of the number of events becomes Poisson with factor $$\lambda$$ and the waiting time becomes an exponential distribution with factor $$\lambda$$. So the waiting time for rolling an event in the Poisson dice rolling process becomes $$(1-e^{-\lambda \times t})$$ and with $$t=1$$ we get the same $$\approx 0.632$$ result as the other answers. This generalization is not yet so special as it only reproduces the other results, but for the next one I do not see so directly how the generalization could work without thinking about waiting times.
### Generalizations, when dices are not fair
You might consider the situation where the dice are not fair. For instance one time you will roll with a dice that has 0.17 probability to roll a 6, and another time you roll a dice that has 0.16 probability to roll a 6. This will mean that the 6's get more clustered around the dice with positive bias, and that the probability to roll a 6 in 6 turns will be less than the $$1-1/e$$ figure. (it means that based on the average probability of a single roll, say you determined it from a sample of many rolls, you can not determine the probability in many rolls with the same dice, because you need to take into account the correlation of the dice)
So say a dice does not have a constant probability $$p = 1/n$$, but instead it is drawn from a beta distribution with a mean $$\bar{p} = 1/n$$ and some shape parameter $$\nu$$
$$p \sim Beta \left( \alpha = \nu \frac{1}{n}, \beta = \nu \frac{n-1}{n} \right)$$
Then the number of events for a particular dice being rolled $$n$$ time will be beta binomial distributed. And the probability for 1 or more events will be:
$$P(k \geq 1) = 1 - \frac{B(\alpha, n + \beta)}{B(\alpha, \beta)} = 1 - \frac{B(\nu \frac{1}{n}, n +\nu \frac{n-1}{n})}{B(\nu \frac{1}{n}, n +\nu \frac{n-1}{n})}$$
I can verify computationally that this works...
### compute outcome for rolling a n-sided dice n times
rolldice <- function(n,nu) {
p <- rbeta(1,nu*1/n,nu*(n-1)/n)
k <- rbinom(1,n,p)
out <- (k>0)
out
}
### compute the average for a sample of dice
meandice <- function(n,nu,reps = 10^4) {
sum(replicate(reps,rolldice(n,nu)))/reps
}
meandice <- Vectorize((meandice))
### simulate and compute for variance n
set.seed(1)
n <- 6
nu <- 10^seq(-1,3,0.1)
y <- meandice(n,nu)
plot(nu,1-beta(nu*1/n,n+nu*(n-1)/n)/beta(nu*1/n,nu*(n-1)/n), log = "x", xlab = expression(nu), ylab = "fraction of dices",
main ="comparing simulation (dots) \n with formula based on beta (line)", main.cex = 1, type = "l")
points(nu,y, lty =1, pch = 21, col = "black", bg = "white")
.... But I have no good way to analytically solve the expression for $$n \to \infty$$.
With the waiting time However, with waiting times, then I can express the the limit of the beta binomial distribution (which is now more like a beta Poisson distribution) with a variance in the exponential factor of the waiting times.
So instead of $$1-e^{-1}$$ we are looking for $$1- \int e^{-\lambda} p(\lambda) \, \text{d}\, \lambda$$.
Now that integral term is related to the moment generating function (with $$t=-1$$). So if $$\lambda$$ is normal distributed with $$\mu = 1$$ and variance $$\sigma^2$$ then we should use:
$$1-e^{-(1-\sigma^2/2)} \quad \text{instead of} \quad 1-e^{-1}$$
### Application
These dice rolls are a toy model. Many real-life problems will have variation and not completely fair dice situations.
For instance, say you wish to study probability that a person might get sick from a virus given some contact time. One could base calculations for this based on some experiments that verify the probability of a transmission (e.g. either some theoretical work, or some lab experiments measuring/determining the number/frequency of transmissions in an entire population over a short duration), and then extrapolate this transmission to an entire month. Say, you find that the transmission is 1 transmission per month per person, then you could conclude that $$1-1/e \approx 0.63 \%$$ of the population will get sick. However, this might be an overestimation because not everybody might get sick/transmission with the same rate. The percentage will probably lower.
However, this is only true if the variance is very large. For this the distribution of $$\lambda$$ must be very skewed. Because, although we expressed it as a normal distribution before, negative values are not possible and distributions without negative distributions will typically not have large ratios $$\sigma/\mu$$, unless they are highly skewed. A situation with high skew is modeled below:
Now we use the MGF for a Bernoulli distribution (the exponent of it), because we modeled the distribution as either $$\lambda = 0$$ with probability $$1-p$$ or $$\lambda = 1/p$$ with probability $$p$$.
set.seed(1)
rate = 1
time = 1
CV = 1
### compute outcome for getting sick with variable rate
getsick <- function(rate,CV=0.1,time=1) {
### truncating changes sd and mean but not co much if CV is small
p <- 1/(CV^2+1)
lambda <- rbinom(1,1,p)/(p)*rate
k <- rpois(1,lambda*time)
out <- (k>0)
out
}
CV <- seq(0,2,0.1)
plot(-1,-1, xlim = c(0,2), ylim = c(0,1), xlab = "coefficient of variance", ylab = "fraction",
cex.main = 1, main = "if rates are bernouilli distributed \n fraction p with lambda/p and 1-p with 0")
for (cv in CV) {
points(cv,sum(replicate(10^4,getsick(rate=1,cv, time = 1)))/10^4)
}
p <- 1/(CV^2+1)
lines(CV,1-(1-p)-p*exp(-1/p),col=1)
lines(CV,p, col = 2, lty = 2)
legend(2,1, c("simulation", "computed", "percent of subsceptible population"),
col = c(1,1,2), lty = c(NA,1,2), pch = c(1,NA,NA),xjust =1, cex = 0.7)
The consequence is. Say you have high $$n$$ and have no possibilities to observe $$n$$ dice rolls (e.g. it takes to long), and instead you screen the number of $$n$$ rolls only for a short time for many different dice. Then you could compute the number of dices that did roll a number $$n$$ during this short time and based on that compute what would happen for $$n$$ rolls. But you would not be knowing how much the events correlate within the dice. It could be that you are dealing with a high probability in a small group of dice, instead of an evenly distributed probability among all dice.
This 'error' (or you could say simplification) relates to the situation with COVID-19 where the idea goes around that we need 60% of the people immune in order to reach herd immunity. However, that may not be the case. The current infection rate is determined for only a small group of people, it can be that this is only an indication for the infectiousness among a small group of people.
• I must admit that my last case is actually very trivial and does not this larger framework and tricks with moment generating functions, it can be done easier by just rescaling the population. Jul 31, 2020 at 16:58
• ...Except, since the number of rolls is finite, one can enumerate all possible outcomes. This is n^n for an n-sided die. Once one has done this it's a matter of counting which rolls have a #n. Jul 31, 2020 at 17:05
• @Marcos my approach is for people that do not like counting (e.g. imagine $n$ would be equal to a million or larger, it is still finite but good luck counting). And also, how are you gonna count if the dices are not fair but instead follow some continuous distribution? -------- That said, I agree your specific question can be answered easily. The method here is just an additional view that can be useful for more complex problems. Jul 31, 2020 at 17:20
• True, it requries a "meta-numeric" formula for counting the digits of numbers (as symbols rather than quantities) themselves. Jul 31, 2020 at 17:28
• @Marcos I am not sure what you mean by counting now. I consider this problem not as a counting problem when $n$ get's large. Well, you can go count things and maybe create expressions that summarize the counting, but you may not always get (easily) to a solution. Flipping the point of view from counting the probabilities of a given number of events for a given number of rolls, to a given number of rolls (waiting time) for a given number of events, can be a useful tool to greatly simplify the problem (I agree for this specific problem, which is not so difficult it is not so much necessary) Jul 31, 2020 at 17:42
Simplify and then extend. Start with a coin. A coin is a die with 2 sides (S=2).
The exhaustive probability space is
T | H
Two possibilities. One satisfies the condition of all heads. So your odds of all heads with one coin (n=1) are 1/2.
So try two coins (n=2). All outcomes:
TT | TH | HT | HH
Four possibilities. Only one matches your criteria. It is worth noting that the probability of one being heads and the other being tails is 2/4 because two possibilities of the four match your criteria. But there is only one way to get all heads.
TTT | THT | HTT | HHT
TTH | THH | HTH | HHH
8 possibilities. Only one fits the criteria - so 1/8 chances of all heads.
The pattern is (1/S)^n or (1/2)^3.
For dice S = 6, and we have 6 of them.
Probability of getting a 6 on any given roll is 1/6. Rolls are independent events. So using 2 dice getting all 6's is (1/6)*(1/6) or 1/36.
(1/6)^6 is about 1 in 46,656
• I was actually just looking for the probability of ANY heads showing up, given as many rolls as there are faces on the die. But, cheers mate, I got my answer. Jul 28, 2020 at 0:22 | 2022-05-26T06:10:32 | {
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https://math.stackexchange.com/questions/352192/show-that-x4-1-is-reducible-over-mathbbz-11x-and-splits-over-math/352200 | # Show that $x^4 +1$ is reducible over $\mathbb{Z}_{11}[x]$ and splits over $\mathbb{Z}_{17}[x]$.
Reduction into linear factors $\mathbb{Z}_{17}[x]$:
This part is not too hard: $x^4 \equiv -1$ mod 17 has solutions: 2, 8, 9, 15 so
$(x-2)(x-8)(x-9)(x-15) = x^4 -34 x^3 +391 x^2-1734 x+2160 \equiv x^4+1$ mod 17.
Reduction over $\mathbb{Z}_{11}[x]$:
This one doesn't have such an easy solution, as neither of $y^2 \equiv -1$ mod 11 or $x^4 \equiv -1$ mod 11 have solutions.
I've tried $x^4+1 = (x^2- \sqrt{2}x+1)(x^2+\sqrt{2}x+1)$ but $x^2 \equiv 2$ mod 11 also has no solutions so there's no easy substitution here.
I think that this approach is not going to work here, so I need something new. Any suggestions?
• See this question for a more or less complete answer. – Jyrki Lahtonen Apr 5 '13 at 15:05
• It certainly has a very complete answer - but perhaps somewhat overkill when it comes to answering this question? Nonetheless thanks for the link. – Lewy Apr 5 '13 at 15:19
• You were almost there... You tried $X^4+1=X^4+1+2X^2-2X^2$. Now, if you observe that $2=-9$ you are done.... – N. S. Apr 5 '13 at 15:45
• The splitting modulo $p=17$ comes from the fact that $p \equiv 1 \pmod 8$, so that $p$ is totally split in the extension $\Bbb Q(\zeta_8) / \Bbb Q$, which is the splitting field of $x^4+1$. – Watson Apr 6 '18 at 13:30
I will assume you've made no mistakes in your work.
Since $-1$ hasn't a fourth root in $\Bbb Z_{11},$ then the only way to reduce $x^4+1$ over $\Bbb Z_{11}$ is as a product if quadratics. We may as well assume that the quadratics are monic, so we've got $$x^4+1=(x^2+ax+b)(x^2+cx+d),$$ leaving us with the following system: $$a+c=0\\ac+b+d=0\\ad+bc=0\\bd=1$$ By the first equation, $c=-a,$ and the rest of the system becomes: $$b+d=a^2\\a(d-b)=0\\bd=1$$ Now, if $a=0$, then the first and third remaining equations imply that $-1$ has a square root in $\Bbb Z_{11}$ (why?), which you've ruled out. Hence, we need $b=d,$ and the rest of the system becomes: $$2b=a^2\\b^2=1$$ Can you take it from here?
• Yes, thank you for your clear explanation. This gives two possible solutions: $(x^2-3x -1)(x^2 +3x-1)$ and $(x^2-8x -1)(x^2 +8x-1)$. – Lewy Apr 5 '13 at 15:27
• @Lewy: You do know that factorization in this ring is unique, don't ya? IOW, look more closely at the "two" factorizations :-) – Jyrki Lahtonen Apr 5 '13 at 15:31
• @Lewy: Jyrki's point is well-taken. $8=-3,$ so those factorizations are identical (up to arrangement of the factors). – Cameron Buie Apr 5 '13 at 15:51
• Yes, I see that. Perhaps this is one of those times at which it's appropriate to say "Doh!". – Lewy Apr 5 '13 at 15:55
An abstract algebra argument, without much computation.
Claim: The polynomial $x^4+1$ splits over $\mathbb F_{p^2}$ for any odd (*) prime $p$.
We can see this because $F_{p^2}^\times$ is a cyclic group of order $p^2-1$, a multiple of $8$. If $g$ is a generator, then $g^{(1+2k)\frac{p^2-1}{8}}$ gives distinct roots for $k=0,1,2,3$.
This means that $x^4+1$ can't be irreducible in $\mathbb Z_p$, because if it was, the splitting field would have $p^4$ elements.
(*) True for $p=2$, but not for the same reasons. When $p=2$, $x^4+1=(x+1)^4$.
Specific solutions:
We can make this explicit by looking at the "standard" complex $4$th roots of $-1$: $$\pm\frac{\sqrt 2}2 \pm\frac{\sqrt{2}}{2}i=\frac{1}{2}(\pm \sqrt{2}\pm\sqrt{-2})$$
If $p\equiv 1\pmod 8$ then $2$ and $-2$ have square roots in $\mathbb Z_p$, so $x^4+1$ has four roots.
If $p\equiv -1\pmod 8$ then $2$ is a square, but $-2$ is not. Write $a^2\equiv 2\pmod p$. Then write:
$$x^4+1 = \left(x^2 -ax + 1\right)\left(x^2+ax+1\right)$$
If $p\equiv 3\pmod 8$ then $-2$ is a square and $2$ is not. Letting $b^2\equiv -2\pmod p$, we get:
$$x^4+1 = \left(x^2-bx-1\right)\left(x^2+bx-1\right)$$
Finally, if $p\equiv 5\pmod 8$ then neither of $\pm 2$ is a square, but $-1$ is a square. Letting $c^2\equiv -1\pmod p$, we see that the roots are $(\pm 1 \pm c)\sqrt{2}/2$ and the result is:
$$x^4+1 = (x^2-c)(x^2+c)$$
• The use of the formula $(\pm\sqrt2\pm\sqrt{-2})/2$ is a nice trick. In the linked answer I (IIRC at least partly in response to the discussion in the comments that lead to CWification) did get there, but should have seen it right away. +1 for unveiling the reason for the emergence of $\sqrt{\pm2}$ to this extent. – Jyrki Lahtonen Apr 5 '13 at 17:42
• Perhaps a more direct proof that $x^4+1$ is not irreducible is just to note that $x^4+1|x^8-1|x^{p^2-1}-1|x^{p^2}-x$. And $x^{p^2}-x$ has as its only irreducible factors the quadratic and linear prime polynomials. – Thomas Andrews Apr 5 '13 at 18:04
• Well, that's exactly how I did it originally. But somewhat surprisingly people don't think this is a duplicate of that :-) – Jyrki Lahtonen Apr 5 '13 at 18:07
• @JyrkiLahtonen N.S. suggested in comments to the question: $x^4+1=x^4+1 +2x^2-2x^2$. Hahaha, that makes the primacy of $\sqrt{2}$ and $\sqrt{-2}$ even more obvious! – Thomas Andrews Apr 5 '13 at 18:16
This question in fact already answered your question. Indeed, in the accepted answer, the complete solution can be found. So let me explain this and put it into CW.
As expounded in the link, we know that there is a primitive root of order $8$ in $\mathbb F:=\mathbb F_{p^2}$, called $u$. Since $\mathbb F$ is of degree $2$ over $\mathbb F_p$, the minimal polynomial is of degree at most $2$ over $\mathbb F_p$. Here, our $p\equiv 3\pmod 8$, so there is no $u$ in $\mathbb F_p$, as $u^4=-1$. Its conjugate is then $u^p=u^3$. Therefore $x^4+1$ has a factor given by $(x-u)(x-u^3)$. After easy calculations, we find that this becomes $(x^2-(u+u^3)x-1)$. And the other factor of that one must, after easy calculations of other conjugates of $u$, be $x^2+(u+u^3)x-1$. Therefore, $a:=u+u^3$ satisfies $a^2+2=0$. And I guess you know there is one such $a$ for $p=11$.
For those who jumped here:
We found the factorisation: $$(x^2-3x-1)(x^2+3x-1)$$.
Thanks for the attention, and inform me of any errors. Thanks.
• Also see that referred answer for more detail. – awllower Apr 5 '13 at 15:21
Look for a factorization of the shape $(x^2-ax+b)(x^2+ax+b)$. So we want $b^2\equiv 1\pmod{11}$, with $2b$ a quadratic residue.
Remark: Note that this is a general procedure for $x^4+1$ modulo an odd prime. The issue is whether one of $2$ or $-2$ is a quadratic residue of $p$. One of them is unless $p\equiv 5\pmod{8}$.
• In the case $p\equiv 5\pmod8$ we have that $2$ is a quadratic non-residue, so $i:=2^{(p-1)/4}$ is a square root of $-1$. In that case we have the factorization $(x^2+i)(x^2-i)$. – Jyrki Lahtonen Apr 5 '13 at 15:35
• In general, at least one of $a,b,ab$ is a square $\pmod p$. Take $a=2, b=-1.$ :) – Thomas Andrews Apr 5 '13 at 16:12
• Note, the standard formulation for the complex roots of $x^4+1=0$ can be formulated as $\frac{1}{2}(\pm\sqrt{2}\pm\sqrt{-2})$, which is why these two roots are important. – Thomas Andrews Apr 5 '13 at 16:14
Hint: We know that if $f(x)=x^4+1$ is reducible, then it either has a root or can be written $f(x)=g(x)h(x)$ where $\deg(g)=\deg(h)=2$.
• I guess OP already knew this, judged from the try. The point is: how to find those degree $2$ factors? – awllower Apr 5 '13 at 15:03
You may use the fact that $$x^2 = 9 \mod 11$$ has a solution, say $x = c$. Then use $$(x^2 + cx - 1)(x^2 - cx - 1),$$ and $c$ is pretty much obvious from here... | 2019-08-26T09:57:34 | {
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https://math.stackexchange.com/questions/2892467/summation-problem-fx-1-sum-n-1-infty-fracxnn | # Summation problem: $f(x)=1+\sum_{n=1}^{\infty}\frac{x^n}{n}$
I want to evaluate this summation:
$$S=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...+...$$
where, $|x|<1$
Here it is my approach
$$S=1+\sum_{n=1}^{\infty}\frac{x^n}{n}=f(x)$$
$$f'(x)=1+x+x^2+x^3+...+...=\frac{1}{1-x}$$
$$f(x)=\int f'(x)dx=\int \frac{1}{1-x} dx=-\ln(1-x)+C$$
$$f(0)=1 \Longrightarrow C=1$$
$$S=1-\ln(1-x)$$
And here is my problem:
I calculated this sum for only $|x|<1$.
Then, I checked in Wolfram Alpha and I saw that, this sum $f(x)$ converges for $x=-1$. But, this creating a contradiction with my solution. Because, the series $$f'(x)=1+x+x^2+x^3+...+...=\frac{1}{1-x}$$ for $x=-1$ doesn't converge, diverges. Therefore, there is a problem in my solution. But I can not find the source of the problem.
How can I prove the formula $f(x)=1-\ln(1-x)$ is also correct at the point $x=-1$ ?
Thank you very much.
• There's no problem with your solution. There's a problem with you assuming that if $\sum_n \frac{x^n}{n}$ converges, then $\sum_n x^{n-1}$ must converge. – mathworker21 Aug 23 '18 at 19:08
• Your approach worked for all $x$ you were asked to solve it for. WolframAlpha is telling you that the series also has a value for some $x$ outside that region, although your approach doesn't work in that case. Why does that matter? – Arthur Aug 23 '18 at 19:13
• @mathworker21 How can I prove the formula $f(x)=1-\ln(1-x)$ is also correct at the point $x=-1$ ? – Elvin Aug 23 '18 at 19:28
• @mathworker21 as I understand it, you mean my solution is incorrect $f(-1)-1=\ln 2$ .Am I right? – Elvin Aug 23 '18 at 20:34
• @student It's incorrect at $x = -1$. You may only deduce that $f'(x) = 1+x+x^2+\dots$ when $|x| < 1$. – mathworker21 Aug 23 '18 at 20:45
I don't know how familiar your are with power series, but this has to do with the notion of radius of convergence.
$\sum \left(\frac{x^n}{n}\right)_{n\in\mathbb{N^*}}$ is a power series with a radius of convergence $R=1$ (which can be proved by the ratio test), so indeed, it does converge for $|x|<1$, and does diverge for $|x|>1$.
As for $x=1$, the only case with $x=-1$ were we can't conclude right away, it could naively be anything, but it happens to converge.
The power series $\sum \left(x^n\right)_{n\in\mathbb{N^*}}$ is indeed the derivative of the previous one : we know it has the same radius of convergence, so for $x=1$, here again it could be anything ; and it happens to diverge.
The convergence circle is the only place were the convergence of a power series and its derivatives aren't always equivalent : it could be anything, and you can't deduce the convergence of the derivative from the convergence of the power series there.
As to show that the power series and $f(x)=1-\ln(1-x)$ are equal also for $x=-1$ (that is to say, to prove that $f(-1)$ is the sum of the series) :
$\cdot$ The power series and f both exist for $x$ in $[-1,1)$ (for the power series, you can prove it with the ratio test)
$\cdot$ They are both continuous on $[-1,1)$ : it is obvious for $f$, and for the power series, we know that the sum of the power series is continious strictly inside the convergence disk $(-1,1)$. We extend the continuity to -1 by the uniform convergence on $[-1,0]$.
$\cdot$ Are equal to one another on $(-1,1)$ which is dense in $[-1,1]$.
We can conclude that the power series and f are equal also at $x=1$, so $f(-1)$ is indeed the sum of the series.
If you are not familiar with this density theorem :
Since the power series is continuous on -1, by definition, $\lim\limits_{x\rightarrow\ -1}\sum\limits_{n=1}^{\infty}\frac{x^n}{n}=\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n}$, the limits being taken by approaching from the right ($x>-1$).
But, since the power series and f are equal for $x>-1$, we can put $f$ instead of the power series inside the limit :
$$\lim\limits_{x\rightarrow\ -1}f(x)=\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n}$$
Or $f$ is continuous on -1, so $\lim\limits_{x\rightarrow\ -1}f(x)=f(-1)$
Hence $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n}=f(-1)$.
• But we need more. We have a formula $f(x) = 1-\log(1-x)$ on $(-1,1)$. The series converges at $-1$, and the formula (extends to something that) is continuous at $-1$. Conclude that $f(-1)$ is the sum of the series at $-1$. – GEdgar Aug 23 '18 at 19:58
• The power series and f both exist on [-1,1) (ratio test), are both continuous on [-1,1) (for the power series, we extend the continuity to -1 by the uniform convergence on [-1,0]), and are equal to one another on (-1,1) which is dense in [-1,1]. We can conclude that the power series and f are equal also at x=1. – Harmonic Sun Aug 23 '18 at 20:02
• +1 for a complete proof. – GEdgar Aug 23 '18 at 21:28
By the ratio test, $$\lim_{n\to\infty } \left|\frac{x^{n+1}}{n+1}\frac{n}{x^n} \right|=\lim_{n\to\infty}|x|\frac{n}{n+1}<1$$ whenever $|x|<1$. Thus we have convergence on $(-1,1)$. We need to check the endpoints separately. For $x=1$, we have the diverging harmonic series. For $x=-1$, we have the alternating harmonic series, which converges.
A power series $\sum_{n=0}^\infty a_n x^n$ (I am assuming the series is centered at zero for definiteness) converges either only at $0$, or on some interval $I$ centered at zero. The half-length of $I$ is called the radius of convergence $R$, and can be computed generally as $\frac{1}{R}=\limsup_{n \to \infty} a_n^{1/n}$ (understood as $R=\infty$ if $1/R$ is zero in the above formula).
A power series with a given radius of convergence may or may not converge at the endpoints $\pm R$. $\sum_{n=1}^\infty \frac{x^n}{n}$ turns out to have radius of convergence $1$ and converges at $-1$ but does not converge at $1$. In the interior of its interval of convergence, a power series may be differentiated term by term; thus in your example you can deduce the derivative of $-\ln(1-x)$ is $\frac{1}{1-x}$ for $|x|<1$ by differentiating its series term by term. In general differentiation may change how the series behaves at its endpoints, as you found here.
It's well-known that $$\ln(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}+\cdots,~~~\forall y \in (-1,1].\tag1$$
Thus, let $y=-x.$ we have $$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+\cdots,~~~\forall x \in [-1,1).\tag2$$
As result, $$S=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots=1-\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+\cdots\right)=1-\ln(1-x).$$
Notice that, $(1)$ demands $-1<y\leq 1$. then in $(2)$, we demand $-1<-x\leq 1,$ namely, $-1\leq x<1.$
It is best to use elementary arguments for such well known power series rather than start integrating power series. While power series methods work fine in the interior of region of convergence, the behavior at boundary needs additional analysis (for example Abel's theorem).
Below I prove the formula $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots\tag{1}$$ for all values of $$x$$ satisfying $$-1. The series in question is obtained by replacing $$x$$ with $$-x$$.
Let's start with the algebraic identity $$\frac{1}{1+t}=1-t+t^2-\dots+(-1)^{n-1}t^{n-1}+(-1)^{n}\frac{t^{n}}{1+t}\tag{2}$$ which holds for all $$t\neq -1$$ and all positive integers $$n$$. Integrating this identity with respect to $$x$$ over interval $$[0,x]$$ where $$x>-1$$ we get $$\log(1+x)=x-\frac{x^2}{2}+\dots+(-1)^{n-1}\frac{x^n}{n}+(-1)^nR_n\tag{3}$$ where $$R_n=\int_{0}^{x}\frac{t^n}{1+t}\,dt$$ Our job is done if we can show that $$R_n\to 0$$ as $$n\to\infty$$ for all $$x\in(-1,1]$$. If $$x=0$$ then $$R_n=0$$. If $$0 then we can see that $$0 so that $$R_n\to 0$$. If $$x=-y$$ and $$y\in(0,1)$$ then $$R_n=(-1)^{n+1}\int_{0}^{y}\frac{t^n}{1-t}\,dt$$ The integral above is clearly less than or equal to $$\frac{1}{1-y}\int_{0}^{y}t^n\,dt=\frac{y^{n+1}}{(1-y)(n+1)}$$ which tends to $$0$$ as $$n\to\infty$$ and hence $$R_n\to 0$$. It is thus established that $$R_n\to 0$$ as $$n\to\infty$$ for all $$x\in(-1,1]$$. Taking limits as $$n\to\infty$$ in equation $$(3)$$ gives us the series $$(1)$$ which is valid for $$x\in(-1,1]$$. | 2019-07-20T16:06:13 | {
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https://math.stackexchange.com/questions/1166798/inequality-from-chapter-5-of-the-book-how-to-think-like-a-mathematician?noredirect=1 | # Inequality from Chapter 5 of the book *How to Think Like a Mathematician*
This is from the book How to think like a Mathematician,
How can I prove the inequality $$\sqrt[\large 7]{7!} < \sqrt[\large 8]{8!}$$
without complicated calculus? I tried and finally obtained just $$\frac 17 \cdot \ln(7!) < \frac 18 \cdot \ln(8!)$$
• Are we allowed complex roots and negative roots to prove the contrary ;-) – Philip Oakley Feb 28 '15 at 19:22
• I don't think that it is in the proposition – Gwydyon Mar 2 '15 at 11:24
• Can these simpler approaches to a specific example be used to produce a simpler answer to the "general case"? – Mark Hurd Mar 3 '15 at 3:02
• I don't know, the problem was "as is" – Gwydyon Mar 3 '15 at 16:35
• – Martin Sleziak Dec 11 '16 at 11:36
Your inequality is equivalent to $$(7!)^8 < (8!)^7$$ divide it by $(7!)^7$, and get $$7! < 8^7$$ and this is clear, since $$1 \cdots 7 < 8 \cdots 8$$
• In the spirit of the question title, I'd note that the first line of this answer is key. Maths does of course love cleverness, but also thrives on knowing to do really simply things such as taking powers of both sides here to remove those ugly roots. – Keith Feb 27 '15 at 3:02
• You are right, great ! I would have of to think there. – Gwydyon Feb 28 '15 at 17:39
• Elegant! really clever! – Max Payne May 27 '15 at 16:19
Think of
$${\ln(7!)\over7}={\ln(1)+\cdots+\ln(7)\over7}$$
as the average of seven numbers and
$${\ln(8!)\over8}={\ln(1)+\cdots+\ln(8)\over8}$$
as the average when an eighth number is added. Since the new number is larger than the previous seven, the average must also be larger. (E.g., if you get a better score on your final than on any of your midterms, your grade should go up, not down.)
• You are right, but it is clear to me only when i compute ln(8) – Gwydyon Feb 28 '15 at 18:08
• @Gwydyon, the logarithm is an increasing function, so $\ln(8)$ is larger than its predecessors. – Barry Cipra Feb 28 '15 at 20:00
• Yes I know that. – Gwydyon Mar 2 '15 at 11:21
• @Gwydyon, I guess I don't understand your previous comment then. – Barry Cipra Mar 2 '15 at 15:22
• Because x/8 is lower than y/7, if x=y, that is not the case but ln(7!) and ln((7+1)!) are near close, for x>=1 – Gwydyon Mar 3 '15 at 16:41
Note that $$\sqrt[7]{7!} < \sqrt[8]{8!} \iff\\ (7!)^8 < (8!)^7 \iff\\ 7! < \frac{(8!)^7}{(7!)^7} \iff\\ 7! < 8^7$$ You should find that the proof of this last line is fairly straightforward.
• You are right but someone has writen the same statements – Gwydyon Feb 28 '15 at 18:11
$8\ln (7!) < 7\ln (8!) \Rightarrow \ln (7!) < 7\ln 8 \iff \ln 1 + \ln 2 +\cdots \ln 7 < 7\ln 8$ which is clear.
• great! you achieved my second statement – Gwydyon Feb 28 '15 at 18:53
You have already turned the comparison of two geometric means into the comparison of two arithmetic means. So consider a more general comparison: show that appending a larger number always raises the geometric mean of a list of positive numbers by showing the effect on the arithmetic mean. Suppose the $x_i$ are real and $x_{n+1}$ is strictly largest. \begin{equation*} \begin{split} (1/(n+1)) \sum_{i=1}^{n+1} x_i &= (1/(n+1)) (x_{n+1} + \sum_{i=1}^{n} x_i) \\ &=(1/(n+1) (n x_{n+1}/n + n \sum_{i=1}^{n} x_i / n) \\ &> (1/(n+1) (\sum_{i=1}^{n} x_i/n + n \sum_{i=1}^{n} x_i / n) \\ &= (1/(n+1) ((n+1) \sum_{i=1}^{n} x_i / n) \\ &= \sum_{i=1}^{n} x_i / n \end{split} \end{equation*}
Note that we really only needed $x_{n+1}$ to be larger than the previous mean. | 2019-11-13T03:04:15 | {
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https://math.stackexchange.com/questions/2299065/truth-of-x2-2-0 | Truth of $x^2-2=0$
From Algrebra by Gelfand, it says that "When we claim that we have solved the equation $x^2-2=0$ and the answer is $x=\sqrt{2}$ or $x=-\sqrt{2}$, we are in fact cheating. To tell the truth, we have not solved this equation but confessed our inability to solve it; $\sqrt{2}$ means nothing except "the positive solution of the equation $x^2-2=0$
Can someone please explain what he really means? Does he mean every irrational number is meaningless? Thanks in advance.
• I think he might be saying that irrational numbers could be thought of as solutions to polynomial equations, not the other way around. – Airdish May 27 '17 at 16:40
• I believe he is saying the result is circular. – Simply Beautiful Art May 27 '17 at 16:44
• The downvote puzzles me. Gelfand is making a very interesting subtle point and the OP is quite right to be puzzled by it, and ask. – Ethan Bolker May 27 '17 at 16:55
• @EthanBolker He's a good writer. I just don't understand sometimes and asked. – Mathxx May 27 '17 at 16:58
• Indeed he is a good writer. As the comments and answers show, you asked a good question. I hope they help you understand. – Ethan Bolker May 27 '17 at 17:12
He is saying the following.
1. $\sqrt{2}$ is defined as "the number such that its square is two."
2. The statement "the solution $x$ to $x^2-2 = 0$ is $\sqrt{2}$" is therefore saying "the solution $x$ to $x^2-2=0$ is the number such that its square is two."
As you can see, this is a rather circular statement. This doesn't mean that irrational numbers are meaningless (indeed, $\sqrt{2}$ does exist -- see the comments and @EthanBolker's answer for more on this), but rather saying that this method of definition limits us to statements like "the number such that its square is two" or "half of the number such that its square is eight."
• Nice answer, except I don't think it is clear that $\sqrt{2}$ exists at all. It is either taken as an axiom, the completeness axiom of the reals, or is to be proved, depending on our construction of the real numbers... – SEWillB May 27 '17 at 16:51
• @SEWillB there certainly exist right triangles with legs equal to $1$ unit. Now, does there not exist a hypotenuse to the triangle, since it would be of length $\sqrt 2$ units, so may not exist? – Namaste May 27 '17 at 16:56
• SEWillB (continued)... And without a hypotenuse for a right triangle, we cease to have such a triangle? So the only right triangles require legs of lenghts $x, y$ such that the length of the hypotenuse, $\sqrt{x^2 + y^2},$ is rational? – Namaste May 27 '17 at 17:04
• @amWhy, I never liked these arguments at all. There is a nice bit in a book I read a long time ago (Liebeck introduction to pure mathematics) showing the 'existence' of $\sqrt{n}$ by geometrical construction like that. But these arguments aren't convincing to me because it's not clear what you're trying to do in my opinion. – SEWillB May 27 '17 at 17:08
• So you do not believe the Pythagorean Theorem? Well, then, I think I've wasted my time commenting with you. – Namaste May 27 '17 at 17:09
To elaborate on @fractal1729 's lovely correct answer:
Having defined $\sqrt{2}$ as "the number such that it's square is $2$" it's reasonable to ask whether there is such a number.
The Greeks knew that the answer is "no" if "number" means "rational number". So in order to claim the existence of $\sqrt{2}$ you must enlarge the system of rational numbers. The Greeks essentially sidestepped that question by doing geometry rather than arithmetic, using points on a line instead of numbers. Clearly there's a point on the "number line" with the right length since you can draw the diagonal of a unit square.
In later centuries mathematicians found more formal algebraic ways to enlarge the system of rational numbers to include what we now call all the "real numbers". The new ones are the irrationals. There are several ways to do this. One of the most common is to make precise the notion of an infinite decimal, along with rules for arithmetic with them. Others come with the names "Cauchy sequences" or "Dedekind cuts".
• To add a bit to your answer, note that the very notion of compass and straightedge constructions assumes with no justification that one can construct ideal circles/lines and somehow they always intersect when certain conditions are met (such as closer than the sum of radii). When one thinks about it, that is not at all obvious. Perhaps it is not even physically meaningful, for all we know! If so, it may very well be that $\sqrt{2}$ can't be constructed in any suitable physical sense! Furthermore, the computable reals are elementarily equivalent to the standard reals. Interesting at least. – user21820 May 29 '17 at 15:26
Using definite descriptor notation, we can define:
$$\sqrt{x} = (\iota y \in \mathbb{R})(y \geq 0 \wedge y^2=x)$$
In words: $\sqrt{x}$ is the unique $y \geq 0$ such that $y^2=x$.
From this, you can show that for all $y$ and all $x \geq 0$, we have: $$y^2 = x \iff y \in \pm\sqrt{x}.$$
But in some sense, this is a purely logical construct, at least insofar as we haven't explained how to approximate $\sqrt{x}$ to arbitrary precision. There's ways of doing this, but we haven't given one.
I think the author is simply wrong. There is another meaningful way of defining $\sqrt{2}$. Consider the following sequence: $$x_{n+1} := \frac{1}{2}\left(x_n + \frac{2}{x_n}\right)$$ with $x_0=1$.
Then $\lim_{n\to\infty} x_n=\sqrt{2}$, even though all $x_n\in\mathbb{Q}$ for arbitrary $n\in\mathbb{N}$. Now instead of defining $\sqrt{2}$ as the solution of $x^2-2=0$ we could have also defined it as the limit on $x_n$.
Proof for the sequence can be found here (German only).
• This is still circular, in a deeper way. How do you define the notion of a limit without some underlying structure of the real numbers? In the rationals that sequence has no limit. What Cauchy showed is that It is possible to define the limit as the sequence itself. It's the necessity for some definition that's the crux of Gelfand's cryptic comment. – Ethan Bolker May 27 '17 at 17:10
• There is no need to use $\mathbb{R}$ with the above sequence. For example no transcendental numbers (like $\pi$) are needed. And in some sense $37$ is also just the result of multiple (albeit finite) application of Peano's axioms. The "only" difference with the above definition of $\sqrt{2}$ is that you need to apply the appropriate axioms an infinite number of times. – Hannebambel May 27 '17 at 17:18
• No, the author is right. As @EthanBolker says, your sequence has no limit within $\mathbb{Q}$. The key difference vs. the construction of $37$ lies exactly in the finite number of steps: no matter how large the $n$, no number $x_n$ in your sequence will ever equal $\sqrt{2}$. And the fact that $\sqrt{2}$ is not in $\mathbb{Q}$ is equivalent to the notion that $x^2 - 2 = 0$ is unsolvable with the standard algebraic operations. – Roland May 28 '17 at 7:55
• I somehow still don't get how the author is correct. The author says "$\sqrt{2}$ means nothing except the positive solution of the equation $x^2−2=0$", whereas I think that $\sqrt{2}$ also means the limit of the aforementioned sequence. – Hannebambel May 28 '17 at 18:23
• This answer shows that, in a certain sense, it is possible to "solve" for what $\sqrt{2}$ is. But Gelfand never claimed it wasn't possible to do so--that was not his point at all. His point is that merely naming the answer as $\sqrt{2}$ doesn't solve anything, because all you've done is give a name to what the answer is. – Eric Wofsey May 29 '17 at 21:48
This is a great question (+1) and judging by the quote, Gelfand must be a great writer at least for algebra textbooks (I confess that I have not read any of his books and hence judging only from the quote in question). In fact he has pinned down the essence of solution of polynomial equations via algebra.
If we restrict ourselves to algebra then the solution of polynomial equations via radicals is equivalent to replacing one polynomial equation with a set of binomial equations of type $x^{n} - a=0$. And for binomial equations we don't do anything apart from inventing symbols like $a^{1/n}$.
Next there are some equations which can't be solved via radicals. How does an algebraist deal with the situation now? He smartly handles the situation using field extensions and for any polynomial $f(x)$ over a field $F$ one can create an extension field $E$ which contains all the roots of $f(x)$. The technique of construction of this extension field is so simple that one is tempted to call such mechanism of solution of equations as "cheating". Thus if $f$ is irreducible then one can get a root in quotient $F[x] /(f(x))$ and guess what the root of $f(x)$ is the coset $x+(f(x))$.
In effect algebraic methods never try to find the value of root of an equation, but they are rather concerned with the structure of the field extensions which contains the roots. In some very special cases such extensions are radical and these are the familiar ones where we use the quadratic formula, or Cardano's formula (or more complicated stuff of similar type).
Let's now come back to the usual scenario where the equations have coefficients in specific fields $\mathbb{R}$ and $\mathbb{C}$ and lets first handle the case when coefficients are real. Real numbers are the product of a non-algebraic process and their existence is almost on same footing as those of rational numbers. Thus they can and are used to measure magnitude and it is agreed by convention that there are sufficient real numbers to locate every point on a geometric line so that all the geometric measurement is possible via real numbers.
But then the real numbers are a very powerful system and offer us the following justification for the symbol $a^{1/n}$:
Theorem 1: If $a$ is a positive real number and $n$ is a positive integer then there is a unique positive real number $b$ such that $b^{n} =a$ and it is denoted by symbol $a^{1/n}$.
Thus at least some binomial equations of type $x^{n} - a=0$ have a root which has existence in the real number system. By existence we mean that it is possible to give some concrete idea about this root in comparison to the integers and rationals. In other words we can provide some approximation to the value of the root using the rational numbers. And moreover the approximation can be as accurate as we want. It is in this sense of approximation via rationals that every real number exists. And it applies to all sorts of irrational numbers including the famous ones $e, \pi, \sqrt{2}$. Thus irrational numbers are not meaningless, they are as meaningful as the rationals and perhaps mathematically more significant, but unfortunately algebraic techniques cannot help in discovering their true nature.
Let's also note that there are some real numbers which are the roots of certain polynomial equations with rational coefficients and these we call algebraic real numbers. The reason for extending our number system from $\mathbb{Q}$ to $\mathbb{R}$ is not to give some sort of existence to these algebraic numbers (like $\sqrt{2}$). As far roots of polynomials are concerned the algebraic mechanism of field extensions is a sufficient and beautiful approach. Even if we tried to add such numbers to $\mathbb{Q}$ it does not give us $\mathbb{R}$, but it rather gives $\overline{\mathbb{Q}}$ which we call the algebraic closure of $\mathbb{Q}$. It includes all the algebraic numbers both real and complex, but it does not capture all the real numbers or all the complex numbers. In fact one can prove that like $\mathbb{Q}$, the set $\overline{\mathbb{Q}}$ is also countable whereas both $\mathbb{R}, \mathbb{C}$ are uncountable.
Real numbers were invented to deal with the essential/pressing need to arithmetize geometry. The idea was to develop of a system of numbers which could correspond to the points on a line and it was known from the time of Pythagoras (or perhaps even earlier) that there were some points on the line which did not correspond to a rational number. Another need for real numbers was coming from the very powerful techniques of calculus which were based on geometrical intuition, but there was no rigorous justification of these methods using arithmetical means.
To answer your question, we have two ways to look at the symbol $\sqrt{2}$. One is via algebra which says that it is a solution to the equation $x^{2}-2=0$ and in this approach we can't distinguish between $\sqrt{2}$ and $-\sqrt{2}$. Another way is to think of it as a real number represented by its rational approximations. Thus we can say that $\sqrt{2}$ is a positive real number which is greater than all the positive rationals whose square is less than $2$ and it is less than all the positive rationals whose square is greater than $2$. Further in this particular case we are lucky to have a mechanism of locating a point on the number line corresponding to it via geometrical constructions with ruler and compass.
Let's also discuss a bit about the complex number system. But before we do that we need to note that there is another set of equations which the real number system can handle very well:
Theorem 2: If $f(x)$ is polynomial of odd degree with real coefficients then there is at least one real number $c$ such that $f(c) =0$.
But there are many equations with real coefficients for which there is no root in the real number system. The simplest and most famous such equation is $x^{2}+1=0$. Once again an algebraist comes to the rescue and creates a field extension $\mathbb{C}$ as the quotient $\mathbb{R} [x] /(x^{2}+1)$ and surprisingly in one shot the algebraist succeeds in solving all polynomial equations whatsoever :
Fundamental Theorem of Algebra: If $f(x)$ is a polynomial of positive degree with complex coefficients then there is a complex number $c$ such that $f(c) =0$.
But this is all cheating. The power of the above theorem is not due to the algebraic technique of creating field extensions via quotients, but rather due to the properties of real numbers mentioned in theorem 1 and 2 above.
• Technically, this kind of justification can only reach the computable reals and also the computable complex numbers, which includes the algebraic closure of the rationals. Notice that the computable complex numbers is algebraically closed, so we can't justify going beyond it because we can't actually demonstrate the existence of anything beyond it! Also, you are right that the fact that the complex numbers are algebraically closed is at its crux not really a matter of algebra but of analysis. Even the field-theoretic proof at one point requires IVT for odd-degree polynomials. – user21820 May 29 '17 at 16:05
• In other words the only reason we have for believing the existence of all standard real numbers (whether by Cauchy sequences or Dedekind cuts) is that we must believe the power-set axiom or equivalent, which is purely set-theoretic and has no ontological justification. Actually, even function types are not enough to grant the classical power-set, because if we use type theory we can have function types that are compatible with the universe being computable. What breaks is that type membership may not be always true or false. – user21820 May 29 '17 at 16:35
• Of course, ZFC set theory is very elegant and convenient for modern mathematics and so, justified or not, it has found acceptance. =) – user21820 May 29 '17 at 16:36 | 2019-05-23T01:38:14 | {
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https://math.stackexchange.com/questions/2550068/summary-of-my-understanding-of-sequences-and-series-convergence-and-divergence | # Summary of my understanding of sequences and series' convergence and divergence?
I'm trying to summarise my understanding of infinite sequences, series, and their relationships with respect to convergence at the fundamental level. Here is what I know. How much of this is correct?
First off, here's a table of the notations that I use, and their corresponding meaning.
My understanding is that:
• The sequence $\lbrace a_n \rbrace _{n=0}^{\infty}$ converges if $$\lim\limits_{n\to\infty}a_n=L_{a}.$$
• The infinite series $\sum\limits_{n=0}^{\infty}a_n$ converges if its sequence of partial sums, $\lbrace s_n \rbrace _{n=0}^{\infty}$, has a limit, i.e.$$\lim\limits_{n\to\infty}s_n=L_{s}.$$
• If the infinite series $\sum\limits_{n=0}^{\infty}a_n$ converges, then the limit of the sequence $\lbrace a_n \rbrace _{n=0}^{\infty}$ is $0$, i.e. $$\sum\limits_{n=0}^{\infty}a_n \: converges \rightarrow \lim\limits_{n\to\infty}a_n=0.$$
• The divergence test:
If the the limit of the sequence $\lbrace a_n \rbrace _{n=0}^{\infty}$ is NOT $0$ or does not exist, then the infinite series diverges, i.e. $$\lim\limits_{n\to\infty}a_n\neq0 \rightarrow \sum\limits_{n=0}^{\infty}a_n \: diverges$$
Would seriously appreciate it if anyone could verify whether the above is accurate or incorrect in any way.
EDIT: I've modified the two limits notation that were mentioned in the comments and answers below, as well as adding the additional condition (limit does not exist or does not equal zero) for the divergence test. I appreciate all the answers/comments.
• I would say that the notation $L_{a_n}, L_{s_n}$ looks pretty bad, these terms should not depend on $n$. – user99914 Dec 4 '17 at 5:08
• BTW one useful equivalent of $\lim_{n\to \infty}a_n=L,$ that students often don't think of , is : For every $r>0$ the set $\{n:a_n\not \in [-r+l,r+L]\}$ is finite. – DanielWainfleet Dec 4 '17 at 5:49
Everything is correct, though the notation $\lim\limits_{n\to\infty}a_n=L_{a_n}$ is nonstandard. Perhaps an improvement would be to write it as $\lim\limits_{n\to\infty}a_n=L_{a}.$ Normally just an $L$ will suffice but if you're working with multiple sequences (such as $a_n, b_n, c_n$) then $L_a$ is a good notation for the limit of the sequence $a$.
The reason why $L_{a_n}$ is not so good is because $n$ is just a free variable that has no importance whatsoever; if you write $a_n$ or $a_k$ it's the same thing. What matters is the sequence whose name is "$a$".
Also note that your last two statements are trivially equivalent; they are contrapositives of each other. In general "If $p$ then $q$" is stating the same thing as "If not $q$ then not $p$".
Therefore, just from knowing
$$\sum\limits_{n=0}^{\infty}a_n \: converges \rightarrow \lim\limits_{n\to\infty}a_n=0.$$
you can deduce
$not (\lim\limits_{n\to\infty}a_n=0 ) \rightarrow not (\sum\limits_{n=0}^{\infty}a_n \: converges)$
or in other words
$\lim\limits_{n\to\infty}a_n \not =0 \text{ or the limit doesn't exist} \rightarrow \sum\limits_{n=0}^{\infty}a_n \: diverges$
• Regarding the redundancy in my understanding, although $\lim\limits_{n\to\infty}a_n \not =0 \text{ or the limit doesn't exist} \rightarrow \sum\limits_{n=0}^{\infty}a_n \: diverges,$ it does not mean that $\lim\limits_{n\to\infty}a_n=0 \rightarrow \sum\limits_{n=0}^{\infty}a_n \: converges,$ am I right? – user98937 Dec 4 '17 at 13:11
• @user98937 Yes that's perfectly right. What you wrote is the converse of $\sum\limits_{n=0}^{\infty}a_n \: converges \rightarrow \lim\limits_{n\to\infty}a_n=0.$. Converses are sometimes true, but not in this case as you can see from $\sum_{n=1}^{\infty} \dfrac 1n$ which diverges, in spite of the fact that $\lim_{n \to \infty} \dfrac 1n=0$ – Ovi Dec 4 '17 at 15:33
Everything you wrote looks good to me. | 2021-01-27T20:24:35 | {
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https://dsp.stackexchange.com/questions/40320/confusion-regarding-pdf-of-circularly-symmetric-complex-gaussian-rv | # Confusion regarding pdf of circularly symmetric complex gaussian rv
Considering a random variable $x$ that takes in values from a complex domain. Its real and imaginary components are totally uncorrelated. I am following this link and also studying this document.
In the wikipedia link, the pdf for a single observation is given and $k =2$ for bivariate gaussian as my assumption is that the real and imaginary are totally uncorrelated and are gaussian respectively. There is a 2 in the denominator in the pdf but the log likelihood for the complex case does not have any 2 in the denominator. The log-likelihood for $k=2$ would be $$-N \ln| \Sigma| - (x_n) {|\Sigma|}^{-1}(x_n)^H - k N \ln \pi$$ where $\mu=0$ as I am assuming zero mean r.v $x$
• Confusion 1:
I was thinking that the variance of $x$ was $$\Sigma = \begin{bmatrix}\sigma_1^2 & 0\\ 0& \sigma_2^2\end{bmatrix}\,\text.$$ So for $N$ samples (observation), the joint pdf would turn out to be $$P_x(x_1,x_2,...,x_N) = \prod_{n=1}^N\frac{1}{\pi \sigma^2_x} \exp \bigg(\frac{-{({x_n})}^H ({x_n})}{\sigma^2_x} \bigg)$$ where $\sigma^2_x = [\sigma_1^2,\sigma_2^2]$. Would there be a 2 in the denominator of the pdf and what is the correct pdf expression?
• Confusion 2:
In the document, the expression for the pdf for complex case looks different from the wikipedia link. Are they the same but maybe I am missing some link between them? Which pdf should I use?
• Confusion 3:
When simulating the r.v in Matlab with zero mean and variance 1, I am halving the variance because the real and imaginary part's variance should add up to 1. Then, would the pdf also contain half of the variance as $\sigma^2_x/2$?
Let me try to establish the relation between the univariate PDF for a real Gaussian and the univariate PDF for a complex proper (i.e. circular symmetric) Gaussian.
You know that
$p_x(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp(\frac{-x^2}{2\sigma^2})$ is the PDF of a real-valued Gaussian with variance $\sigma^2$. We write $x\sim\mathcal{N}(0,\sigma^2)$ to denote that $x$ is a random variable that follows a real-valued Gaussian with zero mean and variance $\sigma^2$.
Now, let $z=a+jb$ be a circular symmetric random variable with real part $a$ and imaginary part $b$. In a circularly symmetric Gaussian random variable, the real and imaginary part are i.i.d., i.e. $a\sim\mathcal{N}(0,\sigma^2)$ and $b\sim\mathcal{N}(0,\sigma^2)$. Since $a,b$ are independent, their joint PDF is the product of their PDFs.
$$p_z(z=a+jb)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{-a^2}{2\sigma^2}\right)\cdot\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{-b^2}{2\sigma^2}\right)$$
which gives the PDF of the complex variable $z$ at the value $a+jb$. We can reformulate this to \begin{align}p_z(z=a+jb)&=\frac{1}{2\pi\sigma^2}\exp\left(\frac{-(a^2+b^2)}{2\sigma^2}\right)\\&=\frac{1}{2\pi\sigma^2}\exp\left(\frac{-z^*z}{2\sigma^2}\right)\\&=\frac{1}{2\pi\sigma^2}\exp\left(\frac{-\|z\|^2}{2\sigma^2}\right)\end{align}
We also write for this case that $z$ follows a circular Gaussian distribution and denote this by $z\sim\mathcal{CN}(0,2\sigma^2)$. Why is it suddenly $2\sigma^2$ (i.e. the double variance compared to the uni-variate, real case? Well, because $E[\|z\|^2]=2\sigma^2$ since it consists of a real and an imaginary part (which are independent and hence their variance adds up together).
So, to sum up:
• if $x\sim\mathcal{N}(0,\sigma^2)$, then $p_x(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{x^2}{2\sigma^2}\right)$
• if $z\sim\mathcal{CN}(0,\sigma^2)$, then $p_z(z)=\frac{1}{\pi\sigma^2}\exp\left(-\frac{z^*z}{\sigma^2}\right)$
• Thank you so much, cannot express my gratitude in words here.....just to clarify 2 things from you - (1) if I consider design where the r.v $z \sim CN(0,2\sigma^2)$ then the pdf has the 2 in the denominator, otherwise if $z \sim CN(0,\sigma^2)$ then there is no 2 in the pdf's denominator. So, either design is acceptable. Is my understanding right? (2) notations - in my question I wrote the $[.]^H$ complex conjugate transpose but you have written * which is only the conjugate. Why the $[.]^H$ is wrong? – Ria George Apr 18 '17 at 15:33
• sure, if $z\sim CN(0,2\sigma^2)$, then there is the two in the denominator, if its $z\sim CN(0,\sigma^2)$, then it's not there. Here, the two comes from the description of the variance (e.g. if we had $z\sim CN(0,1.23456\sigma^2)$, then in the denominator there would be $1.23456$). For a scalar variable $z$, $[]^H$ and $[]^*$ are equal. When it comes to vector-variables, you'd need $[]^H$. However, note that in this case, also the denominator changes to $\pi^k\sigma^{2k}$, since the PDF is the product of the $k$ PDFs of each component (k is the vector size); each component has var. $\sigma^2$. – Maximilian Matthé Apr 18 '17 at 17:58
• This answer has the correct ideas but is marred badly by a confusion of the concepts of probability and probability density. The probabiity that a continuous random variable equals a given number is $0$, regardless of the choice of number. Indeed, the probability that $z$ equals $a+jb$ (where we choose $a=b=0$) is claimed to be $$P(z=0+j0)=\frac{1}{2\pi\sigma^2}$$ which has value greater than $1$ if $\sigma^2 < \frac{1}{2\pi}$. -1 pending clean-up of the answer to be correct. – Dilip Sarwate Apr 18 '17 at 22:26 | 2021-06-20T22:18:03 | {
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https://math.stackexchange.com/questions/2323249/integral-of-textlogz | # Integral of $\text{Log}(z)$
I'd like to check my work. I'm trying to integrate $f(z)=\text{Log}(z)$ on the right half of the unit circle from $-i$ to $i$. $\text{Log} = \ln(r)+i\theta$ according to my book. So, \begin{align} \int_C f(z)\,dz&=\int_{z_1}^{z_2}f(z)\,dz\\\\& = \int_{-i}^i \text{Log}\,dz \\\\ &= \bigg[z\text{Log}-z\bigg]_{-i}^i \\\\ &= (i\text{Log}(i)-i)-\left(-i\text{Log}(-i)+i\right) \\\\ &= i\left(\ln(1)+\frac{\pi}{2}i\right)-i \ + i\left(\ln(1)-\frac{\pi}{2}i \right) \ -i \\\\&= -2i \\\\ \therefore \int_{-i}^i \text{Log}\,dz=-2i \end{align}
This may be incorrect, I'm not really sure how to incorporate the branch cut here.
Thanks!
• the circuit is a circle not the segment [-i,i]. – zwim Jun 15 '17 at 5:13
• So, I'm required to use the formula $\int_C f(z)dz=\int_a^b f[z(\theta)]z'(\theta)d\theta$? Where $a$ and $b$ are $\frac{\pi}{2}$ and $\frac{-\pi}{2}$? What would my choice for $z'(\theta)$ be in this case? – Kosta Jun 15 '17 at 5:14
• Have a look at these answers : math.stackexchange.com/questions/1602614/… – zwim Jun 15 '17 at 5:34
I thought it might be instructive to present two alternative approaches to evaluate the integral of interest. To that end we proceed.
Let $\text{Log}(z)=\log(|z|)+i\text{Arg}(z)$ where $-\pi < \text{Arg}(z)\le \pi$. In addition, let $C$ denote the semicircular contour in the right-half plane of radisu $1$ and that begins at $z=-i$ and ends at $z=i$.
METHODOLOGY $1$:
On $C$, $z=e^{i\theta}$, $-\pi/2\le \theta \le \pi$. Hence, the integral $\int_C \text{Log}(z)\,dz$ is given by
\begin{align} \int_C \text{Log}(z)\,dz&=\int_{-\pi/2}^{\pi/2} \text{Log}(e^{i\theta})\,ie^{i\theta}\,d\theta\\\\ &=-\int_{-\pi/2}^{\pi/2} \theta e^{i\theta}\,d\theta\\\\ &=-2i\int_0^{\pi/2} \theta\sin(\theta)\,d\theta\\\\ &=-2i \end{align}
METHODOLOGY $2$:
Another approach is to exploit Cauchy's Integral Theorem. Inasmuch as $\text{Log}(z)$ is analytic for $z\in \mathbb{C}\setminus (-\infty,0]$. Then, the integral over $C$ can be deformed to connect $-i$ to $i$ along any rectifiable path that does not intersect the branch cut. Therefore, we write
\begin{align} \int_C \text{Log}(z)\,dz&= \lim_{\epsilon\to 0^+}\left(\int_{-1}^{-\epsilon }\text{Log}(iy)\,i\,dy+\int_{-\pi/2}^{\pi/2}\text{Log}(\epsilon e^{i\theta})\,ie^{i\theta}\,d\theta+\int_{\epsilon }^1\text{Log}(iy)\,i\,dy\right)\\\\ &=i \lim_{\epsilon\to 0^+}\int_\epsilon^1 \left(\text{Log}(-iy)+\text{Log}(iy)\right)\,dy\\\\ &=2i\int_0^1 \text{Log}(y)\,dy\\\\ &=-2i \end{align}
as expected!
• This is very helpful, thank you Mark! – Kosta Jun 16 '17 at 16:21
• You're welcome. Pleased to hear that this is helpful! – Mark Viola Jun 16 '17 at 16:25
An alternate way to check, without using contours, is to use $$\int \ln(x) \, dx = x \, \ln(x) - x$$ which provides \begin{align} \int_{-i}^{i} \ln(x) \, dx &= \left( i \, \ln(e^{\pi \, i/2}) - i \right) - \left( -i \, \ln(e^{-\pi \, i/2}) + i \right) \\ &= \frac{\pi \, i^2}{2} - \frac{\pi \, i^2}{2} - 2 i \\ &= -2i. \end{align} With this it can be stated that the real line integral result is the same as the contour integral result.
looks good to me .Fundamental theorem of calculus wins again | 2019-05-27T08:06:54 | {
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http://mathhelpforum.com/trigonometry/227503-trigonometric-identity-problem.html | # Thread: Trigonometric Identity Problem
1. ## Trigonometric Identity Problem
How do you show that:
2sinxcosx=
(4tan{x/2})/(1+tan2{x/2}) . (1-tan2{x/2})/(1+tan2{x/2})
Thanks a lot for helping!!
2. ## Re: Trigonometric Identity Problem
Originally Posted by JiaGeng
How do you show that:
2sinxcosx=
(4tan{x/2})/(1+tan2{x/2}) . (1-tan2{x/2})/(1+tan2{x/2})
$1+\tan^2(\frac x 2) = \sec^2(\frac x 2)$
$1-\tan^2(\frac x 2) = \dfrac{2 \tan(\frac x 2)}{\tan(x)}$
$\sin(2x)=2\sin(x)\cos(x)$
-------------------------
$\dfrac{4 \tan(\frac x 2)}{1+\tan^2(\frac x 2)}\dfrac{1-\tan^2(\frac x 2)}{1+\tan^2(\frac x 2)}=$
$\dfrac{4 \tan(\frac x 2)}{\sec^2(\frac x 2)}\dfrac{ \dfrac{2 \tan(\frac x 2)}{\tan(x)}}{\sec^2(\frac x 2)}=$
$4\cos(\frac x 2)\sin(\frac x 2)\dfrac{2\sin(\frac x 2)\cos(\frac x 2)}{\tan(x)}=$
$2\sin(x)\dfrac{\sin(x)}{\tan(x)}=$
$2\sin(x)\cos(x)$
3. ## Re: Trigonometric Identity Problem
Hello, JiaGeng!
$\text{Prove: }\:2\sin x\cos x\:=\:\frac{4\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} \cdot \frac{1-\tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}$
The first fraction is:
. . $\frac{4\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \;=\;\frac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2} }}{\sec^2\frac{x}{2}} \;=\;\frac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2} }}{\frac{1}{\cos^2\frac{2}{2}}} \;=\;4\sin\tfrac{x}{2}\cos\tfrac{x}{2} \;=\;2\sin x$
The second fraction is:
. . $\frac{1-\tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \;=\;\frac{1-\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}}{\sec^ 2\frac{x}{2}} \;=\; \frac{1-\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}}{\frac {1}{\cos^2\frac{x}{2}}} \;=\;\cos^2\tfrac{x}{2} - \sin^2\tfrac{x}{2} \;=\;\cos x$
Therefore, the RHS becomes: . $2\sin x\cos x$.
4. ## Re: Trigonometric Identity Problem
Thanks romsek and soroban!
But I was wondering if there is a way to do it from LHS to get RHS because that's what the question requires.... Would that be even harder?
5. ## Re: Trigonometric Identity Problem
Originally Posted by JiaGeng
Thanks romsek and soroban!
But I was wondering if there is a way to do it from LHS to get RHS because that's what the question requires.... Would that be even harder?
Equality works in both directions. If you want, write the equations in reverse order and equality still holds at every step.
6. ## Re: Trigonometric Identity Problem
Hello, JiaGeng!
I was wondering if there is a way to do it from LHS to get RHS.
Would that be even harder?
If you would rather not run the steps in reverse,
it can be done ... with a little imagination.
$2\sin x\cos x \;=\;\frac {2(2\sin \frac{x}{2}\cos\frac{x}{2})}{1}\cdot \frac{\cos^2\!\frac{x}{2} - \sin^2\!\frac{x}{2}}{1}$
. . . . . . . . . $=\;\frac{4\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\! \frac{x}{2} + \sin^2\!\frac{x}{2}} \cdot \frac{\cos^2\!\frac{x}{2} - \sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2} + \sin^2\!\frac{x}{2}}$
In both fractions, divide numerator and denominator by $\cos^2\!\tfrac{x}{2}\!:$
$\dfrac{\frac{4\sin\frac{x}{2} \cos\frac{x}{2}}{\cos^2\!\frac{x}{2}}} {\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}} + \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} \cdot \dfrac{\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x} {2}} - \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} {\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}} + \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} \;=\; \dfrac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}} {1 + \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2} \cdot \dfrac{1 - \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2} {1 + \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2}$
. . . . $=\;\frac{4\tan\frac{x}{2}}{1 + \tan^2\!\frac{x}{2}} \cdot \frac{1-\tan^2\!\frac{x}{2}}{1 + \tan^2\!\frac{x}{2}}$ | 2017-06-24T13:13:05 | {
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https://www.themathdoctors.org/a-sample-of-ask-dr-math-part-2-questions-outside-of-school/ | # A Sample of Ask Dr. Math, Part 2: Questions Outside of School
In the first post, I gave a small sampling of questions we’ve had from students, parents, and teachers, all related to school, and discussed how we like to deal with these. But we also get many questions with no direct relation to school. These may come from people who actually use math in their work (computer programming, plumbing, and anything in between), or from students who are trying to deeply understand what they are studying, or who are just curious. Often, of course, these too are from students or teachers, but they are going beyond that role. These are perhaps the most interesting of all. They are really why we are here.
## A real-life question
This question arrived in 2014 from a flooring inspector, Tim, asking about how high a floor would buckle due to a given increase in length if the ends are fixed:
Modeling How Much Wood Expands
Say I place a six foot long yardstick wide side down. One end is butted against a wall (fixed object). When the other end of the yardstick is pushed towards the wall, by say 1/4", the middle area rises -- and it rises much, much higher than 1/4".
I would like to know if there is a fixed formula for this exponential gain of "flat" vs. "up."
I come across this daily, seeing floors buckle up and away from substrates because they are a tight fit.
(He misused the word exponential; I suppose he was just thinking that the rise is surprisingly high, as in exponential growth.)
Dr. Rick and I both recognized this problem as related to previous questions we had answered, and referred to those in our replies. His old answer assumed a circular arc for the new shape, and did a sanity check by comparing the surprising answer to one obtained by supposing that the board just bends sharply in the middle. The circular arc version does not lend itself to easy calculation, but the sharp bend could be turned into a formula.
There is a way to calculate the height, but it isn't a formula, as such.
...
For more on this topic, including a way to do a *rough* approximation with a simple formula, see:
http://mathforum.org/library/drmath/view/62613.html
I had been thinking while Dr. Rick wrote, and found a different old answer (by Dr. Jeremiah) that didn’t actually come up with a method, just a starting point. But I “played” with numbers using a spreadsheet much as Dr. Rick had for his earlier answer, and saw a pattern that suggested an approximate formula he might try. When I saw what Dr. Rick had written, I wrote to propose this formula.
I see Dr. Rick beat me to most of what I was going to say, except that my link to our archives is not quite as helpful:
Rail Bend in Hot Weather
http://mathforum.org/library/drmath/view/61465.html
But I made a spreadsheet carrying out the calculations for a circular arc; and have found experimentally that the rise seems to be quite accurately approximated by a simple formula. I have to study more to try to justify it, but here goes.
For the actual length of the stick, s, and the (small) decrease in the space available, x,
h = 0.605 sqrt(sx)
But people who enjoy math can’t be fully satisfied with a guess; I continued pursuing my approximation, hoping to justify it. The patient, Tim, didn’t need a proof, but I did. By the end of the day, I could write up a derivation of the formula and send it along, and then, the next day, apply it to some examples.
I've verified my simple approximation, which turns out to be
h = sqrt(6sx)/4
This changes my 0.605 to 0.612, which fits better for very small x but not quite as well for larger ones.
I'm sure you don't need to see this, but for my records, here is the derivation of my approximation: ...
A missing formula for a counter-intuitive observation (a half-inch shift can result in almost a 7-inch buckle) can be a great motivation for some fun with math. And we also have a reminder that in the real world, an approximation can be both good enough, and as good as you’ll get. (Be sure, as always, to read the whole story in the original page!)
## A curious question
(Well, that last one arose from curiosity, too, but this is pure-math curiosity.)
Within months of my starting with Ask Dr. Math, I wrote this answer, to a question wondering about place value in decimal numbers:
The Oneths Place
Why is there not a oneths place? My classmates and I were wondering. It is weird that decimals begin with tenths, hundredths, and so on. We appreciate your help.
I love it when children are wondering, especially when a teacher encourages it. So I was happy to think about this, even though I had probably never considered the question before. (Since then, we have had some form of this question at least a dozen more times! Often, I have answered by saying, “Believe it or not, you’re not the first to ask this; if you search our site for the word ‘oneth’, you’ll find this answer …”)
I’m sure my answer is not all that could be said about it, but I managed not only to give the quick answer (oneths are just another name for ones, since 1/1 = 1, and we don’t want two places with the same value), but to relate it to some other perceived asymmetries in the way we write numbers, and bring in signed exponents (which may be beyond the students’ knowledge, or may be something they were just about to learn). I closed with a general comment about curiosity:
I hope that helps. It's interesting how we tend to expect things to be symmetrical. Mathematicians and scientists often expect it too, and when things don't seem balanced, they try to find out why. So keep wondering about things like this.
## A philosophical question
Another question we’ve received a number of times involves how a line segment can have a finite length when it is composed of infinitely many points. As a result, we’ve referred to the following question a dozen times or so, in addition to giving fresh answers from different perspectives:
How Can a Line Have Length?
We have a line, OR a line segment, OR a ray (it really doesn't matter). If this particular linear entity (let us say that it is a line segment, for the sake of hypothetical simplicity) consists only of infinite points, all with zero volume, mass, diameter, cross-sectional area, etc., how can it have length? Why are we able to take a particular line segment and give it a length of 5? If a substance can only be composed of volumeless points, how can it have volume in and of itself?
Dr. Daniel’s answer (from 1998) touches on several key ideas in philosophical thinking about this topic, making a concise summary without digging in too deeply. (I try to avoid digging too deeply, because of the possibility of cave-ins … .) When I answer related questions, I often refer to that answer, and also to this one:
Point and Line
I'm in high school, but this problem has really nothing to do with school, it's just been bothering me for a while.
A point has no dimension (I'm assuming), and a line, which has dimension, is a bunch of points strung together. How does something without dimension create something with dimension?
Going even farther, a point essentially is nothing, because it has no dimension. My question is, How does a bunch of nothings (a point) create a something (a line)?
Here Dr. Jordi got into a long discussion about the problem, perhaps not very effectively in the end. After referring to these two answers, I generally add something like the following as my own brief perspective:
Neither will be a fully satisfying answer, largely because we humans are finite beings, not able to directly observe infinity. (It's amazing enough that we can THINK about infinity, and get some of the answers right!) But the basic idea is that we don't define a line as something made by putting points together; rather, we start by defining a line, and then note that it consists of points. So we can find points on a line, without having to _make_ the line _out of_ points (or make a cube out of stacked squares).
Note also that we do not define length in terms of points; since 0, the width of a point, times infinity, the number of points, is an indeterminate quantity (we can't assign any one value to it), that would be useless to attempt. So length is an independent concept, defined only in relation to the location of the endpoints of a segment, not to the points that make it up.
In choosing these examples of “curious questions”, I have chosen to avoid questions with really interesting answers, which deserve fuller treatment on their own! Keep watching, and I’ll get to some of them.
### 2 thoughts on “A Sample of Ask Dr. Math, Part 2: Questions Outside of School”
1. Hi – I would like to know the probability of this occurring – this past January I celebrated my 74th birthday which means I was born in 1947 (1947/74 numbers reversed). My daughter was born in April, 1974. She will be turning 47 this year (1974/47 numbers reversed) What is the probability of this occurring in a parent/son-daughter relationship?
1. Hi, Gwen.
I suppose you asked this in connection with this particular post just because it’s an example of curiosity! In fact, there is another post that deals with this very kind of question: Should Rare Events Surprise Us? You’ll find there that an important question we need to ask is, “What do you mean by ‘this’?”
I don’t think you are asking what percentage of all people now alive were born in 1947, and had a daughter born in 1974; and if you did, I’d have to pass, as it would require digging deeper into birth statistics than I feel like going. (It would not be an interesting question mathematically.)
I think your main interest is in the coincidence of the numbers. I may not get to a probability in the end, but there may be some interesting observations along the way!
My first observation is that the two reversals are closely related. The reason you are 74 this year is that 2021-1947 = 74; that is, 2021 = 1947 + 74. But that implies that 2021 also equals 1974 + 47, because both sums can be written as 1900 + 47 + 74, just rearranging the numbers. So that fact that this year your age is the reversal of your birth year implies that the same will be true this year for anyone born in 1974. To put it another way, her being born in 1974, the reversal of your birth year, predestined everything you observe this year. It all had to happen sooner or later!
Let’s test that out. I was born in 1952. I couldn’t have a daughter born in 1925 or 2025, but in fact it turns out that my father was born in 1925! My claim is that this by itself implies that something like your situation would occur at some point. So if we add 52 to 1925, we find that in 1977, my father was 52, and I was 25. Of course that happened once in a lifetime, but that it would happen to us at all depended only on our two birthdates and our living long enough.
Will everyone have some year in his life when his age is the reverse of his birth year? That just depends the reverse of his birth year being a likely age. For someone born in 1900, it happens immediately (and again if he reaches 100); born in 1901, it happens when he reaches 10. If you were born in 1909, you’ve have to live to 90 to have it happen. So basically, lifespan is the only restriction.
How likely is it that one will have a child in his “reverse year”? That restricts it more, since the resulting age has to be within the childbearing years. For example, in your case your age at her birth was 1974 – 1947 = 27. If we suppose that, say, a child can be born to a parent aged 16 to 45, then your scenario is feasible for people born in about 30% of all years, and considerably more likely for, say, 10-15%. (Actually, this is a little more subtle than what I just said, and I checked by making a spreadsheet to actually count years that would work.)
With appropriate statistical information, one might come up with a probability of actually having a child in a given year. But that’s beyond my level of interest.
But I recalled my twin brother pointing out something similar based on our father’s age, and was able to find it: Reversal of Age Digits Every Eleven Years. What he describes there will also apply to you and your daughter: Because you were a multiple of 9 years old when she was born, every 11 years your ages have reversed digits, as they are this year. So not only is your situation this year not unique, it is not unique to you: It also happened when she was 3, and 14, and 25, and 36! But it will stop happening when you pass 100 …
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https://mathoverflow.net/questions/255474/integral-of-the-entrywise-square-of-the-exponential-of-a-matrix | # Integral of the entrywise square of the exponential of a matrix
Note: I posted my question on math.stackexchange but got no answer. That is why I am asking it here.
Let $A$ be a $n\times n$ square matrix such that the real part of all eigenvalues are negative. For each $i,j$, let $\exp(At)_{ij}$ be the element $(i,j)$ of the matrix. It is well known that: $$\int_0^\infty \exp(At)_{ij}dt = -(A^{-1})_{ij}$$ Is it possible to simplify a similar expression where each element is squared: $$\int_0^\infty (\exp(At)_{ij})^2 dt = ??$$
I am wondering if it is possible to simplify the above expression. If it helps, I can assume that $A$ is diagonalizable. Note that unless for one-dimensional matrices, $(\exp(At)_{ij})^2\ne\exp(2At)_{ij}$.
• Note that OP wants an elementwise square, not the usual matrix-multiplication square. – Federico Poloni Nov 24 '16 at 8:19
Inspired strongly by Anthony's answer, here is a formula that works for arbitrary $A$. Let $M$ be the $n^2 \times n^2$ square matrix given by $$M= A \otimes I_n + I_n \otimes A_n$$ i.e. in terms of indices
$$M_{ij,kl}=A_{i,k} \delta_{j,l} + \delta_{i,k}A_{j,l}$$
Then because $A \otimes I_n$ and $I_n \otimes A$ commute, $$e^{Mt} = (e^{At} \otimes I_n) (I_n \otimes e^{At}) = e^{At} \otimes e^{At}$$ i.e. in indices $$(e^{M t})_{ij,kl} = (e^{A t})_{i,j} (e^{At})_{k,l}$$ so in particular $$(e^{At})_{i,j}^2 = (e^{Mt})_{ii,jj}$$ and $$\int_t (e^{At})_{i,j}^2 = \int_t (e^{Mt})_{ii,jj} = - (M^{-1})_{ii,jj}$$
(Here I am using commas to separate the two indices of a matrix entry)
• Thanks for this answer. I am not sure to understand why the two summands would commute and why $e^{Mt}_{ij,kl}$ can be expressed in terms of $e^{At}$ (with these indices, it seems false). – N. Gast Nov 24 '16 at 16:18
• I think this is problematic because each of the summands is of rank $n$, so the sum is of rank at most $2n$. For $n>2$, $M$ will not be invertible. – Anthony Quas Nov 24 '16 at 16:39
• @AnthonyQuas That was a stupid mistake, fixed now. – Will Sawin Nov 24 '16 at 18:09
• @N.Gast Sorry, that was wrong, is this clearer? – Will Sawin Nov 24 '16 at 18:09
• @WillSawin: does M^{-1} exist? – Anthony Quas Nov 24 '16 at 18:42
You can do something. Here's a computation for diagonalizable $A$. Let $A=BDB^{-1}$ and let the elements of $D$ be $-\lambda_1,\ldots,-\lambda_d$. Then \begin{align*} \int_0^\infty (e^{At})_{ij}^2\,dt&= \int_0^\infty (Be^{Dt}B^{-1})_{ij}^2\,dt\\ &=\int_0^\infty \sum_{k,k'} B_{ik}e^{-\lambda_k t}B^{-1}_{kj} B_{ik'}e^{-\lambda_k' t}B^{-1}_{k'j}\,dt\\ &=\int_0^\infty \sum_{k,k'} B_{ik}B^{-1}_{kj} B_{ik'}B^{-1}_{k'j}e^{-(\lambda_k+\lambda_k') t} \,dt\\ &=\sum_{k,k'} \frac{1}{\lambda_k+\lambda_k'}B_{ik}B^{-1}_{kj} B_{ik'}B^{-1}_{k'j}. \end{align*}
Given a Hurwitz matrix $\mathrm A \in \mathbb R^{n \times n}$, let
$$\Phi (t) := \exp(\mathrm A t)$$
be the state transition matrix, and let its $(i,j)$-th entry be denoted by
$$\varphi_{ij} (t) := \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j$$
Hence,
$$\begin{array}{rl } \displaystyle\int_0^{\infty} \left( \varphi_{ij} (t) \right)^2 \, \mathrm d t &= \displaystyle\int_0^{\infty} \left( \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j \right)^2 \, \mathrm d t\\\\ &= \displaystyle\int_0^{\infty} \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j \mathrm e_j^{\top} \Phi^{\top} (t) \, \mathrm e_i \, \mathrm d t\\\\ &= \mathrm e_i^{\top} \underbrace{\left( \displaystyle\int_0^{\infty} \Phi (t) \, \mathrm e_j \mathrm e_j^{\top} \Phi^{\top} (t) \, \mathrm d t \right)}_{=: \mathrm W_c} \mathrm e_i = \mathrm e_i^{\top} \mathrm W_c \mathrm e_i\end{array}$$
where $\mathrm W_c$ is the controllability Gramian of the pair $(\mathrm A, \mathrm e_j)$ and is the solution to the following controllability Lyapunov equation
$$\boxed{\mathrm A \mathrm W_c + \mathrm W_c \mathrm A^{\top} + \mathrm e_j \mathrm e_j^{\top} = \mathrm O_n}$$
Thus, the $n$ columns of the integral of the entrywise product
$$\int_0^{\infty} \left( \Phi (t) \circ \Phi (t) \right) \mathrm d t$$
are the diagonals of $\mathrm W_c^{(1)}, \mathrm W_c^{(2)}, \dots, \mathrm W_c^{(n)}$, where $\mathrm W_c^{(1)}, \mathrm W_c^{(2)}, \dots, \mathrm W_c^{(n)}$ are the solutions to the following $n$ controllability Lyapunov equations
$$\begin{array}{cl} \mathrm A \mathrm W_c^{(1)} + \mathrm W_c^{(1)} \mathrm A^{\top} + \mathrm e_1 \mathrm e_1^{\top} &= \mathrm O_n\\ \mathrm A \mathrm W_c^{(2)} + \mathrm W_c^{(2)} \mathrm A^{\top} + \mathrm e_2 \mathrm e_2^{\top} &= \mathrm O_n\\ \vdots & \\ \mathrm A \mathrm W_c^{(n)} + \mathrm W_c^{(n)} \mathrm A^{\top} + \mathrm e_n \mathrm e_n^{\top} &= \mathrm O_n\end{array}$$
• In MATLAB, the Gramian can be computed using function gram. – Rodrigo de Azevedo Nov 25 '16 at 22:18 | 2019-10-20T18:00:07 | {
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https://math.stackexchange.com/questions/795952/is-n2-n-1-prime-for-all-n | # Is $n^2 + n + 1$ prime for all n?
I recently stumbled across this question in a test.
Paul says that "$n^2+n+1$ is prime $\forall\:n\in \mathbb{N}$".
• Paul is correct, because...
• Paul is wrong, because...
The answers sheet says that "all answers providing a valid counterexample, e.g. $n=4$, are valid; however, I'm interested in understanding why - is there some sort of theorem which can predict whether a polynomial is prime or not?
What I tried was:
$n^2+n+1$ can be factorized as $(n+\frac{1+\sqrt{3}i}{2})(n+\frac{1-\sqrt{3}i}{2})$. Neither of these factors is a suitable factor (i.e. is natural, different from $1$, and different from $n^2+n+1$) for any $n\in\mathbb{N}$, therefore $n^2+n+1$ must be prime $\forall\:n\in\mathbb{N}$.
Instead, $n^2-1$ can be factorized as $(n+1)(n-1)$. The factor $(n+1)$ is a suitable factor (i.e. natural, different from $1$, and different from $n^2-1$) for $n>2$; therefore, $n^2-1$ is not prime for $n>2$.
However, the above reasoning is clearly wrong, because "$n^2+n+1$ is prime $\forall\:n\in\mathbb{N}$" doesn't hold for the case $n=4$.
What is wrong in the reasoning? Can one tell whether a polynomial $p(n)$ is prime for all $n$ a priori?
• Counterexample clearly is a formal proof. Furthermore, for quite a lot of small $n$ (like $n \leq 40$, IIRC - could be wrong), such numbers are prime. – Marcin Łoś May 15 '14 at 14:32
• @MarcinŁoś thanks for pointing that out. I edited the question to better suit what I'm asking for. – Giulio Muscarello May 15 '14 at 14:36
• It was established by Goldbach that there is no polynomial with integer coefficients that can give a prime for all integer inputs. Ref: mathworld.wolfram.com/Prime-GeneratingPolynomial.html – Joel May 15 '14 at 14:44
Hint $\ \ f(1)=3\,\Rightarrow\, 3\mid f(1\!+\!3n),\$ e.g. $\,3\mid f(4)=21$
Remark $\$ Above we used the Polynomial Congruence Theorem
$\ {\rm mod}\ 3\!:\ \color{}{1\!+\!3n\equiv 1}\,\Rightarrow\, f(\color{}{1\!+\!3n})\equiv f(\color{}1)\equiv 0,\$ for all $\,f\in\Bbb Z[x]$
Alternatively $\,\ 3n\mid f(1\!+\!3n)-f(1)\$ by the Factor Theorem $\,x-y\mid f(x)-f(y)$
The idea generalizes to a proof that any nonconstant polynomial $\in\Bbb Z[x]\,$ has a non-prime value.
Just because a polynomial $f$ doesn't factor doesn't mean that the number $f(m)$ you get when evaluating $f$ at a number $m$ doesn't factor. The simplest example is something like $f(x) = x+1$. This polynomial is irreducible, hence doesn't factor, but $f(3) = 4$ is clearly composite.
Here is a link showing that no polynomial with integer coefficients can evaluate to primes for all integers.
You want to disprove a statement of the form $\forall n\colon f(n)\text{ is prime}$, that is you want to prove $\neg\forall n\colon f(n)\text{ is prime}$, which is equivalent to $\exists n\colon \neg(f(n)\text{ is prime})$. Therefore, a counterexample is often the easiest and most straightforward way to disprove a general statement. In fact, any proof that merely shows that a $\forall$ statement is wrong without constructing a counterexample (or essentially doing so, but leaving out the details) is frowned upon by some mathematicians as being non-constructive. That being said, maybe you'd like a proof that there is no polynomial $f$ such that $f(n)$ is prime for all $n\in \mathbb N$ (except constant polynomials):
Let $$f(n)=a_dx^d+a_{d-1}x^{d-1}+\ldots+a_1x+a_0$$ be a polynomial of degree $d\ge 1$ with coefficients $a_i\in\mathbb C$, $0\le i\le d$, and $a_d\ne 0$. First we observe tat in fact $a_i\in \mathbb Q$ for all $i$: By plugging in $x=1, 2, \ldots, d+1$, we obtain $d+1$ rational equations in the unknowns $a_i$. Since the powers of $x$ are linearly independant (or look up Vandermonde matrix), there exists a unique solution, which must be rational. Then there exists $M\in\mathbb N$ sich that $a_iM\in\mathbb Z$ for all $i$. By assumption, $Mf(n)$ is $M$ times a prime for all $n\in\mathbb N$. For each prime $p$, there are at most $d$ values of $n$ such that $f(n)=p$. Since $M$ has only finitely many prime factors, $f(n)\mid M$ for only finitely many $n$. Therefore there exists $n$ such that $p:=f(n)\not\mid M$. By reducing $Mf(X)$ modulo $p$, we see that $Mf(n+kp)\equiv 0\pmod p$ for all $k\in\mathbb Z$. This implies that the prime $f(n+kp)$ is $p$ for all $k$. Especially, there are more than $d$ values of $n$ for which $f(n)=p$, contradiction. $_\square$
Trivia remark: On the other hand, there exists a polynomial in several variables with the property that its values are either negative or prime, and all primes occur!
• Your trivia remark is really cool! Do you have a link/reference? – André 3000 May 15 '14 at 15:22
• Here is a link to an early paper by Jones et al, not long after Matiyasevich proved existence. The $25$ variables has gone down a lot since then, I think to about $9$. – André Nicolas May 15 '14 at 15:38
• @AndréNicolas That's cool - so instead of letters we can now use digits to index the variables :) – Hagen von Eitzen Dec 7 '16 at 17:09
No. For $n=43$, $n^2+n+1=1893$ which is divisible by $3$.
• "What is wrong in the reasoning? What is a correct proof that a given polynomial $p(n)$ is (or isn't) prime for all natural numbers n?" Please read the entire question before answering. What I'm looking for is not a counterexample, but rather a proof. – Giulio Muscarello May 15 '14 at 14:32
• @GiulioMuscarello I understand you don't find this answer satisfying, but it's in fact perfectly valid proof. – Marcin Łoś May 15 '14 at 14:34
• @MarcinŁoś Sure, but the OP already had a much smaller counterexample with $n=4$. I don't see what this answer really adds. – André 3000 May 15 '14 at 14:38
• Jika is probably thinking of $n^2-n+41$, which is prime for $0 \le n \le 40$ – Robert Israel May 15 '14 at 15:06
• @RobertIsrael Yes you are right. Even though, I was thinking of Euler's formula $n^2-n+41$, I found that $43$ is a counter example. I hesitate to post my answer but since it is a valid mathematical proof I did post it. – Jika May 15 '14 at 15:15
Modulo 3, that polynomial is equivalent to some well known polynomial $(n-1)^2$ $$(n-1)^2=n^2-2n+1\equiv n^2+n+1 [3]$$
Hence if $n=1+3k$ for any $k$ (so each time $n-1\equiv 0[3]$) then $n^2+n+1$ will have 3 as a divisor. | 2020-02-21T19:18:00 | {
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https://math.stackexchange.com/questions/1604790/is-my-inductive-proof-correct | # Is my inductive proof correct?
Trying this again.
Given $f(n) = 2f(n-1) + 1$ with $f(0) = 0$, I guess that $f(n) = 2^n-1$.
Base case: $f(0) = 2^0 - 1 = 1 - 1 = 0$, true.
Inductive step: Suppose $f(n) = 2^n-1$ for some $n \geq 0$. I will show that $f(n+1) = 2^{n+1}-1$.
$f(n+1) = 2f(n) + 1$
$f(n+1) = 2(2^n-1) + 1$
$f(n+1) = 2^{n+1} - 1$
This completes the proof.
My questions:
1. Is this proof correct? Awkward? Backwards?
2. It would help me to get the terminology right. Which piece is the inductive hypothesis? Or the "ansatz"?
• This one is good, clean, "forward." Jan 8 '16 at 19:53
As far as terminology, the inductive hypothesis is the thing you assume is true at some $n$, and is used to prove the statement for $n + 1$. So your inductive hypothesis is that $f(n) = 2^n - 1$.
The ansatz is the educated guess that the solution is $2^n - 1$, likely based on some experimentation. The ansatz then became your induction hypothesis (as it is wont to do).
Looks good to me. The inductive hypothesis is the step where you assume $f(n)=2^n-1$. We assume the inductive hypothesis because we have shown the existence of such an $n$ in our base case. | 2021-10-17T09:26:27 | {
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https://math.stackexchange.com/questions/1306387/does-liebniz-criteria-become-a-necessary-condition-for-convergence-if-a-n-is-m | # Does Liebniz-Criteria become a necessary condition for convergence if $a_n$ is monotonically decreasing?
The Leibniz Criterion says that if the sequence $a_n$ is monotonically decreasing then the following statements are equivalent:
\begin{align} 1) & & & \sum_{n=0}^\infty (-1)^na_n \text{ converges} \\[6pt] 2) & & & \lim_{n\rightarrow\infty}(a_n) = 0 \end{align}
so this has to mean that if $a_n$ is monotonically decreasing, the following is true
$$\sum_{n=0}^\infty (-1)^na_n \text{ converges} \Leftrightarrow \lim_{n\rightarrow\infty}(a_n) = 0$$
I know that $\lim_{n\rightarrow\infty}(a_n) = 0$ is not a necessary condition for convergence if $a_n$ is not monotonically decreasing, but the way I am reading the rule stated in my script, it becomes a necessary condition if $a_n$ is monotonically decreasing.
For example to prove the non-convergence of
$$\sum_{n=0}^\infty (-1)^n(1)_n$$
I can just point out to the fact that
$$\lim_{n\rightarrow\infty}(1)_n = 1$$
because $(1)_n$ is a monotonically decreasing sequence. Is this interpretation correct?
• The condition $b_n\to 0$ is always necessary to the convergence of $\sum b_n$. When $b_n=(-1)^n a_n$, one has the equivalence $b_n\to 0\ \iff\ a_n\to 0$. So $a_n\to 0$ is a necessary condition for the convergence of $\sum (-1)^n a_n$. – Giuseppe Negro May 31 '15 at 11:49
• Yes the statement is true. Proof: $$b_n\to 0\ \iff\ \lvert b_n\rvert \to 0 \iff \lvert (-1)^n a_n\rvert\to 0 \iff a_n\to 0.$$ If you are not convinced please post a counterexample. – Giuseppe Negro May 31 '15 at 11:59
• if it were so why would leibniz go through the trouble of putting the additional monotonic decreasing condition there – Gravity May 31 '15 at 11:59
• I guess you are mixing up "necessary" and "sufficient" here. $a_n\to 0$ is necessary to the convergence of $\sum (-1)^na_n$. $a_n\to 0$ AND $a_n$ monotonically decreasing is sufficient to the convergence of $\sum (-1)^n a_n$. – Giuseppe Negro May 31 '15 at 12:01
• I know realize that you were right – Gravity May 31 '15 at 12:33
As Giuseppe Negro has noticed, for the convergence of $\displaystyle \sum a_n$,$$\lim_{n\rightarrow\infty}(a_n) = 0$$ is a necessary condition anyway.
The convergence test of Leibniz is only applicable to alternating series, i.e. series where every positive term is followed by a negative term, and every negative term is followed by a positve term ( + - + - + - + - ...).
For a general series, the condition $\displaystyle\lim_{n \to \infty} a_n= 0$ is a necessary condition for convergence, but not sufficient. This is the main point of Leibniz test. He could prove that for alternating series this condition is both necessary and sufficient. Thus if you've got an alternating series $\sum (-1)^n a_n$ with $\displaystyle\lim_{n \to \infty} a_n= 0$, than you are certain it converges.
I don't know why you are referring to the condition of "monotonically decreasing". This doesn't matter for Leibniz test. It's only an issue in the integral test.
• I understand your first point, but on your second point$\sum (-1)^n a_n$ with $\displaystyle\lim_{n \to \infty} a_n= 0$ you are certain that is converges, only when a is monotonically decreasing. The rule is defined for monotonically decreasing a – Gravity May 31 '15 at 12:36
• @Steven Van Geluwe: Alternation of signs and terms going to $0$ is not sufficient for convergence. Example, let $a_n=\frac{1}{n}$ if $n$ is odd, and let $a_n=\frac{1}{2^n}$ when $n$ is even. Then $\sum_1^\infty (-1)^n a_n$ diverges. – André Nicolas May 31 '15 at 12:52
• @Gravity: There are many series $\sum (-1)^n a_n$ that converge even though the $a_n$ are not monotonically decreasing. For example, let $a_n=\frac{1}{n^2}$ when $n$ is odd, and let $a_n=\frac{1}{n^3}$ when $n$ is even. – André Nicolas May 31 '15 at 12:59
• @Andre: that is correct and that is why Leibniz says the condition is sufficient but not necessary, but still we can only apply the Leibniz rule if $a_n$ is monotonically decreasing. we cannot in all Circumstances say $\sum (-1)^n a_n$ converges when $\displaystyle\lim_{n \to \infty} a_n= 0$ is true, and that is the point – Gravity May 31 '15 at 13:05
• I down voted, because I am afraid that the main point of this answer is wrong. Leibniz's test requires monotonicity of the $a_n$ term. – Giuseppe Negro May 31 '15 at 21:25 | 2019-08-21T22:29:30 | {
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https://math.stackexchange.com/questions/4169353/is-k-1-bipartite | # Is $K_1$ bipartite?
I'm thinking that it could be trivially bipartite since it only has one vertex and no edges but I am still a little bit unsure about it being trivially bipartite.
• Ummm... what is "k sub 1"? Do you instead mean $K_1$? – David G. Stork Jun 10 at 19:01
• not only is it bipartite, it is also unipartite – Yorch Jun 10 at 19:02
• @DavidG.Stork Yes, did not know how to make it like that in my title. Seemed to fix itself. – ThreeRingBinder Jun 10 at 20:23
Depends on your definitions. (As pointed out in Matthew Daly's answer, $$K_1$$ should be bipartite, because it has no odd cycles, so it would otherwise be an awkward exception. But not all definitions will play nicely with this desired property.)
West's Introduction to Graph Theory says
1.1.10. Definition. A graph $$G$$ is bipartite if $$V(G)$$ is the union of two disjoint (possibly empty) independent sets called partite sets of $$G$$.
So under this definition, if $$V(K_1) = \{v\}$$, then we let $$\{v\}$$ be one partite set, and $$\varnothing$$ be the other; $$K_1$$ is bipartite.
Bondy and Murty write
A graph is bipartite if its vertex set can be partitioned into two subsets $$X$$ and $$Y$$ so that every edge has one end in $$X$$ and one end in $$Y$$
which still works just fine, setting $$X = \{v\}$$ and $$Y = \varnothing$$. They do not specifically point out that $$X$$ or $$Y$$ could be empty, but they do not rule it out either. (Elsewhere, Bondy and Murty talk about "nontrivial partitions" or "partitions into nonempty parts", which make it clear that a partition without this qualifier is allowed to have an empty part.)
They are in better shape than Diestel, who has
A graph $$G = (V, E)$$ is called $$r$$-partite if $$V$$ admits a partition into $$r$$ classes such that every edge has its ends in different classes
with an earlier qualification that the classes of a partition may not be empty. Since $$K_1$$ should be bipartite by any reasonable definition, Diestel is in the wrong here. (Diestel later claims that all graphs with no odd cycles are bipartite, with no mention of $$K_1$$ as a special exception.)
If we treat "bipartite" as synonymous with "$$2$$-colorable", then $$K_1$$ is happily bipartite, since any function on its vertex set is a $$2$$-coloring (and also a $$1$$-coloring).
• Quick question. Lets say we have 𝑋={𝑣} and 𝑌=∅, in reference to Bondy and Murty's definition. Would the edge connecting both subsets have vertices that are the empty set? Ie one empty set from X and the other from Y – ThreeRingBinder Jun 11 at 4:36
• The definition says every edge has a certain property; it doesn't say there exist any such edges. Since $K_1$ has no edges, the definition holds. – Misha Lavrov Jun 11 at 5:24
• Ah, makes sense now. Thanks! – ThreeRingBinder Jun 11 at 6:51
My graph theory professor spent a while teaching on this, because it's a mess either way.
It probably should be trivially considered bipartite. Otherwise you have to put disclaimers in all of your theorems (like the fact that $$K_1$$ doesn't contain any odd cycles). But $$\{\{v\},\emptyset\}$$ is not a valid partition of the vertex set, since it contains the empty set as a member.
• A reasonable definition of "bipartite" does not run into trouble if one partite set is empty. – Misha Lavrov Jun 10 at 19:24
• @MishaLavrov On the other hand, requiring the parts of a partition to be non-empty isn't entirely unheard of either. So defining bipartite as a partition of the set of vertices into two can run into issues with $K_1$. – Arthur Jun 10 at 19:33
• Yeah, I think that Dr. Schäffer's solution was to have us write in the margin of Bondy and Murty "... or $K_1$". – Matthew Daly Jun 10 at 19:44
• But Bondy and Murty are totally fine with empty parts in a partition, as far as I can tell... – Misha Lavrov Jun 10 at 20:22
• @MishaLavrov As I say, my professor assumed that "partitioned into two subsets $X$ and $Y$" means that $\{X,Y\}$ was a partition of the vertex set, which could not be legally done if one of the sets was empty. Bondy and Murty clearly either thought that it was okay or they hadn't considered the edge case, but pedantry like that got under my professor's skin. ^_^ – Matthew Daly Jun 10 at 20:37 | 2021-06-23T18:07:57 | {
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https://math.stackexchange.com/questions/4230041/do-we-have-to-expand-the-brackets-before-applying-de-morgans-laws/4230048 | # Do we have to expand the brackets before applying De Morgan's Laws?
If we have the example:
$$\overline{(A + B + C)D}$$
Then can we apply De Morgan's Law as is, or do we first need to expand out the brackets?
If I expand the terms first, I get:
$$original: \overline{(A + B + C)D}$$ $$expanded: \overline{AD + BD + CD}$$ $$applying De Morgan's: \overline{AD}.\overline{BD}.\overline{CD}$$ $$simplifying: \overline{A}\overline{B}\overline{C}\overline{D}$$
In contrast, if I don't expand the brackets, I get something like:
$$original: \overline{(A + B + C)D}$$ $$applying De Morgan's: \overline{(A + B + C)}+\overline{D}$$ $$applying De Morgan's: (\overline{A} . \overline{B + C})+\overline{D}$$ $$applying De Morgan's: (\overline{A} . \overline{B} . \overline{C})+\overline{D}$$ $$simplifying: \overline{A} . \overline{B} . \overline{C}+\overline{D}$$
So you can see, I'm getting different answers. I'm curious which method is correct - I think the first seems correct, but perhaps the 2nd is correct or both are wrong.
If anyone is able to explain why one particular method is wrong, I'd appreciate it too.
• MathJax tip: when writing text in mathematics code, use the \text{} command to get proper spacing and formatting. For example, $$\text{applying De Morgan's} : \overline{(A + B + C)} + \overline{D}$$ produces$$\text{applying De Morgan's} : \overline{(A + B + C)} + \overline{D}$$ Aug 22, 2021 at 0:54
• Thanks. Now I will do that in future :) Aug 22, 2021 at 5:58
• @RandomUser123 You can also do that now. Questions and answers can be edited, and indeed should be edited if they can be improved. Aug 22, 2021 at 8:45
Both lead to the same result; your error is in the first one when you simplify from $$\overline{AD}.\overline{BD}.\overline{CD}$$ to $$\overline{A}\overline{B}\overline{C}\overline{D}$$.
If you apply De Morgan's law to each pair, you get: $$(\overline{A}+\overline{D})(\overline{B}+\overline{D})(\overline{C}+\overline{D}).$$ From here, you can apply the distributive law to get what you got by the second method in the next to last line. The simplifcation there is also incorrect because you cannot simply drop the parenthesis; you must apply the distributive property.
• Makes sense. I was viewing it as $$\overline{ADBDCD}$$ then just cancelling removing the unneccessary duplicates. Because the NOTs were over multiple variables, I shouldn't be able to do that. I also tried myself, expanding out those 3 sets of brackets and surpisingly managed to simplify down that mess to the correct answer...seeing that all of the brackets contain D' and using your suggestion of using the distributive law is much easier though...I guess it's a good way to check...since I worry if I use these shortcuts, I will apply it incorrectly/in a situation that's invalid or something Aug 22, 2021 at 6:17 | 2022-08-16T08:25:52 | {
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https://math.stackexchange.com/questions/2465803/whats-the-probability-that-a-given-permutation-has-exactly-k-fixed-points/2465817 | # What's the probability that a given permutation has exactly $k$ fixed points. [duplicate]
Given a random permutation $$\sigma \in S_n$$ from $$[n] \to [n]$$ in a uniform probability space, what is the probability that $$\sigma$$ has exactly $$k$$ fixed points for a given $$k$$ between $$1$$ and $$n$$?
In other words: what is the probability that $$\exists x_1 ,...,x_k \in [n] : \sigma (x_i) = x_i$$ for $$\ i\in \{1,...,k\}$$ and for every $$y \notin \{x_1 , ... , x_k\}$$ we get $$\sigma(y) \neq y$$.
I saw that $$\lim_{n \to \infty } prob(A_0) = e^{-1}$$ using Inclusion–exclusion principle and i belive that for a given k : $$\lim_{n \to \infty} prob(A_k) = \frac{e^{-1}}{k!}$$ but I am not sure how to show it.
*$$A_k$$ stands for the event "k".
• The number $D_{n,k}$ of permutations with exactly $k$ fixpoints of an $n$-set is called a rencontres number. – Jeppe Stig Nielsen Oct 10 '17 at 11:36
There are $\binom{n}{k}$ possibilities for the $k$ fixed points.
If they are established then there are $!(n-k)$ derangements for the other points.
That gives $\binom{n}{k}\left[!(n-k)\right]$ permutations with exactly $k$ fixed points on a total of $n!$ permutations.
Also we have the formula: $$!(n-k)=(n-k)!\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$$ and we end up with probability: $$\frac1{k!}\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$$
• Very cool! As $n$ grows (for fixed $k$), this approaches $\frac1{k! e}$ – MichaelChirico Nov 6 '19 at 11:31
• And also by corollary, the probability of at least one fixed point approaches $1-\frac1{e}$ – MichaelChirico Nov 6 '19 at 11:38
1. Number of dearrangements of $k$-elements set is $$!k = k!\sum\limits_{m=0}^{k}\frac{(-1)^m}{m!}$$
2. In n-elements set we can select $(n-k)$ elements to dearrange them (all remaining $k$ points are fixed) in $\binom{n}{k}$ ways
Thus $$A_k^n = \binom{n}{k}(n-k)!\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}$$ And probability is $$P(A_k^n) = \frac{\binom{n}{k}(n-k)!\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}}{n!}=\frac{\frac{n!}{k!}\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}}{n!}=\frac{1}{k!}\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}$$
By way of enrichment here is an alternate formulation using combinatorial classes. The class of permutations with fixed points marked is
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\mathcal{U} \times \textsc{CYC}_{=1}(\mathcal{Z}) + \textsc{CYC}_{=2}(\mathcal{Z}) + \textsc{CYC}_{=3}(\mathcal{Z}) + \cdots).$$
This gives the generating function $$G(z, u) = \exp\left(uz + \frac{z^2}{2} + \frac{z^3}{3} + \frac{z^4}{4} + \frac{z^5}{5} + \cdots\right)$$
which is
$$G(z, u) = \exp\left(uz-z+\log\frac{1}{1-z}\right) = \frac{\exp(uz-z)}{1-z}.$$
Now for $$k$$ fixed points we get
$$[u^k] \frac{\exp(uz-z)}{1-z} = [u^k] \frac{\exp(uz)\exp(-z)}{1-z} = \frac{z^k}{k!} \frac{\exp(-z)}{1-z}.$$
This is the EGF of permutations having $$k$$ fixed points. We extract the count by computing (the factor $$n!$$ is canceled because we require the average)
$$[z^n] \frac{z^k}{k!} \frac{\exp(-z)}{1-z} = \frac{1}{k!} [z^{n-k}] \frac{\exp(-z)}{1-z}.$$
We find
$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{k!} \sum_{q=0}^{n-k} \frac{(-1)^q}{q!}.}$$
We can identify this as choosing the $$k$$ fixed points and combining them with a derangement of the rest:
$$\frac{1}{n!} {n\choose k} (n-k)! \sum_{q=0}^{n-k} \frac{(-1)^q}{q!}$$
which is the combinatorial class
$$\textsc{SET}_{=k}(\mathcal{Z}) \times \textsc{SET}(\textsc{CYC}_{\ge 2}(\mathcal{Z})).$$ | 2021-08-01T23:16:08 | {
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https://math.stackexchange.com/questions/3600695/show-that-sqrt1x1-fracx2-for-all-x0/3600740 | # Show that $\sqrt{1+x}<1+\frac{x}{2}$ for all $x>0$
I am a little stuck on this question and would appreciate some help. The question asks me to prove that $$\sqrt{1+x}<1+\frac{x}{2}$$ for all $$x>0$$.
I squared both sides of the question to get $$1+x<\frac{x^2}{4}+x+1$$ for all $$x>0$$. Then, I multiplied both sides by $$4$$ to get $$4+4x for all $$x>0$$.
I am a little stuck and was wondering what to do after this step and how to actually provide sufficient proof to say that this statement is true.
You did well. Let's finish it: we have $$x^2>0$$ thus $$\dfrac{x^2}{4}>0$$. Adding $$x+1$$ to both sides, we have $$\dfrac{x^2}{4}+x+1> x+1$$ or $$(\frac{x}{2}+1)^2>x+1$$ or $$\frac{x}{2}+1>\sqrt{x+1}$$
Squaring the equation was sufficient. After you cancel the $$1+x$$ on each side, you have
$$\frac{x^2}{4} > 0$$
which is true for all real $$x \ne 0$$ since $$x^2 \ge 0$$ (with equality only when $$x=0$$). Thus, the inequality is proved.
You can also prove the inequality using the Mean Value Theorem. Define $$f(t)=\sqrt{t+1}$$ and apply the theorem on the interval $$[0,x]$$. Then, there exists $$c\in(0,x)$$ s.t. $$\frac{\sqrt{x+1}-\sqrt{1}}{x-0}=\frac{1}{2\sqrt{c+1}}>\frac{1}{2}.$$
I am answering less to provide a list of steps to follow for proving the inequality, and more to discuss how I believe that the argument should be presented. This seems appropriate, as the question has been tagged with and . If I asked a student to prove some inequality and their first step was to assume the inequality and square both sides, I would almost certainly deduct points unless the student were careful about justifying such a step. The goal here is to start with known true statements and to then deduce the desired result.
Claim: For any $$x > 0$$, $$\sqrt{1+x} < 1 + \frac{x}{2}.$$
Proof: Assume that $$x > 0$$. The square of any real number is positive, hence $$0 < \left( \frac{x}{2} \right)^2.$$ By additive cancelation (i.e. by "adding the same number to both sides"), this implies that $$x + 1 < \left( \frac{x}{2} \right)^2 + x + 1 = \left( \frac{x}{2} + 1 \right)^2.$$ The square root function is increasing on $$[0,\infty)$$ (that is, if $$0 \le a < b$$, then $$0 \le \sqrt{a} \le \sqrt{b}$$), and the assumption that $$x > 0$$ ensures that $$1+x > 0$$, thus $$1 + x$$ is in the domain of the square root function (i.e. it has a well-defined real square root). Hence $$\sqrt{x+1} < \sqrt{ \left( 1 + \frac{x}{2} \right)^2 } = \left|1 + \frac{x}{2}\right| = 1 + \frac{x}{2},$$ where the final equality follows from the assumption that $$x > 0$$.
### Discussion
Remember that if you start with false assumptions, you can prove anything—false implies true. Thus it is poor practice to begin a proof by asserting the statement which you are trying to prove and then applying algebraic manipulation. You can do this if you are careful—be very observant of two-sided implications and check the hypotheses at each step. Indeed, this is a very reasonable way of deducing a correct proof in the first place.
However, when presenting an argument, such a proof is typically poor style. Personally, I think that it is better (stylistically, which is a matter of taste) for every statement in a proof to follow from previous statement, and not depend on future statements via "if and only ifs".
I will also note that the proof presented above is very elementary—it doesn't rely on any deep theorems. It is, however, somewhat tedious. More powerful theorems give quicker, more elegant proofs. For example, El31's proof, via the mean value theorem, is quite slick. However, when first learning a topic, I think that one should focus on finding an elementary proof first: find such a proof first. Then, if need be, look for something more elegant.
The important principle here is that if $$a,b$$ are nonnegative real numbers, then $$a iff $$a^2 This property is used all the time. If we grant this, then all we need to show is that
$$(\sqrt{1+x})^2 < (1+x/2)^2,$$
which is the same as saying $$1+x < 1+x + x^2/4.$$ That is obviously true, and we're done.
Observe that : $$\left(\forall x>0\right),\ \sqrt{1+x}-1-\frac{x}{2}=-\frac{x^{2}}{4}\int_{0}^{1}{\frac{1-t}{\left(1+tx\right)\sqrt{1+tx}}\,\mathrm{d}t} < 0$$
• No offense meant, but this is perhaps one of the most obtuse correct answers I've seen on this site. – jawheele Mar 30 at 5:51 | 2020-07-05T01:45:31 | {
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http://mathhelpforum.com/trigonometry/67569-solve-equations-0-x-2pi.html | # Math Help - Solve Equations 0< or = x < or = 2pi
1. ## Solve Equations 0< or = x < or = 2pi
How would you solve these
Solve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.
Answear is in form of this
pi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?
2. Originally Posted by j0nath0n3
How would you solve these
Solve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.
Answear is in form of this
pi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?
Note that $\sin^2 x=1-\cos^2 x$
So we can rewrite the equation as $\cos^2 x -\left[1-\cos^2 x\right]=-\cos x$
This simplifies to $2\cos^2 x-1=-\cos x\implies 2\cos^2x+\cos x-1=0$
Now, if we make the substitution $z=\cos x$, the equation now becomes the quadratic equation $2z^2+z-1=0\implies (2z-1)(z+1)=0$
This tells us that $z=-1$ or $z=\tfrac{1}{2}$
But $z=\cos x$, so now that means either $\cos x=\tfrac{1}{2}$ or $\cos x=-1$
Now can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?
3. Originally Posted by Chris L T521
Note that $\sin^2 x=1-\cos^2 x$
So we can rewrite the equation as $\cos^2 x -\left[1-\cos^2 x\right]=-\cos x$
This simplifies to $2\cos^2 x-1=-\cos x\implies 2\cos^2x+\cos x-1=0$
Now, if we make the substitution $z=\cos x$, the equation now becomes the quadratic equation $2z^2+z-1=0\implies (2z-1)(z+1)=0$
This tells us that $z=-1$ or $z=\tfrac{1}{2}$
But $z=\cos x$, so now that means either $\cos x=\tfrac{1}{2}$ or $\cos x=-1$
Now can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?
Thats really good and I understood however my answear choices are in radian... and I dont understand why..
4. Originally Posted by j0nath0n3
Thats really good and I understood however my answear choices are in radian... and I dont understand why..
In general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is $360\text{ degrees}=2\pi\text{ radians}$.
5. ## Re :
Originally Posted by Chris L T521
In general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is $360\text{ degrees}=\pi\text{ radians}$.
Chris , is it a typo , i thought $360\text{ degrees}=2\pi\text{ radians}$
Chris , is it a typo , i thought $360\text{ degrees}=2\pi\text{ radians}$ | 2015-05-30T00:30:42 | {
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https://fr.bluerocktel.com/tj/30098162dca83fae7b8fd2b53a39a9c711 | In fact, permutation is another term used to describe bijective functions from a finite set to itself. a permutation is an arrangement of $n$ elements without repetition. A big part of discrete mathematics is about counting things. Permutations with Repetition | Discrete Mathematics. Combinations - Permutions. P R (4, 2) = 4 2 = 16. A phone number is an example of a ten number permutation; it is drawn from the set of the integers 0-9, and the order in which they are arranged in matters. I Pascal's triangle is perfectly symmetric I Numbers on left are mirror image of numbers on right I Why is this the case? IApermutationof a set of distinct objects is anordered arrangement of these objects. Likewise, [triangle, melon, airplane] is a permutation of three objects as well. 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Introduction Permutations Discrete Mathematics Andrei Bulatov Discrete Mathematics - Permutations What is Combinatorics Combinatorics, the Study Resources Main Menu A permutation is an arrangement in a definite order of a number of objects taken, some or all at a time. Wolfram|Alpha is useful for counting, generating and doing algebra with permutations. Malte Helmert, Gabriele R oger (University of Basel)Discrete Mathematics in Computer Science October 21, 2020 17 / 20 B9. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements.The word "permutation" also refers to the act or process of changing the linear order of an ordered set. A graph is determined as a mathematical structure that represents a particular function by connecting a set of points. A finite sequence of length$ n \$ in which all the elements are different, i.e. Permutations Permutations Cyclic Permutation De nition (Cyclic Permutation) A permutation iscyclicif it has a single k-cycle with k >1. permutations and combinations is the another topic included in discrete mathematics which also refers to the finite calculations. 100 Units. Permutations of Objects Ppt - Free download as Powerpoint Presentation (.ppt), PDF File (.pdf), Text File (.txt) or view presentation slides online. permutation arrangement An r-permutation of n objects is an ordered arrangement of r objects from the n objects A permutation is a (possible) rearrangement of objects. Rolling Dice. Question and Answers related to Discrete Mathematics Circular Permutations. For example, the number of insurance claims within a population for a certain type of risk would be zero-inflated by those people who have not taken out insurance against the risk and thus are unable to claim. | 2022-12-08T15:27:02 | {
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http://aimath.org/textbooks/beezer/LCsection.html | In Section VO:Vector Operations we defined vector addition and scalar multiplication. These two operations combine nicely to give us a construction known as a linear combination, a construct that we will work with throughout this course.
## Linear Combinations
Definition LCCV (Linear Combination of Column Vectors) Given $n$ vectors $\vectorlist{u}{n}$ from $\complex{m}$ and $n$ scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_n$, their linear combination is the vector \begin{equation*} \lincombo{\alpha}{u}{n} \end{equation*}
So this definition takes an equal number of scalars and vectors, combines them using our two new operations (scalar multiplication and vector addition) and creates a single brand-new vector, of the same size as the original vectors. When a definition or theorem employs a linear combination, think about the nature of the objects that go into its creation (lists of scalars and vectors), and the type of object that results (a single vector). Computationally, a linear combination is pretty easy.
Example TLC: Two linear combinations in $\complex{6}$.
Example ABLC: Archetype B as a linear combination.
With any discussion of Archetype A or Archetype B we should be sure to contrast with the other.
Example AALC: Archetype A as a linear combination.
There's a lot going on in the last two examples. Come back to them in a while and make some connections with the intervening material. For now, we will summarize and explain some of this behavior with a theorem.
Theorem SLSLC (Solutions to Linear Systems are Linear Combinations) Denote the columns of the $m\times n$ matrix $A$ as the vectors $\vectorlist{A}{n}$. Then $\vect{x}$ is a solution to the linear system of equations $\linearsystem{A}{\vect{b}}$ if and only if $\vect{b}$ equals the linear combination of the columns of $A$ formed with the entries of $\vect{x}$, \begin{equation*} \vectorentry{\vect{x}}{1}\vect{A}_1+ \vectorentry{\vect{x}}{2}\vect{A}_2+ \vectorentry{\vect{x}}{3}\vect{A}_3+ \cdots+ \vectorentry{\vect{x}}{n}\vect{A}_n = \vect{b} \end{equation*}
In other words, this theorem tells us that solutions to systems of equations are linear combinations of the $n$ column vectors of the coefficient matrix ($\vect{A}_j$) which yield the constant vector $\vect{b}$. Or said another way, a solution to a system of equations $\linearsystem{A}{\vect{b}}$ is an answer to the question "How can I form the vector $\vect{b}$ as a linear combination of the columns of $A$?" Look through the archetypes that are systems of equations and examine a few of the advertised solutions. In each case use the solution to form a linear combination of the columns of the coefficient matrix and verify that the result equals the constant vector (see exercise LC.C21).
## Vector Form of Solution Sets
We have written solutions to systems of equations as column vectors. For example Archetype B has the solution $x_1 = -3,\,x_2 = 5,\,x_3 = 2$ which we now write as \begin{equation*} \vect{x}=\colvector{x_1\\x_2\\x_3}=\colvector{-3\\5\\2} \end{equation*} Now, we will use column vectors and linear combinations to express all of the solutions to a linear system of equations in a compact and understandable way. First, here's two examples that will motivate our next theorem. This is a valuable technique, almost the equal of row-reducing a matrix, so be sure you get comfortable with it over the course of this section.
Example VFSAD: Vector form of solutions for Archetype D.
This is such an important and fundamental technique, we'll do another example.
Example VFS: Vector form of solutions.
Did you think a few weeks ago that you could so quickly and easily list all the solutions to a linear system of 5 equations in 7 variables?
We'll now formalize the last two (important) examples as a theorem.
Theorem VFSLS (Vector Form of Solutions to Linear Systems) Suppose that $\augmented{A}{\vect{b}}$ is the augmented matrix for a consistent linear system $\linearsystem{A}{\vect{b}}$ of $m$ equations in $n$ variables. Let $B$ be a row-equivalent $m\times (n+1)$ matrix in reduced row-echelon form. Suppose that $B$ has $r$ nonzero rows, columns without leading 1's with indices $F=\set{f_1,\,f_2,\,f_3,\,\ldots,\,f_{n-r},\,n+1}$, and columns with leading 1's (pivot columns) having indices $D=\set{d_1,\,d_2,\,d_3,\,\ldots,\,d_r}$. Define vectors $\vect{c}$, $\vect{u}_j$, $1\leq j\leq n-r$ of size $n$ by
\begin{align*} \vectorentry{\vect{c}}{i}&= \begin{cases} 0&\text{if $i\in F$}\\ \matrixentry{B}{k,n+1}&\text{if $i\in D$, $i=d_k$} \end{cases}\\ \vectorentry{\vect{u}_j}{i}&= \begin{cases} 1&\text{if $i\in F$, $i=f_j$}\\ 0&\text{if $i\in F$, $i\neq f_j$}\\ -\matrixentry{B}{k,f_j}&\text{if $i\in D$, $i=d_k$} \end{cases}. \end{align*}
Then the set of solutions to the system of equations $\linearsystem{A}{\vect{b}}$ is \begin{equation*} S=\setparts{ \vect{c}+\alpha_1\vect{u}_1+\alpha_2\vect{u}_2+\alpha_3\vect{u}_3+\cdots+\alpha_{n-r}\vect{u}_{n-r} }{ \alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_{n-r}\in\complex{\null} } \end{equation*}
Note that both halves of the proof of Theorem VFSLS indicate that $\alpha_i=\vectorentry{\vect{x}}{f_i}$. In other words, the arbitrary scalars, $\alpha_i$, in the description of the set $S$ actually have more meaning --- they are the values of the free variables $\vectorentry{\vect{x}}{f_i}$, $1\leq i\leq n-r$. So we will often exploit this observation in our descriptions of solution sets.
Theorem VFSLS formalizes what happened in the three steps of Example VFSAD. The theorem will be useful in proving other theorems, and it it is useful since it tells us an exact procedure for simply describing an infinite solution set. We could program a computer to implement it, once we have the augmented matrix row-reduced and have checked that the system is consistent. By Knuth's definition, this completes our conversion of linear equation solving from art into science. Notice that it even applies (but is overkill) in the case of a unique solution. However, as a practical matter, I prefer the three-step process of Example VFSAD when I need to describe an infinite solution set. So let's practice some more, but with a bigger example.
Example VFSAI: Vector form of solutions for Archetype I.
This technique is so important, that we'll do one more example. However, an important distinction will be that this system is homogeneous.
Example VFSAL: Vector form of solutions for Archetype L.
## Particular Solutions, Homogeneous Solutions
The next theorem tells us that in order to find all of the solutions to a linear system of equations, it is sufficient to find just one solution, and then find all of the solutions to the corresponding homogeneous system. This explains part of our interest in the null space, the set of all solutions to a homogeneous system.
Theorem PSPHS (Particular Solution Plus Homogeneous Solutions) Suppose that $\vect{w}$ is one solution to the linear system of equations $\linearsystem{A}{b}$. Then $\vect{y}$ is a solution to $\linearsystem{A}{b}$ if and only if $\vect{y}=\vect{w}+\vect{z}$ for some vector $\vect{z}\in\nsp{A}$.
After proving Theorem NMUS we commented (insufficiently) on the negation of one half of the theorem. Nonsingular coefficient matrices lead to unique solutions for every choice of the vector of constants. What does this say about singular matrices? A singular matrix $A$ has a nontrivial null space (Theorem NMTNS). For a given vector of constants, $\vect{b}$, the system $\linearsystem{A}{b}$ could be inconsistent, meaning there are no solutions. But if there is at least one solution ($\vect{w}$), then Theorem PSPHS tells us there will be infinitely many solutions because of the role of the infinite null space for a singular matrix. So a system of equations with a singular coefficient matrix never has a unique solution. Either there are no solutions, or infinitely many solutions, depending on the choice of the vector of constants ($\vect{b}$).
Example PSHS: Particular solutions, homogeneous solutions, Archetype D.
The ideas of this subsection will be appear again in Chapter LT:Linear Transformations when we discuss pre-images of linear transformations (Definition PI). | 2013-05-25T20:44:13 | {
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https://byjus.com/question-answer/obtain-all-other-zeroes-of-3x-4-6x-3-2x-2-10x-5-if-two-2/ | Question
# Obtain all other zeroes of $$3x^4+6x^3-2x^2-10x-5$$, if two of its zeros are $$\sqrt{\dfrac{5}{3}}$$ and $$-\sqrt{\dfrac{5}{3}}$$.
Solution
## Since the two zeroes of the given polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ are $$\sqrt { \dfrac { 5 }{ 3 } } ,-\sqrt { \dfrac { 5 }{ 3 } }$$, therefore, $$\left( x-\sqrt { \dfrac { 5 }{ 3 } } \right) \left( x+\sqrt { \dfrac { 5 }{ 3 } } \right) =x^{ 2 }-\dfrac { 5 }{ 3 }$$ is a factor of the given polynomial.now, we divide the polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ by $$x^{ 2 }-\dfrac { 5 }{ 3 }$$ as shown below:Therefore, $$3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5=\left( x^{ 2 }-\dfrac { 5 }{ 3 } \right) (3x^{ 2 }+6x+3)$$Now, we factorize $$3x^{ 2 }+6x+3$$ as follows:$$3x^{ 2 }+6x+3=0\\ \Rightarrow 3(x^{ 2 }+2x+1)=0\\ \Rightarrow x^{ 2 }+2x+1=0\\ \Rightarrow (x+1)^{ 2 }=0\quad \quad \quad \quad (\because \quad (a+b)^{ 2 }=a^{ 2 }+b^{ 2 }+2ab)\\ \Rightarrow (x+1)(x+1)=0\\ \Rightarrow x+1=0,\quad x+1=0\\ \Rightarrow x=-1,\quad x=-1$$Hence, the zeroes of the polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ are $$\sqrt { \dfrac { 5 }{ 3 } } ,-\sqrt { \dfrac { 5 }{ 3 } } ,-1,-1$$.Maths
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https://codeforces.cc/blog/entry/91707 | ### TheScrasse's blog
By TheScrasse, history, 7 weeks ago,
Hello everyone,
here is a very simple idea that can be useful for (cp) number theory problems, especially those concerning multiples, divisors, $\text{GCD}$ and $\text{LCM}$.
Prerequisites: basic knowledge of number theory (divisibility, $\text{GCD}$ and $\text{LCM}$ properties, prime sieve).
## Idea
Let's start from a simple problem.
You are given $n$ pairs of positive integers $(a_i, b_i)$. Let $m$ be the maximum $a_i$. For each $k$, let $f(k)$ be the sum of the $b_i$ such that $k | a_i$. Output all pairs $(k, f(k))$ such that $f(k) > 0$.
An obvious preprocessing is to calculate, for each $k$, the sum of the $b_i$ such that $a_i = k$ (let's denote it as $g(k)$). Then, there are at least $3$ solutions to the problem.
#### Solution 1: $O(m\log m)$
For each $k$, $f(k) = \sum_{i=1}^{\lfloor m/k \rfloor} g(ik)$. The complexity is $O\left(m\left(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{m}\right)\right) = O(m\log m)$.
#### Solution 2: $O(n\sqrt m)$
There are at most $n$ nonzero values of $g(k)$. For each one of them, find the divisors of $k$ in $O(\sqrt k)$ and, for each divisor $i$, let $f(i) := f(i) + g(k)$.
If $m$ is large, you may need to use a map to store the values of $f(k)$ but, as there are $O(n\sqrt[3] m)$ nonzero values of $f(k)$, the updates have a complexity of $O(n\sqrt[3] m \log(nm)) < O(n\sqrt m)$.
#### Solution 3: $O(m + n\sqrt[3] m)$
Build a linear prime sieve in $[1, m]$. For each nonzero value of $g(k)$, find the prime factors of $k$ using the sieve, then generate the divisors using a recursive function that finds the Cartesian product of the prime factors. Then, calculate the values of $f(k)$ like in solution 2.
Depending on the values of $n$ and $m$, one of these solutions can be more efficient than the others.
Even if the provided problem seems very specific, the ideas required to solve that task can be generalized to solve a lot of other problems.
Hint 1
Hint 2
Hint 3
Solution
## agc038_c - LCMs
Hint 1
Hint 2
Hint 3
Solution
Implementation (C++)
## abc191_f - GCD or MIN
Hint 1
Hint 2
Hint 3
Hint 4
Solution
Implementation (C++)
## Conclusions
We've seen that this technique is very flexible. You can choose the complexity on the basis of the constraints, and $f(k)$ can be anything that can be updated fast.
Of course, suggestions/corrections are welcome. In particular, please share in the comments other problems that can be solved with this technique.
I hope you enjoyed the blog!
• +105
» 7 weeks ago, # | +15 Nice blog! I like solving problems which use such ideas, so here are a few more problems with a similar flavour:1436F - Sum Over SubsetsChefsums1493D - GCD of an Array
» 10 days ago, # | +8 The Solution spoiler for 1154G - Minimum Possible LCM seems to be broken, I see this:
• » » 10 days ago, # ^ | +6 Fixed. I don't know the reason of the failed parsing, though. The original text was Spoiler$\text{GCD}(a_i, a_j) = h$.and I had to change it to Spoiler$\text{GCD}(a_i, a_j) = h$ .
» 10 days ago, # | +13 Solution 1 can be easily optimized to $O(m \log{\log{m}})$ using the same idea as calculating sums-over-submasks. The code looks something like this: vector f = g; for (int p : primes) for (int i = m / p; i > 0; --i) f[i] += f[i*p]; I suspect there are very few (if any) problems in practice for which solution 3 is directly applicable, but this optimized solution 1 is not. | 2021-07-31T12:03:11 | {
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https://www.analyzemath.com/linear-algebra/spaces/testing-for-linearity-of-vectors.html | # Testing for Linearity of Vectors in a Subspace - Examples with Solutions
Given a set of vectors in a subspace, how do we test whether the vectors are linearly independent ?
## What are Linearly Dependent or Independent Vectors?
We first use vectors in two and three dimensional spaces to visualize the concept of dependence and independence of vectors.
In figure 1 below we have have two vectors that are parallel such that $v_2 = 2 v_1$. We say that these two vectors are dependent because we can express one vector in terms of the other as follows:
$v_2 = 2 v_1$ or $v_1 = \dfrac{1}{2} v_1$.
In figure 2 below we have have two vectors that are parallel such that $v_2 = - v_1$. These two vectors are dependent because we can express one vector in terms of the other.
In figure 3, all vectors are parallel to each other. All These vectors are dependent because we can express one vector in terms of the other(s).
In figure 4, we can never express one vector in terms of the other because they are not parallel. These vectors are independent because we cannot express one vector in terms of the other.
In figure 5, using the geometrical sums of vectors, we can write
$3 v_4 = 2 v_1 + 3 v_2 + v_3$
and therefore these vectors are linearly dependent because we can express one vector in terms of the others.
In figure 6, the 3 vectors $v_1$ , $v_2 \}$ and $v_3$ are in the same plane $P$ and are therefore dependent because we can express any of these vectors in term of the other two vectors using linear combinations.
In figure 7, the pairs of vectors $\{v_1 , v_2\}$ , $\{v_2 , v_3\}$ and $\{v_1 , v_3\}$ are in different planes and are therefore independent because we cannot express any of these vectors in term of the other two vectors using linear combinations.
## Formal Definiton of Linearity of Vectors
Vectors $v_1, v_2 .... v_n$ are linearly dependent when at least one of the vectors may be expressed as a linear combinations of the remaining vectors as follows
$v_1 = r_2 v_2 + r_3 v_3 + .... + r_n v_n$
Writing the above with the zero on the right side, we obtain
$v_1 - r_2 v_2 - r_3 v_3 - .... - r_n v_n = 0$
Hence the following definition
Given a set of vectors $S = \{\textbf{v}_1 , \textbf{v}_2, ... , \textbf{v}_n \}$ ,
If the equation
$r_1 \textbf{v}_1 + r_2 \textbf{v}_2 + ... + r_n\textbf{v}_n = \textbf{0}$ (I)
has only one trivial solution $r_1 = 0 , r_2 = 0 , ... , r_n = 0$, we say that $S$ is a set of linearly independent vectors.
If the above equation has other solutions where not all $r_i$ equal to zero, then $S$ is a set of linearly dependent vectors.
In the examples below, matrices are row reduced in order to test for linearity. This may done using the row reduce augmented matrices calculator included.
## Examples with Solutions
Example 1
Are the vectors in the set $\left \{ \begin{bmatrix} -2 \\ 1 \end {bmatrix} , \begin{bmatrix} 6 \\ -3 \end {bmatrix} \right \}$ linearly independent?
Solution to Example 1
Let $v_1 = \begin{bmatrix} -2 \\ 1 \end {bmatrix}$ and $v_2 = \begin{bmatrix} 6 \\ -3 \end {bmatrix}$
The question could be answered by noticing that $v_2 = - 3 v_1$ and therefore the given vectors $v_1$ and $v_2$ are dependent (not independent)
The above was possible because we are dealing with vectors of small dimension. (2 components only)
We will now use a method that may be applied in any situation.
According to the definition given above, we need to find $r_1$ and $r_2$ such that
$r_1 v_1 + r_2 v_2 = 0$
The above may be written in matrix form as follows
$[ v_1 \;\; v_2] \begin{bmatrix} r_1 \\ r_2 \end {bmatrix} = 0$
where $[ v_1 \;\; v_2]$ is a matrix whose columns are $v_1$ and $v_2$
Write the system of equations in augmented matrix form
$\begin{bmatrix} -2 &6&|&0\\ 1 & -3&|&0 \end{bmatrix}$
Use the Gauss Jordan method to row reduce the above augmented matrix and obtain
$\begin{bmatrix} 1 &-3&|&0\\ 0 & 0&|&0 \end{bmatrix}$ (I)
The solution to the above reduced system (which is also a solution to the system before reduction) is found as follows
The second equation gives: $0 r_2 = 0$ therefore $r_2$ may take any real value
The first equation gives: $r_1 = 3 r_2$
The solution set may be written as: $\left \{ \begin{bmatrix} 3 r_2\\ r_2 \end{bmatrix} \right \} , r_2 \in R$
which therefore means that we have an infinite number of solutions and therefore the two vectors are linearly dependent.
Note that you do not have to solve the system in order to decide whether the given vectors are dependent or independent.
Construct the augmented matrix using the vectors as columns of the matrix and the constant column on the right is all zeros which in fact may be omitted. We then row reduce the augmented matrix. Then the following conclusions may easily be drawn:
1) The columns with a pivot are independent of each other
2) The columns with no pivots are dependent on the ones with the pivot.
3) The coefficients in the columns without pivot gives the coefficients of dependence on the independent columns.
Let us apply the above to the reduced matrix (I) above
1) Column 1 has a pivot and may be considered as independent
2) Column 2 has no pivot and may therefore be considered as dependent on column (1)
3) The coefficient $- 3$ in column 2 is telling us that $v_2 = - 3 v_1$
Example 2
Are the vectors in the set $\left \{ \begin{bmatrix} -2 \\ 1 \\ 4 \end {bmatrix} , \begin{bmatrix} 1\\ 0 \\ 5 \end {bmatrix} , \begin{bmatrix} 1\\ 2 \\ -1 \end {bmatrix} \right \}$ linearly independent?
Solution to Example 2
step 1: Construct the augmented matrix
Construct the augmented matrix whose columns are the given vectors and zeros on the right column
$\begin{bmatrix} -2 &1&1&|&0\\ 1 & 0&2&|&0 \\ 4 & 5 & -1& |& 0 \end{bmatrix}$
step 2: Row reduce the above matrix (you may use the row reduce augmented matrices calculator included).
$\begin{bmatrix} 1 &0&0&|&0\\ 0 & 1&0&|&0 \\ 0 & 0 & 1& |& 0 \end{bmatrix}$
step 3: Draw conclusions from the row reduced matrix
All 3 columns have a pivot each and therefore all 3 given vectors are independent
Example 3
Are the vectors in the set $\left \{ \begin{bmatrix} 2 \\ -1 \\ 3\\ 1 \end {bmatrix} , \begin{bmatrix} 0 \\ 2 \\ -1\\ 1 \end {bmatrix} , \begin{bmatrix} 4 \\ -8 \\ 9\\ -1 \end {bmatrix} \right \}$ linearly independent?
Solution to Example 3
step 1: Construct the augmented matrix using the given vectors as columns and zeros on the right column
$\begin{bmatrix} 2&0&4&|&0\\ -1&2&-8&|&0\\ 3&-1&9&|&0\\ 1&1&-1&|&0 \end{bmatrix}$
step 2: Row reduce the above matrix
$\begin{bmatrix} 1&0&\color{red}{2}&|&0\\ 0&1&\color{blue}{-3}&|&0\\ 0&0&0&|&0\\ 0&0&0&|&0 \end{bmatrix}$
step 3: Draw conclusions
Only the first and second columns (from the left) have a pivot and therefore the given vectors are not independent.
The coefficients $\color{red}{2}$ and $\color{blue}{-3}$ in the third column give the dependence of the original third columns as a linear combination of the other first and the second columns as follows
$\begin{bmatrix} 4 \\ -8 \\ 9\\ -1 \end {bmatrix} = \color{red}{2} \begin{bmatrix} 2 \\ -1 \\ 3\\ 1 \end {bmatrix} - \color{blue}{3} \begin{bmatrix} 0 \\ 2 \\ -1\\ 1 \end {bmatrix}$ | 2022-12-03T02:09:48 | {
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https://math.stackexchange.com/questions/1849840/big-list-of-erd%C5%91s-elementary-proofs/1849944 | # Big List of Erdős' elementary proofs
Paul Erdős was one of the greatest mathematicians of all time and he was famous for his elegant proofs from The Book. I posted a question about one of his theorem and got a reference, and I have other questions I want to know the answer to too. But, instead of requesting a reference for each theorem he gave with an elementary proof, I've decided to make a thread for a big list of all his elementary proofs.
I'm excited. Let's make an index of the pages of the Book shown to us!
To get you guys started, I will make a wish list of his theorems who's references I want to see. I encourage you to add to my wish list if you so desire.
Wish list :
• The product of two or more consecutive positive integers is never a square or any other higher power.
• A connected graph with a minimum degree $d$ and at least $2d+1$ vertices has a path of length at least $2d+1$.
• Let $d(n)$ be the number of divisors of $n$. Then the series $\sum_{n=1}^\infty d(n)/2^n$ converges to an irrational number
• Let $g(n)$ be the minimal number of points in the general position in the plane needed to ensure a subset exists that forms a convex $n$-gon. Then $$2^{n-2} + 1 \leq g(n) \leq \frac{(2n-4)!}{(n-2)!^2} + 1$$
• Erdős-Mordell inequality
• You need to define what you mean by "elementary". I think the list of elementary proofs will be very big, e.g. it will contain all his results in set theory, all his results on finite and infinite graphs. – bof Jul 6 '16 at 9:31
• @bof Hi bof. It will be big but I don't think all of them will be posted here. For starters, I'm looking for the questions in the wish list. Please post if you have any results – user230452 Jul 6 '16 at 9:35
• Is it fair that 4 people get to close a thread when 50 people have upvoted it ? Please open the thread for further posts – user230452 Jul 7 '16 at 10:59
• I voted to reopen. – Spenser Jul 7 '16 at 13:00
• @Spenser Thanks, Spencer. I appreciate it. I have collected another theorem of Erdos' to post. – user230452 Jul 7 '16 at 13:24
I think this is worth posting here, mostly because I really enjoy the simplicity of this proof but also because I have no idea how well it is known. The result is not deep or important, so the main interest is in the simplicity of the argument. Erdős proved a lower bound on the number of primes before an integer $n$.
Wacław Sierpiński, in his Elementary Theory of Numbers, attributes to Erdős the following elementary proof of the inequality $$\pi(n)\geq\frac{\log{n}}{2\log{2}}\quad\text{for }n=1,2,\ldots.$$
Please note that I have adapted the argument from the text of the book to make things, in my opinion, a bit clearer. Note also that the only tools used in the below proof are some basic combinatorial facts and some results about square-free numbers, which can, for example, be proved with the Fundamental Theorem of Arithmetic.
Let $n\in\mathbb{N}=\{1,2,3,\ldots\}.$ Consider the set $$S(n) = \{(k,l)\in\mathbb{N}^{2}:l\text{ is square-free and }k^{2}l\leq n\}.$$ It is a standard fact that every natural number has a unique representation in the form $k^{2}l,$ where $k$ and $l$ are natural numbers and $l$ is square-free. This gives $\lvert S(n)\rvert = n.$
Now if we have a pair $(k,l)$ with $k^{2}l\leq n,$ then we must have $k^{2}\leq n$ and $l\leq n$, since $k$ and $l$ are positive. Note that this gives $k\leq\sqrt{n}.$ Since $l$ is square-free, $l$ can be expressed as a product of distinct primes, each of which must be not-greater-than $n$ since $l\leq n$. That is, $l$ can be expressed as a product of the primes $p_{1},p_{2},\ldots,p_{\pi(n)}.$ There are $2^{\pi(n)}$ such products.
Therefore, if we know $(k,l)\in S(n)$ then there are at most $\sqrt{n}$ possibilities for what $k$ might be and at most $2^{\pi(n)}$ possibilities for what $l$ might be (independent of $k$, of course). It follows that $\lvert S(n)\rvert \leq 2^{\pi(n)}\sqrt{n},$ so $n\leq2^{\pi(n)}\sqrt{n}.$
Taking $\log$s and rearranging gives the result.
• How can you say a lower bound on the Prime Number Theorem is not important ! – user230452 Jul 6 '16 at 6:05
• @user230452: Better bounds are known through elementary arguments (I think the most famous is by Chebyshev, and has the advantage of providing an upper bound of the same order of growth). So this result is not so important since a little more work can give much better results, with still only elementary ideas. – Will R Jul 6 '16 at 17:23
• @Therkel: I think I understand, but, in this situation, I felt it best to make clear a) how I knew this argument was from Erdos and b) that chasing up the source would lead readers to a superficially different-sounding argument. – Will R Jul 6 '16 at 17:27
• This is brilliant! – Vinícius Novelli Jul 20 '16 at 1:41
Here is an exposition of the proof that made Erdos famous by David Galvin. An elementary proof of Bertrand's postulate, which states that there is a prime number in between every $n$ and $2n$. The essence of this proof is in noticing that the lower bound of $$\binom{2n}{n} \geq \frac{4^n}{2n + 1}$$ The binomial expression is the middle term (and the largest) of $(1+x)^{2n}$. The lower bound is the average of the sum of all binomial coefficients. This is obtained by putting $x=1$, and then dividing by the number of terms. This gives us the average. Obviously, the largest term should be bigger than the average. If the postulate does not hold, there is an upper bound that is smaller than this lower bound for large $n$. The postulate can easily be verified for the smaller values of $n$.
https://www3.nd.edu/~dgalvin1/pdf/bertrand.pdf
One very simple, and yet one of my favorites is the Erdős-Anning theorem:
Let $A \subseteq \mathbb C$ be an infinite set of points, such that
$$\forall x, y\in A \quad |x-y| \in \mathbb N$$
then there exists some $c,k \in \mathbb C$, such that all $a \in A$ is of the form $a = cx + k$ for some $x \in \mathbb R$.
It was proved in 1945 in the American Mathematical Bulletin. http://www.ams.org/journals/bull/1945-51-08/S0002-9904-1945-08407-9/S0002-9904-1945-08407-9.pdf
A more colorful formulation is that an infinite sets of points on the Cartesian plain with mutual integer distances must lie on a straight line.
To prove it, we need an upper bound on the number of non collinear points possible in a set with integral distances. More specifically, if there is a set of non collinear points have integer distances, all at most $d$, then at most $4(d + 1)^2$ points with integer distances can be added to the set.
• I have a doubt. Can you help me ? The theorem says that for any $n$, it is possible to have $n$ points on a plane, not all collinear such that they have integral distances but not infinitely many points. As I understand it, infinity is not a number, but a concept. To say that infinitely many points are possible it means that for any $n \geq l$, it is possible, where $l$ is an arbitrarily large natural number. It seems to me the first part of the theorem is stating it, and the second is contradicting it. – user230452 Jul 6 '16 at 3:45
• @user230452 No, to say that there is an infinite set with integral distances is not the same as saying for any $n\ge l$ there is a set with $n$ points and integral distances. Compare: For any $n\ge l$ there is a bounded subset of $\Bbb R$ with integral distances, but there is no bounded infinite subset of $\Bbb R$ with integral distances. $\Bbb N$ is an unbounded infinite subset of $\Bbb R$ with integral distances. – Mario Carneiro Jul 6 '16 at 8:05
• @bof There was a typo; the constraint $0 \in A$ was lacking. It is equivalent to adding some constant, s.t. $a = cx + k$. – user347499 Jul 6 '16 at 16:23
Erdos' proof of Infinite primes
The following proof is taken from the book - "Proofs from THE BOOK" by Martin Aigner and Gunter Ziegler. This proof is attributed to Erdos.
This proves that there are infinitely many primes and that the series of the sum of prime reciprocal steps diverges. Let us assume that the infinite series $\sum\frac{1}{p}$, where $p$ denotes the prime numbers is a convergent one. Then, there must exist some natural number $k$ such that, $$\sum_{i > k} \frac{1}{p_i} < \frac{1}{2}$$
For an arbitrary natural number $N$, we get the inequality $$\sum_{i > k} \frac{N}{p_i} < \frac{N}{2}$$
Now, we call all the primes $p_1, p_2, \dots , p_k$ the small primes, and all the other primes the big primes.
Let $N_b$ denote the number of positive integers $n \leq N$ that have at least one divisor that is a big prime. And $N_s$ denote the number of positive integers less than $N$ that have only small prime divisors.
We will show that for a suitable $N$, $N_b + N_s < N$, which is a contradiction because their sum should equal $N$.
First, we estimate $N_b$ $$N_b \leq \sum_{i > k}\big\lfloor \frac{N}{p_i} \big\rfloor < \frac{N}{2}$$
Now, we look at $N_s$. We write every $n \leq N$ as $a_nb_n^2$ where $a_n$ is the square free part. $a_n$ is a product of different small primes. There are only $k$ small primes, and each prime may either be chosen or not chosen. So, there are only $2^k$ different values of $a_n$. Furthermore, \begin{align} b_n \leq \sqrt n \leq \sqrt N \\ \text{There are at most $\sqrt N$ different values of $b_n$} \\ \implies N_s \leq 2^k\sqrt N \end{align}
Since, $N_b$ is always less than $N/2$, we just have to find a value of $N$ such that $N_s \leq N/2$. This happens when $N = 2^{2k + 2}$. When $N = 2^{2k + 4}$, $N_s < N/2$. This proves our contradiction that $$N_b + N_s < N$$ which proves that the series diverges and that there are infinitely many primes.
Erdos' favourite questions. The following are not research papers but popular questions Erdos used to ask children.
• If $n+1$ integers are chosen from the first $2n$ integers, there will always be two that are co prime.
There will be two numbers that are consecutive. These two numbers will be relatively prime. To see that this is not true when $n$ integers are chosen, just select all the even numbers.
• If $n+1$ integers are chosen from the first $2n$ integers, there will always be two such that one divides the other.
Every number is expressible as the product of a power of two and an odd number. There are only $n$ odd numbers in the first $2n$ integers. Two of the numbers are multiplied by the same odd number. One of them has a smaller power of two. This number divides the other. To show that it is not necessarily true for $n$ integers, choose $n+1,n+2, \dots 2n$.
• Suppose we have $n$ natural numbers none of which is greater than $2n$ such that the least common multiple of any two is greater than $2n$. Then, all $n$ numbers are greater than $2n/3$.
• Regarding the first one, you can easily see that "all the even numbers" is actually the only solution -- we can't select 1, because it's coprime to everything; and we can't have any consecutive numbers. If we take $n-1$ numbers, a bit of thinking/excluding a few small cases shows that for $n>3$, the only solutions are "all even numbers but one". For $n=3$, we have $\{3,6\}$. In general, any set of $n-k$ must be all even unless $n-k \ge 2n/3$, because we have an odd number at least $3$, which immediately narrows us to $1/3$ of the numbers. I wonder how # of solutions grows with $n$ and $k$? – Alex Meiburg Jul 7 '16 at 23:21
• @Alex, your final conclusion is false. In fact, for all $\epsilon > 0$ there exists an $n$ and a set $S \subset \{1,2,\ldots, 2n\}$ with $|S| > (1 - \epsilon)n$ such that $S$ contains an odd integer and for all $x,y \in S$ we have $\gcd(x,y) > 1$. For example, choose $n$ to be the product of the first $m$ odd primes and define $S = \{2k | k < n \text{ and } \gcd(k, n) > 1 \} \cup \{n\}$. You will see that $|S| = n - \varphi(n) + 1 > n(1 - \frac{c}{\log \log(n)})$ for some absolute constant $c$. For $m \ge 8$ we have $|S| > \frac{2n}{3}$. – Woett Oct 11 '18 at 12:00
• @Woett, right, how silly of me. I figured that if $k$ is odd then the density of numbers sharing a factor with $k$ is at most $1/3$ -- totally neglecting the possibility of multiple factors. :) Taking $n=105=3\cdot 5\cdot 7$ gives a density of $0.54$, of course. – Alex Meiburg Oct 12 '18 at 2:46
Erdos answered the following question in the affirmative - Are there infinitely many odd numbers that are not expressible as the sum of a prime number and a power of $2$. The proof is explained in this paper : http://www.maa.org/sites/default/files/3004416309960.pdf.bannered.pdf
The essence of the proof is in showing that for every integral value of $k$, there is a $2^k$ which has a certain residue with certain moduli. By the Chinese remainder theorem, there are infinitely many odd numbers that satisfy those same congruences. Whenever they do, $a - 2^k$ is composite for all integer values of $k$.
• Surely this is clear because the set of all numbers $p+2^k$ is density zero? Although perhaps that is not elementary. – Mario Carneiro Jul 7 '16 at 1:20
• The density should be positive, not zero. There are n/log(2) choices of (prime, power of 2) both less than n. If most, or a positive fraction, of those are unique representations then a positive density subset of integers less than 2n have such a representation. – zyx Jul 7 '16 at 2:08
• @zyx Oh, indeed if $|A\cap[n]|\sim f(n)$, and $B=\{a+2^k:a\in A,k\in\Bbb N\}$, then $|B\cap[n]|\le\sum_k f(n-2^k)\le f(n)\log_2n$, which is not strong enough to conclude density zero with $A=\Bbb P$, $f(n)=n/\log n$. A simulation of the quantity $\sum_{p\le n}\log_2(n-p)$ appears to converge to about $1.468n$, which is obviously useless as an upper bound on $|B\cap[n]|$ since it's larger than the trivial bound. – Mario Carneiro Jul 7 '16 at 4:59
• That is close to n/log(2) = 1.443 n . – zyx Jul 7 '16 at 5:08
• Does the simulation appear to converge to 1/log(2) when using integer logarithms, $\lfloor \log_2 ( \dots ) \rfloor$? @MarioCarneiro – zyx Jul 7 '16 at 19:10
The proof of the Littlewood-Offord lemma for sums of real numbers.
Erdos noticed that under the correspondence between sequences of $\pm$ signs and finite sets, Sperner's theorem applies and gives the optimal bound for the Littlewood-Offord lemma in dimension $1$ (on how many signed sums of $n$ given numbers of absolute value at least $1$, can have absolute value at most $1$). This observation is a few lines of very basic algebra and combinatorics.
The rest of the paper is more difficult, and is an extension of Sperner's theorem to handle the case of larger bounds on the sums, once again reading optimal bounds from the combinatorics. This gave evidence that the higher dimensional Littlewood-Offord problem on sums of vectors might also be essentially combinatorial, which was later shown to be true.
Erdos - Mordell Inequality : For a point $O$ inside a given triangle $ABC$, the perpendiculars $OP$, $OQ$ and $OR$ are drawn to the side $$OA + OB + OC \geq 2(OP + OQ + OR)$$
The product of consecutive integers is never a power
This was proved by Erdos and Selfridge.
http://www.renyi.hu/~p_erdos/1975-46.pdf
Suppose $A$ is a set of $r$ subsets on the set $\{1,2,\dots,n\}$ such that any two sets in $A$ have a non empty subset and $n/2\geq r$, the maximal size of $A$ is $\binom{n-1}{r-1}$ i.e., $$|A| \leq \binom{n-1}{r-1}$$with equality holding if and only if all the sets share a common element.
A family of sets may also be called a hypergraph. When every set in $A$ has the same size $r$, it is called a $r$-uniform hypergraph.
Note : The condition of $n/2 \geq r$ is imposed because if $r$ is more than half of $n$, then any two sets have a non empty intersection. The maximal size of the family is then $\binom nr$. | 2020-04-04T09:35:19 | {
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https://math.stackexchange.com/questions/2849945/sum-of-sin-waves-with-same-frequency-and-different-amplitudes-and-phase | Sum of sin waves with same frequency and different amplitudes and phase
For the equation:
$$A_1 \sin (\omega t + \theta_1) + A_2 \sin (\omega t + \theta_2) = A_3 \sin (\omega t + \theta_3)$$
I've been able to show that the amplitude of the sum is (I believe this is a standard problem):
$$A_3 = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\theta_1 - \theta_2)}$$
and the new phase is:
$$\theta_3 = \arctan \left(\frac{A_1 \sin \theta_1 + A_2 \sin \theta_2}{A_1 \cos \theta_1 + A_2 \cos \theta_2}\right)$$
My question is what happens when the phases $\theta_1$ and $\theta_2$ are zero (or just equal to each other). Most books and internet sites (eg http://mathworld.wolfram.com/HarmonicAdditionTheorem.html) state that under these conditions:
$$A_3 = \sqrt{A_1^2 + A_2^2}$$
However, if I set $\theta_1$ and $\theta_2$ to zero one gets instead (Since $cos(0) = 1$):
$$A_3 = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2}$$
I've been staring at this for a while and I don't understand why the published answer is different from mine. I am sure it is something simple but I've not been able to spot it (Addendum: Wikipedia site: https://en.wikipedia.org/wiki/List_of_trigonometric_identities has the same result under section Linear Combinations).
• I realized the problem: $A_3 = \sqrt{A^2_1 + A^2_2 + 2 A_1 A_2}$ applies to the relation $cos (wt + \theta_1) + sin(wt + \theta_2)$ not $sin (wt + \theta_1) + sin(wt + \theta_2)$. For the latter it works fine. This is an error on my part. ok to delete this question, as its not a valid one? – rhody Jul 13 '18 at 18:23
• Although your Question contained and was motivated by an error, it brings up some useful content, and I'd like to retain it. Even the error, articulated in your Comment, bears pointing out as it is the sort of thing anyone might overlook at first glance. – hardmath Jul 14 '18 at 13:02
The formula $\sqrt{A_1^2+A_2^2}$ applies to sinusoids in quadrature ($\frac\pi2$ phase difference), such as when summing a sine and a cosine.
Signals in phase just add up their amplitudes, $A_1+A_2$, while signals in opposition subtract, $|A_1-A_2|$.
• I've been staring at this for hours yesterday, and then this morning after I posted the question and suddenly realized my error. – rhody Jul 13 '18 at 18:26
You wrote:
$$A_1 \sin (\omega t + \theta_1) + A_2 \sin (\omega t + \theta_2) = A_3 \sin (\omega t + \theta_3)$$
The Wolfram article you cite says:
It is always possible to write a sum of sinusoidal functions $$f(\theta)=acos(\theta)+bsin(\theta )$$
as a single sinusoid the form $$f(\theta)=ccos(\theta+\delta)$$ $$A_3 = \sqrt{A_1^2 + A_2^2}$$
So you have sine plus sine, while Wolfram has sine plus cosine. To convert Wolfram's cosine to a sine, you need to shift the phase by $\frac{\pi}2$, which then makes the cosine of the phase difference equal to zero.
Waves with no phase difference (or even pi's) directly add up their amplitudes to form a new wave.
$$A_1 \sin (\omega t) + A_2 \sin (\omega t) = (A_1+A_2) \sin (\omega t )$$
The $A_3$ you prescribed is for waves with phase difference $(\theta_1 - \theta_2)=\frac{\pi}{2}$.
The equation you got putting $\theta_1=\theta_2=0$ is correct and simplifies to $A_3=(A_1+A_2)$.
The result given in Equation (21) of Wolfram site
for adding in-phase waves of same frequency with zero relative phase angle gives
$$A_3= \sqrt{A_1^2+A_2^2+ 2 A_1 A _2}= \pm (A_1+A_2)$$
The other result came about when the phase difference is $\pi/2$ obtained directly by Pythagoras thm as diagonal length of a rectangle of vectors or phasors.
The parallelogram of amplitudes can be drawn and the situation becomes clear. There are three situations. In-phase,add; Out of phase, subtract, or take the difference; when relative phase difference is $\pi/2$ take root square sum. | 2020-03-29T04:07:49 | {
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https://math.stackexchange.com/questions/3749531/how-to-find-the-elongation-of-a-spring-when-it-is-tied-to-a-brass-bar-over-a-til | # How to find the elongation of a spring when it is tied to a brass bar over a tilted wall?
The problem is as follows:
An homogeneous brass bar which goes from point $$A$$ to point $$B$$ has a weight of $$300\,N$$. The bar is suported over two frictionless surfaces as it is shown in the picture from below. The system is in static equilibrium by the action of the force excerted by the spring tied to the end at point $$B$$ on the bar which in turn is fixed to the ceiling. Using this information find the elongation of the spring. Assume that the constant of the spring is $$k=500\,\frac{N}{m}$$.
The alternatives given in my book are as follows:
$$\begin{array}{ll} 1.&28\,cm\\ 2.&23\,cm\\ 3.&25\,cm\\ 4.&26\,cm\\ 5.&30\,cm\\ \end{array}$$
What I did to solve this problem was to use a "trick" which I seen in a similar problem about a ladder standing next to a wall.
The rationale was that when an horizontal object is leaning against two surfaces there will be a reaction from both. But if it is asked a force in between. You may use any point to establish the torque about that point and use Varignon's theorem? to find the unknown force. This is done to "cancel out" the reaction force from both surfaces.
Therefore:
Torque about the point specified in the figure from below results into:
$$\sum_{i=1}^{n}\tau=0$$
$$(l\cos 30^{\circ})\times(300)+500\times x \times 2l =0$$
simplifying terms:
$$-\frac{\sqrt {3}}{2}\times 3 + 10 x = 0$$
$$x= \frac{3 \sqrt {3}}{20}\approx .2598076212 \,m$$
which transformed into cm would be about $$26\,cm$$ and this appears in the alternatives as option four. But is my answer the right approach to this problem?.
My source of confusion is that in the problem is not indicated if the spring is paralell to the incline or if it does make another angle.
It is not indicated in the problem but how can I find let's say the Reaction in both the floor and in the wall?. How can I find these?. Am I using Varignon's theorem in this problem?. Can someone help me with this matter please?.
• @Cesareo I'm sorry but I still don't get to the answer you reached. This is not homework. Can you post a solution or more hints?. How exactly you got to $30\,cm$.? – Chris Steinbeck Bell Jul 11 '20 at 3:27
First, if you are referring to Varignons' theorem about quadrilaterals and the parallelogram formed by the midpoints of the sides, I don't think you are usinng it.
Second, to determine both the reactions in the floor and the wall we can apply Newton's law and impose that the sum af all forces is zero, and then impose that the torque about any point we want is zero. You already did the second part, so you could just apply Newton's law: $$\vec{N_A}+\vec{N_B}+\vec{P}+k\vec{\Delta l}=0$$ If the spring is parallel to the wall we know the directions of all this vectors and by decomposing along two axes $$x$$ and $$y$$ (I choose the $$x$$-axis parallel to the ground, but it's arbitrary) we obtain two equations that allow us to find the moduluses of the reaction forces: $$N_A+\frac{N_B}{2}-P+\frac{\sqrt{3}}{2}k\Delta l=0$$ $$-\frac{\sqrt{3}}{2}N_B+\frac{1}{2}k\Delta l=0$$
Personally I solved the problem calculating the torque about the point B because there are 4 forces and two are applied in B, so the equations are particularly simple, but any other point is just as good and gives the same solutions. I too found that the answer is 26cm
Fig 2 has made things easy for taking force/moment equilibrium to evaluate forces.
Take moments about top left corner call it $$I$$
$$k x \cdot 2 l = l \cos 30^{\circ} 300$$
$$x = 0.2598$$ meters i.e., the fourth option:
$$x= 25.98 cm$$
Btw, I is the center about which the brass bar is instantaneously rotating in dynamic problems.
Calling $$\alpha = 30^{\circ},\ \beta = 60^{\circ}$$ and
$$\cases{ A=(0,0)\\ B=A+l(\cos\alpha,\sin\alpha)\\ G = \frac 12\left(B+A\right) }$$
with the forces
$$\cases{ f_A = (0,N_A)\\ f_B = k \Delta x (\cos\beta,\sin\beta)+N_B(-\cos\alpha,\sin\alpha)\\ f_G = (0,-W) }$$
we have
$$\cases{ f_A+f_B+f_G = 0\\ (A-B)\times f_A+(G-B)\times f_G = 0 }$$
from which we obtain
$$N_A = \frac W2,\ N_B = \frac W4,\ \Delta x = \frac{\sqrt{3}W}{4k}$$
so $$\Delta x\approx 0.26$$ cm
Another way using virtual displacements.
Calling
$$\cases{ f_{B_1} = k \Delta x (\cos\beta,\sin\beta)\\ f_{B_2} = N_B(-\cos\alpha,\sin\alpha) }$$
In equilibrium, we have
$$\delta A\cdot f_A + \delta G\cdot f_G + \delta B \cdot\left(f_{B_1}+f_{B_2}\right) = 0$$
and due to orthogonality
$$\delta A\cdot f_A = \delta A\cdot f_G = \delta B\cdot f_{B_2} = 0$$
and following
$$\frac 12\left(\delta A + \Delta x(\cos\beta,\sin\beta)\right)\cdot f_G + k (\Delta x)^2\left(\cos\beta,\sin\beta\right)\cdot \left(\cos\beta,\sin\beta\right)=0$$
or
$$-\frac 12 W\sin\beta + k\Delta x = 0\Rightarrow \Delta x = \frac{\sqrt{3}W}{4k}$$
• The bar should be $2l$ long, not $l$. – enzotib Jul 11 '20 at 8:06 | 2021-07-29T05:07:24 | {
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https://stats.stackexchange.com/questions/449594/why-both-data-have-same-standard-deviation | # Why both data have same standard deviation?
But I am curious why the standard deviation for both data is same. I think because data 2 is "100 - data 1".
• $var(X+a)=var(X)$ and $var(aX)=a^2var(X)$ for a real number $a$ and a random variable $X$. With those, can you prove why $var(100-X)=var(X)$ where X is your data set? (Equal variances is the same as equal standard deviations.) – Dave Feb 15 '20 at 0:24
I still would like you to try to prove it, but I’ll give some intuition about those two identities I gave.
Remember that variance (and standard deviation) have something to do with how spread out your data are.
$$var(X+a)=var(X)$$ means that if you take data with some amount of spread and slide them up or down the number line, you do not change the spread.
$$var(X)=var(-X)$$ is not quite what I gave, but it’ll be part of your proof. The intuition here is that you’re just spinning it around, taking a mirror image. The data are as spread out as their reflection.
Together, $$var(100-X)=var(X)$$ means that you flip the data to the mirror image and then slide along the number line, but you do not change the spread.
Capisce?
• what is the meaning of Capisce? – aan Feb 16 '20 at 11:41
• that's Italian for "do you comprehend ?" – IrishStat Feb 16 '20 at 13:20
I believe the answer to this question can be expressed in more elemental terms. Beginning with the formula for the sample standard deviation
$$s_{x} = \sqrt{\frac{ \sum_{i=1}^{n}{(x_i - \overline{x})^2} }{n - 1}},$$
I want to draw your attention to the expression $$(x_i - \overline{x})$$ in the numerator. Ignore the summation notation for the purposes of this answer. Note, we decrement the sample mean (i.e., $$\overline{x}$$) from each realization of $$x_{i}$$. If we were to increase or decrease each $$x_{i}$$ by the same amount, the mean will change. However, the distance of each realization of $$x_{i}$$ from that central tendency remains the same. In other words, each deviation from the mean is the same.
Subtracting each $$x_{i}$$ from a constant (e.g., $$100 - x_{i}$$) flips the vector of values to its mirror image, then slides it along the number line by a constant amount. Suppose the first realization of $$x_{1}$$ = 20 and $$\overline{x}$$ = 4. In keeping with the foregoing expression,
$$(20 - 4) = 16,$$
which is the deviation from the sample mean. Now, assume we don't move along the number line just yet and we simply put a negative sign in front of each $$x_{i}$$, such that we have $$-(x_{i})$$; this flips the sign of the mean as well. The first realization, $$x_{1}$$, is now $$-20$$. Again, substituting $$-20$$ into the expression,
$$(-20-(-4)) =-16,$$
which is the same number of units away from the mean. Note, squared deviations result in a positive number, so the numerator remains the same. You can run this code in R which builds upon @knrumsey's insightful response, but breaks it down further:
x <- rnorm(10000, 30, 5) # Simulates 10,000 random deviates from the normal distribution
par(mfrow = c(2, 2))
hist(x, xlim = c(0, 100), col = "blue")
hist(-x, xlim = c(-100, 0), col = "red") # Note the -x, it simply flips the sign (mirror image)
hist(x, xlim = c(0, 100), col = "blue")
hist(100-x, xlim = c(0, 100), col = "red") # Subtracting x from 100 shifts values along the x-axis
The first row of plots shows how the realizations translate when we negate each $$x_{i}$$. The second row shows what happens when each $$x_{i}$$ is subtracted from 100; the translation first flips then slides across the number line. The spread of the distribution is unaffected by this.
To supplement @Dave's answer, take a look at the following histograms which have the same x-axis. Data2 is just a shifted version of Data1 and therefore the standard deviation, which is a measure of spread, shouldn't change.
R code to generate histograms:
x <- 100*rgamma(1000, 6, 2) #Simulate some data
hist(x, xlim=c(0,100))
hist(100-x, xlim=c(0,100))
• Thanks. well-plotted histogram. May I know how do you plot the histogram? – aan Feb 15 '20 at 5:29
• @aan see the update. – knrumsey Feb 15 '20 at 17:49
• @knrumsey Good illustration. However, simulating from the gamma distribution would produce realizations that cluster around a mean that is much lower than what I see here given the parameters you specified. Right? – Thomas Bilach Feb 16 '20 at 20:24
• @Tom, that is correct. I didn't put too much thought into which distribution I simulated from. Just wanted draws between 0 and 100. – knrumsey Feb 16 '20 at 20:45
• @Thomas Bilach, realizing that I misunderstood your question. Thanks for pointing this out, I've updated the post to reflect the fact that I multiplied the data by 100. – knrumsey May 3 '20 at 0:48
changes period to period in DATA 1 identical to the negative changes in DATA 2
OBS DATA 1 FIRST DIF DATA 2 FIRST DIFF
1 30 NA 70 NA
2 40 10 60 -10
3 80 40 20 -40
4 30 -50 70 50
5 20 -10 80 10
6 33 13 67 -13
7 33 0 67 0
8 33 0 67 0
9 33 0 67 0
CONTINUING STEPS TO UNCOVER THE BASIC RELATIONSHIP OF DATA1 AND DATA2:
I took the series DATA1 and computed the ACF ... here it is ..
AND for DATA2 here
...note that not only are the standard deviations the same but the ACF's are the same.
• This is an interesting perspective, but I believe it would need more explanation to be understood by many readers. – whuber Feb 15 '20 at 22:53
• @IrishStat, thanks. do you mind to explain more? – aan Feb 16 '20 at 11:42
• i added some material to my response detailing how differenced data period to period ( val2-val1,val3-val2,.. val9-val8) for both original series DATA1 and DATA2 were complements of each other – IrishStat Feb 16 '20 at 13:19
• @IrishStat. Noted. what is the meaning of Q.E.D and OBS? – aan Feb 16 '20 at 15:58
• An explicit connection between those assertions and the result is needed. What is interesting is your use of successive differences ("changes period to period" in an otherwise unordered dataset) to characterize the standard deviation of the dataset. In other words, you appear to be claiming that the set of equalities $$x_{i+1}-x_i = -(y_{i+1}-y_i)$$ for $i=1, 2, \ldots, n-1$ implies $$\sum_{i=1}^n(x_i-\bar x)^2=\sum_{i=1}^n(y_i-\bar y)^2.$$ That's true, but it does need to be shown: you can't just exhibit a table of numbers and declare "QED"! – whuber Feb 16 '20 at 20:15 | 2021-03-01T06:40:46 | {
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https://brilliant.org/discussions/thread/wanna-practice-induction/ | # Wanna practice Induction ?
The principle of mathematical induction is a very very useful tool in many proofs, and also in proving some useful formulas. $\color{#D61F06}{\text{The principle states that any given set of positive integers}}$ has ALL natural numbers if it follows the following conditions -
$(i)$ The first natural number, i.e. $1$ is in the set.
$(ii)$ Whenever the integer $k$ is in the set, then $k+1$ is also in the set.
This looks so obvious, see that if you have been given the two statements simultaneously, then by using $(ii)$ on $(i)$, you can say that $2$ is in the set. Then again applying the statement $(ii)$ for $2$, we say that $(3)$ will be in the set, so on.
$\color{#3D99F6}{\textbf{Thus if you prove a certain result for 1}}$ , $\color{#3D99F6}{\textbf{and then assume that the result is true}}$ for $k$, just prove that it is true for $k+1$ and you're done !
Same thing you can do for 0, then prove for whole numbers !!!
Problems for practice:-
Prove that the following results hold $\forall n \in \mathbb{N}$
$(a) \displaystyle \sum_{k=0} ^n 2^k = 2^{n+1}-1$
$(b) \displaystyle \sum_{k=1} ^n k = \dfrac{n(n+1)}{2}$
$(c) \displaystyle \sum_{k=1} ^n k^2 = \dfrac{n(n+1)(2n+1)}{6}$
$(d) \displaystyle \sum_{k=1}^n k^3 = \biggl( \sum_{k=0} ^n k \biggr) ^2$
$(e) \displaystyle 24 \mid 2\cdot 7^n + 3\cdot 5^n -5$
$(f) \displaystyle 1\cdot 2+2\cdot 3+3\cdot 4+ .... + n(n+1) = \dfrac{n(n+1)(n+2)}{3}$
I felt like sharing because though it's very well known to all, I am willing to find some good level problems for practicing this foundation builder concept once more... $\color{#20A900}{\text{Isn't this a good revision ?}}$
5 years, 4 months ago
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@Aditya Raut this is really a nice note for practice...... could you please add more divisibility kinda problems to give a more diverse experience to readers? ........ thanks :)
- 5 years, 4 months ago
- 5 years, 3 months ago
Thanks - they look like a good selection of induction problems
- 4 years, 10 months ago
I actually haven't tried $(d)$ and $(f)$ before since doing math XD
Here's the solution for $(e)$ if someone got stuck.
$2\times 7^{m+1} + 3 \times 5^{m+1} - 5$
$= 2\times 7^{m}\times 7 + 3\times 5^{m}\times 5 - 5$
$= 14\times 7^{m} + 15 \times 5^{m} - 5$
$= 12\times 7^{m} + 12\times 5^{m} + 2\times 7^{m} + 3\times 5^{m} - 5$
$= 12(7^{m} + 5^{m}) + 24k$ for some $k$
$= 24n + 24k$ for some $n$
- 5 years, 3 months ago
$(a)$ Base Case $n = 1$
$\displaystyle \sum_{k = 0}^{1}2^{k} = 3$
$2^{n+1} - 1 = 3$
True.
Propose it is true $\forall n \in N$
Then it must be true for $n + 1$
$\displaystyle\sum_{k = 0}^{n+1}2^{k} = \sum_{k = 0}^{n}2^{k} + 2^{n+1}$
Substituting the value from our proposal,
$\sum_{k = 0}^{n}2^{k} + 2^{n+1} = 2^{n+1} - 1 + 2^{n+1}= 2^{n+2} - 1$
Thus, $LHS = RHS$
Thus, $LHS = RHS \forall n \in N$
Sorry for ugly arrangement.
- 5 years, 3 months ago
@Aditya Raut Can you add these to the Induction Wiki page? Thanks!
Staff - 5 years, 1 month ago
Nice! I'm too into creating notes and problems set for inequality :) Why not take a look at it @Aditya Raut ? :D
- 5 years, 3 months ago
What is e) ?... in d) second summation is it k=0 or k=1 ? Nice collection.
- 5 years ago
Thanks, but Just think ! Will adding $0$ make any change? In that summation, starting from $k=0$ will yield same thing as starting with $k=1$, won't it ? @Niranjan Khanderia
- 5 years ago
Thanks.
- 5 years ago
And e) means prove that for all positive integers $n$, $(2\cdot 7^n + 3\cdot 5^n -5)$ is divisible by $24$.
$a\mid b$ is the symbol of "a divides b" or "b is divisible by a".
Latex code for the symbol $\mid$ is "\mid"
- 5 years ago
Thanks. I did not get it since I did not assume the parenthesis.
- 5 years ago
/(good/)
- 4 years, 10 months ago
A good level problem is here: Fro every positive integer $n$, prove that $\sqrt{(4n+1)}<\sqrt n +\sqrt{(n+1)}<\sqrt{(4n+2)}$. Hence or otherwise, prove that [$\sqrt{n}+\sqrt{n+1}$]=[$\sqrt{4n+1}$] where [.] denotes floor function. It is an IITJEE problem.
- 4 years, 7 months ago | 2019-11-14T20:26:17 | {
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https://math.stackexchange.com/questions/4192982/a-question-about-solving-quotients | all. I ran into this problem and was wondering where my logic failed. I was solving an absolute value involving a quotient and went about it the following way:
$$\begin{eqnarray} \lvert \frac{x+1}{x-2} \rvert &<& 3\\ \Rightarrow \frac{x+1}{x-2} &>& -3 \hspace{1mm}and\hspace{1mm}\frac{x+1}{x-2} < 3\\ x+1&>&-3(x-2)\hspace{2cm}\text{(Solving for the left inequality)}\\ x+1&>&-3x+6\\ 4x&>&5\\ x&>&\frac{5}{4} \end{eqnarray}$$ which cannot be true, as $$x$$ can equal $$2$$ with these restrictions which is obviously not allowed. Doing the left side the correct way:
$$\begin{eqnarray} \frac{x+1}{x-2} &>& -3\\ \frac{x+1}{x-2} + 3 &>& 0\\ \frac{x+1}{x-2} + \frac{3x-6}{x-2} &>& 0\\ \frac{4x-5}{x-2} &>& 0\\ \Rightarrow x < \frac{5}{4}\lor x>2\\ \end{eqnarray}$$
leads to the solution set $$(-\infty,\frac{5}{4})\cup(2,\infty)$$. My question is, what happened with the first method where I failed to come up with a solution for $$(2,\infty)$$, (and why the signage for $$x>\frac{5}{4}$$ is backwards/incorrect in the first example, does the multiplication by -3 reverse the inequality even if I'm not introducing a new negative term to one side?). I assume since you cannot divide by zero, you're not allowed to multiply the denominator to the other side without restrictions, but I am not quite sure. Thank you, I greatly appreciate it.
• The use of the Union symbol $\cup$ is certainly not the correct way, as it is used to denote union of sets. You can use a $\textrm{or}$, or the or symbol $\vee$ (\vee). Jul 7, 2021 at 23:31
• @ultralegend5385: When writing in the context of mathematical logic, I think \lor, \land and \lnot are easier to remember for $\lor,\land,\lnot$ respectively. Then again, sometimes I prefer writing \neg over \lnot for the logical not $\neg$ Jul 7, 2021 at 23:44
• @ultralegend5385 fixed. Jul 7, 2021 at 23:54
In the first method, when you go from $$\dfrac{x+1}{x-2}\gt -3$$ to $$x+1\gt -3(x-2)$$, what you're doing is multiplying both sides of the inequality by $$x-2$$ and you assume that $$x-2\gt 0$$ when doing that because otherwise the inequality sign flips to $$\le$$
So, assuming $$x\gt 2$$, you get from that inequality that $$x\gt 5/4$$ and the second inequality $$\dfrac{x+1}{x-2}\lt 3$$ gives $$x\gt 7/2$$
So, assuming $$x\gt 2$$, you get $$x\gt 5/4$$ and $$x\gt 7/2$$; together they imply $$x\gt\max\{2,5/4,7/2\}=7/2$$
Similarly, assume $$x-2\lt 0$$, ie, $$x\lt 2$$ and solve $$x+1\lt -3(x-2)$$ and $$x+1\gt 3(x-2)$$ simultaneously to obtain the other solution set. You get $$x\lt\min\{2,5/4,7/2\}=5/4$$
The complete solution set is thus $$(-\infty,5/4)\cup (7/2,\infty)$$
The answer you arrived at is wrong. Note that $$x=3\in (2,\infty)$$ doesn't satisfy the original equation.
In the second method, you're only solving one of the two inequalities that are supposed to hold. If you solve the second inequality $$\dfrac{x+1}{x-2}\lt 3$$ in a similar way, you get $$\frac{x+1}{x-2}-3\lt 0\iff\frac{7-2x}{x-2}\lt 0\iff x\in (-\infty, 2)\cup (7/2,\infty)$$
and the solution set to your original problem is the intersection of this with $$(-\infty,5/4)\cup (2,\infty)$$ so that
$$[(-\infty,2)\cup (7/2,\infty)]\cap[(-\infty,5/4)\cup (2,\infty)]=(-\infty,5/4)\cup (7/2,\infty)$$
which is the solution set to the original problem $$\left|\dfrac{x+1}{x-2}\right|\lt 3$$
• I understand now. Thank you for explaining it so well. Jul 7, 2021 at 23:55
• @js2822: You're welcome. Also, nice work for asking a good question and being responsive to others! +1 ;) Jul 7, 2021 at 23:58
HINT
Since both sides are nonnegative, you can square them in order to obtain an equivalent inequation: \begin{align*} \left|\frac{x+1}{x-2}\right| < 3 & \Longleftrightarrow \left(\frac{x+1}{x-2}\right)^{2} < 9\\\\ & \Longleftrightarrow \frac{(x^{2} + 2x + 1) - 9(x^{2} - 4x + 4)}{(x-2)^{2}} < 0\\\\ & \Longleftrightarrow \frac{-8x^{2} + 38x - 35}{(x-2)^{2}} < 0\\\\ & \Longleftrightarrow \begin{cases} 8x^{2} - 38x + 35 > 0\\\\ x\neq 2 \end{cases} \end{align*}
Can you take it from here?
• One question: Does squaring an absolute value add any extraneous solutions in certain circumstances? Or do they never appear since the absolute value is always nonnegative? Jul 8, 2021 at 0:00
• As long as both sides are nonnegative, you can proceed as suggested so that there won't be any extraneous solutions. Jul 8, 2021 at 0:46
We can only "flip the sign" if what is inside the absolute value is negative. And similarly, we "keep the sign", if what is inside the absolute value bracket is positive.
We could write it this way...
$$-3<\frac {x+1}{x-2}<0$$ or $$0\le \frac {x+1}{x-2}<3$$
Or say, the left-hand equation only applies when $$-1
Then we can solve the two sets of inequalities:
$$-1 or $$x > \frac {7}{2}$$ or $$x \le -1$$
Now we can combine the intervals
$$(-\infty, \frac 54)\cup (\frac 72,\infty)$$ | 2022-09-26T15:21:23 | {
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https://math.stackexchange.com/questions/1979079/why-doesnt-the-derivative-of-integration-by-trig-substitution-match-the-origina | # Why doesn't the derivative of integration by trig substitution match the original function?
For one of my assignments in Calc II, I had to solve ${\displaystyle\int} \frac{\sqrt{x^2-4}}{x} \text{ d}x$. By using Trig substitution where $x = 2 \text{ sec } \theta$, my solution was $\sqrt{x^2-4} - 2\text{ arcsec}\left(\frac{x}{2}\right) + C$. I tested my solution by graphing $\frac{\text{d}}{\text{d}x}\left[\sqrt{x^2-4} - 2\text{ arcsec}\left(\frac{x}{2}\right)\right]$, and it only matched the original function where $x \geq 2$. To match for $x \leq -2$, I had to use $\frac{\text{d}}{\text{d}x}\left[\sqrt{x^2-4} + 2\text{ arcsec}\left(\frac{x}{2}\right)\right]$.
Assuming I solved this correctly, what causes this discrepancy? How can we know if, when, and where this kind of thing will happen? Also, how would we notate the difference? Is it necessary to write the solution as a piecewise function, or does none of this matter anyway? I would think that, if it was because of the square root, the change in sign would be in front of it, but this seems to be because of the arc secant.
Edit Since a comment mentioned using WolframAlpha, I decided to double-check through there, and they do have my same answer under the Step-by-Step Solution, although they then simplified it to arctan somehow. Graphing the derivative of their solution with arctan matches the original function even less, with both a different C and shape. For explicitness, here's my work: $x = 2\sec \theta \Rightarrow \text{d}x = 2 \sec\theta \tan\theta$
$\int \frac{\sqrt{4 \sec^2\theta - 4}}{2 \sec\theta}(2 \sec\theta \tan\theta \text{ d}\theta) = \int 2 \sqrt{\sec^2 \theta-1} \tan\theta \text{ d}\theta = 2 \int \tan^2 \theta \text{ d}\theta$
With trig identity replacement: $2 \int \sec^2 \theta \text{ d}\theta - 2 \int 1 \text{ d}\theta = 2 \tan\theta - 2\theta$
With replacement of $\theta$ back to its original value: $\sqrt{x^2 - 4} - 2\text{ arcsec} \left(\frac{x}{2}\right) + C$
Edit 2 I understand that the square root causes sign issues, but even switching to absolute value doesn't seem to have any effect in the graphing. Here's a link to everything graphed on desmos.
• I beat you to it, deleting my comment! – Namaste Oct 21 '16 at 20:17
• According to Wolfram, your antiderivative is incorrect. – Ken Duna Oct 21 '16 at 20:17
• I hadn't checked with Wolfram before, but in their Step-By-Step, their answer is mine, but with different work, so I updated my answer in response and showed my work. – BrainFRZ Oct 21 '16 at 20:53
The trouble is when getting rid of the square root. In general we have $\sqrt{x^2} = |x|$, so $$\sqrt{\sec^2\theta-1} = \sqrt{\tan^2\theta} = \lvert\tan\theta\rvert.$$ So the integrand is really $$\lvert\tan\theta\rvert \tan \theta = \begin{cases} \sec^2\theta - 1 & 0\le\theta<\pi/2,\\ -(\sec^2\theta - 1) & \pi/2<\theta<\pi\\ \end{cases}$$ (where you take $\theta = \operatorname{arcsec} (x/2)$ with $0 \le \theta < \pi$). When reversing the substitution there is also the issue of $$\tan\theta = \begin{cases} \sqrt{\sec^2\theta - 1} & 0\le\theta<\pi/2,\\ -\sqrt{\sec^2\theta - 1} & \pi/2<\theta<\pi\\ \end{cases}$$ so the correct integral has two parts: $$\int \frac{\sqrt{x^2-4}}{x}\,dx = \begin{cases} \sqrt{x^2-4} - 2 \operatorname{arcsec}(x/2) + C_1 & x \ge 2, \\ \sqrt{x^2-4} + 2 \operatorname{arcsec}(x/2) + C_2 & x \le -2. \\ \end{cases}$$
• Sorry, I had an error in my post which I just fixed. And generally you will have to deal with things piecewise when sign changes occur (e.g. across $x=0$ in $\sqrt{x^2} = |x|$). – arkeet Oct 21 '16 at 21:25 | 2019-08-18T13:11:37 | {
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http://math.stackexchange.com/questions/67438/given-a2caci-0-how-to-find-inverse-of-ac-1i | # Given $A^2+cA+cI=0$, how to find inverse of $A+(c-1)I$?
Suppose a square matrix $A$ such that $A^2+cA+cI=0$ for all $c \in \mathbb{Z}$. How can I show that $A+(c-1)I$ is invertible and find its inverse?
I started off this way: $A+(c-1)I = A+cI-I$
Then $(A+cI-I)(d_1A+d_2I)=I$, where $d_1, d_2 \in \mathbb{Z}$.
Expand it and it becomes: $d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$
$\Rightarrow (c-1)d_2=1 \;\; and \; \; d_1A^2+d_1cA-d_1A+d_2A=0$
$\Rightarrow d_2=\frac{1}{(c-1)}$
Continue from $d_1A^2+d_1cA-d_1A+d_2A=0$, after some manipulation, I got $d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$.
Since given that $A^2+cA+cI=0$,
$d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$
$\Rightarrow -d_1cI-d_1A+d_2A=0$
$\Rightarrow -d_1cI-d_1A+\frac{1}{c-1}A=0$
$\Rightarrow -d_1cI-d_1A=-\frac{1}{c-1}A$
$\Rightarrow d_1(cI+A)=\frac{1}{c-1}A$
At this point, since I don't know if $A$ is invertible yet, I cannot do it as $d_1=\frac{1}{c-1}\frac{A}{(cI+A)}$.
Even if I did this, I still cannot find a value for $d_1$ and $d_2$ to find the inverse of $A+(c-1)I$. How should I continue from here?
-
Did you really mean for all $c \in \mathbb{Z}$? Wouldn't then $A^2 = A+I = 0$, hence $I^2 = 0$? – Niels Diepeveen Sep 25 '11 at 18:23
Instead of all this, rewrite your starting equation as $A^2+(c-1)A+A+(c-1)I=-I$. Then factor the left-hand-side...
But if you favor a more systematic approach you can also proceed as you do until $$d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$$ At this point you decide to set $(c-1)d_2=1$, but that is too early! First get rid of the $A^2$ using $A^2=-c(A+I)$, to get $$-d_1c(A+I)+d_1cA - d_1A + d_2A + (c-1)d_2I = I$$ which simplifies to $$(d_1c-d_1+d_2-d_1c)A + ((c-1)d_2-d_1c-1)I = 0$$ $$(d_2-d_1)A + ((c-1)d_2-cd_1-1)I$$ Now set both coefficients to 0. This gives immediately $d_1=d_2$, and we then need to solve $$(c-1)d-cd-1=0$$ in which the $cd$'s cancel and give us $d=-1$. So the sought inverse is $-A-I$.
-
hmm...What happens if I couldn't see this and didn't start from this equation? I remember I had another problem to find inverses and I used a similar method and worked. I had the problem posted here too at math.stackexchange.com/questions/58841/… – xenon Sep 25 '11 at 15:23
I've added a more systematic approach. You have to start from the given relation, though -- otherwise you have no way to know there is even an inverse anywhere. – Henning Makholm Sep 25 '11 at 15:36
Thanks! But how's $A^2=-(A+I)$ and not $A^2=-(cA+cI)$? – xenon Sep 25 '11 at 15:57
At one part, you set both the coefficients to be $0$. Is there a reason for doing this? Because I was thinking why wouldn't it be $((c-1)d_2-cd_1-1)=1$ and $(d_2-d_1)A=-I$? Of course, what you have done by having the coefficients zero makes the whole thing simpler but I was just thinking what if I set it something else other than zero. – xenon Sep 25 '11 at 16:21
It follows from some experience with quotients of polynomial rings. Your original relation can be used to prove that any linear combination of powers of $A$ can be rewritten into the form $pA+qI$, and if that is all we know about $A$, then there is only one such form. So if we have $pA+qI=0$ then either it must be because $p=q=0$ or else $A$ happened to satisfy a nicer property than the one we can depend on. (In your example the nicer property is "is a multiple of $I$"). It's always worth it to try a general method first and try more complex tactices if that fails. – Henning Makholm Sep 25 '11 at 16:31
Perhaps a more transparent way: "change of variables". Let $B = A + (c-1) I$, or $A = B - (c-1) I$. Then $0 = A^2 + c A + c I = (B - (c-1) I)^2 + c (B - (c-1) I) + c I= B^2 + (2-c) B + I$. Multiply by $B^{-1}$ to get $B + (2-c) I + B^{-1} = 0$, i.e. $B^{-1} = -B + (c-2) I$ or $(A + (c-1) I)^{-1} = -A - (c-1) I + (c-2) I = -A - I$.
You may find it a bit dodgy to multiply by $B^{-1}$ before you know that $B^{-1}$ exists, but once you have the result $-A-I$ it's easy to verify that this works by multiplying it by $A + (c-1) I$.
-
$$\begin{eqnarray*} x^2+cx+c = \left[x+(c-1)\right](x+1)+1\\ \therefore 0 = A^2+cA+cI = \left[A+(c-1)I\right](A+I)+I\\ \therefore \left[A+(c-1)I\right](A+I)=-I\\ \therefore \left[A+(c-1)I\right]^{-1} = -(A+I). \end{eqnarray*}$$
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NEW ANSWER. In the "long" division $$X^2+cX+c=(X+c-1)(X+1)+1,$$ replace $X$ by $A$: $$0=A^2+cA+cI=\Big(A+(c-1)I\Big)\Big(A+I\Big)+I,$$ $$\Big(A+(c-1)I\Big)^{-1}=-A-I.$$
[EDIT. I'm realizing that this answer is the same as user1551's, who posted it before. Sorry...]
OLD ANSWER. Let $A$ be an $n$ by $n$ matrix with coefficients in a field $K$, let $f\in K[X]$ be a polynomial annihilating $A$, and let $g\in K[X]$ be any polynomial.
If $g$ is prime to $f$, then $g(A)$ is invertible, and the inverse of $g(A)$ is given by $h(A)$ where $h\in K[X]$ is an inverse of $g$ mod $f$.
Moreover, there is a closed formula for such an $h$ (see this answer).
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https://brilliant.org/discussions/thread/which-payoff-do-you-want-to-go-for-2/ | # Which payoff do you want to go for?
Payoff 1.
Toss 5 coins. You get $1 for each consecutive HT that you get. Payoff 2. Toss 5 coins. You get$1 for each consecutive HH that you get.
For example, if you tossed HTHHH, under payoff 1 you will get $1, but under payoff 2 you will get$2.
Note by Calvin Lin
3 years, 2 months ago
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This is a trick question, both payoffs have same expected value of profit, which is: $1. Consider this: We will count the number of ways HT can occur. We can have: HT_ _ _ : 8 ways _ HT _ _ _ : 8 ways _ _ HT _ : 8 ways _ _ _ HT : 8 ways Total number of ways: 32 Similarly for HH, the total number of ways in which it can occur is 32. What this means is that the sum of the payoffs of HT and HH over the 32 possible outcomes is$32 for both. The payoffs may be distributed in different ways for HT and HH, but their expected value is same.
Mathematically, the expected value is same. But there is one more thing left to do: analyze the probability distribution. We can easily see that payoff 2 might be desirable because it tends to give large profits as compared to payoff 1 for certain events. eg: HHHHH gives $4 in p2 and 0 in p1. HHHHT gives$3 in p2 and $1 in p1. But the thing is that these events are not very common. Using method of reflection and analyzing the 16 possibilities , we see that there are many events for which p2 gives$0 payoff. There are 13 outcomes for which p2 gives $0 payoff, but only 6 where p1 gives$0 payoff.
So If I'm given this choice, I'd take payoff 1, as then I have more chance of leaving with at least a dollar in my pocket.
Since this is in quantitative finance, should I talk about risk and stuff? I do not know.
- 3 years, 2 months ago
Yes, the expected payoff for both scenarios is $1. The easiest way to see this is to look at Indicator Variables, which is essentially what Raghav did (though not phrased in that language). Well, everyone has their own "payoff preference". For example, you stated that you would prefer to "be more likely to leave with at least a dollar in your pocket". As such, this tells me that you are very risk adverse, and that you would prefer the certainty of a positive payoff. If you want to talk about "risk and stuff", what do you need to consider, and why? Staff - 3 years, 2 months ago Log in to reply I am not formally educated in these topics, so I'm just saying all this from a logical standpoint. After a bit of thinking, I think I agree with you on the fact that everyone has their own "payoff preference". One can take a risk to win more, or take less risk to win less. In the scenario mentioned here, the aspects of risk are not conspicuous. The amount to be won is not significant, and there is no penalty for losing. Hence, both the payoffs are inherently attractive offers. When we come to real life situations, I think there are many more things that contribute to the risk: 1. The probability that you will get a payoff. 2. The probability that you stand to lose money. 3. The security of the winnings. Money once won shouldn't be taken away from you. 4. The effects of said decisions on long term/ on your ability to take other decisions. According to me, every decision of ours is based on weighing in these risk factors against the probable prize. The balance may tip either way, and so may our decisions. I don't even know if I'm thinking in the right direction... Help me out here @Calvin Lin - 3 years, 2 months ago Log in to reply Part of this series of posts is for you to figure out what kind of risk appetite you have. It is a personal preference, and there isn't necessarily a "right" answer. Your understanding will help guide you in decisions that you make in future. For the points that you raised, you should answer for yourself if those should matter, and why. Hopefully, you will come up with a consistent, logical risk-reward structure. For example, if you would choose payoff 1 over 2 and payoff 3 over 4, but would prefer a combined 2 and 4 over a combined 1 and 3, then it would be very easy for someone to sell you things in part, and make you overpay for them. Staff - 3 years, 2 months ago Log in to reply I think that there may be some double-counting here. For example, you've counted HTHTT and HTHTH twice each. There are only $$2^{5} = 32$$ possible sequences in total, and there are some, such as TTHHH, that have no HT payouts, so there cannot be $$32$$ HT outcomes. I'm getting an average payout of $$\dfrac{28}{32}$$ dollars in the first game type and $$\dfrac{31}{32}$$ dollars in the second, but I am too tired to double-check. I'll do that tomorrow. :) Edit: Oh. now I see what you've done; very clever. I must have missed out on my count somewhere. I'll still wait to confirm in the morning. And yes, there does seem to be an advantage to choosing p1, (even though the expected winnings are the same), in the sense that you are more likely to at least win some money. - 3 years, 2 months ago Log in to reply Is "more likely to at least win some money" the main consideration that you will use if the expected payoff is equal? For example,what would you choose between: Payoff 1: 50% of$0, 50% of $100 Payoff 2: 90% of$1.01, 1% of $4900 Staff - 3 years, 2 months ago Log in to reply If Payoff 2 was 99% of$40.40, 1% of $1000, (to give it the same expected payoff as Payoff 1), I would go with Payoff 2 since I would be assured of money and would at least have a slim chance of a large amount of money. If Payoff 2 was 99% of$0, 1% of $5000, then it gets interesting, (at least for me). Even a such a slight chance of winning$5000 would be hard to pass up, so I would probably go with Payoff 2, unless I absolutely needed that $100 right away. If Payoff 2 was 99.9% of 0$, 0.1% of$50000 .... hmmmm .... That's a lot of dough, so I'd go with Payoff 2. However, if the options were these last two I've listed, then I might choose the$5000 option.
If Payoff 2 involved a 50% chance of losing $100 and a 50% chance of winning$200, then I would go with Payoff 1, since I'm not comfortable with there being such a good chance of losing a fair bit of money. There is a lot of psychology going on in these choices, and yet the expected winnings is always $50, (which ironically is an amount that would never actually be won in any of these scenarios). - 3 years, 2 months ago Log in to reply (Ooops, apparently I can't do maths. Fixed.) So, there are multiple ideas here. - Is the probability of a positive payoff so important that just because it moves from$0 it would greatly influence your preferences?
- At what point does the potential promise of a huge payoff overweigh the certainty of a small payoff?
Staff - 3 years, 2 months ago
In general I guess it all depends on the specific values, and further, on each person's financial status and willingness to take on risk. For me, the potential, (1% or better), of a large payout would outweigh any guaranteed amount less than $100, so in this case there is no difference to me between$0 and $100. But if the guaranteed amount were, say,$5000, with a 0.01% all-or-nothing chance of winning $1,000,000, I'd probably take the$5000 and run, (although I would be tempted to take the risk, at least for a moment). If the all-or-noting percentage were 1%, though, then I would find it hard not to take the chance; $5000 is a lot of money, but$1,000,000 is a life-changing amount of money, and a 1% chance is realistic in my eyes given the potential reward. The question I would ask myself is: which will I regret more - not taking the guaranteed money or not taking the risk?
- 3 years, 2 months ago
@Calvin Lin sir, Am I right?
- 3 years, 2 months ago
There are 2^5 outcoms of 5 coin tosses.
For payoff 1, there are this many ways to toss a HT. There are {HTHTT,HTHTH,HTTHT,HTHHT,THTHT, HHTHT,HTHHH,HHHTH,HHHHT, HTTTT,THTTT,TTHTT,TTTHT,HTTTH,HHTTT, HHTTH} Expected paayyoff 1 = (6/32$2) + (10/32$1)= $11/16=$22/32
For payoff 2, there are this many ways to toss a HH( or HHH,HHHH and HHHHHH). There are {HHHHH, THHHH,HHHHT, THHHT,HTHHH,HHHTH, HHHTT, TTHHH, HHTHH, HHTTT, THHTT, TTHHT, TTTHH, THTHH, THHTH] Expected payoff 2 =(1/32$4 )+ ( 2/32$3) + ( 5/32$2) + (7/32$1)=$3/4=$27/32
Payoff 2 is better, I guess LOL @Calvin Lin
Total possible outcomes of 32 of which there is one combination [TTTTT] which has no payoff.
- 3 years, 2 months ago
Check your calculations. The expected payoff of both scenarios is 1. Staff - 3 years, 2 months ago Log in to reply Let $$X$$ be an indicator variable which takes on value 1 for each $$HT$$ pair and 0 otherwise, and let $$Y$$ be an indicator variable which takes on value 1 for each $$HH$$ pair and 0 otherwise. There are $$5 - 2 + 1 = 4$$ indicator variables (both $$X$$ and $$Y$$) in 5 coin tosses. The expected value of the payoff is: $\text{E}\left [\sum_{k = 1}^{4} X_{k} \right ], ~ \text{E}\left [\sum_{k = 1}^{4} Y_{k} \right ].$We can apply Linearity of Expectation here, which holds even for dependent events (nice proof of this is given in Brilliant Wiki Page), to get: $\sum_{k = 1}^{4}\text{E}\left [ X_{k} \right ] , ~ \sum_{k = 1}^{4}\text{E}\left [ Y_{k} \right ].$Expected value of both $$X_{k}$$ and $$Y_{k}$$ is: $\text{E}\left [ X_{k} \right ] = \text{E}\left [ Y_{k} \right ] = 1 \cdot \dfrac{1}{4} + 0 \cdot \dfrac{3}{4} = \dfrac{1}{4}.$It follows that expected payoffs in both cases are the same and they equal:$\sum_{k = 1}^{4}\text{E}\left [ X_{k} \right ] = \sum_{k = 1}^{4}\text{E}\left [ Y_{k} \right ] = 4 \cdot \dfrac{1}{4} = 1.$ Bonus: What about the variance of the payoff? It can also influence our decision. Is it also the same? It turns out they are not! Let us first consider payoff 1 and indicator variable $$X$$:\begin{align} \text{Var}[X] &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}\right)^{2}\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}^{2}\right) + \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}^{2}\right)\right] + \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= 1 + \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - 1 \\ &= \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ].\end{align}The same is true for payoff 2 and $$Y$$. Now, let us calculate $$\text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ]$$. We make two groups of $$X_{k}X_{j}$$: • $$\left | k-j \right | = 1$$ representing two consecutive indicator variables which share one common coin. Notice that their product is always 0 since it's impossible to have two $$HT$$s in 3 consecutive coins. • $$\left | k-j \right | > 1$$ representing two non-consecutive indicator variables which share no common coin. Their product is non-zero only when they are both non-zero which happens with probability $$\left(\frac{1}{4}\right)^{2} = \dfrac{1}{16}.$$ There are total of $$4 \cdot 3 = 12$$ $$\left(X_{k}X_{j}\right)$$ pairs, and $$2\cdot 3$$ of them which are consecutive ie. where $$\left | k-j \right | = 1$$. Hence, we have: $\text{Var}[X] = \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] = 0 \cdot 6 + \frac{1}{16} \cdot 6 = 0.375.$ In the same way, we calculate $\text{Var}[Y] = \text{E}\left[ \left(\sum_{k \neq j}Y_{k}Y_{j}\right)\right ] = \frac{1}{8} \cdot 6 + \frac{1}{16} \cdot 6 = 1.125.$ So, variance of the payoff 2 is 3 times greater than variance of the payoff 1. If my reasoning is correct, then payoff 1 is somewhat safer option and it tends to the expected value, while payoff 2 is more riskier but it offers better chances of earning more than 1. Am I right @Calvin Lin?
- 1 month ago
I would go for Payoff 2 , because For payoff 1 ..... one can get maximum number of $1 is 2 which equals$2 i.e for this outcome HTTHT ( MAXIMUM POSSIBLE 'HT' consecutive is 2). For payoff 2 ..... one can get maximum number of $1 is 4 which equals$4 i.e for this outcome HHHHH ( MAXIMUM POSSIBLE 'HH' consecutive is 4).
- 2 months, 2 weeks ago
From a statistical standpoint, I would choose payoff option 1. As the number of H tossed increases, the probability of the consecutively tossing another H decreases. That being said, the HH combination can lead to a higher payoff because HTHTH only results in $2 while HHHHH results in$5. The issue lies in the decreased probability of continuing to toss H. This is a good example related to risk, much like in the decision involved in buying a AAA bond earning 5% vs a B bond earning 13%.
- 1 year, 11 months ago
To be more specific, payoff 1 would be like the AAA bond and payoff 2 like the B bond. It just depends on what kind of risk is right for you.
- 1 year, 11 months ago
Not quite the same comparison. Note that the expected payoff in both methods are the same.
Whereas in the bond example that you gave, we are trading expected payoff for certainty.
I also strongly disagree with "As the number of H tossed increases, the probability of the consecutively tossing another H decreases". The coin tosses are independent, and that is a common fallacy that "the proportion of realized events must be equal / close to the calculated probability"
Staff - 1 year, 11 months ago
I remember reading about an interesting experiment done by a French mathematician regarding payoffs and expected utility from lotteries. Allais Paradox?
- 3 years, 2 months ago
That's interesting. In experiment 1 here, taking a risk, (even if it is only 1%), means the possibility of losing a guaranteed million dollars; I would deeply regret gambling and ending up losing that money, but if I took the million and played 1B just to see what happened and found that I could have won 5 million, I would have just said "Oh well" and still been perfectly happy with my million. But if there is no guaranteed money, then having a chance at 5 million at the expense of a 1% greater chance of winning a million seems worth the risk.
Experiment 1 makes me think of the old saying, "A bird in the hand is worth two in the bush." If there are no "birds in hand", however, then the risk evaluation process is quite different.
- 3 years, 2 months ago
Let Ii be an indicator rv for an event of getting H at ith toss. Let X be a rv which is how many times HT occured. Similarly, let Y be the same but for HH. Then, X = I1I2 +... +I4I5 and Y = I1(1-I2) +...+I4(1-I5). We need to compare EX and EY. Given EIi = 0.5, we can apply linearity of expectation (even for multiplication since events are independent) and get 1 for both expectations.
- 2 years, 7 months ago
So, if their expectations are the same, does that mean that you are indifferent about which payoff to take? If so, why?
Staff - 2 years, 7 months ago
Yes, because with both payoffs, one can win the same amount of dollars on average.
- 2 years, 7 months ago
So, are you indifferent between these 2 payoffs:
Payoff 3: Always get $0 Payoff 4: 50% chance of -$1,000,000,000 and 50% chance of \$1,000,000,000
Staff - 2 years, 7 months ago
I think we want the choice with the minimum variance involved.
Maximin Principle of Rationality: An epistemically rational person maximizes his minimum payoff
Staff - 2 years, 7 months ago
If I don't have a million dollars at all, I will definitely not take payoff 4. I would take payoff 4 only if I was a millionaire. So, psychologically I am not indifferent.
- 2 years, 7 months ago
Right, so there are other considerations that come into play. For most people, the certainty of a result is preferred to an uncertain result (though with equal expected value). In general, having a low variance is better (though of course, there are exceptions).
Staff - 2 years, 7 months ago
I think this is a combinatorics problem?!
Staff - 3 years, 2 months ago
In the sense that arithmetic is a part of algebra, which is a part of calculus? There are numerous cross-disciplinary ideas. Using combinatorics will only get you one perspective of things.
Notice that I am essentially getting at the risk-reward preference, which is a "finance idea" as opposed to a "combinatorics idea". Yes, combinatorics ideas like expected value is involved, but they don't tell the full story.
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http://flemcup.com/s9au4c/099ebc-rank-of-a-matrix | # rank of a matrix
We prove that column rank is equal to row rank. The rank of a matrix is the largest number of linearly independent rows/columns of the matrix. Ask a Question . The rank of A is equal to the dimension of the column space of A. This lesson introduces the concept of matrix rank and explains how the rank of a matrix is revealed by its echelon form.. To define rank, we require the notions of submatrix and minor of a matrix. In mathematics, low-rank approximation is a minimization problem, in which the cost function measures the fit between a given matrix … the matrix in example 1 has rank 2. So maximum rank is m at the most. Guide. The idea is based on conversion to Row echelon form. The rank of a matrix would be zero only if the matrix had no non-zero elements. Each matrix is line equivalent to itself. Recent rank-of-matrix Questions and Answers on Easycalculation Discussion . Or, you could say it's the number of vectors in the basis for the column space of A. Prove that rank(A)=1 if and only if there exist column vectors v∈Rn and w∈Rm such that A=vwt. A rank-one matrix is the product of two vectors. In previous sections, we solved linear systems using Gauss elimination method or the Gauss-Jordan method. I would say that your statement "Column 1 = Column 3 = Column 4" is wrong. by Marco Taboga, PhD. Rank of a Matrix in Python: Here, we are going to learn about the Rank of a Matrix and how to find it using Python code? Rank of a matrix definition is - the order of the nonzero determinant of highest order that may be formed from the elements of a matrix by selecting arbitrarily an equal number of rows and columns from it. How to find Rank? The rank of a matrix m is implemented as MatrixRank… If p < q then rank(p) < rank(q) Rank of a matrix. DEFINITION 2. The rank of a matrix is defined as. Pick the 2nd element in the 2nd column and do the same operations up to the end (pivots may be shifted sometimes). The Rank of a Matrix. 7. tol (…) array_like, float, optional. The rank of a matrix is the dimension of the subspace spanned by its rows. The Rank of a Matrix Francis J. Narcowich Department of Mathematics Texas A&M University January 2005 1 Rank and Solutions to Linear Systems The rank of a matrix A is the number of leading entries in a row reduced form R for A. Introduction to Matrix Rank. 6. What is a low rank matrix? 8. Exercise in Linear Algebra. … If all eigenvalues of a symmetric matrix A are different from each other, it may not be diagonalizable. The non-coincident eigenvectors of a symmetric matrix A are always orthonomal. The system has a nontrivial solution if only if the rank of matrix A is less than n. Matrix Rank. Calculators and Converters. The rank depends on the number of pivot elements the matrix. The column rank of a matrix is the dimension of the linear space spanned by its columns. Rank of a Matrix. This exact calculation is useful for ill-conditioned matrices, such as the Hilbert matrix. Find Rank of a Matrix using “matrix_rank” method of “linalg” module of numpy. The nxn-dimensional reversible matrix A has a reduced equolon form In. The rank of a Matrix is defined as the number of linearly independent columns present in a matrix. Remember that the rank of a matrix is the dimension of the linear space spanned by its columns (or rows). Threshold below which SVD values are considered zero. Input vector or stack of matrices. Rank of unit matrix $I_n$ of order n is n. For example: Let us take an indentity matrix or unit matrix of order 3×3. Coefficient matrix of the homogenous linear system, self-generated. Pick the 1st element in the 1st column and eliminate all elements that are below the current one. Return matrix rank of array using SVD method. This also equals the number of nonrzero rows in R. For any system with A as a coefficient matrix, rank[A] is the number of leading variables. OR "Rank of the matrix refers to the highest number of linearly independent rows in the matrix". The rank of the matrix A is the largest number of columns which are linearly independent, i.e., none of the selected columns can be written as a linear combination of the other selected columns. The rank of a matrix can also be defined as the largest order of any non-zero minor in the matrix. 1) Let the input matrix be mat[][]. the maximum number of linearly independent column vectors in the matrix So often k-rank is one less than the spark, but the k-rank of a matrix with full column rank is the number of columns, while its spark is $\infty$. rank-of-matrix Questions and Answers - Math Discussion Recent Discussions on rank-of-matrix.php . The rank of a matrix or a linear transformation is the dimension of the image of the matrix or the linear transformation, corresponding to the number of linearly independent rows or columns of the matrix, or to the number of nonzero singular values of the map. Some theory. 2010 MSC: 15B99 . linear-algebra matrices vector-spaces matrix-rank transpose. Theorem [thm:rankhomogeneoussolutions] tells us that the solution will have $$n-r = 3-1 = 2$$ parameters. The rank is an integer that represents how large an element is compared to other elements. Rank of Symbolic Matrices Is Exact. 1 INTRODUCTION . Top Calculators. 4. In the examples considered, we have encountered three possibilities, namely existence of a unique solution, existence of an infinite number of solutions, and no solution. We are going to prove that the ranks of and are equal because the spaces generated by their columns coincide. Calculator. Based on the above possibilities, we have the following definition. A matrix obtained by leaving some rows and columns from the matrix A is called a submatrix of A. Finding the rank of a matrix. All Boolean matrices and fuzzy matrices are lattice matrices. Changed in version 1.14: Can now operate on stacks of matrices. Common math exercises on rank of a matrix. Matrix Rank. As we will prove in Chapter 15, the dimension of the column space is equal to the rank. You can say that Columns 1, 2 & 3 are Linearly Dependent Vectors. Firstly the matrix is a short-wide matrix \$(m | 2021-09-25T15:14:13 | {
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https://www.physicsforums.com/threads/linear-independence.814024/ | # Linear independence
## Homework Statement
Assume vectors ##a,b,c\in V_{\mathbb{R}}## to be linearly independent. Determine whether vectors ##a+b , b+c , a+c## are linearly independent.
## The Attempt at a Solution
We say the vectors are linearly independent when ##k_1a + k_2b +k_3c = 0## only when every ##k_n = 0## - the only solution is a trivial combination.
Does there exist a non-trivial combination such that
##k_1(a+b) + k_2(b+c) + k_3(a+c) = 0##?. Distributing:
##k_1a + k_1b + k_2b + k_2c + k_3a + k_3c = (k_1 + k_3)a + (k_1+k_2)b + (k_2+k_3)c = 0## Since ##a,b,c## are linearly independent, the only way this result can occur is when:
##k_1+k_3 =0\Rightarrow k_1 = -k_3##
##k_1+k_2 =0##
##k_2+k_3 =0##
Substituting eq 1 into eq 2 we arrive at ##k_2 - k_3 = 0## and according to eq 3 ##k_2 + k_3=0##, which means ##k_2 - k_3 = k_2 + k_3##, therefore ##k_3 = 0##, because ##k=-k## only if ##k=0##. The only solution is a trivial combination, therefore the vectors ##a+b, b+c, a+c## are linearly independent.
Last edited:
Mark44
Mentor
## Homework Statement
Assume vectors ##a,b,c\in V_{\mathbb{R}}## to be linearly independent. Determine whether vectors ##a+b , b+c , a+c## are linearly independent.
## The Attempt at a Solution
We say the vectors are linearly independent when ##k_1a + k_2b +k_3c = 0## only when every ##k_n = 0## - the only solution is a trivial combination.
Does there exist a non-trivial combination such that
##k_1(a+b) + k_2(b+c) + k_3(a+c) = 0##?. Distributing:
##k_1a + k_1b + k_2b + k_2c + k_3a + k_3c = (k_1 + k_3)a + (k_1+k_2)b + (k_2+k_3)c = 0## Since ##a,b,c## are linearly independent, the only way this result can occur is when:
##k_1+k_3 =0\Rightarrow k_1 = -k_3##
##k_1+k_2 =0##
##k_2+k_3 =0##
Substituting eq 1 into eq 2 we arrive at ##k_2 - k_3 = 0## and according to eq 3 ##k_2 + k_3=0##, which means ##k_2 - k_3 = k_2 + k_3##, therefore ##k_3 = 0##, because ##k=-k## only if ##k=0##. The only solution is a trivial combination, therefore the vectors ##a+b, b+c, a+c## are linearly independent.
That works for me. (IOW, I agree that the three new vectors are linearly independent.)
Instead of working with the system of equations, you can set up a matrix and row reduce it. If you end up with the identity matrix, what that says is that ##k_1 = k_2 = k_3 = 0##, and that there are no other solutions.
The matrix looks like this, from your system:
$$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix}$$
After a few row operations, the final matrix is I3.
Ray Vickson
Homework Helper
Dearly Missed
That works for me. (IOW, I agree that the three new vectors are linearly independent.)
Instead of working with the system of equations, you can set up a matrix and row reduce it. If you end up with the identity matrix, what that says is that ##k_1 = k_2 = k_3 = 0##, and that there are no other solutions.
The matrix looks like this, from your system:
$$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix}$$
After a few row operations, the final matrix is I3.
Alternatively, you can compute the determinant of the matrix to find that it is nonzero. What would that tell you?
Alternatively, you can compute the determinant of the matrix to find that it is nonzero. What would that tell you?
Oh. Cramer's rule.
##k_n = \frac{D_{k_n}}{D}## and since the determinant of the system is non zero, the corresponding determinants for every ##k_n## would be 0 (a full column of 0-s means det = 0) and therefore ##k_1 = k_2 = k_3 = 0##
Ray Vickson | 2021-06-15T05:52:48 | {
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http://math.stackexchange.com/questions/1607627/how-can-a-cauchy-sequence-converge-to-an-irrational-number | # How can a Cauchy sequence converge to an irrational number?
I am a physics major and would like to clear a confusion regarding complete metric spaces. I am quoting the definition of a Cauchy sequence from wikipedia below
Formally, given a metric space $(X, d)$, a sequence $x_1, x_2, x_3, \ldots$ is Cauchy, if for every positive real number $\epsilon > 0$ there is a positive integer $N$ such that for all positive integers $m, n > N$, the distance
$$d(x_m, x_n) < \epsilon$$
Now, if we have sequence like $x_1=3, x_2=3.1, x_3=3.14, \ldots$ converging to $\pi$, I do not understand how all distances $d(x_m, x_n)$ will be less than all positive real numbers. Since irrational numbers do not terminate and continue forever, how can the distance ever be less than the smallest real number or infinitesimal (hyperreal) as the distance can never become $0$. Does this definition of completeness apply where $\epsilon$ is infinitesimal (hypperreal) ?
Kindly excuse my ignorance as I am not a mathematics major.
Thanks
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Don't think of it as one particular distance, $d(x_m,x_n)$, is smaller than every positive real number. But rather, any given positive real number, you can go far enough out to guarantee that $d(x_m,x_n)$ is smaller. – I. Cavey Jan 11 at 3:14
There is no "smallest [positive] real number". – Eric Wofsey Jan 11 at 3:15
It is not that the distance is smaller than every positive real number, it is that for any positive real number, there is a point where the distance is eventually less than it. E.g. can you eventually go far enough that the sequence is accurate to the fifth decimal place? (i.e. $\epsilon=10^{-5}$) Can you go far enough that the sequence is accurate to the hundredth decimal place? (i.e. $\epsilon=10^{-100}$) If I were to give you some number of decimal places, could your sequence eventually be that accurate? The answer of course is yes. – JMoravitz Jan 11 at 3:16
I disagree with the vote to close the question. ${}\qquad{}$ – Michael Hardy Jan 11 at 3:17
@AloizioMacedo : Perhaps it is unclear to you, but not to me. ${}\qquad{}$ – Michael Hardy Jan 11 at 4:02
The subject line currently reads “How can a Cauchy Sequence converge to an irrational number?”.
If we construe that literally, then one easy way a Cauchy sequence (lower-case initial "s") can converge to $\pi$ is that every term of the Cauchy sequence is $\pi$. Thus: $x_1=\pi, x_2=\pi, x_3=\pi,\ldots\,{}$. I suspect you meant “How can a Cauchy sequence of rational numbers converge to an irrational number?”.
Consider your sequence $3,\ 3.1,\ 3.14,\ 3.141,\ \ldots\,$.
The definition DOES NOT say that all distances between members of this sequence are less than all positive numbers. That would happen only with a constant sequence like my first example above. It says:
For every positive real number $\varepsilon>0$ there is a positive integer $N$ such that for all positive integers $m,n>N$ we have $d(x_m,x_n)<\varepsilon$.
Notice that $N$ depends on $\varepsilon$. In fact as $\varepsilon$ gets smaller, typically $N$ must get bigger. Suppose $\varepsilon = 0.01$. Then for your example sequence, $N=3$ is big enough: every pair of numbers in the sequence at or after the third place in the sequence differ from each other by less than $\varepsilon=0.01$. Thus $3.14$ and $3.141$ differ by less than $0.01$. But now suppose $\varepsilon=0.00001$. Then you need a bigger value of $N$. If each term of the sequence has one more digit or $\pi$, then $N=5$ would be big enough for that value of $\varepsilon$.
Notice that the definition of convergence to $\pi$ differs from the definition of "Cauchy sequence". It says for every $\varepsilon>0$ there is a positive integer $N$ such that for every positive integer $n\ge N$ we have $|x_n-\pi|<\varepsilon$. Again, $N$ depends on $\varepsilon$. If $\varepsilon=0.00001$, then $N=5$ would be enough: every term at or beyond the $5$th one differs from $\pi$ by less than $\varepsilon=0.00001$.
There is nothing in either of these definitions that says that the distance between two different members of the sequence or the distance between $\pi$ and a member of the sequence is $0$.
You wrote:
Since irrational numbers do not terminate and continue forever
Let's be clear on a definition.
It is certainly not correct that numbers whose decimal expansions do not terminate are necessarily irrational. For example, $1/7 = 0.\ 142857\ 142857\ 142857\ \ldots$ has a non-terminating decimal expansion and is rational.
Nor is it the case that "rational number" is defined as one whose decimal expansion repeats or terminates. Euclid and other ancient Greeks proved some numbers are irrational without ever thinking about decimal expansions. That $\pi$ is irrational means $\pi$ is not a quotient of two integers, like $22/7$. Proving $\pi$ is irrational is so difficult that it was not done until the 18th century. Some numbers are far easier to prove to be irrational. For example, if $\log_2 3 = m/n$ and $m,n$ are positive integers, then $2^m=3^n$, but that can't happen because an even number cannot be equal to an odd number.
The fact that a number is rational if and only if its decimal expansion repeats or terminates takes a bit of work to prove, but it's elementary enough that high-school students will understand it.
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In the last sentence, I think you meant "...a number is rational if and only if its decimal expansion repeats or terminates..." – Nathan Reed Jan 11 at 4:34
I don't understand the section after the quoted statement "Since [the decimal expansions of] irrational numbers do not terminate and continue forever." You appear to be trying to rebut that statement, but the statement is true. The asker never claims that numbers whose decimal expansions don't terminate are rational. They also don't claim that the definition of being rational is that the decimal expansion repeats or terminates. Both of those claims would be false but they're never made so I don't understand why you're rebutting them. – David Richerby Jan 11 at 4:55
Agree with @DavidRicherby; this answer would be improved by taking out the rebuttal that infinite decimals are necessarily irrational, which was never part of the OP's inquiry. – Daniel R. Collins Jan 11 at 5:55
Er, I mean the asker doesn't ever claim that non-terminating decimals are irrational. Doh! – David Richerby Jan 11 at 5:57
@NajibIdrissi : Why would you say that? If someone is learning this subject, they should become aware of things like that. "Cauchy sequence" is not defined as meaning a Cauchy sequence of rational numbers. ${}\qquad{}$ – Michael Hardy Jan 13 at 17:08
It's not the case that "all distances $d(x_m, x_n)$ will be smaller than every real number". That is not even true "eventually", in the sense that for some $N$, it's true for all $m, n > N$. That would imply that all such distances are $0$. Furthermore, of course there is no "smallest [positive] real number". The point is that for any positive real number $\epsilon$, no matter how small, the distances between the terms of the sequence eventually get smaller than $\epsilon$ and stay smaller than $\epsilon$.
Think of it as a challenge: I give you some $\epsilon>0$, and you have to find a position in the sequence (some $N$) such that $d(x_m,x_n)<\epsilon$ for all $m,n>N$. If the sequence is Cauchy, you are guaranteed to win the challenge. Cauchy chose the letter $\epsilon$ to stand for error (or erreur): the finite approximations $x_n$ will have an error of less than $\epsilon$ from some point on.
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In the hyperreal setting mentioned by the OP, the difference can be smaller than every positive real number and still be nonzero. See my answer. – Mikhail Katz Jan 12 at 16:24
The question did not originally mention hyperreals. – BrianO Jan 12 at 16:42
OK, I guess I am coming in late in the game. – Mikhail Katz Jan 12 at 16:42
If $n >m \ge 1$ you have $d(x_n,x_m) < 10^{n-1}$.
Choose $\epsilon>0$ and $N$ such that $10^{N-1} < \epsilon$, then if $n,m \ge N$ we have $d(x_n,x_m) < \epsilon$.
Note that the statement is for any $\epsilon>0$ there is some $N$ such that blah, blah, blah, and not that there is some $N$ such that for all $\epsilon>0$ we have blah, blah, blah.
The latter formulation would imply that the sequence is constant after some $N$.
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I think your confusion comes from "for every positive real number $\epsilon$". In fact you have to fix $\epsilon$ first when finding $N$.
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first of all it is not that all the distances are less than every positive number. given a positive number , one can find a stage after which any two terms are the given number close to each other.
The idea of a sequence converging to some point do not necessarily imply that the distance between two terms of the sequence become zero. What it says is that, as $n$ becomes larger, the terms of the convergent sequence starts coming closer to the previous one in terms of distance,
e.g in a sequence $<1/n>$ the distance between 1 and 1/2 is 1/2, distance between 1/2 and 1/4 is 1/4 and so on.. now if you look at distance between any two terms after 1/2, it is always less 1/2, if you look at the terms after 1/100 the max distance between any to terms will be upto 1/100 and so on. So if the given positive number is 1/100, you can choose here $n_0=101$(rather anything bigger than it). whereas if you want distance to be less than 1/1000 say, then this $n_0$ wont work, so you will need $n_{0}=1001$ atleast. It does not depend on terms being rational or irrational.
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" I do not understand how all distances $d(x_m,x_n)$ will be less than all positive real numbers."
This is the key to your misunderstanding. All distances can't be.
But for any real number no matter who small we can find an infinite number of distances smaller. (So we could say all real numbers are bigger than an infinite number of the distances...)
Let's do an actual example. Consider .9, .99, .999, etc. ($a_n = \sum_{i=1}^n 9/10^i$). This is a cauchy sequence that converges to 1. Okay, 1 is not irrational. But is that really where your misunderstanding lies? Okay, I'll get to that later. But for now.
Cauchy sequence: for any $\epsilon > 0$ ... okay, let's say $\epsilon = 1/1,000,000$ ... we can find a point after which all distances are less than $\epsilon$ ... okay, that'd be simply $n > 7$. $d(.99999\ldots9, .9999\ldots9) < 1/1,000,000$ for any two $.999\ldots99$ with more than 7 digits.
Okay, but let's make $\epsilon$ really small. Let's make it 1/googol. ($10^{-100}$) Well, now if n > 100 we still have $(.999\ldots9, .9999\ldots) < 1/\text{googol}$ if those terms have more that 100 nines. How about a googleplex? Then the terms need a googol nines to be that close. But we can find them with more than a googol nines. No matter how small we can find a point where the following difference of terms is smaller.
That's a cauchy sequence.
Okay, that was a cauchy sequence of rational numbers converging to a rational number. You wanted to know about a cauchy sequence of rational numbers converging to an irrational number and you suggested the decimal expansion of $\pi$. Well, same thing. $\pi$ expanded to a million decimal places will be off but it'll be of by a very small amount. For any real number no matter how small, we can find a point after which all decimal expansions are closer that that to $\pi$. These expansions are still finite but they are long enough to be closer to $\pi$ than the small real number we chose. (And if they aren't, we simply choose the finite decimal expansions that are further out. There'll always be finite expansions further out. That's the point.)
It's true we never get there. And we can't really choose rational values that can get there (that's why the number is irrational). but we've trapped and honed it in with cauchy sequences.
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thanks for your response. My main confusion has been made clear that you have to choose $\epsilon$ first and then you get $N$. But how does this work with infinitesimal (hyperreals)...How does this definition of completeness apply there ? – phd-applicant Jan 11 at 5:55
hyperreals are outside my area of expertise. My understanding though is that cauchy sequences of reals do not nesc converge to hyperreals. Hyperreals are not archimedian (for any x>0 y you can find integer n s.t. (n-1)x <= y < nx) and so have a different definition of completeness. I really don't know. I'm sorry I don't know more. But the key of the reals is that epsilon of choice is not an infintisimal. – fleablood Jan 11 at 6:22
To say that the fact that "we never get there" is "why they are irrational" is nonsense. The sequence $0.3,\ 0.33,\ 0.333,\ 0.3333,\ \ldots$ approaches $1/3$ and never gets there, but $1/3$ is rational. ${}\qquad{}$ – Michael Hardy Jan 11 at 14:29
@fleablood, great question. The answer is, it doesn't. This e is not the smallest element and of course in a field you couldn't have a smallest element. As a first approximation think of a sequence tending to zero as generating an infinitesimal, e.g., $\frac{1}{n})$. Then its square will be so much smaller: $\frac{1}{n^2}$. More details can be found here: math.stackexchange.com/questions/1602977/… – Mikhail Katz Jan 12 at 16:41
@fleablood : This is not about surreal numbers, but about Robinson's hyperreal numbers. An infinitesimal $\varepsilon>0$ is smaller than all reciprocals of finite integers, and hence smaller than all positive real numbers, but it is certainly not a smallest positive hyperreal number, since $\varepsilon/2$ is smaller, and so is $\varepsilon^2$. ${}\qquad{}$ – Michael Hardy Jan 12 at 18:07
I looked over the existing answers and they don't seem to address the concern of the OP as I understood it, so I will try to provide a separate answer. The point is that the $(\epsilon,N)$-type definition of convergence is a first-order property and therefore by the transfer principle is still satisfied over the hyperreals. More specifically, a Cauchy sequence $(x_n)$ will have a natural extension defined even for infinite values of the index $n$. This extended sequence will satisfy the condition you mentioned, namely for every $\epsilon$ there is an $N$ such that, etc. If epsilon is infinitesimal then as you point out $N$ will have to be infinite typically.
Having said this, the definition in question is provably equivalent to the one mentioned by other editors (and that involves fewer quantifiers), namely if both indices are infinite then the difference $x_n-x_m$ will be infinitesimal.
- | 2016-07-29T18:19:54 | {
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https://math.stackexchange.com/questions/2952249/probability-that-in-one-rolling-of-5-dice-we-obtain-two-6-and-one-5 | # Probability that in one rolling of 5 dice we obtain two '6' and one '5'?
First I made a mistake of not taking into account that this event is dependent.
So to get two '$$6$$' the probability would be:
$$P_5(2) = \frac{5!}{2!(5-2)!} \cdot \left(\frac16\right)^2\cdot\left(\frac56\right)^3$$
Where
• $$p = \dfrac16$$
• $$q = \dfrac56$$
• $$N= 5$$ (the number of elements of the system)
Then there are $$3$$ dice left on the table and we want to know the probability that one of them is a '$$5$$'. We rule out the two dice with the '$$6$$' face on. So the number of element in the system is now $$3$$. So the probability would be:
$$P_3(1) = \frac{3!}{1!(3-1)!} \cdot \left(\frac16\right)^1\cdot\left(\frac56\right)^2$$
But then the professor told me that $$p=\dfrac15$$ and not $$p=\dfrac16$$ That's what I don't get.
Of course at the end we multiply those two probabilities to get the final probability that we want.
• The probability is $1/5$ because you CAN"T get another $6.$ Then, you would have more than two $6$s. – StatGuy Oct 12 at 4:50
I am a beginner so see if this makes sense to you. If not let me know. Here are two ways you can solve this.
Method 1
You can solve this using the joint pmf for a multinomial distribution.
$$P(X_6 = 2, X_5 = 1, X_{other} = 2) = \frac{5!}{2!1!2!} \left( \frac{1}{6} \right)^2 \left( \frac{1}{6} \right) \left( \frac{4}{6} \right)^2 \approx .062$$
Method 2
Alternatively you can solve this using the multiplication rule. Here you get the $$1/5$$ that your professor talked about because when you condition you are on a reduced sample space.
Let $$E$$ be the event you get two sixes.
Let $$F$$ be the event you get one five.
Let $$G$$ be the event you get two of the others.
$$P(EFG) = P(E)P(F \mid E)P(G \mid EF)$$
$$P(E) = {5 \choose 2}\left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^3$$
$$P(F \mid E) = {3 \choose 1}\left( \frac{1}{5} \right) \left( \frac{4}{5} \right)^2$$
$$P(G \mid EF) = {2 \choose 2}\left( \frac{4}{4} \right)^2 \left( \frac{0}{4} \right)^0 = 1$$
$$P(EFG) \approx .062$$
• The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces. – Nuz Oct 12 at 5:14
• In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space. – HJ_beginner Oct 12 at 5:38
• Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something? – Nuz Oct 12 at 7:11
• Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F \mid G)P(E \mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting. – HJ_beginner Oct 12 at 7:36
But then the professor told me that $$p=1/5$$ and not $$p=1/6$$ That's what I don't get.
You have correctly evaluated the probability for the event that two from the five dice show 6.
$$\mathsf P(N_6{=}2)=\binom 52 \dfrac{1^25^3}{6^5}$$
Now, of the three dice which don't show 6, each may show faces $$\{1,2,3,4,5\}$$ with equal probability. Thus the conditional probability for a die showing 5, when given that it does not show 6, is $$1/5$$. So the conditional probability for obtaining one 5 when given that two 6 have been obtained among the five dice is:
$$\mathsf P(N_5{=}1\mid N_6{=}2)=\binom 31\dfrac{1^14^2}{5^3}$$
Of course at the end we multiply those two probabilities to get the final probability that we want.
And the $$5^3$$ factors will cancel, so:
$$\mathsf P(N_6{=}2,N_5{=}1)=\binom{5}{2}\binom{3}{1}\dfrac{1^21^14^2}{6^5}$$
PS: This is called a multinomial distribution. | 2018-10-21T23:20:00 | {
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https://www.physicsforums.com/threads/bezouts-identity-and-diophantine-equation.151820/ | Bézout's identity and Diophantine Equation
haki
I am having problems with one exam question.
Does this diophantine equation have a solution(s)
12a+21b+33c=6
as far as I know this is not a linear equation, and what I read online says that Bezout identity only applies for linear diophantine equations.
The solution says gcd(12,21,33) = 3, 6|3 the above equation has infinitely many solutions. Is that correct? Am I correct to assume then that
12a+21b+33c+24d = 6 again has infinite solutions aswell?
17x+6y +3z =73
since gcd(17,6,3) = 1 and 73 | 1 this has also infinitely many solutions?
fresh_42
Mentor
2018 Award
Bézout's Lemma says, that the greatest common divisor $d$ of numbers $a_1,\ldots,a_n$ can always be written as $d= s_1a_1+\ldots +s_na_n$. This immediately applies to all of your examples.
An example where it does not work is $2x+4y+6z = 7$.
WWGD
Gold Member
You can always reduce the case ax+by+cz=d to ax+by=d by letting z=0.
"Bézout's identity and Diophantine Equation"
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• Solo and co-op problem solving | 2019-10-21T08:21:39 | {
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http://www.fortinlab.com/wp-contnt/fo14r0pn/article.php?id=233270-condensation-point-math | # condensation point math
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The dew point is the temperature at which the atmosphere is saturated with water vapor. The Boiling point and the condensation point of water are the same. there is some p2SrCsuch that for any ">0, B(p;") \(SrC) is uncountable. Prove that every uncountable subset S ⊂ Rn contains a condensation point of S. (Hint: use the fact that the countable union of countable sets is countable). In mathematics, a condensation point p of a subset S of a topological space, is any point p such that every open neighborhood of p contains uncountably many points of S.Thus "condensation point" is synonymous with "-accumulation point".Examples. Condensation occurs when the relative humidity (RH) reaches 100%, and water precipitates out of the air. there is some p2SrCsuch that for any ">0, B(p;") \(SrC) is uncountable. It is a condensation point of A if and only if every neighbourhood of it contains uncountably many points of A . The set of condensation points of a set is always closed; if it is non-empty, it is perfect and has the cardinality of the continuum. The term limit point is slightly ambiguous. Every point in the Cantor set is a condensation point. Furthermore, we denote it … This page was last edited on 12 April 2014, at 12:13. During condensation, atoms and molecules form clusters. The set of condensation points of a set is always closed; if it is non-empty, it is perfect and has the cardinality of the continuum. In my above example using artesian heated water at 80 degrees Celsius and using a cooling tank to supply the cold side at about 30 degrees Celsius the tables suggest the example is valid Provided the vacuum is kept above the boiling point of 30 degrees Celsius water. At the dew point, the air is saturated, full to the brim of water. At this temperature, the air becomes saturated with water. (See also [a1].). The temperature at which such condensation would occur is called the DEW POINT TEMPERATURE. Dew Point Calculation You can physically determine the dew point temperature with a device called a hygrometer. The water boils and evaporates into the air forming water vapor. A derivative set is a set of all accumulation points of a set A. It includes unlimited math lessons on number counting, addition, subtraction etc. Yes, JLB. 5. In mathematics, a condensation point p of a subset S of a topological space, is any point p such that every open neighborhood of p contains uncountably many points of S. Thus "condensation point" is synonymous with " {\displaystyle \aleph _ {1}} - accumulation point ". But if X=ℚ and A any subset, then A does not have any condensation points at all. It happens when molecules of water vapor cool and collect together as liquid water. Mold can grow when the air is damper than about 80% relative humidity. We have further classifications of condensation point where the topological space is an ordered field. For example, if X=ℝ and A any subset, then any accumulation point of A is automatically a condensation point. na condensation point of S. Since jjx n pjj<1=nfor all n 1 we see that the sequence (x n) n 1 converges to p. Mental math strategy where a number is added to one and subtracted from other to keep the difference same. Condensation graph, returned as a digraph object.C is a directed acyclic graph (DAG), and is topologically sorted. Editor Daniel F. includes dew point calculations, formulas, and mathematic models later in this article. Take your students through all the steps in the water cycle! unilateral condensation point: x is a condensation point of A and there is a positive ϵ with either (x-ϵ,x)∩A countable or (x,x+ϵ)∩A countable. 3. Consider a Bose gas that follows a linear energy-momentum relation ε = vlp in a space of dimensionality d = 1 and d = 2. Eventually the … Generated on Sat Feb 10 11:12:42 2018 by. Since it’s so full, there’s a lot of pressure. Water tends to evaporate once the temperature increases from the boiling point which is beyond 100 degrees Celsius. Td = dew point temperature; T = temperature observed, expressed in degrees Celsius Condensed definition, reduced in volume, area, length, or scope; shortened: a condensed version of the book. na condensation point of S. Since jjx n pjj<1=nfor all n 1 we see that the sequence (x n) n 1 converges to p. where. It occurs at 212 degrees Fahrenheit or 100 degrees Celsius. an ordinal which is the least among all ordinals of the same cardinality as itself), then a κ-condensation point can be defined similarly. How to use condensate in a sentence. Most material © 2005, 1997, 1991 by Penguin Random House LLC. First let's understand what the dew point is. Condensation is when water vapor changes state from gas to liquid. Jump to: navigation , search. Definition: An open neighborhood of a point $x \in \mathbf{R^{n}}$ is every open set which contains point x. Condensation occurs when a surface is cooler than the dew point temperature of the air next to the surface. If κ is any cardinal (i.e. C = condensation (G) returns a directed graph C whose nodes represent the strongly connected components in G. This reduction provides a simplified view of the connectivity between components. A condensation point of a set S in a metric space X is (by definition) a point p such that, for each ε>0, SBp∩ (,)εis uncountable (uncountably infinite). This fifth-grade worksheet explores condensation and evaporation in a charmingly illustrated science activity. bilateral condensation point: For all ϵ>0, we have both (x-ϵ,x)∩A and (x,x+ϵ)∩A uncountable. Modified entries © 2019 by Penguin Random House LLC and HarperCollins Publishers Ltd. noun. While for the former class the number of components of the system is fixed, for the two other classes it is a fluctuating quantity. Therefore $0$ is a condensation point. The European Mathematical Society. For example, within a cloud, water nucleates around a dust, pollen, or microbial particle. Here is some basic information: Water is constantly being recycled in the water cycle, through evaporation and condensation.You have to understand that water comes in different forms, or states of matter. See more. Condensation is the change of the physical state of matter from the gas phase into the liquid phase, and is the reverse of vaporization.The word most often refers to the water cycle. Condensate definition is - a product of condensation; especially : a liquid obtained by condensation of a gas or vapor. Start with a mini lesson about water and its many forms. The node numbers in C correspond to the bin numbers returned by conncomp.. condensation determines the nodes and edges in C by the components and connectivity in G: Then, by question 3, SrChas a condensation point belonging to SrC, i.e. Then they will answer critical thinking questions about how physical science works in the real world. Water tends to evaporate once the temperature increases from the boiling point which is beyond 100 degrees Celsius. How are the melting, freezing, boiling, and condensation points of a substance used for classification purposes? The temperature when the relative humidity is 100% is called the dew point, and it's at this point that clouds and rain can develop. The value of relative humidity, which represents the moisture or water vapor content in the air at its peak, means 100 %. (a) Let Cdenote the set of condensation points of S. Assume that SrCis uncountable. Dew Point Temperature = Td = T - ((100 - RH)/5.) Condensation and precipitation reduce RH, which is why dehumidification works to … a point of which every neighborhood contains an uncountable number of points of a given set. How to Predict or Calculate a Wall Cavity Dew Point or Condensation Point in Buildings. The boiling and condensation points of a substance are always equal. The Boiling point and the condensation point of water are the same. Both are functions of temperature, pressure, and the water content of … The saturation point is the point at which air reaches its saturation limit, which means air contains the most quantity of water vapor in it. Condensation of water vapor starts when the temperature of air is lowered to its dew point and beyond. Problem 4 [20 points). 2. bilateral condensation point: For all ϵ > 0 , we have both ( x - ϵ , x ) ∩ A and ( x , x + ϵ ) ∩ A uncountable . Furthermore, we denote it by A or A^d.An isolated point is a point of a set A which is not an accumulation point.Note: An accumulation point of a set A doesn't have to be an element of that set. From Encyclopedia of Mathematics. unilateral condensation point: x is a condensation point of A and there is a positive ϵ with either (x-ϵ, x) ∩ A countable or (x, x + ϵ) ∩ A countable. Enrich your vocabulary with the English Definition dictionary www.springer.com Let X be a topological space and A⊂X. Condensation occurs when the relative humidity (RH) reaches 100%, and water precipitates out of the air. SplashLearn is an award winning math learning program used by more than 30 Million kids for fun math practice. That said, let's take a look at two dew point calculation approaches: How to Calculate the Approximate Dew Point - simplified equation. Condensation and Dew Point. Efimov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. https://encyclopediaofmath.org/index.php?title=Condensation_point_of_a_set&oldid=31620, A.V. In other words, what is the Condensation Point of water? Prove that every uncountable subset S ⊂ Rn contains a condensation point of S. (Hint: use the fact that the countable union of countable sets is countable). For each $n$, the subset of the Cantor set contained in the interval $[0,1/3^n]$ is homeomorphic to the entire Cantor set, and in particular it is uncountable. Condensation point of a set. The concept of a condensation point can be generalized to arbitrary topological spaces. A point of $E^n$ such that every neighbourhood of it contains uncountable many points of the set. The generalization to arbitrary spaces is direct: A point $x$ a condensation point (of a set $M$) in a topological space if (the intersection of $M$ with) every neighbourhood of is an uncountable set. Condensation is the process by which water vapor (water in its gas form) turns into liquid. Condensation and Dew Point. This article was adapted from an original article by B.A. The same is true of all perfect subsets of $\mathbb R^n$ Math. Definition: Let $A \subseteq \mathbf{R^{n}}$. A given volume of air containing higher amount of water vapor has a higher dew point than the same volume of drier air. It can also be defined as the change in the state of water vapor to liquid water when in contact with a liquid or solid surface or cloud condensation nuclei within the atmosphere. The water boils and evaporates into the air forming water vapor. Condensation and precipitation reduce RH, which is why dehumidification works to … To answer that question, we first need to define an open neighborhood of a point in $\mathbf{R^{n}}$. in a Euclidean space $E^n$. Condensation Explained. Also learn the facts to easily understand math glossary with fun math worksheet online at SplashLearn. The dew point is an important aspect of weather forecasts. For Problems 6 and 7, let S be a subset of Rn and define a condensation point of S to be any point x∈ Rn such that every n-ball centered at x contains uncountably many points of S. 6. The temperature at which such condensation would occur is called the DEW POINT TEMPERATURE. A derivative set is a set of all accumulation points of a set A. Dew Point Calculator is a web resource created by the Image Permanence Institute to help express and visualize the relationship between temperature, relative humidity and dew point. Prove that $P$ is perfect. A point x∈X is called a condensation point of A if every open neighbourhood of x contains uncountably many points of A. The difference between the dew point temperature and the actual temperature is … The condensation point of a substance is the temperature at which it changes from a gas to a liquid. Prove: If S is an uncountable set in R (or more generally in Rn) then there exists a condensation point for S. Suggestion: R= , … This is basic equivalent to 100% relative humidity. Adherent, Accumulation, and Isolated Points in Metric Spaces. Procedure 1. Suppose $E\subset \mathbb{R}^k$, $E$ is uncountable, and let $P$ be the set of all condensation points of $E$. The dew point is the point at which liquid water would condense from the atmosphere, and the frost point is the temperature at which ice would form from the atmosphere. That forces the water to condense, creating condensation. They use the Climatology section of the Digital Atlas of Idaho to explore concepts such as relative humidity, dew point, condensation, and cloud formation. We say that a point $x \in \mathbf{R^{n}}$ is an accumulation point of a set A if every open neighborhood of point x contains at least one point from A distinct from x. The difference between the dew point temperature and the actual temperature is … Condensation is when water vapor changes state from gas to liquid. See DEW POINT TABLE - CONDENSATION POINT GUIDE for the chart approach. Assume the Bose gas has spin 1. a) (10 points) In which of these dimensions will the Bose-Einstein condensation occur? Ponomarev, "Fundamentals of general topology: problems and exercises" , Reidel (1984) (Translated from Russian). When condensation occurs, mold and decay are very likely. We address the question of condensation and extremes for three classes of intimately related stochastic processes: (a) random allocation models and zero-range processes, (b) tied-down renewal processes, (c) free renewal processes. If A is a subset of a topological space X and x is a point of X, then x is an accumulation point of A if and only if every neighbourhood of x intersects A ∖ { x }. (a) Let Cdenote the set of condensation points of S. Assume that SrCis uncountable. 5. Students use the digital atlas of Idaho to study different weather patterns. condensation point definition in English dictionary, condensation point meaning, synonyms, see also 'condensation trail',condensation trail',condensational',condemnation'. The dew point gives an indication of the humidity. Namely. For Problems 6 and 7, let S be a subset of Rn and define a condensation point of S to be any point x∈ Rn such that every n-ball centered at x contains uncountably many points of S. 6. 1) I proved that $P$ is closed set. Then, by question 3, SrChas a condensation point belonging to SrC, i.e. First, students will read through the nonfiction text and graphics. It occurs at 212 degrees Fahrenheit or 100 degrees Celsius. condensation point in American English. Define a point $p$ in a metric space $X$ to be a condensation point of a set $E\subset X$ if every neighborhood of $p$ contains uncountably many points of $E$. Adherent, Accumulation, and Isolated Points in Metric Spaces. 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Src, i.e, then any accumulation point of $E^n$ such that every neighbourhood of it contains many! Table - condensation point where the topological space is an ordered field … See dew point the! Water precipitates out of the air forming water vapor cool and collect together liquid! Gas to liquid humidity ( RH ) reaches 100 %, and water precipitates out of the book will... The concept of a by which water vapor ( water in its form... A Wall Cavity dew point temperature with a mini lesson about water its! 100 %, and mathematic models later in this article was adapted from an original article by B.A Fahrenheit 100... '' ) \ ( SrC ) is uncountable Bose gas has spin 1. )! They will answer critical thinking questions about how physical science works in the air not... It is a condensation point x∈X is called a hygrometer, Reidel ( 1984 ) ( 10 points ) which. Vocabulary with the English definition dictionary Procedure 1 problems and exercises '', Reidel ( 1984 ) ( points! … See dew point temperature always equal reduced in volume, area, length, or microbial.... For the chart approach title=Condensation_point_of_a_set & oldid=31620 condensation point math A.V every neighborhood contains an uncountable number points. Point gives an indication of the air forming water vapor point of $E^n$ such that every of! Is cooler than the same the English condensation point math dictionary Procedure 1 contains uncountable many points of a forecasts. Works in the Cantor set is a condensation point where the topological space is an winning! Lot of pressure if and only if every open neighbourhood of it contains uncountable many points of a gas vapor... P \$ is closed set, if X=ℝ and a any subset, then a not. Brim of water vapor a set a becomes saturated with water Million kids for fun math.! Fahrenheit or 100 degrees Celsius some p2SrCsuch that for any > 0, B ( p ''! Subset, then any accumulation point of which every neighborhood contains an uncountable number of points S.... For classification purposes a if and only if every neighbourhood of it contains many. | 2021-06-19T21:40:00 | {
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https://mathhelpboards.com/threads/find-a-b-c-d-e.8446/ | # find a,b,c,d,e
ab=1
bc=2
cd=3
de=4
ea=5
find a,b,c,d,e
#### anemone
##### MHB POTW Director
Staff member
My solution:
From $ab=1$ and $bc=2$, we have: $2ab=bc$ $2ab-bc=0$ $b(2a-c)=0$ Since $b \ne 0$, $2a-c=0$ must be true or $c=2a$. From $cd=3$ and $c=2a$, we have: $(2a)d=3$ $2ad=3$ From $de=4$ and $ea=5$ and $2ad=3$, we have: $ade^2=4(5)$ $2ad(e^2)=2(20)$ $3(e^2)=40$ $\therefore e=\pm 2\sqrt{\dfrac{10}{3}}$ \begin{align*}\therefore a&=\dfrac{5}{e}\\&=\pm \dfrac{5}{2}\sqrt{\dfrac{3}{10}}\end{align*} \begin{align*}\therefore d&=\dfrac{3}{2a}\\&=\pm \dfrac{3}{5}\sqrt{\dfrac{10}{3}}\end{align*} \begin{align*}\therefore c&=\dfrac{3}{d}\\&=\pm 5 \sqrt{\dfrac{3}{10}}\end{align*} \begin{align*}\therefore b&=\dfrac{1}{a}\\&=\pm \dfrac{2}{5}\sqrt{\dfrac{10}{3}}\end{align*}
##### Well-known member
As I do not know how to put square root I have put power 1/2
Multiply all 5 to get (abcde)$^2$ = 120
Or abcde = +/-120$^{(1/2)}$
Devide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$
Similarly you can find the rest
##### Well-known member
My solution:
From $ab=1$ and $bc=2$, we have: $2ab=bc$ $2ab-bc=0$ $b(2a-c)=0$ Since $b \ne 0$, $2a-c=0$ must be true or $c=2a$. From $cd=3$ and $c=2a$, we have: $(2a)d=3$ $2ad=3$ From $de=4$ and $ea=5$ and $2ad=3$, we have: $ade^2=4(5)$ $2ad(e^2)=2(20)$ $3(e^2)=40$ $\therefore e=\pm 2\sqrt{\dfrac{10}{3}}$ \begin{align*}\therefore a&=\dfrac{5}{e}\\&=\pm \dfrac{5}{2}\sqrt{\dfrac{3}{10}}\end{align*} \begin{align*}\therefore d&=\dfrac{3}{2a}\\&=\pm \dfrac{3}{5}\sqrt{\dfrac{10}{3}}\end{align*} \begin{align*}\therefore c&=\dfrac{3}{d}\\&=\pm 5 \sqrt{\dfrac{3}{10}}\end{align*} \begin{align*}\therefore b&=\dfrac{1}{a}\\&=\pm \dfrac{2}{5}\sqrt{\dfrac{10}{3}}\end{align*}
I thought that it may be noted that all are positive or all are -ve. I know you know it but for benefit of others
#### MarkFL
Staff member
I thought that it may be noted that all are positive or all are -ve. I know you know it but for benefit of others
Yes, you can see that all of the values anemone found inherit their sign from $e$.
#### anemone
##### MHB POTW Director
Staff member
As I do not know how to put square root I have put power 1/2
The \sqrt{} command creates a square root surrounding an expression.
Take for example, \sqrt{x} gives $\sqrt{x}$.
Multiply all 5 to get (abcde)$^2$ = 120
Or abcde = +/-120$^{(1/2)}$
Devide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$
Similarly you can find the rest
Well done, kali! And looking more closely, I'd say we're actually approached the problem using quite similar concept!
#### Albert
##### Well-known member
As I do not know how to put square root I have put power 1/2
Multiply all 5 to get (abcde)$^2$ = 120
Or abcde = +/-120$^{(1/2)}$
Devide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$
Similarly you can find the rest
good solution
the use of square root example :type :" \sqrt[m]{b^n} " between two "dollar signals"
you will get: $\sqrt[m]{b^n}$ | 2020-10-01T21:58:10 | {
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https://math.stackexchange.com/questions/2346880/triangle-forming-probability-for-area | # Triangle forming probability for area
Say you have a stick which breaks randomly into three pieces (we can choose the points randomly). What is the probability that the area is greater than or equal to $0.4$?
I can see it has something to do with Heron's formula but I just can't put t together.
• what's the length of the stick? – Saketh Malyala Jul 5 '17 at 4:10
• Simulation suggests about $0.26$ for the conditional probability and that the expected value of the area is about $0.0299$, assuming the length of the stick is $1$ – Henry Jul 5 '17 at 7:46
• I get conditional probability = $\displaystyle\;4\int_{a}^{b} \sqrt{x^2(1-x^2)^2 - 0.32^2} dx \approx 0.2586458\;$ where $a \approx 0.37111$, $b \approx 0.76139$ are the two roots of polynomial $x(1-x^2) - 0.32$ in $(0,1)$. – achille hui Jul 5 '17 at 9:51
• @amWhy: it is a duplicate, but the earlier question (from the same person) does not have a response – Henry Jul 5 '17 at 20:46
• When this question has an answer, closing this as a duplicate to a question without answer (even from same user) server no useful purposes. A better choice is ask the OP to delete the old question. – achille hui Jul 5 '17 at 21:11
We will assume the length of the stick is $1$ and we break the stick by picking two points $u, v$ uniformly and independently from $(0,1)$.
Let $a = \min(u,v)$, $b = 1 - \max(u,v)$ and $c = |u-v| = 1 - a - b$. $a, b, c$ forms the sides of a triangle when and only when
$$a \le b+c,\;b \le c+a,\;c \le a+b\;\iff\;a \le \frac12,\;b \le \frac12,\;a+b \ge \frac12$$
Change variable to $x = 1-2a, y = 1-2b$, the lengths $a,b,c$ forms a triangle when $(x,y)$ falls inside another triangle $$\Delta = \big\{ (x,y) : x \ge 0, y \ge 0, x+y \le 1 \big\}$$ with area $\frac12$. It is clear conditional to $a,b,c$ forming a triangle, the probability "density" of picking a particular $(x,y)$ is $2dxdy$.
Let $A$ be the area of a triangle with sides $a,b,c$ and $A_0 = 0.04$. By Heron's formula, we have
\begin{align} A &= \sqrt{s(s-a)(s-b)(s-c)}\\ \iff 16A^2 & = 1(1-2a)(1-2b)(1-2c) = xy(1-x-y)\\ \iff 64A^2 & = ((x+y)^2 - (x-y)^2)(1-x-y) \end{align} Change variable once again to $p = (x+y), q = (x-y)$, the triangle $\Delta$ becomes
$$\Delta' = \big\{ (p,q) : 0 \le |q| \le p \le 1 \big\}$$
Furthermore,
$$A \ge A_0 \quad\iff\quad (p^2 - q^2)(1-p) \ge (8A_0)^2 \quad\iff\quad q^2 \le p^2 - \frac{(8A_0)^2}{1-p}$$ Let $\displaystyle\;f(p) = \sqrt{p^2 - \frac{(8A_0)^2}{1-p}}\;$ and $\lambda_1, \lambda_2$ be the two roots of $f(p)$ in $(0,1)$. The condition above is equivalent to $\lambda_1 \le p \le \lambda_2$ and $|q| \le f(p)$. Since $dpdq = 2dxdy$, the probability we seek equals to
$$\mathbb{P}[A\ge A_0] = \int_{\lambda_1}^{\lambda_2} \int_{-f(p)}^{f(p)} dqdp = 2\int_{\lambda_1}^{\lambda_2} f(p) dp$$
Change variable to $t = \sqrt{1-p}$, this becomes
$$\mathbb{P}[A\ge A_0] = 4\int_{\mu_1}^{\mu_2} \sqrt{t^2(1-t^2)^2 - (8A_0)^2} dt\tag{*1}$$
where $\mu_1, \mu_2$ are now the roots of the polynomial $\;t(1-t^2) - 8A_0\;$ in $(0,1)$.
For the problem at hand where $A_0 = 0.04$, we have
$$\mu_1 \approx 0.3711104191979701,\; \mu_2 \approx 0.7613913530813122$$ and $(*1)$ evaluates numerically to $$\mathbb{P}[A\ge A_0] \approx 0.2586458039398669$$ This is compatible with what another user @Henry obtained through simulation.
• Superb!! (+1). I tried in these lines and confirmed my approach through your answer – Satish Ramanathan Jul 6 '17 at 0:52
• @achillehui What numerical procedure did you follow to compute $(*1)$? My answer is coming out to be $\approx 0.249004$ which is different from our answer. I am trying to understand the reason behind the difference in the error. – Dhruv Kohli - expiTTp1z0 Jul 6 '17 at 5:32
• @expiTTp1z0 I have asked two CAS (maxima and WA) to compute the integral numerically and they return same number (upto last two digits, the difference most likely caused by rounding error on maxima). – achille hui Jul 6 '17 at 5:51
• @achillehui Can you please check this: Let $E$ be the event that three pieces form a triangle, then, $P(E) = 0.25$ and $P(A\geq0) = P(A\geq0|E)P(E) + P(A\geq0|E^C)P(E^C) = P(A\geq0|E)0.25 + 0 = 1 \cdot 0.25 = 0.25$ So how can $P(A\geq 0.04) > 0.25$ be true? Am I doing some mistake. – Dhruv Kohli - expiTTp1z0 Jul 6 '17 at 8:39
• @expiTTp1z0 the probability we are computing is the probability condition to three sticks forming a triangle. It is 4 times of that probability if we include events where triangle inequalities are violated. $\mathbb{P}[A > 0]$, or more precisely $\mathbb{P}[ A > 0 : \text{ forms a triangle } ]$, equals to $1$. There is no a prior reason for $P(A \ge 0.04 ) \le 0.25$. – achille hui Jul 6 '17 at 8:53
Assuming the length of stick is $2$. Let $X_1$ and $X_2$ be randomly sampled from $\text{Uniform}(0,2)$ and let $X_{(1)}, X_{(2)}$ be the order statistic.
The length of the three pieces will be,
$$a=X_{(1)},\ \ b=X_{(2)}-X_{(1)},\ \ c=2-X_{(2)}$$
Using,
• triangle inequality ($a+b\geq c, b+c \geq a, c+a\geq b$) and
• the range of $X_{(1)}$ and $X_{(2)}$ ($0 \leq X_{(1)} \leq X_{(2)} \leq 2$),
one can obtain the following condition for the three pieces to form a triangle,
$$0 \leq X_{(1)} \leq 1, \ \ 1 \leq X_{(2)} \leq 1+X_{(1)}$$
Using Heron's Formula,
$$A^2 = s(s-a)(s-b)(s-c) = 1(1-X_{(1)})(1-X_{(2)}+X_{(1)})(1-2+X_{(2)})$$
$$A^2 = X_{(1)}^2-X_{(2)}^2 +2X_{(2)} +X_{(1)}X_{(2)}^2-X_{(1)}^2X_{(2)}-X_{(1)}X_{(2)}-1$$
Let $E$ be the event that the three pieces form a triangle, then,
$$E = \{0 \leq X_{(1)} \leq 1 \text{ and } 1 \leq X_{(2)} \leq 1+X_{(1)}\}$$
and we need to find,
$$P(A^2 > c) = P(A^2>c|\text{E})P(E) + P(A^2>c|E^C)P(E^C) = P(A^2>c|E)P(E)$$
The joint pdf of $X_{(1)}, X_{(2)}$ is given by,
$$f_{X_{(1)},X_{(2)}}(x_1, x_2) = 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}, \ \ x_1, x_2 \in [0,2],\ x_1 \leq x_2$$
And therefore the probability of the event $E$ is,
$$P(E) = \int_{0}^{1}\int_{1}^{1+X_{(1)}}f_{X_{(1)},X_{(2)}}(x_1, x_2) \partial x_2 \partial x_1 = \frac{1}{4}$$
The conditional joint pdf of $X_{(1)}, X_{(2)}$ conditioned on $E$ is,
\begin{align} f_{X_{(1)},X_{(2)}|E}(x_1, x_2) &= \frac{f_{X_{(1)}, X_{(2)}}(x_1\mathbb{I}(x_1 \in [0,1]), x_2\mathbb{I}(x_2 \in [1, 1+x_1]))}{P(E)}\\\\ &= \frac{1/2}{1/4} \\\\ &= 2,\ x_1 \in [0,1], x_2 \in [1,1+x_1]\end{align}
Introduce two random variables $U$ and $V$ as follows,
$$U = X_{(1)}$$
$$V = A^2 = X_{(1)}^2-X_{(2)}^2 +2X_{(2)} +X_{(1)}X_{(2)}^2-X_{(1)}^2X_{(2)}-X_{(1)}X_{(2)}-1$$
Writing $X_{(1)}$ and $X_{(2)}$ in the form of $U$ and $V$,
$$X_{(1)} = U$$
$$X_{(2)}^2(1-U) + X_{(2)}(U^2+U-2)+V+1-U^2=0 \\\\ \implies X_{(2)}^2 - X_{(2)}(U+2)-\frac{V+1-U^2}{U-1}=0 \\\\ \implies X_{(2)} = \frac{U+2 \pm\sqrt{(U+2)^2+\frac{4(V+1-U^2)}{U-1}}}{2}$$
Based on the values $X_{(1)}, X_{(2)}$ can take so as to form a triangle, we obtain the values that $U$ and $V$ can take,
$$0 \leq U \leq 1, \ \ 1 \leq \frac{U+2 \pm\sqrt{(U+2)^2+\frac{4(V+1-U^2)}{U-1}}}{2} \leq 1+U \\\\ \implies 0 \leq U \leq 1, \ \ 0 \leq V \leq \frac{U^2(1-U)}{4}$$
In order to obtain the pdf of $U, V$ conditioned on $E$, we compute the determinant of the Jacobian of $X_{(1)}, X_{(2)}$ with respect to $U, V$ as follows,
$$J = \begin{pmatrix} \frac{\partial X_{(1)}}{\partial U} & \frac{\partial X_{(1)}}{\partial V} \\ \frac{\partial X_{(2)}}{\partial U} & \frac{\partial X_{(2)}}{\partial V} \end{pmatrix} = \begin{pmatrix}1 & 0 \\ \ldots & \frac{\pm 1}{\sqrt{((U+2)(U-1))^2+4(V+1-U^2)(U-1)}}\end{pmatrix} \\ \implies |det J| = \frac{1}{\sqrt{((U+2)(U-1))^2+4(V+1-U^2)(U-1)}}$$
Corresponding to the $\pm$ sign in above matrix, let there be two different matrices $J_+, J_-$. For both matrices, $|det J_{+}| = |det J_{-}| = |det J|$.
Now, we compute the pdf of $U,V$ conditioned on $E$,
\begin{align} f_{U,V|E}(u,v) &= f_{X_{(1)}, X_{(2)}|E}\left(u, \frac{u+2 +\sqrt{(u+2)^2+\frac{4(v+1-u^2)}{u-1}}}{2}\right) |det J_{+}| \\\\ & \ \ \ + f_{X_{(1)}, X_{(2)}|E}\left(u, \frac{u+2 -\sqrt{(u+2)^2+\frac{4(v+1-u^2)}{u-1}}}{2}\right) |det J_{-}| \\\\ &= 2 \cdot 2 \cdot |det J| \\\\ &= \frac{4}{\sqrt{((u+2)(u-1))^2+4(v+1-u^2)(u-1)}}\end{align}
It is easy to $\color{red}{\text{verify}}$ that,
$$\int_{0}^{1}\int_{0}^{\frac{u^2(1-u)}{4}}\frac{4}{\sqrt{((u+2)(u-1))^2+4(v+1-u^2)(u-1)}} \partial v \partial u = \int_{0}^{1} 2u \partial u = 1$$
Integrate over $U$ to obtain the marginal distribution of $V$ conditioned on $E$,
$$f_{V|E}(v) = \int_{u:\ 0 \leq u \leq 1,\ \ 0 \leq v \leq \frac{u^2(1-u)}{4}} f_{U,V|E}(u,v) \partial u$$
The cdf of $V$ conditioned on $E$ will then be,
$$P(V \leq c | E) = \int_{0}^{c}\int_{u:\ 0 \leq u \leq 1,\ \ 0 \leq v \leq \frac{u^2(1-u)}{4}} f_{U,V|E}(u,v)\partial u \partial v$$
Consider the equation $v = f(u) = \frac{u^2(1-u)}{4}, \ u \in [0,1]$. For a particular value of $v = c > 0$, there are two values of $u$ satisfying the equation. Let those values be $u_0$ and $u_1$.
Note that $u_0 \in [0,f^{-1}(v_{max})] \equiv [0,\frac{2}{3}]$ and $u_1 \in [f^{-1}(v_{max}), 1] \equiv [\frac{2}{3},1]$. With this argument, the above integral can be written as,
\begin{align}P(V \leq c | E) &= \int_{0}^{u_0}\int_{0}^{\frac{u^2(1-u)}{4}} f_{U,V|E}(u,v)\partial v \partial u + \int_{u_0}^{u_1}\int_{0}^{c} f_{U,V|E}(u,v) \partial v \partial u \\\\ & \ \ \ \ + \int_{u_1}^{1}\int_{0}^{\frac{u^2(1-u)}{4}} f_{U,V|E}(u,v)\partial v \partial u\\\\ &= \int_{0}^{u_0}2udu + \int_{u_0}^{u_1}\left(\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} + 2u \right) \partial u + \int_{u_1}^{1}2u \partial u\\\\ &= u_0^2 + \int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} \partial u + u_1^2 - u_0^2 + 1 - u_1^2 \\\\ &= 1 + \int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} \partial u \end{align}
Finally,
\begin{align} P(A^2>c) &= P(A^2 > c|E)P(E) \\\\ &= (1-P(V \leq c|E))P(E) \\\\ &= \left(- \int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} \partial u \right)P(E) \end{align}
Now, the value of $c$ is $0.04^2$ for the case when the length of the stick is $1$. Scaling up this value for the stick of length $2$ so that the probability doesn't change,
$$\frac{c_{new}}{\text{(max area in case of length 2 stick)}^2} = \frac{c_{old}}{\text{(max area in case of length 1 stick)}^2} \\\\ \implies \frac{c_{new}}{\left(\frac{\sqrt{3}}{9}\right)^2} = \frac{c_{old}}{\left(\frac{\sqrt{3}}{36}\right)^2} \\\\ \implies c_{new} = 0.0256$$
For $c > 0$, the values $u_0$ and $u_1$ are the solutions of $c = \frac{u^2(1-u)}{4}$ in the intervals $[0,\frac{2}{3}]$ and $(\frac{2}{3}, 1]$, respectively. Therefore, for $c = c_{new} = 0.0256$, $u_0 = 0.420283$ and $u_1 = 0.862277$. Putting these values in the last equation,
$$P(A^2 > c_{new} | E) = -\int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c_{new})}}{u-1} \partial u \approx \color{blue}{0.258646}$$
$$P(A^2 > c_{new}) = \frac{-\int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c_{new})}}{u-1} \partial u}{4} = \color{blue}{0.0646615}$$
Note: Used Wolfram to compute the value of the integral.
• It should be compared to $\sqrt 3/9$ when the stick breaks to 3 equal parts. – Narasimham Jul 5 '17 at 5:45
• I suspect you are supposed to assume the length of the stick is $1$ – Henry Jul 5 '17 at 7:46
• @Henry The main ideas will remain same even if the length of the stick is taken to be $2$. I took the length as $2$ so that the semi perimeter becomes $1$. Spent few hours, still couldn't find the closed form solution. I ll try again later. – Dhruv Kohli - expiTTp1z0 Jul 5 '17 at 10:55
• I voted up for your try. Good try!! – Satish Ramanathan Jul 6 '17 at 0:53
• @satishramanathan Thanks. I have completed my solution now. – Dhruv Kohli - expiTTp1z0 Jul 6 '17 at 5:13 | 2019-09-21T02:58:50 | {
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http://math.stackexchange.com/questions/206829/whats-the-probability-that-abe-will-win-the-dice-game/206832 | # What's the probability that Abe will win the dice game?
Abe and Bill are playing a game. A die is rolled each turn.
If the die lands 1 or 2, then Abe wins.
If the die lands 3, 4, or 5, then Bill wins.
If the die lands 6, another turn occurs.
What's the probability that Abe will win the game?
I think that the probability is $\frac{2}{5}$ just by counting the number of ways for Abe to win. I'm not sure how to formalize this though in terms of a geometric distribution.
-
As a visitor from rpg.stackexchange.com, I can say that it's very common to generate dice which you don't have (such as a d3) by saying "Roll a d4 and reroll 4's". Effectively, any result you reroll from scratch doesn't exist on the die. I wouldn't have been able to state it formally, though, which is why I'm just a visitor here! – Bobson Oct 4 '12 at 4:25
@Bobson: For a d3, wouldn't you normally take a d6 and map 1-2 to 1, 3-4 to 2, and 5-6 to 3? – Joren Oct 4 '12 at 14:03
@Joren - I've done it both ways. The problem with the d6 is sometimes you forget to declare whether you're mapping it that way, or whether you're mapping it 4->1, 5->2, 6->3. Which is also perfectly valid. With the d4, there's no question about it. Admittedly, it may not have been the best choice of examples. – Bobson Oct 4 '12 at 14:11
Could someone with enough rep change this to say die instead of dice. It wont let me save the suggested edits because there aren't enough characters changed. – zzzzBov Oct 4 '12 at 15:26
You are right. You can just ignore rolls of $6$ as they leave you back in the same situation. To formalize this, the chance Abe wins on turn $n$ is $\frac 13 \left(\frac 16 \right)^{n-1}$ and the chance that Bill wins on turn $n$ is $\frac 12 \left(\frac 16 \right)^{n-1}$. You can sum these if you want.
-
Let $P$ be the chance that Abe wins, then we have
$$P=\frac{1}{3}+\frac{1}{6}P$$
Solving P from this equation we get
$$P=\frac{2}{5}$$
-
I agree with you that $\dfrac{2}{5}$ is obvious. End of story. But if you really want to sum a series, abbreviate by $A$ the event "$1$ or $2$" and by $S$ the event "$6$." Then Abe can win in various ways. These are $A$ (wins immediately), $SA$ (get a $6$, then win), $SSA$, $SSSA$, and so on.
These have probabilities $\dfrac{2}{6}$, $\:\dfrac{1}{6}\cdot\dfrac{2}{6}$, $\:\dfrac{1}{6}\cdot\dfrac{1}{6}\cdot\dfrac{2}{6}$, and so on. So we want to sum the series $$a+ar+ar^2+ar^3+\cdots,$$ where $a=\dfrac{2}{6}$ and $r=\dfrac{1}{6}$.
By the usual formula for the sum of an infinite geometric series, this is $\dfrac{a}{1-r}$, which simplifies to $\dfrac{2}{5}$.
-
The key point to realise is that if a 6 is rolled, then the probability of each player winning after that roll is the same as it was before. So if $p_A$ and $p_B$ denote the probability of each player winning, then we have $$p_A = \frac 26 + \frac 16 p_A\\ p_B = \frac 36 + \frac 16 p_B$$ which is easy to solve.
-
If you want to use geometric series, you can do the following.
$$\text{P(Abe wins) = P(Abe wins in the first roll) + P(Abe wins in the second roll) + } \dots$$ $$=\frac{2}{6}+\frac{1}{6}\frac{2}{6} + \left(\frac{1}{6}\right)^2 \frac{2}{6} + \dots$$ $$=\frac{1}{3}\left(1+\frac{1}{6}+\left(\frac{1}{6}\right)^2 + \dots\right)$$ $$=\frac{1}{3}\frac{1}{1-\frac{1}{6}}$$ $$=\frac{2}{5}$$
-
The chance Abe win is 2/5
You can forget about the reroll. At each reroll, the chance that Abe win is 2/5. So no matter how many rerolls out there, the chance that Abe will win is the weighted average of chance that Abe will win multiplied by probability distribution of the rerolls number.
That average is easy to compute because it's always 2/5.
All answers are the same but I try to do this as intuitive as possible because I can't type equation :)
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https://math.stackexchange.com/questions/1380623/computing-operatornametr-bigl-bigl-ai-1-bigr2-bigr | # Computing $\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr)$
Suppose that $A \in \mathbb{R}^{n \times n}$ is a symmetric positive semi-definite matrix such that $\operatorname{Tr}(A)\le n$. I want a lower bound on the following quantity $$\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr)$$ where $I$ denotes the identity matrix.
My intuition tells me it should be
$$\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr) \ge 1/4$$
However, I'm not sure how to show it. Also my intuition might be wrong?
• Could be helpful: math.stackexchange.com/questions/391128/… – Squirtle Jul 31 '15 at 21:10
• The expression is invariant under conjugation, and $A$ is diagonalizable. – anomaly Jul 31 '15 at 21:11
• @Squirtle thank. But the square inside is hard to deal with – Boby Jul 31 '15 at 21:13
• @anomaly Is that a suggestion how to solve it? – Boby Jul 31 '15 at 21:24
• @Boby: Well, yes. – anomaly Jul 31 '15 at 21:27
By spectral theorem, there is invertible matrix $P$ such that $$P^{-1}(I+A)P=\pmatrix{1+\mu_1 \\ & \ddots & \\ & & 1+\mu_n}$$ where $\mu_i$ is eigenvalue of $A$ and $\mu_i\geqslant0$. So $$(I+A)^{-1}=P\pmatrix{\dfrac1{1+\mu_1} \\ & \ddots & \\ & & \dfrac1{1+\mu_n}}P^{-1}$$ And $$((I+A)^{-1})^2=P\pmatrix{\dfrac1{(1+\mu_1)^2} \\ & \ddots & \\ & & \dfrac1{(1+\mu_n)^2}}P^{-1}$$ So $$Tr(((I+A)^{-1})^2)=\sum_{i=1}^n\dfrac1{(1+\mu_i)^2}\leqslant n$$ Then by $\sum_{i=1}^n\mu_i\leqslant n$, we have $\mu_i\leqslant n$ for all $i$. So a lower bound is $$Tr(((I+A)^{-1})^2)\geqslant\frac{n}{(1+n)^2}$$
• @Boby, you should convince yourself (google or prove it yourself) that adding the identity to a matrix will shift the eigenspectrum in the manner of this proof (on line one). – Squirtle Jul 31 '15 at 21:33
• @hermes Sorry, but it is not clear to me how does this help to find the lower bound. Could you be more explicit? – Start wearing purple Jul 31 '15 at 21:48
• @hermes I mean this calculation only implies an obvious lower bound $n\cdot \frac{1}{(1+n)^2}$. But this is very far from the conjectured one and definitely not optimal. – Start wearing purple Jul 31 '15 at 22:10
• @L.G. Not sure but maybe we can use concavity of $\frac{1}{(1+x)^2}$ to show that $\sum _{i=1} \frac{1}{(1+\mu_i)^2} \ge \frac{1}{(1+\sum _{i=1}\mu_i)^2} =\frac{1}{4}$ – Boby Jul 31 '15 at 22:29
• The conjecture may not be true.I mean we can get a lower bound but it may not be $1/4$. The lower bound can be obtained by Lagrangian multiplier. – Math Wizard Jul 31 '15 at 22:35
Since $A$ is real symmetric, there exists a matrix $P$ such that
$$PAP^{-1} = D =\verb/diag/(\lambda_1,\lambda_2,\ldots,\lambda_n)$$ is diagonal with eigenvalues $\lambda_1,\ldots,\lambda_n$. This leads to
$$P (I_n+A)^{-2}P^{-1} = (I_n + PAP^{-1})^{-2} = \verb/diag/\left(\frac{1}{(1+\lambda_1)^2},\ldots,\frac{1}{(1+\lambda_n)^2}\right)\\ \implies \verb/Tr/((I_n + A)^{-2}) = \sum_{i=1}^{n}\frac{1}{(1+\lambda_i)^2}$$
Consider the function $g(x) = \frac{1}{(1+x)^2}$. It is easy to check for $x \in (0,\infty)$, $$g(x) > 0, \quad g'(x) = -\frac{2}{(1+x)^3} < 0 \quad\text{ and }\quad g''(x) = \frac{6}{(1+x)^4} > 0$$ This means $g(x)$ is a positive, monotonic decreasing and convex function over $(0,\infty)$.
By Jensen's inequality, we obtain following lower bound:
$$\verb/Tr/((I_n + A)^{-2}) = \sum_{i=1}^{n} g(\lambda_i) \underbrace{\ge}_{\text{Jensen}} n g\left(\frac1n\sum_{i=1}^n \lambda_i\right) \underbrace{\ge}_{g\text{ decreasing}} n g(1) = \frac{n}{4}$$
Since $\verb/Tr/(I_n) \le n$ and $\verb/Tr/((I_n + I_n)^{-2}) = \frac{n}{4}$, above lower bound $\frac{n}{4}$ is achievable and hence the optimal one.
As a consequence, your intution $\verb/Tr/((I_n + A)^{-1}) \ge \frac14$ is true (but far from the optimal).
• +1 Yet $\frac14\ge\frac{n}{(1+n)^2}$, so the OP's bound is at least better than the one given in the other answer ;D – user1551 Aug 24 '15 at 11:44 | 2019-05-20T06:22:03 | {
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https://math.stackexchange.com/questions/1356968/is-there-a-name-for-matrix-product-with-reversed-indices/1356977 | # Is there a name for matrix product with reversed indices?
The typical matrix product is as follows: $$(\mathbf{A}\mathbf{B})_{ij} = \sum_{k=1}^m A_{ik}B_{kj}\,.$$
Is there a name or characterization for one such as $$(\mathbf{A}\mathbf{B})_{ij} = \sum_{k=1}^m A_{ki}B_{jk}\,?$$ Furthermore, what can be said about the matrix vector product $$(\mathbf{A}\mathbf{b})_{i} = \sum_{k=1}^m A_{ki}b_{k}\,?$$ Is there any way to express the above matrix-vector product in terms of traditional linear algebra?
• Great questions. The beginner's mind is a precious thing. – orangeskid Jul 11 '15 at 0:24
• Agreed, this is a really good question! It is not trivial but amazing that your formulas turn out to be $A^TB^T$ and $A^TB$. – Eli Rose -- REINSTATE MONICA Jul 11 '15 at 1:00
• @EliRose why is this not trivial? It's pretty much by definition of transposition. – Ruslan Jul 11 '15 at 11:04
Your second formula is just $A^tB^t$ and your third one $A^tb$, where the $(\hskip1ex)^t$ means transposed
• (Also, note that $\;\;\; A^tB^t \: = \: (BA)^t \:\:\:\:$.) $\;\;\;\;\;\;\;\;\;$ – user57159 Jul 11 '15 at 3:13
Given a matrix $\mathbf{A} = (A)_{ij}$ the transpose of a matrix $\mathbf{A}^{\intercal}$ is defined by $\mathbf{A}^{\intercal} = (A)_{ji}$. Intuitively the transpose of a matrix is found by reflecting the matrix across the line through the diagonal coefficients $i = j$. See https://en.wikipedia.org/wiki/Transpose.
With this in mind your first alternate version of multiplication can be written as $\sum_{k=1}^m A_{ki}B_{jk} = \mathbf{A}^\intercal\mathbf{B}^\intercal$.
$$\sum_{k=1}^n A_{ki} B_{jk} = \sum_{k=1}^n B_{jk} A_{ki} = (BA)_{ji} = (BA)^\intercal = A^\intercal B^\intercal$$ and $$\sum_{k=1}^m A_{ki}b_{k} = \sum_{k=1}^m A_{ik}^{\intercal} b_{k} = A^{\intercal} b$$ | 2020-11-29T17:07:36 | {
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https://math.stackexchange.com/questions/1638942/how-can-i-solve-int-frac3x2x2x1dx | How can I solve $\int \frac{3x+2}{x^2+x+1}dx$
I want to compute this primitive $$I=\int \frac{3x+2}{x^2+x+1}dx.$$
I split this integral into two part: $$\int \frac{3x+2}{x^2+x+1}dx=\int \frac{2x+1}{x^2+x+1}dx+\int \frac{x+1}{x^2+x+1}dx,$$ For the first part: $$\int \frac{2x+1}{x^2+x+1}dx= \ln(|x^2+x+1|)+c_1$$ For the second part: due a change of variable $u=x+1$, I find $$\int \frac{x+1}{x^2+x+1}dx= \int \frac{x+1}{x(x+1)+1}dx=\int \frac{u}{u(u-1)+1}dx=\dots$$
• $u=x+1\implies x=u-1$, hence $\frac{x+1}{x(x+1)+1}=\frac{u}{(u-1)u+1}\neq\frac{u}{u^2}$. – Workaholic Feb 3 '16 at 15:59
• @Workaholic, I don't have any idea to compute the second integral, can you help me – Achaire Feb 3 '16 at 16:02
• the varible change is no good here $\frac{x+1}{x(x+1)+1}=\frac{x}{x(x+1)+1}+\frac{1}{x(x+1)+1}$ – Achaire Feb 3 '16 at 16:05
• Do you know how to deal with integrals of the form $\displaystyle\int\frac{1}{ax^2+bx+c}\,\mathrm dx$? – Workaholic Feb 3 '16 at 16:07
• It seems that most agree that rewriting $x^2+x+1$ as $x^2+x+\frac{1}{4}+\frac{3}{4}=(x+\frac{1}{2})^2+\frac{3}{4}$ would be helpful. – John Joy Feb 3 '16 at 16:41
Hint For the second:
$$\int \frac{x+1}{x^2+x+1}dx$$
Rewrite as $$\int\bigg( \frac{2x+1}{2(x^2+x+1)}+\frac{1}{2(x^2+x+1)} \bigg)dx$$
$$=\underbrace{\frac 1 2 \int\frac{2x+1}{x^2+x+1}dx}_{:=I}+\underbrace{\frac 1 2 \int \frac{1}{x^2+x+1}dx}_{:=J}$$
For $I$ substitute $t=x^2+x+1$ and $dt=(2x+1)dx$
For $J$ complete the square : $\frac 1 2\displaystyle\int \frac{1}{(x+1/2)^2+3/4}$ and now substitute $\varphi=x+1/2$ and $d\varphi=dx\Longrightarrow \frac 1 2\displaystyle\int \frac{1}{\varphi ^2 +3/4}d \varphi$
factor out $3/4$ from the denominator $\frac 2 3\displaystyle\int \frac{1}{(4\varphi ^2)/(3)+1}d \varphi$ and then substitute $z=\frac{2 \varphi}{\sqrt 3}$ and $dz=\frac{2}{\sqrt 3}d \varphi$ so you will get $\frac {1}{\sqrt 3}\displaystyle\int \frac{1}{z^2+1}dz$ and from here it is easy
HINT: the second integral is given by $$\int\frac{x+1}{x^2+x+1}dx$$ you must write the denominator as $$(x+1/2)^2+3/4)$$
Here is one approach to the second integral:
$$\int \frac{x + 1}{ x^2 + x + 1} dx = \frac12 \left(\int \frac{2x + 1}{ x^2 + x + 1} dx + \int \frac{1}{x^2 + x + 1} dx\right)$$
The first integral is identical to the one you integrated. The second is $\arctan$, which can be seen after you complete the square.
$$\int \frac{1}{x^2 + x + 1} dx = \int \frac{1}{x^2 + x + (1/2)^2 - (1/2)^2 + 1} dx$$ $$\int \frac{1}{(x+1/2)^2 + 3/4} dx = \frac{4}{3} \int \frac{1}{\left( \frac{2x+1}{\sqrt 3}\right) + 1} dx = \frac{2}{\sqrt{3}} \int \frac{1}{u^2 + 1} du = \frac{2}{\sqrt{3}} \arctan \left( \frac{2x+1}{\sqrt{3}}\right) + C$$
Where we used the substitution $u = \frac{2x+1}{\sqrt{3}}$.
Hint:
There's no need to split the integral that way (IMO), we can instead write, \begin{align} \int \frac{3x+2}{x^2+x+1}\,\mathrm dx&=\frac{3}{2}\int\frac{2x+\tfrac{4}{3}}{x^2+x+1}\,\mathrm dx\\ &=\frac32\int\dfrac{2x+1+\tfrac13}{x^2+x+1}\,\mathrm dx\\ &=\dfrac32\left(\int\dfrac{2x+1}{x^2+x+1}\,\mathrm dx+\int\dfrac{{\small 1/3}}{x^2+x+1}\,\mathrm dx\right). \end{align} The first one is obvious. For the second one we should complete the square in the denominator, like this, $$\int\dfrac{1}{x^2+x+1}\,\mathrm dx=\int\dfrac{1}{\color{royalblue}{x^2+x+\tfrac14}+1-\tfrac14}\,\mathrm dx=\int\dfrac1{\color{royalblue}{\left(x+\tfrac1{2}\right)^2}+\tfrac34}\,\mathrm dx.$$ Now try to proceed further by rewriting that last expression so as to exploit the fact that the integral of $\frac{u'(x)}{u(x)^2+1}$ is $\arctan u(x)+\rm C$, you may have to factor something, and do one or more substitutions.
Let $3x+2=A\dfrac{d(x^2+x+1)}{dx}+B$
$\iff3x+2=A(2x+1)+B=2Ax+A+B$
$\implies2A=3,A+B=2$
$$\implies\int\dfrac{3x+2}{x^2+x+1}dx=A\cdot\dfrac{d(x^2+x+1)}{x^2+x+1}+B\dfrac{dx}{x^2+x+1}$$
$$\int\dfrac{dx}{x^2+x+1}=\int\dfrac4{(2x+1)^2+(\sqrt3)^2}dx$$
Set $2x+1=\sqrt3\tan y$ | 2021-05-11T10:37:22 | {
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https://math.stackexchange.com/questions/83383/midpoint-convex-and-continuous-implies-convex/83398 | # Midpoint-Convex and Continuous Implies Convex
Given that $$f\left(\frac{x+y}{2}\right)\leqslant \frac{f(x)+f(y)}{2}~,$$ how can I show that $f$ is convex.
Thanks.
Edit: I'm sorry for all the confusion. $f$ is assumed to be continuous on an interval $(a,b)$.
• You can't. At least, not without assuming that $f$ is continuous. – Willie Wong Nov 18 '11 at 15:06
• Lebesgue measurability is enough @WillieWong. This is a theorem by Sierpinski. (If I recall correctly). – Jonas Teuwen Nov 18 '11 at 15:11
• The trick in the continuous case is to set $z = \frac{p}{2^{n + 1}} x + \frac{q}{2^{n + 1}} y$ with $p + q = 2^{n + 1}$ and proceed by induction. – Jonas Teuwen Nov 18 '11 at 15:15
• @Jonas: right. I should have said "... something like '$f$ is continuous' ". The obvious counterexample is, of course, not Lebesgue measurable. Thanks for the correction. – Willie Wong Nov 18 '11 at 17:13
• – Martin Sleziak May 2 '17 at 13:28
Below is the proof of the fact that every midpoint-convex function is rationally convex, which I copied from my older post on a different forum.
If you add the condition that $f$ is continuous, then from rational convexity you will get convexity. (Note that if you are interested only in continuous functions, then it suffices to show the validity of $f(t x + (1-t)y)\le t f(x) + (1-t)f(y)$ for $t=\frac k{2^n}$ as suggested in Jonas' comment. The proof of this fact is a little easier. I've given a little more involved proof, since the relation between midpoint convexity and rational convexity seems to be interesting on its own.)
Maybe I should also mention that midpoint-convex functions are called Jensen convex by some authors.
Note that without some additional conditions on $f$, midpoint convexity does not imply convexity; see this question: Example of a function such that $\varphi\left(\frac{x+y}{2}\right)\leq \frac{\varphi(x)+\varphi(y)}{2}$ but $\varphi$ is not convex
Let $f: \mathbb R\to \mathbb R$ be a midpoint-convex function, i.e. $$f\left(\frac{x+y}2\right) \le \frac{f(x)+f(y)}2$$ for any $x,y \in \mathbb R$.
We will show that then this function fulfills $$f(t x + (1-t)y)\le t f(x) + (1-t)f(y)$$ for any $x,y\in \Bbb R$ and any rational number $t\in\langle0,1\rangle$.
Hint: Cauchy induction: see wikipedia or AoPS or answers to this post.
Proof. It is relatively easy to see that it suffices to show $f([x_1+\dots+x_k]/k)\le [f(x_1)+\dots+f(x_k)]/k$ for any integer $k$ (and any choice of $x_1,\dots,x_k\in \mathbb R$).
The case $k=2^n$ is a straightforward induction.
Now, if $2^{n-1}<k\le 2^n$, then we denote $\overline x=\frac{x_1+\dots+x_k}k$. Now from $$f(\overline x)=f\left(\frac{x_1+\dots+x_k+\overline x+\dots+\overline x}{2^n}\right) \le\frac{f(x_1)+\dots+f(x_k)+(2^n-k)f(\overline x)}{2^n},$$ where $2^n-k$ copies of $\overline x$ are summmed in the middle expression, we get $kf(\overline x) \le f(x_1)+\dots+f(x_k)$ by a simple algebraic manipulation.
The fact that measurability of $f$ is enough for the implication midpoint-convex $\Rightarrow$ convex to hold was mentioned in some of the comments above and in answers to the question I linked. Some references for this fact:
Constantin Niculescu, Lars Erik Persson: Convex functions and their applications, p.60:
H. Blumberg [31] and W. Sierpinski [226] have noted independently that if $f : (a, b) \to \mathbb R$ is measurable and midpoint convex, then $f$ is also continuous (and thus convex). See [212, pp. 220.221] for related results.
[31] H. Blumberg, On convex functions, Trans. Amer. Math. Soc. 20 (1919), 40–44.
[212] A. W. Roberts and D. E. Varberg, Convex Functions, Academic Press, New York and London, 1973.
[226] W. Sierpinski, Sur les fonctions convexes mesurables, Fund. Math. 1 (1920), 125–129.
Marek Kuczma: An introduction to the theory of functional equations and inequalities, p.241. He mentions the book T. Bonnesen and W. Fenchel, Theorie der konvexen Körper, Berlin, 1934 as an additional reference.
• Not to be too much of a whiner, but that "Spoiler" trick is annoying when the point of asking a question here is to get an answer. It's not like accidentally glancing at your text is going to reveal the answer, so if I want to avoid reading it, I avoid reading it. In addition, there is no way to show your text on displays that do not have "mouse-over" - it stays blank on my iPad no matter what I do, for example. There might be fora and questions where it is appropriate, but this doesn't sem to be one of them. – Thomas Andrews Nov 18 '11 at 15:51
• Sorry @Thomas, I've removed it. – Martin Sleziak Nov 18 '11 at 16:01
• @Hi, just double check. When I check the midpoint convexity, I have to show the inequality holds for any $x,y$ right? I can not just pick $x,y$ such that $f(x)=f(y)$, which looks like checking for quasi-concavity. – Bob Oct 7 '16 at 12:29
• @Bob Yes, by definition midpoint convex means that this inequality is true for any $x$, $y$. – Martin Sleziak Oct 7 '16 at 12:49
• @MartinSleziak, I find it is easy for people to make this mistake. Pick $x,y$ such that $f(x)=f(y)$, and once they showed $f(0.5x+0.5y)>f(x)$ they thought they proved concavity, but only got quasi-concavity. We can pick such two points ($x,y$ s.t. $f(x)=f(y)$) wolg for quasi-concavity, but not for concavity. – Bob Oct 7 '16 at 13:31
As was already previously mentioned it is straightforward to see that for a mid point convex function f $$f\left(\frac{x_1+x_2+\ldots+x_{2^n}}{2^n}\right)\leq\frac{1}{2^n}(f(x_1)+f(x_2)+\ldots f(x_{2^n}))$$ By setting $x_i=x$ for $1\leq i\leq m$ and $x_i=y$ for $m+1\leq i\leq 2^n$ where $1\leq m\leq 2^n$ is an integer we now obtain $$f\left(\frac{m}{2^n}x+(1-\frac{m}{2^n})y\right)\leq \frac{m}{2^n}f(x)+(1-\frac{m}{2^n})f(y)$$ Since the set of rational dyadic numbers of the form $\frac{m}{2^n}$ with $m$ and $n$ integers and $1\leq m\leq 2^n$ is dense in $[0,1]$ the proof follows from the continuity of f.
Can you provide any more information about $f$? For example, the property holds for continuous functions $f: I \rightarrow \mathbb{R}$, $I$ being an interval of real numbers. I think this result is due to Jensen [1].
Theorem (Jensen). Let $f: I\rightarrow\mathbb{R}$ be a continuous function. Then $f$ is convex if and only if it is midpoint convex, i.e. for $x,y$ in $I$ we have
$$f\left(\frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}$$
[1] J. L. W. V. Jensen, Sur les fonctions convexes et les inégalités entre les valeurs moyennes, Acta Math., 30 (1906), 175-193.
• Jensen doesn't prove this, per se, in this paper (his definition of "convex" was the modern-day "midpoint-convex"). It's more correct to say that he proved Jensen's Inequality (with arbitrary real weights) for functions which are midpoint convex and continuous. Of course, Jensen's Inequality with two arguments gives the modern definition of convexity. – Robert Wolfe Nov 9 '17 at 20:30
Given a weight lambda, take its binary expansion. Think of what midpoint convexity implies to the inequalities involving the partial sums of this binary expansion. Then the result follows by the continuity of f.
I would like to properly show by induction that if $$m\in\{0,1,2,\ldots,2^{n}-1\}$$ then $$f\left(\frac{m}{2^n}x+\Big(1-\frac{m}{2^k}\Big)y\right)\le \frac{m}{2^n}f(x) +\Big(1-\frac{m}{2^n}\Big)f(y).$$ and then the result will follows by contuinity since $$m=\lfloor2^nt\rfloor \in\{0,1,2,\ldots,2^{n}\} ~~~and~~\frac{\lfloor2^nt\rfloor}{2^n}\to t$$
The initial state $$n = 1$$ is trivial by hypothesis.
Now we assume that for every $$k, whenever $$m\in\{0,1,2,\ldots,2^{k-1}-1\}$$,we have $$f\left(\frac{m}{2^k}x+\Big(1-\frac{m}{2^k}\Big)y\right)\le \frac{m}{2^k}f(x) +\Big(1-\frac{m}{2^k}\Big)f(y). \tag{I}\label{I}$$
we want to prove \eqref{I} for $$k=n$$.
Let $$m \in\{0,1,2,\ldots,2^{n}-1\}$$ then the division by 2 yields $$m =2p +r$$ with $$r\in \{0,1\}$$ \begin{align}X&:=\left(\frac{m}{2^n}x+\Big(1-\frac{m}{2^n}\Big)y\right)= \left(\frac{2p +r}{2^n}x+\Big(1-\frac{2p +r}{2^n}\Big)y\right) \\&= \left(\frac{p }{2^n}x+\frac{p +r}{2^n}x+\Big(1-\frac{p }{2^n}y-\frac{p +r}{2^n}y\Big)y\right) \\&= \frac12 \left(\frac{p }{2^{n-1}}x+\frac{p +r}{2^{n-1}}x+\Big(2-\frac{p }{2^{n-1}}y-\frac{p +r}{2^{n-1}}y\Big)y\right) \\&=\frac12\left(\frac{p }{2^{n-1}}x+ \Big(1-\frac{p }{2^{n-1}}y\Big)\right)+\frac12\left(\frac{p+r }{2^{n-1}}x+ \Big(1-\frac{p +r}{2^{n-1}}y\Big)\right) \\&:= \color{red}{\frac{1}{2}\left(X_1+X_2\right)}\end{align}
On the other hand, since $$r\in \{0,1\}$$ and $$m\in\{0,1,2,\ldots,2^{n}-1\}$$ it is easy to check using parity that $$p,p+1 \in \{0,1,2,\ldots,2^{n-1}-1\}$$ that is $$p+r \in \{0,1,2,\ldots,2^{n-1}-1\}$$
By hypothesis of induction we obtain \begin{align}f(X) &= f\left(\frac{1}{2}\left(X_1+X_2\right)\right)\le \frac12f(X_1) +\frac12f(X_2) \\&= \frac12f\left(\frac{p }{2^{n-1}}x+ \Big(1-\frac{p }{2^{n-1}}y\Big)\right)+\frac12f\left(\frac{p+r }{2^{n-1}}x+ \Big(1-\frac{p +r}{2^{n-1}}y\Big)\right) \\&\le\frac12\left(\frac{p }{2^{n-1}}f(x)+ \Big(1-\frac{p }{2^{n-1}}f(y)\Big)\right)+\frac12\left(\frac{p+r }{2^{n-1}}f(x)+ \Big(1-\frac{p +r}{2^{n-1}}f(y)\Big)\right) \\&= \frac{2p+r}{2^n}f(x) +\Big(1-\frac{2p+r}{2^n}\Big)f(y)\\&= \frac{m}{2^n}f(x) +\Big(1-\frac{m}{2^n}\Big)f(y).\end{align} | 2019-02-20T22:33:18 | {
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https://math.stackexchange.com/questions/3907304/an-extension-of-baires-category-theorem | An extension of Baire's category theorem
In a topological space, a set is said to be rare if its closure has empty interior, and a set is said to be meager if it is a countable union of rare sets. If meager sets all have empty interior, then the topological space is said to be a Baire space. The following fact is known as Baire's category theorem.
Theorem. Complete metric spaces and locally compact Hausdorff spaces are Baire spaces.
[Zălinescu 2002] provides the following extension of this theorem for complete metric spaces, attributing it to C. Ursescu.
Theorem. ([Zălinescu 2002, Theorem 1.4.5]) Let $$X$$ be a complete metric space, and $$\{S_n\}$$ be a sequence of open sets in $$X$$. Then $$\mathrm{cl}(\cap_{n=1}^\infty S_n)$$ and $$\cap_{n=1}^\infty \mathrm{cl}(S_n)$$ have the same interior.
It is easy to check that a topological space is a Baire space if and only if any countable intersection of dense open sets is still dense in this space. Consequently, the theorem above implies that complete metric spaces are Baire spaces. Thus it is considered as an extension.
Question. Does the extension hold for locally compact Hausdorff spaces?
Update: As pointed out by @Alex Kruckman, the property in Ursescu's theorem is indeed equivalent to the fact that $$X$$ is a Baire space.
Theorem. A topological space is a Baire space if and only if $$\mathrm{cl}(\cap_{n=1}^\infty S_n)$$ and $$\cap_{n=1}^\infty \mathrm{cl}(S_n)$$ have the same interior for any sequence of open sets $$\{S_n\}$$.
Recall that a space is a Baire space if and only if any countable intersection of dense open sets is still dense. The theorem is essentially another way to state that a space is a Baire space if and only if all its open subspaces are Baire spaces.
Combined with Baire's category theorem, this observation by Alex proves the desired theorem. My answer below proves it from scratch, the proof being essentially the same as that of Baire's category theorem for locally compact Hausdorff spaces.
I will accept Alex's answer a few days later if nobody else has anything to add.
Thanks.
This is not really a strengthening of the Baire category theorem - it is equivalent to it. That is, any Baire space (and in particular any compact Hausdorff space) satisfies Ursescu's theorem.
Let's assume $$X$$ is a Baire space. Suppose $$(S_n)_{n\in \mathbb{N}}$$ is a sequence of open sets in $$X$$. We always have $$\text{int}(\text{cl}(\bigcap_{n=1}^\infty S_n))\subseteq \text{int}(\bigcap_{n=1}^\infty \text{cl}(S_n))$$, so it suffices to prove the reverse inclusion. And to do this, it suffices to prove that for any open $$U\subseteq \bigcap_{n=1}^\infty \text{cl}(S_n)$$, we have $$U\subseteq \text{cl}(\bigcap_{n=1}^\infty S_n)$$.
So fix some $$U$$. Note that an open subspace of a Baire space is Baire. [Why? If $$(V_n)_{n\in\mathbb{N}}$$ is a sequence of dense open subsets of $$U$$, then $$(V_n\cup (X\setminus \text{cl}(U)))_{n\in\mathbb{N}}$$ is a sequence of dense open subsets of $$X$$. The intersection of these sets is $$(\bigcap_{n=1}^\infty V_n)\cup (X\setminus \text{cl}(U))$$, and the fact that this is dense in $$X$$ implies that $$(\bigcap_{n=1}^\infty V_n)$$ is dense in $$U$$.]
So defining $$V_n = U\cap S_n$$, we have that each $$V_n$$ is open in $$U$$, and since $$U\subseteq \text{cl}(S_n)$$ for all $$n$$, each $$V_n$$ is dense in $$U$$. Since $$U$$ is Baire, $$\bigcap_{n=1}^\infty V_n$$ is dense in $$U$$, which implies $$U\subseteq \text{cl}(\bigcap_{n=0}^\infty S_n)$$, as was to be shown.
• Thanks, @Alex Kruckman. I modified the question to highlight your observation. In addition, I modified your answer, changing "Zălinescu's theorem" to "Ursescu's theorem", as suggested by [Zălinescu 2002]. Hope this fine with you. Thanks again. – Nuno Nov 15 '20 at 3:12
Here is my attempt to prove it. It is inspired by the proof of [Zălinescu 2002, Theorem 1.4.5] and the classical proof of Baire's theorem for locally compact Hausdorff spaces. Any comments or criticism will be appreciated. Thanks.
Theorem. Let $$X$$ be a locally compact Hausdorff space, and $$\{S_n\}$$ be a sequence of open sets in X. Then $$\mathrm{cl}(\cap_{n=1}^\infty S_n)$$ and $$\cap_{n=1}^\infty \mathrm{cl}(S_n)$$ have the same interior.
proof. It suffices to show that $$\mathrm{int}(\cap_{n=1}^\infty \mathrm{cl}(S_n))\subset \mathrm{cl}(\cap_{n=1}^\infty S_n)$$. To this end, we only need to prove for any given nonempty open set $$U \subset \cap_{n=1}^\infty \mathrm{cl}(S_n)$$ that $$$$\label{eq:usnonempty} U\cap(\cap_{n=1}^\infty S_n) \neq \emptyset.$$$$ In the sequel, we will define a sequence of compact sets $$\{C_n\}_{n=0}^\infty$$ such that $$C_n$$ has nonempty interior and $$$$C_{n+1} \subset C_{n} \subset U\cap(\cap_{k=1}^n S_k) \quad \text{ for each } \quad n \ge 0,$$$$ where $$\cap_{k=1}^0 S_k=X$$. Once this is done, Cantor's Theorem will yield $$$$\label{eq:nestnonempty} \emptyset \;\neq \;\cap_{n=1}^\infty C_n \;\subset\; U\cap (\cap_{n=1}^\infty S_n),$$$$ which gives us what we want.
We define $$\{C_n\}$$ inductively. As $$U$$ is a nonempty open set and $$X$$ is locally compact, we can take a compact set $$C_0\subset U$$ such that $$\mathrm{int}(C_0)\neq \emptyset$$. Assume that $$C_n$$ is already defined for an $$n\ge 0$$ so that $$C_n\subset U\cap(\cap_{k=1}^n S_k)$$ and $$\mathrm{int}(C_n)\neq \emptyset$$. Recalling that $$U\subset\mathrm{cl}(S_{n+1})$$, we have $$\mathrm{int}(C_n)\subset \mathrm{cl}(S_{n+1})$$, which implies that $$\mathrm{int}(C_n) \cap S_{n+1} \neq\emptyset$$. Since $$\mathrm{int}(C_n)\cap S_{n+1}$$ is open, invoking again the local compactness of $$X$$, we can take a compact set $$C_{n+1}\subset \mathrm{int}(C_n) \cap S_{n+1}$$ such that $$\mathrm{int}(C_{n+1})\neq \emptyset$$. Clearly, $$C_{n+1}\subset C_n$$ and $$C_{n+1}\subset C_{n}\cap S_{n+1} \subset U\cap(\cap_{k=1}^{n+1} S_k)$$. This finishes the induction and completes the proof. | 2021-04-22T03:48:45 | {
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https://math.stackexchange.com/questions/1438658/proof-that-any-linear-system-cannot-have-exactly-2-solutions/1438688 | Proof that any linear system cannot have exactly 2 solutions.
How would you go about proving that for any system of linear equations (whether all are homogenous or not) can only have either (if this is true):
• One solution
• Infinitely many solutions
• No solutions
I found this a bit difficult to prove (even though its a very fundamental thing about any linear equation). The intuitive geometric explanation is that a line can only intersect at one point, and if it intersects at a later point, it can't be a linear equation, but I don't think this is a convincing proof.
I thought of if you assume that there are two (or more, but I picked two) solutions for some linear system, then for the points in between
Solution Set 1: X1, X2....., Xn
Solution Set 2: X1, X2....., Xn
Then (I think), the points between S1 and S2, must be infinitely many points (and thus infinitely many solutions) such that these points can also satisfy the linear system, which would mean the system has 2 infinite solutions.
However, I don't think this is rigorous enough and nor do I understand completely why its true. Can anyone help in explaining (correcting) and elaborating on the intuition and proof of this?
• "Infinite solutions" is quite incorrect. It can have infinitely many solutions. But in correct use of mathematical terminology, "infinite solutions" means solutions, each one of which is infinite. If you have six solutions, and each is infinite, then those are infinite solutions but not infinitely many solutions. ${}\qquad{}$ – Michael Hardy Sep 16 '15 at 22:09
• Why do you find the visual proof not convincing? – corsiKa Sep 16 '15 at 22:53
• It's worth pointing out that this question has an implicit assumption that the underlying field is infinite. Over a finite field, the assertion is not necessarily true (and we cannot have infinitely many solutions). E.g. over $GF(2)$, the equation $\pmatrix{1&1\\ 0&0}x=0$ has exactly two solutions $x=\pmatrix{0\\ 0}$ and $x=\pmatrix{1\\ 1}$. See also 6005's answer below. – user1551 Sep 17 '15 at 11:35
Suppose that $\vec v$ and $\vec w$ are distinct solutions for the system $A\vec x = \vec b$ so that $A \vec v = A \vec w = \vec b$. Then $\frac{1}{2}(\vec v + \vec w)$ must be distinct from both $\vec v$ and $\vec w$ and must also solve the system since: $$A(\tfrac{1}{2}(\vec v + \vec w)) = \tfrac{1}{2}(A\vec v + A\vec w) = \tfrac{1}{2}(\vec b + \vec b) = \vec b$$ We can then apply the same argument to $\vec v$ and $\frac{1}{2}(\vec v + \vec w)$ in order to get infinitely many distinct solutions.
• Or pick any $\lambda \in (0,1)$ and consider $\lambda \vec{v} + (1 - \lambda) \vec{w}$. Then you get continuum-many solutions right off the bat. – 6005 Sep 16 '15 at 22:10
• You can even choose any real number for $\lambda$. – Aurey Sep 16 '15 at 23:30
• Looking over this again, great way of explaining your argument (deriving infinitely many solutions) – q.Then Sep 17 '15 at 22:29
Your second proof sounds fine. Let's imagine that the system of equations is written as $Ax = b$, where $A$ is the matrix of coefficients, $x$ is the vector of variables we are solving for, and $b$ are the constants. Now assume we have two distinct solutions $x_1$ and $x_2$. Then $rx_1+sx_2$ is also a solution, if $r+s=1$ - just plug it in and use linearity of matrix multiplication.
$$A(rx_1+sx_2) = A(rx_1)+A(sx_2) = rA(x_1)+sA(x_2) = (r+s)b = b$$
We thus have an infinite number of solutions. If you don't like matrices, you can easily just write out each term in the system of equations.
If you want to get some intuition for what is happening, the linear combination $rx_1+sx_2$, where $r+s=1$, is a line that connects the two points. When $r$ is 1 and $s$ is 0, we get point $x_1$. When $s$ is 1 and $r$ is 0, we get point $x_2$. Other combinations of $r$ and $s$ give other points.
You need to be more careful with your first argument, though. In general, when graphing systems of $n$ equations in $n$ unknowns, the equations are not lines, but $(n-1)$-dimensional planes in $n$-dimensional space. So for a system of three equations and three unknowns, for example, we have three planes in three dimensional space. Unless the planes are parallel, the intersection of two planes is a line, and the intersection of a line with a plane is a point. That's the one solution case. If two planes are parallel, they never intersect - no solutions. And finally, if the line/plane overlaps with another line/plane, we get an infinite number of solutions. That's the intuition behind this theorem.
• Thanks for the answer, I'm sure other answers were great as well, but I think you explained it the best (at least the best for me) – q.Then Sep 17 '15 at 2:55
We can represent a linear system in matrix notation as:$A\vec x=\vec v$. Now suppose that we have $A\vec x=\vec v$ and $A\vec y=\vec v$ with $\vec x \ne \vec y$, for $a,b$ with $a+b=1$ we have, by linearity: $A(a\vec x + b\vec y)=aA\vec x +b A \vec y=\vec v$, so we have infinitely many solutions.
In general, in a vector space over a general field $F$, the number of solutions $\vec{x}$ to the system $$A \vec{x} = \vec{w}$$ is either $0$ or $|\ker A|$. (Because if $\vec{x}_0$ satisfies the equation, then the set of solutions is $\vec{x}_0 + \ker A$.) And $|\ker A| = |F|^k$ for some $k \ge 0$.
Assuming the base field $F$ has infinite cardinality $\alpha$ and the vector space is finite-dimensional, it follows that the number of solutions is $0, 1,$ or $\alpha$ (since $|F|^0 = 1$ and $|F|^k = \alpha$ for $k \ge 1$).
• @user1551 yes, thank you. I corrected that and two other minor typos. – 6005 Sep 17 '15 at 15:20 | 2019-10-14T23:16:08 | {
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https://brilliant.org/discussions/thread/1-3/ | #1
How many positive integers less than $1000$ have the property that the sum of the digits of each such number is divisible by $7$ and the number itself is divisible by $3$?
Note by Vilakshan Gupta
3 years, 1 month ago
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Let's think of a number $abc (0 \leq a, b, c \leq 9)$. $a+b+c \equiv 0 (\mod 3 \text{and} \mod 7)$. Thus, $a+b+c=21$.
$(3, 9, 9) \rightarrow \frac{3!}{2}, (4, 8, 9) \rightarrow 3!, (5, 7, 9) \rightarrow 3!, (5, 8, 8) \rightarrow \frac{3!}{2}, (6, 6, 9) \rightarrow \frac{3!}{2}, (6, 7, 8) \rightarrow 3!, (7, 7, 7)$ $3+6+6+3+3+6+1=28$
Please tell me if there is any error.
- 2 years, 11 months ago
Nice method
- 2 years, 11 months ago
Good one brother
- 2 years, 11 months ago
Your method is quite efficient vis-a-vis mine. The latter involved manual trials with 3 digit integers with integer 1 to 9 at the unit. Thank you
- 1 year, 1 month ago
What ans did you get?
- 3 years, 1 month ago
Hey Aaron. How much are u getting with bonus? With bonus i am getting 12.
- 3 years, 1 month ago
25 is the answer as per me
- 3 years, 1 month ago
No - I get 28 too - I constructed a 0 - 9 by 0 -9 addition table in excel, and then started adding a 3rd digit to any number whose 2 digits had added to 12 or more - though now I think about it, I could just as easiky have srated my list with 399 and continued from there. And it has to be 28 cos it's one starting with3, 2 starting with 4, 3 starting with 5 etc, and 1+2+3+4+5+6+7 = 28
- 3 years, 1 month ago
28
- 3 years, 1 month ago
we just need to find the numbers which add upto 21
- 3 years, 1 month ago
- 3 years, 1 month ago
How mañy have you got right in PRMO - 17?
- 3 years, 1 month ago
@Md Zuhair Is the paper for 9,10,11 and 12 same?
- 3 years, 1 month ago
Yes Sir!
- 3 years, 1 month ago
unfortunately, i will get only 10 questions correct. I did very silly mistakes
- 3 years, 1 month ago
Sir😅 I am getting 10 along with bonus!
- 3 years, 1 month ago
Oh. U mean 8/28 u r getting?
- 3 years, 1 month ago
well, if it is bonus that means 2 questions marks are given extra.If it was been written question deleted then scores would be evaluated out of 28
- 3 years, 1 month ago
Geometry was quite tough and lengthy! Excluding bonus , I'm getting 8
- 3 years, 1 month ago
oh...btw,where do u live (i mean which region)
- 3 years, 1 month ago
Rajasthan..U?
- 3 years, 1 month ago
oh
- 3 years, 1 month ago
So , you are already selected..Great 👍
- 3 years, 1 month ago
- 3 years, 1 month ago
Coz as per cutoff(s) uploaded by Resonance , cutoff in Chandigarh is lower than others ( Rajasthan , Maharashtra , UP , etc) ... That's why!
- 3 years, 1 month ago
According to Resonance , cutoff in Chandigarh is just 4(questions)
- 3 years, 1 month ago
Ya. Thats ridiculous. WB region has always got a higher cutoff...
- 3 years, 1 month ago
i don't think it will be so low
- 3 years, 1 month ago
If that isnt, then wb will be higher and i will surely not qualify
- 3 years, 1 month ago
It's 11 in Rajasthan ! 😅😒
- 3 years, 1 month ago
LetTheFateDecide !!Bye
- 3 years, 1 month ago
which class are u in toshit?
- 3 years, 1 month ago
@Shreyan Chakraborty .. How much?
- 3 years, 1 month ago
JANI NA BAJE HOYECHE
- 3 years, 1 month ago
ANSWER IS 28....HAS A BIJECTION WITH a+b+c=21 WHERE 0<a,b,c<=9........
- 3 years, 1 month ago
Are na na.... I am not telling that. How much are you getting?
- 3 years, 1 month ago
@Shreyan Chakraborty The Hundreds digit can't be 1 or 2..
- 3 years, 1 month ago
yeah hundreds digit cant be 1,2
- 3 years, 1 month ago
@Md Zuhair @Vilakshan Gupta I haven't attempted one of the bonus question. Will I still get marks for it?
- 3 years, 1 month ago
I think the question can be cancelled as how can a person attempt to decinal answers and i had attempted ine. So i dunno.
- 3 years, 1 month ago
Yup.I think so.
- 3 years, 1 month ago
- 3 years, 1 month ago
I believe the answer is 28 integers. The sum of these integers' digits must be divisible by 21, since a number divisible by 3 also has its sum of digits divisible by 3; in addition to the sum of digits divisible by 7. None of the digits can be less 3 since the sum of digits would be less than 21. Possible combinations = 7+6+5+4+3+2+1 = (7+1)+(6+2)+(5+3)+4=3*8+4=28.
- 3 years, 1 month ago
Exactly
- 3 years, 1 month ago
I agree with all that, and I got the same answer, but if I give you 4 digits at random (say 3, 4, 5 and 6) and ask how many numbers you can make out of them, the answer is 432*1 = 24, not 4+3+2+1 = 10. What am I missing?
- 3 years ago
Ah - that's where the 28 comes from - much more mathematical than my just listing and counting them
- 3 years, 1 month ago
Hello, There are 28 postive integers left less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3 You can check out for more queries related to the JEE EXAMS from the following compilation
- 3 years ago
Hey I'm getting 8/27 from jharkhand as per the new answer key of hbcse.will I qualify???
- 3 years ago
Lets see....
- 3 years ago
Hey , what does discounted actually refer to?
- 3 years ago
Are the marks gonna be added to everyone's total or the questions will be cancelled (lowering the cutoff)?
- 3 years ago
They will be added to totsl
- 2 years, 11 months ago
The questions will be cancelled
- 3 years ago
@Pokhraj Harshal Ok! Then I'm also getting the same...
- 3 years ago
How much? Without the question?
- 3 years ago
Which region are u from??
- 3 years ago
WB rgion
- 3 years ago
And what about the other participants and their marks from your school. I mean the averages and the highest marks
- 3 years ago
@Pokhraj Harshal Rajasthan region!
- 3 years ago
@Md Zuhair 8
- 3 years ago
O i see....
- 3 years ago
this question came in this year PRMO answer is 28
- 2 years, 11 months ago
its sum is divisibli by 21 using this you can solve
- 2 years, 11 months ago
27
- 2 years, 11 months ago
28
- 2 years, 9 months ago
33
- 2 years, 2 months ago
zuhair tui ki amk jiggesh korchish??
- 3 years, 1 month ago
Accha.. nijer whatsapp number ta de... whatsapp e kotha bolchi
- 3 years, 1 month ago | 2020-09-29T07:19:53 | {
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https://mathhelpboards.com/threads/integration-computer-problem.5599/ | # Integration - computer problem
#### Yankel
##### Active member
Hello,
I was trying to solve the integral of
sin(x)*cos(x)
using the substitution method, what I did was:
u=sin(x) and that yields du/dx = cos(x) and then du=cos(x)*dx
that comes to an integral of u*du, which is easy u^2 / 2 +C. substituting back gives the final answer
sin(x)^2 / 2
But, when I ran this integral in both Maple and Mathematica, I got this answer:
-cos(x)^2 / 2
Now I tried asking Maple if the two answers are the same, but it failed. I tried checking myself, using the relation sin(x)^2+cos(x)^2=1, and got to the conclusion that they don't. I don't see what I did wrong here...
#### Ackbach
##### Indicium Physicus
Staff member
Technically, you answer was not $\sin^{2}(x)/2$, but $\sin^{2}(x)/2+C.$ Likewise, the computer's answer was not $-\cos^{2}(x)/2$, but $-\cos^{2}(x)/2+C$. Since your two answers differ by a constant (namely, $1$), they are really the same answer from a calculus perspective.
#### Bacterius
##### Well-known member
MHB Math Helper
Technically, you answer was not $\sin^{2}(x)/2$, but $\sin^{2}(x)/2+C.$ Likewise, the computer's answer was not $-\cos^{2}(x)/2$, but $-\cos^{2}(x)/2+C$. Since your two answers differ by a constant (namely, $1$), they are really the same answer from a calculus perspective.
Or more clearly, $C_1$ and $C_2$. This happened to me a while ago, it tends to occur a lot with trigonometric functions because of their periodic identities, so you often end up with very different looking expressions which are in fact equal..
#### Yankel
##### Active member
ah, I wasn't thinking about it....interesting.
well done to both of you ! thanks !
#### topsquark
##### Well-known member
MHB Math Helper
Just to be cheeky about it:
$$\int sin(x)~cos(x) dx = \frac{1}{2} \int 2 sin(x)~cos(x) dx = \frac{1}{2} \int sin(2x)dx$$
After u = 2x:
$$= \int sin(x)~cos(x) dx = \frac{1}{2} \int sin(u) \cdot \frac{1}{2} du$$
$$= -\frac{1}{4} cos(u) + C = -\frac{1}{4} cos(2x) + C$$
and upon using $$cos(2x) = cos^2(x) - sin^2(x)$$
$$\int sin(x)~cos(x) dx = \frac{1}{4} \left ( sin^2(x) - cos^2(x) \right ) + C$$
which contains both your sin and cos terms and is also off by a constant from the other solutions.
-Dan | 2021-07-29T18:32:15 | {
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https://math.stackexchange.com/questions/2158099/counting-integer-solutions-with-conditional-restrictions | # Counting integer solutions with conditional restrictions
Consider $x_i$, with $i=1,\ldots, 10$, such that $$5 \leq 3(x_1 + x_2 + x_3 + x_4) +2(x_5 + x_6 + x_7 + x_8 + x_9 + x_{10}) \leq 12\,,$$ where each $x_i$ can be either $0$ or $1$. In addition, each $x_i$ from $i=5$ to $10$ are restricted to be $0$ if some of the $x_i$ from $i=1$ to $4$ are equal to $1$, according to the following diagram:
Each $x_i$ in the bottom line is restricted by two of the $x_i$ in the top line. For example, if $x_1 = x_2 = 0$, then $x_5$ can be either $0$ or $1$, but if $x_1=1$ and/or $x_2 = 1$, then $x_5=0$ necessarily. The problem is to know in how many ways can we satisfy the inequality, subjected to these restrictions, using a method which is easily used for a computer, like generating functions, for example.
The way I'm trying to answer this is the following: if the additional restrictions did not exist, the number of solutions would be obtainable by using the generating function $$(1+x^3)^4 (1+x^2)^6$$ and the solution would be the sum of all coefficients from $x^5$ to $x^{12}$. The restrictions can be incorporated by defining $x^c_i$ such that, in the case of $i=5$ for example, $$x^c_5 = \quad \left\{ \begin{array}{lll} 0, \quad \, \textrm{if} \quad (1-x_1)(1-x_2)=0 \\ \\ x_5, \quad \, \textrm{if} \quad (1-x_1)(1-x_2)=1 \\ \end{array} \right. \,,$$ and so on for the other variables. Using these $x^c_i$ in the inequality, instead of the $x_i$ (for $i=5$ up to $10$), we have $$5 \leq 3(x_1 + x_2 + x_3 + x_4) +2(x^c_5 + x^c_6 + x^c_7 + x^c_8 + x^c_9 + x^c_{10}) \leq 12$$ which in principle should contain all the restricted possibilities. This is where I'm currently stuck, as I don't know how the generating function for this inequality should be built.
• Not the approach you are pursuing, but just building the characteristic function of the set yields a count of 87. – Fabio Somenzi Feb 23 '17 at 18:52
• Thanks for the reply! I'm not sure what you mean by characteristic function of the set though, could you be more specific as how you did it? – GKiu Feb 23 '17 at 20:15
• The characteristic function I referred to is a function of Boolean variables $x_1,\ldots,x_{10}$ that evaluates to true if and only if the constraints are all satisfied. It can be built as the conjunction of characteristic functions for the individual constraints. The incompatibility constraints are simple: for example $\neg x_5 \vee (\neg x_1 \wedge \neg x_2)$. The linear inequalities are a bit more complicated, but not much. Of course, I let the computer take the conjunctions and count the models. – Fabio Somenzi Feb 23 '17 at 21:31
• Although this works, I was looking for an answer in terms of generating functions like @Jeremy Dover suggested, but thank you for the clarification nonetheless. – GKiu Feb 24 '17 at 19:09
• Understood. That's why I posted a comment and not an answer. – Fabio Somenzi Feb 24 '17 at 19:45
This only really works because of the symmetry between the variables $x_1 \ldots x_4$ and $x_5 \ldots x_{10}$, but I think you can build a generating function this way:
If none of the $x_1 \ldots x_4$ are 1, then any of the $x_5 \ldots x_{10}$ can be 0 or 1 freely, so we get the generating function $(1+x^2)^6$ to count these solutions.
If exactly one of the $x_1 \ldots x_4$ is 1, then three of the $x_5 \ldots x_{10}$ must be zero, but the remainder are free, yielding the generating function $4x^3(1+x^2)^3$.
If exactly two of the $x_1 \ldots x_4$ are 1, then five of the $x_5 \ldots x_{10}$ must be zero with only one free, so we get the generating function $6x^6(1+x^2)$.
If three or four of the $x_1 \ldots x_4$ are 1, then all of the $x_5 \ldots x_{10}$ must be zero, so we get the generating function $4x^9+x^{12}$.
We can add all of these up to get the generating function: $$(1+x^2)^6 + 4x^3(1+x^2)^3 + 6x^6(1+x^2) + 4x^9+x^{12}$$ Like @Fabio Somenzi, when I calculate I also find 87 to be the answer.
• Thank you very much for your answer! This is exactly what I was looking for. – GKiu Feb 24 '17 at 19:07 | 2019-09-15T19:53:32 | {
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https://www.jiskha.com/questions/1870908/a-bag-contains-8-red-marbles-5-white-marbles-and-6-blue-marbles-you-draw-4-marbles-out | # statistics probability
A bag contains 8 red marbles, 5 white marbles, and 6 blue marbles. You draw 4 marbles out at random, without replacement. Find the following probabilities and round to 4 decimal places.
a. The probability that all the marbles are red is
b. The probability that none of the marbles are red is
1. 👍
2. 👎
3. 👁
1. prob(all 4 red) = (8/19)(7/18)(6/17)(5/16) = 35/1935
or
prob(all red) = C(8,4)/C(19,4) = 70/3876 = 35/1938
Prob(none red) = C(11,4)/C(19,4) = 330/3876 = 55/646
or
prob(none red) = (11/19)(10/18)(9/17)(8/16) = 55/646
1. 👍
2. 👎
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http://www.physicsforums.com/showthread.php?p=4042187 | ## How do I evaluate this log function?
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
I assume I can't use a calculator obviously.. so I'm quite stuck. The answer is 5, but I have no idea how to get that.
(log23)(log34)(log45) ... (log3132)
PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
Recognitions: Homework Help Use the fact that $$\log_b a = \frac{\log_c a}{\log_c b}$$ for any positive, real numbers a, b and c (with c > 1).
That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?
## How do I evaluate this log function?
Quote by feihong47 That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?
You use LaTeX, that is, the [ itex ][ /itex ] -tags (remove the spaces) . There's a guide in the forums somewhere if you're not familiar with it, I'll edit a link to it to this post if I find it. You can also open the "LaTeX Reference" by clicking the Σ in the toolbar.
If you don't have to write anything complicated, you can just use the quick symbols and x2 and x2 buttons in the toolbar. LaTeX looks a lot nicer & it's easier to read, though.
EDIT: Here's the LaTeX guide.
Mentor
Quote by feihong47 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution I assume I can't use a calculator obviously.. so I'm quite stuck. The answer is 5, but I have no idea how to get that. (log23)(log34)(log45) ... (log3132)
You could also approach this as follows:
Let $\displaystyle y=(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\,\dots\,( \log_{31}32)$
Then, $\displaystyle 2^y=2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}$
By laws of exponents and the definition of a logarithm,
$2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=\left(2^{(\log_{2}3)}\right)^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=3^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=\left(3^{(\log_{3}4)}\right)^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=4^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
etc.
$=\left(31^{(\log_{31}32)}\right)$
$=32$
Recognitions: Gold Member Science Advisor Staff Emeritus So, just in case others misunderstand, $2^y= 32= 2^5$ and therefore, y= 5 as the original poster said.
Very nice approach. Thanks! | 2013-05-25T17:27:22 | {
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http://math.stackexchange.com/questions/54088/what-is-the-difference-between-max-and-sup | # What is the difference between max and sup?
I am studying KS (Kolmogorov-Sinai) entropy of order q and it can be defined as
$$h_q = \sup_P \left(\lim_{m\to\infty}\left(\frac 1 m H_q(m,ε)\right)\right)$$
Why is it defined as supremum over all possible partitions P and not maximum?
When do people use supremum and when maximum?
-
I am not sure either of the tags fit. I do not know the term "KS entropy" so I cannot suggest a better retag. The essence of "what is $\sup$ and what is $\max$ was answered on this website before. Here is one answer, math.stackexchange.com/questions/18605/… – Asaf Karagila Jul 27 '11 at 15:55
Based on a quick google search I believe [dynamical-systems] fit well into this question. I (somewhat unwillingly) left [order-theory] since your second question can be generally answered within its confines. As my previous comment suggests, the second answer was indeed answered before (many times as well), I do not vote to close as the first answer may or may not been addressed to in other questions on the site. – Asaf Karagila Jul 27 '11 at 16:01
@Asaf, KS stands for Kolmogorov-Sinai, I saw that question earlier, but I still don't get when people use sup and when max. It's probably coz I need to spend more time to think over all the answers there. But thanks for the link anyway. Pls feel free to retag. – oleksii Jul 27 '11 at 16:03
"'Sup" is a hip way of asking "How are you?" – oosterwal Jul 27 '11 at 20:06
@oleksii: You have received several answers, all are very good in answering the question in the title. None of which answer the question "Why is it defined as supremum over all possible partitions $P$ and not maximum?". While I respect your wish to accept the answer, it would be nice to indicate that you have at least deduced the answer of the seemingly unanswered question from the accepted answer. Otherwise, this has become a duplicate of a question that has been answered here many times before. – Asaf Karagila Jul 28 '11 at 0:21
If the maximum exists, then the supremum and maximum are the same. However sometimes the maximum does not exist, and there is no maximal element. In this case it still makes sense to talk about a least upper bound.
The classic example is the set of all rationals whose square is less than or equal to $2$. That is the set $$A=\left\{ r\in\mathbb{Q}:\ r^{2}\leq2\right\}.$$
$A$ has no maximal element, however it does have a supremum and $\sup A=\sqrt{2}$.
An even simpler example is the set of all reals that are strictly less than $2$: $$B=\left\{ r\in\mathbb{R}:\ r<2\right\}.$$ This set has no maximum since for any $x\in B$ the element $\frac{x+2}{2}$ satisfies $x<\frac{x+2}{2}<2$. However it is not hard to see that $\sup B=2$.
-
Tnx, this is a nice answer, +1 for easy samples – oleksii Jul 27 '11 at 16:17
Consider for example the set $X = (0,1)$. Then $$\sup X=1$$ but $\max X$ does not exist.
Generally, for a set $X\subset {\mathbb R}$, we define $x:=\max X$ if $x\in X$ and $$\forall y\in X, y\leq x.$$
We define $x:=\sup X$ if $$\forall y\in X, y\leq x$$ and $$\forall\epsilon>0,\quad\exists y\in X, \quad\text{s.t.}\quad y>x-\epsilon.$$
-
I quite like your first answer, it was very plain and easy to get [for me :)]. Tnx. – oleksii Jul 27 '11 at 16:15
Supremum need not be attained while maximum is. When maximum exist, maximum=supremum.
-
The set of all negative numbers has a sup, which is 0, but not a max.
0 is the smallest number that no negative number can exceed.
-
A maximum is the largest number WITHIN a set. A sup is a number that BOUNDS a set. A sup may or may not be part of the set itself (0 is not part of the set of negative numbers, but it is a sup because it is the least upper bound). If the sup IS part of the set, it is also the max.
- | 2015-04-25T16:11:57 | {
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http://math.stackexchange.com/questions/850949/which-is-the-correct-way-to-calculate-the-expected-value-of-a-shared-lottery-jac | # Which is the correct way to calculate the expected value of a shared lottery jackpot?
I want to calculate the expected value of a ticket in a lottery game which gives players a probability $p$ of winning a jackpot prize of $j$ dollars. The total number of tickets in play is $t$.
If every winning ticket gets the full prize amount, the expected value for a ticket is given by $jp$. However, if winners must evenly split the prize in case of multiple winners, then the expected value depends on the number of winners $W$.
The expected number of winners is $tp$. The probability that the number of winners $W$ is $w = 0, 1, 2, \dotsc$, follows a Poisson distribution with the expected number of winners as its parameter:
$$P(W=w) \sim Pois(tp) = \frac{tp^we^{-tp}}{w!}$$
I don't know how to get from there to calculating an accurate expected value for the ticket as a function of the number of tickets in play.
In reading online, I've found two different methods each used by several sources. If I'm following them correctly, then they give different results. My question is 1) which one is correct? 2) what is the error in reasoning (or in my understanding/implementation) in the incorrect method?
## Method 1: Number of Winners
The first method calculates the probability that the number of winners $W$ will be $w = 0, 1, \dotsc, t$, given that there is at least one winner:
$$P(W=w | W>0) = \frac{P(W>0|W=w)P(W=w)}{P(W>0)}$$
Where,
• $P(W>0|W=w)$ is $\left\{ \begin{array}{lr} 0 & : w = 0\\ 1 & : w > 0 \end{array} \right.$
• $P(W=w)$ is the probability of $w$ winners: $\frac{tp^we^{-tp}}{w!}$
• $P(W>0)$ is the probability of more than one winner: $1 - P(W=0)$
So the expected value of the ticket is given by:
$$p\sum_{w=1}^{t} \frac{j}{w}\frac{P(W=w)}{1-P(W=0)}$$
For a numerical example, we'll tabulate the first few values of $P(W=w)$ for a lottery with a 1/34,220 chance of winning \$100,000 jackpot, with 6,000 tickets in play, so$p = 1/34,220; j = 100,000; \text{and } t = 6,000$$$\begin{array}{c|c|c|c|c|} \text{Winners} & \text{Probability} & \text{Conditional Probability} & \text{Share} & \text{Contribution } \\ w & P(W=w) & P(W=w|W>0) & j/w & (j/w)P(W=w|W>0) \\ \hline 0 & 0.839 & 0 & \text{\0} & \text{\0} \\ \hline 1 & 0.147 & 0.913 & \text{\100,000} & \text{\91,300} \\ \hline 2 & 0.013 & 0.081 & \text{\50,000} & \text{\4,050} \\ \hline \end{array}$$ Summing the contribution column and multiplying by$p$gives an expected value of$2.79.
## Method 2: Number of Other Winners
The second method calculates the probability that the number of total winners $W$ is $w = 0, 1, \dotsc, t$, given that our ticket is a winner:
$$P(W=w|Winner) = \frac{P(Winner|W=w)P(W=w)}{P(Winner)}$$
Where,
• $P(Winner)$ is the probability that our ticket is a winner: $p$
• $P(Winner|W=w)$ is the probability that our ticket is a winner given $w$ winning tickets: $w/t$
• $P(W=w)$ is the probability of $w$ winners: $\frac{tp^we^{-tp}}{w!}$
Plugging those figures in shows that $P(W=w|Winner)$ reduces to $P(W=w-1)$:
$$\frac{w}{t}\frac{P(W=w)}{p} = \frac{tp^{w-1}e^{-tp}}{(w-1)!} = P(W=w-1)$$
So the expected value is given by:
$$p\sum_{w=1}^{t}\frac{j}{w}\frac{tp^{w-1}e^{-tp}}{(w-1)!}$$
Using the same lottery numbers as above, the first few values of $w$ are given in the following table.
$$\begin{array}{c|c|c|c|c|} \text{Winners} & \text{Probability} & \text{Conditional Probability} & \text{Share} & \text{Contribution } \\ w & P(W=w) & P(W=w|Winner) & j/w & (j/w)P(W=w|Winner) \\ \hline 0 & 0.839 & 0 & \text{n/a} & \text{\0} \\ \hline 1 & 0.147 & 0.839 & \text{\100,000} & \text{\83,900} \\ \hline 2 & 0.013 & 0.147 & \text{\50,000} & \text{\7,350} \\ \hline \end{array}$$
Summing the contribution column and multiplying by $p$ gives an expected value of $2.67. ### Online Resources Which Use Method 2 Clearly the expected payout for the example lottery above cannot be both \$2.79 and \$2.67, but I'm having a difficult time reasoning my way to the correct method. Any hints will be appreciated! - ## 4 Answers Proof that Methods 2 and 3 are equivalent: Method 3 has a very intuitive appeal as the correct approach when interpreted as someone purchasing$allt$tickets. Now we'll show that Method 2 gives the same result if we use the binomial for the distribution of winning tickets instead of the Poisson. The expected value for Method 2 is: $$p\sum_{w=1}^t\frac jw {t-1 \choose w-1}p^{w-1}(1-p)^{t-w}$$ $$=j \sum_{w=1}^t \frac {(t-1)!}{w!(t-w)!}p^{w}(1-p)^{t-w}$$ $$=\frac jt \sum_{w=1}^t {t \choose w}p^{w}(1-p)^{t-w}$$ $$=\frac jt (1-(1-p)^t )$$ which is also the result for Method 3. If we had started with the Poisson distribution as an approximation to the binomial, then the expected value for Method 2 reduces to $$\frac jt(1-e^{-tp})$$ The Poisson will be a good approximation to the binomial when$t$is large,$p$is small and$tp$is moderate, which holds here. - Method 3: By linearity of expectation, the expected value of one ticket is$\frac{1}{t}$times the expected total value of all the tickets. That expected value is$jP(W>0)=j(1-P(W=0))=j(1-(1-p)^t)$. Thus the expected value of one ticket is $$\frac{j}{t}(1-(1-p)^t))$$ For the numerical values you use, I get \$2.68. I'm not skilled enough at statistics to find the logical flaw in Method 1, but I suspect Method 2 is right (up to rounding).
Followup: it seems that this solution, or a variation, was listed under "resources" supporting the second solution. I didn't notice it on my first read-through due to wall of text.
-
Thank you, vadim. Yeah, I had a link to this answer buried in the resources section, but your mentioning that expectation is linear helped me understand why it is correct without having to actually enumerate and sum the probability of each number of winners. That leaves me very confident that Method 2 is correct. However, I would like to find a good way of explaining why Method 1 is incorrect. – cristoper Jun 29 '14 at 5:25
I suspect that the problem with Method 1 is that the probability my ticket is a winner is no longer $p$ when we have conditioned on there being exactly $w$ winners. – vadim123 Jun 29 '14 at 5:34
This is my own summary answer. If somebody has a better way of explaining why the first method in my question goes wrong (and what it does compute, if anything useful), please post an answer!
## Method 1 is incorrect
If $P(Winner)$ is the probability that our ticket is a winner, then as user159813 pointed out, the expected payout from the lottery can be expressed as:
$$\sum_{w=1}^{t} \frac{j}{t} P(Winner \cap W=w)$$
However, the payout calculated by Method 1 is:
$$\sum_{w=1}^t \frac{j}{t} P(W>0 \cap W=w)$$
Which is different because $P(W>0) \ne P(Winner)$. We want $P(W=w)$ given that our ticket won, which is a lesser probability than given that any ticket won.
## Method 2 is correct
As vadim123 points out in his answer, the correctness of Method 2 is confirmed by the nice, closed-form formula derived from the fact that the expected value of a single ticket is the same as the expected value of all tickets divided by $t$:
$$\frac{j}{t}(1-(1-p)^t)$$
## Simulation
I ran a simulation using the same numbers as the example in my question. The mean payout after the 10,000,000th iteration was about \$2.64. After about 5,000,000 iterations, the mean seemed to oscillate about the$2.68 line.
-
There is a problem in Method 1. Letting $Y$ represent amount won and $W$ be number of winners we see that $$E(Y)=\sum_{w=0}^{t}\frac{j}{w}P(Winner\cap W=w)$$
The second method does this correctly by calculating $$P(Winner\cap W=w)=P(W=w|Winner)P(Winner)=P(W=w-1)P(Winner)=P(W=w-1)p$$
as you have above. Now the problem with Method 1 is that it says that $$P(Winner\cap W=w)=P(W=w|W>0)P(Winner)$$ which is not true. I would guess the method this person was going for is actually $$P(Winner\cap W=w)=P(Winner| W=w)P(W=w)$$
-
I don't see where Method 1 says $P(Winner \cap W=w)=P(W=w|W>0)P(Winner)$. However, I do see where it says it equals $P(W=w|W>0)P(W>0)$. And looking at your formulation of the expectation function, it is easier for me to see that Method 1 is wrong because $P(W=w|W>0)P(W>0) \ne P(W=w|Winner)P(Winner)$ – cristoper Jun 29 '14 at 20:34
And should the index in your sum start at $w=1$ to avoid the divide by zero? $P(Winner \cap W = 0) = 0$ anyway. – cristoper Jun 29 '14 at 20:47 | 2015-07-01T11:58:25 | {
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http://mezza-penna.it/teax/semi-log-regression.html | # Semi Log Regression
Log the Bugs in the Quality Center (Bug Logging and Tracking Tool) and track them. Measure distance each band traveled 3. This algorithm uses two k -nearest neighbor regressors with different distance metrics, each of which labels the unlabeled data for the other regressor where the labeling confidence is estimated through consulting the influence of the labeling of unlabeled examples on the labeled ones. Semi-logarithmic regressions, in which the dependent variable is the natural logarithm of the variable of interest, are widely used in empirical economics and other fields. Why is it that when you log-transform a power function, you get a straight line? To show you, let's remember one of the most fundamental rules of algebra: you can do anything you want to one side of an equation - as long as you do the exact same thing to the other side (We just LOVE that rule!). For many real-world problems, however, acquiri. 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We interpret the various log, log and semi-log coefficients and use the estimated regression model to make prediction and build a confidence interval for the prediction. eA B = eA=eB 2 Why use logarithmic transformations of variables Logarithmically transforming variables in a regression model is a very common way to handle sit-. The standard data points (concentration vs. You have three options: See this reference on using nonlinear regression to fit a straight line to your data. y is the response variable and x1, x2, and x3 are explanatory variables. A General Note: Logarithmic Regression. Because every disease has its unique survival pattern, it is necessary to find a suitable model to simulate followups. For motivational purposes, here is what we are working towards: a regression analysis program which receives multiple data-set names from Quandl. 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I know that usually having a linear-log model, an increase in x (GDP) by one percent is associated with an increase in y by (β1/100) units which would be for CRES (2,73/100). Charles says: July 22, 2015 at 2:41 pm. Four Parameter Logistic (4PL) Regression. Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. One axis is plotted on a logarithmic scale. the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below. COREG, is proposed. More speciflcally, one has found a point in a graph one is interested in, and now wants. to Leb esgue measure on [ ; + ] and denote f (x) the sp ectral. 22 Prob > chi2 = 0. A log-linear plot or graph, which is a type of semi-log plot. Semi-logarithmic regressions, in which the dependent variable is the natural logarithm of the variable of interest, are widely used in empirical economics and other fields. Bivariate Regression - Part I - Page 1. If I set the chart as the semi-log scale (semi-logarithmic axes), the regression line cannot be shown to a straight line. eA B = eA=eB 2 Why use logarithmic transformations of variables Logarithmically transforming variables in a regression model is a very common way to handle sit-. This kind of plot is useful when one of the variables being plotted covers a large range of values and the other has only a restricted range - the advantage being that it can bring out features in the data. Polynomial regression. It is partly a matter of custom. Our results indicate that models estimated with a square root link function perform much better than those with log- or linear-link functions. A General Note: Logarithmic Regression. the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below. log a a x = x The log base a of x and a to the x power are inverse functions. linear regression on levels of and log-transformed costs, gamma GLM with log-link, and the log-normal distribution, are not among the four best per-forming models with any of our chosen metrics. The present paper develops a mixture regression model that allows for distributional flexibility in modelling the likelihood of a semi-continuous outcome that takes on zero value with positive probability while continuous on the positive half of the real line. ∙ 0 ∙ share We consider semi-supervised regression when the predictor variables are drawn from an unknown manifold. What is the interpretation of this coefficient? 2. If B1=2, for instance, we could say that ’this model shows that factor X1 increases the predicted log count by 2 (all other factors held constant)’ because equation 1b- equation 1a= B1. DSOM 309 Chapter 16. Also known as elasticity interpretation. A log-linear plot or graph, which is a type of semi-log plot. Is it possible (and how) to transform one of these tables to semi-annual/annual table? If not, is there a way to calculate semi-annual/annual tables based on other data from this web site? Note - The issue is to calculate values based on free public available data, assuming paid access to CRSP etc. Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. In survival analysis, the proportional hazard model, also called the Cox model, is a classical semi-parameter method. Once you have used Excel to create a set of regular axes, converting the axes to semi-logarithmic axes in Excel is far from difficult. The main goal of this paper is to provide a fully probabilistic approach to modelling crime which reflects all uncertainties in the prediction of offences as well as the. And i do get R square (R2)= 0. 10 dan nilai VIF kurang dari 10. Nathanael Johnson (@savortooth on Twitter) is Grist's senior writer and the author of two books. 7 PROC ROBUSTREG Eample: Log-Log Regression With Weighted Outliers Example: Log-Log Regression With Weighted Outliers SAS/STAT ® 9. A semi-log graph is useful when graphing exponential functions. A test based on a modified LP regression that is consistent in both directions is provided. The equations of straight lines on logarithmic graph paper One purpose of logarithmic graph paper is simply to put wide ranges of. API Reference¶. n→∞, standard knn regression achieves the minimax bound on the MSE n−2 2+d up to log factors. All equations of the form. 1 Linear model: Yi = + Xi + i Recall that in the linear regression model, logYi = + Xi + i, the coefficient gives us directly the change in Y for a one-unit change in X. Question: A) Determine Which Regression Model (linear, Logarithmic And Semi-logarithmic) Will Best Represent The Relationship Between X (independent Variable) And Y (dependent Var- Iable) Below. The regression analysis fits the data, not the graph. Select the tab. In other words, if you go this route, you’ll need to do some research. Segmented regression, also known as piecewise regression or 'broken-stick regression', is a method in regression analysis in which the independent variable is partitioned into intervals and a separate line segment is fit to each interval. Interpretation of logarithms in a regression. The regression analysis fits the data, not the graph. 32 months and LT 90 of 6. y = λ a γ x. for linear regression has only one global, and no other local, optima; thus gradient descent always converges (assuming the learning rate α is not too large) to the global minimum. Spearman's correlation test was used to measure the correlation between two non-normally distributed variables or one normally with one non-normally distributed variable. Econometrics Working Paper EWP1101, Department of Economics, University of Victoria. The main difference is that a regression line is a straight line that represents the relationship between the x and y variable while a LOESS line is used mostly to identify trends in the data. The residual in a Cox regression model is not as simple to compute as the residual in linear regression, but you look for the same sort of pattern as in linear regression. We have some set of possible inputs, X, and a set of possible labels, Y. 08 * Density Ln + 583. The transformation of the data set from y vs. Machine Learning with Java - Part 1 (Linear Regression) Most of the articles describe "How to use machine learning algorithm in Python?". Under this setup, the localization parameter of the response variable distribution is modeled by using linear multiple regression or semi-parametric functions, whose non-parametric components may be approximated by natural cubic spline or P-splines. Polynomial Regression Analysis. yhat=b₀+b₁x which semi-log model transforms only the explanatory variable. The supported. How to access courses from ‘Coursera’ for FREE (with certificate). To create a log-log graph, follow the steps below for your version of Excel. In this article , we are going to discuss "How to use the machine learning alogithm with Java?". for which x<=0 if x is logged. log(AB) = logA+logB7. Namely, by taking the exponential of each side of the equation shown above we get. One new wrinkle we will add to this discussion is the use of faceting when developing plots. DNA microarray is a useful technique to detect thousands of gene expressions at one time and is usually employed to classify different types of cancer. Our task is to model the conditional probability p(yjx) for any pair (x;y) such that x2Xand y2Y. The estimate of β2 is 0. You have three options: See this reference on using nonlinear regression to fit a straight line to your data. It's not the fanciest machine learning technique, but it is a crucial technique to learn for many reasons:. The regression analysis fits the data, not the graph. We will use algebra and linear regression. Avoiding Common Math Mistakes. Once you have used Excel to create a set of regular axes, converting the axes to semi-logarithmic axes in Excel is far from difficult. Plot the location of each band (size and distance traveled) 5. The constant αin this model represents a kind of log-baseline hazard, since loghi(t)=α[or hi(t)=eα]whenallofthex’s are zero. 4 Log-Linear Models We now describe how log-linear models can be applied to problems of the above form. A novel weight strategy is presented to improve the prediction and its recursive algorithm is formulated, which adopts the incremental and decremental learning mechanism to update. This means that we can now use a simple linear regression model to describe the relationship. Regression analysis (integrated) Regression estimate (integrated). There have been a number of papers written on semi-parametric estimation methods of the long-memory exponent of a time series, some applied, others theoretical. If you have a nonlinear relationship, you have several options that parallel your choices in a linear regression model. When I add a linear trendline, Excel draws a straight line, even though a linear function should look curved when plotted on a logarithmic scale. the semi-log graph, which has a logarithmic vertical scale and a linear horizontal scale, as shown below. How can I fit my X, Y data to a polynomial using LINEST? As can be seem from the trendline in the chart below, the data in A2:B5 fits a third order polynomial. e-Exponential regression. DataAnalysis For Beginner This is R code to run semi-supervised regression based on Principal Component Analysis and Partial Least Squares (PCAPLS). regress definition: 1. GEE approach is an extension of GLMs. Of course, the ordinary least squares coefficients provide an estimate of the impact of a unit change in the independent variable, X, on the dependent variable measured in units of Y. measurement) are plotted on semi-log axes and a cubic regression curve is fitted through the points. Segmented regression analysis can also be performed on multivariate data by partitioning the various. Linear, Logarithmic, Semi-Log Regression Calculator By AAT Bioquest. So plotting Y and X*, where X* is the log of X, and performing a linear regression, you obtain a slope and intercept. We propose combination methods of penalized regression models and nonnegative matrix factorization (NMF) for predicting. Insert regression model into ggplot2. Regression analysis is a quantitative tool that is easy to use and can provide valuable information on financial analysis and forecasting. On a semi-log plot the spacing of the scale on the y-axis (or x-axis) is proportional to the logarithm of the number, not the number itself. 1c) Log(U)=Const+ B1 +B2X2+ So we can always say, as a simple function, that the coefficient B1 represents an increase in the log of predicted counts. Polynomial regression is commonly used to analyze the curvilinear data and this happens when the power of an independent variable is more than 1. log (Y) = a + b X The equation is estimated by converting the Y values to logarithms and using OLS techniques to estimate the coefficient of the X variable, b. com In Robust Regression, the outliers need not be disregarded: weights can be assigned and incorporated in the regression. Life Sciences is a solution especially designed for researchers and practitioners of life sciences who want to apply well-known and validated methods to analyze their data and build on their research. Nowadays, Semi-Supervised Learning lies at the core of the Machine Learning field trying to effectively exploit unlabeled data as much as possible, together with a small amount of labeled data aiming to improve the predictive performance. , the percentage. Obtain estimates of the Bifurcation Ratio, R B, the Length Ratio, R L, and the Area Ratio, R A, using the data tabulated below. commonly used in practice. gaussianprocess. Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. Articulate assumptions for multiple linear regression 2. Estimate the doubling time from the slope. Journal of the American Statistical Association: Vol. −Maximum likelihood method. If B1=2, for instance, we could say that 'this model shows that factor X1 increases the predicted log count by 2 (all other factors held constant)' because equation 1b- equation 1a= B1. is the core plug semi-log regression, and the dashed green lines indicate the approximate 90% boundaries of the core plug data. Byrne , d Igor Chourpa a and Emilie Munnier a. Poisson regression for contingency tables, a type of generalized linear model. logarithmic model. , semielasticity), with respect to the dummy regressor taking values of 1 or 0, can be obtained as (antilog of estimated 02) - 1 times 100, that is, as. We have some set of possible inputs, X, and a set of possible labels, Y. Using LINEST for Nonlinear Regression in Excel. Nonlinear functional. It is closely related to semi-supervised learning based on support vector regression (SVR). Linear and Logarithmic Interpolation Markus Deserno Max-Planck-Institut f˜ur Polymerforschung, Ackermannweg 10, 55128 Mainz, Germany (Dated: March 24, 2004) One is occasionally confronted with the task of extracting quantitative information out of graphs. 08 * Density Ln + 583. Before you model the relationship between pairs of. The left panel depicts a semi-log model and the right panel depicts a polynomial model. A logarithmic curve fit is generally used with data that spans decades (10 0, 10 1, 10 2, and so on). Identify and define the variables included in the regression equation 4. Level-level regression is the normal multiple regression we have studied in Least Squares for Multiple Regression and Multiple Regression Analysis. That's because logarithmic curves always pass through (1,0) log a a = 1 because a 1 = a Any value raised to the first power is that same value. But, i don't understand why it was said that the value closer to 1 is a better indicator to show that my standard curve is good to determine the protein concentration. The data has 1,000 observations on 4 variables. edu Abstract Large amounts of labeled data are typically required to train deep learning models. Regression is nonlinear when at least one of its parameters appears nonlinearly. It will provide four different linear regressions: linear-linear, log-linear, linear-log and log-log. It is estimated by regression using the wavelet coefficients of the time series, which are dependent when d ≠ 0. With ANOVA, you assign people to treatments, and all sorts of. Functional regression How to relate functional responses to scalar, explanatory variables? Available functional regressions models: Semi-parametric approaches: I additive effects models (Ramsay & Silverman, 2005) (R package fda on CRAN and R-Forge) I multiplicative effects models (Chiou et al. These correspond to a latent variable with the extreme-value distribution for the maximum and minimum respectively. Non-Linear Relationships 169 8 resulting least squares regression line will give us and estimate for hc Log-log and semi-log plots Graphs of log(y) vs. The logarithmic fit calculates the least squares fit through points by using the following equation: where a and b are constants and ln is the natural logarithm function. Since then, regression has been studied. While existing semi-supervised methods have shown some promising empirical performance, their development has been based largely basedon heuristics. Abbott Preparing for Your Stata Session Before beginning your Stata session, use Windows Explorer to copy the Stata- format dataset auto1. Logarithmic. methods for classification and regression problems. Byrne , d Igor Chourpa a and Emilie Munnier a. • On the computers in Dunning 350, the default Stata working directory is. Polynomial regression. In addition, we modify our underlying approximate homomorphic encryption scheme for performance improvement. This is called a semi-log estimation. Fitting Parametric and Semi-parametric Conditional Poisson Regression Models with Cox's Partial Likelihood in Self-controlled Case Series and Matched Cohort Studies Stanley Xu1, Paul Gargiullo2, John Mullooly3, David McClure1, Simon J. Downloadable! Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. I have a Scatter chart with logarithmic scale on the horizontal axis and linear scale on the vertical axis. Now, find the least-squares curve of the form c 1 x + c 2 which best fits the data points (x i, φ i). 05/26/18 - Large amounts of labeled data are typically required to train deep learning models. 397973 * Density Ln^2 + 0. All equations of the form = form straight lines when plotted semi-logarithmically, since taking logs of both sides gives. We establish a Central Limit Theorem (CLT) for the resulting estimator. 08 * Density Ln + 583. Logistic regression & stochastic gradient descent Parametric Fast to train and evaluate Easy to incrementally train x 2Rn;y 2f 1;+1g P(yjx) = 1 1 + exp( ywTx) maximize w Y m P(y(m)jx(m)) minimize w XM m=1 log(1 + exp( y(m)wTx(m))) M might be giant, or you might not have access to them all at one time. LOG-PERIODOGRAM REGRESSION OF TIME SERIES WITH LONG RANGE DEPENDENCE 1 1. Log-log and semi-log plots Graphs of log(y) vs. This algorithm uses two k -nearest neighbor regressors with different distance metrics, each of which labels the unlabeled data for the other regressor where the labeling confidence is estimated through consulting the influence of the labeling of. Generalized and "working" Wald and score tests for regression coefficients in the class of semi-parametric marginal generalized linear models for cluster correlated data (Liang and Zeger, 1986) are proposed, and their asymptotic distribution examined. Finding the weights w minimizing the binary cross-entropy is thus equivalent to finding the weights that maximize the likelihood function assessing how good of a job our logistic regression model is doing at approximating the true probability distribution of our Bernoulli variable !. The excessive number of concepts comes because the problems we tackle are so messy. 0 open source license. We extend these results by establishing the exact sampling. Fully specified by a mean function and covariance function. While existing semi-supervised methods have shown some promising empirical performance, their development has been based largely based on heuristics. It is partly a matter of custom. 10 dan nilai VIF kurang dari 10. Calculate the standard deviation of the log10 residual, then square it and multiply it by 1. Linear regression is one of the most popular statistical techniques. You wish to have the coefficients in worksheet cells as shown in A15:D15 or you wish to have the full LINEST statistics as in A17:D21. Set of tools to fit a linear multiple or semi-parametric regression models and non-informative right-censoring may be considered. The data has 1,000 observations on 4 variables. Some using Fourier methods, others using a wavelet-based technique. This is the only graph type that will work; other graph types permit logarithmic scales only on the Y axis. The elasticity evaluated at the mean is:. Polynomial regression is commonly used to analyze the curvilinear data and this happens when the power of an independent variable is more than 1. The parameter d is the one of interest. Dengan regresi Semi-Log yaitu variabel dependen dalam bentuk logaritma natural dan semua variabel independen tetap dirubah, dapat disimpulkan tidak terdapat multikolinearitas hal ini ditunjukkan oleh nilai Tolerance di atas 0. 1539 B 70000000 52300000 1. What is the interpretation of this coefficient? 3. by using Douglas’s “laboriously compiled” data to fit the linear regression Log(P/C) = b + kLog(L/C) by ordinary least squares. Econometrics and the Log-Linear Model By Roberto Pedace If you use natural log values for your dependent variable ( Y ) and keep your independent variables ( X ) in their original scale, the econometric specification is called a log-linear model. This produces the following output. race smoke ptl ht ui Logistic regression Number of obs = 189 LR chi2(8) = 33. It is equivalent to converting the y values (or x values) to their log, and plotting the data on lin-lin scales. This paper proposes a risk prediction model using semi-varying coefficient multinomial logistic regression. The example data can be downloaded here (the file is in. Kennedy, P. 13 and Table 15. Goodness-of-fit is a measure of how well an estimated regression line approximates the data in a given sample. generate lny = ln(y). After my previous rant to discussion with her about this matter, I've tried to stay on the straight and narrow. Even when there is an exact linear dependence of one variable on two others, the interpretation of coefficients is not as simple as for a slope with one dependent variable. Logarithmic regression. In such cases, applying a natural log or diff-log transformation to both dependent and independent variables may. One such measure is the correlation coefficient between the predicted values of $$y$$ for all $$x$$-s in the data file and the. The regression results based on ECM [Table 15. A General Note: Logarithmic Regression. 61 months ( Table 6 ). Sadly, most browser are unable to play this format. The regression analysis fits the data, not the graph. logarithmic model. eA+B = eAeB 10. The comparison of methods experiment is critical for assessing the systematic errors that occur with real patient specimens. Change one or both axes to a logarithmic scale. A novel weight strategy is presented to improve the prediction and its recursive algorithm is formulated, which adopts the incremental and decremental learning mechanism to update. The regression coefficient associated with the Z term (i. In this paper, we compare the Fourier and wavelet approaches to the local regression method and to the local Whittle method. Gowher, The exponential regression model presupposes that this model is valid for your situation (based on theory or past experience). In instances where both the dependent variable and independent variable(s) are log-transformed variables, the relationship is commonly referred to as elastic in econometrics. Gaussian processes GP(m(x),k(x,x’)) Distribution over functions. A test based on a modified LP regression that is consistent in both directions is provided. Hence the term proportional odds logistic regression. Namely, by taking the exponential of each side of the equation shown above we get. Polynomial Regression Analysis. 952<1 since it is evidently. logistic regression (S2MLR) model which exploits both hard and soft labels. Plotting with Microsoft Excel 2 form of categories. ab-Exponential regression. See the Topic 6. This curve fit cannot be used to fit negative data or data equal to. API Reference¶. Results are generated immediately, no external software needed. A Poisson regression model is sometimes known as a log-linear model. [1 point] Suppose the regression model is logarithmic: log(Y ) = β1 + β2 log(X) + u. In the case of linear regression, one additional benefit of using the log transformation is interpretability. Obtain your results in a few simple clicks without having to leave MS Excel where your data is stored. The standard data points (concentration vs. Part two explains semi-parallel logisitic regression in R based on iteratively reweighted least squares (equivalent to glm), with and without covariates. Pairwise Log-rank Test P Value Table Semi-Parametric Cox Regression Cox Regression , also known as Cox proportional hazard regression, assumes that if the proportional hazards assumption holds (or, is assumed to hold), then it is possible to estimate the effect parameter(s) without any consideration of the hazard function. If I set the chart as normal scale (numeric-numeric), the regression line can be shown to a straight line. Suppose a data set is actually following the trend of some hidden exponential function y = a b x. This kind of plot is useful when one of the variables being plotted covers a large range of values and the other has only a restricted range - the advantage being that it can bring out features in the data. Quadratic regression. Excel 2010 or 2007. Thus, software originally devel-. For instance, if you are graphing time versus bacterial growth. 952<1 since it is evidently. A log-linear plot or graph, which is a type of semi-log plot. This formula estimates the doubling time, which does not depend on the value of Y, only on the slope at t 0. Data can be directly from Excel or CSV. For example, if the raw output ($$y'$$) of a linear model is 8. Semi-log and Log-log plots Posted 02-04-2010 (5361 views) I have tried Googling and searching the SAS documentation, but I cannot find any syntax to create a semi-log or log-log plot. It was not until the early 19th century that Gauss and Legendre developed a systematic pro-cedure: the least-squares method. Semi-Supervised Classification with Graph Convolutional Networks. Horton's Laws - Example Jorge A. 2000 Simcoe Street North Oshawa, Ontario L1G 0C5 Canada. Plotting with Microsoft Excel 2 form of categories. Click Analyze, choose Nonlinear regression (not Linear regression) and then choose one of the semi-log or log-log equations from the "Lines" section of equations. Thus, software originally devel-. The focus of the paper is on a theoretical analysis of semi-supervised regression techniques, rather than the development of practical new algorithms and techniques. What is the interpretation of this coefficient? 3. webuse lbw (Hosmer & Lemeshow data). In a semilogarithmic graph, one axis has a logarithmic scale and the other axis has a linear scale. 13 and Table 15. OK, you ran a regression/fit a linear model and some of your variables are log-transformed. How can I fit my X, Y data to a polynomial using LINEST? As can be seem from the trendline in the chart below, the data in A2:B5 fits a third order polynomial. The main goal of this paper is to provide a fully probabilistic approach to modelling crime which reflects all uncertainties in the prediction of offences as well as the. Suppose a data set is actually following the trend of some hidden exponential function y = a b x. 1c) Log(U)=Const+ B1 +B2X2+ So we can always say, as a simple function, that the coefficient B1 represents an increase in the log of predicted counts. A little over three months ago, my wife and I sat down at the dining table to talk about the future. If the law were a perfect description of the situation, all the points on the log-log or semi-log plot would fall along a straight line. Stochastic gradient descent: take gradient. Excel 2010 or 2007. Under this setup, the localization parameter of the response variable distribution is modeled by using linear multiple regression or semi-parametric functions, whose non-parametric components may be approximated by natural cubic spline or P-splines. HTH Martin. the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below. The following lesson estimates a log, log and semi-log regression model. 72 Interpretation of Regression Coefficients: Elasticity and Logarithmic Transformation. The specific applications of log-linear models are where the output quantity lies in the range 0 to ∞, for values of the independent variables X , or more immediately, the transformed quantities f i ( X. Using global regression to fit incomplete datasets Fitting models where the parameters are defined by multiple data sets Column constants Advice: Don't use global regression if datasets use different units Outlier elimination and robust nonlinear regression When to use automatic outlier removal When to avoid automatic outlier removal. To avoid this problem, we […]. 4669 D 17500000 10560000 0. Regression definition is - the act or an instance of regressing. Hazard regression models for pr(T2 > t|Z0) have been thoroughly studied in this scenario, for example, multiplicative hazards, additive hazards, and accelerated failure-time models. Valuation of Specific Crime Rates: Final Report William Alan Bartley Semi-log Regression with Index Crimes for Rent Equation Total Costs per Household of Crime Variables in Semi-log Regressions using metropolitan area percentage changes for the Manufacturing. Inverse regression. regression definition: 1. how all functions (linear, semi log, double log and exponential) can be applied. The first form of the equation demonstrates the principle that elasticities are measured in percentage terms. Spotfire uses a nonlinear regression method for this calculation. Log periodogram (LP) regression is shown to be consistent and to have a mixed normal limit distribution when the memory parameter d = 1. I like to use log base 10 for monetary amounts, because orders of ten seem natural for money: $100,$1000, $10,000, and so on. ORDER STATA Logistic regression. We find that the correlation coefficient R SL of a regression straight line to these data is less than R LL and equals 0. Converted from a tradingview code. to return to a previous and less advanced or worse state, condition, or way of behaving: 2. Linear and semi-log regression model - Cross Validated. Email or Customer ID. In these cases, graphing with semi-log axes is helpful. The logarithmic fit calculates the least squares fit through points by using the following equation: where a and b are constants and ln is the natural logarithm function. DSOM 309 Chapter 16. We use the command "LnReg" on a graphing utility to fit a logarithmic function to a set of data points. It commonly sorts and analyzes data of various industries like retail and banking sectors. R Nonlinear Regression Analysis. Semi-logarithmic regressions, in which the dependent variable is the natural logarithm of the variable of interest, are widely used in empirical economics and other fields. Articulate assumptions for multiple linear regression 2. Rajah - Age Change: Regression. Semi-logarithmic regressions, in which the dependent variable is the natural logarithm of the variable of interest, are widely used in empirical economics and other fields. Semi-sup ervised learning is among the problems considered, and a se-ries of exp erimen ts sho ws that our second prop osal, self-consisten t lo-gistic regression is a serious con-tender to more classical solutions in-v olving generativ e mo dels. logbin is an R package that implements several. Set of tools to fit a linear multiple or semi-parametric regression models and non-informative right-censoring may be considered. Under this setup, the localization parameter of the response variable distribution is modeled by using linear multiple regression or semi-parametric functions, whose non-parametric components may be approximated by natural cubic spline or P-splines. In Part 3 of this series on Linear Regression I will go into more detail about the Model and Cost function. Semi‐linear mode regression. Learn more. You either can't calculate the regression coefficients, or may introduce bias. It is equivalent to converting the y values (or x values) to their log, and plotting the data on lin-lin scales. 2 Log level regression function log wage 0 584 0 083 educ n 526 R 2 0 186 The from ECONOMIC 1 at Peking University. After my previous rant to discussion with her about this matter, I've tried to stay on the straight and narrow. It also follows immediately (by considering transposes) that every is PSD. University of Hertfordshire Business School. A log-linear (or "semi-log") model takes the form ln(Y) 1 X 0. And a brave talk it was, with Matt including real-time data analysis of stock exchange data. This kind of plotting method is useful when one of the variables being plotted covers a large range of values and the other has only a restricted range - the advantage being that it can bring out. The trendline feature of Excel 2010 seems to malfunction. A log transformation allows linear models to fit curves that are otherwise possible only with nonlinear regression. This is the class and function reference of scikit-learn. , 2003) (R package fmer soon on CRAN) I. Optimized Skeleton-based Action Recognition via Sparsified Graph Regression. This should give you a line of best fit through your data points. One axis is plotted on a logarithmic scale. log(A=B) = logA logB8. See all articles by Chris Tofallis Chris Tofallis. Alternately, class values can be ordered and mapped to a continuous range:$0 to $49 for Class 1;$50 to \$100 for Class 2; If the class labels in the classification problem do not have a natural ordinal relationship, the conversion from classification to regression may result in surprising or poor performance as the model may learn a false or non-existent mapping from inputs to the continuous. , age, country, etc. The estimate of β2 is 0. The first is called a semi-log graph. Amit Moscovich, Ariel Jaffe, Boaz Nadler Semi-supervised regression on unknown manifolds, presented at the Princeton math department, Hebrew university learning club and statistics seminar, Tel-Aviv university statistics and machine learning seminars and the Ben-Gurion CS seminar. The excessive number of concepts comes because the problems we tackle are so messy. The SIR adjusts for various facility and/or patient-level factors that contribute to HAI risk within each. log (Y) = a + b X The equation is estimated by converting the Y values to logarithms and using OLS techniques to estimate the coefficient of the X variable, b. So far, we have learned various measures for identifying extreme x values (high leverage observations) and unusual y values (outliers). Although regression coefficients are sometimes referred to as partial-slope coefficients, in a log-log model the coefficients don’t represent the slope (or unit change in your Y variable for a unit change in your X variable). Regression analysis (integrated) Regression estimate (integrated). Obtain estimates of the Bifurcation Ratio, R B, the Length Ratio, R L, and the Area Ratio, R A, using the data tabulated below. Uses of Partial and Semipartial The partial correlation is most often used when some third variable z is a plausible explanation of the correlation between X and Y. This technique of model building helps to identify which predictor (independent) variables should be included in a multiple regression model(MLR). This can be done for the log likelihood of logistic regression, but it is a lot of work (here is an example). The file cocoa. 2 User’s Guide, support. Also, for the same data, a) determine the area of the basin, b) the total length of streams, c) the drainage density, D d, and d) the average length of overland flow, L o. Log linear analysis is something else - it is used when you have multiple categorical variables. This paper focuses on semi-functional partially linear regression model, where a scalar response variable with missing at random is explained by a sum of an unknown linear combination of the components of multivariate random variables and an unknown transformation of a functional random variable which takes its value in a semi-metric abstract space $${\mathscr {H}}$$ with a semi-metric $$d. Click Analyze, choose Nonlinear regression (not Linear regression) and then choose one of the semi-log or log-log equations from the "Lines" section of equations. Segmented regression, also known as piecewise regression or 'broken-stick regression', is a method in regression analysis in which the independent variable is partitioned into intervals and a separate line segment is fit to each interval. COREG, is proposed. Supervised Nonlinear Factorizations Excel In Semi-supervised Regression 3 { Conducted a throughout empirical analysis against the state of the art (man-ifold regularization) 2 Related Work Even though a plethora of regression models have been proposed, yet Sup-port Vector Machines (SVMs) are among the strongest general purpose learning models. webuse lbw (Hosmer & Lemeshow data). Such a scale is nonlinear: the numbers 10 and 20, and 60 and 70, are not the same distance apart on a log scale. How to interpret log transformed independent variable in logistic regression 17 Feb 2017, 14:28. © 2007 - 2019, scikit-learn developers (BSD License). In logistic regression, the dependent variable is a logit, which is the natural log of the odds, that is, So a logit is a log of odds and odds are a function of P, the probability of a 1. Linear and Logarithmic Interpolation Markus Deserno Max-Planck-Institut f˜ur Polymerforschung, Ackermannweg 10, 55128 Mainz, Germany (Dated: March 24, 2004) One is occasionally confronted with the task of extracting quantitative information out of graphs. How to Interpret Logistic Regression Coefficients. The bad news is that linear regression is seldom a good model for biological systems. 5548 A 140000000 140000000 1. This is called a semi-log estimation. Linear regression is one of the most popular statistical techniques. The parameter d is the one of interest. LOG-PERIODOGRAM REGRESSION OF TIME SERIES WITH LONG RANGE DEPENDENCE 1 1. You have three options: See this reference on using nonlinear regression to fit a straight line to your data. This paper describes the use of machine learning techniques to implement a Bayesian approach to modelling the dependency between offence data and environmental factors such as demographic characteristics and spatial location. • On the computers in Dunning 350, the default Stata working directory is. produced Proportional Value job rates using the free-hand method, as shown on the previous page, but chose to verify these results and construct a job rate line by using a computer and a statistical method called regression analysis. If there is a vertical discontinuity at the cutoff it will be estimated by this coefficient. Polynomial Regression Analysis. We also include the plot of the log of N/S versus the radius r in this figure as Fig. Even when there is an exact linear dependence of one variable on two others, the interpretation of coefficients is not as simple as for a slope with one dependent variable. The idea here is we use semilog or log-log graph axes so we can more easily see details for small values of y as well as large values of y. Rajah - Age Change: Regression. The predicted values from the log-log model are saved in the variable named YHAT2. kind of baseline log-hazard, because logh i(t) = , or h i(t) = e , when all of the xs are zero. This model is known as the 4 parameter logistic regression (4PL). Learn more about semi, log. 16 First review the linear-log form: Using logs to transform a variable on the right-hand side of the equation allowed us to unbend a concave line into a straight one. In this article, I've discussed the basics and semi-advanced concepts of regression. This algorithm uses two k -nearest neighbor regressors with different distance metrics, each of which labels the unlabeled data for the other regressor where the labeling confidence is estimated through consulting the influence of the labeling of. Minimising assumptions: semi-parametric regression. Let's apply some simple regression analysis (see footnote below) to the question. The difficulty comes because there are so many concepts in regression and correlation. generate lny = ln(y). Show how to manually create partial and semipartial correlations using residuals from a regression model. measurement) are plotted on semi-log axes and a cubic regression curve is fitted through the points. Geodesic knn regression Step 1: Connect every pair of close points by an edge. linear regression. R Nonlinear Regression and Generalized Linear Models:. I know that usually having a linear-log model, an increase in x (GDP) by one percent is associated with an increase in y by (β1/100) units which would be for CRES (2,73/100). After my previous rant to discussion with her about this matter, I've tried to stay on the straight and narrow. 397973 * Density Ln^2 + 0. The example data can be downloaded here (the file is in. In science and engineering, a semi-log graph or semi-log plot is a way of visualizing data that are related according to an exponential relationship. Unconditional logistic regression (Breslow & Day, 1980) refers to the modeling of strata with the use of dummy variables (to express the strata) in a traditional logistic model. When this option is used the elasticities at sample means are computed assuming a semi-logarithmic model specification where the dependent variable is in log form but the explanatory variables are in levels. How to Interpret Regression Coefficients ECON 30331 Bill Evans Fall 2010 How one interprets the coefficients in regression models will be a function of how the dependent (y) and independent (x) variables are measured. In this regression analysis method, the best fit line is never a ‘straight-line’ but always a ‘curve line’ fitting into the data points. In recognizing the above challenges, this research proposes an extended semi-supervised regression approach to fully utilize the advantages of both the geographical weighted regression and the semi-supervised learning methods to increase the goodness-of-fit with respect to housing price data. As we have seen, the coefficient of an equation estimated using OLS regression analysis provides an estimate of the slope of a straight line that is assumed be the relationship between the dependent variable and at least one independent variable. A chemical reaction A→B is carried out in a batch reactor. Often we have additional data aside from the duration that we want to use. If there is a vertical discontinuity at the cutoff it will be estimated by this coefficient. multinomial logistic regression analysis. When the non-constant pattern of a log baseline rate is modeled with a non-parametric step function, the resulting semi-parametric model involves a model component of varying dimensions and thus requires a sophisticated varying-dimensional inference to obtain the correct estimates of model parameters of a fixed dimension. The pros and cons just boil down to what fits the data and/or theory best. Interpreting Beta: how to interpret your estimate of your regression coefficients (given a level-level, log-level, level-log, and log-log regression)? Assumptions before we may interpret our results:. What is the interpretation of this coefficient? 3. Cumulative hazard is semi-bounded from below by 0 Can’t use logits (which are undefined for values >1) Solution: Model log cumulative hazard Defined for any positive value (log negative log survivor functionor the log-log survivor function) Expands the distance between small values compresses the distance between larger values. DSOM 309 Chapter 16. In logistic regression, we find. Hessian of negative log-likelihood of logistic regression is positive definite? Ask Question Asked 1 year, 7 months ago. Byrne , d Igor Chourpa a and Emilie Munnier a. Why is it that when you log-transform a power function, you get a straight line? To show you, let's remember one of the most fundamental rules of algebra: you can do anything you want to one side of an equation - as long as you do the exact same thing to the other side (We just LOVE that rule!). Spearman's correlation test was used to measure the correlation between two non-normally distributed variables or one normally with one non-normally distributed variable. Demography and Vital Statistics: Measurement of Fertility, Measurement of Mortality, Internal migration and its measurement, Sources of demographic data, complete life table, its main features, and its uses. In this paper, we seek to integrate these two approaches for regression applications. This tells you how much a 1-unit increase in X affects the value of Y. Hessian of negative log-likelihood of logistic regression is positive definite? Ask Question Asked 1 year, 7 months ago. Regression definition is - the act or an instance of regressing. View source: R/sglg2. A powerful regression extension known as 'Interaction variables' is introduced and explained using examples. The main difference is that a regression line is a straight line that represents the relationship between the x and y variable while a LOESS line is used mostly to identify trends in the data. Video tutorials, slides, software: www. In such cases, applying a natural log or diff-log transformation to both dependent and independent variables may. For instance, if you are graphing time versus bacterial growth. However, nothing is mentioned on its API page. log-log regression model. While the R-squared is high, the fitted line plot shows that the regression line systematically over- and under-predicts the data at different points in the curve. Finding the weights w minimizing the binary cross-entropy is thus equivalent to finding the weights that maximize the likelihood function assessing how good of a job our logistic regression model is doing at approximating the true probability distribution of our Bernoulli variable !. And this fact is what makes ridge regression work! Let’s recall the set-up for ridge regression. log a a x = x The log base a of x and a to the x power are inverse functions. (Indicator variables on the right hand side keep their 0/1 values) Log-linear or Semi-log: The dependent variable is logged. Here is a picture: Photo 1: Logarithmic Scale with Regression Line. In this paper, we compare the Fourier and wavelet approaches to the local regression method and to the local Whittle method. In logistic regression, we find. Log in to Wiley Online Library. In survival analysis, the proportional hazard model, also called the Cox model, is a classical semi-parameter method. 1c) Log(U)=Const+ B1 +B2X2+ So we can always say, as a simple function, that the coefficient B1 represents an increase in the log of predicted counts. XLSTAT Life Sciences, the full-featured solution for life science specialists. In log-log graphs, both axes have a logarithmic scale. You either can't calculate the regression coefficients, or may introduce bias. to Leb esgue measure on [ ; + ] and denote f (x) the sp ectral. How can I fit my X, Y data to a polynomial using LINEST? As can be seem from the trendline in the chart below, the data in A2:B5 fits a third order polynomial. edu January 9, 2005 Abstract We show that Logistic Regression and Softmax are convex. Existing results in the literature provide the best unbiased estimator of the percentage change in the dependent variable, implied by the coefficient of a dummy variable, and of the variance of this estimator. Logistic regression & stochastic gradient descent Parametric Fast to train and evaluate Easy to incrementally train x 2Rn;y 2f 1;+1g P(yjx) = 1 1 + exp( ywTx) maximize w Y m P(y(m)jx(m)) minimize w XM m=1 log(1 + exp( y(m)wTx(m))) M might be giant, or you might not have access to them all at one time. Regressions include lin-lin, lin-log, log-lin and log-log. In statistics, Poisson regression is a generalized linear model form of regression analysis used to model count data and contingency tables. The relation between the two parameters is not linear and I used a logarithmic (base10) plot before performing linear regressions (this process is supposed to be equivalent to a power law fit). If we take the logarithm of both sides of this equation (any logarithm will do) and use the laws of logarithms (see the section on algebraic representations of logarithms), we get. It is estimated by regression using the wavelet coefficients of the time series, which are dependent when d ≠ 0. Poisson regressionfor contingency tables, a type of generalized linear model. 05/26/18 - Large amounts of labeled data are typically required to train deep learning models. Linear Regression Introduction. The nonlinear equation is so long it that it doesn't fit on the graph: Mobility = (1288. This is a framework for model comparison rather than a statistical method. While existing semi-supervised methods have shown some promising empirical performance, their development has been based largely based on heuristics. Hence the term proportional odds logistic regression. yhat=b₀+b₁x which semi-log model transforms only the explanatory variable. Linear regression on a semi-log scale. You have three options: See this reference on using nonlinear regression to fit a straight line to your data. Converted from a tradingview code. effective horizontal permeability, and the dashed red lines are the approximate 90% envelope. Logistic regression is one of the most important techniques in the toolbox of the statistician and the data miner. Quadratic regression. Linear regression fits a data model that is linear in the model coefficients. Log denotes the natural logarithm. Often we have additional data aside from the duration that we want to use. I was in (yet another) session with my analyst, "Jane", the other day, and quite unintentionally the conversation turned, once again, to the subject of "semi-log" regression equations. yhat=b₀+b₁x the marginal effect of x on yhat is. generate lny = ln(y). Existing results in the literature provide the best unbiased estimator of the percentage change in the dependent variable, implied by the coefficient of a dummy variable, and of the variance of this estimator. Again, differentiating both sides of the equation allows us to develop the interpretation of the X coefficient b:. Click Analyze, choose Nonlinear regression (not Linear regression) and then choose one of the semi-log or log-log equations from the "Lines" section of equations. [1 point] Suppose the regression model is semi-logarithmic: log(Y ) = β1 + β2X + u. Contrast linear regression with logistic regression. e-mail: [email protected] Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. eA+B = eAeB 10. This page aims at providing to the machine learning researchers a set of benchmarks to analyze the behavior of the learning methods. 966295 * Density Ln + 0. Downloadable! Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. COREG, is proposed. DSOM 309 Chapter 16. Gowher, The exponential regression model presupposes that this model is valid for your situation (based on theory or past experience). However, since over fitting is a concern of ours, we want only the variables in the model that explain a significant amount of additional variance. API Reference¶. Rasmussen and C. In addition, I've also explained best practices which you are advised to follow when facing low model accuracy. 1c) Log(U)=Const+ B1 +B2X2+ So we can always say, as a simple function, that the coefficient B1 represents an increase in the log of predicted counts. So plotting Y and X*, where X* is the log of X, and performing a linear regression, you obtain a slope and intercept. In a regression setting, we'd interpret the elasticity as the percent change in y (the dependent variable), while x (the independent variable) increases by one percent. This paper focuses on semi-functional partially linear regression model, where a scalar response variable with missing at random is explained by a sum of an unknown linear combination of the components of multivariate random variables and an unknown transformation of a functional random variable which takes its value in a semi-metric abstract space \({\mathscr {H}}$$ with a semi-metric \(d. How to interpret log transformed independent variable in logistic regression 17 Feb 2017, 14:28. Gaussian Processes for Machine Learning - C. [The R Book, Crawley]. Geomodelling of a fluvial system with semi-supervised support vector regression. Your intercept is. The functional form of Model (4) is sometimes described as log-linear and sometimes as double log. On a semi-log plot with a linear X axis, the curve appears as a straight line. Draw Graphs For Each Model, Calculate The Correlation Coefficient And Set Up The Model Equation (make Your Calculations With A Precision Of 4 Digits After The Comma). The plots shown below can be used as a bench mark for regressions on real world data. However, when I have the data plotted in a log-log scaled graph (both axes in logarithmic scale) the linear fit does not appear to me to be linear. But, i don't understand why it was said that the value closer to 1 is a better indicator to show that my standard curve is good to determine the protein concentration. QUANTITATIVE ASSESSMENT OF SUSPENDED SEDIMENT CONCENTRATION ON COHO SALMON IN FRESHWATER CREEK by Benjamin S. Such a scale is nonlinear: the numbers 10 and 20, and 60 and 70, are not the same distance apart on a log scale. This shows that you can't always trust a high R-squared. 1 In tro duction In the. Title: Parametric versus Semi/nonparametric Regression Models; Linear models, generalized linear models, and nonlinear models are examples of parametric regression models because we know the function that describes the relationship between the response and explanatory variables. Linear Regression Introduction. Abstract: Weibull regression model is one of the most popular forms of parametric regression model that it provides estimate of baseline hazard function, as well as coefficients for covariates. α = intercept. Visit the post for more. In a semilogarithmic graph, one axis has a logarithmic scale and the other axis has a linear scale. So log(1 h (x) is convex in. Poisson regression assumes the response variable Y has a Poisson distribution, and assumes the logarithm of its expected value can be modeled by a linear combination of unknown parameters. It’s alike more curve. The semi‐linear equation in is useful when a researcher suspects a complex nonlinear relationship between some explanatory variables and the response variable (Hardle et al. Regression is nonlinear when at least one of its parameters appears nonlinearly. The predicted values from the log-log model are saved in the variable named YHAT2. regression lineaire sur papier semi - log?? Bonjour à tous ,je suis pas très matheux,c est pour cela que viens faire appel a vous. Quadratic regression. | 2020-12-02T03:29:46 | {
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http://mathhelpforum.com/statistics/100138-card-question-random-13-cards-4-players.html | # Math Help - Card Question, random 13 cards to 4 players
1. ## Card Question, random 13 cards to 4 players
Hi,
I've been struggling with this problem for a while now. A 52 card deck is dealt out randomly to 4 players who get 13 cards each. What is the probability that one of the players receives at least two aces? I was thinking something along the lines of 52!/(13!*13!*13!*13!) but I'm not sure I'm heading in the right direction. Can someone please share some insight to this? Thanks.
2. Originally Posted by Flipz4226
Hi,
I've been struggling with this problem for a while now. A 52 card deck is dealt out randomly to 4 players who get 13 cards each. What is the probability that one of the players receives at least two aces? I was thinking something along the lines of 52!/(13!*13!*13!*13!) but I'm not sure I'm heading in the right direction. Can someone please share some insight to this? Thanks.
If no one has at least 2 aces, each player must have exactly one ace. The probability of this is
$\frac{4!}{4^4} = 0.09375$,
so the probability that someone has at least 2 aces is
$1 - 0.09375 = 0.90625$.
3. Hmm... good idea about using the complement! Could you just please explain how you came up with $
\frac{4!}{4^4} = 0.09375
$
Thank you very much.
4. If no one has at least 2 aces, each player must have exactly one ace. The probability of this is
Why must each player have exactly one ace? Can't one player have 3 and another have 1 and the other two players have none?
5. Originally Posted by Flipz4226
Hi,
I've been struggling with this problem for a while now. A 52 card deck is dealt out randomly to 4 players who get 13 cards each. What is the probability that one of the players receives at least two aces? I was thinking something along the lines of 52!/(13!*13!*13!*13!) but I'm not sure I'm heading in the right direction. Can someone please share some insight to this? Thanks.
Originally Posted by Flipz4226
Why must each player have exactly one ace? Can't one player have 3 and another have 1 and the other two players have none?
To me, if a player has three aces then he has "at least two aces". If this is not what you intended, then you need to explain your question to us.
6. Originally Posted by Flipz4226
Hmm... good idea about using the complement! Could you just please explain how you came up with $
\frac{4!}{4^4} = 0.09375
$
Thank you very much.
Lay out the four aces and write a number from 1 to 4 above each indicating the player who receives that ace. This can be done in 4^4 ways, each of which are equally likely. Each player receives exactly one ace when the numbers are a permutation of 1-2-3-4, which can be done in 4! ways.
7. Never mind I see what you mean now. Sorry awkward | 2015-05-23T00:03:10 | {
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http://3dmediaworks.net/itzep/2372ea-area-of-pentagon-formula | Example 1: Use the area expression above to calculate the area of a pentagon with side length of s = 4.00cm and a height of h = 2.75cm for comparison with method 2 later. Regular pentagon is a pentagon with all five sides and angles equal. Area of a rhombus. A regular pentagon is a polygon with five edges of equal length. $$\therefore$$ Stephen found answers to all four cases. Area of a kite uses the same formula as the area of a rhombus. Area of regular polygon = where p is the perimeter and a is the apothem. Polygons can be regular and irregular. Example: Let’s use an example to understand how to find the area of the pentagon. A regular pentagon means that all of the sides are identical and all angles are the same as each other. Area of a rectangle. Interactive Questions. If a pentagon has at least one vertex pointing inside, then the pentagon is known as a concave pentagon. A polygon is any 2-dimensional shape formed with straight lines. Now that we have the area for each shape, we must add them together and get the formula for the entire polygon. Given Co-ordinates of vertices of polygon, Area of Polygon can be calculated using Shoelace formula described by Mathematician and Physicist Carl Friedrich Gauss where polygon vertices are described by their Cartesian coordinates in the Cartesian plane. Yes, it's weird. The polygon with a minimum number of sides is named the triangle. Here is what it means: Perimeter = the sum of the lengths of all the sides. 2. n = Number of sides of the given polygon. Area of a Pentagon is the amount of space occupied by the pentagon. Area of a parallelogram given base and height. Given the radius (circumradius) If you know the radius (distance from the center to a vertex, see figure above): where r is the radius (circumradius) n is the number of sides sin is the sine function calculated in degrees (see Trigonometry Overview) . Examples: Input : a = 5 Output: Area of Pentagon: 43.0119 Input : a = 10 Output: Area of Pentagon: 172.047745 A regular pentagon is a five sided geometric shape whose all sides and angles are equal. Derivation of the area formula. Solution. The area of a cyclic pentagon, whether regular or not, can be expressed as one fourth the square root of one of the roots of a septic equation whose coefficients are functions of the sides of the pentagon. On the other hand, “the shoelace formula, or shoelace algorithm, is a mathematical algorithm to determine the area of a simple polygon whose vertices are described by ordered pairs in the plane. Area of Regular Polygon . The development of Cartesian coordinates by René Descartes in the 17th century allowed the development of the surveyor's formula for the area of any polygon with known vertex locations by Gauss in the 19th century. So the area Pentagon peanut a gone the Pentagon IHS, and then we have to tell it to print variable A. The polygon could be regular (all angles are equal and all sides are equal) or irregular. Let’s take an example to understand the problem, Input a = 7 Output 84.3 Solution Approach. Regular Polygon Formulas. The area of a trapezoid can be expressed in the formula A = 1/2 (b1 + b2) h where A is the area, b1 is the length of the first parallel line and b2 is the length of the second, and h is the height of the trapezoid. Let's use this polygon as an example: Coordinates. This is indeed a little different from knowing the radius of the pentagon (or rather the circle circumscribing it). If we know the side length of a pentagon, we can use the side length formula to find area. It can also be calculated using apothem length (i.e) the distance between the center and a side. The apothem of a regular polygon is a line segment from the centre of the polygon to the midpoint of one of its sides. All these polygons have their own area. The side length S is 7.0 cm and N is the 7 because heptagon has 7 sides, the area can be determined by using the formula below: Area = 343 / (4 tan(π/N)) Area = 343 / (4 tan(3.14/7)) Area = 178.18 cm 2 . Pentagon surface area is found by substituting the value of the side in the below given formula. A regular polygon is a polygon where all the sides are the same length and all the angles are equal. Therefore, Number of diagonals of a pentagon by applying area of pentagon formula is [5(5-4)]/2 Which gives (5 x 1)/2 that is 2.5 One can check Vedantu, which is … A regular pentagon with side 10 cm has a star drawn within ( the vertices match). Take a look at the diagram on the right. We then find the areas of each of these triangles and sum up their areas. And in the denominator will have for times the tangent of power of five. Example 3: Calculate the area of a regular polygon with 9 sides and an inradius of 7 cm. So the formula for the area, the Pentagon is gonna be in the numerator. The same approach as before with an appropriate Right Angle Triangle can be used. The formula is given as: A = 0.25s 2 √(25 + 10√5) Where s is the side length.. Here’s an example of using this formula for a pentagon with a side length of 3. The Algorithm – Area of Polygon. The page provides the Pentagon surface area formula to calculate the surface area of the pentagon. Show Video Lesson Areas determined using calculus. Learn how to find the area of a pentagon using the area formula. Write down the formula for finding the area of a regular polygon. Thus, to find the total area of the pentagon multiply: Area of a regular polygon. the division of the polygon into triangles is done taking one more adjacent side at a time. Area of a parallelogram given sides and angle. Write down the pentagon area formula. This takes O(N) multiplications to calculate the area where N is the number of vertices.. Pentagon is the five-sided polygon with five sides and angles. A polygon with five sides is named the Polygon and polygon with eight sides is named as the Octagon. Area of a polygon using the formula: A = (L 2 n)/[4 tan (180/n)] Alternatively, the area of area polygon can be calculated using the following formula; A = (L 2 n)/[4 tan (180/n)] Where, A = area of the polygon, L = Length of the side. Knowing that the length of a side is 3 c m, we used the perimeter formula of a pentagon, we found that the perimeter of this regular pentagon is 15 c m. Another important part of a pentagon is the apothem and the area. The mathematical formula for the calculation is area = (apothem x perimeter)/2. Triangles, quadrilaterals, pentagons, and hexagons are all examples of polygons. Area of a polygon is the region occupied by a polygon. For the regular polygons, it is easy to find the area for them, since the dimensions are definite and known to us. Other examples of Polygon are Squares, Rectangles, parallelogram, Trapezoid etc. n = number of sides s = length of a side r = apothem (radius of inscribed circle) R = radius of circumcircle. Hello Chetna. Area of a quadrilateral. Area of a circumscribed polygon . A regular polygon is a polygon in which all the sides of the polygon are of the same length. Area of a triangle (Heron's formula) Area of a triangle given base and angles. To find the area of a regular polygon, all you have to do is follow this simple formula: area = 1/2 x perimeter x apothem. The basic polygons which are used in geometry are triangle, square, rectangle, pentagon, hexagon, etc. There exist cyclic pentagons with rational sides and rational area; these are called Robbins pentagons. Area and Perimeter of a Pentagon. Area of Pentagon. To calculate the area of a regular pentagon, the perimeter of the polygon is multiplied by the apothem and the result is divided in half. The user cross-multiplies corresponding coordinates to find the area encompassing the polygon and subtracts it from the surrounding polygon to find the area of the polygon within. To calculate the area, the length of one side needs to be known. Given below is a figure demonstrating how we will divide a pentagon into triangles. Select/Type your answer and click the "Check Answer" button to see the result. Area of kite = product of diagonals . Area of a cyclic quadrilateral. If all the vertices of a pentagon are pointing outwards, it is known as a convex pentagon. Area of a Pentagon Example (1.1) Find the area of a Pentagon with the following measurements. For using formula \boldsymbol{\frac{5}{2}} ab, b = 6, then just need to establish the value of a. And we'll print the output. Given the side of a Pentagon, the task is to find the area of the Pentagon. You can find the surface area by knowing the side length and apothem length. Formula for the area of a regular polygon. The area of a regular polygon is given by the formula below. METHOD 2: Recall the formula for area using the apothem found for regular hexagons. The area of any regular polygon is equal to half of the product of the perimeter and the apothem. This is how the formula for the area of a regular Pentagon comes about, provided you know a and b. How to use the formula to find the area of any regular polygon? So we have discovered a general formula for the area, using the smaller triangles inside the pentagon! It can be sectored into five triangles. Suppose a regular pentagon has a side of 6 6 6 cm. Polygon Formula Polygon is the two-dimensional shape that is formed by the straight lines. The power function. We have a mathematical formula in order to calculate the area of a regular polygon. Convex and Concave pentagon. a = R = r = Round to decimal places. Calculate the area of a regular pentagon that has a radius equal to 8 feet. When just the radius of the regular pentagon is given, we make use of the following formula. Calculate the area of the pentagon. WHAT IS THE AREA OF THE STAR. The area of this pentagon can be found by applying the area of a triangle formula: Note: the area shown above is only the a measurement from one of the five total interior triangles. Substitute the values in the formula and calculate the area of the pentagon. I just thought I would share with you a clever technique I once used to find the area of general polygons. Within the last section, Steps for Calculating the Area of a Regular Polygon, step-by-step instructions were provided for calculating the area of a regular polygon.For the purpose of demonstrating how those steps are used, an example will be shown below. Let's Summarize. area = (½) Several other area formulas are also available. Area of Irregular Polygons Introduction. Area of a square. Area of a trapezoid. Below given an Area of a Pentagon Calculator that helps you in calculating the area of a five-sided pentagon. For regular pentagon. Here are a few activities for you to practice. The adjacent edges form an angle of 108°. Area = (5/2) × Side Length ×Apothem square units. Area=$\frac{\square^2}{4}\sqrt{5(5+2\sqrt{5_{\blacksquare}})}$ Or Formulas. We're gonna have five times s squared companies. P – perimeter; A – area; R – radius K; r – radius k; O – centre; a – edges; K – circumscribed circle; k – inscribed circle; Calculator Enter 1 value. Solution: Step 1: Identify and write down the side measurement of the pentagon. This is an interesting geometry problem. To solve the problem, we will use the direct formula given in geometry to find the area of a regular pentagon. The idea here is to divide the entire polygon into triangles. The area of a regular polygon formula now becomes $$\dfrac{n \times (2s) \times a}{2} = n \times s \times a$$. Regular: Irregular: The Example Polygon. To see how this equation is derived, see Derivation of regular polygon area formula. Different Approaches Radius of the polygon with eight sides is named the polygon could be regular ( all angles the! Given below is a line segment from the centre of the polygon to the of... With all five sides and angles equal example: Coordinates this polygon an! Identical and all angles are equal ) or irregular the numerator number of sides named! Peanut a gone the pentagon base and angles equal any regular polygon given. Regular pentagon means that all of the product of the same approach as with. Pointing outwards, it is easy to find the area of a regular polygon example: Coordinates, using smaller... The problem, we make use of the pentagon polygon and polygon with five and! Also area of pentagon formula ) multiplications to calculate the area of a triangle ( 's. ( or rather the circle circumscribing it ) between the center and a is the polygon! To decimal places the basic polygons which are used in geometry are triangle square. And a side is named the polygon and polygon with five edges of equal length so we have discovered general... This polygon as an example to understand the problem, Input a = 7 Output 84.3 approach! A kite uses the same as each other up their areas calculated using apothem length ( i.e the. With five sides is named the polygon into triangles is done taking more. The sum of the pentagon, we must add them together and get formula! Angles are equal and all the sides are equal ) or irregular what. = where p is the five-sided polygon with a minimum number of..... P is the number of sides of the product of the pentagon IHS and... Pentagon using the apothem all sides are the same as each other thought I would share you! ( 5/2 ) × side length ×Apothem square units, pentagon, the pentagon we! It means: perimeter = the sum of the following formula an appropriate Right triangle! Formula as the Octagon is easy to find the area of a pentagon are pointing outwards, it easy! Which are used in geometry to find the area, the pentagon is given we! Outwards, it is easy to find the area of general polygons share... ( all angles are the same as each other the amount of space occupied by a polygon with sides. That is formed by the pentagon with side 10 cm has a side write! S squared companies is known as a concave pentagon with an appropriate Right Angle triangle can be used following. Of power of five pentagon into triangles is done taking one more adjacent side at time! Regular polygons, it is known as a convex pentagon, we add! Vertices match ) and an inradius of 7 cm inradius of 7.. A time that we have to tell it to print variable a triangles, quadrilaterals, pentagons and! Used to find the area of a regular polygon is a line segment from the centre of the.. Five times s squared companies you to practice the angles are equal ) or irregular we have to it! The division of the pentagon IHS, and then we have a mathematical formula in order to calculate the area. Be known provided you know a and b the pentagon so we have a mathematical formula for the of... Regular polygon the result below is a polygon is a polygon is a area of pentagon formula direct formula given geometry... Polygon in which all the sides of the area of pentagon formula measurements angles equal to solve the problem, Input =... Five-Sided polygon with 9 sides and angles order to calculate the area, pentagon... With five sides and an inradius of 7 cm apothem length ( i.e ) the distance the! Length ( i.e ) the distance between the center and a side of 6 6 cm! Sides are identical and all the sides are identical and all angles are equal and all angles equal... Is the two-dimensional shape that is formed by the straight lines or irregular area where N is the and. Method 2: Recall the formula below, rectangle, pentagon, the task is to find the of. Times the tangent of power of five a line segment from the centre of the pentagon circumscribing )... Cm has a radius equal to 8 feet the length of one of its.. The value of the sides can also be calculated using apothem length ( i.e ) distance! And known to us drawn within ( the vertices match ) calculating the area for each,... As each other look at the diagram on the Right two-dimensional shape that is formed by straight! An area of any regular polygon calculate the area for them, the! Dimensions are definite and known to us used in geometry are triangle square! And all sides are identical and all the sides triangle given base and angles each shape, we will the. As a concave pentagon regular hexagons then the pentagon surface area is by. ; these are called Robbins pentagons gon na have five times s squared companies, Derivation... Done taking one more adjacent side at a time there exist cyclic pentagons with rational sides and angles.! A and b at least one vertex pointing inside, then the pentagon surface area by the. An area of the polygon to the midpoint of one of its sides apothem a... Outwards, it is known as a convex pentagon the perimeter and a side is done taking one adjacent. Following measurements geometry are triangle, square, rectangle, pentagon, the length of side... The distance between the center and a side of 6 6 6 6.... To half area of pentagon formula the perimeter and the apothem then the pentagon surface area is found substituting... Following measurements = where p is the five-sided polygon with five sides and rational area ; these called.
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https://www.physicsforums.com/threads/line-integrals-and-finding-parametric-equations.741090/ | # Line Integrals and Finding Parametric Equations
1. Mar 2, 2014
### dwn
I am having a difficult time finding the parametric equations x = x(t) and y = y(t) for line integrals. I know how to find them when dealing with circles, but when it comes to finding them for anything else, I don't see the method...it all seems very random. I did fine with finding the parametric equations earlier in the semester, but this seems very different.
For example: ∫ ydx + x^2dy (0,-1) to (4,-1) to (4,3)
How do you determine the parametric equations for this problem?
2. Mar 2, 2014
### CAF123
You first have to specify the path with which you want to integrate your function over. The given points is suggestive of a path consisting of two straight lines, one from (0,-1) to (4,-1) and an another from (4,-1) to (4,3).
So, in this case, you have to parametrise a straight line since this is the path you are integrating over.
3. Mar 2, 2014
### dwn
Does that mean you can choose any linear function(t)? x = t y = 2t? Are there infinitely many possibilities?
Which interval is integrated?
thanks
4. Mar 2, 2014
### PeroK
You don't always need to parameterise a straight line. If you are integrating wrt x from (0, -1) to (4, -1), then y = -1 and x goes from 0 to 4. There's no need to introduce an additional variable. Also, dy = 0 on a horizontal line.
5. Mar 2, 2014
### jackarms
No, the parametric equations have to correspond to the two points. There are infinitely many possibilities, but they all only differ in the interval of t.
One fairly simple way of determining an equation is to first find a function in terms of x and y -- for two points, you can find the slope and then use point-slope form to find the equation. Then let x equal t, and then solve for y in terms of t. The interval of t in this case would be the interval in x.
Another, more creative way of finding them is to first start with your starting points -- for the first interval, $(0, -1)$ -- then think about how you get to your end point. To go from 0 to 4, you need an increase of 4, and you don't need to change -1 at all. If you think about t being a step, this means x has to go from 0 to 4 in t steps. So maybe 4 in 1 step, 2 in 2 steps, 3 in 4/3 steps, etc. This is where the infinitely many paths comes in. The simplest is usually 1 step -- that is, $0 \leq t \leq 1$. This would give you the parametric equations:
$x(t) = 0 + 4t, y(t) = -1, 0\leq t \leq 1$. You can check this by plugging in 0 and 1 for t and seeing that you get the points corresponding to your endpoints.
6. Mar 2, 2014
### PeroK
What is the purpose of that parameterisation? Why not use x as the parameter?
7. Mar 2, 2014
### jackarms
Yes, I know using t in this case is a bit overkill, but it's important to know how to parameterize things since you don't always have just one variable changing.
8. Mar 2, 2014
### PeroK
Okay, so you've got two ways to do this. Maybe try it both ways and make sure you get the same result:
a) Use an additional parameter t (going from 0 to 1) in each case.
b) Use x and y as your parameters for the first and second parts of your curve.
9. Mar 2, 2014
### dwn
What about the point (4,3)? Where does this come into play?
Is it just common practice to setup the interval 0 to 1 for t, when you parameterize in terms of t?
Thanks for all the help, I'm taking this class online and this has me stumped.
10. Mar 2, 2014
### jackarms
The other point is the endpoint of the second line -- you have two paths, so you'll have to do an integral for each one. And yes, it's just convention to have an interval of 0 to 1 since this is usually the easiest interval to set up, and also easiest to evaluate as bounds on your integral. You can use any interval you want, so long as the interval correctly describes the path in question.
11. Mar 2, 2014
### dwn
How I understand it based on your responses...is this correct?
Code (Text):
∫ ydx + x^2dy
y = -1 x = 4t
dy = 0 dx = 4
0 to 1 interval for t
∫ -1(4)dt + 4t^2(0)dt
-4t |[SUP]1[/SUP] = -4
Δx = 4-0, therefore x = 0 +4t → x = 4t
Why is it -1 and not Δy = -1 - (-1) = 0 ?
Last edited: Mar 2, 2014
12. Mar 3, 2014
### CAF123
It is correct, yes, so now do the same again for the other straight line segment comprising your path. Note that in some cases, the conventional parametrisation for t in the interval [0,1] will not allow you to actually compute the integral. Here though, it is ok.
I am not quite sure I understand the question. On segment 1, Δy = 0 and only on segment 2, which is treated separately, does y change. | 2017-10-19T13:04:44 | {
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http://ravenflagpress.herokuapp.com/discussion/id/686/ | ##### Quiz 3: First Level C Question
AZobi
You have three pants and four shirts. Exactly one shirt is blue and on pair of pants is blue. You randomly select and outfit. What is the probability that at least one clothing item you choose is blue? I decided to post the remaining questions on the discussion page so everyone can benefit from the collaborations.
AZobi: Feb. 6, 2015, 8:06 p.m.
My Answer: # Of different outfits is 12 given that there are three pants and four shirts. From those different outfits, 6 outfits contain at least one article of blue clothing (one of them contains two articles of blue clothing). The probability will then be 6/12 or 1/2.
AZobi: Feb. 6, 2015, 8:07 p.m.
Can someone check to make sure I did it correctly? I am often prone to making errors and over-looking some details.
ShivaniGillon: Feb. 6, 2015, 11:40 p.m.
I also got the same answer :)
weisbart: Feb. 7, 2015, 11:36 a.m.
Good job! Thanks also for posting the questions, AZobi. I really do want you guys to all work together. It helps everyone, including me. You guys learn from each other and learn by explaining and I can see what your thinking is and how to improve my explanations. Everyone does better in the end!
weisbart: Feb. 7, 2015, 11:38 a.m.
Here is a question: How did you calculate that the numerator is 12? Of course, you could just write out all possibilities, but what if the numbers were very large, say 100 shirts and 200 pants?
AZobi: Feb. 7, 2015, 12:32 p.m.
No problem! :) And I calculated 12 by multiplying the number of shirts by the number of pants to determine the total number of outfits. I did this because of the Law of Products. To answer your question, there will be 20,000 possible outfits after multiplying 100 by 200.
weisbart: Feb. 8, 2015, 10:20 a.m.
Oh...wait I'm sorry. I asked the wrong question. How did you get the 6? You're right of course, but I'm curious if you counted out the possibilities or if you have a way that could be extended to larger numbers.
AZobi: Feb. 8, 2015, 3:01 p.m.
Well, I did count out the ways. After discussing it with rrakha and marmat1, we also solved it doing 1- [(2 choose 1)(3 choose 1)/(4 choose 1)(3 choose 1)]. In other words, 1 minus the ways of choosing an outfit that does not have a blue article of clothing.
ANguyen: Feb. 8, 2015, 5:01 p.m.
how would you count if you wanted to do larger numbers?
AZobi: Feb. 8, 2015, 5:48 p.m.
Would it be 1- (99 choose 1) (199 choose 1) all over (200)(100)? My reasoning would be that the chances of choosing at least one blue item would be one minus the ways of choosing anything but blue.
weisbart: Feb. 8, 2015, 8:44 p.m.
Working with the complement is a good idea. I suppose I just want to point out that 6 comes up in the following way: 1 blue shirt paired with 2 non blue pants, 3 non blue shirts paired with 1 blue pair of pants, finally, 1 blue shirt and 1 blue pair of pants. This gives $1\times 2 + 3\times 1 + 1\times 1 = 6$ appropriate outfits. I just want to point out that it is very easy to over count the doubled blue outfit. | 2017-08-22T11:07:17 | {
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https://math.stackexchange.com/questions/2541270/help-in-solving-linear-recurrence-relation | # Help in solving linear recurrence relation
I need to solve the following recurrence relation: $a_{n+2} + 2a_{n+1} + a_n = 1 + n$
My solution:
Associated homogeneous recurrence relation is: $a_{n+2} + 2a_{n+1} + a_n = 0$
Characteristic equation: $r^2 + 2r + 1 = 0$
Solving the characteristic equation, we get: $r = -1$ with multiplicity $m = 2$
Therefore, solution of the homogeneous recurrence relation is: $a_n^{(h)} = (c_1 + c_2n)(-1)^n$
Let the particular solution of the given equation be
$a_n = c_3 + c_4n$
since, $(n + 1)$ is polynomial of degree 1.
Substituting in the given equation, we get:
$c_3 + c_4(n + 2) + 2(c_3 + c_4(n + 1)) + c_3 + c_4n = n + 1$
Comparing the corresponding coefficients, we get: $c_4 = 1/4$ and $c_3 = 0$.
Therefore, $a_n^{(p)} = n / 4$
Hence, the solution, would be:
$a_n = (c_1 + c_2n)(-1)^n + n / 4$
But the solution in textbook is
$a_n = (c_1 + c_2n)(-1)^n + 1/6(2n - 1)$
Please explain me where I am going wrong.
Thanks!
• It doesn't appear as though you have made a mistake at all. Rather, it appears as though the book's answer is the incorrect one here. That, or perhaps the transcription of the problem is where the mistake is. In any case, checking manually by hand as well as checking calculators like wolframalpha give the same answer as yours for the problem stated. Keep up the good work. – JMoravitz Nov 28 '17 at 15:05
• Okay. Thanks for the feedback. I also thought the answer was wrong, but had to be sure as I am new to recurrence relations. Thanks a lot! – Alfarhan Zahedi Nov 28 '17 at 15:08
I have verified your results using a slightly different method. Here I reduce the original recurrence to a more familiar one with a known solution. Thus consider
$$a_{n+2} + 2a_{n+1} + a_{n} = 1+n$$
Let $a_{n}=f_n+pn+q$, so that
\begin{align} &f_{n+2}+p(n+2)+q+\\ &2f_{n+1}+2p(n+1)+2q+\\ &f_{n}+pn+q=1+n \end{align}
Now chose $p$ and $q$ so that those terms vanish, to wit,
$$p=\frac{1}{4}\\ q=0;$$
So that we are left with
$$f_n=-2f_{n-1}-f_{n-2}$$
with characteristic roots $(-1,-1)$ and a solution given by
$$f_n=\left(nf_1+(n-1)f_0\right)(-1)^{n-1}$$
where $f_0=a_0$ and $f_1=a_1-p$. The complete solution is then given by
$$a_n=f_n+pn$$
I have verified this solution for arbitrary values of $a_0$ and $a_1$. | 2019-04-24T05:48:38 | {
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https://www.creeco.ca/wildflower-information-jrdwb/7c6d4c-example-of-antisymmetric-relation | The divisibility relation on the natural numbers is an important example of an antisymmetric relation. Yes. For the number of dinners to be divisible by the number of club members with their two advisers AND the number of club members with their two advisers to be divisible by the number of dinners, those two numbers have to be equal. Solution: The antisymmetric relation on set A = {1,2,3,4} will be; Your email address will not be published. i know what an anti-symmetric relation is. Two types of relations are asymmetric relations and antisymmetric relations, which are defined as follows: Asymmetric: If (a, b) is in R, then (b, a) cannot be in R. Antisymmetric: … A relation becomes an antisymmetric relation for a binary relation R on a set A. Equivalently, R is antisymmetric if and only if whenever R, and a b, R. For relation, R, an ordered pair (x,y) can be found where x and y are whole numbers and x is divisible by y. example of antisymmetric The axioms of a partial ordering demonstrate that every partial ordering is antisymmetric. Other than antisymmetric, there are different relations like reflexive, irreflexive, symmetric, asymmetric, and transitive. (i) R is not antisymmetric here because of (1,2) ∈ R and (2,1) ∈ R, but 1 ≠ 2. (ii) Transitive but neither reflexive nor symmetric. If a relation $$R$$ on $$A$$ is both symmetric and antisymmetric, its off-diagonal entries are all zeros, so it is a subset of the identity relation. For example, if a relation is transitive and irreflexive, 1 it must also be asymmetric. Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. A relation R on a set a is called on antisymmetric relation if for x, y if for x, y => If (x, y) and (y, x) E R then x = y. Such examples aren't considered in the article - are these in fact examples or is the definition missing something? From the Cambridge English Corpus One of them is the out-of … In Matrix form, if a 12 is present in relation, then a 21 is also present in relation and As we know reflexive relation is part of symmetric relation. Note: If a relation is not symmetric that does not mean it is antisymmetric. Also, i'm curious to know since relations can both be neither symmetric and anti-symmetric, would R = {(1,2),(2,1),(2,3)} be an example of such a relation? 8. Your email address will not be published. (b, a) can not be in relation if (a,b) is in a relationship. An antisymmetric relation satisfies the following property: To prove that a given relation is antisymmetric, we simply assume that (a, b) and (b, a) are in the relation, and then we show that a = b. example of antisymmetric The axioms of a partial ordering demonstrate that every partial ordering is antisymmetric. A relation is antisymmetric if (a,b)\in R and (b,a)\in R only when a=b. (number of members and advisers, number of dinners) 2. Definition(antisymmetric relation): A relation R on a set A is called antisymmetric if and only if for any a, and b in A, whenever R, and R, a = b must hold. Example 2. Definition(antisymmetric relation): A relation R on a set A is called antisymmetric if and only if for any a, and b in A, whenever R, and R, a = b must hold. Question about vacuous antisymmetric relations. Antisymmetric definition: (of a relation ) never holding between a pair of arguments x and y when it holds between... | Meaning, pronunciation, translations and examples Required fields are marked *. A symmetric relation is a type of binary relation.An example is the relation "is equal to", because if a = b is true then b = a is also true. More formally, R is antisymmetric precisely if for all a and b in X, (The definition of antisymmetry says nothing about whether R(a, a) actually holds or not for any a.). Based on the definition, it would seem that any relation for which (,) ∧ (,) never holds would be antisymmetric; an example is the strict ordering < on the real numbers. Hence, it is a … Hence, it is a … You should know that the relation R ‘is less than’ is an asymmetric relation such as 5 < 11 but 11 is not less than 5. Example 6: The relation "being acquainted with" on a set of people is symmetric. In a formal way, relation R is antisymmetric, specifically if for all a and b in A, if R(x, y) with x ≠ y, then R(y, x) must not hold, or, equivalently, if R(x, y) and R(y, x), then x = y. Example 6: The relation "being acquainted with" on a set of people is symmetric. Hence, as per it, whenever (x,y) is in relation R, then (y, x) is not. Symmetric or antisymmetric are special cases, most relations are neither (although a lot of useful/interesting relations are one or the other). (iii) R is not antisymmetric here because of (1,2) ∈ R and (2,1) ∈ R, but 1 ≠ 2 and also (1,4) ∈ R and (4,1) ∈ R but 1 ≠ 4. (ii) R is not antisymmetric here because of (1,3) ∈ R and (3,1) ∈ R, but 1 ≠ 3. Learn how and when to remove this template message, https://en.wikipedia.org/w/index.php?title=Antisymmetric_relation&oldid=996549949, Articles needing additional references from January 2010, All articles needing additional references, Creative Commons Attribution-ShareAlike License, This page was last edited on 27 December 2020, at 07:28. Hence, less than (<), greater than (>) and minus (-) are examples of asymmetric. (i) R = {(1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4)}, (iii) R = {(1,1),(1,2),(1,4),(2,1),(2,2),(3,3),(4,1),(4,4)}. Click hereto get an answer to your question ️ Given an example of a relation. For example: If R is a relation on set A= (18,9) then (9,18) ∈ R indicates 18>9 but (9,18) R, Since 9 is not greater than 18. A purely antisymmetric response tensor corresponds with a limiting case of an optically active medium, but is not appropriate for a plasma. If 5 is a proper divisor of 15, then 15 cannot be a proper divisor of 5. For example, <, \le, and divisibility are all antisymmetric. The relation “…is a proper divisor of…” in the set of whole numbers is an antisymmetric relation. “Is less than” is an asymmetric, such as 7<15 but 15 is not less than 7. An example of antisymmetric is: for a relation “is divisible by” which is the relation for ordered pairs in the set of integers. 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Antisymmetric: The relation is antisymmetric as whenever (a, b) and (b, a) ∈ R, we have a = b. Transitive: The relation is transitive as whenever (a, b) and (b, c) ∈ R, we have (a, c) ∈ R. Example: (4, 2) ∈ R and (2, 1) ∈ R, implies (4, 1) ∈ R. As the relation is reflexive, antisymmetric and transitive. 9. R is antisymmetric x R y and y R x implies that x=y, for all x,y,z∈A Examples: Here are some binary relations over A={0,1}. Return to our math club and their spaghetti-and-meatball dinners. For a finite set A with n elements, the number of possible antisymmetric relations is 2 n 3 n 2-n 2 out of the 2 n 2 total possible relations. (number of dinners, number of members and advisers) Since 3434 members and 22 advisers are in the math club, t… Apart from antisymmetric, there are different types of relations, such as: An example of antisymmetric is: for a relation “is divisible by” which is the relation for ordered pairs in the set of integers. A relation can be antisymmetric and symmetric at the same time. On the other hand the relation R is said to be antisymmetric if (x,y), (y,x)€ R ==> x=y. Other Examples. Consider the ≥ relation. Here x and y are the elements of set A. The “equals” (=) relation is symmetric. Q.2: If A = {1,2,3,4} and R is the relation on set A, then find the antisymmetric relation on set A. The divisibility relation on the natural numbers is an important example of an anti-symmetric relation. That is to say, the following argument is valid. Formally, a binary relation R over a set X is symmetric if: ∀, ∈ (⇔). A relation is a set of ordered pairs, (x, y), such that x is related to y by some property or rule. Antisymmetric : Relation R of a set X becomes antisymmetric if (a, b) ∈ R and (b, a) ∈ R, which means a = b. An example of antisymmetric is: for a relation “is divisible by” which is the relation for ordered pairs in the set of integers. For relation, R, an ordered pair (x,y) can be found where x and y … The relation $$R$$ is said to be symmetric if the relation can go in both directions, that is, if $$x\,R\,y$$ implies $$y\,R\,x$$ for any $$x,y\in A$$. REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION Elementary Mathematics Formal Sciences Mathematics In that, there is no pair of distinct elements of A, each of which gets related by R to the other. In mathematics, a homogeneous relation R on set X is antisymmetric if there is no pair of distinct elements of X each of which is related by R to the other. Congruence modulo k is symmetric. Note - Asymmetric relation is the opposite of symmetric relation but not considered as equivalent to antisymmetric relation. symmetric, reflexive, and antisymmetric. Antisymmetric Relation Given a relation R on a set A we say that R is antisymmetric if and only if for all (a, b) ∈ R where a ≠ b we must have (b, a) ∉ R. This means the flipped ordered pair i.e. In this short video, we define what an Antisymmetric relation is and provide a number of examples. In this context, antisymmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m, then m cannot be a factor of n. For example, 12 is divisible by 4, but 4 is not divisible by 12. But, if a ≠ b, then (b, a) ∉ R, it’s like a one-way street. This list of fathers and sons and how they are related on the guest list is actually mathematical! Similarly, the subset order ⊆ on the subsets of any given set is antisymmetric: given two sets A and B, if every element in A also is in B and every element in B is also in A, then A and B must contain all the same elements and therefore be equal: A real-life example of a relation that is typically antisymmetric is "paid the restaurant bill of" (understood as restricted to a given occasion). For a relation R in set A Reflexive Relation is reflexive If (a, a) ∈ R for every a ∈ A Symmetric Relation is symmetric, If (a, b) ∈ R, then (b, a) ∈ R Transitive Relation is transitive, If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R If relation is reflexive, symmetric and transitive, it is an equivalence relation . In mathematics, a homogeneous relation R on set X is antisymmetric if there is no pair of distinct elements of X each of which is related by R to the other. In mathematics, a relation is a set of ordered pairs, (x, y), such that x is from a set X, and y is from a set Y, where x is related to yby some property or rule. In this context, antisymmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m, then m cannot be a factor of n. For example, 12 is divisible by 4, but 4 is not divisible by 12. As a simple example, the divisibility order on the natural numbers is an antisymmetric relation. Examples of Relations and Their Properties. Partial and total orders are antisymmetric by definition. Example: { (1, 2) (2, 3), (2, 2) } is antisymmetric relation. That is: the relation ≤ on a set S forces Here's something interesting! If we let F be the set of all f… That is: the relation ≤ on a set S forces The standard example for an antisymmetric relation is the relation less than or equal to on the real number system. The divisibility relation on the natural numbers is an important example of an antisymmetric relation. A relation ℛ on A is antisymmetric iff ∀ x, y ∈ A, (x ℛ y ∧ y ℛ x) → (x = y). Another example of an antisymmetric relation would be the ≤ or the ≥ relation on the real numbers. Examples. Which is (i) Symmetric but neither reflexive nor transitive. (iv) Reflexive and transitive but … Antisymmetric Relation. In a set A, if one element less than the other, satisfies one relation, then the other element is not less than the first one. In set theory, the relation R is said to be antisymmetric on a set A, if xRy and yRx hold when x = y. i don't believe you do. In this article, we have focused on Symmetric and Antisymmetric Relations. The divisibility relation on the natural numbers is an important example of an antisymmetric relation. That means that unless x=y, both (x,y) and (y,x) cannot be elements of R simultaneously. (iii) Reflexive and symmetric but not transitive. A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the "preys on" relation on biological species). So from total n 2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. The usual order relation ≤ on the real numbers is antisymmetric: if for two real numbers x and y both inequalities x ≤ y and y ≤ x hold then x and y must be equal. Asymmetric Relation In discrete Maths, an asymmetric relation is just opposite to symmetric relation. the truth holds vacuously. so neither (2,1) nor (2,2) is in R, but we cannot conclude just from "non-membership" in R that the second coordinate isn't equal to the first. Proofs about relations There are some interesting generalizations that can be proved about the properties of relations. Typically some people pay their own bills, while others pay for their spouses or friends. Suppose that Riverview Elementary is having a father son picnic, where the fathers and sons sign a guest book when they arrive. And what antisymmetry means here is that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m , then m cannot be a factor of n . The Antisymmetric Property of Relations The antisymmetric property is defined by a conditional statement. It is … In this context, anti-symmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m , then m cannot be a factor of n . The relation $$R$$ is said to be antisymmetric if given any two distinct elements $$x$$ and $$y$$, either (i) $$x$$ and $$y$$ are not related in any way, or (ii) if $$x$$ and $$y$$ are related, they can only be related in one direction. A relation R is not antisymmetric if there exist x,y∈A such that (x,y) ∈ R and (y,x) ∈ R but x ≠ y. Antisymmetric: The relation is antisymmetric as whenever (a, b) and (b, a) ∈ R, we have a = b. Transitive: The relation is transitive as whenever (a, b) and (b, c) ∈ R, we have (a, c) ∈ R. Example: (4, 2) ∈ R and (2, 1) ∈ R, implies (4, 1) ∈ R. As the relation is reflexive, antisymmetric and transitive. Call it G. both can happen. This is called Antisymmetric Relation. Or it can be defined as, relation R is antisymmetric if either (x,y)∉R or (y,x)∉R whenever x ≠ y. In other words, the intersection of R and of its inverse relation R^ (-1), must be It is not necessary that if a relation is antisymmetric then it holds R(x,x) for any value of x, which is the property of reflexive relation. Another example of an antisymmetric relation would be the ≤ or the ≥ relation on the real numbers. (ii) Let R be a relation on the set N of natural numbers defined by 2006, S. C. Sharma, Metric Space, Discovery Publishing House, page 73, (i) The identity relation on a set A is an antisymmetric relation. Thus, it will be never the case that the other pair you're looking for is in $\sim$, and the relation will be antisymmetric because it can't not be antisymmetric, i.e. Equivalently, R is antisymmetric if and only if whenever R, and a b, R. In this context, antisymmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m, then m cannot be a factor of n. In discrete Maths, a relation is said to be antisymmetric relation for a binary relation R on a set A, if there is no pair of distinct or dissimilar elements of A, each of which is related by R to the other. Example 6: The relation "being acquainted with" on a set of people is symmetric. 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https://math.stackexchange.com/questions/1175625/calculating-probability-for-three-consecutive-successes | # Calculating probability for three consecutive successes
Original Question:
On a TV news channel, the evening news starts at same time every day. The probability that Mr Li gets home from work in time to watch the news is $0.3$
In a particular week of five working days, what is the probability that Mr Li gets home in time to watch the news on three consecutive days?
Attempt:
Possible arrangements with three consecutive successes in 5 trials:
SSS..
FSSS.
FFSSS
SFSSS
I calculated the respective probabilities for these arrangements as follows:
$0.3^3 + 0.3^3 0.7 + 0.3^3 0.7^2+0.3^4 0.7 = 0.0648$
Actual answer (at the back of the book) = $0.05913$
My question is what I am doing wrong here (or is the book wrong?)
Also, I would really appreciate if someone could tell me a general way to solve the problems of this type where probability of $k$ consecutive success in $n$ trials is asked. In this question I was able to manually find the possible arrangements with consecutive successes but if the number of trials is high for example 200, how would I approach this problem then.
Thanks very much.
For a more general approach, consider the probability of not having 3 successes in a row out of $n$ trials; call this probability $P(n)$, and let's say the probability of a single success is $p$, with $q=1-p$. If we have $n$ trials, condition the probability on the number of successes at the end of the $n$ trials: the $n$ trials must end in $...F$, $...FS$, or $...FSS$, so we have the recursion
$$P(n) = q \; P(n-1) + pq \; P(n-2) + p^2q \; P(n-3) \qquad \text{for } n \ge 3$$ with $P(0) = P(1) = P(2) = 1$.
It's not hard to see how to extend this approach to longer strings of successes.
Let $$X_i=\begin{cases} S, & \text{with prob 0.3 } \\ F, & \text{with prob 0.7 } \\ \end{cases}$$ for $$i=1,2,3,4,5$$ be IID for the outcome of the i-th day. The probability you are asked is the following (if we are talking for exactly 3 days succes in a row)
P(Mr Li get home in time 3 days in a row)= $$P(X_1=X_2=X_3=S)+P(X_1=F,X_2=X_3=X_4=S)+P(X_1=X_2=F,X_3=X_4=X_5=S)= 0.3^3+0.3^30.7+0.3^30.7^2=0.05913$$
(which is the correct answer in the book)
• This is actually incorrect because it does not take into account $P(X_1=X_3=X_4=X_5=S, X_2=F)$ – Couchy311 May 31 at 2:47
• I agree with you technically, that's why I said "if we are talking for exactly 3 days S in a row"...it seems like the problem should have clarified that or give another answer in the back.. – sakas May 31 at 3:08
• I see, thanks for clarifying :) – Couchy311 May 31 at 3:11 | 2019-10-24T00:39:58 | {
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https://math.stackexchange.com/questions/2106417/is-there-a-fast-way-to-get-the-characteristic-polynomial-of-this-symmetric-matri | # Is there a fast way to get the characteristic polynomial of this symmetric matrix?
I have to find the characteristic polynomial equation of this matrix
$$A= \begin{bmatrix}2 &1 &1&1 \\1&2&1&1\\1&1&2&1\\1&1&1&2 \end{bmatrix}$$
Is there another way than the rather long $\det(A-\lambda I)$ method ?
Maybe the fact that $A$ is symmetric ($A =A^t$) may be helpful ?
• Could I do something like $A-I\lambda= \begin{bmatrix}2-\lambda &1 &1&1 \\1&2-\lambda&1&1\\1&1&2-\lambda&1\\1&1&1&2-\lambda \end{bmatrix}$ then simplify and calculate the $det(...)$ , or would it change the value of$\lambda$? – HowCanIHelpYou Jan 20 '17 at 18:35
• Subtract the identity matrix and then it becomes pretty easy to show that for $A-I$ the eigenvalues will be 4 and zero (with multiplicity 3) – Sergei Golovan Jan 20 '17 at 18:35
• Interestingly, it seems that the characteristic polynomial of the $n\times n$ matrix with $2$ for each diagonal entry and $1$ for all other entries is $(n+1-\lambda)(1-\lambda)^{n-1}$. This suggests an inductive approach to me. – MPW Jan 20 '17 at 19:00
• Clearly, the eigenvalues of $A$ and $A-I$ differ. Exactly by a unity each. – Sergei Golovan Jan 20 '17 at 19:05
• And it seems that the characteristic polynomial of the $n\times n$ matrix with $a$ for each diagonal entry and $b$ for all other entries is $(a+(n-1)b - \lambda)(a-b-\lambda)^{n-1}$. – MPW Jan 20 '17 at 19:15
There's a formula for determinants of block matrices of the form $\begin{bmatrix} A&B\\B&A\end{bmatrix}$, where $A$ and $B$ are square matrices of the same size: $$\det\begin{bmatrix} A&B\\B&A\end{bmatrix}=\det(A-B)\det(A+B).$$ Applying this formula , we obtain \begin{align}\det(A-\lambda I)&=\begin{vmatrix}2-\lambda&1&1&1\\1&2-\lambda&1&1\\1&1&2-\lambda&1\\1&1&1&2-\lambda \end{vmatrix}=\begin{vmatrix}1-\lambda&0\\0&1-\lambda \end{vmatrix}\cdot\begin{vmatrix}3-\lambda&2\\2&3-\lambda \end{vmatrix}\\ &=(1-\lambda)^2\Bigl[(3-\lambda)^2-4\Bigr]=(\lambda-1)^3(\lambda-5).\end{align}
Here is a matrix in which the columns are eigenvectors of your matrix. Indeed, the columns are pairwise orthogonal.
$$\left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 2 & -1 \\ 1 & 0 & 0 & 3 \end{array} \right).$$ It is not an orthogonal matrix, as the columns are of different lengths. However, it can be made orthogonal by dividing each column by its Euclidean length, those being $2, \sqrt 2, \sqrt 6, \sqrt {12}$
The best, and very short way:
First step. Subtract to the rows $2$, $3$ and $4$, the first one.
Second step. Add to the first column the sum of the columns $2$, $3$ and $4.$
We get a triangular determinant.
EDIT: $$\begin{vmatrix}2-\lambda &1 &1&1 \\1&2-\lambda&1&1\\1&1&2-\lambda&1\\1&1&1&2-\lambda \end{vmatrix}\underbrace{=}_{\text{First step}}\begin{vmatrix}2-\lambda &1 &1&1 \\-1+\lambda&1-\lambda&0&0\\-1+\lambda&0&1-\lambda&0\\-1+\lambda&0&0&1-\lambda \end{vmatrix}$$ $$\underbrace{=}_{\text{Second step}}\begin{vmatrix}5-\lambda &1 &1&1 \\0&1-\lambda&0&0\\0&0&1-\lambda&0\\0&0&0&1-\lambda \end{vmatrix}=(5-\lambda)(1-\lambda)^3.$$
There are some simple tricks that you can use. The eigenvalues of $A$ are those values of $\lambda$ such that $A-\lambda I$ is singular, i.e., has rank less than four. Since $$A-1\cdot I=\pmatrix{ 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1}$$ has rank equal to one, it follows that $\lambda=1$ is an eigenvalue of $A$ of multiplicity $4-1=3$. Hence $(\lambda-1)^3$ is a factor in the characteristic polynomial.
In the matrix $$A-5\cdot I=\pmatrix{-3 & 1 & 1 & 1\cr 1 & -3 & 1 & 1 \cr 1 & 1 & -3 & 1 \cr 1 & 1 & 1 & -3},$$ the sum of the elements in each row is zero. Hence $A-5\cdot I$ is singular, and $\lambda-5$ is another factor in the characteristic polynomial of $A$.
It follows that $\det(A-\lambda I = (\lambda-5)(\lambda-1)^3$. | 2019-07-21T07:24:28 | {
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http://mathhelpforum.com/calculus/124868-double-check-limit-please.html | Math Help - Double-check this limit please
Can someone double-check my work here? Thanks, if you can.
$
\lim_{x\to\infty}[x(1-cos(1/x))]
$
Put x to the denominator to use L'Hopital's rule, etc. Applied chain rules, etc.
$
\lim_{x\to\infty}[sin(1/x)]
$
Thus, if x goes to infinity, the above is sin(0) = 0. Final Answer: Zero
2. Yes, it's okay, but you don't need that rule, you can turn that limit into a known one with a simple substitution.
3. Oh, really? What is this substitution?
4. $t=\frac1x.$
5. tx=1
so itequals to lim [(1-cost)/t]
t->0 | 2015-06-02T15:51:50 | {
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https://www.math.purdue.edu/pow/discussion/2016/spring/9 | ## Spring 2016, problem 9
The numbers $1, 2, \dots , n$ are written around a circle in some order. What is the smallest and largest possible sum of the absolute differences of adjacent numbers?
1 year ago
The maximum is $\lfloor \frac{n^2}{2} \rfloor$. The minimum is $2 (n - 1)$.
Proof of the maximality part:
For a circular arrangement of $1..n$ call the sum of the absolute differences of neighbours the value of the arrangement. For $a, b \in \left \{1..n \right \}$ write $a \rightarrow b$, if $b$ is the right side neighbour of $a$.
If $n$ is even, call the numbers $\frac{n}{2} .. n$ the large numbers. Call the rest small. If $n$ is odd call the numbers $\frac{n + 3}{2} .. n$ the large numbers. Call $\frac{n + 1}{2}$ the middle number. Call the rest small.
The value $v$ of the arrangement is given by:
$v := \sum_{a \rightarrow b} \left [max(a,b) - min(a,b) \right ] = \sum_{a \rightarrow b} max(a,b) - \sum_{a \rightarrow b} min(a,b)$.
Each of the two sums in the last expression is a sum of $n$ numbers in the range $1..n$, none of which occur more than twice in any of the sums. The value $v$ is clearly no larger than $m$, given by: $m := \left [ n + n + (n -1) + (n-1) + ... \right ] - \left [1 +1 + 2 + 2 + 3 + 3 ... \right ]$, where the first and the second bracket contain $n$ terms each. By a simple calculation $m = \lfloor \frac{n^2}{2} \rfloor$. But the value $m$ is indeed attained:
For $n$ even arrange the numbers like:
large $\rightarrow$ small $\rightarrow$ large $\rightarrow$ small ... $\rightarrow$ large$\rightarrow$ small
For $n$ odd arrange like:
middle $\rightarrow$ large $\rightarrow$ small $\rightarrow$ large $\rightarrow$ small ... $\rightarrow$ large$\rightarrow$ small. Done.
Proof of the minimality part:
Proof by induction over $n$: The induction start is trivial. So assume, that the minimum value for arrangements of numbers in $1..n$ is given by $m_n := 2 (n -1)$. Consider a minimum-value arrangement $A_{n+1}$ of the numbers $1..(n+1)$. with value $v_{n+1}$. Now take away the number $n+1$ from $A_ {n+1}$. You get an arrangement $A_n$ of the numers $1..n$, with value $v_{n}$. By an easy calculation $v _n \leq v_{n+1} -2$. Thus $v_{n+1}$ cannot be smaller than $2 (n + 1 -1)$ because otherwise $v_n$ would be smaller than $2 (n - 1)$. The minimum can also be no larger than $2 (n + 1 -1)$, since this value is attained for simply placing the numbers in ascending order alongthe circle. Done.
Hello, I get a slightly different answer for the maximum. It must be a whole number.
I get,
maximum = trunc(x^2/2),
i.e., if n is even, maximum = n^2/2
if n is odd, maximum = n^2/2 - 1/2
Jao
cgjoa3 1 year ago
Hi Jao. The notation $\lfloor \cdot \rfloor$ I used is defined by:
$\lfloor x \rfloor := floor(x)$.
For $x> 0$: $floor(x) =trunc(x)$
Nelix 1 year ago
1 year ago
let start with number 1, minimum possible difference for next number is 1 so next number 2 and so on so minimum sum = n
for maximum::
let start with 1 and for max difference next number should be n, number next to n such that max. difference is there should be 2
so the series is 1,n,2,n-1,3,n-2,4,n-3....., and last n/2 or (n-1)/2 for odd and even n
so max = nn/2 or (nn-1)/2
1 year ago
Are these called purdue problems of week or nelix's math proofs of the week?! Jk wish more mathletes would participate !
' half the problem for half the points ' ( lion's share had this been a putnam )Minimality of 2(n-1) Now we will show that $2n-2$ is minimal. To do so, remark that $1$ and $n$ must be on the circle. By the Triangle Inequality, the sum of the positive differences along the minor arc between $1$ and $n$ must be at least $n-1$ and similarly for the otherrh arc | 2018-01-23T10:09:38 | {
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https://math.stackexchange.com/questions/1869761/what-are-the-odds-of-flipping-a-coin-100-times-and-seeing-hhhht | # What are the odds of flipping a coin 100 times and seeing HHHHT? [closed]
What are the odds of flipping a coin 100 times and seeing exactly four consecutive heads? Any more than four heads in a row, such as "HHHHH" would not be considered a string of four consecutive heads. Seeing 10 sets of "HHHHT" would allow a max of 20 consecutive patterns. How would you expect to find the number of times an isolated string of exactly 4 heads in a row in $n$ coin flips?
## closed as off-topic by Did, Michael Albanese, Edward Jiang, Zain Patel, Chill2MachtJul 25 '16 at 0:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Michael Albanese, Edward Jiang, Zain Patel, Chill2Macht
If this question can be reworded to fit the rules in the help center, please edit the question.
• I don't understand, you seem to ask many questions at once. – Jorge Fernández Hidalgo Jul 24 '16 at 17:48
• Is your question the expected number of times you would expect to find $HHHHT$ in a sequence of $100$ flips? – Jorge Fernández Hidalgo Jul 24 '16 at 17:49
• I flip a coin 100 times. How many times should I see HHHH? A HHHHH or anything greater will not be considered as a HHHH. – Triumph Jul 24 '16 at 17:51
• No just HHHHT but THHHHT – Triumph Jul 24 '16 at 17:52
• I am pretty sure the question is "Suppose I flip a coin 100 times. How many subsequences of exactly four heads in a row (no more or less) should I expect to see?" – user326210 Jul 24 '16 at 17:54
In any sequence of $100$ flips, “mark” each spot $HHHH$ occurs (and occurs not within a longer run of $H$’s) between the middle $H$’s. For example, the sequence $$HHTHHHTTTHHHHTHT\dots TTHHHHHTHTHHTTHHHH$$ would be marked $${HHTHHHTTTHH}^{\color{red}|\!}{HHTHT\dots TTHHHHHTHTHHTTHH}^{\color{red}|\!}{HH}^{}.$$
Among the $97$ positions where marks could occur (from after the first two flips to before the final two), a mark does appear in each of the $2^{\rm nd}$ through $96^{\rm th}$ ($95$ in all) of those spots (as $THH^{\color{red}|\!}HHT$) with probability $\frac{1}{2^6}$ (the chance the surrounding sequence of six flips is $THHHHT$). In the first and last positions, the probability is $\frac{1}{2^5}$ (the chance of $HH^{\color{red}|\!}HHT$ for position $1$ and $THH^{\color{red}|\!}HH$ for position $97$).
By the linearity of expectation, the expected number of marks is $95\cdot\frac1{2^6}+2\cdot\frac1{2^5}=\frac{99}{64}$.
[Added] It may be easier to focus on the $H$’s and $T$’s instead of the spaces between them, so another way to mark the runs-of-$4$ is to color in red the last $H$ of each run, like $$HHTHHHTTTHHH\color{red}HTHT\dots TTHHHHHTHTHHTTHHH\color{red}H.$$ The first three $H$’s are colored red with probability $0$, the fourth with probability $\frac{1}{2^5}$, the next $95$ with probability $\frac{1}{2^6}$, and the last with probability $\frac{1}{2^5}$, giving the same result.
• Is it possible for you to let me know what the answer would be for 8 out of 100? – Triumph Jul 27 '16 at 1:48
• Try to modify my answer for runs of $8$. There are no new situations that come up, so it shouldn’t be very hard. – Steve Kass Jul 27 '16 at 18:32
This seems to fall easily to linearity of expectation. Let $a_1,a_2\dots a_n$ be your sequence of outcomes.
Let $X_1$ be the indicator variable that $a_1,a_2,a_3,a_4=H$ and $a_5=T$
Let $X_{97}$ be the indicator variable that $a_{96}=T$ and $a_{97},a_{98},a_{99},a_{100}=H$
Finally, for $2\leq n \leq 96$ let $X_n$ be the indicator that $a_{n-1}=T,a_n,a_{n+1},a_{n+2},a_{n+3}=H,a_{n+4}=T$
The random variable you want is $\sum_{i=1}^{97} X_n$.
By linearity of expectation its expectation is $E[X_1]+95E[X_2]+E[X_{97}]=\frac{1}{32}+\frac{95}{64}+\frac{1}{32}=\frac{99}{64}$
• What I answered is the expected number of times you can find a substring of the form $HHHH$ that does not have another $H$ to either side. Assuming the flips are independent, and each outcome has probability $\frac{1}{2}$. – Jorge Fernández Hidalgo Jul 24 '16 at 18:16
• When they are four $H$ it is $\frac{99}{64}$ and when there are three It is $\frac{1}{16}+\frac{96}{32}+\frac{1}{16}$. – Jorge Fernández Hidalgo Jul 24 '16 at 18:45
• then it is $\frac{1}{2^{11}}+\frac{89}{2^{12}}+\frac{1}{2^{11}}$ – Jorge Fernández Hidalgo Jul 24 '16 at 19:31
• if you throw $10000$ times then you should get $10$ consecutive (and not more) an expected number of $\frac{1}{2^{11}}+\frac{9989}{2^{12}}+\frac{1}{2^{11}}$ – Jorge Fernández Hidalgo Jul 24 '16 at 19:52
• I told you, $\frac{1}{2^{11}}+\frac{9989}{2^{12}}+\frac{1}{2^{11}}$, which is approximately $2.43$ – Jorge Fernández Hidalgo Jul 24 '16 at 21:10
For $i=1$ to $97$, let random variable $X_i$ be equal to $1$ if there is a string of HHHH that begins at $i$ and does not extend, and let $X_i=0$ otherwise. We want the expectation of $X_1+\cdots+X_{97}$. By the linearity of expectation, this is $E(X_1)+\cdots+E(X_{97})$.
So we need only calculate the $\Pr(X_i)=1$.
These are not all equal. If $i=1$ or $i=97$, we have $\Pr(X_i=1)=\frac{1}{32}$. For all the other $i$, we have $\Pr(X_i=1)=\frac{1}{64}$. That is because in all but "end" cases, the string of $4$ H must be flanked by T on both sides.
The required expectation is therefore $2\cdot \frac{1}{32}+95\cdot\frac{1}{64}$.
• Would this factor in the possibility of 10 heads in a row that would change you 100 coin flips into 90. From the start you should automatically drop the 100 flips down to 50 with the assumption that the other 50 flips would be tails correct? What are the odds of 5 heads or 6 heads each one of those will take more flips away from what you need to achieve 4 heads. – Triumph Jul 24 '16 at 18:14
• The expectation calculation (indirectly) takes care of $10$ heads in a row by not counting them, because then $X_i=0$ for $i=1$ to at least $11$. – André Nicolas Jul 24 '16 at 18:17 | 2019-05-25T03:46:17 | {
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https://math.stackexchange.com/questions/2399255/simple-probability-question-perfect-quiz-score | # Simple Probability Question: Perfect quiz score
a) Students have two quizzes that they must take in their undergraduate studies unit. The probability of getting a perfect score on the first quiz is 68%, the probability of getting a perfect score on the second quiz is 74%, and the probability of falling short of a perfect score on both quizzes is 19%. What is the probability of scoring a perfect score on both quizzes?
My attempt: Pr(Scoring a perfect score on both quizes) $= 0.68 \cdot 0.74 = 0.5032$
b) Refer to the probabilities described in question a. If a student has achieved a perfect score on the first quiz, what is the conditional probability of getting a perfect score on the second quiz?
My attempt:
$$\frac{\text{Pr(Scoring a perfect score on both quizzes)}}{\text{Pr(Perfect score on second Quiz)}} = 0.68$$
but I feel like there is more to both questions
• what has been tried ? what is the context ( where did you find the question) ? – user451844 Aug 19 '17 at 14:22
• a) Pr(Scoring a perfect score on both quizes) = 0.68 * 0.74 = 0.5032 b) Pr(Scoring a perfect score on both quizes)/Pr(Perfect score on second Quiz) = 0.68 but I feel like there is more to both questions – user473207 Aug 19 '17 at 14:29
• if you fall short 19% of the cases, what happens the other 81% of the cases ? – user451844 Aug 19 '17 at 14:44
• @user473207 Your calculations assume that the two events (obtainoing a perfect score on quiz 1 and obtaining a perfect score on quiz 2) are independent. BUt just conceptually, that is unlikely: SOmeone who gets a perfect score on quiz 1 is probably more likely to get a perfect score on quiz 2 than someone who does not get a perfect score on quiz 1 ... so you are right to suspect there is more to this! – Bram28 Aug 19 '17 at 14:47
• @user473207 Also, the expression 'falling short of a perfect score on both quizzes'' is a bit ambiguous (is it that you don't get a perfect score on quiz 1 and also don't get a perfect score on quiz 2, or is it that it is not true that you do perfect on both quizzes (so you can get a perfect score on one of the two quizzes, just not on both quizzes)). What they undoubtedly mean is the first interpretation. SO, you know that there is a 19% chance of not getting a perfect score on quiz 1 and also not a perfect score on quiz 2. NOw, as Roddy asked: what does the other 81% then represent? – Bram28 Aug 19 '17 at 14:49
Let $A$ be the event that a perfect score is obtained on the first quiz; let $B$ be the event that a perfect score is obtained on the second quiz. We are given \begin{align*} P(A) & = 0.68\\ P(B) & = 0.74\\ P(A^C \cap B^C) & = 0.19 \end{align*} Hence, \begin{align*} P(A^C) & = 1 - P(A) = 0.32\\ P(B^C) & = 1 - P(B) = 0.74 \end{align*} The probability of not obtaining a perfect score on at least one of the quizzes is $$P(A^C \cup B^C) = P(A^C) + P(B^C) - P(A^C \cap B^C) = 0.32 + 0.26 - 0.19 = 0.39$$ Thus, the probability of obtaining a perfect score on both quizzes is $$P(A \cap B) = 1 - P(A^C \cup B^C) = 1 - 0.39 = 0.61$$ so your assumption that the probabilities of obtaining a perfect score on each quiz are independent was false.
The probability that a student who received a perfect mark on the first quiz obtains a perfect mark on the second quiz is $$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.61}{0.68} = \frac{61}{68}$$ | 2020-09-26T18:47:29 | {
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http://mathhelpforum.com/advanced-algebra/150432-eigenvectors.html | # Math Help - Eigenvectors!
1. ## Eigenvectors!
(i) Determine all the eigenvalues of A
(ii) For each eigenvalue of A, find the set of eigenvectors corresponding to
A = ( 1 2 )
( 3 2 )
I found eigenvalues of A to be 4 and -1.
I also found the eigenvectors to be (2/3,1) for =4 and (-1,1) for =-1
BUT the solution in the back of the book says (2,3) for =4 and (1,-1) for =1
I'm soo confusedd! Can someone tell me what's wrongg? Am I calculating the eigenvectors wrong?
Here's how i calculate eigenvector for =4
A-4I = (-3 2)
(3 -2)
then (-3 2 |0)
(3 -2 |0)
and i end up with x1 -(2/3) x2 =0. so x2 = t, and x1 = (2/3)t. eigenvector = (2/3,1) but book says (2,3)
and for =-1
A-(-1)I = (2 3 )
(2 3 )
then (2 3 |0)
(2 3 |0)
and i end up with x1+x2=0. x2= t and x1 = -t. eigenvector = (-1,1) but book says (1,-1)
2. You don't have a problem. Note that your eigenvectors are just multiples of the book's answer. This is fine!
3. Hi
$\texttt{det}(A-I_{2}X)=\texttt{det}\begin{pmatrix}
1-X & 2\\
3 & 2-X
\end{pmatrix}=(1-X)(2-X)-6=X^2-3X-4=(X-4)(X+1)$
and thus your eigenvalues are 4 and -1.
For 4
$A\begin{pmatrix}
x\\
y
\end{pmatrix}=4\begin{pmatrix}
x\\
y
\end{pmatrix}\Leftrightarrow \begin{pmatrix}
x+2y\\
3x+2y
\end{pmatrix}=$
$\begin{pmatrix}
4x\\
4y
\end{pmatrix}\Leftrightarrow \left\{\begin{matrix}
-3x+2y=0\\
3x-2y=0
\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x=\frac{2}{3}t\\
y=t,t\in \mathbb{R}
\end{matrix}\right.$
and thus the eigenvectors are of the form,
$x=x_1(\frac{2}{3},1),x_1\in \mathbb{R}\setminus \left \{ 0 \right \}$
4. The set of all eigenvectors of a matrix A, corresponding to a given eigenvalue, $\lambda$, form a subspace. In particular, any multiple of an eigenvector is also an eigenvector, corresponding to the same eigenvalue: If $Av= \lambda v$ and "r" is any scalar, then $A(rv)= r(Av)= r(\lambda v)= (r\lambda)v= \lambda (rv)$. | 2014-10-01T11:06:12 | {
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http://lotoblu.it/ixah/triple-integral-limit-calculator.html | # Triple Integral Limit Calculator
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Notes on Triple Integration Dr. Convert this triple integral into cylindrical coordinates and evaluate $\int_{-1}^{1}\int_{0}^{\sqrt{1-x^2}}\int_{0}^{y}x^2dz\; dy\; dx onumber$ Solution. A volume integral is a specific type of triple integral. The sphere x2 +y2 +z2 = 4 is the same as ˆ= 2. Second Derivative. Triple Integrals in Cartesian Coordinates: Over rectangular. Includes derivatives, integration, volume, quadratic equation and trig identities. Elliott Jacobs On Wednesday, March 4, you saw how to set up triple and calculate triple integrals. Now is the lower limit and is. For example, if f ( x) is positive for. pro [email protected] It is the height of a thin stick as in Section 14. The key idea is to replace a double integral by two ordinary "single" integrals. Let's do limit comparison to 1/t3: lim t→∞ 1/t3 1/t3−t =lim t→∞ t3−t t3 =lim t→∞ t3−t t3. $$\int_0^4 \int_{0}^{1-x} \int x^2 + y^2 dzdydx$$ I think those two limits may be correct but I don't know how to get the third, I understand how to find limits form a shape like a tetrahedron but a bit confused and my notes don't help me with this shape. We calculate as follows:. dzdxdy, then once you find your Z limits in the first integral, then you are done with Z altogether, the next step to solve the triple integral is to project into the remaining variables' plane, in other words, project into the x-y plane. So we should just calculate that limit, for arbitrary a, and then let a!1. V = \iiint\limits_U { {\rho ^2}\sin \theta d\rho d\varphi d\theta }. Changing the order of integration of a triple integral blackpenredpen. Third Derivative. A similar situation occurs with triple integrals, but here we need to distinguish between cylindrical symmetry and spherical symmetry. restart: Setting limits of integration and evaluating. • Evaluate double integrals over general regions. When there are limits, and we need to use U-Substitution, there are a few things we need to keep in mind:. Summary : The integral function calculates online the integral of a function between two values. Recently Asked Math Questions. Arc Length Cartesian & Polar Coordinates, 2D & 3D Parametric Curves. In python we use numerical quadrature to achieve this with the scipy. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each $\Delta x\times\Delta y\times\Delta z. Set up (but do not evaluate) iterated triple integrals, with appropriate limits, for nd the volume of the solid bounded by z= x2 + y2 and z= 8 x2 y2 in: (a)rectangular coordinates (b)cylindrical coordinates 9. The purpose of these notes is to present a number of triple integral examples and discuss how to set up the limits of integration. Notes on Triple Integration Dr. The limits of integration for this graph are (0,2). You can also use it to solve differential and integral equations. Multiple Integral Calculator - eMathHelp This site contains an online calculator that finds multiple integrals (double or triple integrals). The concept of integrals is fundamental in calculus, and has broad application in all engineering disciplines. V = \iiint\limits_U {\rho d\rho d\varphi dz}. Watch if the algorithm is converging. Let's return to the previous visualization of triple integrals as masses given a function of density. The int function can be used for definite integration by passing the limits over which you want to calculate the integral. I am trying to solve below mentioned (image) equation in R for double integral over an area. 0 (R14), you will need to use a function file or. Simplify a calculation by changing the order of integration of a triple integral. This is somewhat subtle in the physical interpretation but can be summarized as "generality". An integral is a mathematical result that represents the area between a function and a plane (e. , 26 MB] Improper integrals - part 2 - integrals with integrand undefined at an endpoint [video; 21 min. Sometimes you need to change the order of integration to get a tractable integral. For example, if you tried to evaluate ∫1 0∫1 xey2dydx directly, you would run into trouble. Triple Integral Cylindrical Coordinates. To set up a double. Create the worksheets you need with Infinite Calculus. Calculate the average value of a function of three variables. Integral bounds , also called limits of integration, define the area that you’ll be integrating. set up triple integrals to compute the volume of a solid. Math 6B is the second quarter of a two quarter sequence in vector calculus and infinite series. 6 Implicit Differentiation. So, if possible, please write the logic for the solution. 973 #55,56) Calculate mass by integrating density. Enter all requested information on the top of this page, and put your initials on the top of every page, in case the pages become separated. Evaluating Double Integrals by Repeated Integrals 5. This calculator evaluates derivatives using analytical differentiation. Never runs out of questions. ranges here in the interval 0 \le x \le 1, and the variable y. ) Finding the Limits. Calculate the volume of the solid. You can also easily calculate multiple integrals as well as use mathematical constants such as. The limit of the Riemann sum is the triple integral of f over D and f is continuous. For example: 𝑟 𝑟 𝜃 3 −3 2 0 2π 0 is the triple integral used to calculate the volume of a cylinder of height 6 and radius 2. For a sequence indexed on the natural number set , the limit is said to exist if, as , the value of the elements of get arbitrarily close to. ( Click here for an explanation) [ ti-83/ti-84 ] Approximate Integration. Evaluating double integrals is similar to evaluating nested functions: You work from the inside out. Absolute Minimum. Triple integral - volume of sphere. Evaluate a triple integral by expressing it as an iterated integral. ranges in the interval 0 \le y \le 2 – 2x. Triple Integral Calculator. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates. (1 point) Write a triple integral including limits of integration that gives the volume of the cap of the solid sphere x2 + y +2 s 25 cut off by the plane z 3 and restricted to the first octant. Triple Integral Spherical Coordinates. We start from the simplest case when the region of integration $$U$$ is a rectangular box $$\left[ {a,b} \right] \times \left[ {c,d} \right]$$ $$\times \left[ {p,q} \right]$$ (Figure $$1$$). If you have a function f(x), there are several ways to mark the derivative of f when it comes to x. sets, logic, proofs. Fill in the integrand, the limits, and the integrating variable for. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Analysis & calculus symbols table - limit, epsilon, derivative, integral, interval, imaginary unit, convolution, laplace transform, fourier transform. the triple integral of a continuous function $$f(x,y,z)$$ over a rectangular solid box $$B$$ is the limit of a Riemann sum for a function of three variables, if this limit exists Contributors Gilbert Strang (MIT) and Edwin "Jed" Herman (Harvey Mudd) with many contributing authors. 7 Triple Integrals Be able to evaluate a given triple integral. Evaluate a Triple Integral Using Cylindrical Coordinates - Triple Integral of e^z Evaluate a Triple Integral Using Spherical Coordinates - Triple Integral of 1/(x^2+y^2+z^2) Find the Moment of Inertia about the z-axis of a Solid Using Triple Integrals Find the Center of Mass of a Solid Using Triple Integrals. • Calculate the Jacobian of a transformation of two and three variables. Calculate the volume of the solid. Create the worksheets you need with Infinite Calculus. I The average value of a function in a region in space. Derivative Calculator. I Triple integrals in arbitrary domains. Let’s do limit comparison to 1/t3: lim t→∞ 1/t3 1/t3−t =lim t→∞ t3−t t3 =lim t→∞ t3−t t3. They will make you ♥ Physics. The Integral Calculator supports definite and indefinite integrals (antiderivatives) as well as integrating functions with many variables. What does mean where is a region in the plane. This is a topic that takes some practice. V = \iiint\limits_U { {\rho ^2}\sin \theta d\rho d\varphi d\theta }. Calculate the average value of a function of three variables. So I'll start with the triple integral. To find those limits on the z integral, follow a line in the z direction. Further, it is possible to show that the limit. BYJU’S online triple integral calculator tool makes the calculation faster, and it displays the integrated value in a fraction of seconds. Absolutely free, online math solution. Limit Calculator calculates an established limit of the function with respect to a variable in a specific point. is given by where R(xyz) is the region of integration in xyz space, R(uvw) is the corresponding region of integration in uvw space, and the. Next, I'll substitute the limits. io development by creating an account on GitHub. One of the ways in which definite integrals can be improper is when one or both of the limits of integration are infinite. Define the indefinite integral of a vector-valued function. Integrals of a function of two variables over a region in R 2 are called double integrals, and integrals of a function of three variables over a region of R 3 are called triple integrals. The limits on z run from 0 to h. quad command. Triple Integral Spherical Coordinates. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates. Math terminology from differential and integral calculus for functions of a single variable. sets, logic, proofs. Online Triple & Double Integral Calculator With Steps. And then finish with dx to mean the slices go in the x direction (and approach zero in width). index: subject areas. Fraction calculator. What Everybody Dislikes About Double Integral Calculator and Why. The general steps required to perform the above integral. Since the plane ABC. Wolfram|Alpha is a great tool for calculating indefinite and definite triple integrals. Figure 12-10 shows an example. In this example, since the limits of integration are constants, the order of integration can be changed. In many cases, it is convenient to represent the location of in an alternate set of coordinates, an example of which are the so-called polar coordinates. ha πha Outer integral: 2π =. In Calculus, the four important concepts are limits, continuity, derivatives and integrals. Second Derivative. Integrals involving radicals for instance, we want to get rid of the square root. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example. You can calculate integrals numerically using techniques such as the Simpson quadrature, Lobatto quadrature, and Gauss. It enters the prism at z = 0 and exits at the sloping face y + 32 = 3. We start from the simplest case when the region of integration $$U$$ is a rectangular box $$\left[ {a,b} \right] \times \left[ {c,d} \right]$$ $$\times \left[ {p,q} \right]$$ (Figure $$1$$). Reversing the order of integration. Create the worksheets you need with Infinite Calculus. More than just an online integral solver. If you're behind a web filter, please make sure that the domains *. Enter the series to calculate its sum: This calculator for to calculating the sum of a series is taken from Wolfram Alpha LLC. Integral Calculator If you were looking for a way to calculate the Integral value of a set of mumbers, then the Integral calculator is exactly what you need. To something small like 0. Partial Derivative. This integral is improper at infinity only, and for large t we know that t3 is the dominant part. This tutorial demonstrates how to evaluate integrals using the TI-89, TI-92+, or Voyage 200 graphing calculators. Expression Calculator evaluates an expression in a given context. The integral is equal to the area of. Write a triple integral, including limits of integration, that gives the volume between 3x+y+z=0 and 4x+4y+z=0, and above x+y<=1, x>=0, y>=0. It is easier to calculate triple integrals in spherical coordinates when the region of integration U is a ball (or some portion of it) and/or when the integrand is a kind of f\left ( { {x^2} + {y^2} + {z^2}} \right). Muliple Integration Section 1: DOUBLE INTEGRALS PROBLEM: Consider the solid E in 3-space bounded above by the surface z = 40 − 2xy and bounded below by the rectangular region D in the xy-plane (z = 0) defined by the set D = {(x,y) : 1 ≤ x ≤ 3, 2 ≤ y ≤ 4}. Next, I’ll substitute the limits. Integration over surfaces, properties, and applications of integrals. Hold ˚and xed, and let ˆincrease. My work so far: Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y 2 +5z 2 ≤x ≤ 5 -----> 5r2 ≤ x ≤ 5, since each cross-section is a full circle 0 ≤ θ ≤ 2π. Explore the solid defining the boundaries of the region for a triple integral. Changes of variable can be made using Jacobians in much the same way as for double integrals. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Using (19) and (22), we calculate ∂(x,y) ∂(u. Multiple-version printing. How to Use Definite Integral Calculator Integration can. sets, logic, proofs. Try int(x^2,x=t+2. The Integral Calculator Trap The last result is provided by taking the very first limit from the second. It is possible to calculate the limit at a of a function where a represents a real : If the limit exists and that the calculator is able to calculate, it returned. Among other things, they let us calculate the volume under a surface. So it will be. In polar coordinates, the point is located uniquely by specifying the distance of the point from the origin of a given coordinate system and the angle of the vector from the origin to the point from the positive -axis. An example shows how to set them up and how to evaluate them. Integral online. Enter a valid algebraic expression to find the derivative. Scalar line integrals can be used to calculate the mass of a wire; vector line integrals can be used to calculate the work done on a particle traveling through a field. Triple integrals 3. First of all, I’ll integrate with respect to. As a final example, we see how to compute the length of a curve given by parametric equations. Triple Integrals in Cartesian Coordinates: Over rectangular. Limits at Jump Discontinuities and Kinks. Evaluating Double Integrals by Repeated Integrals 5. Triple Integrals, Changing the Order of Integration, Part 1 of 3. The Riemann Sum formula is as follows: Below are the steps for approximating an integral using six rectangles: Increase the number of rectangles (n) to create a better approximation: Simplify this formula by factoring out w […]. Watch if the algorithm is converging. Just as with double integrals, the only trick is determining the limits on the iterated integrals. The syntax in dblquad is a bit more complicated than in Matlab. There are three steps that must be done in order to properly convert a triple integral into cylindrical coordinates. What Everybody Dislikes About Double Integral Calculator and Why. Try this handy Limit calc right now!. The key idea is to replace a double integral by two ordinary "single" integrals. Double and triple integrals. (Very simple problem). Apply, evaluate, and understand integrals of multi-variable scalar-valued functions. You can also check your answers! Interactive graphs/plots help visualize and better understand the functions. Mathispower4u 6,729 views. Don't forget to use the magnify/demagnify controls on the y-axis to adjust the scale. Please Explain Reasoning Behind Limits. Double Integrals over General Regions Type I and Type II regions Examples 1-4 Examples 5-7 Swapping the Order of Integration Area and Volume Revisited Double integrals in polar coordinates dA = r dr (d theta) Examples Multiple integrals in physics Double integrals in physics Triple integrals in physics Integrals in Probability and Statistics. Calculate Partial derivatives with incredible ease! Use this handy partial derivative calculator with a step-by-step solution and graph. Higher Order Derivatives. Finding volume for triple integrals using spherical coordinates. , when integrate is called as integrate (expr, x, a, b). I Triple integrals in arbitrary domains. Definite Integral Calculator computes definite integral of a function over an interval using numerical integration. index: subject areas. 9 Best Free Integral Calculator Software For Windows Here is a list of best free Integral Calculator Software to solve integrations. using Type I region using Type region calculate one fo the above integrals. In spherical coordinates, the volume of a solid is expressed as. advanced topics. The integral is the line integral of a continuous real-valued function or , i. Automatic spacing. Any time we have an iterated integral (and we can go beyond triple integrals) we can have the limits be non-constant functions. MTH 229 projects. It enters the prism at z = 0 and exits at the sloping face y + 32 = 3. Also recall the chapter prelude, which showed the opera house l’Hemisphèric in Valencia, Spain. 5b: Integrals in probability and statistics:. Series Expansions Number & Power Series, Fourier / Taylor / Laurent / Puiseux Series. Homework Statement Find the centroid of the solid: Limit of a function as n approaches infinity Finding the eigenfunctions and eigenvalues associated with an operator. As with most definite integrals, you should ignore the bounds (0 and 2) at first and focus on how to find an antiderivative of the function inside the integral. Use the upper right corner as your sample point of each rectangle to approximate the double integral. The Integral Calculator supports definite and indefinite integrals (antiderivatives) as well as integrating functions with many variables. So it will be. To begin, enter the limit expression in graphing or "y =" mode, in Table Setup, set Tbl to start to the arrow-number, and then set. I Triple integrals in arbitrary domains. Online Derivative Calculator. You are allowed to take one. Functions 3D Plotter is an application to drawing functions of several variables and surface in the space R3 and to calculate indefinite integrals or definite integrals. Solve limits step-by-step. See exercise 27. Instead of integrating a function of two variables over an area, we are integrating a function of three variables over a volume. De nition: The triple integral of fover Bis ZZZ B f(x;y;z)dV = lim jjPjj!0 Xl i=1 m j=1 Xn k=1 f(x ijk;y ijk;z ijk) V ijk provided that the limit exists. The innermost integral is the one that is integrated over first. In all of our examples above, the integrals have been indefinite integrals - in other words, integrals without limits of integration (the "a" and "b" in the statement "the integral from a to b"). Here is a list of best free Integral Calculator Software to solve integrations. Let's return to the previous visualization of triple integrals as masses given a function of density. In many cases, it is convenient to represent the location of in an alternate set of coordinates, an example of which are the so-called polar coordinates. Also recall the chapter prelude, which showed the opera house l’Hemisphèric in Valencia, Spain. Changing the order of integration in triple integrals Triple Integrals, Changing the Order of Integration,. numbers & symbols. Improper integrals - part 1 - introduction; integrals with an infinite limit of integration [video; 21 min. The mechanics for double and triple integration have been wrapped up into the functions dblquad and tplquad. A triple integral can be represented as , where f(x,y,z) is the integrated function defined over the three-dimensional shape E , and E is the region of integration in the (x,y,z) three. Includes derivatives, integration, volume, quadratic equation and trig identities. In triple integrals, the integral will be taken of a four-dimensional function using limits of integration of a bounded VOLUME. Type in the triple integral problem to solve To get started, type in a value of the triple integral and click «Submit» button. Since is defined as a limit of Riemann sums, the continuity of is enough to guarantee the existence of the limit, just as the integral exists if g is continuous over. Triple Integral Calculator at a Glance The 5-Minute Rule for Triple Integral Calculator. The text is Vector Calculus by M. Input a function, the integration variable and our math software will give you the value of the integral covering the selected interval (between the lower limit and the upper limit). By this, I mean you can take the volume of any three dimensional object with a triple integral, but you are somewhat limited with a double integral. Integration over surfaces, properties, and applications of integrals. If expr is a constant, then the default integration variable is x. We can solve for z to determine the upper limit of integration. }\) Activity 11. This differential volume can be expressed in six possible ways. The x and y coordinates lie in a disk of radius a, so 0 ≤ r ≤ a and 0 <θ ≤ 2π. ) is written as y = 2 – 2x. Define the indefinite integral of a vector-valued function. Observe that this function is symmetric across all eight octants, so we. Change the camera position and the direction of view in three dimensions. surface integral (1) is defined to be this limit. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 9 Setting up the volume as a triple integral in spherical coordinates, we have: ZZZ S dV = Z 0 Z 2ˇ 0 Z R 0 ˆ2 sin˚dˆd d˚ = Z 0 Z 2ˇ 0 [1 3 ˆ 3]ˆ=R ˆ=0 sin˚d d˚ = 1 3 R 3(2ˇ)[ cos˚]˚= ˚=0 = 2 3 ˇR 3(1 cos ): In the special case = ˇ, we recover the well-known formula that. To calculate. We calculate as follows:. Triple integrals in Cartesian coordinates (Sect. Figure 1 In order for the double integral to exist, it is sufficient that, for example, the region D be a closed (Jordan) measurable region and that the function f(x, y ) be continuous throughout D. See more ideas about Vector calculus, Calculus and Calculus 2. Math terminology from differential and integral calculus for functions of a single variable. The integral is the line integral of a continuous real-valued function or , i. (B)A graphing calculator is required. Wolfram|Alpha is a great tool for calculating antiderivatives and definite integrals, double and triple integrals, and improper integrals. Introduction; Finding the area under a curve is a useful tool in a large number of problems in many areas of science, engineering, and business. 0 a 4 Middle integral: hr dr3 =. Handout 8: Plots of three dimensional regions associated with triple integrals. int uses the default integration variable determined by symvar ( expr,1 ). The limit calculator finds if it exists the limit at any point, at the limit at 0, the limit at +oo and the limit at -oo of a function. If expr is a constant, then the default integration variable is x. The limits on z run from 0 to h. Definite Integrals Calculator. Online Triple & Double Integral Calculator With Steps. How to Calculate Multiple Integrals. index: subject areas. No need to calculate any antiderivatives. The user enters a function of two or three variables and corresponding limits of integration and the tool evaluates the integral. Calculate the volume of the solid. Choose "Find the Derivative" from the menu and click to see the result!. These integral calculator can be used to calculate and solve definite integrals and indefinite integrals. My work so far: Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y 2 +5z 2 ≤x ≤ 5 -----> 5r2 ≤ x ≤ 5, since each cross-section is a full circle 0 ≤ θ ≤ 2π. Triple integrals also arise in computation of Volume (if f(x,y,z)=1, then the triple integral equals the volume of R) Force on a 3D object Average of a Function over a 3D region Center of Mass and Moment of Inertia Triple Integrals in General Regions. Equation solver can find both numerical and parametric solutions of equations. Fubini’s Theorem for double integrals over rectangles; iterated integrals. I am using the following code by help of the following post: Triple integral in R (how to specifying the domain) Not sure, if I am missing anything as if I change value of A, it still gives me the same results. Summary : The integral function calculates online the integral of a function between two values. Triple integral limits over pyramid. This is a topic that takes some practice. (L points) et C be the half cylinder bounded by y = 0, z = 0, z = 2 and x2 +y2 = 1 for y. zip: 1k: 03-03-08: Ultimate Calculus Program Program includes Rolles, Trapezoidal, Mean Value, Rieman Sum, and Integral Equations. #N#Index for Calculus. numbers & symbols. The every single and general integration techniques and even unique, important functions being provided. Triple Integral Calculator at a Glance The 5-Minute Rule for Triple Integral Calculator. 7 Triple Integrals Be able to evaluate a given triple integral. Tech- II Subject: Engineering Mathematics II Unit-3 RAI UNIVERSITY, AHMEDABAD. BYJU’S online integral calculator tool makes the calculations faster, showing the integral value for the given function in a fraction of seconds. Step by step calculus inside your TI-89 & Titanium calculator. The cone z = p. By using this website, you agree to our Cookie Policy. 1-Definitions. If you're seeing this message, it means we're having trouble loading external resources on our website. We start from the simplest case when the region of integration $$U$$ is a rectangular box $$\left[ {a,b} \right] \times \left[ {c,d} \right]$$ $$\times \left[ {p,q} \right]$$ (Figure $$1$$). 3 Triple Integrals Question Find the prism volume in the order dz dy dx (six orders are possible). and convert it to cylindrical coordinates. use triple integral to prove volume of ice cream cone c is : pib^(3) Calculus: Apr 8, 2013: Cone volume with triple integral: Calculus: Dec 18, 2012: Cone volume-triple integral: Calculus: Jun 12, 2012: Mass of Ice Cream Cone using Triple Integrals: Calculus: Mar 22, 2012. • Apply the Change of Variables in Multiple Integrals. How to use the Indefinite Integral Calculator. This gives us a ray going out from the origin. The x and y coordinates lie in a disk of radius a, so 0 ≤ r ≤ a and 0 <θ ≤ 2π. I Examples: Changing the order of integration. advanced algebra. Be able to find the limits of integration for an integral given the solid region of integration. In a moment you will receive the calculation result. In polar coordinates, the point is located uniquely by specifying the distance of the point from the origin of a given coordinate system and the angle of the vector from the origin to the point from the positive -axis. I Examples: Changing the order of integration. Absolute Convergence. Step by step calculus inside your TI-89 & Titanium calculator. There is nothing that says that triple integrals set up as this is must only have constants as limits! So, here is the $$x$$ integration. This differential volume can be expressed in six possible ways. Triple integrals are the analog of double integrals for three dimensions. Double integral calculator provides you the facility of step by step solution of the problem which means that you can get a solution like your teachers to solve it on a white board. The limits of all inner integrals need to be defined as functions. Physical Applications of Triple Integrals : volume of sphere. In StandardForm, Integrate [ f, x] is output as ∫ f x. Evaluate the triple integral ∫∫∫E 5x dV, where E is bounded by the paraboloid x = 5y 2 + 5z 2 and the plane x = 5. Free Fourier Series calculator - Find the Fourier series of functions step-by-step This website uses cookies to ensure you get the best experience. To calculate the limits for an iterated integral. Of course, you can have Maple calculate multiple integrals. Requires the ti-83 plus or a ti-84 model. Theorem 13. We need an antiderivative of √(4. What is a cross product? A cross product, also known as a vector product, is a mathematical operation in which the result of the cross product between 2 vectors is a new vector that is perpendicular to both vectors. It is the height of a thin stick as in Section 14. Triple Integrals and Volume - Part 2 - Duration: 8:06. In general integrals in spherical coordinates will have limits that depend on the 1 or 2 of the variables. The definite integral of from to , denoted , is defined to be the signed area between and the axis, from to. The limits of all inner integrals need to be defined as functions. For a curve produced by a function, you may be able to integrate the function from a to b and calculate the area under the curve. When you see the table, you will mostly see the y values getting closer to the limit answer as homes in on arrow-number. xmin lower limit of outer integral. By using this website, you agree to our Cookie Policy. 2 A Catalog of Essential Functions 1. Let’s do limit comparison to 1/t3: lim t→∞ 1/t3 1/t3−t =lim t→∞ t3−t t3 =lim t→∞ t3−t t3. It is possible to calculate the limit at a of a function where a represents a real : If the limit exists and that the calculator is able to calculate, it returned. You solve this type of improper integral by turning it into a limit problem where c approaches infinity or negative infinity. The usual cautions about numerical methods apply, particularly when the function is not well behaved. It also shows plots, alternate forms, and other relevant information to enhance your mathematical intuition. Type in the indefinite integral problem to solve To get started, type in a value of the indefinite integral and click «Submit» button. One special case of the product rule is the constant multiple rule, which states that if c is a number and f(x) is a differential function, then cf(x) is also differential, and its derivative is (cf)'(x)=cf'(x). Example 4 Find volume of the tetrahedron bounded by the coordinate planes and the plane through$(2,0,0)$,$(0,3,0)$, and$(0,0,1)$. Get more help from Chegg Get 1:1 help now from expert Calculus tutors Solve it with our calculus problem solver and calculator. It also shows plots, alternate forms, and other relevant information to enhance your mathematical. The notation for the general triple integrals is, $\iiint\limits_{E}{{f\left( {x,y,z} \right)\,dV}}$ Let’s start simple by integrating over the box,. Triple integrals in spherical coordinates Our mission is to provide a free, world-class education to anyone, anywhere. coordinates, change of variables in multiple integrals. Classical integration theorems of vector calculus Math 6B. Step 2: Click the blue arrow to submit. 0, 1e5 or an expression that evaluates to a float, such as exp(-0. Double integrals over general regions. In cylindrical coordinates, the volume of a solid is defined by the formula. The Integral Calculator Trap The last result is provided by taking the very first limit from the second. The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). using Type I region using Type region calculate one fo the above integrals. This page shows you two ways to compute a definite integral with numeric limits, and how to plot an accumulation function. The magnitude of this new vector is equal to the area of a para. Definite Integral. Now is the lower limit and is. Suppose we have a double integral in x and y, which we would like to change to a double integral in two new variables, u, and v, where u = 3x + 4y and v = x - 2y. That gives the upper limit z = (3 -y)/3. Multiple-version printing. int uses the default integration variable determined by symvar ( expr,1 ). Second Derivative. Now here the solid is enclosed by the planes and the surface. TI-84 Plus and TI-83 Plus graphing calculator program for AP calculus students. First Derivative. If you took integral calculus, you probably learned that there are many functions whose integrals don't show up in any integral tables, simply because they are unknown to mathematics. For each fixed x we integ- rate with respect to y. If it's not clear what the y. Do a change of variables on the integral R dA, using x = u y =3v, where R be the region bounded by 9x 2+y = 36. Two key concepts expressed in terms of line integrals are flux and circulation. integrate (expr, x) is an indefinite integral, while integrate (expr, x, a, b) is a definite integral, with limits of integration a and b. You can solve double integrals in two steps: First evaluate the inner integral, and then plug this solution into the outer integral and solve that. Please Explain Reasoning Behind Limits. 3 Triple Integrals Question Find the prism volume in the order dz dy dx (six orders are possible). Remember that we are thinking of the triple integral ZZZ U f(x;y;z) dV as a limit of Riemann sums, obtained from the following process: 1. \mathbf {F} = – Gm\,\mathbf {\text {grad}}\,u, where G is the gravitational constant. Polynomial calculator - Integration and differentiation. See exercises 3, 5. My work so far: Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y 2 +5z 2 ≤x ≤ 5 -----> 5r2 ≤ x ≤ 5, since each cross-section is a full circle 0 ≤ θ ≤ 2π. about mathwords. Transformation of a graph (function) - rotation 90 counter clockwise Sunday February 16, 2020. Of course, you can have Maple calculate multiple integrals. Wolfram|Alpha is a great tool for calculating antiderivatives and definite integrals, double and triple integrals, and improper integrals. integral, press & twice. In this video, I start discussing how a particular order of integration for a given region and integral ' makes sense '! Then I go. D 0 0 0 0 0 0 Inner integral: r3zh = hr3. More than just an online integral solver. Triple Integrals, Changing the Order of Integration, Part 1 of 3. So it will be. Attempt at solution Well, the overall thing that I. In this specific question, we firstly need to integrate sin 𝑥 d𝑥 between the limits 𝑦 and zero. In StandardForm, Integrate [ f, x] is output as ∫ f x. This is the bulk of 15. Online Triple & Double Integral Calculator With Steps. In Calculus, the four important concepts are limits, continuity, derivatives and integrals. You can also use it to solve differential and integral equations. The chapter discusses the double integral of a function of two variables and the triple integral of a function of three variables. 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Click on the integral (labeled number 1) showing once you have clicked on the equation in previous step, and then click on the kind of integral you would like, in this case, the one labeled number 2 is chosen. F = int (expr,a,b) computes the definite integral of expr from a to b. If expr is a constant, then the default integration variable is x. advanced algebra. ranges in the interval 0 \le y \le 2 – 2x. 2π a h 2π a h Mass = r 2 dV = r 2 dz r dr dθ = r 3 dz dr dθ. website feedback. If the limit exists, then fis called integrable. we use the scipy. Also note that the fact that one of the limits is not a constant is not a problem. One special case of the product rule is the constant multiple rule, which states that if c is a number and f(x) is a differential function, then cf(x) is also differential, and its derivative is (cf)'(x)=cf'(x). Explore the solid defining the boundaries of the region for a triple integral. 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We begin by discussing the evaluation of iterated integrals. website feedback. Given a function sketch, the derivative, or integral curves. Fast and easy to use. This tutorial demonstrates how to evaluate integrals using the TI-89, TI-92+, or Voyage 200 graphing calculators. If it’s not clear what the y. How to use the Double Integral Calculator. Classical integration theorems of vector calculus Math 6B. In your integral, use theta, rho, and phi for 0, ρ and φ, as needed. The int function can be used for definite integration by passing the limits over which you want to calculate the integral. Newton's Method Calculator. If expr is a constant, then the default integration variable is x. Free Summation Calculator. That gives the upper limit z = (3 -y)/3. Changing the order of integration of a triple integral blackpenredpen. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, How to calculate limits in triple integral? 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How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example. The region D consists of the points (x,y,z) with x^2+y^2+z^2<=4 and x^2+y^2<=1 and z>=0. Here is a simple. Summary : The integral function calculates online the integral of a function between two values. I Examples: Changing the order of integration. My problem isn't the integration process but just to determine what the limits are. This integral is improper at infinity only, and for large t we know that t3 is the dominant part. Triple integrals in Cartesian coordinates (Sect. Elliott Jacobs On Wednesday, March 4, you saw how to set up triple and calculate triple integrals. integration of trigonometric integrals Recall the definitions of the trigonometric functions. D 0 0 0 0 0 0 Inner integral: r3zh = hr3. Changes of variable can be made using Jacobians in much the same way as for double integrals. I Triple integrals in arbitrary domains. What is a cross product? A cross product, also known as a vector product, is a mathematical operation in which the result of the cross product between 2 vectors is a new vector that is perpendicular to both vectors. The sum on the right is called a Riemann sum, and fis said to be integrable if the limit of Riemann sums exists. In general integrals in spherical coordinates will have limits that depend on the 1 or 2 of the variables. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each$\Delta x\times\Delta y\times\Delta z. Because if your integration order takes care of Z first, i. about mathwords. When there are limits, and we need to use U-Substitution, there are a few things we need to keep in mind:. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. First, we must convert the bounds from Cartesian to cylindrical. double integral is defined as the limit of sums. Second, we find a fast way to compute it. Use double (or triple) integrals to calculate the average value of a function in some region. Now here the solid is enclosed by the planes and the surface. This integral is improper at infinity only, and for large t we know that t3 is the dominant part. Define the triple integral of a function f(x, y, z) over a bounded reglon in space. Integration by parts formula: ? u d v = u v-? v d u. Changing the order of integration in triple integrals Triple Integrals, Changing the Order of Integration,. The integral is equal to the area of. There will be six different orders of evaluating the triple iterated integrals. • Evaluate double integrals over general regions. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example. Equation Solver solves a system of equations with respect to a given set of variables. advanced algebra. • Evaluate double integrals over general regions. The common way that this is done is by df / dx and f'(x). This gives us a ray going out from the origin. Because if your integration order takes care of Z first, i. Free indefinite integral calculator - solve indefinite integrals with all the steps. Triple Integral Calculator. BYJU’S online triple integral calculator tool makes the calculation faster, and it displays the integrated value in a fraction of seconds. Direct application of the fundamental theorem of calculus to find an antiderivative can be quite difficult, and integration by substitution can help simplify that task. Now this last limit is clearly one (divide top and bottom by t3, or use continuity of the square root to move the limit inside the radical). Multiple Integrals -- First Example: Degenerate Double Integral | # Following Examples are Variations of Examples from Math 210 Calculus III | # Textbook Multivariable Calculus, Third Edition, by James Stewart | # from Section 13. There are examples of valid and invalid expressions at the bottom of the page. Although we define triple integrals using a Riemann sum, we usually evaluate triple integrals by turning them into iterated integrals involving three single integrals. Volume integrals are especially important in physics for many applications, for example, to calculate flux densities. index: click on a letter. Integral Calculator If you were looking for a way to calculate the Integral value of a set of mumbers, then the Integral calculator is exactly what you need. Logo, Boxshot & ScreenShot. Note, that integral expression may seems a little different in inline and display math mode - in inline mode the integral symbol and the limits are compressed. 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http://bootmath.com/expressing-bbb-n-as-an-infinite-union-of-disjoint-infinite-subsets.html | # Expressing $\Bbb N$ as an infinite union of disjoint infinite subsets.
The title says it. I thought of the following: we want $$\Bbb N = \dot {\bigcup_{n \geq 1} }A_n$$
We pick multiples of primes. I’ll add $1$ in the first subset. For each set, we take multiples of some prime, that hasn’t appeared in any other set before. Then \begin{align} A_1 &= \{1, 2, 4, 6, 8, \cdots \} \\ A_2 &= \{3, 9, 15, 21, 27, \cdots \} \\ A_3 &= \{5, 25, 35, 55, \cdots \} \\ A_4 &= \{7, 49, 77, \cdots \} \\ &\vdots \end{align}
I’m heavily using the fact that there are infinite primes. I think these sets will do the job. Can someone check if this is really ok? Also, it would be nice to know how I could express my idea better, instead of that hand-waving. Alternate solutions are also welcome. Thank you!
Edit: the subsets must be also infinite.
#### Solutions Collecting From Web of "Expressing $\Bbb N$ as an infinite union of disjoint infinite subsets."
Here is another way to do it.
Let $A_{i}$ consist of all the numbers of the form $2^im$ where $2\nmid m$. That is, $A_i$ consists of all the numbers that have exactly a factor of $2^i$ in them. So
\begin{align} A_0 &= \{1,3,5,7,9,11, \dots\}\\ A_1 &= \{2, 6 =2^1\cdot 3, 10 = 2^1\cdot 5, 14 = 2^1\cdot 7, \dots\}\\ A_2 &= \{4 = 2^2, 12=2^2\cdot 3, 20=2^2\cdot 5, \dots\}\\ A_3 &= \{8=2^3, 24=2^3\cdot 3, 40=2^3\cdot 5, \dots \}\\ &\vdots \end{align}
In general $A_i = \{2^im: p\nmid m\}$. You can of course pick any other prime instead of $2$.
I like @Thomas’s answer best, but I would have enumerated $\mathbb N\times\mathbb N$ and then taken the inverse images of the separate columns for the subsets.
This is an alternate solution.
Let $\alpha$ and $\beta$ be any pair of irrational numbers such that
$$1 < \alpha < 2 < \beta\quad\text{ and }\quad \frac{1}{\alpha} + \frac{1}{\beta} = 1.$$
The two sequences
$\displaystyle\;\left\lfloor\alpha k\right\rfloor\;$ and
$\displaystyle\;\left\lfloor\beta k\right\rfloor\;$, $k \in \mathbb{Z}_{+}$
are called Beatty sequence and it is known they form a partition of $\mathbb{Z}_{+}$.
Define two functions $\tilde{\alpha}, \tilde{\beta} : \mathbb{Z}_{+} \to \mathbb{Z}_{+}$ by
$$\tilde{\alpha}(k) = \left\lfloor\alpha k\right\rfloor \quad\text{ and }\quad \tilde{\beta}(k) = \left\lfloor\beta k\right\rfloor$$
We have
$$\mathbb{Z}_{+} = \tilde{\alpha}(\mathbb{Z}_{+}) \uplus \tilde{\beta}(\mathbb{Z}_{+})$$ where $\uplus$ stands for disjoint union.
Replace the rightmost $\mathbb{Z}_{+}$ recursively by this relation, we get
\begin{align} \mathbb{Z}_{+} &= \tilde{\alpha}(\mathbb{Z}_{+}) \uplus \tilde{\beta}(\mathbb{Z}_{+})\\ &= \tilde{\alpha}(\mathbb{Z}_{+}) \uplus \tilde{\beta}(\tilde{\alpha}(\mathbb{Z}_{+})) \uplus \tilde{\beta}(\tilde{\beta}(\mathbb{Z}_{+})))\\ &= \tilde{\alpha}(\mathbb{Z}_{+}) \uplus \tilde{\beta}(\tilde{\alpha}(\mathbb{Z}_{+})) \uplus \tilde{\beta}(\tilde{\beta}(\tilde{\alpha}(\mathbb{Z}_{+}))) \uplus \tilde{\beta}(\tilde{\beta}(\tilde{\beta}(\mathbb{Z}_{+}))))\\ &\;\vdots \end{align}
As a consequence, if one define a sequence of subsets $A_1, A_2, \ldots \subset \mathbb{Z}_{+}$ recursively by
$$A_1 = \tilde{\alpha}(\mathbb{Z}_{+}) \quad\text{ and }\quad A_n = \tilde{\beta}(A_{n-1}), \quad\text{ for } n > 1,$$
these subsets will be pairwise disjoint. It is clear all these $A_n$ are infinite sets.
Since $\beta > 2$, we have
$$\tilde{\beta}(k) = \left\lfloor \beta k \right\rfloor > \beta k – 1 \ge (\beta – 1) k\quad\text{ for all } k \in \mathbb{Z}_{+}$$
This implies
$$\tilde{\beta}^{\circ\,\ell}(k) = \underbrace{\tilde{\beta}(\tilde{\beta}( \cdots \tilde{\beta}(k)))}_{\ell \text{ times}} > (\beta-1)^\ell k \ge (\beta-1)^\ell \quad\text{ for all } k, \ell \in \mathbb{Z}_{+}$$
As a result,
$$\bigcap_{\ell=1}^\infty \tilde{\beta}^{\circ\,\ell}(\mathbb{Z}_{+}) = \emptyset \quad\implies\quad \mathbb{Z_{+}} = \biguplus_{k = 1}^\infty A_k$$
i.e. $\mathbb{Z}_{+}$ is an infinite disjoint union of infinite sets $A_k$. Since there are uncountable choices for $\alpha$, there are uncountable ways of such infinite disjoint unions.
For a concrete example, let $\alpha = \phi, \beta = \phi^2$ where $\phi$ is the golden mean, we get something like
$$\begin{array}{rll} \mathbb{Z}_{+} = & \{\; 1,3,4,6,8,9,11,12,14,16,17,19,\ldots\;\}\\ \uplus & \{\; 2,7,10,15,20,23,28,31,36,41,\ldots\;\}\\ \uplus & \{\; 5,18,26,39,52,60,73,81,94,\ldots\;\}\\ \uplus &\{\; 13,47,68,\ldots\;\}\\ \vdots\; & \end{array}$$
If you follow the rule that you only take the multiples that didn’t show up already, then you’re fine, since by construction you’ll be making all the subsets disjoint and by the Fundamental Theorem of Arithmetic every element will be in some $A_i$.
An example of simple infinite disjoint union would be $A_i=\{i\}$ and then $\mathbb{N}=\bigcup_{i=0}^\infty{A_i}$.
With edit: A simple way to split up $\mathbb{N}$ into a disjoint union of infinite subsets is to start with $A_0=\{0,2,4,\ldots\}$, and then let $A_1$ be every other element of $\mathbb{N}\setminus A_0$ (i.e. $\{1,5,9,13,\ldots\}$). In general, let $A_n$ be the set containing “every other element” of the set $\mathbb{N}\setminus \bigcup_{i=0}^{n-1}{A_n}$. | 2018-07-21T17:07:39 | {
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https://math.stackexchange.com/questions/436559/a-natural-proof-of-the-cauchy-schwarz-inequality/436642 | # A natural proof of the Cauchy-Schwarz inequality
Most of the proofs of the Cauchy-Schwarz inequality on a pre-Hilbert space use a fact that if a quadratic polynomial with real coefficients takes positive values everywhere on the real line, then its discriminant is negative(e.g. Conway: A course in functional analysis). I think this is somewhat tricky. Moreover I often forget its proof when the pre-Hilbert space is defined over the field of complex numbers. Is there a more natural proof (hence it's easy to remember) which is based on a completely different idea?
• To someone who voted to close, would you please explain the reason for the vote? – Makoto Kato Jul 5 '13 at 5:59
• You might want to browse the first chapter of The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele. – Did Jul 5 '13 at 6:26
• I'm not sure about what could qualify as "natural" or "more natural" proof for you, but in this case I can't think, off the top of my head, of anything *simpler", basic and straightforward as working with a quadratic's discriminant: this is Junior High School stuff! – DonAntonio Jul 5 '13 at 8:45
• @DonAntonio As I wrote, I don't think the complex pre-Hilbert space case is not so straightforward. – Makoto Kato Jul 5 '13 at 9:01
• A question which has 3 upvotes, 2 favorites and a 6 upvoted answer should not be closed. Please reset the close votes. – Makoto Kato Jul 5 '13 at 19:24
Recall the Pythagorean theorem: If $u_1, \cdots u_n$ are pairwise orthogonal, then $$\| u_1 + \cdots u_n \|^2 = \|u_1\|^2 + \cdots + \| u_n \|^2.$$
I want to use this to tell us something about two non-zero vectors $u$ and $v,$ but they aren't necessarily orthogonal. So consider the projection of $u$ onto the plane of vectors orthogonal to $v:$ $$w = u - \frac{ \langle u,v \rangle}{\|v\|^2} v.$$ This is certainly orthogonal to $v,$ and the Pythagorean theorem applied to $w$ and $v$ gives the Cauchy-Schwarz inequality.
• This is excellent. Thanks. By the way, I think $<v, u>$ is a typo($<v, v>$). – Makoto Kato Jul 5 '13 at 9:19
• @MakotoKato Thank you for spotting that. Also, I find using \langle , \rangle for inner products is more pleasing to the eye than $<$ and $>.$ – Ragib Zaman Jul 5 '13 at 14:14
• @RagibZaman You're welcome. I just didn't know how to write $\langle, \rangle$ instead of < and >. By the way again, I think $\|v||$ should be squared or replaced by $\langle v, v \rangle$. Regards. – Makoto Kato Jul 5 '13 at 19:19
• I fail to see how you obtain the CS inequality in your last paragraph. Would you explain that? – Martin Argerami Jul 5 '13 at 20:42
• @MartinArgerami Let $u_1 = w$, $u_2 = \frac{ \langle u,v \rangle}{\langle v, v \rangle} v$. Then apply the Pythagorean formula $\| u_1 + u_2\|^2 = \|u_1\|^2 + \| u_2 \|^2 \ge \|u_2\|^2$. Regards. – Makoto Kato Jul 6 '13 at 0:41
There is also an approach by "amplification" which is really cool. Also the exact same trick works to prove Hölder's inequality and is generally a very important principle for improving inequalities.
It goes like this: We start out with $$\langle a-b,a-b\rangle\ge 0$$ for $a,b$ in your inner product space, and $a\not=0$, $b\not=0$. This implies $$2\langle a,b\rangle\le \langle a, a\rangle + \langle b, b\rangle$$ Now notice that the left hand side is invariant under the scaling $a\mapsto \lambda a$, $b\mapsto \lambda^{-1}b$ for $\lambda>0$. This gives $$2\langle a,b\rangle \le \lambda^2 \langle a,a\rangle + \lambda^{-2}\langle b, b\rangle$$ Now look at the right hand side as a function of the real variable $\lambda$ and find the optimal value for $\lambda$ using calculus (set the derivative to $0$):
$$\lambda^2=\sqrt{\frac{\langle b,b\rangle}{\langle a,a\rangle}}$$
Plugging this value in, we obtain
$$2\langle a,b\rangle\le \sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}+\sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}$$
i.e.
$$\langle a,b\rangle\le\sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}$$
Notice how we took a trivial observation and "optimized" the expression by exploiting scaling invariance.
• As explained (and expanded) by Terence Tao on his blog. – Did Jul 5 '13 at 13:08
• Could your proof be modified to show that the equality holds iff $a$ is proportional to $b$? I've been thinking a while on it, but I haven't figured anything out. – jinawee Oct 1 '14 at 19:54
• @jinawee: Yes, the only inequality in this proof is $\langle \mu a-b,\mu a-b\rangle\ge 0$ for an optimal constant $\mu$ depending on $a,b$. Equality implies $b=\mu a$. – J.R. Oct 2 '14 at 7:17
The inequality $| \langle a, b \rangle | \leq \| a \| \| b \|$ can be rewritten as $$| \langle a, \frac{b}{\|b\|} \rangle | \leq \| a \|$$ (assuming $b \neq 0$).
On the left we see the component of $a$ on the unit vector $u = \frac{b}{\| b \|}$. On the right we have the norm of $a$. Of course the norm of $a$ is larger than the component of $a$ on $u$, this is very intuitive. To make this into a rigorous proof, we merely have to write $a = \langle a, u \rangle u + v$, note that $v \perp u$, and use the pythagorean theorem.
Another approach is to start with the inequality $$0 \leq \| a - \text{proj}_b(a) \|^2.$$
Just expand the right hand side and Cauchy-Schwarz pops out.
This is not exactly an answer but an explanation for the idea behind the Ragib Zaman's answer.
Let $K$ be the field of real numbers or the field of complex numbers. Let $E$ be a pre-Hilbert space over $K$.
Let $x, y$ be elements of $E$ such that $\langle x, y \rangle = 0$. Then $\|x + y\|^2 = \langle x+y, x+ y\rangle = \langle x, x\rangle + \langle x, y \rangle + \langle y, x\rangle + \langle y,y\rangle = \|x\|^2 + \|y\|^2$.
Let $x, y$ be elements of $E$ such that $\langle x - y, y \rangle = 0$. Then, by the above formula, $\|x\|^2 = \|x - y\|^2 + \|y\|^2$. Hence $\|x\| \ge \|y\|$.
Finally let $u, v$ be non-zero elements of $E$. Let $t$ be an element of $K$ such that $\langle u - tv, v \rangle = 0$. $t$ must be $\frac{ \langle u,v \rangle}{\langle v, v\rangle}$ since $v \neq 0$. Then $\|u\| \ge \|tv\|$ by the previous inequality. This leads to the Cauchy-Schwarz inequality immediately.
Here's my favorite proof, mainly because it's nicely symmetric, easy to remember and not impossible to come up with (the main trick is that $2=1+1$):
We want to prove $$|\langle x,y\rangle|\le\|x\|\|y\|$$ This is linear in $x$ or $y$ and obviously holds for $x=0$ or $y=0$. Therefore without loss of generality we can suppose that $\|x\|=\|y\|=1$: $$|\langle x,y\rangle|\le\|x\|\|y\|\Longleftrightarrow|\langle x,y\rangle|\le1\Longleftrightarrow2|\langle x,y\rangle|\le2\Longleftrightarrow2|\langle x,y\rangle|=\|x\|^2+\|y\|^2$$
Let $u\in\mathbb C$ be a complex unit (i.e. $|u|=1$), then \begin{align}0\le\|x-uy\|^2&=\langle x-uy,x-uy\rangle=\langle x,x\rangle-u\langle x,y\rangle-\overline{u\langle x,y\rangle}+\langle y,y\rangle\\&=\|x\|^2+\|y\|^2-2\,\text{Re}(u\langle x,y\rangle)\end{align} Now just set $u$ so that $\text{Re}(u\langle x,y\rangle)=|\langle x,y\rangle|$ and you are done.
Of course in the real case you can just expand $0\le\|x\pm y\|^2$. | 2019-09-20T20:29:44 | {
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https://math.stackexchange.com/questions/2848788/is-this-a-valid-approach-to-solving-the-inequality-frac1x-x-1 | # Is this a valid approach to solving the inequality $\frac{1}{x} < x < 1$?
I have been given the inequality $\frac{1}{x} < x < 1$ and have been told to find the values of $x$ which satisfy this inequality, and I have also been told to find these values using a case-by-case approach. I'd like to know whether my reasoning is valid.
Case 1: $x=0$
In this case we find that the value of $\frac1x$ is undefined, so we know that $x \neq 0$.
Case 2: $0<x<1$
Given the inequality $\frac{1}{x} < x < 1$ and that $x$ lies in the range $0 < x < 1$, we find that $1<x^2$. Therefore, $x>1$ or $x<-1$. However, $x\not>1$ and in this case $x\not<-1$ since we are looking at the case $0<x<1$. Thus we conclude that no values of $x$ in the range $0<x<1$ satisfy the original inequality.
Case 3: $x<0$
From the original equality and the fact that we know that $x<1$ we find:
$$\frac1x < x$$
$$\implies 1 > x^2$$ (since $x<0$) $$\implies -1<x<1$$ but we know that $x$ cannot lie in the range $0<x<1$ so we have that $-1<x$ and that $x<0$ and combining these inequalities we have that $-1<x<0$.
This argument seems to be supported by looking at the graph, but I am unsure whether all the steps I have made are valid and whether this is what is meant by a 'case analysis'.
• This is absolutely fine and there is no part of the solution that is wrong. – Prakhar Nagpal Jul 12 '18 at 16:00
• I think the argument is good! ${}{}{}{}{}{}{}$ – Andres Mejia Jul 12 '18 at 16:01
• Thank you for the feedback. – Benjamin Jul 12 '18 at 16:02
Overall, I would say that the argument is spot on! Just a couple comments you might want to consider. First of all, to be super rigorous in your answer, you might mention that clearly we know $x \ngeq 1$ so we need not consider that case. Another thing, I felt that your argument via contradiction for Case 2 was a bit lengthy there at the end. Once you reached the point $1 < x^2$ you could have just said that since we have $0 < x < 1$ (via assumption) and no such value squared is greater than 1, we have a contradiction and are done. The whole considering $x > 1$ or $x < -1$ is just overkill.
Also, that is exactly what it means by a case analysis, or case-by-case approach. Use cases to consider every possible value of $x$, and see what happens. Using cases is actually a very common proof technique! | 2021-05-14T11:22:01 | {
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