# Repeated Eigenvalues

General Form for Solutions in the Case of Repeated Eigenvalues For the di erential equation x_(t) = Ax(t); assume that we have found the single eigenvalue and a corresponding eigenvec-. 1 Find the eigenvalues and associated eigenspaces of each of the following matrices. then every eigenvalue of X is an eigenvalue of A, and the associated eigenvector is in V = R(M) if Xu = λu, u 6= 0 , then Mu 6= 0 and A(Mu) = MXu = λMu so the eigenvalues of X are a subset of the eigenvalues of A more generally: if AM = MX (no assumption on rank of M), then A and X share at least Rank(M) eigenvalues Invariant subspaces 6-6. Input the components of a square matrix separating the numbers with spaces. edu It is well known that a function can be decomposed uniquely into the sum of an odd and an even function. So the good case is when the geometric multiplicity of each eigenvalue equals its algebraic multiplicity because then. @Star Strider: Thanks for the suggestion, I was unaware of this function. The repeated eigenvalue λ2= corresponds to the eigenvectors v2,1= and v2,2=. distinct real eigenvalues, different signs saddle point –4 –2 0 2 4 y –4 –2 2 4 x distinct real eigenvalues, both negative node sink –4 –2 0 2 y distinct real roots, both positive node source –2 0 2 4 y –4 –2 2 4 x repeated eigenvalues, positive improper node, source –4 –2 0 2 4 y –4 –2 2 4 x repeated eigenvalues. It is called complete if there are m linearly independent eigenvectors corresponding to it and defective otherwise. Next, everything is repeated with the second eigenvalue and the second eigenvector - the 2nd pr. They are not unique when there are repeated eigenvalues, as in the example above of a disk rotating about any of its diameters (§ B. The eigenspace corresponding to one eigenvalue of a given matrix is the set of all eigenvectors of the matrix with that eigenvalue. So the first thing we do, like we've done in the last several. If an eigenvalue algorithm does not produce eigenvectors, a common practice is to use an inverse iteration based algorithm with μ set to a close approximation to the eigenvalue. As noted above, if λ is an eigenvalue of an n × n matrix A, with corresponding eigenvector X, then (A − λIn)X = 0, with X 6= 0, so det(A−λIn) = 0 and there are at most n distinct eigenvalues of A. The general solution is Y(t) = e^(a*t)*V0 + t*e^(a*t)*V1 such that a is the repeated eigenvalue and V1=(A-a*I)V0 (A=2x2 matrix such that A*V1=a*V1, I=2x2 identity matrix) or V1=0 vector. Section PEE Properties of Eigenvalues and Eigenvectors The previous section introduced eigenvalues and eigenvectors, and concentrated on their existence and determination. Defective matrices A are similar to the Jordan (canonical) form A = XJX−1. Costco wholesale business plan how to write a compare and contrast essay pdf how do i assign a drive letter to a new ssd drive writing argumentative essays middle school essay on video game violence, art institute essay examples thinking critically with psychological science answers. Horn Department of Electrical Engineering and Computer Science, MIT and CSAIL, MIT, Cambridge, MA 02139, USA e-mail: [email protected] (c) First of all, by part (b), we know A has at least a complex eigenvalue. 1 - Eigenvalue Problem for 2x2 Matrix Homework (pages 279-280) problems 1-16 The Problem: • For an nxn matrix A, find all scalars λ so that Ax x=λ GG has a nonzero solution x G. Now is the next step. Thanks for your reply. Eigenvalues and Eigenvectors of a 3 by 3 matrix Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. This video series will not get side tracked by special cases with embedded pole/zero cancellations, repeated poles and non-simple Jordan forms and the like. If eigenvalue stability is established for each component individually, we can conclude that the original (untransformed) system will also be eigenvalue stable. Computing its characteristic polynomial. component is being recorded, and then "removed". All the eigenvectors corresponding to of contain components with , where represents the position of each nonzero weights associated with and. When there are two linearly independent eigenvectors k 1 and k 2. The spectral decomposition of x is returned as components of a list with components. In this section we discuss the possibility that the eigenvalues of A are not distinct. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. This lack of eigenvalues and eignevectors will not occur if we use F = C, the field of complex numbers. Description. If is a repeated eigenvalue, only one of repeated eigenvalues of will change. Unfortunately I'm not good at theory and mathematics. I have used eig2image. An eigenvector of a matrix is a vector that, when left-multiplied by that matrix, results in a scaled version of the same vector, with the scaling factor equal to its eigenvalue. It means that some nonzero vector is mapped to zero times itself - that is, to the zero vector. Since the geometric multiplicity j for j is the dimension of E j, there will be exactly j vectors in this basis. The same situation applies, if Ais semi-simple, with repeated eigenvalues. i s are the repeated eigenvalues of M. the eigenvalues of A) are real numbers. Nonhomogeneous Systems - Solving nonhomogeneous systems of differential. When there are two or more resonant modes corresponding to the same natural frequency'' (eigenvalue of ), then there are two further subcases: If the eigenvectors corresponding to the repeated eigenvalue (pole) are linearly independent, then the modes are. When there is a basis of eigenvectors, we can diagonalize the matrix. 1 Introduction to Eigenvalues Linear equationsAx D bcomefrom steady stateproblems. We prove it by contradiction. distinct real eigenvalues, different signs saddle point -4 -2 0 2 4 y -4 -2 2 4 x distinct real eigenvalues, both negative node sink -4 -2 0 2 y distinct real roots, both positive node source -2 0 2 4 y -4 -2 2 4 x repeated eigenvalues, positive improper node, source -4 -2 0 2 4 y -4 -2 2 4 x repeated eigenvalues. But you can find enough independent eigenvectors -- Forget the "but. The following graph shows the Gershgorin discs and the eigenvalues for a 10 x 10 correlation matrix. Given a matrix A, recall that an eigenvalue of A is a number λ such that Av = λ v for some vector v. The matrix norm for an n × n matrix is defined. Next, everything is repeated with the second eigenvalue and the second eigenvector - the 2nd pr. Some regular eigenvectors might not produce any non-trivial generalized eigenvectors. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. Equal eigenvalues. Math Vids offers free math help, free math videos, and free math help online for homework with topics ranging from algebra and geometry to calculus and college math. Expert Answer 100% (1 rating). Let Abe a square (that is, n n) matrix, and suppose there is a scalar and a. We start by finding the eigenvalues and eigenvectors of the upper triangular matrix T from Figure 3 of Schur's Factorization (repeated in range R2:T4 of Figure 1 below). In general, nonlinear differential equations are required to model actual dynamic systems. (c) First of all, by part (b), we know A has at least a complex eigenvalue. 2 Harmonic Oscillators 114 6. (ii) Then will be an orthonormal basis of and the required matrix P is given by Further , a diagonal matrix and diagonal entries of D are , each repeated as many times as the dimensions of. i can easily find the eigenvector for eigenvalue 1, but i don't understand the method for the other two. Characteristic equations. 0 along the global X-axis. It decomposes matrix using LU and Cholesky decomposition The calculator will perform symbolic calculations whenever it is possible. That's what we want to do in PCA, because finding orthogonal components is the whole point of the exercise. 1 Eigenvalues and Eigenvectors The product Ax of a matrix A ∈ M n×n(R) and an n-vector x is itself an n-vector. • The pattern of trajectories is typical for two repeated eigenvalues with only one eigenvector. It is a \repeated eigenvalue," in the sense that the characteristic polynomial (T 1)2 has 1 as a repeated root. This matrix may be either a Vandermonde matrix or a modal matrix. In that case, one can give explicit algebraic formulas for the solutions. Deﬁnition 5. The eigenvalues may be chosen to occur in any order along the diagonal of T and for each possible order the matrix U is unique. However, the power method can find only one eigenvector, which is a linear combination of the eigenvectors. Eigenvalues and Eigenvectors. The phase portrait thus has a distinct star. is an eigenvector for , then. The center has trigonometric solutions that are the parametric representations of closed curves. De nition If Ais a matrix with characteristic polynomial p( ), the multiplicity of a root of pis called the algebraic multiplicity of the eigenvalue. Background We will now review some ideas from linear algebra. Consider a ﬁrst order diﬀerential equation of the form. (d) Show that A has no real eigenvalue if D<0. joan on December 24th, 2018 @ 2:26 pm I can now understand how this works! really interisting and I now see things way more clear!. So, the system will have a double eigenvalue, $$\lambda$$. A System of Differential Equations with Repeated Real Eigenvalues Solve = 3 −1 1 5. Despite their utility, students often leave their linear algebra courses with very little intuition for eigenvectors. 1 A Complete Eigenvalue We call a repeated eigenvalue complete if it has two distinct (linearly independent) eigenvectors. In general there will be as many eigenvalues as the rank of matrix A. For any x ∈ IR2, if x+Ax and x−Ax are eigenvectors of A ﬁnd the corresponding eigenvalue. Read "An iterative method for finite dimensional structural optimization problems with repeated eigenvalues, International Journal for Numerical Methods in Engineering" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. We found: e igenvalue λ = 4, and the associated eigenvectors are: x = x3 × 1 2 1 2 1 Geometric interpretation of eigenvalues and eigenvectors IX a repeated eigenvalue λ = −2. As noted above, if λ is an eigenvalue of an n × n matrix A, with corresponding eigenvector X, then (A − λIn)X = 0, with X 6= 0, so det(A−λIn) = 0 and there are at most n distinct eigenvalues of A. If is a diagonal matrix with the eigenvalues on the diagonal, and is a matrix with the eigenvectors as its columns, then. Unfortunately I'm not good at theory and mathematics. An eigenvector is a nonzero vector that, when multiplied against a given square matrix, yields back itself times a multiple. vector w is called a generalized eigenvector corresponding to the eigenvalue 1. The constituent matrices are then found to be equal to the product of two matrices with elements partly from a certain matrix and its inversion. (b) Find the eigenvalues of A. Find the characteristic polynomial and the eigenvalues. how can i find eigenvectors for repeated eigenvalues and complex eigenvalues. Since they appear quite often in both application and theory, lets take a look at symmetric matrices in light of eigenvalues and eigenvectors. And we're asked to find the general solution to this differential equation. Firstly we look at matrices where one or more of the eigenvalues is repeated. , λi with multiplicity k. a vector containing the $$p$$ eigenvalues of x, sorted in decreasing order, according to Mod(values) in the asymmetric case when they might be complex (even for real matrices). This happens when the dimension of the nullspace of A−λI (called the geometric multiplicity of λ) is strictly less than the arithmetic multiplicity m. So far we have covered what the solutions look like with distinct real eigenvalues and complex (nonreal) eigenvalues. Show that A and AT do not have the. It is called complete if there are m linearly independent eigenvectors corresponding to it and defective otherwise. 3 Eigenvalues and Eigenvectors. Such matrices arise, for example, in invariant subspace decomposition approaches to the symmetric eigenvalue problem. Since some eigenvalues may be repeated roots of the characteristic polynomial, there may be fewer than neigenvalues. eigenvalues of A = · a h h b ¸ and constructs a rotation matrix P such that PtAP is diagonal. (2018) A new method for computation of eigenvector derivatives with distinct and repeated eigenvalues in structural dynamic analysis. 1 Find the eigenvalues and associated eigenspaces of each of the following matrices. Recall that given a symmetric, positive de nite matrix A we de ne R(x) = xTAx xTx: Here, the numerator and denominator are1 by 1matrices, which we interpret as numbers. If eigenvalue stability is established for each component individually, we can conclude that the original (untransformed) system will also be eigenvalue stable. where the eigenvalues are repeated eigenvalues. So far we have considered the diagonalization of matrices with distinct (i. This is all part of a larger lecture series on differential equations here on educator. , λi 6= λj for i 6= j, then A is diagonalizable (the converse is false — A can have repeated eigenvalues but still be diagonalizable) Eigenvectors and diagonalization 11–22. Suppose the 2 2 matrix Ahas repeated eigenvalues. Repeated Eigenvalues 1 Section 7. EIGENVALUES AND EIGENVECTORS 5 Similarly, the matrix B= 1 2 0 1 has one repeated eigenvalue 1. In order to determine the eigenvectors of a matrix, you must first determine the eigenvalues. Penny [ + - ] Author and Article Information. So our strategy will be to try to find the eigenvector with X=1, and then if necessary scale up. SIAM Journal on Matrix Analysis and Applications 37:4, 1581-1599. • The pattern of trajectories is typical for two repeated eigenvalues with only one eigenvector. Matrix Calculator applet. 1 A Complete Eigenvalue We call a repeated eigenvalue complete if it has two distinct (linearly independent) eigenvectors. Solution motivation from 1D: x0= axhas solution x(t) = eatx(0);. non-zero vectors v, w satisfying: Av = av Aw = dw which brings us to your second question. Moreover, these methods are surely going to have trouble if the matrix has repeated eigenvalues, distinct eigenvalues of the same magnitude, or complex eigenvalues. one repeated eigenvalue. I'm setting up my program to solve for x, y and z (it's a 3x3 matrix) using Gauss elimination. 1 (Eigenvalue and eigenvector). In the case where the 2X2 matrix A has a repeated eigenvalue and only one eigenvector, the origin is called an improper or degenerate node. Let's say we have the following second order differential equation. The matrix norm for an n × n matrix is defined. Diagonalisation of 2 x 2 and 3 x 3 matrices. Figure 1 – Eigenvectors of a non-symmetric matrix. m to find the eigen vectors at each point on the image (in my image, there is grey values on the concentric circular region and background is black ). However, ker(B I 2) = ker 0 2 0 0 = span( 1 0 ): Motivated by this example, de ne the geometric multiplicity of an eigenvalue. Eigenvalues and Eigenvectors. 5 Nonautonomous Linear Systems 130. If is a diagonal matrix with the eigenvalues on the diagonal, and is a matrix with the eigenvectors as its columns, then. Input the components of a square matrix separating the numbers with spaces. (If there is no such eigenvector,. 3 Introduction In this Section we further develop the theory of eigenvalues and eigenvectors in two distinct directions. For real asymmetric matrices the vector will be complex only if complex conjugate pairs of eigenvalues are detected. i don't really understand how to find the corresponding eigenvectors to repeated eigenvalues. Eigenvalues have a number of convenient properties. The calculator will find the eigenvalues and eigenvectors of the given square matrix, with steps shown. Let A be an n´ n matrix over a field F. Recall that in the case of a repeated eigenvalue (of algebraic multiplicity m) we might not have m linearly independent eigenvectors. Still assuming 1 is a real double root of the characteristic equation of A , we say 1 is a complete eigenvalue if there are two linearly independent eigenvect ors ~ 1 and ~ 2. A defective matrix has at least one multiple eigenvalue that does not have a full set of linearly independent eigenvectors. 6 Genericity 101 CHAPTER 6 Higher Dimensional Linear Systems 107 6. In the previous cases we had distinct eigenvalues which led to linearly independent solutions. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. In this form, the state matrix is a diagonal matrix of its (non-repeated) eigenvalues. Find the eigenvalues and eigenvectors of the matrix. Plugging in 1 0 to A, we see that solutions go counterclockwise. 22 Matrix exponent. For real asymmetric matrices the vector will be complex only if complex conjugate pairs of eigenvalues are detected. Given an eigenvalue λ i of an n×n matrix M, its geometric multiplicity is the dimension of Ker(M −λ iI n), and it is the number of Jordan blocks corresponding to λ i. repeated eigenvalues. The Effect of Close or Repeated Eigenvalues on the Updating of Model Parameters from FRF Data M. Proofs of the theorems are either left as exercises or can be found in any standard text on linear algebra. Start at the top of the leftmost column, and use the vectors as you go down the column. Find the characteristic polynomial and the eigenvalues. Computing its characteristic polynomial. The sum of the sizes of all Jordan blocks corresponding to an eigenvalue λ i is its algebraic multiplicity. We do not normally divide matrices (though sometimes we can multiply by an inverse). Clarence Wilkerson In the following we often write the the column vector " a b # as (a;b) to save space. The rest are similar. The matrix norm for an n × n matrix is defined. and suppose that Bhas ndistinct eigenvalues. one repeated eigenvalue. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. Under this matrix norm, the inﬁnite series converges for all A and for all t, and it deﬁnes the matrix exponential. So, once again: eigenvectors are direction cosines for principal components, while eigenvalues are the magnitude (the variance) in the principal components. 3 Repeated Eigenvalues 119 6. The vector v is called an eigenvector corresponding to the eigenvalue λ. Slope field for y' = y*sin(x+y) System of Linear DEs Real Distinct Eigenvalues #1. 2 (i) It is easy to see that eigenvalues and eigenvectors of a linear transformation are same as those of the associated matrix. J has the eigenvalues of A on its main diagonal, is upper triangular, and has 0’s and 1’s in the upper triangle. Then we can find the largest integer so that are linearly independent, but are not. We prove it by contradiction. The effectiveness. EigenValues is a special set of scalar values, associated with a linear system of matrix equations. This presents us with a problem. General Form for Solutions in the Case of Repeated Eigenvalues For the di erential equation x_(t) = Ax(t); assume that we have found the single eigenvalue and a corresponding eigenvec-. To prove this, we start with a general lemma on. An eigenvector is a nonzero vector that, when multiplied against a given square matrix, yields back itself times a multiple. Using the previous problem, show that either V is an eigenvector for Aor else (A I)V is an eigenvector for A. for each eigenvalue. Intuition: If the eigenvalues of A are all zero, then for arbitrary vector x, we have Ax=0. Deﬁnition 5. 1 Distinct eigenvalues a matrix might have repeated eigenvalues and still be diagonalizable. The main diagonal of T contains the eigenvalues of A repeated according to their algebraic multiplicities. y(t) = eaty0:. If this is the situation, then we actually have two separate cases to examine, depending on whether or not we can find two linearly independent eigenvectors. [We say that a sign pattern matrix B requires k repeated eigenvalues if every A E Q(B) has an eigenvalue of algebraic multiplicity at least k, and k is a minimum with respect to this requirement. The vector v is called an eigenvector corresponding to the eigenvalue λ. Whereas the fundamental eigenfrequency of a structure usually is of great importance, some situations may require different dynamic objectives such as maximum eigenvalue separation [17],. The proposed approach drastically reduces the coherence time requirements, enhancing the potential of quantum resources available today and in the near future. Subsection 3. The spectrum of eigenvalues is found by solving for the roots of the characteristic polynomial or secular equation det(A- I)=0. Still assuming 1 is a real double root of the characteristic equation of A , we say 1 is a complete eigenvalue if there are two linearly independent eigenvect ors ~ 1 and ~ 2. Now I’ll use Gaussian elimination method to simplify these equations. Recall that given a symmetric, positive de nite matrix A we de ne R(x) = xTAx xTx: Here, the numerator and denominator are1 by 1matrices, which we interpret as numbers. If we further assume, as in §3, that the matrix H is Hermitian,14 with its eigenvalues 1h real and its eigenvectors xh forming a base of m-space and orthonormal, 15 (4. 3 COMPLEX AND REPEATED EIGENVALUES 15 A. Supplementary notes for Math 265 on complex eigenvalues, eigenvectors, and systems of di erential equations. Pointing out the eigenvalue again is just repeating something you already said. In solving an eigenvalue problem, the eigenvalues will be determined as well as the corresponding configurations of system. We do not normally divide matrices (though sometimes we can multiply by an inverse). So, once again: eigenvectors are direction cosines for principal components, while eigenvalues are the magnitude (the variance) in the principal components. 1 (Eigenvalue and eigenvector). The relation for finding Eigenvalue corresponds to the Eigenvector x is. y(t) = eaty0:. It decomposes matrix using LU and Cholesky decomposition The calculator will perform symbolic calculations whenever it is possible. We start by finding the eigenvalues and eigenvectors of the upper triangular matrix T from Figure 3 of Schur’s Factorization (repeated in range R2:T4 of Figure 1 below). As before, perturbations in. The eigenvalues and eigenvectors of a matrix are essential in many applications across the sciences. Slides by Anthony Rossiter 3 ) (t) L 1[( sI A) 1] Poles come from determinant of (sI-A) which are clearly the same as the. 3 power method for approximating eigenvalues 551 Note that the approximations in Example 2 appear to be approaching scalar multiples of which we know from Example 1 is a dominant eigenvector of the matrix. Continuation: Repeated Real Eigenvalues, Complex Eigenvalues -- Lecture 26. Such a block has one repeated eigenvalue and only one eigenvector regardless of its dimension. We can nd the eigenvalue corresponding to = 4 using the usual methods, and nd u. 2 Harmonic Oscillators 114 6. The spectral decomposition of x is returned as components of a list with components. Indeed, BAv = ABv = A( v) = Av since scalar multiplication commutes with matrix multiplication. The complete case. Conclusion: there is one degree of freedom to determine the eigenvector itself and therefore also the derivative contains a degree of freedom. Finding Eigenvectors with repeated Eigenvalues. and when eigenvalues 0 and how can i draw phase portrait. Which eigenvectors do MATLAB/numpy display when eigenvalues are repeated. In fact, it is easy to see that this happen if and only if we have more than one equilibrium point (which is (0,0)). Finding Eigenvectors with repeated Eigenvalues. (c) Show that A has one repeated real eigenvalue if D =0. 2 Solving Systems with Repeated Eigenvalues If the characteristic equation has only a single repeated root, there is a single eigenvalue. We prove it by contradiction. where the eigenvalues are repeated eigenvalues. Repeated Eigenvalues We studyhomogeneous autonomous system: dx dt = Ax In Lecture 10 and 11, we learn how to solve this system wheneigenvalues 1 and 2 of matrix A arereal and di erentandcomplex conjugate, respectively. m in MATLAB to determine how the solution curves (trajectories) of the system Az behave A. However, if a matrix has repeated eigenvalues, it is not similar to a diagonal matrix unless it has a full (independent) set of eigenvectors. An eigenvector is determined uniquely in case of distinct eigenvalues up to a constant. This multiple is a scalar called an. Condition numbers, returned as a vector. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): The spectral stability problem for periodic traveling water waves on a two–dimensional fluid of infinite depth is investigated via a perturbative approach, computing the spectrum as a function of the wave amplitude beginning with a flat surface. Option 3: Generally, an (n x n) matrix with repeated eigen values can be diagonalised if we obtain n linearly independent eigen vectors for it. and when eigenvalues 0 and how can i draw phase portrait. Eigenvalues and Eigenvectors 1. Repeated eigenvalues indicate linear dependence within the rows and columns of A. Repeated Eigenvalues 1. 3 Introduction In this Section we further develop the theory of eigenvalues and eigenvectors in two distinct directions. The Rayleigh Principle for Finding Eigenvalues April 19, 2005 1 Introduction Here I will explain how to use the Rayleigh principle to nd the eigenvalues of a matrix A. The repeated eigenvalue λ2= corresponds to the eigenvectors v2,1= and v2,2=. General Form for Solutions in the Case of Repeated Eigenvalues For the di erential equation x_(t) = Ax(t); assume that we have found the single eigenvalue and a corresponding eigenvec-. Of course v is then called the eigenvector of A corresponding to λ. model to be determined by the eigenvalues of the A matrix. Penny [ + - ] Author and Article Information. Where there is not, we can't. 2 (Spectrum and spectral radius). So our strategy will be to try to find the eigenvector with X=1, and then if necessary scale up. This is a coupled eigenvalue problem through A and its adjoint A*. Appendix B Advanced topics Lecture B. SUPPLEMENT ON EIGENVALUES AND EIGENVECTORS We give some extra material on repeated eigenvalues and complex eigenvalues. So even though a real asymmetric x may have an algebraic solution with repeated real eigenvalues, the computed solution may be of a similar matrix with complex conjugate pairs of eigenvalues. EigenValues is a special set of scalar values, associated with a linear system of matrix equations. non-zero vectors v, w satisfying: Av = av Aw = dw which brings us to your second question. If eigenvalues are repeated, we may or may not have all n linearly independent eigenvectors to diagonalize a square matrix. (2018) A new method for computation of eigenvector derivatives with distinct and repeated eigenvalues in structural dynamic analysis. Find more Mathematics widgets in Wolfram|Alpha. The data used in this example are from the following experiment. If we further assume, as in §3, that the matrix H is Hermitian,14 with its eigenvalues 1h real and its eigenvectors xh forming a base of m-space and orthonormal, 15 (4. One of Lemma's high quality modules plucked from our universe of content. For real asymmetric matrices the vector will be complex only if complex conjugate pairs of eigenvalues are detected. It is a \repeated eigenvalue," in the sense that the characteristic polynomial (T 1)2 has 1 as a repeated root. An eigenvalue problem is a special type of problem where the solution exists only for special values (i. We prove that the volume of n satis es: j nj jRPN 3j = n 2 ; where N = n+1 2 is the dimension of the space of real symmetric matrices of size n n. We can’t ﬁnd it by elimination. The matrix norm for an n × n matrix is defined. 1 - Eigenvalue Problem for 2x2 Matrix Homework (pages 279-280) problems 1-16 The Problem: • For an nxn matrix A, find all scalars λ so that Ax x=λ GG has a nonzero solution x G. Next, everything is repeated with the second eigenvalue and the second eigenvector - the 2nd pr. A matrix and its transpose both have the same eigenvalues. An eigenvector is determined uniquely in case of distinct eigenvalues up to a constant. Every eigenvalue ß. , it could just be the diagonal matrix with diagonal entries 2. This approach is an extension of recent work by Daily and by Juang et al. Background We will now review some ideas from linear algebra. We have accomplished this by the use of a non-singular modal matrix P i. Show that A and AT do not have the. Mechanical Systems and Signal Processing 107 , 78-92 Online publication date: 1-Jul-2018. Let Abe a square (that is, n n) matrix, and suppose there is a scalar and a. how can i find eigenvectors for repeated eigenvalues and complex eigenvalues. The rest are similar. eigenvalues tell the entire story. Slope field. For 2x2, 3x3, and 4x4 matrices, there are complete answers to the problem. Indeed, BAv = ABv = A( v) = Av since scalar multiplication commutes with matrix multiplication. In particular, if you have repeated eigenvalues, it doesn't automatically mean you have a multidimensional eigenspace: for instance, the matrix 1 1 0 1 has the eigenvalue 1 with multiplicity 2, but its only eigenvector is the vector (1,0). = 1 exactly once from each column (as is done above in Equation 6) results in a new sum of zero for the elements of each column vector. If are the distinct eigenvalues of A ,for each eigenvalues , find an orthonormal basis of the eigensubspace. is at most the multiplicity of. All eigenvalues of A are distinct ⇒diagonalizable There are repeated eigenvalues, e. 5 Repeated Eigenvalues 95 5. I am trying to understand the principle underlying the general solution for repeated eigenvalues in systems of differential equations. In this chapter, we provide basic results on this subject. The eigen value and eigen vector of a given matrix A, satisfies the equation Ax = λx , where, λ is a number, also called a scalar. (solution: x = 1 or x = 5. However, if a matrix has repeated eigenvalues, it is not similar to a diagonal matrix unless it has a full (independent) set of eigenvectors. To tackle the issue of non-smoothness of repeated eigenvalues, we propose an estimator constructed by averaging all repeated eigenvalues. 06 or 18/700. Returns A const reference to the column vector containing the eigenvalues. 2 Intuitive Example. He's also an eigenvector. Find the eigenvalues and eigenfunctions of the Sturm-Liouville problem 00u = u 0 1. Consider a ﬁrst order diﬀerential equation of the form. In general there will be as many eigenvalues as the rank of matrix A. Since some eigenvalues may be repeated roots of the characteristic polynomial, there may be fewer than neigenvalues. Complex eigenvalues and eigenvectors satisfy the same relationships with l 2C and~x 2Cn. , multiplicity of "2". Costco wholesale business plan how to write a compare and contrast essay pdf how do i assign a drive letter to a new ssd drive writing argumentative essays middle school essay on video game violence, art institute essay examples thinking critically with psychological science answers. Note that this sets all entries of the matrices and vectors to 0. and is applicable to symmetric or nonsymmetric systems. An eigenvector is a nonzero vector that, when multiplied against a given square matrix, yields back itself times a multiple. So far we have covered what the solutions look like with distinct real eigenvalues and complex (nonreal) eigenvalues. This is because u lays on the same subspace (plane) as v and w, and so does any other eigenvector. How will I calculate largest eigen values and its correspoinding eigen vector of Hessian matrix to select new seed point as discussed above. y(0) = y0: Of course, we know that the solution to this IVP is given by. I'm setting up my program to solve for x, y and z (it's a 3x3 matrix) using Gauss elimination. The picture is more complicated, but as in the 2 by 2 case, our best insights come from finding the matrix's eigenvectors : that is, those vectors whose direction the. In this section we discuss the possibility that the eigenvalues of A are not distinct.


# How to calcualate how many unique set of 6 can i have in a given set.

Hello my question is quite simple i would think but i just cant seem to find an answer. I have a set of $\{1,2,3,4,5,6,7,8,9,10\}$ and i would like to calculate how many unique given sets of $6$ can i get from this set. In other words for the number $1$ i would end up with $[1,2,3,4,5,6] [1,3,4,5,6,7] [1,4,5,6,7,8] [1,5,6,7,8,9] [1,6,7,8,9,10]$ I would move down the line with the number $2$ to compare to unique sets of $6$ note: when moving to two I would no longer do this $[2,1,3,4,5,6]$ because it repeats my first case above. its there a formula to figure this sort of thing? Thanks in advance.

when I work this out on paper i end up with 15 sets here is how

for 1
[1,2,3,4,5,6]
[1,3,4,5,6,7]
[1,4,5,6,7,8]
[1,5,6,7,8,9]
[1,6,7,8,9,10]

for 2
[2,3,4,5,6,7]
[2,4,5,6,7,8]
[2,5,6,7,8,9]
[2,6,7,8,9,10]
for 3
[3,4,5,6,7,8]
[3,4,6,7,8,9]
[3,5,6,7,8,9,10]
for 4 [4,5,6,7,8,9]
[4,6,7,8,9,10]
for 5 [5,6,7,8,9,10]

after that i cant make any more groups of $6$ thus i end up with $15$ sets.

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I'd be pretty certain there are other questions similar to this on the site that might help you. –  Simon Hayward Dec 2 '12 at 20:30

Yes, there is, it is called the binomial, written $\binom{n}{k}$, read $n$ choose $k$. The value is $$\binom{n}{k}=\frac{n!}{k!(n-k)!}.$$ So, in your case, you have $$\binom{10}{6}=\frac{10!}{6!4!}=210.$$ I hope you find this helpful!

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Forgive my stupidity Clayton I have two problems here. 1 how are we ending up with 210? the "!" operator is what exactly? Also i worked this out on paper and i only end up with 15 sets of 6 so I'm obviously missing something can you help me understand. –  Miguel Dec 2 '12 at 19:19
The "!" stands for the factorial, that is, $n!=n\cdot(n-1)\cdot \cdots\cdot (2)(1)$. So, $$10!=10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1.$$ Dividing by $6!4!$ is what gives you the 210 unique subsets of order 6. Does this make it more clear? –  Clayton Dec 2 '12 at 19:23
it makes it more clear but i just dont see how the resul makes sense. For example i edit my question can you look how i reached 15. Yet this formula is given me 210 im obviously missing something but i dont see what it could be? forgive my ignorance. –  Miguel Dec 2 '12 at 19:28
No worries, it is great to ask questions. Now, looking at the subsets you've listed, where would the set $[1,2,3,4,5,7]$ fit into? Also, we can have the sets $[1,2,3,4,5,8]$, $[1,2,3,4,5,9]$, and $[1,2,3,4,5,10]$, for example. Once we've exhausted one "component," which I'm taking to be the last one, change the next index over, so now we count $[1,2,3,4,6,7]$. So on and so forth. Hope it helps! –  Clayton Dec 2 '12 at 19:36
THANK YOU! :-D i knew there was something i was missing. Thank you! I truly appreciate your response. –  Miguel Dec 2 '12 at 19:40

It exactly the number of ways to choose $6$ elements out of $10$, i,e. the binomial coefficient$$\binom{10}{6}=\frac{10!}{6!4!}$$

-

Binomial coefficients count the number of distinct subsets of $k$ elements from a set containing $n$ elements. The notation for this is $\binom{n}{k}$ which is equal to $\frac{n!}{k!(n-k)!}$

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# A (challenging?) question around the Jacobian matrix

1. Apr 5, 2008

### Aleph-0

Here is the problem:

Suppose that g is a diffeomorphism on R^n. Then we know that its jacobian matrix is everywhere invertible.
Let us define the following matrix valued function on R^n
$$H_{i,j} (x) = \int_0^1 \partial_i g^j(tx) dt$$
where $$g^j$$ are the components of g.
Question : Is $$(H_{i,j}(x))_{i,j}$$ (which could be interpreted as a mean of the Jacobian matrix of g) invertible for any x ?

My guess is that the answer is negative, but I find no counter-examples.
Any Help ?

2. Apr 7, 2008

### ObsessiveMathsFreak

I believe you may have to move into three dimensions to obtains a counterexample. It is possible that this theorem is in fact true.

Edit:
Actually, I believe that a counterexample is provided by;

$$\mathbf{g}((x,y,z))=(x \cos(2\pi z)-y \sin(2\pi z),x \sin(2\pi z)+y \cos(2\pi z),z)$$
with
$$H((0,0,1))$$ being singular.

Can anyone else check this?

Last edited: Apr 7, 2008
3. Apr 7, 2008

### Aleph-0

Indeed, we even have H(x,y,1) singular for any x,y !
Thank you !
I wonder if the theorem is true in dimension 2...
(it is trivially true in dimension 1).

4. Apr 7, 2008

### Haelfix

Interesting problem. I thought about this for awhile, and couldn't come up with a counter in D =2. Can anyone prove this statement? Its not clear to me why it should be true, and there seems to be something interesting (maybe topological) lurking behind it.

Last edited: Apr 7, 2008
5. Apr 14, 2008

### ObsessiveMathsFreak

I believe it is true because of the nature of coordinate lines and their tangents in 2D.

Firstly, assume that we can find a H in 2D that is singular. Therefore the columns of H, which we'll denote $$\mathbf{h}_i$$ are linearly dependant. So there exist non zero $$\alpha_i$$ such that

$$\Sigma \alpha_i \mathbf{h}_i = \mathbf{0}$$

Now by the definition

$$\mathbf{h}_i = \int_0^1 \frac{\partial \mathbf{g}}{\partial x_i}(t \mathbf{x}) dt$$
Therefore

$$\Sigma \alpha_i \mathbf{h}_i =\int_0^1 \Sigma \alpha_i \frac{\partial \mathbf{g}}{\partial x_i}(t \mathbf{x}) dt = \mathbf{0}$$

Now looking at the integral along this curve. We let
$$\mathbf{v}(t)=\Sigma \alpha_i \frac{\partial \mathbf{g}}{\partial x_i}(t(1,0)) = \alpha_1 \frac{\partial \mathbf{g}}{\partial x_1}(t(1,0)) + \alpha_2 \frac{\partial \mathbf{g}}{\partial x_2}(t(1,0))$$
And so we have
$$\int_0^1 \mathbf{v}(t) dt = \mathbf{0}$$
We cannot have $$\mathbf{v}=\mathbf{0}$$, because the jacobian is always fully ranked, and hence any linear combination of its columns cannot be zero.

So $$\mathbf{v}(t)$$ is non zero everywhere, and its integral is zero. But if we let $$\mathbf{v}(t)=\frac{d \mathbf{z}}{dt}(t)$$, we can see that

$$\int_0^1 \mathbf{v}(t) dt = \int_0^1 \frac{d \mathbf{z}}{dt}(t) dt = \mathbf{z}(1)-\mathbf{z}(0) = \mathbf{0}$$

Which means that $$\mathbf{v}$$ is the tangent vector to some closed curve $$\mathbf{z}(t)$$

We now make a change of variables by letting
$$\mathbf{x}(\mathbf{u}) = \left( \begin{array}{cc} \alpha_1 & -\alpha_2 \\ \alpha_2 & \alpha_1 \end{array} \right) \mathbf{u}$$

Then it can be shown that $$\mathbf{v}(t) = \frac{d \mathbf{g}}{d u_1}(t(1,0))$$, i.e. $$v$$ is the tangent vector to the $$u_2=0$$ coordinate curve. But this means that this coordinate curve is closed. Since the mapping from u to x is a diffeomorphism, and the function g is a diffeomorphism, the function from u to g is a diffeomorphism on R^2. This means that it cannot have any closed coordinate curves. Therefore we have a contridiction, and so H must be non singular.

I hope this is OK. I'm not being especially rigorous here.

In higher dimensions, some of the $$\alpha_i$$ can be zero, and hence the mapping from u to x may not be a diffeomorphism. Essentially you have some wiggle room once you move into higher dimensions.


### Author Topic: Can there exists infinite number of solutions given initial conditions.  (Read 969 times)

#### Xinyu Jiao

• Newbie
• Posts: 1
• Karma: 0
##### Can there exists infinite number of solutions given initial conditions.
« on: September 25, 2018, 08:58:39 PM »
In class, we were given an example where a differential equation can have two solutions given some initial condition. Specifically, the equation was $y' = y^\alpha$ with $0<\alpha<1$, and initial condition $y(0) = 0$. This shows that it's not unique, because it does not satisfy some condition which I do not understand.

My question is, can there be a differential equation (of order 1) such that given an initial condition, can acquire an infinite number of solutions? The answer to this question should be able to shed light as to the mechanism through which the equation acquires more than one solution.
« Last Edit: September 25, 2018, 09:18:16 PM by Victor Ivrii »

#### Victor Ivrii

• Elder Member
• Posts: 2511
• Karma: 0
##### Re: Can there exists infinite number of solutions given initial conditions.
« Reply #1 on: September 25, 2018, 09:27:59 PM »
For condition see Section 2.8 of the textbook or this Lecture Note

Yes, this equation $y'=3 y^{2/3}$ (I modified it for simplicity) has a general solution $y=(x-c)^{3}$ but also a special solution $y=0$. Thus problem $y'=3 y^{2/3}$, $y(0)=0$ has an infinite number of solutions. Restricting ourselves by $x>0$ we get solutions y=\left\{\begin{aligned} &0 &&0<x<c,\\ &(x-c)^3 && x\ge c\end{aligned}\right. with any $c\ge 0$ and similarly for $x< 0$.

This happens because this Lipschitz condition is violated at each point of the solution $y=0$.

#### Kathryn Bucci

• Jr. Member
• Posts: 6
• Karma: 2
##### Re: Can there exists infinite number of solutions given initial conditions.
« Reply #2 on: October 06, 2018, 10:59:07 AM »
If 𝑦′=𝑦𝛼 with 0 < 𝛼 < 1 e.g. 𝑦′=3𝑦2/3= f(t,y), then ∂f/∂y=2y-1/3 is not continuous at (0,0).
According to theorem 2.4.2 (existence and uniqueness for 1st order nonlinear equations), both f and ∂f/∂y have to be continuous on an interval containing the initial point (0,0) - ∂f/∂y is not continuous there so you can't infer that there is a unique solution.

#### Victor Ivrii

Continuity of $\frac{\partial f}{\partial y}$ is not required, but "Hölder  property" $|f(x,y)-f(x,z)|\le M$ is.


# What is the difference between a supremum and maximum; and also between the infimum and minimum?

What is the difference between a supremum and maximum; and similarly the infimum and minimum? Also, how does one tell if they exist?

Here is an example:

$$x_n = \frac{n}{2n-1}$$

Determine whether the maximum, the minimum, the supremum, and the infimum of the sequence $$x_n$$ n=1 to n=+∞

My understanding:

• The limit is 1/2.
• $$x_n$$ is decreasing.
• The supremum exists as n goes to +∞.
• The infimum does not exist as limited by n=1, minimum = 1/2.
• The maximum and supremum exists, both equal to 1.
• One distinction I like: If the $\sup A$ is an element of $A$ then $\sup A=\max A$, etc... so they only differ on membership to the set. For example $0=\inf \{1/n:n\in \mathbb{N}\}$ but this set has no minimum since $0 \notin \{1/n:n\in \mathbb{N}\}$. Mar 30 '16 at 1:48

If the max exists, then it is the supremum. If the min exists, then it is the infimum. For an infinite set, it can happen that the max does not exist but the supremum does exist, and/or that the min does not exist but the infimum does exist.

In this case, $\frac{n}{2n-1}$ is never actually equal to $\frac{1}{2}$, but $\frac{1}{2}$ is still the infimum, because it is a lower bound and because one can find $n$ with $\frac{n}{2n-1}<1/2+\epsilon$ whenever $\epsilon>0$ (which follows from the limit observation you made). Since the infimum is not attained, the minimum doesn't exist.

• Hi, is your last sentence a general statement? Since if infimum =1/2; it follows min = 1/2? And I am correct in my assumption max and sup = 1? Mar 30 '16 at 1:51
• The infimum is 1/2, but there is no minimum here. And yes, the max is 1, so the sup is also 1.
– Ian
Mar 30 '16 at 1:52
• Okay, so if the max and min exist then it is the sup inf, but there is no max or min then the sup inf is just the result of the limit? Mar 26 '20 at 0:38
• @ErockBrox In general "the limit" needn't even make sense; not every set of real numbers is countable, and those that are aren't necessarily given in the form of a sequence. For a set with no max or min, there is no cleaner way to describe the sup and inf than just the definition of sup and inf (as least upper bound and greatest lower bound respectively).
– Ian
Mar 26 '20 at 1:02

As Ian pointed out, the difference can only appear with infinite sets. Here is the definition of minimum and infimum:

• For the set $$A$$ to have a minimum $$x_m$$ it means that $$\exists x_m\in A$$ such that $$x\geq x_m$$ $$\forall x\in A$$.
• For the set $$A$$ to have an infimum $$x_i$$ it means that $$\exists x_m$$ such that $$x\geq x_m$$ $$\forall x\in A$$, and $$x_m$$ is the largest possible such number.

Notice that for the minimum, we require $$x_m$$ to still be in the set, while the infimum does not need to be in the set. That is the key point. For example, for $$A=(0,\infty)$$, we have that $$x\geq 0$$ $$\forall x\in A$$, and zero is the largest number that makes this inequality true for all elements of $$A$$. Since $$0\notin A$$, we cannot say that it is a minimum, so it can only be the infimum.

• Great mention about the infimum not having to be in the set; that is indeed a key point. Mar 14 '19 at 18:07

My answer is conceptual rather than an exact solution to your problem. I believe you will solve the problem after you understand the concepts.

Let's start with the definition of an upper bound. $M$ is an upper bound of the set $A$ if every element of $A$ is less than $M$. And the minimum of these $M$'s is called the supremum of $A$, denoted by $\sup A$. That is to say, if $M$ is an upper bound of $A$ and there is not an upper bound of $A$ smaller than $M$, then $\sup A = M$. And there is a fundamental property of real numbers called the least upper bound axiom which states the following:

A set $X$ has a supremum iff $X$ is non-empty and has an upper bound.

You can play with these definitions (use terms like lower bound instead of upper bound) to obtain the cases for the infimum.

And as others indicated, the difference between the supremum and the maximum is that maximum of a set must be contained by the set but supremum of a set may not be contained by the set. When the first case is satisfied, that is, if a set has a maximum, then the maximum is equal to the supremum as you can show easily using the definitions above. For example, if $S = [1,5]$ then $\sup S = \max S = 5$. But if $S = [1,5)$, then $\sup S = 5$ and $S$ has no maximum.

Therefore, one can easily show, for your example, that the set $S =\left\{ {x_n \:\:\: |\:\:\: n \ge 1} \right\}$ has a lower bound (eg. $0$ is a lower bound for the set $S$), and is non-empty ($1 \in S$). Therefore, by the greatest lower bound property (which you should have obtained by playing the definitions above) the set has an infimum.

Remark: For some sets by using induction or other simple methods it is easy to show that whether the set has a supremum or not. But try to show that the set consisting of the numbers $$\bigg(1+ \frac{1}{n}\bigg)^n$$ where $n\ge 1$ has a supremum. You will see that it is not that easy.

• Well-explained!! Best! Oct 2 '17 at 20:53

The difference between supremum and maximum is that for bounded, infinite sets, the maximum may not exist, but the supremum always does. Consider the set $$(0,1)$$. Does this set contain a largest element? The answer is no because for any $$x \in (0,1)$$, $$\tfrac{x+1}{2}$$ is also in $$(0,1)$$ and $$x < \tfrac{x+1}{2}$$. That is, for any $$x \in (0,1)$$, we can find an element in $$(0,1)$$ which is larger than $$x$$. Thus there is no maximal element of $$(0,1)$$. By contrast, the supremum is the least upper bound for the set. The supremum does not need to be in the set. For $$(0,1)$$, the supremum is 1. This means that $$1$$ is an upper bound for $$(0,1)$$ [which is obvious] and that no number smaller than $$1$$ is an upper bound for $$(0,1)$$ [which follows from the above reasoning]. Whenever the maximum exists, it is equal to the supremum. Conversely, if the supremum lies in the set, then the maximum exists and is equal to this supremum.

The difference between minimum and infimum is similar. In my example, $$(0,1)$$ has no minimum, but the infimum is 0.

For your set, we see the sequence goes $$1, \,\,\,\,\, 2/3, \,\,\,\,\, 3/5, \,\,\,\,\, 4/7, \ldots.$$ The sequence is clearly decreasing. The number 1 is in the set and no other member of the set is larger than 1. Hence 1 is a maximum for the set. Since the maximum exists, it is equal to the supremum, so the supremum is also 1. There is no minimum element since for any element $$x_n$$ of the set, the value $$x_{n+1}$$ is less than $$x_n$$. However, the infimum is 1/2. To see this, we note that all members are greater than 1/2, so 1/2 is a lower bound. To be the infimum, it needs to be the greatest lower bound, so you need to prove that $$1/2 + a$$ is not a lower bound for any $$a > 0$$. Indeed, if $$a > 0$$, taking $$n$$ large enough that $$1/(4n-2) < a$$ shows that $$x_n = \frac{n}{2n -1} = \frac{n-1/2 + 1/2}{2n-1} = \frac 1 2 + \frac 1{4n-2} < \frac 1 2 + a.$$ Thus we have found a member of the sequence which is smaller than $$1/2 + a$$ so $$1/2 + a$$ is not a lower bound for the sequence. Hence $$1/2$$ is the infimum.

• Suprema and infima do not necessarily exist, an example is the positive reals which has an infimum of 0 but no supremum Mar 30 '16 at 2:03
• Sure, sorry. I will edit. I meant to say they always exist for bounded sets. Mar 30 '16 at 2:38
• I think I understood what you mean with your example, just making sure :-) Mar 30 '16 at 2:59
• You were absolutely right; I just forgot to include the word "bounded." Thanks for the note. Mar 30 '16 at 3:11

Apparently, you know how to obtain the limits (here: upper bound 1, lower bound 1/2). That is any case is then the supremum (or infimum at the lower end) - or the set is unlimited, e.g. the naturals on top, the reals both ways: Nothing exists then.

Two cases:
(1) The limit is IN the set, e.g. actually attained by the sequence. Then it is maximum AND supremum (or minimum AND infimum), e.g. 1.
(2) The limit is OUTSIDE the set, not attained by the sequence (though only barely missed). That is your case at the lower end, 1/2. Then it is only supremum (or infimum) AND there is no maximum (minimum).

You might have some doubts coming in via the index n (your 3rd suggestion $$n\to\infty$$ and supremum)?
$$n\to\infty$$ has nothing to do with supremum ("large $$n$$"?). Only the $$x_n$$ themselves do.

Take-home bits:

1. No bound - no ..mum, nothing.
2. Minimum = Infimum OR no Minimum. (ditto max=sup or no max). In particular, these can not differ!


# Show discontinuous points of increasing function on $[0,1]$ is $F_\sigma$set

Problem: Let $f$ be an increasing function on $[0,1]$, Show discontinuous points of $f$ on $[0,1]$ is $F_\sigma$ set. But it is not always a $G_\delta$ set.

My thoughts: It is easy to prove that there exist a injective mapping from $A=\{x\in[0,1]|f\text{ is discontinuous at } x\}$ to $\mathbb{Q}$. So $A$ is most a countable set. Since $\mathbb{Q}$ is $F_\sigma$ set, so is subset(Is it right?). And similarly, since $\mathbb{Q}$ is not a $G_\delta$set, so the subset.

Yes, it's true that the set of discontinuities of $f$ is countable. And every countable subset of $\mathbb R$ is clearly an $F_\sigma.$ So we have the first claim.
The second claim is false. For a counterexample, take an $f$ with just one discontinuity. Now just note that a one point set is a $G_\delta.$
• can this discontinuous points set can always be $G_\delta$ set? I think, if we take $A=[0,1]\cap \mathbb{Q}$, then $A$ is not $G_\delta$ set but $F_\sigma$, am I right? – DuFong May 8 '16 at 19:18


Complement, open and closed sets

Definition: Let X be a metric space, and E $$\subseteq X$$, E is closed if it is equal to its closure.

Definition: A metric subset U of X is open if for every point in U there exists an open ball centered around the point that is contained within U.

Theorem: Let X be a metric space. A Subset $$Y \subseteq X$$ is closed iff $$Y^c$$ is open.

Proof:

Assume Y is closed so $$Y= \bar{Y}$$ (so it is equal to is closure). Which means that $$y\in Y \iff \forall$$ $$r>0 B(y,r) \cap Y \neq \emptyset$$.

Let $$a \in Y^c$$ be arbitrary. Since Y is closed $$\exists r>0 : B(a,r) \cap Y = \emptyset$$. This means that $$B(a,r) \subseteq Y^c$$. Since a was arbitrary, $$Y^c$$ is open.

Suppose $$Y^c$$ is open. So $$y\in Y^c$$ $$\iff$$ $$\exists r>0$$ : $$B(y,r) \subseteq Y^c$$. Let x $$\in \bar{Y}$$ be arbitrary. Note that it suffices to show that $$\bar{Y} \subseteq Y$$. So for any possible $$r>0$$ we have $$B(x,r) \cap Y \neq \emptyset$$. Since $$Y^c$$ is open, it follows that $$x \in Y$$.

Is the proof correct? I would very much love feedback, I'd be very very thankful.

• in general, it would be helpful to include your definitions of open and closed, because there are many equivalent ways of approaching a subject. In fact, the theorem you wrote is sometimes taken as the definition. Even though we might be able to hazard a guess as to which definition you're using based on what you've written, including the definitions can help us to provide better answers (so that we don't assume facts unknown to you etc) – peek-a-boo Jun 4 at 2:59
• @peek-a-boo I just did. Thanks for telling me. – topologicalmagician Jun 4 at 3:11
• A metric space is closed by definition. – copper.hat Jun 4 at 4:09
• Closed is usually defined as the complement of open, so you may want to specify what you mean by closed. – copper.hat Jun 4 at 4:11
• @copper.hat I did. A subspace E of X is closed if $E=\bar{E}$, where $\bar{E}$ is defined to be the closure, i.e. the set of all limit/adherent points. – topologicalmagician Jun 4 at 4:12

I'm not concinced by the last part of the $$Y^\complement$$ open implies $$Y$$ closed part. Why "Since $$Y^\complement$$ is open it follows that $$x \in Y$$" is unclear: but can this be expanded:

Assume $$x \notin Y$$ then $$x \in Y^\complement$$ and by openness there is some $$r>0$$ such that $$B(x,r) \subseteq Y^\complement$$ or equivalently, $$B(x,r) \cap Y=\emptyset$$, but that contradicts $$x \in \overline{Y}$$. So $$x \in Y$$ must hold.

• Suppose $Y^c$ is open. So $y\in Y$ $\iff$ $\forall r>0$ : $B(y,r) \cap Y \neq \emptyset$. Why isn't that enough? – topologicalmagician Jun 4 at 4:53
• Your A appeared while I was removing my usual infinite series of typos from mine. – DanielWainfleet Jun 4 at 4:53
• @topologicalmagician I see no (logical direct) link between the two statements $Y^\complement$ open and that equivalence. – Henno Brandsma Jun 4 at 5:55
• @Henno Brandsma, what about the contrapositive of $y\in Y^c$ iff $\exists r>0$ : $B(y,r) \subseteq Y^c$? – topologicalmagician Jun 4 at 13:36
• P iff Q is equivalent to not P iff not Q – topologicalmagician Jun 4 at 13:46

The last sentence is unclear. I think you should add a little, like: $$x\in \bar Y \implies \forall r>0\,(B(x,r)\cap Y\ne \emptyset) \implies$$ $$\implies (\neg (\exists r>0\, (B(x,r)\subset Y^c)\,))\implies$$ $$\implies (\neg (x\in Y^c))\; \text {...[because } Y^c \text { is open...] }\implies$$ $$\implies x \in Y.$$

• Suppose $Y^c$ is open. So $y\in Y$ $\iff$ $\forall r>0$ $B(y,r) \cap Y \neq \emptyset$. Isn't that enough? – topologicalmagician Jun 4 at 4:54
• I would very much appreciate your help – topologicalmagician Jun 4 at 5:03
• It basically is the contrapositive of "So $y\in Y^c \iff ….."$ – topologicalmagician Jun 4 at 5:08
• It's not that you made a mistake. It's just that it's not totally obvious how your last sentence follows from the 2nd last. – DanielWainfleet Jun 4 at 5:10
• but is using the contrapositive logically correct? Is what I used in fact the contrapositive? – topologicalmagician Jun 4 at 13:52


# Proving a set of numbers has arithmetic progressions of arbitrary length, but none infinite

For each real number $x$, let $[x]$ be the largest integer less than or equal to $x$. For example, $$[5] = 5$$

$$[7.9] = 7,$$ and $$[−2.4] = −3.$$

An arithmetic progression of length $k$ is a sequence $a_1, a_2, \dots, a_k$ with the property that there exists a real number $b$ such that $a_{i+1} − a_i = b$ for each $1 \leq i \leq k − 1.$

Let $\alpha > 2$ be a given irrational number. Let $S = \{[n · α] : n ∈ Z\}$, the set of all integers that are equal to $[n \alpha]$ for some integer $n$.

(a) Prove that for any integer $m \geq 3$, there exist m distinct numbers contained in $S$ which form an arithmetic progression of length $m$.

(b) Prove that there exist no infinite arithmetic progressions contained in $S$.

I'm thinking that for (a), some application pigeonhole principle can be considered... but otherwise, I don't really know how to prove this.

Some suggestions on a strategy would be awesome. Thanks.

• How familiar are you with the proof that the set of fractional parts $\{\{n\alpha\}\}_{n \in \mathbb N}$ is dense in the unit interval? – Erick Wong Aug 18 '14 at 3:15
• Not very familiar at all... – user164403 Aug 18 '14 at 14:14
• By the way, how does this relate to Number Theory? I know only very elementary Number Theory, so just wondering if there are some good books that cover these types of Number Theory questions. – user164403 Aug 18 '14 at 15:50

Outline: We want to construct an arithmetic sequence of length $n$ that is a subsequence of the sequence $(\lfloor i\alpha\rfloor)$.

For $i=1,2,3, \dots, 10n+1$, let $b_i$ be the fractional part of $i\alpha$, that is, $i\alpha-\lfloor i\alpha\rfloor$. These fractional parts are all distinct, since $\alpha$ is irrational. By the Pigeonhole Principle, there exist $i$ and $j$ with $i\lt j$ such that $|b_j-b_i|\lt \frac{1}{10n}$. Let $k=j-i$. Then either (i) $b_k\gt 1-\frac{1}{10n}$ or (i) $b_k\lt \frac{1}{10n}$.

We look first at case (i). Consider the numbers $k\alpha$, $2k\alpha$, $3k\alpha$ and so on up to $nk\alpha$. The fractional parts of these are slowly decreasing. Thus the numbers $\lfloor k\alpha\rfloor$, $\lfloor 2k\alpha\rfloor$, $\lfloor 3k\alpha\rfloor$, and so on up to $\lfloor nk\alpha\rfloor$ form an arithmetic progression of length $n$. There is a lot of slack, we can continue well beyond $\lfloor nk\alpha\rfloor$.

Case (ii) is essentially the same, except that the fractional parts of the $ik\alpha$ slowly increase instead of decreasing.

The argument that our sequence has no infinite arithmetic progression uses similar ideas.

• Would it be essential to apply the Pigeonhole Principle? I'm not used to using it, so I'm just wondering if there are other ways. Also, when do you know that a problem uses the Pigeonhole Principle? Is there anything that 'ticks', and you know that you have to use the Pigeonhole Principle? Are there any clues that show this in the question when not explicitly stated? – user164403 Sep 6 '14 at 4:19
• Well, I guess we do not have to mention the Pigeonhole Principle explicitly. But we will have to use it implicitly, since the $\alpha$ is general. It comes up whenever we are interested in the distribution of fractional parts. Here it was integer parts, but the technique was worth trying, and works. An indication that Pigeonhole might be useful is if the problem says that a set has more than $k$ elements, then a certain nice thing happens. The problem above, however, does not really have that shape. – André Nicolas Sep 6 '14 at 5:20
• What type of math is this? I'm very interested in it and would like to learn more by checking out resources pertaining to that aspect of math. – user164403 Sep 30 '14 at 3:42
• It belongs, more or less, to Diophantine Approximation. – André Nicolas Sep 30 '14 at 3:45
• Are there any good books on those kinds of questions? This question was from a Math Challenge, and this challenge stated that to prepare, one should go over such topics as sequences and series, which I thought was to be in real analysis. – user164403 Sep 30 '14 at 16:40


How can this function have two different antiderivatives?

I'm currently operating with the following integral:

$$\int\frac{u'(t)}{(1-u(t))^2} dt$$

But I notice that

$$\frac{d}{dt} \frac{u(t)}{1-u(t)} = \frac{u'(t)}{(1-u(t))^2}$$

and

$$\frac{d}{dt} \frac{1}{1-u(t)} = \frac{u'(t)}{(1-u(t))^2}$$

It seems that both solutions are possible, but that seems to contradict the uniqueness of Riemann's Integral.

So the questions are:

1. Which one of them is the correct integral?
2. If both are correct, why the solution is not unique?
3. The pole at $u(t)=1$ has something to say?
• @SanchayanDutta I have edited the title to fit the question a little bit better. If you have ideas for further improvements, go ahead and edit the post. – Martin Sleziak Sep 7 '15 at 20:13
• "but that seems to contradict the uniqueness of Riemann's Integral" whoever said antiderivatives were unique? – PyRulez Sep 8 '15 at 0:12
• – Dargor Sep 8 '15 at 0:14
• @PabloGalindoSalgado Notice that it says "unique up to a constant" – PyRulez Sep 8 '15 at 0:17
• @PyRulez Yup, yup. I thought that your question were refered to that result, not the constant. Sorry. – Dargor Sep 8 '15 at 0:19

It is not really a contradiction, since difference of the two functions is constant: $$\frac1{1-u(t)} - \frac{u(t)}{1-u(t)} = \frac{1-u(t)}{1-u(t)}=1.$$ (Derivative of a constant function is zero. Primitive function is determined uniquely up to a constant.)

I have not verified that the derivatives are correct, but notice that $$\frac1{1-u(t)}-\frac{u(t)}{1-u(t)}=1$$

that is, the difference between both antiderivatives is constant. This implies that the derivatives of both functions are the same.

$$\frac{d}{dt} \frac{u(t)}{1-u(t)}=\frac{d}{dt} \left(\frac{1}{1-u(t)}-1\right)=\frac{d}{dt} \frac{1}{1-u(t)}$$

Note that $1$ is just a constant so it vanishes. But when computing the antiderivative you will specify the constant according to initial conditions.

An equivalent way to see this, but from a slightly different perspective:

$$\frac{1}{1 - u(t)} = \frac{1 - u(t) + u(t)}{1 - u(t)} = 1 + \frac{u(t)}{1 - u(t)}$$

Perhaps this technique is worth seeing, namely, tinkering with an expression using additive inverses; specifically, by subtracting then adding back $u(t)$ to the numerator, we come to see that the two expressions differ only by a constant (i.e., $1$).

This technique occasionally presents itself when integrating; for example,

$$\int \frac{x}{x+1} dx = \int \frac{x+1-1}{x+1} dx = \int 1 - \frac{1}{x+1} dx$$

where the leftmost integrand is, in my estimation, a bit less friendly than the rightmost.

The other answers are excellent and clearly explain that indeed anti derivatives of a function may differ by a constant. I wanted to add that your "problem" can be introduced both in a "forward" and a "backward" fashion.

You gave a "backward" example, that is when two clearly not equal functions upon differentiation produce the same function. But working the other way around can also introduce the "problem", that is integration of the same function (and omitting the constant) may still produce a constant difference in the resulting functions. To illustrate what I mean consider the following tempting conclusion

$$\text{Let } f(x) = x, g(x) = x \text{ then } f(x) = g(x) \text{ and } \int f(x)dx = \int g(x)dx$$

We now apply the results of integration theory on both sides of the equation (which should result in applying the same operations on the symbols, right?) and decide not to add an extra constant to either of both. If we let the resulting expressions be denoted respectively by $F(x)$ and $G(x)$, we might be tempted to say that $F(x) = G(x) = \frac{1}{2}x^2$, but this conclusion is also not entirely true because the results depend on the operations applied. Moreover, the applied operations might be (unknowingly) forced upon us by the form of the equation!

To see this, consider the differential equation $y' = y$. Suppose I have the solution $z = 2e^x$, now we might be tempted to say

$$\frac{z'}{z} = 1 \rightarrow \int \frac{z'}{z}dx = \int dx \xrightarrow{\text{set both integration constants 0}} \log(z) = x \rightarrow z = e^x \rightarrow 2e^x = e^x$$

which is obviously false.

I have sometimes seen students struggling with this, so I thought I might as well add it. Hope it helps! :)

I will explain the concept using the derivative since you are already pretty familiar with that. Lets define to $\int \:f\left(x\right)dx$ to be some fucntion where $\frac{d}{dx}\left(g\left(x\right)\right)=f\left(x\right)$. (Note this is not the exact definition of an anti-derivative, but an intuitive way of thinking about it.)

So if $g\left(x\right)=x^2,\:\frac{d}{dx}\left(g\left(x\right)\right)=f\left(x\right)=2x$ And in reverse,$\int \:f\left(x\right)dx=x^2$

But what happens if $g\left(x\right)=x^2+5,\:\frac{d}{dx}\left(g\left(x\right)\right)=f\left(x\right)=2x$, Notice the 5 is a constant and disappears when taking the derivative. If we apply the reverse, we have no way of getting back the 5. In reverse you still get $\int \:f\left(x\right)dx=x^2$.

So now here comes the problem, when you are going in reverse you have no idea what the constant is. This is purely because the constant disappears when taking the derivative. For example $\frac{d}{dx}\left(x^2+8\right)=\frac{d}{dx}\left(x^2+5\right)=\frac{d}{dx}\left(x^2+1\right)$! So when finding the Anti-Derivative you will get a function plus or minus some constant.


# Math Help - Simple Integral Question (I think it's simple)

1. ## Simple Integral Question (I think it's simple)

I am supposed to know this by heart and I do:
http://www.wolframalpha.com/input/?i=integral+(1%2F(x^2+%2B+1))

However, on a specific problem I'm having, I need to know to get the following:
http://www.wolframalpha.com/input/?i=integral+1%2F(x^2+%2B+36)

My question is:
Is there a way for me to use what I have memorized to derive the second link? Like how can I make sense of the second link?

Any input would be greatly appreciated!

2. Originally Posted by s3a
I am supposed to know this by heart and I do:
http://www.wolframalpha.com/input/?i=integral+(1%2F(x^2+%2B+1))

However, on a specific problem I'm having, I need to know to get the following:
http://www.wolframalpha.com/input/?i=integral+1%2F(x^2+%2B+36)

My question is:
Is there a way for me to use what I have memorized to derive the second link? Like how can I make sense of the second link?

Any input would be greatly appreciated!
I'll give you a hint. See if you can figure it out yourself. Use the formula you memorized to make a general formula for the derivative of $tan^{-1}(\frac {x}{a})$

3. Eek, its better to just use LaTex (if you know how) and link the problem.

However this isn't a "simple" integral, unless you are familiar with trig substitutions. Lets do a little bit of math trickery:

$x = 6tan(\theta)$
$dx=6sec^{2}(\theta)d\theta$

Thus:

$\int \frac{6sec^{2}(\theta)d\theta}{(6tan{\theta})^{2}+ 36} \Rightarrow \int \frac{6sec^{2}(\theta)d\theta}{36(tan^{2}\theta+1) }$

From here we simplify:
$\int \frac{sec^{2}(\theta)d\theta}{6sec^{2}(\theta)} \Rightarrow \int \frac{1}{6}d\theta$

This integral is easy to evaluate:

$\int \frac{1}{6}d\theta \Rightarrow \frac{1}{6}\theta+C$

However, we need our final answer in terms of x, not theta. So, we go back to our original substitution:

$x = 6tan(\theta)$

. . .and solve for theta. Do not "memorize" formulas or "tricks" (this really isn't a trick though I called it so) - memorize (or KNOW) identities, and familiar algebraic structures.

4. I learned it like this:

$\int \frac{1}{x^2+36} \, dx$

$=~ \frac{1}{36} \int \frac{1}{\frac{x^2}{36}+1} \, dx$

$=~ \frac{1}{36} \int \frac{1}{\left(\frac{x}{6}\right)^2+1} \, dx$

(Then let $u=x/6 ~\implies~ du=dx/6$.)

$=~ \frac{1}{36} \int \frac{6}{u^2+1} \, du$

$=~ \frac{1}{6} \int \frac{1}{u^2+1} \, du$

You can probably complete it from there.

5. Originally Posted by ANDS!
Eek, its better to just use LaTex (if you know how) and link the problem.

However this isn't a "simple" integral, unless you are familiar with trig substitutions. Lets do a little bit of math trickery:

$x = 6tan(\theta)$
$dx=6sec^{2}(\theta)d\theta$

Thus:

$\int \frac{6sec^{2}(\theta)d\theta}{(6tan{\theta})^{2}+ 36} \Rightarrow \int \frac6{sec^{2}(\theta)d\theta}{36(tan^{2}\theta+1) }$

From here we simplify:
$\int \frac{sec^{2}(\theta)d\theta}{6sec^{2}(\theta)} \Rightarrow \int \frac{1}{6}d\theta$

This integral is easy to evaluate:

$\int \frac{1}{6}d\theta \Rightarrow \frac{1}{6}\theta+C$

However, we need our final answer in terms of x, not theta. So, we go back to our original substitution:

$x = 6tan(\theta)$

. . .and solve for theta. Do not "memorize" formulas or "tricks" (this really isn't a trick though I called it so) - memorize (or KNOW) identities, and familiar algebraic structures.
However, there is a MUCH simpler way to do this:

$\frac{d}{dx} tan^{-1}x = \frac {1}{x^2+1}$
$\frac{d}{dx} tan^{-1}(\frac {x}{a}) = \frac {1}{(\frac{x}{a})^2+1}.\frac{1}{a} = \frac {a}{x^2+a^2}
$

Notice how the question is in similar format:
$\int \frac{1}{x^2+36}dx = \frac{1}{6} \int \frac{6}{x^2+36}dx = \frac{1}{6} tan^{-1}\frac{x}{6}+C
$

Hope this helps, s3a

6. Originally Posted by drumist
I learned it like this:

$\int \frac{1}{x^2+36} \, dx$

$=~ \frac{1}{36} \int \frac{1}{\frac{x^2}{36}+1} \, dx$

$=~ \frac{1}{36} \int \frac{1}{\left(\frac{x}{6}\right)^2+1} \, dx$

(Then let $u=x/6 ~\implies~ du=dx/6$.)

$=~ \frac{1}{36} \int \frac{6}{u^2+1} \, du$

$=~ \frac{1}{6} \int \frac{1}{u^2+1} \, du$

You can probably complete it from there.
Sorry, I was typing when you posted; your way works well, too.

And s3u: if you're going to take a timed test like AP Calc AB, memorizing $\frac{d}{dx} tan^{-1}(\frac {x}{a}) = \frac {a}{x^2+a^2}$ could save a lot of time. Of course, you don't need to memorize it, I'm sure, because you saw where that came from, you could re-figure that out in a few seconds.

7. All of these methods are the same. Personally I like the method I used as it shows where it all comes from - but then it is all a matter of taste.

8. Originally Posted by ANDS!
All of these methods are the same. Personally I like the method I used as it shows where it all comes from - but then it is all a matter of taste.
I actually prefer the method you used, but the one I used is more useful in a test perspective, to save a lot of time. And, yes, they are all, in essence, the same method; that's the beautiful consistency of Math.


# posterior probability of bag given ball (evidence)

Question:

Given the distribution of the coloured balls in three different bags:
- Bag A: 1 Red 2 Black 2 Blue
- Bag B: 2 Red 4 Black 4 Blue
- Bag C: 10 Red 2 Black 3 Green

we carry out two independent experiments:
1) pick a bag, then pick a ball from the chosen bag.
2) pick a ball uniformly and random from all of the 30 balls.

We observe that the ball is red (for both experiment).
Now, we want to find the posterior probability of ball was taken from bag C given the ball chosen is Red for each of the experiment (i.e. P(Bag C | Red)).

My attempt:
For Experiment 1:
Using Bayes Rule we have:
P(Bag C | Red) = P(Bag C and Red) / P(Red)

From the distribution, we have P(BagC, Red) = 10/30
Now,P(Red)
= (by law of total probability) sum of conditional probability of
P(Red | Bag X) * P(Bag X)
= P(Red|BagC)*P(BagC) + P(Red|BagB)*P(BagB) + P(Red|BagA)*P(BagA)
= (1/3)(10/15) + (1/3)(2/10) + (1/3)(1/5)
= 16/45

And hence P(BagC|Red) = (10/30) / (16/45) = 15/16 ???

For Experiment 2:
this is the part where i am confuse, because intuitively, i dont not see the difference between both question! But i slept over it and think maybe the difference is in the way we calculate P(red), and thats why the following answer:

So using the same formula above, but with difference calculation of the probability of getting a red ball from the bag:

P(Red) = total number of red / total number of balls = 13/30
Hence, P(BagC|Red)
= P(Red, BagC) / P(Red)
= (10/30) / (13/30)
= 10/13

Intuitively, i thought my answer made sense because if we were to choose a bag first, the probability of the red ball coming from bag C will be larger since the proportion of red balls in bag C is significantly higher.
In comparison, if we pick randomly from the pool of 30 balls, the contribution of C into the pool is higher as well (10/13 red balls from C).

I am not entirely sure if my approach is correct here, and would wish that you can validate my answers.

EDIT:
Rectify the error pointed out in the comments

Your answer to the first question is correct. (ETA: Or was, before you edited $\frac{5}{8}$ to $\frac{15}{16}$. $<$g$>$)

Your answer to the second question is incorrect, because the $P(\text{red}, \text{Bag$C$}) = 10/30$, not $10/45$. (That is, out of the $30$ balls, each equally likely to be selected, exactly $10$ of them are both red and in Bag $C$.) You then obtain the correct answer of $10/13$.

• Hi Brian, good catch. However, if thats the case, then my first answer wouldnt be correct as well, isnt it? P(red, bagC) = 10/30, then for the first experiment, we should have P(bagC | Red) = (10/30) / (16/45) = 15/16 – rockstone Apr 21 '16 at 16:48
• @rockstone: No, it was right before. You should have $P(\text{Bag$C$} \mid \text{red}) = \frac{P(\text{red}, \text{Bag$C$})}{P(\text{red})} = \frac{2/9}{16/45} = \frac{5}{8}$. The difference is that the balls are not equally likely in the first problem; the bags are. – Brian Tung Apr 21 '16 at 17:39
• @brain, not exactly sure how you get 2/9. – rockstone Apr 21 '16 at 19:19
• @brain, after some thought, is $\frac{2}{9} = (\frac{10}{15})(\frac{1}{3}) =$ $P(red | BagC) P(C)$. Okay, then, my next question, if $P(A|B)P(B) = P(A, B)$, then for the second part, if we do use Bayes theorem, $P(BagC|red) = \frac{P(red|BagC) P(BagC)} {P(red)} = \frac{P(red, BagC)}{P(red)}$ why do we use $\frac{10}{30}$ instead of $\frac{10}{45}$? hope you can enlighten me on this issue. – rockstone Apr 21 '16 at 19:31
• @rockstone: Because all $30$ balls are equally likely, of which $10$ are both red and in Bag $C$. – Brian Tung Apr 21 '16 at 19:53

In the pool of all $30$ balls, label the balls by the label of the bag they came from. Then, there are $10$ red 'C' balls. So the second question asks for the probability of a ball being a red 'C' ball, given that it is a red ball. There are totally $13$ red balls, and $10$ of these are 'C', so the probability is $\dfrac{10}{13}$.

You can do this using Bayes' theorem too, but you will get the same answer.

• Yes, the second part should be 10/13! – rockstone Apr 21 '16 at 17:04


Combinatorics - pigeonhole principle question

This is for self-study. This question is from Rosen's "Discrete Mathematics And Its Applications", 6th edition.

An arm wrestler is the champion for a period of 75 hours. (Here, by an hour, we mean a period starting from an exact hour, such as 1 P.M., until the next hour.) The arm wrestler had at least one match an hour, but no more than 125 total matches.

1 - Show that there is a period of consecutive hours during which the arm wrestler had exactly 24 matches.

2 - Is the statement in the previous exercise true if 24 is replaced by

a) 2? b) 23? c) 25? d) 30?

My solution to part 1 is the following, based on Rosen's solution to a similar problem given as example in the text:

1 - If I consider $a_i$ to be the number of competitions until the $i^{th}$ hour, then, $1\leq a_1<a_2<\cdots<a_{75}\leq 125$, because there are no more than 75 hours and the total number of competitions is no more than 125. Now I will add 24 to all the terms of the above inequality: $25\leq a_1+24<a_2+24<\cdots<a_{75}+24\leq 149$.

There are 150 numbers $a_1,\cdots,a_{75},a_1+24,\cdots,a_{75}+24$. By the inequalities above, these numbers range from 1 to 149. Then, by the pigeonhole principle, at least two of them are equal (in a list of 150 integers ranging from 1 to 149, at least two are equal).

Because all the numbers $a_1,\cdots,a_{75}$ are distinct, and all numbers $a_1+24,\cdots,a_{75}+24$ are also distinct, it follows that $a_i = a_j + 24$ for some $i > j$. Therefore, from the $(j+1)^{th}$ hour to the $i^{th}$ hour, there were exactly 24 competitions.

Now, I will show the attempt at a solution for part 2:

2 - a) The same reasoning as above can be used:

$1\leq a_1<a_2<\cdots<a_{75}\leq 125$. Adding 2 to all terms:

$3\leq a_1 + 2<a_2 + 2<\cdots<a_{75} + 2\leq 127$

So, there are 150 numbers that range from 1 to 127. Therefore, by the pigeonhole principle, at least $\left \lceil \frac{150}{127} \right \rceil$ = 2 numbers must be equal. So, there is an $a_i = a_j + 2$. This guarantees that there is a period of consecutive hours during which there were exactly 2 competitions.

b) The same reasoning as above can be used:

$1\leq a_1<a_2<\cdots<a_{75}\leq 125$. Adding 23 to all terms:

$24\leq a_1 + 23<a_2 + 23<\cdots<a_{75} + 23\leq 148$

So, there are 150 numbers that range from 1 to 148. Therefore, by the pigeonhole principle, at least $\left \lceil \frac{150}{148} \right \rceil$ = 2 numbers must be equal. So, there is an $a_i = a_j + 23$. This guarantees that there is a period of consecutive hours during which there were exactly 23 competitions.

c) $1\leq a_1<a_2<\cdots<a_{75}\leq 125$. Adding 25 to all terms:

$26\leq a_1 + 25<a_2 + 25<\cdots<a_{75} + 25\leq 150$

In this case, there are 150 numbers that range from 1 to 150. So, there are not necessarily two equal numbers. Therefore, we can't conclude anything directly.

But it is possible to show that the statement is not true for 25, because an explicit counter-example (suggested below in the comments) can be found. Suppose the number of matches until each one of the 75 hours is, respectively: $\{1,2,\cdots,25,51,\cdots,75,101,\cdots,125\}$. Here, there are no pairs of numbers whose difference is 25.

d) $1\leq a_1<a_2<\cdots<a_{75}\leq 125$. Adding 30 to all terms:

$31\leq a_1 + 30<a_2 + 30<\cdots<a_{75} + 30\leq 155$

In this case, there are 150 numbers that range from 1 to 155. So, we can't apply the pigeonhole principle here, similarly to the above situation. It is not easy to find a counter-example in this case; I think that, for 30, the statement is always true (that is, there is always a period of consecutive hours during which there were exactly 30 matches). But I'm not sure how to prove it.

Edit: I think I found a way to prove it, based on the suggestion given by Lopsy; I've included it as an answer to this question.

Thank you in advance.

-
Just because the pigeonhole principle doesn't apply immediately doesn't mean that it isn't true - you'd need to come up with an explicit counter-example. – Thomas Andrews Jan 8 '12 at 16:07
In the case of $25$, you can come up with a pretty easy counter-example - $\{a_i\}=\{1,2,...,25,51,...,75,101,...125\}$. Can you come up with a counter-example for $30$? – Thomas Andrews Jan 8 '12 at 16:15
@ThomasAndrews: Actually it seems that it's not possible to come up with a counter-example for 30. If I use the same reasoning you used for 25: $\{1,\cdots,30,61,\cdots,90,121,\cdots,125\}$. In this list there are only 65 hours. But there are 75 hours and at least 1 match per hour. If I added more numbers smaller than 125 to this list, I would get 2 numbers whose difference is 30. It seems to mean that the statement is valid for 30. – anonymous Jan 8 '12 at 17:54
Wow, that's a whole lot of arm wrestling. – Nate Eldredge Jan 8 '12 at 20:32
Hint: there can only be three $a_i$ in the set {1, 31, 61, 91, 121}, right? Okay, now keep using this idea... you should be able to finish the problem using this. – Lopsy Jan 9 '12 at 2:15

I'm posting this answer to my own question because I think I found a solution, but I would appreciate feedback on whether the solution below is complete.

The only detail that is missing to complete part d is that I need to show that the statement is true if 24 in the original statement is replaced by 30; that is, I will try to show that there is a period of consecutive hours during which the arm wrestler had exactly 30 matches (since there seems to be no counter-example for this case).

I think I found a way to prove it, based on the suggestion given by Lopsy in the comments to the question. This attempt is very similar to the accepted answer here: Prove that 2 students live exactly five houses apart if.

We build the following sets: the 5-element sets $\{1, 31, 61, 91, 121\}$, $\{2, 32, 62, 92, 122\}$, ..., $\{5, 35, 65, 95, 125\}$, and the 4-element sets $\{6, 36, 66, 96\}$, $\{7, 37, 67, 97\}$, ..., $\{30, 60, 90, 120\}$. The purpose of these sets is to represent every possibility of number of matches played so far. They encompass all numbers from 1 to 125 and, inside each set, the difference between adjacent numbers is 30; however, there are no numbers from two different sets whose difference is 30. There are 5 five-element sets, and 25 four-element sets.

Now, if I want to choose 75 numbers from all sets so that there are no two numbers whose difference is 30, I have to choose at most 3 elements from the sets with 5 elements (for example, I can choose 1, 61 and 121 from the set $\{1, 31, 61, 91, 121\}$; if I choose one more, then clearly two of them will have difference equal to 30), and at most 2 elements from the sets with 4 elements (for example, I can choose 30 and 120 from $\{30, 60, 90, 120\}$, but, if I choose one more, then clearly two of them will have difference 30). There are 30 sets in total; since there are 5 sets with 5 elements and 25 sets with 4 elements, the maximum number of elements that can be chosen so that no two of them have 30 as difference is $3\times 5 + 2\times 25 = 15 + 50 = 65$ numbers. But I need to choose 75 numbers. So, there must be some numbers whose difference is 30.

Does this make sense, or is there some mistake?

-
Looks fine to me. – Peter Taylor Feb 5 '12 at 13:29


# Mean of the residuals in Simple Linear Regression *Beginner Alert*

##### New Member
Hey guys,

First of all, I would like to apologize for the question I'm gonna ask regarding the fact, that I am absolute beginner in statistics.

Currently I'm trying to test some hypotheses in Rstudio using Simple Linear Regression, as I am aware that before interpreting results of my model I have to provide verification of assumptions that should be meet within the model.

Unfortunately, one of the key assumption is not very clear for me, could someone explain please?

"The mean of the residuals should be equal to zero."

I have no clue which of the residuals I should sum and then divide by number of items. I tried every possibility but it's never zero (I tried it also within other models). So basically my conclusion is that I'm doing something wrong.

I tried to plot residuals and it seems fine for me.

I would be very thankful for any advice or explanation.

Thank you very much.

#### Dason

If you take the mean of the residuals for the observations that were used in the model you should get 0. Here is an example using R since that's what you're using.

Code:
> head(mtcars)
mpg cyl disp  hp drat    wt  qsec vs am gear carb
Mazda RX4         21.0   6  160 110 3.90 2.620 16.46  0  1    4    4
Mazda RX4 Wag     21.0   6  160 110 3.90 2.875 17.02  0  1    4    4
Datsun 710        22.8   4  108  93 3.85 2.320 18.61  1  1    4    1
Hornet 4 Drive    21.4   6  258 110 3.08 3.215 19.44  1  0    3    1
Hornet Sportabout 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2
Valiant           18.1   6  225 105 2.76 3.460 20.22  1  0    3    1
> o <- lm(mpg ~ wt, data = mtcars)

resid         residuals     residuals.glm residuals.lm  resizeImage
Mazda RX4     Mazda RX4 Wag        Datsun 710    Hornet 4 Drive
-2.2826106        -0.9197704        -2.0859521         1.2973499
-0.2001440        -0.6932545
> mean(residuals(o))
[1] 2.024748e-16
Note that the final line might look like it's non-zero but... that's zero. When using a computer there is rounding error and the value being displayed as the mean of the residuals is... 0.0000000000000002024748. Which is close to the smallest allowable value when working with floating point data...

##### New Member
If you take the mean of the residuals for the observations that were used in the model you should get 0. Here is an example using R since that's what you're using.

Code:
> head(mtcars)
mpg cyl disp  hp drat    wt  qsec vs am gear carb
Mazda RX4         21.0   6  160 110 3.90 2.620 16.46  0  1    4    4
Mazda RX4 Wag     21.0   6  160 110 3.90 2.875 17.02  0  1    4    4
Datsun 710        22.8   4  108  93 3.85 2.320 18.61  1  1    4    1
Hornet 4 Drive    21.4   6  258 110 3.08 3.215 19.44  1  0    3    1
Hornet Sportabout 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2
Valiant           18.1   6  225 105 2.76 3.460 20.22  1  0    3    1
> o <- lm(mpg ~ wt, data = mtcars)

resid         residuals     residuals.glm residuals.lm  resizeImage
Mazda RX4     Mazda RX4 Wag        Datsun 710    Hornet 4 Drive
-2.2826106        -0.9197704        -2.0859521         1.2973499
-0.2001440        -0.6932545
> mean(residuals(o))
[1] 2.024748e-16
Note that the final line might look like it's non-zero but... that's zero. When using a computer there is rounding error and the value being displayed as the mean of the residuals is... 0.0000000000000002024748. Which is close to the smallest allowable value when working with floating point data...
yes! You're right!

Thank you very much!

#### Dason

Side note: there is never a need to apologize for being a beginner. We all start somewhere. Please let us know if you have other questions!

#### noetsi

##### No cake for spunky
Except dason. He was born (if that is the right word for a bot) with a perfect knowledge of statistics...

I have spent 15 years studying regression and this is the first time I have heard that one raised. Usually you look at the residuals to see if you have heteroskedacity, non-normality, or non-linearity.

#### ondansetron

##### TS Contributor
The nonlinearity could imply the mean error is not 0.

True model: Y= B0 +B1X + B2X^2
Fit: yhat= b0 + b1x

then Y-yhat = residual
implies
E[(B0 +B1X + B2X^2)-(b0 + b1x)] = E[B2X^2] not equal to zero for B2 and X not zero... so the average error would be B2X^2... this could show as curvature in a residual plot of residuals vs x, for example.

#### Dason

True. But that doesn't matter because the sample mean of the residuals will be 0 regardless.

#### ondansetron

##### TS Contributor
True. But that doesn't matter because the sample mean of the residuals will be 0 regardless.
Cause that's how the line of best fit is defined (partially), yaknow?

#### ondansetron

##### TS Contributor
I think that is his point
And mine too, but generally the idea of plotting residuals vs x is to help see if you're misspecified!

#### hlsmith

##### Less is more. Stay pure. Stay poor.
Yeah, i would have been content with the outputted residual median being E-07. Good points though.

#### noetsi

##### No cake for spunky
And mine too, but generally the idea of plotting residuals vs x is to help see if you're misspecified!
That is true, but different than the original posters question. In fact since the mean of the residuals will always be zero you don't need to check for that - it has to be true. Unless you have a software issue


# Probability question about “good” tickets.

The problem:

Out of 20 exam tickets, 16 tickets are "good". Tickets are carefully mixed, and students take turns pulling one ticket. Who has the better chance to draw a "good" ticket the first or the second student in the queue?

My attempt:

Obviously the probability of the first student getting a "good" ticket is $16/20 = 4/5 = 0.8$

Now here's where I'm confused. I considered two cases. If the first student got a "good" ticket then the chances of the second student are $15/19 = 0.789$, so lower than the first student.

However if the first student didn't get a "good" ticket, the chances of the second student are $16/19 =0.842$, so slightly better chances than the first one.

So is the answer "depends on whether the first student gets a "good" ticket"?

• The question is to "predict" the propability of the 2nd student getting a good card not knowing what the first person draws. – Cornman Sep 21 '17 at 20:32
• Hint. Is there any of the 20 tickets that the second student has a different chance of getting from any other? – Henning Makholm Sep 21 '17 at 20:34
• So I have to count the probability of the second student having better chances? That would be equal to $4/20 * 16/19 = 0.168$. But I'm kinda lost after that. – Nick202 Sep 21 '17 at 20:34
• Draw a "probability tree". :) – Cornman Sep 21 '17 at 20:36

You're almost there. You just need to apply conditional probability rules.

Essentially, the unconditional probability that the second student gets a good ticket is equal to a weighted average of the conditional probabilities $\frac{15}{19}$ and $\frac{16}{19}$, where the weights are the probabilities that the first student did or did not get a good ticket, respectively. If you carry out that computation, you get

$$\frac{4}{5} \times \frac{15}{19} + \frac{1}{5} \times \frac{16}{19} = \frac{76}{95} = \frac{4}{5}$$

So the two students have equal probabilities of getting a good ticket.

The same result can be obtained by observing that if you lay the tickets out "face down," so to speak, the only thing distinguishing the first and second tickets is the arbitrary order you laid them out in. Since the first two tickets are already destined for the first two students, they must have the same probability of being good, namely $\frac45$. In fact, the same logic applies to all the students.

• Mild edit after a small math-o. – Brian Tung Sep 21 '17 at 20:37
• I see. I calculated ($4/20 * 16/19$) to find out the probability of the first student getting a bad ticked and second one getting a good ticket. Then I guess I wanted to calculate the opposite(first one getting good and second one getting bad) to see which one would have higher probability. I found out that they are equal. Would that be considered a correct solution? – Nick202 Sep 21 '17 at 20:44
• If I understand you correctly, that would work, but it seems pretty indirect to me. The most straightforward solution to me is the one by symmetry. – Brian Tung Sep 21 '17 at 21:20

If you don't look at the first student's ticket, you should do a weighted average of the two possibilities and you will discover that the second student's chance of getting a good ticket is also $0.8$. The fact that one was already drawn does not matter at all. How can the chance be different from if they each drew, then swapped tickets before looking at them?

Let $A_1$ and $A_2$ denote the Bernoulli random variables of student 1 and student 2 picking a good ticket, respectively. So for example, the probababilty that student one picks a good ticket is $P(A_1 = G)$.

First, lets calculate the probability $P(A_1 = G)$. This is a simple case of the hypergeometric distribution, since the good cards form a "sub-population" within the total "population" of cards:

$$P(A_1 = G) = \frac{{16\choose{1}}{4\choose{0}}}{20\choose{1}} = \frac{16}{20} = .8$$

And thus the probability that the first student chooses a bad card is: $P(A_1 = B) = .2$.

Now for second student: Use the law of total probability.

$$P(A_2 = G) = P(A_2 = G | A_1 = G)P(A_1 = G) + P(A_2 = G | A_1 = B)P(A_1 = B)$$

The conditional distribution of $A_2$ given $A_2$ is also hypergeometric.

$$= \frac{{15\choose{1}}{4\choose{0}}}{19\choose{1}}*.8 + \frac{{16\choose{1}}{3\choose{0}}}{19\choose{1}}*.2$$ $$= \frac{15}{19}*.8 + \frac{16}{19}*.2 = .8$$

A-ha! the probability is the same. Go figures.


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