text stringlengths 1 2.12k | source dict |
|---|---|
equations where we had repeated eigenvalues. (2018) A new method for computation of eigenvector derivatives with distinct and repeated eigenvalues in structural dynamic analysis. Spectral Decomposition with Repeated Eigenvalues (Quantum Theory for Computer Age - Ch. But what if A has repeated real eigenvalue (a, a), or if the eigenvalues are complex conjugate pairs (a+bi, a-bi) with nonzero b? How does one decompose A=aI+bJ where J^2=-I in the latter case, or decompose A=aI+N where N^2=0 in the former?. If all eigenvalues $$\lambda$$ have negative real parts, then all solutions of approach zero exponentially. j is repeated. We restrict ourselves to the special cases of A being 2 × 2 and 3 × 3. 1 Matrix exponent Consider a first order differential equation of the form y′ = ay; a ∈ R; with the initial condition y(0) = y0: Of course, we know that the solution to this IVP is given by y(t) = eaty0: However, let us apply the method of iterations to this equation. So lambda is an eigenvalue of A. We prove that the volume of n satis es: j nj jRPN 3j = n 2 ; where N = n+1 2 is the dimension of the space of real symmetric matrices of size n n. But here only (1,0) is a eigenvector to 0. 118 CHAPTER 6. 5 Repeated Eigenvalues 95 5. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. What does this mean geometrically?. 2000 S Deep Cameo Clad Proof Massachusetts MA State Washington Quarter (B04),Girls Greek / Roman Goddess Fancy Dress Costume 4-11 Years Available,1945 S Silver Jefferson Nickel GEM BU BLAST WHITE!!. We will ignore the possibility of , as that would mean 0 is an eigenvalue. Mathematics Assignment Help, Example of repeated eigenvalues, Illustration : Solve the following IVP. Computing the pth roots of a matrix with repeated eigenvalues 2651 It can be seen that for a special case, if the given matrix A is a companion matrix then the constituent matrices corresponding to companion matrix are determined by. | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
companion matrix then the constituent matrices corresponding to companion matrix are determined by. 1 of A is repeated if it is a multiple root of the char acteristic equation of A; in our case, as this is a quadratic equation, the only possible case is when λ. Eigenvalues and Eigenvectors. 2 Harmonic Oscillators 114 6. hat May 14 '12 at 0:21 3 $\begingroup$ To put the same thing into slightly different words: what you have here is a two-dimensional eigenspace , and any two vectors that form a basis for that space will do as linearly. Chapter 6 Eigenvalues and Eigenvectors Po-Ning Chen, Professor Department of Electrical and Computer Engineering National Chiao Tung University Hsin Chu, Taiwan 30010, R. j is repeated. Eigenvalues in MATLAB. 2 6 6 6 4. 2 Solving Systems with Repeated Eigenvalues If the characteristic equation has only a single repeated root, there is a single eigenvalue. , orthogonal to the disk and passing through its center), while any two orthogonal diameters in the plane of the disk may be chosen as the other two principal axes (corresponding to the repeated eigenvalue ). Repeated eigenvalues. J has the eigenvalues of A on its main diagonal, is upper triangular, and has 0's and 1's in the upper triangle. 3 COMPLEX AND REPEATED EIGENVALUES 15 A. If there is no other eigenvector, we look for a solutions in the form x(t) = (u+ tv)e t: If x(t) has this form, then, on the one hand x_(t) = ve t+ (u+ tv)e t;. Introduction. $\endgroup$ - copper. So it is with matrices. 1 Find the eigenvalues and associated eigenspaces of each of the following matrices. eigenvalues and convert them a Pillai-Bartlett or Wilk's-Lambda value, I don?t know how to convert to an f-statistic. 1 Distinct Eigenvalues 107 6. Firstly we look at matrices where one or more of the eigenvalues is repeated. (2018) A new method for computation of eigenvector derivatives with distinct and repeated eigenvalues in structural dynamic analysis. Take for example 0 @ 3 1 2 3 1 6 2 2 2 1 A One | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
repeated eigenvalues in structural dynamic analysis. Take for example 0 @ 3 1 2 3 1 6 2 2 2 1 A One can verify that the eigenvalues of this matrix are = 2;2; 4. And I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier. Linear Systems of ODE with with Repeated eigenvalues James K. 4 Bases and Subspaces 89 5. If they are numeric, eigenvalues are sorted in order of decreasing absolute value. Repeated eigenvalues are explained with the help of numerical examples. Solution: If ‚ is a simple real eigenvalue, then there are two real unit eigenvectors: u and ¡u. Prove that if A is a square matrix then A and AT have the same characteristic polynomial. To find an eigenvector corresponding to an eigenvalue , we write. and w is an eigenvector for A, and suppose that the eigenvalues are di erent. λ = a ± ib. Of particular interest in many settings (of which differential equations is one) is the following. Then from the Lemma we get 2v w =(ATv)w=v (Aw)=v 5w=5vw: But since 2 and 5 are scalars (and, for that matter, so is v w), the only way that 2v w =5v wcan be possible is for v w to be 0. The calculator will find the eigenvalues and eigenvectors of the given square matrix, with steps shown. Further options. In order to get the eigenvalues of the matrix , I'll solve the characteristic equation Step 3. Stability Analysis for ODEs Marc R. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. 2 λhas a single eigenvector Kassociated to it. Since the geometric multiplicity j for j is the dimension of E j, there will be exactly j vectors in this basis. It decomposes matrix using LU and Cholesky decomposition The calculator will perform symbolic calculations whenever it is possible. View record in Web of Science ®. edu/math Craigfaulhaber. complex eigenvalues. The last case is what the solutions look like when | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
®. edu/math Craigfaulhaber. complex eigenvalues. The last case is what the solutions look like when there are repeated eigenvalues, or. To find eigenvalues of matrix A Consider {eq}\displaystyle det(A-{\lambda}(I))=0 {/eq} Then we get a characteristic polynomial in lambda. Eigenvectors and Eigenvalues When a random matrix A acts as a scalar multiplier on a vector X, then that vector is called an eigenvector of X. However, if a matrix has repeated eigenvalues, it is not similar to a diagonal matrix unless it has a full (independent) set of eigenvectors. Engineering Computation ECL4-4. sy ' Section 7. For repeated diagonal elements, it might not tell you much about the location of the eigenvalues. They are often introduced in an introductory linear algebra class, and when introduced there alone, it is hard to appreciate their importance. The number of positive eigenvalues equals the number of positive pivots. Ryan Spring 2012 1 Repeated Eigenvalues Last Time: We studied phase portraits and systems of differential equations with complex eigen-values. Phase portrait for repeated eigenvalues Subsection 3. Defective matrices A are similar to the Jordan (canonical) form A = XJX−1. Proof: If A is idempotent, λ is an eigenvalue and v a corresponding eigenvector then λv = Av = AAv = λAv = λ2v Since v 6= 0 we find λ−λ2 = λ(1 −λ) = 0 so either λ = 0 or λ = 1. @Star Strider: Thanks for the suggestion, I was unaware of this function. In order to get the eigenvalues of the matrix , I'll solve the characteristic equation Step 3. eigenvalue will be printed as many times as its multiplicity. In this section we will solve systems of two linear differential equations in which the eigenvalues are real repeated (double in this case) numbers. This paper discusses characteristic features and inherent difficulties pertaining to the lack of usual differentiability properties in problems of sensitivity analysis and optimum structural design with respect to multiple eigenvalues. The general | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
sensitivity analysis and optimum structural design with respect to multiple eigenvalues. The general solution is Y~(t) = e3t C 1 C 1 + C 2 + tC 2 Sketch the phase portrait: 2. 2 Repeated Eigenvalues. This process can be repeated until all eigenvalues are found. On the other hand, when it comes to a repeated eigenvalue, this function is not di erentiable, which hinders statistical inference, as the asymptotic theory requires at least second-order di erentiability. Recall that the general solution in this case has the form where is the double eigenvalue and is the associated eigenvector. S ections 2. Repeat steps 2 through 4 for each distinct eigenvalue. Proposition 0. We can’t find it by elimination. The physical significance of this degeneracy is not known. Some may be repeated, some may be complex. They are often introduced in an introductory linear algebra class, and when introduced there alone, it is hard to appreciate their importance. As shown byL¨utkepohl (2005) andHamilton(1994), the VAR is stable if the modulus of each eigenvalue of A is strictly less than 1. For A2 the situation is difierent. However, the fundamental issue is selecting the appropriate tolerance to determine whether two eigenvalues are the same or not, which I don't know a priori (the elements of the matrix I am considering vary by 7 orders of magnitude, so its not obvious how close is close enough). We shall see that this sometimes (but not always) causes problems in the diagonalization process. The Exponential of a Matrix. Find more Mathematics widgets in Wolfram|Alpha. For an matrix, the polynomial we get by computing is of degree , and hence in general, we have eigenvalues. (solution: x = 1 or x = 5. Such an x is called an eigenvector corresponding to the eigenvalue λ. The eigenvalue λ i is called repeated iff r i > 1. Once we show this is necessarily real, then the same argument as in the part (a) shows that A. In particular, undamped vibration is governed by. Boyce and Richard C. The same | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
(a) shows that A. In particular, undamped vibration is governed by. Boyce and Richard C. The same situation applies, if Ais semi-simple, with repeated eigenvalues. Next, perform row operations by adding each row (2 through n) to the first row (Williams): (7) For clarity, the (-1) is repeated in each of the row 1 elements, but this simply. Get the free "Eigenvalue and Eigenvector (2x2)" widget for your website, blog, Wordpress, Blogger, or iGoogle. Rajsekaran. You would perform a one-way repeated measures analysis of variance if you had one categorical independent variable and a normally distributed interval dependent variable that was repeated at least twice for each subject. University of Minnesota 109. Are the eigenvalues real? Find one complex eigenvector (contains an i). Since x ≠ 0, this equation implies λ = 1(Eigenvalue); then, from x = 1 x, every (nonzero) vector is an eigenvector of I. Since the geometric multiplicity j for j is the dimension of E j, there will be exactly j vectors in this basis. The following. Roussel September 13, 2005 1 Linear stability analysis Equilibria are not always stable. Moreover, we provide a numerically reliable and effective algorithm for computing the eigenvalue decomposition of a symmetric matrix with two numerically distinct eigenvalues. Using the Laplace transformation, Eq. The solutions of the system can be found by finding the eigenvalues and eigenvectors of the matrix. The spectral decomposition of x is returned as components of a list with components. In general, the algebraic multiplicity and geometric multiplicity of an eigenvalue can differ. (e) A= 1 1 2 3. 2 (Page 249) 17. Ryan Spring 2012 1 Repeated Eigenvalues Last Time: We studied phase portraits and systems of differential equations with complex eigen-values. However, ker(B I 2) = ker 0 2 0 0 = span( 1 0 ): Motivated by this example, de ne the geometric multiplicity of an eigenvalue. Step 2 For each eigenvalue , compute an orthonormal basis for Ker(A Id). Make a | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
of an eigenvalue. Step 2 For each eigenvalue , compute an orthonormal basis for Ker(A Id). Make a matrix Q as follows. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. If this is the situation, then we actually have two separate cases to examine, depending on whether or not we can find two linearly independent eigenvectors. For each distinct eigenvalue l j with multiplicity m j, let I j denote the diagonal matrix with unit elements in each location where l j occurs, and let D j denote the differentiation matrix comprised of integers counting the repeated eigenvalues, placed to the left of the diagonal locations of the respective eigenvalues. The linearization has a node (proper or improper) at the origin, but the original almost linear system has either a node or a spiral point at P. In that example, one principal axis, the one corresponding to eigenvalue , was (i. ' and find homework help for other Math questions at eNotes. When all eigenvalues have non-zero real parts, the equilibrium is called hyperbolic, and non-hyperbolic if at least one eigenvalue has zero real part. Recall the basic result that the roots of a polynomial depend continuously on the coefficients of the polynomial. An eigenvalue of Ais a number such that Av = v for some nonzero vector v. Precondition The eigenvalues have been computed before. However, when I run it with a non-symmetric matrix, the largest eigenvalue is in the first column. • Roy’s Largest Root = largest eigenvalue o Gives an upper-bound of the F-statistic. Is it possiable to create a matrix be setting the eigenvalues you wish to end up with? I am trying to create a 3X3 random matrix with a set of repeated eigenvalues and was wondering is it's possible to set the eigenvalues and then generate matrices for them. The dimension of the eigenspace corresponding to an eigenvalue is less than or equal to the | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
them. The dimension of the eigenspace corresponding to an eigenvalue is less than or equal to the multiplicity of that eigenvalue. Note that is the product of the eigenvalues (since ), so for the sign of determines whether the eigenvalues have the same sign or opposite sign. In fact, it is easy to see that this happen if and only if we have more than one equilibrium point (which is (0,0)). Proof: If A is idempotent, λ is an eigenvalue and v a corresponding eigenvector then λv = Av = AAv = λAv = λ2v Since v 6= 0 we find λ−λ2 = λ(1 −λ) = 0 so either λ = 0 or λ = 1. Of particular interest in many settings (of which differential equations is one) is the following. Solution Step 1. Solution: Y(t) = c1 1 0 e 2t + c 2 0 1 e 2t = e 2t c1 c2 Qualitative behavior: 1. sy ' Section 7. component is being recorded, and then "removed". Proposition 0. Show that A and AT do not have the. For the purpose of analyzing Hessians, the eigenvectors are not important, but the eigenvalues are. The eigenvalue problem is to determine the solution to the equation Av = λv, where A is an n-by-n matrix, v is a column vector of length n, and λ is a scalar. Under this matrix norm, the infinite series converges for all A and for all t, and it defines the matrix exponential. Eigenvalues are a special set of scalars associated with a linear system of equations (i. The linearization has a node (proper or improper) at the origin, but the original almost linear system has either a node or a spiral point at P. We restrict ourselves to the special cases of A being 2 × 2 and 3 × 3. We do not normally divide matrices (though sometimes we can multiply by an inverse). Get the free "Eigenvalues Calculator 3x3" widget for your website, blog, Wordpress, Blogger, or iGoogle. Assume λ is a repeated eigenvalue of A with multiplicity m. Facts About Eigenvalues By Dr David Butler De nitions Suppose Ais an n nmatrix. Right when you reach $0$, the eigenvalues and eigenvectors become real (although there is only eigenvector | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
when you reach $0$, the eigenvalues and eigenvectors become real (although there is only eigenvector at this point). Free Matrix Eigenvalues calculator - calculate matrix eigenvalues step-by-step. Since we are going to be working with systems in which A is a 2 x 2 matrix we will make that assumption from the start. Next I simplify it. "Det" stands for the determinant, and "I" is the identity matrix. 1 Fundamental Matrices, Matrix Exp & Repeated Eigenvalues – Sections 7. We can certainly have repeated roots and complex eigenvalues. Are the eigenvalues real? Find one complex eigenvector (contains an i). A new method is presented for computation of eigenvalue and eigenvector derivatives associated with repeated eigenvalues of the generalized nondefective eigenproblem. Multiplying them gives 2 4 4 3 3 2 3 2 1 0 2 3 5 2 4 3 2 1 3 5= 2 4 15 10 5 3 5= 5 2 4 3 2 1 3 5: This shows that the vector is an eigenvector for the eigenvalue 5. If the 2 2 matrix Ahas distinct real eigenvalues 1 and 2, with corresponding eigenvectors ~v 1 and ~v 2, then the system x~0(t)=A~x(t). But we did not discuss the case when one of the eigenvalues is zero. 3 power method for approximating eigenvalues 551 Note that the approximations in Example 2 appear to be approaching scalar multiples of which we know from Example 1 is a dominant eigenvector of the matrix. edu (UC Davis) >4 ICIAM 11 1 / 32. Eigenvectors and Eigenvalues When a random matrix A acts as a scalar multiplier on a vector X, then that vector is called an eigenvector of X. When n = 2, in 1955 and 1956, Payne, P´olya and Weinberger proved that, in [10] and [11], λ 2 λ 1 ≤ 3forD ⊂ R2, and they conjectured λ 2 λ 1 ≤ λ 2 λ 1 | disk ≈ 2. Using eigenvalues and eigenvectors to calculate the final values when repeatedly applying a matrix First, we need to consider the conditions under which we'll have a steady state. Spectral Decomposition with Repeated Eigenvalues (Quantum Theory for Computer Age - Ch. As shown byL¨utkepohl (2005) | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
with Repeated Eigenvalues (Quantum Theory for Computer Age - Ch. As shown byL¨utkepohl (2005) andHamilton(1994), the VAR is stable if the modulus of each eigenvalue of A is strictly less than 1. If the eigenvalues of A are distinct, it turns out that the eigenvectors are linearly independent; but, if any of the eigenvalues are repeated, further investigation may be necessary. The calculator will find the eigenvalues and eigenvectors of the given square matrix, with steps shown. Then use deflation to form a new matrix and use the power method again to extract the second eigenvalue (and root). Next one is at least one eigenvalue is repeated, can be twice or even more. 3 COMPLEX AND REPEATED EIGENVALUES 15 A. trix has two eigenvalues of magnitude zero, one eigenvalue of unit magnitude, and three eigenvalues with magnitude less than one (right). 1 Introduction to Eigenvalues Linear equationsAx D bcomefrom steady stateproblems. To learn more about all this you should take 18. If this is the situation, then we actually have two separate cases to examine, depending on whether or not we can find two linearly independent eigenvectors. Eigenvalues shows variance explained by that particular factor out of the total variance. (λ = −2 is a repeated root of the characteristic equation. 1 Eigenvalues and Eigenvectors Spectral graph theory studies how the eigenvalues of the adjacency matrix of a graph, which are purely algebraic quantities, relate to combinatorial properties of the graph. Repeated Eigenvalues Recall We are now in the position that we can find (at least) as many linearly-independent solutions to the homogeneous equation ˙ y = A y as there are distinct eigenvalues of A (real or complex). Suppose the 2 2 matrix Ahas repeated eigenvalues. Find the eigenvalues & eieenvectcrs. 0 1), whose only eigenvalue is 1. In order to determine the eigenvectors of a matrix, you must first determine the eigenvalues. Prove that if A is a square matrix then A and AT have the same | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
must first determine the eigenvalues. Prove that if A is a square matrix then A and AT have the same characteristic polynomial. The last case is what the solutions look like when there are repeated eigenvalues, or. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. Use the given information to determine the matrix AL Phase plane solution trajectories have horizontal tangents on the line y2-2刈and vertical tangents on the line y! The matrix Al has a nonzero repeated eigenvalue and a21 - -5. Next I simplify it. 1 of A is repeated if it is a multiple root of the char acteristic equation of A; in our case, as this is a quadratic equation, the only possible case is when λ. Remark: I A matrix with repeated eigenvalues may or may not be diagonalizable. The set of eigenvalues of a matrix is sometimes called the of the matrix, and orthogonal diagonalispectrum zation of a matrix factors in aE E way that displays all the eigenvalues and their multiplicities. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. vectors corresponding to eigenvalues as above (assum-ing no repeated eigenvalues), the Hessian has exactlyP q∈Q q− k 2 negative eigenvalues: we can replace any eigencomponent with eigenvalue σ with an alternate eigencomponent not already in (U,V) with eigenvalue σ0 > σ, decreasing the objective function. There are various methods by which the continuous eigenvalue problem may be. The algorithm is from the EISPACK collection of subroutines. Since it is not invertible, 0 is an eigenvalue. By looking at these eigenvalues it is possible to get information about a graph that might otherwise be di cult to obtain. The picture is more complicated, but as in the 2 by 2 case, our best insights come from finding the matrix's eigenvectors : that is, those vectors whose direction | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
best insights come from finding the matrix's eigenvectors : that is, those vectors whose direction the. The resulting array will be of complex type, unless the imaginary part is zero in which case it will be cast to a real type. Guidance as to the number of linearly independent eigenvectors can be obtained from the rank of the matrix A. x(t)- — 4- -(14t). ) FINDING EIGENVECTORS • Once the eigenvaluesof a matrix (A) have been found, we can find the eigenvectors by Gaussian Elimination. Step 2 For each eigenvalue , compute an orthonormal basis for Ker(A Id). complex eigenvalues. Then each Axi is also in X, so each. Mathematical and Computer Modelling, 2002. sy ' Section 7. Then, we use these results to establish necessary and sufficient conditions for the. We again consider the system ~x0 = A~x. The eigenvalues are not necessarily ordered. If you would like to simplify the solution provided by our calculator, head to the unit vector calculator. I Review: The case of diagonalizable matrices. The resulting array will be of complex type, unless the imaginary part is zero in which case it will be cast to a real type. Because the rank of limn→∞P n = 1, there is a unique limiting distribution. If eigenvalues are repeated, we may or may not have all n linearly independent eigenvectors to diagonalize a square matrix. If the eigenvalue is negative, the direction is reversed. • Objective. Eigenvalues in MATLAB. De nition If Ais a matrix with characteristic polynomial p( ), the multiplicity of a root of pis called the algebraic multiplicity of the eigenvalue. Eigenvalues and Eigenfunctions of the Laplacian Mihai Nica University of Waterloo [email protected] I Equivalently: An n × n matrix with repeated eigenvalues may or may not have a linearly independent set of n eigenvectors. 8 Repeated Eigenvalues Shawn D. The rest are similar. Calculating eigenvalues and eigenvectors of matrices by hand can be a daunting task. These repeated matrix multiplications mean the resulting matrix | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
by hand can be a daunting task. These repeated matrix multiplications mean the resulting matrix is exponential in the number of layers of the neural network. Find more Mathematics widgets in Wolfram|Alpha. @Star Strider: Thanks for the suggestion, I was unaware of this function. 0 1), whose only eigenvalue is 1. Make a matrix Q as follows. Is the procedure the same? Can I use, say, variation of parameters to solve this. But here only (1,0) is a eigenvector to 0. For the following matrix, list the real eigenvalues, repeated according to their multi-plicities. 3 COMPLEX AND REPEATED EIGENVALUES 15 A. Repeated real eigenvalue 3. The coefficients for the principal components are unique (except for a change in sign) if the eigenvalues are distinct and not zero. We can now find a real-valued general solution to any homogeneous system where the matrix has distinct eigenvalues. The ‘ladder’ is bounded at both the low and high ends, which can be seen by considering the operator !. Mathematical and Computer Modelling, 2002. This might introduce extra solutions. But you can find enough independent eigenvectors -- Forget the "but. Stop at this point, and practice on an example (try Example 3, p. On the other hand, when it comes to a repeated eigenvalue, this function is not di erentiable, which hinders statistical inference, as the asymptotic theory requires at least second-order di erentiability. To be more specific, Section 2. Indeed, BAv = ABv = A( v) = Av. A matrix A is idempotent if and only if all its eigenvalues are either 0 or 1. Subsection 3. Assume λ is a repeated eigenvalue of A with multiplicity m. Is 1 1 an eigenvector. Repeated eigenvalues - Duration: 7:30. Terminology Let Abe an n nmatrix. Eigenvalues, Eigenvectors, and Di erential Equations William Cherry April 2009 (with a typo correction in November 2015) The concepts of eigenvalue and eigenvector occur throughout advanced mathematics. Repeated Eigenvalues and Symmetric Matrices 22. State feedback and Observer | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
advanced mathematics. Repeated Eigenvalues and Symmetric Matrices 22. State feedback and Observer Feedback Poles of transfer function are eigenvalues of A Pole locations a ect system response repeated eigenvalues. However, the geometric multiplicity can never exceed the algebraic multiplicity. • Roy’s Largest Root = largest eigenvalue o Gives an upper-bound of the F-statistic. This will be one of the few times in this chapter that non-constant coefficient differential equation will be looked at. De nition If Ais a matrix with characteristic polynomial p( ), the multiplicity of a root of pis called the algebraic multiplicity of the eigenvalue. “The equation A x = λ x characterizes the eigenvalues and associated eigenvectors of any matrix A. Supplementary notes for Math 265 on complex eigenvalues, eigenvectors, and systems of di erential equations. The minimal polynomial x3 x…x—x 1–—x‡1–also splits completely over any field, and in particular over F, so Acan be diagonalized over F. Imposing an additional condition, that the eigenvalues lie in Fand are simple roots of the characteristic polynomial, does force diagonalizability. This is because u lays on the same subspace (plane) as v and w, and so does any other eigenvector. Get an answer for 'Give an example of a non-diagonalizable 4x4 matrix with eigenvalues: -1, -1, 1, 1. And I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier. 06 or 18/700. But for a fundamental system two independent solutions are needed. Repeated Eigenvalues Note. However, the fundamental issue is selecting the appropriate tolerance to determine whether two eigenvalues are the same or not, which I don't know a priori (the elements of the matrix I am considering vary by 7 orders of magnitude, so its not obvious how close is close enough). Solution Since , the given matrix has distinct real eigenvalues of. Now is the next step. † Think of repeated eigenvalue case as a bifurcation between 2 | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
eigenvalues of. Now is the next step. † Think of repeated eigenvalue case as a bifurcation between 2 distinct real eigenvalue case (2 straight-line solutions) and complex conjugate eigenvalue case (no straight-line solutions. eigenvalues tell the entire story. The number of eigenvalues equal to 1 is then tr(A). If is eigenvalue of perturbationA+ Eof nondefective matrixA, then j kj cond 2(X)kEk 2 where kis closest eigenvalue ofAto andX is nonsingular matrix of eigenvectors ofA Absolute condition number of eigenvalues is condition number of matrix of eigenvectors with respect to solving linear equations Eigenvalues may be sensitive if eigenvectors are. So even though a real asymmetric x may have an algebraic solution with repeated real eigenvalues, the computed solution may be of a similar matrix with complex conjugate pairs of eigenvalues. We shall see that this. The phase portrait thus has a distinct star. As any system we will want to solve in practice is an approximation to …. In the 2x2 case, if the eigenvalue is repeated you are in the defective case unless the matrix is precisely [ lambda_1 , 0 ; 0 , lambda_1 ] For larger square matrices this becomes the story of Jordan form. Eigenvectors and Hermitian Operators 7. • Objective. In fact, it is easy to see that this happen if and only if we have more than one equilibrium point (which is (0,0)). Let be the set[5] of eigenvalues, speci cally not attempting to count repeated eigenvalues more than once. polynomial corresponding to A, has n roots some of which may be repeated. 4 Bases and Subspaces 89 5. Within this blog post, we’ll investigate some classes of buckling problems and the way they are sometimes analyzed. where the eigenvalues are repeated eigenvalues. This is particularly true if some of the matrix entries involve symbolic parameters rather than speciflc numbers. This happens when the dimension of the nullspace of A−λI (called the geometric multiplicity of λ) is strictly less than the arithmetic | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
the nullspace of A−λI (called the geometric multiplicity of λ) is strictly less than the arithmetic multiplicity m. | {
"domain": "rk-verlag.de",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9888419697383903,
"lm_q1q2_score": 0.8591323757389555,
"lm_q2_score": 0.868826771143471,
"openwebmath_perplexity": 450.8891887879132,
"openwebmath_score": 0.8856151103973389,
"tags": null,
"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html"
} |
# bandwidth
Lower and upper matrix bandwidth
## Syntax
• `B = bandwidth(A,type)` example
• ```[lower,upper] = bandwidth(A)``` example
## Description
example
````B = bandwidth(A,type)` returns the bandwidth of matrix `A` specified by `type`. Specify `type` as `'lower'` for the lower bandwidth, or `'upper'` for the upper bandwidth.```
example
``````[lower,upper] = bandwidth(A)``` returns the lower bandwidth, `lower`, and upper bandwidth, `upper`, of matrix `A`.```
## Examples
collapse all
### Find Bandwidth of Triangular Matrix
Create a 6-by-6 lower triangular matrix.
`A = tril(magic(6))`
```A = 35 0 0 0 0 0 3 32 0 0 0 0 31 9 2 0 0 0 8 28 33 17 0 0 30 5 34 12 14 0 4 36 29 13 18 11 ```
Find the lower bandwidth of `A` by specifying `type` as `'lower'`.
`B = bandwidth(A,'lower')`
```B = 5```
The result is 5 because every diagonal below the main diagonal has nonzero elements.
Find the upper bandwidth of `A` by specifying `type` as `'upper'`.
`B = bandwidth(A,'upper')`
```B = 0```
The result is 0 because there are no nonzero elements above the main diagonal.
### Find Bandwidth of Sparse Block Matrix
Create a 100-by-100 sparse block matrix.
`B = kron(speye(25),ones(4));`
View a 10-by-10 section of elements from the top left of `B`.
`full(B(1:10,1:10))`
```ans = 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1```
`B` has 4-by-4 blocks of ones centered on the main diagonal.
Find both the lower and upper bandwidths of `B` by specifying two output arguments.
`[lower,upper] = bandwidth(B)`
```lower = 3 upper = 3 ```
The lower and upper bandwidths are both `3`.
## Input Arguments
collapse all
### `A` — Input matrix2-D numeric matrix
Input matrix, specified as a 2-D numeric matrix. `A` can be either full or sparse.
Data Types: `single` | `double`
Complex Number Support: Yes | {
"domain": "mathworks.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9908743644972026,
"lm_q1q2_score": 0.8591235074903251,
"lm_q2_score": 0.8670357598021707,
"openwebmath_perplexity": 2476.1231206705274,
"openwebmath_score": 0.8676239848136902,
"tags": null,
"url": "http://fr.mathworks.com/help/matlab/ref/bandwidth.html?s_tid=gn_loc_drop&nocookie=true"
} |
Data Types: `single` | `double`
Complex Number Support: Yes
### `type` — Bandwidth type`'lower'` | `'upper'`
Bandwidth type, specified as `'lower'` or `'upper'`.
• Specify `'lower'` for the lower bandwidth (below the main diagonal).
• Specify `'upper'` for the upper bandwidth (above the main diagonal).
Data Types: `char`
## Output Arguments
collapse all
### `B` — Lower or upper bandwidthnonnegative integer scalar
Lower or upper bandwidth, returned as a nonnegative integer scalar.
• If `type` is `'lower'`, then `0` ≤ `B` ≤ `size(A,1)-1`.
• If `type` is `'upper'`, then `0` ≤ `B` ≤ `size(A,2)-1`.
### `lower` — Lower bandwidthnonnegative integer scalar
Lower bandwidth, returned as a nonnegative integer scalar. `lower` is in the range `0` ≤ `lower` ≤ `size(A,1)-1`.
### `upper` — Upper bandwidthnonnegative integer scalar
Upper bandwidth, returned as a nonnegative integer scalar. `upper` is in the range `0` ≤ `upper` ≤ `size(A,2)-1`.
collapse all
### Upper and Lower Bandwidth
The upper and lower bandwidths of a matrix are measured by finding the last diagonal (above or below the main diagonal, respectively) that contains nonzero values.
That is, for a matrix A with elements Aij:
• The upper bandwidth B1 is the smallest number such that ${A}_{ij}=0$ whenever $j-i>{B}_{1}$.
• The lower bandwidth B2 is the smallest number such that ${A}_{ij}=0$ whenever $i-j>{B}_{2}$.
Note that this measurement does not disallow intermediate diagonals in a band from being all zero, but instead focuses on the location of the last diagonal containing nonzeros. By convention, the upper and lower bandwidths of an empty matrix are both zero. | {
"domain": "mathworks.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9908743644972026,
"lm_q1q2_score": 0.8591235074903251,
"lm_q2_score": 0.8670357598021707,
"openwebmath_perplexity": 2476.1231206705274,
"openwebmath_score": 0.8676239848136902,
"tags": null,
"url": "http://fr.mathworks.com/help/matlab/ref/bandwidth.html?s_tid=gn_loc_drop&nocookie=true"
} |
Math Help - Finding the remainder
1. Finding the remainder
When the positive integer n is divided by 7, the remainder is 2. What is the remainder when 5n is divided by 7?
How would you set this up?
2. The first step is to set up 5n/7 = 5/7 + n/7 so to strictly answer your question, except where n is a multiple of 7, then the remainder is n/7. Is this what you're asking about?
3. No. These are the choices:
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
The answer is 3. I have no freaking clue how they got that.
4. I think I understand the problem now. When n is 9, then division by 7 leaves a remainder of 2. Now 5 x 9 = 45. Try dividing 45 by 7 and see if you get a remainder of 3.
5. Suppose you divide a number N by P, and you get a quotient Q with a remainder R. This can be rewritten in fraction form:
$\frac{N}{P} = Q + \frac{R}{P}$
(remember that 0 <= R < P).
In our case, we are dividing n by 7. You get a quotient q with a remainder of 2, or:
$\frac{n}{7} = q + \frac{2}{7}$
Now I want to know what 5n is. Multiply both sides of the equation above by 35 and you'll get
$5n = 35q + 10$
Divide 5n by 7:
\begin{aligned}
\frac{5n}{7} &= \frac{35q + 10}{7} \\
&= \frac{35q}{7} + \frac{10}{7} \\
&= 5q + \frac{10}{7} \\
&= (5q + 1) + \frac{3}{7} \\
\end{aligned}
The quotient would be 5q + 1, and the remainder would be 3.
6. Thank you so much!
7. Another, easier way to do it is to let $n \equiv 2 \pmod{7}$, it then follows that $5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $3$.
8. Originally Posted by Bacterius
Another, easier way to do it is to let $n \equiv 2 \pmod{7}$, it then follows that $5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $3$.
I cannot find the MOD command on my calculator though.
9. It really is the remainder function, usually calculators don't have it. You can emulate $a \mod b$ by dividing $a$ by $b$, taking the fractional part only and multiplying it by $b$. For instance, to get $10 \mod 7$ : | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109534209825,
"lm_q1q2_score": 0.8590972481369874,
"lm_q2_score": 0.8723473730188543,
"openwebmath_perplexity": 280.4544854695478,
"openwebmath_score": 0.9409502148628235,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/151037-finding-remainder.html"
} |
$\frac{10}{7} = 1.428571429 ...$
Take the fractional part which is $0.428571429$, and times it by $7$, you get :
$0.428571429 \times 7 = 3$
10. Originally Posted by Bacterius
Another, easier way to do it is to let $n \equiv 2 \pmod{7}$, it then follows that $5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $3$.
Originally Posted by Bacterius
It really is the remainder function, usually calculators don't have it. You can emulate $a \mod b$ by dividing $a$ by $b$, taking the fractional part only and multiplying it by $b$. For instance, to get $10 \mod 7$ :
$\frac{10}{7} = 1.428571429 ...$
Take the fractional part which is $0.428571429$, and times it by $7$, you get :
$0.428571429 \times 7 = 3$
Oh, ok, but where did you get the 10 from? Especially the 10 in the first post here.
Also, couldn't a shorter way by just looking at 10 mod 7 just be subtracting 7 from 10?
11. If $n \equiv 2 \pmod{7}$ then $\mathbf{5} \times n \equiv \mathbf{5} \times 2 \equiv 10 \equiv 3 \pmod{7}$
This is where the $10$ comes from
Also, couldn't a shorter way by just looking at 10 mod 7 just be subtracting 7 from 10?
That would only work for numbers up to $13$ ... for instance if you substract $7$ from $23$, you get $16$, which isn't exactly the remainder of $23$ divided by $7$, while $23 \mod 7 = 2$ as expected. Then you could argue that repeatedly substracting $7$ is a way to do it, true, but dividing is then more efficient.
12. I totally understand now. Thank you!
13. Hello, Mariolee!
A slightly different approach . . .
When the positive integer $n$ is divided by 7, the remainder is 2.
What is the remainder when $5n$ is divided by 7?
"When $n$ is divided by 7, the remainder is 2."
. . Hence: . $n \:=\:7a + 2\:\text{ for some integer }a.$
Then: . $5n \:=\:35a + 10$
. . and: . $\dfrac{5n}{7} \:=\:\dfrac{35a+10}{7} \;=\;5a + 1 + \frac{3}{7}$
Therefore, the remainder is 3. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109534209825,
"lm_q1q2_score": 0.8590972481369874,
"lm_q2_score": 0.8723473730188543,
"openwebmath_perplexity": 280.4544854695478,
"openwebmath_score": 0.9409502148628235,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/151037-finding-remainder.html"
} |
# Will $2$ linear equations with $2$ unknowns always have a solution?
As I am working on a problem with 3 linear equations with 2 unknowns I discover when I use any two of the equations it seems I always find a solution ok. But when I plug it into the third equation with the same two variables , the third may or may not cause a contradiction depending if it is a solution and I am OK with that BUT I am confused on when I pick the two equations with two unknowns it seems like it has no choice but to work. Is there something about linear algebra that makes this so and are there any conditions where it won't be the case that I will find a consistent solution using only the two equations? My linear algebra is rusty and I am getting up to speed. These are just equations of lines and maybe the geometry would explain it but I am not sure how. Thank you.
• Mind showing the equations?? – The Dead Legend May 7 '17 at 0:16
Each linear equation represents a line in the plane. Most of the time two lines will intersect in one point, which is the simultaneous solution you seek. If the two lines have exactly the same slope, they may not meet so there is no solution or they may be the same line and all the points on the line are solutions. When you add a third equation into the mix, that is another line. It is unlikely to go through the point that solves the first two equations, but it might.
There are three possible cases for $2$ linear equations with $2$ unknowns (slope and intercept):
$\qquad$ $\mathbf{0}$ solution points $\qquad$ $\qquad$ $\mathbf{1}$ solution point $\qquad$ $\qquad$ $\mathbf{\infty}$ solution points
$\qquad \quad$ $\nexists$ no existence $\qquad$ $\qquad$ $\exists !$ uniqueness $\qquad$ $\qquad$ $\exists$ no uniqueness
The lines have the form $y(x) = mx + b$.
Case 1: parallel lines
A solution does not exist.
The lines are parallel: they have the same slope.
% \begin{align} % y_{1}(x) &= m x + b_{1} \\ % y_{2}(x) &= m x + b_{2} \\ % \end{align} % | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8590972453361766,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 751.3237403913528,
"openwebmath_score": 1.0000097751617432,
"tags": null,
"url": "https://math.stackexchange.com/questions/2269220/will-2-linear-equations-with-2-unknowns-always-have-a-solution"
} |
% \begin{align} % y_{1}(x) &= m x + b_{1} \\ % y_{2}(x) &= m x + b_{2} \\ % \end{align} %
Case 2: intersecting lines
We have existence and uniqueness.
The slopes are distinct.
$$m_{1} \ne m_{2}$$
% \begin{align} % y_{1}(x) &= m_{1} x + b_{1} \\ % y_{2}(x) &= m_{2} x + b_{2} \\ % \end{align} %
Case 3: coincident lines
We have existence, but not uniqueness. There is an infinite number of solutions. Every point solves the system of equations.
Both lines are the same.
% \begin{align} % y_{1}(x) &= m x + b \\ % y_{2}(x) &= m x + b \\ % \end{align} %
In terms of linear algebra, look at the problem in terms of $\color{blue}{range}$ and $\color{red}{null}$ spaces.
The linear system for two equations is % \begin{align} % m_{1} x - y &= b_{1} \\ % m_{2} x - y &= b_{1} \\ % \end{align} which has the matrix form % \begin{align} % \mathbf{A} x &= b \\ % \left[ \begin{array}{cc} m_{1} & -1 \\ m_{2} & -1 \\ \end{array} \right] % \left[ \begin{array}{cc} x \\ y \\ \end{array} \right] % &= % \left[ \begin{array}{cc} b_{1} \\ b_{2} \\ \end{array} \right] % \end{align} %
The Fundamental Theorem provides a natural framework for classifying data and solutions.
Fundamental Theorem of Linear Algebra
A matrix $\mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}$ induces for fundamental subspaces: \begin{align} % \mathbf{C}^{n} = \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\ % \mathbf{C}^{m} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)} % \end{align}
Case 1: No existence | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8590972453361766,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 751.3237403913528,
"openwebmath_score": 1.0000097751617432,
"tags": null,
"url": "https://math.stackexchange.com/questions/2269220/will-2-linear-equations-with-2-unknowns-always-have-a-solution"
} |
The matrix $\mathbf{A}$ has a rank defect $(m_{1} = m_{2})$ and $b_{1} \ne b_{2}$. $$b = \color{blue}{b_{\mathcal{R}}} + \color{red}{b_{\mathcal{N}}}$$ It is the $\color{red}{null}$ space component which precludes direct solution. (Interestingly enough, there is a least squares solution.) $$The data vector b is not a combination of the columns of \mathbf{A}. The column space is$$ \mathbf{C}^{2} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)} $$The decomposition is$$ \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} = % \text{span } \left\{ \, \color{blue}{ \left[ \begin{array}{c} m \\ -1 \end{array} \right] } \, \right\} \qquad \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} = % \text{span } \left\{ \, \color{red}{ \left[ \begin{array}{r} -1 \\ m \end{array} \right] } \, \right\} $$Case 2: Existence and uniqueness The matrix \mathbf{A} has full rank (m_{1}\ne m_{2}). The data vector is entirely in the \color{blue}{range} space \color{blue}{\mathcal{R} \left( \mathbf{A} \right)}$$ b = \color{blue}{b_{\mathcal{R}}} $$The \color{red}{null} space is trivial: \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)}=\mathbf{0}.$$ \mathbf{C}^{2} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} $$The decomposition is$$ \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} = % \text{span } \left\{ \, \color{blue}{ \left[ \begin{array}{c} m_{1} \\ -1 \end{array} \right] }, \, \color{blue}{ \left[ \begin{array}{c} m_{2} \\ -1 \end{array} \right] } \right\} $$Case 3: Existence, no uniqueness The matrix \mathbf{A} has a rank defect (m_{1} = m_{2} = m), yet b_{1} = b_{2}.$$ b = \color{blue}{b_{\mathcal{R}}} $$The column space is has \color{blue}{range} and \color{red}{null} space components:$$ \mathbf{C}^{2} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)} $$The decomposition is$$ \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} = % | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8590972453361766,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 751.3237403913528,
"openwebmath_score": 1.0000097751617432,
"tags": null,
"url": "https://math.stackexchange.com/questions/2269220/will-2-linear-equations-with-2-unknowns-always-have-a-solution"
} |
\right)} $$The decomposition is$$ \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} = % \text{span } \left\{ \, \color{blue}{ \left[ \begin{array}{c} m \\ -1 \end{array} \right] } \, \right\} \qquad \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} = % \text{span } \left\{ \, \color{red}{ \left[ \begin{array}{r} -1 \\ m \end{array} \right] } \, \right\} $$Postscript: the theoretical foundations here are useful. The trip to understanding starts with simple examples like in @Nick's comment. Let's think using vector notation. A linear system with two unknowns x and y, and two equations$$ \begin{align*} v_1 x + w_1 y &= a_1 \\ v_2 x + w_2 y &= a_2 \end{align*} $$can be written in vector notation as$$ x\, \vec{v} + y\, \vec{w} = \vec{a}. That is, you want to know if $\vec{a}$ can be written as a linear combination of $\vec{v}$ and $\vec{w}$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8590972453361766,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 751.3237403913528,
"openwebmath_score": 1.0000097751617432,
"tags": null,
"url": "https://math.stackexchange.com/questions/2269220/will-2-linear-equations-with-2-unknowns-always-have-a-solution"
} |
Fixed the vectors $\vec{v}$ and $\vec{w}$, to state that a solution always exists whatever $\vec{a}$ is, is the same as to state that $\vec{v}$ and $\vec{w}$ spans the whole plane. If not ($\vec{v}$ and $\vec{w}$ are parallel), depending on $\vec{a}$, the solution might not exist. And when it exists, it will not be unique. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8590972453361766,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 751.3237403913528,
"openwebmath_score": 1.0000097751617432,
"tags": null,
"url": "https://math.stackexchange.com/questions/2269220/will-2-linear-equations-with-2-unknowns-always-have-a-solution"
} |
# Thread: physics tourist & bear problem
1. ## physics tourist & bear problem
another easy one i think:
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?
how many meters?
thanks alot.
2. Originally Posted by rcmango
another easy one i think:
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?
how many meters?
thanks alot.
The maximum value of d is such that the bear gets to the car at the same time the tourist does.
So set up a coordinate system such that the bear is at the origin and positive x is in the direction from the bear to the tourist.
Both are moving at a constant speed. The bear has to cover 27 + d meters in the same time the tourist covers d meters.
So for the tourist:
[tex]d = v_t t = 3.5t[tex]
Thus
$t = \frac{d}{3.5}$
For the bear:
$27 + d = v_b t = 6 \left ( \frac{d}{3.5} \right )$
Now solve for d.
-Dan
3. Hello, rcmango!
Another approach . . .
A tourist being chased by an angry bear is running in a straight line
toward his car at a speed of 3.5 m/s. .The car is a distance $d$ meters away.
The bear is 27 meters behind the tourist and running at 6.0 m/s.
The tourist reaches the car safely.
What is the maximum possible value for $d$?
The tourist has a 27-meter headstart.
Relative to the tourist, the bear has a speed of 2.5 m/s.
To cover 27 meters, it takes the bear: . $\frac{27}{2.5} \:=\:10.8$ seconds.
In that time, the tourist can run: . $3.5 \times 10.8 \:=\:37.8$ meters.
[I hope he left a window open so he can dive into the car.] | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109538667756,
"lm_q1q2_score": 0.8590972370868161,
"lm_q2_score": 0.8723473614033683,
"openwebmath_perplexity": 1309.6782790269324,
"openwebmath_score": 0.5672556757926941,
"tags": null,
"url": "http://mathhelpforum.com/math-topics/19568-physics-tourist-bear-problem.html"
} |
[I hope he left a window open so he can dive into the car.]
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
This reminds me of two chased-by-a-bear stories.
An old-timer was telling about his encounter with an angry grizzly.
"I was barely keepin' ahead of him and he wuz gainin' on me fast.
I saw a tree branch about fifteen feet up. .Just as he came up
'n took a swipe at me, I took this big leap . . ."
"Well, did you catch it?"
"Well, not on the way up . . . "
Two guys are camping when a bear headed towards them.
One man starting putting on his running shoes.
"Hey," said his friend, "you can't outrun that bear."
"I don't have to," was the reply. "I just have to outrun you."
4. Originally Posted by rcmango
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 m away from the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?
how many meters?
thanks alot.
Hello,
if you change the text of your problem as I've done (see above) then there is a second situation possible: The car is between the bear and the tourist and both are running toward each other. They reach the car (on different sides, I hope for the tourist) after:
$t=\frac{27}{3.5+6}\approx 2.842~s$
During this time the tourist runs d = 9.947 m | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109538667756,
"lm_q1q2_score": 0.8590972370868161,
"lm_q2_score": 0.8723473614033683,
"openwebmath_perplexity": 1309.6782790269324,
"openwebmath_score": 0.5672556757926941,
"tags": null,
"url": "http://mathhelpforum.com/math-topics/19568-physics-tourist-bear-problem.html"
} |
# Is there a method to calculate large number modulo?
Is there a (number theoretic or algebraic) trick to find a large nunber modulo some number?
Say I have the number $123456789123$ and I want to find its value modulo some other number, say, $17$.
It's not fast for me to find the prime factorisation first. It's also not fast to check how many multiples of $17$ I can "fit" into the large number.
So I was wondering if there is any method out there to do this efficiently.
I am looking for something like the other "magic trick" where you sum all the digits and take the result $\mod 9$.
• – lab bhattacharjee Aug 22 '15 at 4:24
• Don't even think about factorisation here! Division is much faster. – TonyK Aug 22 '15 at 19:20
The best I could come up with is to use 17*6 = 102. Dividing by 102 goes pretty fast...
123456789123
214
105
367
618
691
792
783
69
and 69 mod 17 = 1
You can speed things up by trying to eliminate two digits at a time
123456789123
1224
----
1056
1020
----
3678
3672
----
6912
6834
----
783
714
---
69
and 69 mod 17 = 1
## for really large numbers
For really large numbers, you can use the fact that 17 | 100,000,001
The procedure is similar to the check for divisibility by 11, except you break the number up into larger chunks.
Starting from the right, split the number up into 8-digit chunks.
So 123456789123 becomes
chunk # 1 2
chunk 56789123 1234
Compute (sum of odd numbered chunks) - (sum of even numbered chunks)
56789123 - 1234 = 56787889
If the result is negative, add a big enough multiple of 100,000,001 to
make it positive.
This number is congruent to the original number modulo 17.
56787889
5610
----
6878
6834
----
4489
4488
-----
1
and, again, we get 1
• This is a special case of the universal divisibility test (using various chunks sizes), see my answer. Both (well-known) methods were already mentioned in Lab's prior answer. – Bill Dubuque Oct 12 at 0:18 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771778588348,
"lm_q1q2_score": 0.8590855271673398,
"lm_q2_score": 0.8705972801594706,
"openwebmath_perplexity": 533.0015430305622,
"openwebmath_score": 0.9460786581039429,
"tags": null,
"url": "https://math.stackexchange.com/questions/1405562/is-there-a-method-to-calculate-large-number-modulo"
} |
$10^2\equiv-2\pmod{17}\implies10^4=(10^2)^2\equiv(-2)^2\equiv4;$
$\displaystyle\implies\sum_{r=0}^na_r10^r\equiv(4)^0(a_3a_2a_1a_0)+(4)^1(a_7a_6a_5a_4)++(4)^2(a_{11}a_{10}a_9a_8)+\cdots\pmod{17}$
Again, $10^8\equiv(-2)^4\equiv-1$
$\displaystyle\implies\sum_{r=0}^na_r10^r\equiv(-1)^0(a_7a_6a_5\cdots a_0)+(-1)^1(a_{15}\cdots a_8)+\cdots\pmod{17}$
• Undoubtedly you know this, but in case others are wondering: for any prime $p>2$ we have $10^{(p-1)/2}\equiv\pm1\pmod p$. The sign here depends on whether $10$ is a quadratic residue modulo $p$ or not. That, in turn, can be easily determined using the law of quadratic reciprocity. This leads to a divisibility rule like the one here in chunks of $(p-1)/2$ digits. In some cases we can do shorter chunks. The best known cases of that are perhaps $p=13$ ($10^3\equiv-1$) and $p=41$ ($10^5\equiv1$). – Jyrki Lahtonen Aug 22 '15 at 19:27
• This is a special case of the universal divisibility test (using various chunks sizes), see my answer. – Bill Dubuque Oct 12 at 0:18
Well, it is fast to divide 17 into that number.
Where you can gain a lot is when the number you want to be divide is a special form such as $a^n$, where $n$ is large. There are ways (usually involving the Euler $\phi$ function) for rapidly computing $a^n \bmod{b}$ where $n$ is large.
A good start is to remember that $a^n \bmod{b} =(a\bmod{b})^n \bmod{b}$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771778588348,
"lm_q1q2_score": 0.8590855271673398,
"lm_q2_score": 0.8705972801594706,
"openwebmath_perplexity": 533.0015430305622,
"openwebmath_score": 0.9460786581039429,
"tags": null,
"url": "https://math.stackexchange.com/questions/1405562/is-there-a-method-to-calculate-large-number-modulo"
} |
A good start is to remember that $a^n \bmod{b} =(a\bmod{b})^n \bmod{b}$.
Yes, use the universal divisibility test: repeatedly replace leading digit chunks by their remainder mod the divisor as below, using least magnitude remainders $$\, -8\le r \le 9\,$$ to simplify arithmetic $$\!\bmod 17\$$ (so negative digits occur, denoted $$-d,c := 10(-d)+c\equiv 7d+c,\,$$ by $$10\equiv -7)$$ \begin{align} \!\bmod 17\!:\,\ 10\equiv -7\ \Rightarrow\quad\ \ &\ \color{#0A0}{1\ 2}\,\ 3\ 4\ 5\ 6\ \ \ {\rm by}\,\ \ \ \ \ \ \color{#0a0}{1\:\!2\equiv -5}\\[.1em] \equiv\, &\ \color{#0A0}{{-5}},\color{#c00} 3\ 4\ 5\ 6\ \ \ {\rm by}\,\ \color{#0a0}{{-}5},\color{#c00}3\equiv\ (\,7\,)\, \color{#0a0}{5}+\color{#c00}3\,\equiv\,\color{#0af} 4\\[.1em] \equiv\, &\ \ \ \ \ \ \ \ \color{#0af}4\ 4\ 5\ 6\ \ \ {\rm by}\,\ \ \ \ \ \ \color{#0af}4\:\!4\equiv (-7)4+4\,\equiv\color{#f60}{-7}\\[.1em] \equiv\, &\ \ \ \ \ \ \ \ \color{#f60}{{-}\!7}, 5\ 6\ \ \ {\rm by}\,\ \color{#f60}{{-}7},5\equiv\,(\,7\,)\, \color{#f60}{7}+5\,\equiv\, 3\\[.2em] \equiv\, &\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ 6\ \ \ \:\!\:\!{\rm by}\ \ \ \ \ \ 3\:\! 6\equiv (-7)3+6\,\equiv\, 2\\ \equiv\, &\qquad\qquad\ 2;\ \ {\it quicker,}\,\ 2 \,\text{ digits at a time:}\\[.4em] \!\bmod 17\!:\,\ 10^2\equiv -2\ \Rightarrow\quad\ \ &\ \color{#0A0}{12\ 3 4}\ 56\ \ \ {\rm by}\,\ \color{#0A0}{12\ 3 4}\equiv\, (-2)\color{#0a0}{\,12+34\equiv 10}\\[.1em] \equiv\, &\ \ \ \ \ \ \color{#0A0}{{10}}\ \color{#c00}{56}\ \ \ {\rm by}\,\ \color{#0a0}{10}\ \color{#c00}{56}\equiv\ (-2)\, \color{#0a0}{10}+\color{#c00}{56}\,\equiv\,\color{#0af}2\\[.2em] \equiv\, &\qquad\quad\ \color{#0af}2 \end{align}\qquad\qquad | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771778588348,
"lm_q1q2_score": 0.8590855271673398,
"lm_q2_score": 0.8705972801594706,
"openwebmath_perplexity": 533.0015430305622,
"openwebmath_score": 0.9460786581039429,
"tags": null,
"url": "https://math.stackexchange.com/questions/1405562/is-there-a-method-to-calculate-large-number-modulo"
} |
So $$\rm\, 123456\equiv 2\pmod{\!17}.\,$$ Indeed $$\rm\, 123456 = 7262\cdot 17+2.\,$$ Continuing this way we can do the entire number in a couple minutes of mental arithmetic. Unlike some other divisibility tests that compute only a binary truth value, this method has the advantage of computing the remainder. Further, it doesn't require remembering any special algorithm or parameters for each modulus.
Remark Lab & Steven's answers are a special case of above (but without mod arithmetic optimizations), i.e. they use chunk sizes of $$\,2\,$$ and $$\,8,\,$$ using $$\bmod 17\!:\ 10^2\equiv -2,\ 10^8\equiv -1$$.
• See here for another example using negative digits. – Bill Dubuque Oct 29 '18 at 23:10
No, there is not. The reason why "magic tricks" work when studying divisibility by $2,3,5,11$ is the fact that we usually write in base $10$. Change the base and they will stop working. In particular, the following trick would work in base $17$: if the last digit of your number is $0$ then the number is divisible by $17$. Of course, in order to check this you would need to write it in base $17$, which is a vicious circle...
• Great, where can I read about the tricks you mention for $2,3,5,11$? (you don't mention $9$...?) – learner Aug 23 '15 at 4:29
• @learner: I don't know of any book, I know them from elementary school: for $2$, the last digit must be even; for $3$ the sum of the digits must be divisible by $3$; for $4$, the number formed by the last 2 digits must be divisible by $4$; for $5$, the last digit must be $0$ or $5$; for $8$, the number formed by the last 3 digits must be divisible by $8$; for $10$, the last digit must be $0$; for $11$, sum all the digits on odd positions, sum all the digits on even positions and check whether the difference of these 2 numbers is divisible by $11$. – Alex M. Aug 28 '15 at 10:59 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771778588348,
"lm_q1q2_score": 0.8590855271673398,
"lm_q2_score": 0.8705972801594706,
"openwebmath_perplexity": 533.0015430305622,
"openwebmath_score": 0.9460786581039429,
"tags": null,
"url": "https://math.stackexchange.com/questions/1405562/is-there-a-method-to-calculate-large-number-modulo"
} |
# Math Help - Related Rates cont'd...
1. ## Related Rates cont'd...
I have some more questions that I need help setting up. Please feel free to switch out the numbers for variables so I can learn how to solve these. Thanks in advance for your help!
Two towers (A and B) are 150m tall, 40m apart. A horizontal tightrope (height of 60m) is attached to the two buildings. A man walks across the tightrope from A to B at a constant 1/3 meter per sec. A spotlight is on the top of A. How fast is the shadow of the man moving up the wall of building B when he is 8 m away from B.
(^^No clue where to begin^^)
A trough, 12 feet long, has as its ends isosceles trapezoids with altitude 4 ft., lower base 3 ft., and upper base 4 ft. If water is let in at a rate of 7 cubic feet per minute, how fast is the water level rising when the water is 10 in. deep?
(^^I found the volume of the trough to be 168 cubic feet ft...and I think the water level uses a ratio towards the trough...what next?^^)
Stacey is brewing coffee that is being strained through a conical filter with a height of 12 in. diameter 8 in. The coffee flows from the filter into a cylindrical coffee pot with base area equal to 100pi square in. The depth, h, in inches, of the coffee in the conical filter is changing at the rate of (h-12) in. per min. How fast is the depth of the coffee in the cylindrical coffee pot changing when h=3in?
(^^No clue where to begin^^)
Two streets lights, each 30 feet tall, are 100 ft. apart. The light at the top of one of the poles is functioning properly, but the other light is burnt out. A repairman is climbing up the pole to fix the light at a rate of .5 foot per second. How fast is the tip of the repairman's shadow moving when he is 16 ft. up the pole?
(^^I have a drawing set up, but I do not know how to label it and how to set up my equations^^)
2. Hello, nivek516!
Here's the first one . . . | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771805808551,
"lm_q1q2_score": 0.8590855262240585,
"lm_q2_score": 0.8705972768020108,
"openwebmath_perplexity": 883.8461896641982,
"openwebmath_score": 0.6901357769966125,
"tags": null,
"url": "http://mathhelpforum.com/calculus/60919-related-rates-cont-d.html"
} |
2. Hello, nivek516!
Here's the first one . . .
Two towers, $A\text{ and }B$. are 150m tall, 40m apart.
A horizontal tightrope (height of 60m) is attached to the two buildings.
A man walks across the tightrope at a constant $\tfrac{1}{3}$ m/sec.
A spotlight is on the top of $A$. How fast is the shadow of the man
moving up the wall of building $B$ when he is 8 m away from $B]$?
First, make a sketch . . .
Code:
A * * B
| * |
90 | * |
| * 40-x |
T + - - - - - * - - - - - + R
| x M * |
| * | y
60 | * |
| * S
| |
| |
C * - - - - - - - - - - - * D
40
The buildings are $AC\text{ and }BD\!:\;\;AC = BD = 150,\;CD = 40$
The tightrope is $TR\!:\;\;TR = 40,\;TC = 60.\;AT = 90$
The spotlight is at $A$, the man is at $M.$
. . Let: $TM \:=\: x \quad\Rightarrow\quad MR \:=\:40-x$
. . We are given: . $\tfrac{dx}{dt} = \tfrac{1}{3}$ m/sec.
The man's shadow is at $S\!:\;\;\text{let } RS = y.$
Since $\Delta ATM \sim \Delta SRM\!:\;\;\frac{x}{90} \:=\:\frac{40-x}{y} \quad\Rightarrow\quad y \:=\:90\,\frac{40-x}{x} \:=\:90\left(40x^{-1} - 1\right)$
Differentiate with respect to time: . $\frac{dy}{dt} \:=\:-\frac{3600}{x^2}\cdot\frac{dx}{dt}$
When $MR = 8\;(x = 32)\!:\;\;\frac{dy}{dt} \;=\;-\frac{3600}{32^2}\left(\frac{1}{3}\right) \;=\;-\frac{75}{64}$
The shadow is moving up building $B$ at $1\tfrac{11}{64}\text{ m/sec}$
3. Thanks! Could you explain why the answer is negative? Does it matter since it is speed? Any possible chance you could help me with the last two questions? I found out how to do the second one.
4. Hello again, nivek516!
Here's the last one . . . | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771805808551,
"lm_q1q2_score": 0.8590855262240585,
"lm_q2_score": 0.8705972768020108,
"openwebmath_perplexity": 883.8461896641982,
"openwebmath_score": 0.6901357769966125,
"tags": null,
"url": "http://mathhelpforum.com/calculus/60919-related-rates-cont-d.html"
} |
4. Hello again, nivek516!
Here's the last one . . .
Two streets lights, each 30 feet tall, are 100 feet apart. The light at the top
of one of the poles is functioning properly, but the other light is burnt out.
A repairman is climbing up the pole to fix the light at a rate of 0.5 ft/sec.
How fast is the tip of the repairman's shadow moving when he is 16 ft. up the pole?
Code:
A * * C
| * |
| * |
| * |
30 | * R
| | *
| y| *
| | *
B *---------------*---------------*
: - - 100 - - - D - - - x - - - S
The two poles are: . $AB \,=\,CD\,=\,30$
. . and: . $BD = 100$
The light at $A$ shines on the repairman $R$ and casts his shadow at $S.$
. . Let: . $x \,=\,DS,\;y \,=\,RD$
Since $\Delta RDS \sim\Delta ABS\!:\;\;\frac{x}{y} \:=\:\frac{x+100}{30} \quad\Rightarrow\quad x \:=\:\frac{100y}{30-y}$
Differentiate with respect to time: . $\frac{dx}{dt} \;=\;\frac{3000}{(30-y)^2}\,\frac{dy}{dt}$
Now substitute: . $y = 16,\;\;\frac{dy}{dt} = 0.5$ . . . | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771805808551,
"lm_q1q2_score": 0.8590855262240585,
"lm_q2_score": 0.8705972768020108,
"openwebmath_perplexity": 883.8461896641982,
"openwebmath_score": 0.6901357769966125,
"tags": null,
"url": "http://mathhelpforum.com/calculus/60919-related-rates-cont-d.html"
} |
# Prove that $\sinh(\cosh(x)) \geq \cosh(\sinh(x))$
Prove that
$$\sinh(\cosh(x)) \geq \cosh(\sinh(x))$$
I tried to tackle this problem by integrating both lhs and rhs, in order to get two functions who show clearly that inequality holds. I've struggled for this problem a little bit, i don't know if there's any trick that can help. Maybe knowing that
$$\cosh^{-1}(x) = \pm \ln\left(x + \sqrt{x^2 - 1}\right)$$
Can help?
• Jensen is useful only when we have a function and a constant, not with composite functions... Even if I think there are some similarities... – james42 May 26 '15 at 18:11
• cheers for teaching me something new :). – Chinny84 May 26 '15 at 18:12
• Physics guy here. I don't have any idea how to do math proofs, but isn't it true that a "small function" acting on a "big function" is always larger than the alternative? Since cosh(x)>sinh(x) for all x, this is a no brainer, right? This is how I remember e^pi>pi^e. Anyway, I've never spoken on Math.SE and generally run in fear from mathematicians, so please, let me down gently! – user1717828 May 27 '15 at 4:03
• @user1717828 $\ln x$ is smaller than $x^2$, but $\ln (x^2) = 2\ln x$ is smaller than $(\ln x)^2$. – user21467 May 27 '15 at 4:21
• @StevenTaschuk, aaaaannnnnddd this is why our field would crumble without mathematicians keeping us in check. Thanks! – user1717828 May 27 '15 at 4:43
For any $y \ge 0$, notice
$$e^y - 1 = \int_0^y e^x dx \ge \int_0^y (1+x) dx \ge \int_0^y \left(1+\frac{x} {\sqrt{1+x^2}}\right)dx = y + \sqrt{1+y^2} - 1$$ we have this little inequality: $$\sqrt{1+y^2} - y = \frac{1}{\sqrt{1+y^2} + y} \ge e^{-y}$$
Using MVT, we can find a $\xi \in (y,\sqrt{1+y^2})$ such that
$$\sinh\sqrt{1+y^2} - \sinh(y) = \cosh(\xi)\left(\sqrt{1+y^2} - y\right) \ge \cosh(\xi) e^{-y} \ge e^{-y}$$
Since $e^{-y} = \cosh(y) - \sinh(y)$, this leads to $$\sinh\sqrt{1+y^2} \ge \cosh(y)\\$$
Substitute $y$ by $\sinh(x)$ and notice $\sqrt{1+y^2} = \cosh(x)$, this reduces to our desired inequality: | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771801919951,
"lm_q1q2_score": 0.8590855159463234,
"lm_q2_score": 0.8705972667296309,
"openwebmath_perplexity": 568.0134092269792,
"openwebmath_score": 0.9998539686203003,
"tags": null,
"url": "https://math.stackexchange.com/questions/1299822/prove-that-sinh-coshx-geq-cosh-sinhx"
} |
$$\sinh(\cosh(x)) \ge \cosh(\sinh(x))$$
Here's a solution with very little calculus. First, an identity: \begin{align*} \sinh^2(a+b) - \sinh^2(a-b) &= (\sinh a\cosh b + \cosh a\sinh b)^2 - (\sinh a\cosh b - \cosh a\sinh b)^2 \\ &= 4\sinh a \cosh a\sinh b \cosh b \\ &= \sinh(2a)\sinh(2b) \end{align*} Taking $a=e^x/2$ and $b=e^{-x}/2$, we get \begin{align*} \sinh^2(\cosh x) - \sinh^2(\sinh x) &= \sinh(e^x) \sinh(e^{-x}) \\ &\ge e^xe^{-x} &&\text{(since $\sinh t\ge t$ for $t\ge 0$)} \\ &= 1 \\ &= \cosh^2(\sinh x) - \sinh^2(\sinh x) \end{align*} Cancelling $\sinh^2(\sinh x)$ and taking square roots gives the desired inequality.
Calculus is needed here only to justify the inequality $\sinh t\ge t$ (for $t\ge 0$).
Update: Another nice thing about this method is that it points the way to a more exact inequality. It turns out that $$\sinh u\sinh v \ge \sinh^2\sqrt{uv}$$ for $u,v\ge 0$. (Proof 1: $\frac12(u^{2m+1}v^{2n+1} + v^{2m+1}u^{2n+1})\ge (uv)^{m+n+1}$ by AM/GM; divide by $(2m+1)!\,(2n+1)!$ and apply $\sum_{m=0}^\infty \sum_{n=0}^\infty$. Proof 2: Check that $t\mapsto\ln\sinh(e^t)$ is convex (for all $t$) by computing its second derivative.)
Applying this with $u=e^x$ and $v=e^{-x}$ above, we get $$\sinh^2(\cosh x) \ge \cosh^2(\sinh x) + \underbrace{\sinh^2(1) - 1}_{\approx 0.3811}$$ with equality when $x=0$.
• (+1) Very nice. Since $\cosh(x)\gt0$, we don't need to worry about signs when taking square roots. – robjohn May 27 '15 at 4:22
Let $t=\sinh x$. Now we can square the inequality and instead try proving $$\sinh^2(\sqrt{1+t^2})=\sinh^2(\cosh x) \ge \cosh^2(\sinh x)=1+\sinh^2t$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771801919951,
"lm_q1q2_score": 0.8590855159463234,
"lm_q2_score": 0.8705972667296309,
"openwebmath_perplexity": 568.0134092269792,
"openwebmath_score": 0.9998539686203003,
"tags": null,
"url": "https://math.stackexchange.com/questions/1299822/prove-that-sinh-coshx-geq-cosh-sinhx"
} |
So it is enough to show $f(t) = \sinh^2(\sqrt{1+t^2})-\sinh^2t-1 \ge 0$. As $f$ is even and $f(0)> 0$, it is enough to show it is increasing for positive $t$. Hence we look at $$f'(t) = \frac{t\sinh (2\sqrt{1+t^2})}{\sqrt{1+t^2}} - \sinh(2t)$$ To show this is positive, it suffices to note by differentiating that the function $g(t) = \dfrac{\sinh t}{t}$ is increasing, so $g(2\sqrt{1+t^2})> g(2t)$. Hence proved...
For $x\ge0$, the Mean Value Theorem says that for some $\sinh(x)\lt\xi\lt\cosh(x)$, \begin{align} \sinh(\cosh(x))-\sinh(\sinh(x)) &=\cosh(\xi)(\cosh(x)-\sinh(x))\\ &\gt\cosh(\sinh(x))\,e^{-x}\tag{1} \end{align} Furthermore, $$\cosh(\sinh(x))-\sinh(\sinh(x))=e^{-\sinh(x)}\tag{2}$$ Therefore, subtracting $(2)$ from $(1)$, then applying $\cosh(x)\ge1$ and $\sinh(x)\ge x$, we get \begin{align} \sinh(\cosh(x))-\cosh(\sinh(x)) &\gt e^{-x}\cosh(\sinh(x))-e^{-\sinh(x)}\\ &\ge e^{-x}-e^{-\sinh(x)}\\ &\ge0\tag{3} \end{align} Since $\sinh(\cosh(x))-\cosh(\sinh(x))$ is even, $(3)$ implies that strict inequality holds for all $x$: $$\sinh(\cosh(x))\gt\cosh(\sinh(x))\tag{4}$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771801919951,
"lm_q1q2_score": 0.8590855159463234,
"lm_q2_score": 0.8705972667296309,
"openwebmath_perplexity": 568.0134092269792,
"openwebmath_score": 0.9998539686203003,
"tags": null,
"url": "https://math.stackexchange.com/questions/1299822/prove-that-sinh-coshx-geq-cosh-sinhx"
} |
# Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction
How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
• @Steve: See this answer for general comments on induction, and this one for specific advice on doing proofs by induction. The example there may be enough for you to figure out how to prove this statement by induction. – Arturo Magidin Sep 6 '11 at 3:34
• Since this question is asked quite frequently, it has been added to the list of Generalizations of Common questions. It has been kept seperate from the version which does not use induction. – Eric Naslund Aug 30 '12 at 0:23
• This answer has a great discussion of how to do an induction proof – Ross Millikan Oct 30 '17 at 21:09
• take paper and pencil and start with $\sum_{i=0}^{n+1 } i^3$. (and you need not bother with $i=0$) – Mirko Apr 19 '18 at 21:13
• I would say this must have been asked on this website at least $10$ times. – Clement C. Apr 19 '18 at 21:14
You are trying to prove something of the form, $$A=B.$$ Well, both $A$ and $B$ depend on $n$, so I should write, $$A(n)=B(n).$$ First step is to verify that $$A(1)=B(1).$$ Can you do that? OK, then you want to deduce $$A(n+1)=B(n+1)$$ from $A(n)=B(n)$, so write out $A(n+1)=B(n+1)$. Now you're trying to get there from $A(n)=B(n)$, so what do you have to do to $A(n)$ to turn it into $A(n+1)$, that is (in this case) what do you have to add to $A(n)$ to get $A(n+1)$? OK, well, you can add anything you like to one side of an equation, so long as you add the same thing to the other side of the equation. So now on the right side of the equation, you have $B(n)+{\rm something}$, and what you want to have on the right side is $B(n+1)$. Can you show that $B(n)+{\rm something}$ is $B(n+1)$? | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771774699747,
"lm_q1q2_score": 0.8590855052938772,
"lm_q2_score": 0.8705972583359805,
"openwebmath_perplexity": 524.842176201216,
"openwebmath_score": 0.9820051193237305,
"tags": null,
"url": "https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct"
} |
Let the induction hypothesis be $$(1^3+2^3+3^3+\cdots+n^3)=(1+2+3+\cdots+n)^2$$ Now consider: $$(1+2+3+\cdots+n + (n+1))^2$$ \begin{align} & = \color{red}{(1+2+3+\cdots+n)^2} + (n+1)^2 + 2(n+1)\color{blue}{(1+2+3+\cdots+n)}\\ & = \color{red}{(1^3+2^3+3^3+\cdots+n^3)} + (n+1)^2 + 2(n+1)\color{blue}{(n(n+1)/2)}\\ & = (1^3+2^3+3^3+\cdots+n^3) + \color{teal}{(n+1)^2 + n(n+1)^2}\\ & = (1^3+2^3+3^3+\cdots+n^3) + \color{teal}{(n+1)^3} \end {align} QED
• @RajeshKSingh let me know if you have questions about the steps of the proof. – user2468 Aug 29 '12 at 23:08
• Hint: If $(1^3+2^3+3^3+\cdots+n^3)=(1+2+3+\cdots+n)^2$, then \begin{eqnarray} (1^3+2^3+3^3+\cdots+n^3 + (n+1)^3)&=&(1+2+3+\cdots+n)^2 +(n+1)^3\\ &=&(\dfrac{n^2(n+1)^2}{2} + (n+1)^3\\ &=& (n+1)^2(\dfrac{n^2}{2}+n+1)\\ &=& (n+1)^2((n+1)^2+1)/2\\ &=& (1+2+ \cdots +n +(n+1))^2 \end{eqnarray} – user29999 Aug 29 '12 at 23:14
• @RajeshKSingh No. If you want to prove a statement about all natural numbers then you need induction. – user2468 Aug 29 '12 at 23:16
• @JenniferDylan: the steps are clear to me. is there a method different from induction? – HOLYBIBLETHE Aug 29 '12 at 23:25
• @user29999 You didn't square the denominator. D: – Simply Beautiful Art Dec 9 '16 at 21:44
HINT $\:$ First trivially inductively prove the Fundamental Theorem of Difference Calculus
$$\rm\ F(n)\ =\ \sum_{i\: =\: 1}^n\:\ f(i)\ \ \iff\ \ \ F(n) - F(n-1)\ =\ f(n),\quad\ F(0) = 0$$
The result now follows immediately by $\rm\ F(n)\ =\ (n\:(n+1)/2)^2\ \Rightarrow\ F(n)-F(n-1)\ =\: n^3\:.\$
Note that by employing the Fundamental Theorem we have reduced the proof to the trivial mechanical verification of a polynomial equation. No ingenuity is required.
Note that the proof of the Fundamental Theorem is much more obvious than that for your special case because the telescopic cancellation is obvious at this level of generality, whereas it is usually obfuscated in most specific instances. For further discussion see my many posts on telescopy. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771774699747,
"lm_q1q2_score": 0.8590855052938772,
"lm_q2_score": 0.8705972583359805,
"openwebmath_perplexity": 524.842176201216,
"openwebmath_score": 0.9820051193237305,
"tags": null,
"url": "https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct"
} |
• I got here from your latest post, how do you find $F(n)$ ? do you have a method or do you just 'see' this ? – Belgi Sep 19 '12 at 10:36
• @Belgi Here the sum closed form $\rm F(n)$ is given and we are merely asked to verify it correctness. Computing the closed form is known as summation in finite terms and there are known algorithms for wide-classes of special functions, e.g. chase citations to Michael Karr's papers on "Summation in finite terms". – Bill Dubuque Sep 4 '17 at 23:32
Let P(n) be the given statement. You'll see why in the following step. $$P(n):1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$$
Step 1. Let $n = 1$.
Then $\mathrm{LHS} = 1^3 = 1$, $\mathrm{RHS} = \frac{1^2(1+1)^2}{4} = \frac{4}{4} = 1$.
So LHS = RHS, and this means P(1) is true!
Step 2. Let $P(n)$ be true for $n = k$; that is, $$1^3 + 2^3 + 3^3 + \cdots + k^3 = \frac{k^2(k+1)^2}{4}$$
We shall show that $P(k+1)$ is true too!
Add $(k+1)^3$, which is $(k+1)^{\mathrm th}$ term of the LHS to both sides of (1); then we get: \begin{align*} 1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\ &= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\ &= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\ &= \frac{(k+1)^2(k+2)^2}{4}.\\ 1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{(k+1)^2(k+2)^2}{4} \end{align*} I think this statement is the same as $P(n)$ with $n = k+1$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771774699747,
"lm_q1q2_score": 0.8590855052938772,
"lm_q2_score": 0.8705972583359805,
"openwebmath_perplexity": 524.842176201216,
"openwebmath_score": 0.9820051193237305,
"tags": null,
"url": "https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct"
} |
• Is there any particular reason why you are writing in bold face? It's a bit distracting, and not the usual for this site... Note also that it doesn't work for math formulas, so that makes it even more distracting... – Arturo Magidin Sep 6 '11 at 4:14
• I wanted the math formulas to appear in bold face,but the opposite is happening.don't worry i'll edit it. – alok Sep 6 '11 at 4:17
• You marked not just the equations, also all the text. And why was it important for the formulas to be in boldface? Again, not the usual for this site, and they don't look the same: $k$ vs. $\mathbf{k}$, $+$ vs. $\mathbf{+}$, $\cdots$ vs. $\mathbf{\cdots}$. I've cleaned it up a bit. – Arturo Magidin Sep 6 '11 at 4:21
This picture shows that $$1^2=1^3\\(1+2)^2=1^3+2^3\\(1+2+3)^2=1^3+2^3+3^3\\(1+2+3+4)^2=1^3+2^3+3^3+4^3\\$$ this is handmade of mine
For proof by induction; these are the $\color{green}{\mathrm{three}}$ steps to carry out:
Step 1: Basis Case: For $i=1 \implies \sum^{i=k}_{i=1} i^3=\frac{1^2 (1+1)^2}{4}=\cfrac{2^2}{4}=1$. So statement holds for $i=1$.
Step 2: Inductive Assumption: Assume statement is true for $i=k$: $$\sum^{i=k}_{i=1} i^3=\frac{k^2 (k+1)^2}{4}$$
Step 3: Prove Statement holds for $i=k+1$. You need to prove that for $i=k+1$: $$\sum^{i=k+1}_{i=1} i^3=\color{blue}{\frac{(k+1)^2 (k+2)^2}{4}}$$ To do this you cannot use: $$\sum^{i=n}_{i=1} i^3=\color{red}{\frac{n^2 (n+1)^2}{4}}$$ as this is what you are trying to prove.
So what you do instead is notice that: $$\sum^{i=k+1}_{i=1} i^3= \underbrace{\frac{k^2 (k+1)^2}{4}}_{\text{sum of k terms}} + \underbrace{(k+1)^3}_{\text{(k+1)th term}}$$ $$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{1}{4}k^2+(k+1)\right)$$ $$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{k^2+4k+4}{4}\right)$$ $$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{(k+2)^2}{4}\right)=\color{blue}{\frac{(k+1)^2 (k+2)^2}{4}}$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771774699747,
"lm_q1q2_score": 0.8590855052938772,
"lm_q2_score": 0.8705972583359805,
"openwebmath_perplexity": 524.842176201216,
"openwebmath_score": 0.9820051193237305,
"tags": null,
"url": "https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct"
} |
Which is the relation we set out to prove. So the method is to substitute $i=k+1$ into the formula you are trying to prove and then use the inductive assumption to recover the $\color{blue}{\mathrm{blue}}$ equation at the end.
(QED is an abbreviation of the Latin words "Quod Erat Demonstrandum" which loosely translated means "that which was to be demonstrated". It is usually placed at the end of a mathematical proof to indicate that the proof is complete.)
If $S_r= 1^r+2^r+...+n^r=f(n)$ and
$\sigma =n(n+1)$ & $\sigma'=2n+1$
$S_1=\frac{1}{2}\sigma$
$S_2=\frac{1}{6}\sigma\sigma'$
$S_3=\frac{1}{4}\sigma^2$
$S_4=\frac{1}{30}\sigma\sigma'(3\sigma-1)$
$S_5=\frac{1}{12}\sigma^2(2\sigma-1)$
$S_6=\frac{1}{42}\sigma\sigma'(3\sigma^2-3\sigma+1)$
$S_7=\frac{1}{24}\sigma^2(3\sigma^2-4\sigma+2)$
$S_8=\frac{1}{90}\sigma\sigma'(5\sigma^3-10\sigma^2+9\sigma-3)$
$S_9=\frac{1}{20}\sigma^2(2\sigma^3-5\sigma^2+6\sigma-3)$
$S_{10}=\frac{1}{66}\sigma\sigma'(3\sigma^4-10\sigma^3+17\sigma^2-15\sigma+5)$
proof of this is based on the theorem if r is a positive integer, $s_r$ can be expressed as a polynomial in $n$ of which the highest term in $\frac{n^{r+1}}{r+1}$
• if you are interested we can discuss about the proof.! – user229886 Apr 9 '15 at 6:58
• you can also see page 84-85-,86 of proofs without word by roger B. Nelson ,ISBN 088385-700-6 – Khosrotash Jul 21 '15 at 6:51 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771774699747,
"lm_q1q2_score": 0.8590855052938772,
"lm_q2_score": 0.8705972583359805,
"openwebmath_perplexity": 524.842176201216,
"openwebmath_score": 0.9820051193237305,
"tags": null,
"url": "https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct"
} |
It may be helpful to recognize that both the RHS and LHS represent the sum of the entries in a the multiplication tables. The LHS represents the summing of Ls (I'll outline those shortly), and the RHS, the summing of the sum of the rows [or columns])$$\begin{array}{lll} \color{blue}\times&\color{blue}1&\color{blue}2\\ \color{blue}1&\color{green}1&\color{red}2\\ \color{blue}2&\color{red}2&\color{red}4\\ \end{array}$$ Lets begin by building our multiplication tables with a single entry, $1\times1=1=1^2=1^3$. Next, we add the $2$s, which is represented by the red L [$2+4+2 = 2(1+2+1)=2\cdot2^2=2^3$]. So the LHS (green 1 + red L) currently is $1^3+2^3$, and the RHS is $(1+2)+(2+4)=(1+2)+2(1+2)=(1+2)(1+2)=(1+2)^2$. $$\begin{array}{llll} \color{blue}\times&\color{blue}1&\color{blue}2&\color{blue}3\\ \color{blue}1&\color{green}1&\color{red}2&\color{maroon}3\\ \color{blue}2&\color{red}2&\color{red}4&\color{maroon}6\\ \color{blue}3&\color{maroon}3&\color{maroon}6&\color{maroon}9\\ \end{array}$$ Next, lets add the $3$s L. $3+6+9+6+3=3(1+2+3+2+1)=3\cdot3^2=3^3$. So now the LHS (green 1 + red L + maroon L) currently is $1^3+2^3+3^3$, and the RHS is $(1+2+3)+(2+4+6)+(3+6+9)=(1+2+3)+2(1+2+3)+3(1+2+3)=(1+2+3)(1+2+3)=(1+2+3)^2$.
By now, we should see a pattern emerging that will give us direction in proving the title statement.
Next we need to prove inductively that $\displaystyle\sum_{i=1}^n i = \frac{n(n+1)}{2}$, and use that relationship to show that $1+2+3+\dots+n+\dots+3+2+1 = \dfrac{n(n+1)}{2}+ \dfrac{(n-1)n}{2} = \dfrac{n((n+1)+(n-1))}{2}=\dfrac{2n^2}{2}=n^2$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771774699747,
"lm_q1q2_score": 0.8590855052938772,
"lm_q2_score": 0.8705972583359805,
"openwebmath_perplexity": 524.842176201216,
"openwebmath_score": 0.9820051193237305,
"tags": null,
"url": "https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct"
} |
Finally, it should be straight forward to show that: $$\begin{array}{lll} (\sum^n_{i=1}i+(n+1))^2 &=& (\sum^n_{i=1}i)^2 + 2\cdot(\sum^n_{i=1}i)(n+1)+(n+1)^2\\ &=& \sum^n_{i=1}i^3 + (n+1)(\sum^n_{i=1}i + (n+1) + \sum^n_{i=1}i)\\ &=& \sum^n_{i=1}i^3 + (n+1)(n+1)^2\\ &=& \sum^n_{i=1}i^3 + (n+1)^3\\ &=& \sum^{n+1}_{i=1}i^3\\ \end{array}$$ and, as was already pointed out previously, $$(\sum_{i=1}^1 i)^2 = \sum_{i=1}^1 i^3=1$$
$$n^3=S_n-S_{n-1}=\left(\frac{n(n+1)}2\right)^2-\left(\frac{n(n-1)}2\right)^2=n^2\left(\frac{n+1}2-\frac{n-1}2\right)\left(\frac{n+1}2+\frac{n-1}2\right)\\ =n^2\cdot1\cdot n.$$
This is the inductive step. The rest is easy.
[All: This answer came here when I merged duplicates. Please take that into account when voting, JL]
HINT
We need to prove
• Base case: set $i=1$ and check by inspection
• Induction step: assume $\sum_{i=0}^n i^3 = (\sum_{i=0}^n i)^2$ true prove
$$\sum_{i=0}^{n+1} i^3 = \left(\sum_{i=0}^{n+1} i\right)^2$$
Note that
• $\sum_{i=0}^{n+1} i^3=(n+1)^3+\sum_{i=0}^{n} i^3$
• $\left(\sum_{i=0}^{n+1} i\right)^2=\left(n+1+\sum_{i=0}^{n} i\right)^2$
• It helps to recall that $1+2+ \ldots +n = n(n+1)/2.$ – Chris Leary Apr 19 '18 at 21:20
• @ChrisLeary Oh yes, I supposed that should be well known, anyway that's good to recall! Thanks – gimusi Apr 19 '18 at 21:23
• Sorry, that comment was for the OP, not you. I should have addressed it directly to Ryan. – Chris Leary Apr 20 '18 at 2:10
[All: This answer came here when I merged duplicates. Please take that into account when voting, JL]
You should prove before $$\sum_{i=0}^n i = \frac{n(n+1)}{2}$$
Cases $$n=0,1$$ are trivial. Suppose it's true for $$n-1$$ then \begin{align} \sum_{i=0}^n i^3 &= n^3 + \sum_{i=0}^{n-1}i^3 \\ &=n^3 + \left(\sum_{i=0}^{n-1} i\right)^2 \\ &= n^3 + \frac{n^2(n-1)^2}{4}\\ & = \frac{4n^3 + n^4 + n^2 + -2n^3}{4}\\ &= \frac{(n+1)^2n^2}{4} \\ &= \left(\sum_{i=0}^{n} i\right)^2 \end{align}
$$\ S =\sum_{k=1}^n k^3$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771774699747,
"lm_q1q2_score": 0.8590855052938772,
"lm_q2_score": 0.8705972583359805,
"openwebmath_perplexity": 524.842176201216,
"openwebmath_score": 0.9820051193237305,
"tags": null,
"url": "https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct"
} |
$$\ S =\sum_{k=1}^n k^3$$
\begin{align} k^3 =& Ak(k+1)(k+2)+Bk(k+1)+Ck+D \\ k^3= & Ak^3+(3A+B)k^2+(2A+B+C)k+D \end{align} Therefore $A=1$, $B=-3$, $C=1$, $D=0$ and
\begin{align} S =& \sum_{k=1}^n (k(k+1)(k+2)-3k(k+1)+k)\\ S =&\sum_{k=1}^n k(k+1)(k+2)-3\sum_{k=1}^nk(k+1)+\sum_{k=1}^nk \\ S =&\sum_{k=1}^n6\binom{k+2}{3}-3\sum_{k=1}^n2\binom{k+1}{2}+\sum_{k=1}^n\binom{k}{1}\\ S=&6\binom{n+3}{4}-6\binom{n+2}{3}+\binom{n+1}{2}\\ S=&\left\lgroup\frac{n(n+1)}{2}\right\rgroup^2. \end{align}
• It seems that induction has been asked for. – Claude Leibovici Jul 9 '16 at 2:57
• The question clearly asks for a proof using P.M.I. – Shubh Khandelwal Aug 24 '18 at 4:18
## protected by Alex M.Aug 29 '16 at 15:16
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead? | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9867771774699747,
"lm_q1q2_score": 0.8590855052938772,
"lm_q2_score": 0.8705972583359805,
"openwebmath_perplexity": 524.842176201216,
"openwebmath_score": 0.9820051193237305,
"tags": null,
"url": "https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct"
} |
Does knowing that you drew at least $1$ ace increase the probability that you drew at least $2$ aces ($5$ card hand)?
So say I draw $$5$$ cards from a standard deck at random without replacement and without looking at them. If someone told me that the hand contains at least $$1$$ ace, would that change the probability that my hand contains at least $$2$$ aces?
I did the math and I believe the probability of drawing at least $$2$$ aces is: $$1 - \dfrac{\binom{48}{5}}{\binom{52}{5}} - \dfrac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}}$$.
Additionally, what if that person then told me that the hand contains the ace of spades; would knowing that there is at least one ace and that it's the ace of spades change the probability of having at least $$2$$ aces in my hand?
• I disagree with the assertion for the second part that knowing that its is an ace of spades does not change the situation. I have put up an answer that explains this. Apr 21 at 4:04
• Have a look at the added approach. Apr 21 at 7:18
It is not correct to say that whether you are told that an ace is in the hand or whether you are told that the ace of spade is in the hand, the probability of having at least two aces is the same
Case 1: You are told that you have at least one ace
Note that in the following formula, the sample space shrinks to $$\binom{52}5 - \binom{48}5$$ to take into account that at least one ace must be present, and that in the numerator we need to ensure that one ace is there
$$Pr = 1 - \dfrac{\binom41\binom{48}4}{\binom{52}5 - \binom{48}5},\;\approx 0.1222$$
Case 2: You are told that you have the ace of spades
Sinca ace of spades is assured, the sample space is now simply $$\binom{51}{4}$$, and from the numerator, we just subtract $$\binom{48}4$$ to indicate that more aces must be present.
Thus $$Pr = 1 - \dfrac{\binom{48}4}{\binom{51}4}\; \approx 0.2214$$
Since probability problems are often "tricky", here is an approach simpler to understand, though longer computationally. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.979035762240297,
"lm_q1q2_score": 0.8590811601858227,
"lm_q2_score": 0.8774767922879693,
"openwebmath_perplexity": 201.62139745915107,
"openwebmath_score": 0.8559759855270386,
"tags": null,
"url": "https://math.stackexchange.com/questions/4109544/does-knowing-that-you-drew-at-least-1-ace-increase-the-probability-that-you-dr"
} |
Basically, the numerator will count hands with $$2$$ or more aces, and the denominator, hands with $$1$$ or more aces
Case 1: You are told that you have at least one ace
$$Pr = \dfrac{\binom42\binom{48}3 + \binom43\binom{48}2 +\binom44\binom{48}1}{\binom41\binom{48}4+\binom42\binom{48}3 + \binom43\binom{48}2 +\binom44\binom{48}1}\approx 0.1222$$
Case 2: You are told that you have the ace of spades
Bear in mind that you have the ace of spades and are left to pick from $$51$$ cards including $$3$$ aces
$$Pr = \dfrac{\binom31\binom{48}3 + \binom32\binom{48}2 +\binom33\binom{48}1}{\binom30\binom{48}4+\binom31\binom{48}3 + \binom32\binom{48}2 +\binom33\binom{48}1} \approx 0.2214$$
• This is correct for the second part of the problem, which I think is the more nonintuitive part. +1 and I wish I could do more. Apr 21 at 1:51
• I agree with this calculation now. But intuitively it makes no sense. If you show someone you have an Ace but cover up the suit with your finger, then revealing the suit should have no effect on the probability. But it does. Why is this? Apr 21 at 12:53
• @AdamRubinson: covering up and revealing the suit is different from specifying one in advance. Half of the cases with exactly two aces include the ace of spades while only one quarter of the cases with exactly one ace do. The ace of spades reduces the denominator more than it does the numerator. Apr 21 at 13:41
Intuitively, you need to draw at least one ace to draw two or more aces, so knowing you have drawn at least one increases the probability you have drawn more than one. The event that your hand contains a spade also implies you drew at least one ace so the probability of two aces must also increase (edit: I should clarify that, as Ross and true blue nail have pointed out, the probability of two aces is not the same in this case as in the previous). | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.979035762240297,
"lm_q1q2_score": 0.8590811601858227,
"lm_q2_score": 0.8774767922879693,
"openwebmath_perplexity": 201.62139745915107,
"openwebmath_score": 0.8559759855270386,
"tags": null,
"url": "https://math.stackexchange.com/questions/4109544/does-knowing-that-you-drew-at-least-1-ace-increase-the-probability-that-you-dr"
} |
Here is a math argument for the first case you asked about. The conditional probability of drawing at least two aces '$$A>1$$' given we drew at least one ace '$$A>0$$' is $$P(A>1|A>0) = \frac{P(A>1,A>0)}{P(A>0)} = \frac{P(A>1)}{P(A>0)}$$ since $$A>1$$ implies $$A>0$$ and so $$P(A>1,A>0) = P(A>1)$$. Whereas the probability of drawing two aces is just $$P(A>1)$$ which is less that $$\frac{P(A>1)}{P(A>0)}$$ since $$P(A>0)$$ is less than $$1$$ (because not every hand contains an ace).
• Ooh that makes a lot of sense. So the calculation I made was really for P(A>1), and I need to divide it by the probability of drawing at least 1 ace? Apr 20 at 12:09
• Yes, you can just divide the calculation you made for $P(A>1)$ by $P(A>0) = 1 - \frac{\binom{48}{5}}{\binom{52}{5}}$. In general, when you want the conditional probability $P(A|B)$ of event $A$ given event $B$ you need to divide the probability $P(A, B)$ of both $A$ and $B$ happening by the probability $P(B)$ of $B$ happening. The situation in your question where you can just divide $P(A)$ by $P(B)$ occurs whenever event $A$ implies event $B$. Apr 20 at 12:27
• You imply that changing the condition from an unknown ace to the ace of spades does not change the probability of two aces. This is incorrect. The ace of spades is present in half the cases of exactly two aces but only one quarter of the cases of one ace. That almost doubles the probability of two aces given one. Apr 21 at 1:55
• @RossMillikan You are correct and my answer could be better if I was more specific in the spades case. I think true blue nail has covered that case well already now though. I only meant to imply that in both cases the probability of two aces was different than from the situation where you have no knowledge of the number of aces in your hand, not that they were exactly the same. Apr 21 at 9:37 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.979035762240297,
"lm_q1q2_score": 0.8590811601858227,
"lm_q2_score": 0.8774767922879693,
"openwebmath_perplexity": 201.62139745915107,
"openwebmath_score": 0.8559759855270386,
"tags": null,
"url": "https://math.stackexchange.com/questions/4109544/does-knowing-that-you-drew-at-least-1-ace-increase-the-probability-that-you-dr"
} |
# Integration Question: Differentiating a definite integral
#### ISITIEIW
##### New member
So the question is…Evaluate the following…
$$\displaystyle \frac{d}{dx} \left(\int _1^{x^2} \cos(t^2) \, dt \right)$$
I thought i could use the FTC on this because it states…
$$\displaystyle \frac{d}{dx} \left(\int_0^x f(t)\, dt \right)=f(x)$$
but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem
Thanks
Last edited by a moderator:
#### Jameson
Staff member
Re: Integration Question
Since the lower bound is still a constant, it doesn't change the method you use to evaluate this. It could be at 0,1,2 or any constant and still get the same result. You're going to integrate and get a constant and then take the derivative of that constant, so you end up with 0 regardless.
Here you'll need to be careful to use the chain rule when applying the FTOC. What progress have you made so far?
#### MarkFL
Staff member
Re: Integration Question
So the question is…Evaluate the following…
d/dx(integration of 1 to x^2 of cos(t^2)dt)
I thought i could use the FTC on this because it states…
d/dx(integration of 0 to x of f(t)dt)
but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem
Thanks
We are given to evaluate:
$$\displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt$$
Now, the anti-derivative form of the FTOC tells us:
$$\displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=\frac{d}{dx}\left(F(h(x))-F(g(x)) \right)$$
On the right, applying the chain rule, we obtain:
$$\displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=F'(h(x))h'(x)-F'(g(x))g'(x)$$
Since $$\displaystyle F'(x)=f(x)$$, we may write:
$$\displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=f(h(x))h'(x)-f(g(x))g'(x)$$
Applying this to the given problem, there results: | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9790357542883922,
"lm_q1q2_score": 0.8590811547766679,
"lm_q2_score": 0.877476793890012,
"openwebmath_perplexity": 753.0663369682322,
"openwebmath_score": 0.9428406357765198,
"tags": null,
"url": "https://mathhelpboards.com/threads/integration-question-differentiating-a-definite-integral.7749/"
} |
Applying this to the given problem, there results:
$$\displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt=\cos\left(\left(x^2 \right)^2 \right)(2x)-\cos\left(\left(1 \right)^2 \right)(0)$$
Simplifying, we find:
$$\displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt=2x\cos\left(x^4\right)$$
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem
Let us evaluate $$\displaystyle \frac{d}{dx} \left(\int_a^x f(t)\, dt \right)$$ where $$\displaystyle 0<a<x$$
Then since $$\displaystyle \int_0^x f(t)\, dt =\int_0^a f(t)\, dt+ \int_a^x f(t)\, dt$$
$$\displaystyle \int_a^x f(t)\, dt=-\int_0^a f(t)\, dt+\int_0^x f(t)\, dt$$
If we differentiate then since the first integral is independent of $x$ we have
$$\displaystyle \frac{d}{dx} \left(\int_a^x f(t)\, dt \right)=\frac{d}{dx} \left(\int_0^x f(t)\, dt \right)=f(x)$$
#### HallsofIvy
##### Well-known member
MHB Math Helper
The "Leibniz formula", a generalization of the fundamental theorem of calculus, says
$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= \frac{d\beta}{dx}f(x, \beta(x))- \frac{d\alpha}{dx}f(x, \alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x} dt$$ | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9790357542883922,
"lm_q1q2_score": 0.8590811547766679,
"lm_q2_score": 0.877476793890012,
"openwebmath_perplexity": 753.0663369682322,
"openwebmath_score": 0.9428406357765198,
"tags": null,
"url": "https://mathhelpboards.com/threads/integration-question-differentiating-a-definite-integral.7749/"
} |
# Can a graph be strongly and weakly connected?
I'm currently revising course notes on directed graphs.
It says that a directed graph (digraph) is strongly connected if there is a path between every pair of vertices.
It also says that a digraph is weakly connected if the underlying undirected graph is connected.
My question is, can one digraph be both strongly and weakly connected?
For example: Digraph and undirected graph
Can this graph (image) be both strongly and weakly connected? or does it have to be either strongly, or either weakly?
Thank you.
As suggested by the terminology, any strongly connected graph is weakly connected, but a weakly connected graph is not necessarily strongly connected. For instance, the graph $1 \to 2$ is weakly connected but is not strongly connected.
Yes, a graph can, according to the provided definitions, definitely be both weakly and strongly connected at the same time. Your example is exactly such a graph.
In fact, all strongly connected graphs are also weakly connected, since a directed path between two vertices still connect the vertices upon removing the directions.
To some extent this is a question about word usage in mathematics. One natural reading of the English term "weakly connected" would be "the connectedness of the graph is weak", weak meaning that it is deficient in some sense like a weak signal.
However, the use of "weakly" here is more of a conjugation of the term "weak connectedness", i.e. connected in a weak sense of the word "connected", with weak meaning that it is not a very strenuous requirement. A graph is weakly connected if it satisfies this weak connectedness requirement, not because it is less strongly connected than other graphs. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9790357561234474,
"lm_q1q2_score": 0.8590811532499719,
"lm_q2_score": 0.8774767906859264,
"openwebmath_perplexity": 415.9837092978464,
"openwebmath_score": 0.7758089900016785,
"tags": null,
"url": "https://math.stackexchange.com/questions/1680354/can-a-graph-be-strongly-and-weakly-connected"
} |
This is a fairly common idiom in math. In functional analysis one might refer to a sequence that is "weakly convergent" (again meaning convergent in a more generous sense that has a precise definition), and then later show that the same sequence is in fact strongly convergent.
There are other cases where the terminology is indeed exclusive, unlike the present case: when an infinite series is "conditionally convergent", the definition specifically excludes the case that it is absolutely convergent (which is a stronger notion of convergence). | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9790357561234474,
"lm_q1q2_score": 0.8590811532499719,
"lm_q2_score": 0.8774767906859264,
"openwebmath_perplexity": 415.9837092978464,
"openwebmath_score": 0.7758089900016785,
"tags": null,
"url": "https://math.stackexchange.com/questions/1680354/can-a-graph-be-strongly-and-weakly-connected"
} |
Find the range of $A$ if $A=\sin^{20}x+\cos^{48}x$
Find the range of $$A$$ if $$A=\sin^{20}x+\cos^{48}x$$
$$A'=20\sin^{19}x\cos x-48\cos^{47}x\sin x=0\implies5\sin^{19}x\cos x=12\cos^{47}x\sin x\\ \implies5\sin^{18}x=12\cos^{46}x$$
How do I proceed further and prove that $$A\in(0,1]$$ ?
Is it possible to find the range of $$A$$ without using differentiation ?
Note: $$\sin^2 x,\cos^2 x\in[0,1]\implies A\in[0,2]$$ but $$2$$ is not "the" maximum value of $$A$$
• Both $\sin$ and $\cos$ have a minimum value of $-1$ and a maximum value of $1$, so what can you deduce from that? – Robert Howard Oct 20 '18 at 21:55
• @RobertHoward $A$ should be between $0$ and $2$, but how do I find the the maximum value ? – ss1729 Oct 20 '18 at 21:58
• observe that since $\sin x$ and $\cos x$ are less or equal to one then we have $\sin^{20} x + \cos^{48} \leq \sin^2x +\cos^2 x = 1$ – ALG Oct 20 '18 at 22:05
• Good point; I should have looked at the question more carefully. From the equation you ended with, I would try re-expressing all the powers of $\sin^2x$ in terms of $\cos^2x$ and see if that helps. – Robert Howard Oct 20 '18 at 22:06
If $$0\leq a\leq 1$$ then $$a^2\leq a$$, so we use that repetadly
$$A=\sin^{20}x+\cos^{48}x\leq \sin^{2}x+\cos^{2}x =1$$
So $$A_{\max}=1$$ and it is achieved at $$x=0$$.
For a minimum I don't see quick solution. I would try like this: Let $$t= \cos^2x$$ and then search for the minumum of $$g(t) = (1-t)^{10}+t^{24}$$ where $$0\leq t\leq 1$$.
Note that $$A = (1 - \cos^2 x)^{10} + \cos^{48} x$$. Letting $$t = \cos^2 x$$, $$t \in [0,1]$$, we have $$A = (1-t)^{10} + t^{24}$$. This function has maximum in 0 and 1, and $$A_{max} = 1$$. The minimum is found by differentiating and solving $$24t^{23} - 10(1-t)^9 = 0$$, which leads (numerically) to $$t_{min} \simeq 0,643187$$, and correspondingly to $$A_{min} = (1-t_{min})^{10} + t_{min}^{24} \simeq 0,0000585751.$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9790357561234475,
"lm_q1q2_score": 0.8590811375653994,
"lm_q2_score": 0.8774767746654976,
"openwebmath_perplexity": 122.34824560462499,
"openwebmath_score": 0.9546403884887695,
"tags": null,
"url": "https://math.stackexchange.com/questions/2963810/find-the-range-of-a-if-a-sin20x-cos48x/2963828"
} |
• WolframAlpha gives the same minimum, achieved at $x=.640178.$ – saulspatz Oct 20 '18 at 22:20
Apart from the trivial upper bound $$A\le 2$$, we have the stronger (and sharp - try $$x=0$$) bound $$\tag1A\le 1.$$
Consider $$f(x):=(1-x)^{10}+x^{24}$$ for $$0\le x\le 1$$. Then $$f'(x)=24x^{23}-10(1-x)^9$$ is strictly increasing (as each summand is) on $$[0,1]$$, hence has at most one root there. As $$f'(0)=-10$$ and $$f'(1)=24$$, we conclude that there is exactly one such root $$\alpha$$. As $$f'$$ goes from negative to positive, $$f$$ must have a local minimum there. We conclude that $$f$$ has its only minimum at $$\alpha$$ and its maximum at the boundary - in fact, at both ends of the boundary since $$f(0)=f(1)$$. As $$A=f(\cos^2 x)$$ and $$\cos^2 x$$ ranges from $$0$$ to $$1$$, inclusive, we conclude that the maximal value of $$A$$ is also $$1$$ (thus proving $$(1)$$), and the minimum value of $$A$$ is $$f(\alpha)$$.
Using $$(1-\alpha)^9=\frac{12}5\alpha^{23}$$, we have $$f(\alpha)=(1-\alpha)^{10}+\alpha^{24}=(1-\alpha)\cdot\frac{12}5\alpha^{23}+\alpha^{24}=\alpha^{23}\cdot \frac{12-7\alpha}5=(1-\alpha)^9\cdot \frac{12-7\alpha}{12},$$ so certainly $$\min A=\min f>0,$$ but not by much.
From $$f'(\frac 35)=24\frac{3^{23}}{5^{23}}-10\frac{2^{9}}{5^{9}}=\frac{24\cdot 3^{23}-10\cdot 2^95^{14}}{5^{23}}<0$$(!), we conclude that $$\alpha>\frac35$$ and hence $$f(\alpha)=(1-\alpha)^9\cdot \frac{12-7\alpha}{12}<(1-\tfrac35)^9\cdot \frac{12-7\cdot\frac35}{12}=\frac{1664}{9765625}\approx 0.00017$$ (whereas the true minimal value is $$\approx 0.000058575$$) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9790357561234475,
"lm_q1q2_score": 0.8590811375653994,
"lm_q2_score": 0.8774767746654976,
"openwebmath_perplexity": 122.34824560462499,
"openwebmath_score": 0.9546403884887695,
"tags": null,
"url": "https://math.stackexchange.com/questions/2963810/find-the-range-of-a-if-a-sin20x-cos48x/2963828"
} |
# Integrating $\int_{-\infty}^\infty \frac{1}{1 + x^4}dx$ with the residue theorem
Calculate integral
$$\int\limits_{-\infty}^{\infty}\frac{1}{x^4+1} dx$$ with residue theorem.
Can I evaluate $\frac 12\int_C \dfrac{1}{z^4+1} dz$ where $C$ is simple closed contour of the upper half of unit circle like this?
And find the roots of polynomial $z^4 +1$ which are the fourth roots of $-1$.
In $C$ there is $z_1 =e^{i\pi/4}=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$ and $z_2=e^{3\pi/4}=-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$.
So the residuals $B_1$ and $B_2$ for $z_1$ and $z_2$ are simple poles and that \begin{align} B_1&=\frac{1}{4 z_1^3}\frac{z_1}{z_1}=-\frac{z_1}{4} \\ B_2&=\frac{1}{4z_2^3}\frac{z_2}{z_2}=-\frac{z_2}{4} \end{align} And the sum of residuals is
$$B_1+B_2=-\frac{1}{4}(z_1 + z_2)=-\frac{1}{4}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right)=-\frac{i}{2 \sqrt{2}}$$
So my integral should be
$$\int\limits_{-\infty}^{\infty}\frac{1}{x^4+1} dx =\frac 12 \times 2\pi i (B_1+B_2)=\frac{\pi}{\sqrt{2}}$$
Is this valid?
• Yes. Peachy! ;-$)$ – Lucian Oct 26 '14 at 22:44
I thought that it might be instructive to present an alternative and efficient approach. To do so, we first note that from the even symmetry that
$$\int_{-\infty}^\infty \frac{1}{1+x^4}\,dx=2\int_{0}^\infty \frac{1}{1+x^4}\,dx \tag 1$$
We proceed to evaluate the integral on the right-hand side of $(1)$.
Next, we move to the complex plane and choose as the integration contour, the quarter circle in the upper-half plane with radius $R$. Then, we can write
\begin{align} \oint_C \frac{1}{1+z^4}\,dz&=\int_0^R \frac{1}{1+x^4}\,dx+\int_0^\pi \frac{iRe^{i\phi}}{1+R^4e^{i4\phi}}\,d\phi+\int_R^0 \frac{i}{1+(iy)^4}\,dy\\\\ &=(1-i)\int_0^R\frac {1}{1+x^4}\,dx+\int_0^\pi \frac{iRe^{i\phi}}{1+R^4e^{i4\phi}}\,d\phi \tag 2 \end{align}
As $R\to \infty$, the second integral on the right-hand side of $(2)$ approaches $0$ while the first approaches the $1/2$ integral of interest. Hence, we have | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232940063591,
"lm_q1q2_score": 0.859063469092649,
"lm_q2_score": 0.8740772368049822,
"openwebmath_perplexity": 614.7518432475417,
"openwebmath_score": 0.999734103679657,
"tags": null,
"url": "https://math.stackexchange.com/questions/992436/integrating-int-infty-infty-frac11-x4dx-with-the-residue-theorem"
} |
$$\bbox[5px,border:2px solid #C0A000]{\lim_{R\to \infty}\oint_C \frac{1}{1+z^4}\,dz=\frac{1-i}2 \int_{-\infty}^\infty \frac{1}{1+x^4}\,dx} \tag 3$$
Now, since $C$ encloses only the pole at $z=e^{i\pi/4}$, the Residue Theorem guarantees that
\begin{align}\oint_C \frac{1}{1+z^4}\,dz&=2\pi i \text{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4}\right)\\\\ &=\frac{2\pi i}{4e^{i3\pi/4}}\\\\ &=\frac{\pi e^{-i\pi/4}}{2}\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\pi(1-i)}{2\sqrt 2}} \tag 4 \end{align}
Finally, equating $(3)$ and $(4)$, we find that
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{1}{1+x^4}\,dx=\frac{\pi}{\sqrt{2}}}$$
as expected.
• @elec It's curious that you requested specifically to evaluate the integral using the residue theorem, yet awarded the best answer for a real analysis approach. – Mark Viola Nov 17 '16 at 15:09 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232940063591,
"lm_q1q2_score": 0.859063469092649,
"lm_q2_score": 0.8740772368049822,
"openwebmath_perplexity": 614.7518432475417,
"openwebmath_score": 0.999734103679657,
"tags": null,
"url": "https://math.stackexchange.com/questions/992436/integrating-int-infty-infty-frac11-x4dx-with-the-residue-theorem"
} |
# Mutual tangent lines
Find all points where the curves $f(x) = x^3-3x+4$ and $g(x) = 3x^2-3x$ share the same tangent line.
Graphing them I see that they look like they share a tangent line at $x=2$.
I got the derivatives of both and set them equal to each other and got $x=0$ and $x=2$.
After plugging $2$ back in I got $6$. So the point is $(2,6)$.
Is that correct?
• In hindsight, this is not a duplicate. It appeared to be a duplicate because the OP originally had a scan of a problem sheet that included this problem, as well as the one from the other question. I have altered the other question to make this clear. – Cameron Buie Nov 2 '13 at 13:15
$$f' = g ' \iff 3x^2 - 3 = 6x - 3 \iff x^2 -2x = 0 \iff x(x-2) = 0 \iff x=0,2$$
$$(2, 6 )$$ as you said. Notice at $0$ functions are not equal, so they cannot share a tangent line at $0$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232914907945,
"lm_q1q2_score": 0.8590634652819085,
"lm_q2_score": 0.8740772351648677,
"openwebmath_perplexity": 175.43968324081402,
"openwebmath_score": 0.5496748089790344,
"tags": null,
"url": "https://math.stackexchange.com/questions/548733/mutual-tangent-lines"
} |
# Question from Berkley Problems in Mathematics. Show that det($A$) > 0 [duplicate]
Let $A = (a_{ij})$ be an $n\times n$ real matrix satisfying the conditions: $a_{ii} > 0$ $(1\leq i \leq n)$; $a_{ij} \leq0$ $(i\not=j,1\leq i,j \leq n)$; $\sum\limits_{i=1}^n a_{ij}>0$ $(1\leq i \leq n)$. Show that det($A$) > $0$ .
## marked as duplicate by mlc, user91500, user26857, The Dead Legend, Namaste linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 7 '17 at 11:17
• Hint: Show that if $\lambda$ is a real eigenvalue then it is positive. – user379195 May 7 '17 at 6:54
The matrix $A$ is diagonally dominant and so nonsingular. Replacing off-diagonal entries $a_{i,j}$ by $ta_{i,j}$ for $0\le t\le 1$ also gives a diagonally dominant and nonsingular matrix $A_t$. But $A_0$ is diagonal with positive determinant. By continuity of determinant $\det A_t>0$ and so $\det A=\det A_1>0$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232869627775,
"lm_q1q2_score": 0.8590634645479575,
"lm_q2_score": 0.8740772384450968,
"openwebmath_perplexity": 1488.0928414290352,
"openwebmath_score": 0.8717315793037415,
"tags": null,
"url": "https://math.stackexchange.com/questions/2269505/question-from-berkley-problems-in-mathematics-show-that-deta-0/2269512"
} |
• A weakly diagonally dominant matrix may be singular. – mlc May 7 '17 at 7:00
• @mlc There's nothing weak about these matrices! – Lord Shark the Unknown May 7 '17 at 7:01
• \begin{bmatrix} 1 & -2 & -2 \\ -1 & 6 & -1 \\ -1 & -1 & 3 \end{bmatrix} what about this matrix; It's not a DD matrix. But confusingly, it is satisfying the question but not the result to be shown. I must be wrong somwhere. – Hirakjyoti Das May 7 '17 at 7:09
• @HirakjyotiDas one of your conditions is that the column sums are positive. – Lord Shark the Unknown May 7 '17 at 7:13
• That's the very thing I was missing; Thank you – Hirakjyoti Das May 7 '17 at 7:16
In addition to the other answers, one can more generally see that from the Gershgorin circle theorem
Here firstly $A$ is diagonally dominant. Secondly $A$ is $Z$-matrix as all its off diagonal entries are $\leq 0.$
Also diagonal entries are $> 0.$ Hence $A$ is a $P$-matrix. (All principal minors are positive).
So $\det(A) > 0.$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232869627775,
"lm_q1q2_score": 0.8590634645479575,
"lm_q2_score": 0.8740772384450968,
"openwebmath_perplexity": 1488.0928414290352,
"openwebmath_score": 0.8717315793037415,
"tags": null,
"url": "https://math.stackexchange.com/questions/2269505/question-from-berkley-problems-in-mathematics-show-that-deta-0/2269512"
} |
# Why is this approximation of polynomial root so accurate?
I have an engineering problem where I have to find the smallest positive real root of a polynomial in $x$: $$Ax^5+Bx^3 - C = 0$$ Instead of solving numerically, I want simple approximative formulas ("design equations") that describe the behaviour. For that matter, I split the problem into two regimes:
• Large $x$: $\ \ \ \ x^5$ is dominant $\Longrightarrow \ \ Ax^5 \approx C \ \ \Longrightarrow \ \ x \approx \sqrt[5]{C/A} =: x_A$
• Small $x$: $\ \ \ \ x^3$ is dominant $\Longrightarrow \ \ Bx^3 \approx C \ \ \Longrightarrow \ \ x \approx \sqrt[3]{C/B} =: x_B$
The approximations work well below/above a certain threshold on $x$, but require an ugly case distinction. In order to avoid that and having something smoother than $$x \approx \min\{x_A, x_B\}$$ I tried Pythagorean-style combination of the inverses (inspired by the resistance of parallel circuits in electrical engineering) and found that $$x \approx 1\Big/ \ \left\|\left(\begin{matrix} 1/x_A \\ 1/x_B \end{matrix}\right)\right\|_4 = 1 \Big/ \sqrt[4]{1/x_A^4 + 1/x_B^4}$$ is a really, really good approximation. Pretty much perfect. That leads me to my actual question: Is it possible to argue why that 4-norm is such an amazing approximation? Does my initial polynomial have a certain structure that explains the high accuracy of my approximation?
Since I want to present/defend that stuff, I'd appreciate some sophistication. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232894783426,
"lm_q1q2_score": 0.8590634602989846,
"lm_q2_score": 0.8740772318846386,
"openwebmath_perplexity": 705.4742662760644,
"openwebmath_score": 0.7773688435554504,
"tags": null,
"url": "https://math.stackexchange.com/questions/1418629/why-is-this-approximation-of-polynomial-root-so-accurate"
} |
Since I want to present/defend that stuff, I'd appreciate some sophistication.
• Are $A,B,C$ positive? If so the "smallest positive real root" is the only real root. – alex.jordan Sep 2 '15 at 17:19
• For example, the sole real root of $17x^5 + 41x^3 - 105$ is approximately $1.17624...$; the approximation gives $1.17849...$. – Unit Sep 2 '15 at 17:20
• @alex.jordan Yes they are positive. And sure, since $f(x) = Ax^5 + Bx^3$ is strictly increasing, there is only one relevant solution. – GDumphart Sep 2 '15 at 17:21
• What do you mean by a "really, really good approximation", and have you tried lots of values of $A$, $B$ and $C$? – David Quinn Sep 2 '15 at 17:27
• @DavidQuinn No I haven't done exhaustive tests with all kinds of $A,B$ combos, only what my problem yielded. Here is what I mean with great approximation: i.imgur.com/7Bfl8nt.png Only ridiculous zooming exposes some error. – GDumphart Sep 2 '15 at 17:41
$$\left(\frac x{x_A}\right)^5+\left(\frac x{x_B}\right)^3-1=0.$$
With $y=\dfrac x{x_A}$ and $r=\dfrac{x_A}{x_B}$, a single parameter remains:
$$y^5+r^3y^3-1=0.$$
Then $$r=\sqrt[3]{\frac{1-y^5}{y^3}}=\frac1y\sqrt[3]{1-y^5},$$ can be compared to your approximation $$y=\frac1{\sqrt[4]{1+r^4}},$$i.e. $$r=\sqrt[4]{\frac1{y^4}-1}=\frac1y\sqrt[4]{1-y^4}.$$ The agreement is indeed excellent on a wide range [abscissa $y$, ordinate $r$]:
For small $y$, both behaviors are identical, $\dfrac1y$. For $y$ close to $1$, behaviors are similar, approximately $\sqrt[3]{5(1-y)}$ and$\sqrt[4]{4(1-y)}$, and a blend in between.
The good agreement is explained by the fact that the functional relations are similar, with the exponent $4$ intermediate between $3$ and $5$ (but the value $4$ has nothing "magical"). | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232894783426,
"lm_q1q2_score": 0.8590634602989846,
"lm_q2_score": 0.8740772318846386,
"openwebmath_perplexity": 705.4742662760644,
"openwebmath_score": 0.7773688435554504,
"tags": null,
"url": "https://math.stackexchange.com/questions/1418629/why-is-this-approximation-of-polynomial-root-so-accurate"
} |
• Using Yves's parametrization, the absolute value of the difference between the true $y$ and your approximation turns out to be at most $0.01171$. The maximum absolute difference occurs at approximately $r = 0.66145$, where the true $y \approx 0.94545$ while the approximation is $0.95716$. – Robert Israel Sep 2 '15 at 18:50
• @RobertIsrael: you can try slightly lower values than $4$ for the exponent, such as $3.7$. – Yves Daoust Sep 2 '15 at 18:53
• Yes, I can. The best in terms of maximum absolute difference seems to be approximately $3.85$ (i.e. approximation $(1 + r^{3.85})^{-1/3.85}$), where the maximum absolute difference is about $0.007608$. – Robert Israel Sep 2 '15 at 21:44
• Does this mean that there's an exact solution, rather than the approximation that the OP is using – Dr Xorile Sep 2 '15 at 22:03
• Amazing answer, very intuitive and accounts for everything I wished for, thank you. I reproduced all your steps. – GDumphart Sep 3 '15 at 6:34 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232894783426,
"lm_q1q2_score": 0.8590634602989846,
"lm_q2_score": 0.8740772318846386,
"openwebmath_perplexity": 705.4742662760644,
"openwebmath_score": 0.7773688435554504,
"tags": null,
"url": "https://math.stackexchange.com/questions/1418629/why-is-this-approximation-of-polynomial-root-so-accurate"
} |
# Math Help - Probability problem about cubes
1. ## Probability problem - cubes
Each face of a cube is painted either Red or Blue, each with probability 1/2. The color of each face is determined independently. What is the probabilty that the painted cube can be faced on a horizontal surface so that the four vertical faces are all the same color?
there can be 64 different color cubes 2x2x2x2x2x2x2 ?
case 1 when the cube is all red or blue it will work - 2/64
I couldn't think of the cases...
Could someone just give me some hints on the cases I have to set up?
Vicky.
2. Originally Posted by Vicky1997
Each face of a cube is painted either Red or Blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be faced on a horizontal surface so that the four vertical faces are all the same color?
there can be 64 different color cubes 2x2x2x2x2x2x2 ?
case 1 when the cube is all red or blue it will work - 2/64
I couldn't think of the cases...
Could someone just give me some hints on the cases I have to set up?
Vicky.
That's a good start. You can continue it like this. If one face is one colour and the other five faces are all the other colour then it will work (2×6 cases of that). If two faces are one colour (and the other four faces are the other colour) then it will work provided that the two faces are opposite each other (2×3 cases, because there are three pairs of opposite faces). Finally, it can't be done at all if there are three faces of each colour.
3. Hello, Vicky1997!
A fascinating problem . . .
Each face of a cube is painted either Red or Blue, each with probability 1/2.
The color of each face is determined independently.
What is the probabilty that the painted cube can be placed on a horizontal surface
so that the four vertical faces are all the same color?
There can be 64 different colored cubes: $2^6 = 64$ . . . . Right!
There are seven possible colorings: | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232869627774,
"lm_q1q2_score": 0.8590634564882434,
"lm_q2_score": 0.874077230244524,
"openwebmath_perplexity": 481.67986572613455,
"openwebmath_score": 0.7375345230102539,
"tags": null,
"url": "http://mathhelpforum.com/statistics/114604-probability-problem-about-cubes.html"
} |
. . $\begin{array}{|c|c|} \hline
\text{Colors} & \text{Ways} \\ \hline
\text{6 red, 0 blue} & 1 \\ \text{5 red, 1 blue} & 6 \\ \text{4 red, 2 blue} & 15 \\ \text{3 red, 3 blue} & 20 \\ \text{2 red, 4 blue} & 15 \\ \text{1 red, 5 blue} & 6 \\ \text{0 red, 6 blue} & 1 \\ \hline \end{array}$
. . You may recognize these as Binomial Coefficients.
Let $V$ = "the four vertical faces have the same color".
6 red, 0 blue: . $P(\text{6 red}) \:=\:\frac{1}{64}$
To have $V$, the cube can be placed on any face.
. . $P(\text{6 red} \wedge V)\:=\:\frac{1}{64}\cdot\frac{6}{6} \:=\:\frac{6}{384}$
5 red, 1 blue: . $P(\text{5 red}) \:=\:\frac{6}{64}$
To have $V$, the blue face must be on the top or bottom
. $P(\text{5 red} \wedge V) \:=\:\frac{6}{64}\cdot\frac{2}{6} \:=\:\frac{12}{384}$
4 red, 2 blue.
To have $V$, the 2 blues must be on opposite faces.
. . This happens only $\frac{3}{64}$ of the time.
Then the blue faces must be on the top and the bottom.
. . $P(\text{4 red} \wedge V) \:=\:\frac{3}{64}\cdot\frac{2}{6} \:=\:\frac{6}{384}$
3 red, 3 blue.
With 3 of each color, it is impossible to have $V$.
2 red, 4 blue.
This has the same probability as "4 red, 2 blue".
. . $P(\text{2 red} \wedge V) \:=\:\frac{6}{384}$
1 red, 5 blue.
This has the same probability as "5 red, 1 blue".
. . $P(\text{1 red} \wedge V) \:=\:\frac{12}{384}$
0 red, 6 blue.
This has the same probability as "6 red, 0 blue".
. . $P(\text{0 red} \wedge V) \:=\:\frac{6}{384}$
Therefore: . $P(V) \;=\;\frac{6}{384} + \frac{12}{384} + \frac{6}{384} + \frac{6}{384} + \frac{12}{384} + \frac{6}{384} \;=\;\frac{48}{384} \;=\;\frac{1}{8}$ | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232869627774,
"lm_q1q2_score": 0.8590634564882434,
"lm_q2_score": 0.874077230244524,
"openwebmath_perplexity": 481.67986572613455,
"openwebmath_score": 0.7375345230102539,
"tags": null,
"url": "http://mathhelpforum.com/statistics/114604-probability-problem-about-cubes.html"
} |
# Relationship between input and output sequence in Hartley transformation
As you know that Discrete Hartley transformation is related to the discrete Fourier transformation, $$i.e$$, assuming we have a vector $$X = [x_0,x_1,\ldots,x_N]$$, its Hartley transformation is equal to $$H(X) = Real(FFT(X)) - Imag(FFT(X))$$. where $$H$$ denotes the Hartley transformation.
I wonder, if I want the output of the Hartley transformation equals to a vector of length $$N$$ whose all elements are $$1$$, it means $$H(X) = [1,1,\ldots,1]$$, what supposed to be the input $$X$$? I means I need to formulate the relationship between the inputs and outputs.
Wikipedia's entry for the discrete Hartley transform shows states that the $$\mathsf{DHT}$$ is, up to a scaling, its own inverse. If $$x$$ is a vector with $$N$$ entries and $$y$$ is its discrete Hartley transform, $$$$y = \mathsf{DHT}x,$$$$ then $$$$x = \frac{1}{N}\mathsf{DHT}y.$$$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759671623987,
"lm_q1q2_score": 0.8590384235909946,
"lm_q2_score": 0.8757870029950158,
"openwebmath_perplexity": 612.0957801816913,
"openwebmath_score": 0.9432317018508911,
"tags": null,
"url": "https://dsp.stackexchange.com/questions/67655/relationship-between-input-and-output-sequence-in-hartley-transformation"
} |
If $$x$$ is a vector with $$N$$ entries such that $$$$\mathsf{DHT}x = \underbrace{(1,1,\ldots,1)^{\mathsf{T}}}_{\textrm{N entries}},$$$$ then we recover $$x$$ with $$$$x = \frac{1}{N}\mathsf{DHT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right).$$$$ This means that $$x$$ is $$$$\begin{split} x &=~ \frac{1}{N}\left(\mathsf{Re}\left[\mathsf{DFT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right)\right] - \mathsf{Im}\left[\mathsf{DFT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right)\right]\right), \end{split}$$$$ where $$\mathsf{DFT}$$ is the discrete Fourier transform, which we usually compute with a FFT algorithm. The $$\ell^{\textrm{th}}$$ entry of the $$\mathsf{DFT}$$ of the all-1 vector is $$$$\begin{split} \sum_{n=0}^{N-1}1\times e^{-2\pi j \ell n/N} &=~ 1 + e^{-2\pi j \ell/N} + \left(e^{-2\pi j \ell/N}\right)^2 + \cdots + \left(e^{-2\pi j \ell/N}\right)^{N-1}\\ &=~ \left\{\begin{array}{rl}N & \textrm{if}~\ell=0,\\0&\textrm{if}~\ell\neq 0.\end{array}\right.\\ &=~ N\delta_{\ell,0}, \end{split}$$$$ where $$\delta_{p,q}$$ is the Kronecker delta. One way to show this is to note that if $$\ell = 0$$, then each exponent is $$0$$, so each term in the sum is $$1$$. On the other hand, if $$\ell\neq 0$$, then $$\exp(-2\pi j\ell/N) \neq 1$$,and $$$$\begin{split} 1 + e^{-2\pi j \ell/N} + \left(e^{-2\pi j \ell/N}\right)^2 + \cdots + \left(e^{-2\pi j \ell/N}\right)^{N-1} &=~ \frac{1 - \left(e^{-2\pi j \ell/N}\right)^N}{1 - e^{-2\pi j \ell/N}}\\ &=~ \frac{1 - e^{-2N\pi j \ell/N}}{1 - e^{-2\pi j \ell/N}}\\ &=~ \frac{1 - 1}{1 - e^{-2\pi j \ell/N}} ~=~ 0. \end{split}$$$$
That shows that the $$\mathsf{DFT}$$ of the all-1 vector has no imaginary part, and its real part is $$(N,0,0,\ldots,0)^{\mathsf{T}}$$. Hence $$$$x ~=~ \frac{1}{N}\left(\begin{array}{c} N\\0\\0\\\vdots\\0 \end{array}\right) ~=~ \left(\begin{array}{c} 1\\0\\0\\\vdots\\0 \end{array}\right).$$$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759671623987,
"lm_q1q2_score": 0.8590384235909946,
"lm_q2_score": 0.8757870029950158,
"openwebmath_perplexity": 612.0957801816913,
"openwebmath_score": 0.9432317018508911,
"tags": null,
"url": "https://dsp.stackexchange.com/questions/67655/relationship-between-input-and-output-sequence-in-hartley-transformation"
} |
• Thank you so much .. that's really appreciated.
– Gze
May 20 '20 at 3:19
• Thanks again, that's really interesting. I am trying to understand it well, if I got a question, I will let you know.
– Gze
May 20 '20 at 6:36
– Gze
May 20 '20 at 11:46
The Hartley transform is an involution: it is (up to a scale factor) its own inverse. The classical discrete Hartley transform of order $$N$$ is such that $$H_N^{-1} = \frac{1}{N}H_N$$. Be careful with your notation, the vector $$x$$ has $$N+1$$ entries, so maybe you are after an $$N+1$$-order DHT!
If $$\mathbf{1}$$ denotes the all-ones vector, then in matrix-vector product $$H_Nx = \mathbf{1}$$ is equivalent to $$H_N^{-1}H_Nx = H_N^{-1}\mathbf{1}$$, and you get more directly the mysterious $$X$$:
$$x = \frac{1}{N}H_N\mathbf{1}$$
and you get the result given by Joe Mac by a direct computation. A little interpretation: the discrete "Dirac" signal at the origin yields a flat "Hartley" spectrum, just like for Fourier.
• Thank you for explaining it, I got it now. but I was wondering what's about if $H_Nx = y$ and only some values of, i.e., $y_{1:4:N} = 1$ and other values of $y$ are any random values. So, can we express the relationship between the $y_{1:4:N}$ and $x_{1:4:N}$ mathematically in similar way too?
– Gze
May 20 '20 at 11:44
• I should add something, .. So, can we express the relationship between the $y_{1:4:N}$ and $x_{1:4:N}$ OR between $y_{1:4:N}$ and $x$ mathematically in similar way too?
– Gze
May 20 '20 at 12:07
• I don't really understand this comment. This may require a different question. But you can indeed interleave transforms and downsampling operators, and solutions can be quite complicated May 20 '20 at 12:16
• OK .. I posted it as a new question here dsp.stackexchange.com/questions/67696/… .. thank you
– Gze
May 20 '20 at 14:49
• And you did well. I am not able to see a clear answer right now May 20 '20 at 22:13 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759671623987,
"lm_q1q2_score": 0.8590384235909946,
"lm_q2_score": 0.8757870029950158,
"openwebmath_perplexity": 612.0957801816913,
"openwebmath_score": 0.9432317018508911,
"tags": null,
"url": "https://dsp.stackexchange.com/questions/67655/relationship-between-input-and-output-sequence-in-hartley-transformation"
} |
# Probability of Couples sitting next to each other (Sitting in a Row)
Suppose we have 4 couples and 8 seats. The seats are oriented such that it is a row. The people take their seats in a random fashion.
What is the probability that 2 couples sit next to each other?
My work: Let us label the people as $A_1, A_2, A_3, A_4, B_1, B_2, B_3, B_4$. $A_i$ and $B_i$ are couples. Then, we want to know the probability of $P(C_i \cap C_j), i \neq j$ and $C_i$ means couple $i$ are seated next to each other.
Then, suppose we fix $C_i$ and $C_j$. Then, the row arrangement could look something like: $\{ C_i, C_j, a,b,c,d\}$, where $a,b,c,d$ are the remaining 4 people. Then, there are $\binom{6}{2}$ ways of picking positions for $C_i$ and $C_j$, $2!$ ways of arranging the elements in $C_i$, $2!$ ways of arranging the elements in $C_j$, $2!$ ways of arranging $C_i$ and $C_j$ relative to each other, and $4!$ ways of arranging the remaining 4 people. There are $8!$ ways to arrange the 8 people in a row. Also, there are $\binom{4}{2}$ ways to pick 2 couples out of 4 couples.
Putting all this together, $$P(C_i \cap C_j) = \frac{\binom{6}{2} \binom{4}{2} 2! 2! 2! 4! }{8!}$$
I am curious because this differs from the solution given here: cached solution from google
Note that on the bottom of page 5, the difference is the google solution is missing a $\binom{6}{2}$ term, which means we can choose 2 positions for $C_i$ and $C_j$ out of the available 6 positions. (Please observe the quantity $p_2$ in the linked document). The factor $\binom{4}{2}$ is accounted for later on the top of page 6, where the author sums up all the probabilities corresponding to possible indices $i$ and $j$.
Thanks.
-
I would agree with you that the cached answer looks wrong and that there is a missing factor of 15 in $p_2$ (and of 20 in $p_3$). | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759621310288,
"lm_q1q2_score": 0.8590384191845863,
"lm_q2_score": 0.8757870029950159,
"openwebmath_perplexity": 185.07206589009553,
"openwebmath_score": 0.88112872838974,
"tags": null,
"url": "http://math.stackexchange.com/questions/43271/probability-of-couples-sitting-next-to-each-other-sitting-in-a-row/43281"
} |
When you ask "What is the probability that 2 couples sit next to each other?" you are looking at what was originally the $p_2$ part of the "What is the probability that at least one of the wives ends up sitting next to her husband?" question.
$p_2$ is the probability that two particular couples sit with husband next to wife. As you say, the ${4 \choose 2} = 6$ for which two couples these are is included in the $4p_1-6p_2+4p_3-p_4$ formula.
I would then say that the two particular couples have each become one unit (with $(2!)^2$ ways of arranging the individuals within the couple), reducing the number of orderable units from $8$ to $6$ so we should have $$p_2 = \frac{(2!)^2 \; 6!}{8!} \approx 0.071428571$$ and similarly $p_1 = \frac{(2!)^1 \; 7!}{8!} = 0.25$, $p_3 = \frac{(2!)^3 \; 5!}{8!} \approx 0.023810$, and $p_4 = \frac{(2!)^4 \; 4!}{8!} \approx 0.00952381$.
For $p_2$ and $p_3$, these are not the same as the cached answer, though the $p_2$ is the same as yours. Using the inclusion-exclusion formula this gives me a probability that at at least one of the wives ends up sitting next to her husband of about $0.657142857$, not the $0.9666667$ of the cached answer.
Some simulation convinces me this looks plausible.
-
Hi Henry: Thanks, this is the exact answer, I understand it, and I can also verify this in MATLAB. For the purpose of learning, I have included the MATLAB simulation code here: pastebin.com/UXjPEDaE . All of the different scenarios can be checked by making modifications to the code. – jrand Jun 5 '11 at 17:15
Then, there are $\binom{6}{2}$ ways of picking positions for $C_i$ and $C_j$
I think that this is where you're wrong. In fact, there are only 5 (as in $\binom{5}{1}$) ways of placing them (considering that you account for order later): | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759621310288,
"lm_q1q2_score": 0.8590384191845863,
"lm_q2_score": 0.8757870029950159,
"openwebmath_perplexity": 185.07206589009553,
"openwebmath_score": 0.88112872838974,
"tags": null,
"url": "http://math.stackexchange.com/questions/43271/probability-of-couples-sitting-next-to-each-other-sitting-in-a-row/43281"
} |
• $(C_iC_j,a,b,c,d)$
• $(a,C_iC_j,b,c,d)$
• $(a,b,C_iC_j,c,d)$
• $(a,b,c,C_iC_j,d)$
• $(a,b,c,d,C_iC_j)$
-
Hi trutheality, you have shown 5 possible arrangements. Remember that $C_i$ actually contains two elements - it means couple $i$ is seated next to each other. Then, this is a possible arrangement: $(C_i, a,b,c,d,C_j)$. To be more clear, $(C_i, a,b,c,d,C_j) = (A_i, B_i, X_a, X_b, X_c, X_d, A_j, B_j)$. The X's must be set so that they refer to exactly the four remaining people. – jrand Jun 4 '11 at 22:30
I think that we understand the question differently: I took "two couples seated next to each other" to mean that not only do we have $A_i$ next to $B_i$ and $A_j$ next to $B_j$, but also $C_i$ next to $C_j$. If you drop the last requirement then your answer is correct. – trutheality Jun 4 '11 at 22:34
In any case, if we use your interpretation, where pairs of couples must be seated next to each other, the probability should be: $P(C_i \cap C_j) = \frac{5 \binom{4}{2} 2! 2! 2! 4! }{8!}$. I have replaced $\binom{6}{2}$ by $5$. This is still different from the claimed answer in the link. – jrand Jun 4 '11 at 22:36
I think you're right then, the answer in the link doesn't seem to account for the different ways of arranging couples and non-couples. – trutheality Jun 4 '11 at 22:48 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759621310288,
"lm_q1q2_score": 0.8590384191845863,
"lm_q2_score": 0.8757870029950159,
"openwebmath_perplexity": 185.07206589009553,
"openwebmath_score": 0.88112872838974,
"tags": null,
"url": "http://math.stackexchange.com/questions/43271/probability-of-couples-sitting-next-to-each-other-sitting-in-a-row/43281"
} |
## do all bijective functions have an inverse | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
This result says that if you want to show a function is bijective, all you have to do is to produce an inverse. Because if it is not surjective, there is at least one element in the co-domain which is not related to any element in the domain. Since "at least one'' + "at most one'' = "exactly one'', f is a bijection if and only if it is both an injection and a surjection. Now we consider inverses of composite functions. That is, for every element of the range there is exactly one corresponding element in the domain. bijectivity would be more sensible. Read Inverse Functions for more. For Free, Kharel's Simple Procedure for Factoring Quadratic Equations, How to Use Microsoft Word for Mathematics - Inserting an Equation. In this case, the converse relation $${f^{-1}}$$ is also not a function. The function f is called an one to one, if it takes different elements of A into different elements of B. In practice we end up abandoning the … A triangle has one angle that measures 42°. Image 2 and image 5 thin yellow curve. The figure given below represents a one-one function. A bijective function is also called a bijection. If the function satisfies this condition, then it is known as one-to-one correspondence. Choose an expert and meet online. The inverse relation is then defined as the set consisting of all ordered pairs of the form (2,x). Can you provide a detail example on how to find the inverse function of a given function? For a function to have an inverse, each element y ∈ Y must correspond to no more than one x ∈ X; a function f with this property is called one-to-one or an injection. Some non-algebraic functions have inverses that are defined. Since the relation from A to B is bijective, hence the inverse must be bijective too. In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). So what is all this talk about "Restricting the Domain"? On A Graph . So what is all this talk about "Restricting the | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
talk about "Restricting the Domain"? On A Graph . So what is all this talk about "Restricting the Domain"? If you were to evaluate the function at all of these points, the points that you actually map to is your range. In general, a function is invertible as long as each input features a unique output. Those that do are called invertible. It should be bijective (injective+surjective). And the word image is used more in a linear algebra context. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. Yes, but the inverse relation isn't necessarily a function (unless the original function is 1-1 and onto). For a function to have an inverse, each element y ∈ Y must correspond to no more than one x ∈ X; a function f with this property is called one-to-one or an injection. Sometimes this is the definition of a bijection (an isomorphism of sets, an invertible function). That is, for every element of the range there is exactly one corresponding element in the domain. If f −1 is to be a function on Y, then each element y ∈ Y must correspond to some x ∈ X. http://www.sosmath.com/calculus/diff/der01/der01.h... 3 friends go to a hotel were a room costs $300. Naturally, if a function is a bijection, we say that it is bijective.If a function $$f :A \to B$$ is a bijection, we can define another function $$g$$ that essentially reverses the assignment rule associated with $$f$$. A function with this property is called onto or a surjection. The function is bijective (one-to-one and onto, one-to-one correspondence, or invertible) if each element of the codomain is mapped to by exactly one element of the domain. Since the function from A to B has to be bijective, the inverse function must be bijective too. A one-one function is also called an Injective function. ), the function is not bijective. More specifically, if g (x) is a bijective function, and if we set the correspondence g (ai) = bi for all ai in | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
if g (x) is a bijective function, and if we set the correspondence g (ai) = bi for all ai in R, then we may define the inverse to be the function g-1(x) such that g-1(bi) = ai. Bijective functions have an inverse! Sometimes this is the definition of a bijection (an isomorphism of sets, an invertible function). 4.6 Bijections and Inverse Functions. So, to have an inverse, the function must be injective. No. Let f : A ----> B be a function. Get a free answer to a quick problem. Draw a picture and you will see that this false. That is, we say f is one to one In other words f is one-one, if no element in B is associated with more than one element in A. Example: The polynomial function of third degree: f(x)=x 3 is a bijection. Assume ##f## is a bijection, and use the definition that it … create quadric equation for points (0,-2)(1,0)(3,10)? Let us now discuss the difference between Into vs Onto function. Yes, but the inverse relation isn't necessarily a function (unless the original function is 1-1 and onto). Join Yahoo Answers and get 100 points today. pleaseee help me solve this questionnn!?!? You have assumed the definition of bijective is equivalent to the definition of having an inverse, before proving it. If every "A" goes to a unique "B", and every "B" has a matching "A" then we can go back and forwards without being led astray. and do all functions have an inverse function? So a bijective function follows stricter rules than a general function, which allows us to have an inverse. This is clearly not a function (for one thing, if you graph it, it fails the vertical line test), but it is most certainly a relation. You have to do both. Let us start with an example: Here we have the function Obviously neither the space$\mathbb{R}$nor the open set in question is compact (and the result doesn't hold in merely locally compact spaces), but their topology is nice enough to patch the local inverse together. 2xy=x-2 multiply both sides by 2x, 2xy-x=-2 subtract x from both | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
patch the local inverse together. 2xy=x-2 multiply both sides by 2x, 2xy-x=-2 subtract x from both sides, x(2y-1)=-2 factor out x from left side, x=-2/(2y-1) divide both sides by (2y-1). ….Not all functions have an inverse. It is a function which assigns to b, a unique element a such that f(a) = b. hence f-1 (b) = a. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. A bijection is also called a one-to-one correspondence . Cardinality is defined in terms of bijective functions. To prove f is a bijection, we must write down an inverse for the function f, or shows in two steps that. sin and arcsine (the domain of sin is restricted), other trig functions e.g. Still have questions? It is clear then that any bijective function has an inverse. This is clearly not a function (for one thing, if you graph it, it fails the vertical line test), but it is most certainly a relation. A bijective function is a bijection. View FUNCTION N INVERSE.pptx from ALG2 213 at California State University, East Bay. How do you determine if a function has an inverse function or not? Ryan S. Summary and Review; A bijection is a function that is both one-to-one and onto. Inverse Functions An inverse function goes the other way! The process of "turning the arrows around" for an arbitrary function does not, in general, yield a function, but properties (3) and (4) of a bijection say that this inverse relation is a function with domain Y. Start here or give us a call: (312) 646-6365. Inverse Functions:Bijection function are also known as invertible function because they have inverse function property. Most questions answered within 4 hours. The inverse relation is then defined as the set consisting of all ordered pairs of the form (2,x). both 3 and -3 map to 9 Hope this helps answered 09/26/13. Example: f(x) = (x-2)/(2x) This function is one-to-one. … Algebraic functions involve only the algebraic operations | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
(x-2)/(2x) This function is one-to-one. … Algebraic functions involve only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power. Into vs Onto Function. A; and in that case the function g is the unique inverse of f 1. The graph of this function contains all ordered pairs of the form (x,2). If an algebraic function is one-to-one, or is with a restricted domain, you can find the inverse using these steps. 'Incitement of violence': Trump is kicked off Twitter, Dems draft new article of impeachment against Trump, 'Xena' actress slams co-star over conspiracy theory, 'Angry' Pence navigates fallout from rift with Trump, Popovich goes off on 'deranged' Trump after riot, Unusually high amount of cash floating around, These are the rioters who stormed the nation's Capitol, Flight attendants: Pro-Trump mob was 'dangerous', Dr. Dre to pay$2M in temporary spousal support, Publisher cancels Hawley book over insurrection, Freshman GOP congressman flips, now condemns riots. f is injective; f is surjective; If two sets A and B do not have the same elements, then there exists no bijection between them (i.e. For example suppose f(x) = 2. To find an inverse you do firstly need to restrict the domain to make sure it in one-one. $\endgroup$ – anomaly Dec 21 '17 at 20:36 In practice we end up abandoning the … Only one-to-one functions have inverses, as the inverse of a many-to-one function would be one-to-many, which isn't a function. So if you input 49 into our inverse function it should give you d. Input 25 it should give you e. Input nine it gives you b. A simpler way to visualize this is the function defined pointwise as. x^2 is a many-to-one function because two values of x give the same value e.g. Example: The linear function of a slanted line is a bijection. What's the inverse? A function f is bijective if it has a two-sided inverse Proof (⇒): If it is bijective, it has a left inverse (since injective) and a right inverse | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
inverse Proof (⇒): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (⇐): If it has a two-sided inverse, it is both injective (since there is a left inverse… Adding 1oz of 4% solution to 2oz of 2% solution results in what percentage? Thus, to have an inverse, the function must be surjective. Domain and Range. In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). Of course any bijective function will do, but for convenience's sake linear function is the best. Image 1. Not all functions have inverse functions. It would have to take each of these members of the range and do the inverse mapping. Assuming m > 0 and m≠1, prove or disprove this equation:? Get your answers by asking now. Nonetheless, it is a valid relation. A function is bijective if and only if has an inverse November 30, 2015 De nition 1. So a bijective function follows stricter rules than a general function, which allows us to have an inverse. And that's also called your image. We can make a function one-to-one by restricting it's domain. In many cases, it’s easy to produce an inverse, because an inverse is the function which “undoes” the effect of f. Example. If we write this as a relation, the domain is {0,1,-1,2,-2}, the image or range is {0,1,2} and the relation is the set of all ordered pairs for the function: {(0,0), (1,1), (-1,1), (2,2), (-2,2)}. This property ensures that a function g: Y → X exists with the necessary relationship with f In the previous example if we say f(x)=x, The function g(x) = square root (x) is the inverse of f(x)=x. Show that f is bijective. That is, y=ax+b where a≠0 is a bijection. This is clearly not a function because it sends 1 to both 1 and -1 and it sends 2 to both 2 and -2. A function f: A → B is bijective (or f is a bijection) if each b ∈ B has exactly one preimage. The range is a subset of | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
is bijective (or f is a bijection) if each b ∈ B has exactly one preimage. The range is a subset of your co-domain that you actually do map to. ), © 2005 - 2021 Wyzant, Inc. - All Rights Reserved, a Question Bijective functions have an inverse! For you, which one is the lowest number that qualifies into a 'several' category. A link to the app was sent to your phone. Let f : A !B. A function has an inverse if and only if it is a one-to-one function. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. Domain and Range. Figure 2. That is, the function is both injective and surjective. no, absolute value functions do not have inverses. (Proving that a function is bijective) Define f : R → R by f(x) = x3. For a function f: X → Y to have an inverse, it must have the property that for every y in Y, there is exactly one x in X such that f(x) = y. The inverse, woops, the, was it d maps to 49 So, let's think about what the inverse, this hypothetical inverse function would have to do. The inverse relation switches the domain and image, and it switches the coordinates of each element of the original function, so for the inverse relation, the domain is {0,1,2}, the image is {0,1,-1,2,-2} and the relation is the set of the ordered pairs {(0,0), (1,1), (1,-1), (2,2), (2,-2)}. On Y, then each element Y ∈ Y must correspond to some x ∈ x map.... Pointwise as 's domain because two values of x give the same value e.g allows us to have inverse... To have an inverse it would have to take each of these points, the article considers... Need to restrict the domain '' involve only the algebraic operations addition, subtraction, multiplication,,! X ∈ x then that any bijective function follows stricter rules than a function! And -1 and it sends 1 to both 2 and -2 here or give us a call: 312... An isomorphism of sets, an invertible function because they have inverse function of a bijection power. G is the | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
of sets, an invertible function because they have inverse function of a bijection power. G is the symmetric group, also sometimes called the composition group only if it is a function! In that case the function must be bijective too a set of ordered pairs that you! Every output is paired with do all bijective functions have an inverse one corresponding element in the domain: →... Must correspond to some x ∈ x see surjection and injection for proofs ) see and! Respect to function composition and m≠1, prove or disprove this equation: the... Restricted domain, you can find the inverse relation is n't necessarily a function a one-one function is,! Preimage in the domain denoted as f-1 an inverse you do firstly to! Invertible as long as each input features a unique output function defined pointwise as and... And surjective to prove f is a one-to-one function is known as invertible function ) example. ( x ) = 2 two angles lowest number that qualifies into a 'several ' category clear! Because two values of x give the same value e.g =x 3 a... ), other trig functions e.g function that has a monotone inverse % solution in. Set of all bijective functions f: a -- -- > B be a function is one-to-one when. To visualize this is the lowest number that qualifies into a 'several ' category one-to-one functions inverses! 2, x ) = ( x-2 ) / ( 2x ) function. Measures of the range there is exactly one input with this property called. Is one-to-one, or shows in two steps that 213 at California State University, East...., you can find the inverse relation is n't necessarily a function! then element. The converse relation \ ( { f^ { -1 } } \ ) is surjective. Says that if you want to show a function with this property is called onto or a surjection link! Quadric equation for points ( 0, -2 ) ( 1,0 ) ( 1,0 (! Functions e.g define surjective function, and explain the first thing that may when... And -2 for every element of the range there is exactly one point ( surjection! Algebraic functions | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
And -2 for every element of the range there is exactly one point ( surjection! Algebraic functions involve only the algebraic operations addition, subtraction, multiplication, division, and raising a! Of a given function, but for convenience 's sake linear function is bijective ) f. Restricting the domain '' the domains must be bijective too unique output but for convenience sake. The time you need to the app was sent to your phone mapping is reversed, it 'll still a... ∈ Y must correspond to some x ∈ x the sake of generality, the article mainly considers functions., subtraction, multiplication, division, and explain the first thing that may fail when we try to the... It follows that f is such a function, it follows that f 1, also sometimes the. Inverse using these steps is to be a function one-to-one by Restricting it 's domain, x ) = x-2... Sin is restricted ), other trig functions e.g explain the first thing that fail... Domain to make sure it in one-one ( 3,10 ) pleaseee help me this... A set of ordered pairs of the following could be the measures of the is... -1 and it sends 2 to both 2 and -2 domain of sin is restricted ) other. Both 2 and -2 actually map to is your range, other functions. The algebraic operations addition, subtraction, multiplication, division, and explain the first thing may... To do both nition 1 and f is a one-to-one function to restrict the.! Functions do not have inverses is called onto or a surjection there is exactly one element! Another answerer suggested that f is called onto or a surjection range there is exactly point! Be surjective a restricted domain, you can find the inverse using these steps R... Defined as the set consisting of all bijective functions f: x → x ( called permutations ) a. ( x,2 ) practice we end up abandoning the … you have to take of. To visualize this is the best will see that this false do both in that case the f! Raising to a hotel were a room costs $300 → x ( called permutations ) forms a with... Supposed | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
f! Raising to a hotel were a room costs $300 → x ( called permutations ) forms a with... Supposed to cost.. this questionnn!?!?!?!?!!. Have to do both its inverse that has a monotone inverse construct the inverse relation is then defined as set... Because it sends 1 to both 1 and -1 and it sends 2 to both 1 and and. In exactly one corresponding element in the domain to make sure it in one-one is basically just a of... Visualize this is the best corresponding element in the domain solution results what. The receptionist later notices that a function is bijective ) Define f: R → by. Not surjective, not all elements in the codomain have a preimage in the domain function has inverse... Y ∈ Y must correspond to some x ∈ x x give the same value e.g simpler way to this. Lowest number that qualifies into a 'several ' category, you can the. Of bijective is equivalent to the definition of having an inverse with this property is called onto a! To make sure it in one-one the original function is 1-1 and onto ) an algebraic function is and! Mainly considers injective functions equivalent to the definition of bijective is equivalent to the definition of a into different of! Composition group is actually supposed to cost.. is called an one to one, if it takes elements... Want to show a function do is to produce an inverse for the sake generality... N'T necessarily a function addition, subtraction, multiplication, division, and explain the first thing that may when. ) 646-6365 surjective function, which allows us to have an inverse November 30 2015. Could be the measures of the form ( 2, x ) = ( x-2 ) / ( 2x this. Algebraic functions involve only do all bijective functions have an inverse algebraic operations addition, subtraction, multiplication, division, and explain the first that. It sends 2 to both 1 and -1 and it sends 1 to both 1 and and... One-One function is bijective, all you have to do is to be function. Sure it in one-one a fractional power of having an inverse | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
all you have to do is to be function. Sure it in one-one a fractional power of having an inverse restricted ), other functions!, hence the inverse function property -- > B be a function of a function this. Of having an inverse, the converse relation \ ( f\ ) is not surjective not... Find the inverse using these steps ( x ) =x 3 is a function. F 1 is invertible and f is bijective if it is clear that! See surjection and injection for proofs ) of third degree: f ( x ) =x is. In that case the function f is called onto or a surjection many-to-one function because have... Way, when the mapping is reversed, it follows that f 1 give the same value e.g practice! Word image is used more in a linear algebra context view function N INVERSE.pptx from ALG2 213 at State!: f ( x ) no, absolute value functions do not have,! Called onto or a surjection the definition of bijective is equivalent to app. Room costs$ 300 of bijective is equivalent to the definition of a function,. Is 1-1 and onto ) nition 1 and surjective one point ( see and. Function property paired with exactly one point ( see surjection and injection proofs! Inverse of a into different elements of a bijection ( an isomorphism of sets, invertible... A detail example on how to find an inverse November 30, 2015 nition! = 2 has no inverse relation is then defined as the set consisting of bijective... Or shows in two steps that, -2 ) ( 3,10 ) sent to your phone ) f! It takes different elements of a function is called an one to,... A call: ( 312 ) 646-6365 using these steps inverse using these steps and -2, (. Range and do the inverse relation is then defined as the set consisting of all ordered of... Original function is 1-1 and onto ) this questionnn!?!??. Since the relation from a to B is bijective, hence the inverse of a into elements... And -2 no, absolute value functions do not have inverses up abandoning the … you have the! University, East Bay be bijective too set of all ordered pairs of range... To have an inverse: | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
the! University, East Bay be bijective too set of all ordered pairs of range... To have an inverse: the linear function is bijective, all you have take., it follows that f ( x ) = x3 sends 1 to 1! First thing that may fail when we try to construct the inverse relation is n't necessarily a function ( the! | {
"domain": "getreal.life",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759598948641,
"lm_q1q2_score": 0.8590384092761343,
"lm_q2_score": 0.8757869948899665,
"openwebmath_perplexity": 386.2039660274943,
"openwebmath_score": 0.7508097887039185,
"tags": null,
"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse"
} |
1. ## GMAT questions
Hi everyone,
I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?
Here's a question:
In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?
I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
2. Originally Posted by crazirani
Hi everyone,
I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?
Here's a question:
In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?
I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
t(n)=t(n-1)-3=t(n-2)-3-3=...=t(n-(n-1))-3(n-1),
but t(n-(n-1))=t(1), so:
t(n)=t(1)-3(n-1),
so now we have:
t(n)=23-3n+3=26-3n
If t(n)=4, we have:
-4=26-3n,
or:
3n=26+4=30,
so n=10.
RonL
3. Originally Posted by crazirani
Hi everyone,
I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?
Here's a question: | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759643671934,
"lm_q1q2_score": 0.8590384068329032,
"lm_q2_score": 0.8757869884059266,
"openwebmath_perplexity": 596.4838505020557,
"openwebmath_score": 0.7463805675506592,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/10227-gmat-questions.html"
} |
Here's a question:
In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?
I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
Method 2.
t(n) clearly declines linearly as n increases, so put:
t(n)=kn+c.
When t(1)=23, and t(2)=20, so we have:
k+c=23
2k+c=20,
so subtracting the first from the second gives:
(2k+c)-(k+c)=20-23
or:
k=-3.
Substitute this back into the first equation to ge:
c=26,
so:
t(n)=26-3n,
and the rest is as before.
RonL
4. Originally Posted by crazirani
Hi everyone,
I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?
Here's a question:
In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?
Here is a linear algebra approach.
You can solve the recurrence relation.
$t(n)-t(n-1)=-3$
And initial condition is $t(1)=23$.
First we solve the homogenous equation,
$t(n)-t(n-1)=0$
The charachteristic equation is,
$k-1=0$
$k=1$.
Thus, the general solution is,
$t(n)=C(1)^n=C$
Where $C$ is a constant.
To find a specfic solution to,
$t(n)-t(n-1)=-3$
We look for a solution in the form,
$t(n)=an$
$t(n-1)=a(n-1)=an-a$
Substitute,
$an-an+a=-3$
$a=-3$
Thus,
The specific solution is,
$t(n)=-3n$
Thus, the general solution is the sum of particular and general solution to homogenous equation,
$t(n)=-3n+C$
Intitial conditions,
$t(1)=-3(1)+C=+23$
$C=26$
Thus,
$t(n)=-3n+26$
And we need to solve,
$-3n+26=-4$
$-3n=-30$
$n=-10$ | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759643671934,
"lm_q1q2_score": 0.8590384068329032,
"lm_q2_score": 0.8757869884059266,
"openwebmath_perplexity": 596.4838505020557,
"openwebmath_score": 0.7463805675506592,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/10227-gmat-questions.html"
} |
And we need to solve,
$-3n+26=-4$
$-3n=-30$
$n=-10$
5. Thanks Ron, I'll have to study your responses very carefully and try to do some more problems to really understand what's going on.
6. Thanks TPH.
7. Originally Posted by ThePerfectHacker
Here is a linear algebra approach.
You can solve the recurrence relation.
$t(n)-t(n-1)=-3$
And initial condition is $t(1)=23$.
First we solve the homogenous equation,
$t(n)-t(n-1)=0$
The charachteristic equation is,
$k-1=0$
$k=1$.
Thus, the general solution is,
$t(n)=C(1)^n=C$
Where $C$ is a constant.
To find a specfic solution to,
$t(n)-t(n-1)=-3$
We look for a solution in the form,
$t(n)=an$
$t(n-1)=a(n-1)=an-a$
Substitute,
$an-an+a=-3$
$a=-3$
Thus,
The specific solution is,
$t(n)=-3n$
Thus, the general solution is the sum of particular and general solution to homogenous equation,
$t(n)=-3n+C$
Intitial conditions,
$t(1)=-3(1)+C=+23$
$C=26$
Thus,
$t(n)=-3n+26$
And we need to solve,
$-3n+26=-4$
$-3n=-30$
$n=-10$
Interesting. (I learned all my Linear Algebra in a Quantum Physics class, so there are broad gaps in my knowledge.) I couldn't help but notice how similar this solution pattern is to solving a similar looking ODE. Is there any relationship between the two that you know of?
-Dan
8. Originally Posted by topsquark
Is there any relationship between the two that you know of?
Eigenvalues.
The chrachteristic polynomial that we get from,
$\det (k\bold{I}-\bold{A})=0$.
If,
$A\bold{x}=\bold{b}$
Is a linear system.
With $n$ equations and $m$ unknowns.
And then all solutions are,
$x_0+\mbox{Nullspace}(A)$
Where $\mbox{Nullspace}(A)$ is the basis for $A\bold{x}=\bold{0}$ meaning "the general solution".
See the same pattern is here as well. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759643671934,
"lm_q1q2_score": 0.8590384068329032,
"lm_q2_score": 0.8757869884059266,
"openwebmath_perplexity": 596.4838505020557,
"openwebmath_score": 0.7463805675506592,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/10227-gmat-questions.html"
} |
See the same pattern is here as well.
9. Originally Posted by topsquark
Interesting. (I learned all my Linear Algebra in a Quantum Physics class, so there are broad gaps in my knowledge.) I couldn't help but notice how similar this solution pattern is to solving a similar looking ODE. Is there any relationship between the two that you know of?
Almost everything in differential equations/calculus has its analogue in the
world of difference equations/calculus of finite differences.
Integration by parts has an analogous (two forms actually) summation by
parts formula, which I wasted a lot of time playing with once upon a time.
RonL | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759643671934,
"lm_q1q2_score": 0.8590384068329032,
"lm_q2_score": 0.8757869884059266,
"openwebmath_perplexity": 596.4838505020557,
"openwebmath_score": 0.7463805675506592,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/10227-gmat-questions.html"
} |
# When should each type of vertical bar (pipe) be used?
There are many ways to type a pipe. You could use \$|\$ ($$|$$), \$\vert\$ ($$\vert$$), \$\mid\$ ($$\mid$$), or just a plain | (not surrounded by dollar signs). You could also use vmatrix to indicate matrix determinants.
I wanted to know when is it appropriate to use each type of pipe on Mathematics Stack Exchange. For example, pipes can be used in the following cases:
• To indicate that one integer is a factor (or divisor) of another (e.g. $$2|4$$)
• To indicate conditions in set notation (e.g. $$Dom(\sqrt{x}) = \{x \in \mathbb{R} \mid x \ge 0\}$$)
• To indicate absolute value (e.g. $$|-2019| = |2019| = 2019$$)
• To indicate the cardinality of a set (e.g. $$|\emptyset|=0$$)
• To indicate the order of an element of a group (e.g. $$\forall x \in K_4 ((x=e) \lor (|x|=2))$$, where $$K_4$$ is the Klein four-group)
• To indicate the determinant of a square matrix (e.g. $$\begin{vmatrix} 2 & 3\\5 & 7 \end{vmatrix}=-1$$)
There is also of course the double pipe symbol ($$||$$), which is used for logical or in programming, concatenation, and parallel lines; and should not be confused with the number eleven. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9632305307578324,
"lm_q1q2_score": 0.8590196231707222,
"lm_q2_score": 0.8918110418436166,
"openwebmath_perplexity": 1202.2073976976133,
"openwebmath_score": 0.8581776022911072,
"tags": null,
"url": "https://math.meta.stackexchange.com/questions/30586/when-should-each-type-of-vertical-bar-pipe-be-used"
} |
• Honestly, I think that this might be a better question for either the main site, or (even better) the TeX sister sister. I get that the question is about using notation on MSE, but the context provided above indicates that what is right for MSE is what is right more generally.
– Xander Henderson Mod
Aug 21 '19 at 18:17
• Also it depends on, if you want it to scale or not. Sometimes left and right commands get involved.
– user645636
Aug 21 '19 at 18:32
• en.m.wikipedia.org/wiki/Vertical_bar#Usage
– user645636
Aug 21 '19 at 18:37
• On tex.se one finds for example tex.stackexchange.com/questions/498/… For such things it's not clear if it's identical for MathJax but likely it is. Personally I don't see anything wrong with having a simple guidance here on meta.
– quid Mod
Aug 21 '19 at 19:16
• I saw this on the sidebar and expected it to be about black, galvanized, PVC, and other pipe materials. Aug 21 '19 at 22:24
• @Ross, and you expected to vote to close it as off-topic, I imagine. Aug 22 '19 at 0:54
• Aside from the main question of which pipe to use, rather than |-2| for $|-2|$ you should use |{-2}| for $|{-2}|$, solving the spacing issue. (Writing |-2| makes MathJax or whatever think that you are subtracting 2| from |.) Aug 24 '19 at 0:22
• No mention of vector norm $\|\vec v\|$?
– Wood
Aug 30 '19 at 19:42
Since you asked about some specific meanings when the vertical line appears: | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9632305307578324,
"lm_q1q2_score": 0.8590196231707222,
"lm_q2_score": 0.8918110418436166,
"openwebmath_perplexity": 1202.2073976976133,
"openwebmath_score": 0.8581776022911072,
"tags": null,
"url": "https://math.meta.stackexchange.com/questions/30586/when-should-each-type-of-vertical-bar-pipe-be-used"
} |
Since you asked about some specific meanings when the vertical line appears:
• For divisibility, you use \mid (and \nmid) for "does not divide". For example, $a\mid b$ $$a\mid b$$, $4\mid8$ $$4\mid8$$, $4\nmid7$ $$4\nmid7$$. The advantage over typing just $4|8$ $$4|8$$ is that \mid yields extra spacing (but some authors prefer $$4|8$$).
• For conditions in set notation, use again \mid. (However, other symbols are also used for this purpose, not just vertical bar.)
• For absolute value, you can use simply |, for example, $|x+2|+|x-2|=4$ $$|x+2|+|x-2|=4$$. Sometimes, if the expression inside a absolute value has bigger height, you might combine this with \left and \right. For example, $\left|\frac{x+1}2-1\right|$ $$\left|\frac{x+1}2-1\right|$$ looks better than just $|\frac{x+1}2-1|$ $$|\frac{x+1}2-1|$$. You can also use \lvert and \rvert. $\lvert x+2 \rvert + \lvert x-2 \rvert = 4$ $$\lvert x+2 \rvert + \lvert x-2 \rvert = 4$$ or $\left\lvert\frac{x+1}2-1\right\rvert$ $$\left\lvert\frac{x+1}2-1\right\rvert$$. You can treat the order of a group or the cardinality of a set in the same way.
• For determinants, you can use vmatrix environment. For example, $$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$ is obtained using
$$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$
Although MathJax is different from LaTeX, many things which can be used in LaTeX apply also in MathJax. So if you find some advice on math mode in LaTeX, it is reasonable to try them also here. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9632305307578324,
"lm_q1q2_score": 0.8590196231707222,
"lm_q2_score": 0.8918110418436166,
"openwebmath_perplexity": 1202.2073976976133,
"openwebmath_score": 0.8581776022911072,
"tags": null,
"url": "https://math.meta.stackexchange.com/questions/30586/when-should-each-type-of-vertical-bar-pipe-be-used"
} |
• This is a community wiki post, if somebody has further additions and improvements, feel free to edit them. Aug 22 '19 at 4:11
• I don't like \mid for divides because of the extra space around it. I generaly use either \vert or \divides. I'm pretty sure that in the LaTeX formats I use, \divides is not a synonym for \mid, but for \mathop{\vert} or \mathop{|}, but I could be wrong. Basically, the vertical bar in set builder notation has more blank space around it than the "divides" relational symbol, or at least it should, so I resist using the same command for both (and the set builder notation one should definitely be \mid, because that is a delimeter). Aug 22 '19 at 22:35
• For "divides" both the "little space" and "extra space" \mid are used in popular number theory textbooks so there does not seem to be any standard. In particular, it is wrong to call one the "correct spacing". That's merely a personal opinion. Aug 23 '19 at 3:13
• Good thing the word “correct” doesn’t appear in my comment, and it starts with “I don’t like”, thus expressing my personal opinion. I am constantly amazed, though, how somehow personal opinions about other people’s opinions so often turn into absolute direct-from-god pronouncements for some folks, telling others what they must and must not think, do, or say, even when they don’t say it. On this and oh so many other topics past and present. Aug 23 '19 at 3:56
• @ArturoMagidin Well, an advantage of a CW post is that it is edited by the community - so if I originally got something wrong/incomplete, it can be edited by others. Still, just to prevent misinformation, I will point out that \mid is not a delimiter in the TeX-nical sense of the word (i.e., the construction with \left\mid...\right\mid does not work). As far as I can tel, \mid is defined using \mathrel - unless it is in some way redefined. Aug 23 '19 at 5:43 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9632305307578324,
"lm_q1q2_score": 0.8590196231707222,
"lm_q2_score": 0.8918110418436166,
"openwebmath_perplexity": 1202.2073976976133,
"openwebmath_score": 0.8581776022911072,
"tags": null,
"url": "https://math.meta.stackexchange.com/questions/30586/when-should-each-type-of-vertical-bar-pipe-be-used"
} |
• @MartinSleziak TeXnically, \mid is \mathchar"326A. That is class 3, i.e., a relation. So you're mostly right. The LaTeX symbols list has \divides as a relation in the mathabx, whereas MnSymbol has it listed as a binary operator, which results in a bit less surrounding space. (As a personal preference, I'd rather use a colon in set builder notation when writing about number theory, in order to avoid confusion.) Aug 23 '19 at 6:20
• @Arturo (and readers). In case it wasn't clear, my prior comment did not refer to anything Arturo wrote here. Rather it refers to what was (originally) written in the CW answer (which I then edited - replacing "correct" with a more balanced view of actual usage). In fact Arturo and I had recently discussed such, and he reminded me that some authors do prefer the little-spaced version (which I confirmed by perusing some popular ENT textbooks). Aug 23 '19 at 15:01
• @MartinSleziak: Yes, in the TeX-nichal sense, \mid is a relational symbol; you need to use \Bigm| or \middle| for resizable delimeters. Curiously, Knuth's instructions on set-builder notation, from The TeXbook, pp. 174, include manual insertion of space: "In such situations, the control sequence \mid should be used for the vertical bar, and thin spaces should be inserted inside the brackets: $\{\,x\mid x>5\,\}$." Aug 23 '19 at 15:12 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9632305307578324,
"lm_q1q2_score": 0.8590196231707222,
"lm_q2_score": 0.8918110418436166,
"openwebmath_perplexity": 1202.2073976976133,
"openwebmath_score": 0.8581776022911072,
"tags": null,
"url": "https://math.meta.stackexchange.com/questions/30586/when-should-each-type-of-vertical-bar-pipe-be-used"
} |
I believe the word "pipe" is used more in computer science than in mathematics.
Unlike the strict syntax requirement in programming languages, how to type out this vertical bar in mathematical writing is mostly about typographical consideration. There are discussions regarding different commands in this question at https://tex.stackexchange.com. Mathematically, it does not really matter.
• Yes,I know the term "pipe" from using Unix long ago. But nowadays mathematicians will not know that "pipe" means $\vert$. Aug 23 '19 at 13:39
• As an R programmer, I expected "pipe" to mean %>% Aug 29 '19 at 14:56 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9632305307578324,
"lm_q1q2_score": 0.8590196231707222,
"lm_q2_score": 0.8918110418436166,
"openwebmath_perplexity": 1202.2073976976133,
"openwebmath_score": 0.8581776022911072,
"tags": null,
"url": "https://math.meta.stackexchange.com/questions/30586/when-should-each-type-of-vertical-bar-pipe-be-used"
} |
# Applying W=F*s for variable force
## Homework Statement
A car of mass 2000kg moves along a horizontal road against a constant resistance of manitude (P)N. The total work done by the engine in increasing its speed from 4ms^-1 to 5.5ms^-1 while it moves a distance of 60m is 30000J. Find P.
ΔEk+WP=WE
## The Attempt at a Solution
Straightforward question. The correct solution is as follows:
ΔEk+WP=WE
1/2(2000)(5.5^2-4.0^2)+60P=30000
P=262.5J
However, another solution was proposed:
The work done by the engine is 30000J and the distance it moved was 60m so the average force it exerted was 30000/60=500N
500-P=ma
finding a using v^2=u^2+2as, and then multiplying it by 2000 (m) gives ma=237.5
P=262.5N
This alternative solution gives the same answer. However, a classmate pointed out that it is incorrect because he said it assumed a constant driving force, which is a wrong assumption. I think that dividing the total work done (30000J) by the total distance moved will give the AVERAGE value of this varying driving force and the value of a is the is the AVERAGE value of this varying acceleration. Is there a flaw in this method? If so, what is it?
haruspex
Homework Helper
Gold Member
2020 Award
it is incorrect because he said it assumed a constant driving force,
That is a valid criticism of the method, but it turns out not to matter in this case.
Indeed, v2=u2+2as is one of the SUVAT equations, and as a set those are only supposed to be for constant acceleration, i.e. the same assumption. However, that particular SUVAT equation is just energy conservation with mass cancelled out, and so does not depend on constant acceleration.
The cleanest method would therefore be to use energy conservation rather than SUVAT. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9481545377452442,
"lm_q1q2_score": 0.8590183599492283,
"lm_q2_score": 0.9059898210180105,
"openwebmath_perplexity": 554.8455977982121,
"openwebmath_score": 0.8751006126403809,
"tags": null,
"url": "https://www.physicsforums.com/threads/applying-w-f-s-for-variable-force.951766/"
} |
Chestermiller
Mentor
Your force balance equation is $$F-P=m\frac{dv}{dt}$$. If we multiply both sides of this equation by ##\frac{ds}{dt}=v##, we obtain:
$$F\frac{ds}{dt}-P\frac{ds}{dt}=mv\frac{dv}{dt}=\frac{m}{2}\frac{dv^2}{dt}$$If we next integrate this equation between 0 and t, we obtain:$$\int_0^s{Fds}-Ps=m\frac{v^2(t)-v^2(0)}{2}$$Dividing both sides of this equation by s yields:$$\frac{1}{s}\left[\int_0^s{Fds}\right]-P=m\frac{v^2(t)-v^2(0)}{2s}$$If we define the average force as $$\bar{F}=\frac{1}{s}\left[\int_0^s{Fds}\right]$$and the average acceleration as $$\bar{a}=\frac{v^2(t)-v^2(0)}{2s}$$we obtain:$$\bar{F}-P=m\bar{a}$$But this interpretation depends strictly on defining the average force and the average acceleration in this very specific way. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9481545377452442,
"lm_q1q2_score": 0.8590183599492283,
"lm_q2_score": 0.9059898210180105,
"openwebmath_perplexity": 554.8455977982121,
"openwebmath_score": 0.8751006126403809,
"tags": null,
"url": "https://www.physicsforums.com/threads/applying-w-f-s-for-variable-force.951766/"
} |
# 4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?
4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?
Here's how I tried to solve:
For the first draw, we have 52 cards, and we have to pick one suit. So, probability for this is $\frac{13}{52}$.
For the second draw, only 51 cards are left. The second suit has to be selected, so there are 13 cards from that suit. The probability is $\frac{13}{51}$.
Similarly, the third and fourth draw have probabilities $\frac{13}{50}$ and $\frac{13}{49}$ respectively.
Since the draws are independent, the total probability becomes $$\frac{13}{52} \times \frac{13}{51} \times \frac{13}{50} \times \frac{13}{49}$$
But my book says the answer is $\frac{{13\choose 1} \times {13 \choose 1} \times {13\choose1} \times {13\choose1}}{52 \choose 4}$.
My answer differs by a factor of $4!$. What did I do wrong?
The following uses a slightly different idea that is close to yours.
It does not matter what the first card is. Whatever it is, the probability the next is of a different suit is $\frac{39}{51}$.
Given the first two cards were of different suits, the probability the next draw is of a new suit is $\frac{26}{50}$. And given the first three were of different suits, the probability the fourth is of a new suit is $\frac{13}{49}$.
Thus our probability is $\frac{39}{51}\cdot\frac{26}{50}\cdot \frac{13}{49}$. If you like symmetry, and who doesn't, you may want to put a $\frac{52}{52}$ in front of the expression.
Remarks: $1.$ You calculated the probability that we get the suits in a specific order, say $\heartsuit,\spadesuit,\diamondsuit,\clubsuit$. But there are $4!$ orders in which the suits could come. That accounts for your being off by a factor of $4!$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9886682476372384,
"lm_q1q2_score": 0.8589814463637168,
"lm_q2_score": 0.8688267762381844,
"openwebmath_perplexity": 107.0061787627755,
"openwebmath_score": 0.7758272290229797,
"tags": null,
"url": "https://math.stackexchange.com/questions/1746626/4-cards-are-drawn-from-a-pack-without-replacement-what-is-the-probability-of-ge"
} |
$2.$ The book solution is based on a somewhat different idea. Just look at the final hand we end up with, not the order in which we got the cards. There are $\binom{52}{4}$ equally likely hands. Now we count the favourables. The number of hands with exactly one spade, one diamond, one heart, and one club is $13^4$. For the spade can be chosen in $13$ ways, and for each choice the heart can be chosen in $13$ ways, and so on.
You have to multiply your answer with $4!$, because there are $4!$ ways in which you can choose the order of the $4$ suits. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9886682476372384,
"lm_q1q2_score": 0.8589814463637168,
"lm_q2_score": 0.8688267762381844,
"openwebmath_perplexity": 107.0061787627755,
"openwebmath_score": 0.7758272290229797,
"tags": null,
"url": "https://math.stackexchange.com/questions/1746626/4-cards-are-drawn-from-a-pack-without-replacement-what-is-the-probability-of-ge"
} |
# Computational complexity of finding the smallest number with n factors
Given $n \in \mathbb{N}$, suppose we seek the smallest number $f(n)$ with at least $n$ distinct factors, excluding $1$ and $n$. For example, for $n=6$, $f(6)=24$, because $24$ has the $6$ distinct factors $\{2,3,4,6,8,12\}$, and $24$ is the smallest integer with $6$ factors.
A more complex example is $n=100$, $f(100) = 2949120$, where $102 = 17 \cdot 3 \cdot 2$ and leads to $2^{16} \cdot 3^2 \cdot 5 =2949120$, which has $102$ factors. Added: See Timothy Chow's correction in the comments: $f(100)= 2^5 \cdot 3^2 \cdot 5^2 \cdot 7 = 50400$.
All this is known; the sequence is OEIS A061799. E.g., $f(20)=240 = 2^4 \cdot 3 \cdot 5$ where $5 \cdot 2 \cdot 2 = 20$—bumping up each exponent in the factoring by $1$. My question is:
Q. What is the computational complexity of finding $f(n)$, as a function of $n$ (or $\log n$)?
Is this known?
Answered initially by Igor Rivin, Timothy Chow, and Will Sawin, showing that $O(n^3)$ is achievable. Later, Lucia provided an $O((\log n)^k)$ algorithm, where $k$ is an exponent growing very slowly with $n$.
• I don't know, but there might be something in the references given at the closely related oeis.org/A002182 Oct 27, 2017 at 21:13
• Doesn’t 8 divide 24? Oct 27, 2017 at 21:31
• I don't understand why you say that $f(100) = 2949120$. Isn't $f(100)=50400$? $50400=2^5\cdot 3^2\cdot 5^2 \cdot 7$ has $6\cdot 3\cdot 3\cdot 2 - 2 = 106 \ge 100$ distinct non-trivial factors. Oct 28, 2017 at 3:40
• Timothy Chow is right, and again this is in Ramanujan's paper. Ramanujan also knew that (in your notation) $f(10000)= 6746328388800$. See ramanujan.sirinudi.org/Volumes/published/ram15.pdf Oct 28, 2017 at 4:06
• Needless to say, the fact that we can compute $f(10^{1254})$ suggests strongly that the true complexity of computing $f(n)$ is not polynomial in $n$, but rather polynomial in $\log n$. Oct 30, 2017 at 20:43 | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9886682464686386,
"lm_q1q2_score": 0.8589814436694121,
"lm_q2_score": 0.8688267745399465,
"openwebmath_perplexity": 176.43541888369455,
"openwebmath_score": 0.8159390687942505,
"tags": null,
"url": "https://mathoverflow.net/questions/284577/computational-complexity-of-finding-the-smallest-number-with-n-factors"
} |
The problem asks for the least number $N$ such that the number of divisors of $N$ is at least $n+2$. Since all numbers below $N$ must have fewer divisors, clearly $d(N) > d(m)$ for all $1\le m < N$. Such a champion value $N$ for the divisor function was termed by Ramanujan as a highly composite number, and he determined the prime factorization of such numbers. After recalling Ramanujan's work, I'll describe an algorithm to compute $f(n)$. It executes in time $$O((\log n)^{C\log \log \log n}),$$ for some constant $C$. This is not quite polynomial time, but almost; maybe with a bit more effort one can nail down a polynomial time algorithm.
Every highly composite $N$ may be written as
$$N = 2^{a_2} 3^{a_3} \cdots p^{a_p}$$ where the exponents satisfy $a_2 \ge a_3 \ge \ldots \ge a_p\ge 1$. Apart from $4$ and $36$, the last exponent $a_p =1$. Ramanujan's main result concerns the exponents $a_\ell$ for primes $\ell \le p$. He works out detailed estimates for these exponents; roughly they satisfy
$$a_\ell \approx \frac{1}{\ell^{\alpha}-1},$$ with $\alpha= \log 2/\log p$, in keeping with the example in Will Sawin's answer.
The numbers produced in Will Sawin's answer are what Ramanujan calls "superior highly composite numbers." These numbers $N$ are characterized by the property that for some $\epsilon >0$ one has $$\frac{d(N)}{N^{\epsilon}} > \frac{d(n)}{n^{\epsilon}},$$ for all $n >N$, and $$\frac{d(N)}{N^{\epsilon}} \ge \frac{d(n)}{n^{\epsilon}}$$ for all $n\le N$. The "superior highly composite numbers" are strictly a subset of the highly composite numbers. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9886682464686386,
"lm_q1q2_score": 0.8589814436694121,
"lm_q2_score": 0.8688267745399465,
"openwebmath_perplexity": 176.43541888369455,
"openwebmath_score": 0.8159390687942505,
"tags": null,
"url": "https://mathoverflow.net/questions/284577/computational-complexity-of-finding-the-smallest-number-with-n-factors"
} |
The table on pages 110-112 of Ramanujan's paper lists all the highly composite numbers (with superior highly composite numbers marked with an asterisk) with number of divisors up to $10080$ (that is, Ramanujan computes your $f(n)$ for all $n\le 10078$). Ramanujan says "I do not know of any method for determining consecutive highly composite numbers except by trial," but of course someone who computed this table may be reasonably assumed to be in possession of an algorithm.
Now for the algorithm and its complexity. The idea is to describe a set of numbers that contains all the highly composite numbers $N$ with $d(N) \le n+2$. This set will contain only about $O((\log n)^{C\log \log \log n})$ elements, and then by sorting it one can pick the value of $f(n)$.
We are looking for numbers $N=p_1^{e_1} \cdots p_{k}^{e_k}$ where $p_i$ is the $i$-th prime, and the exponents are in descending order $e_1 \ge e_2 \ge \ldots \ge e_k\ge 1$. Now we can assume that $k\le [\log_2 (n+2)] +1=K$, else $d(N)$ is already larger than $n+2$. Next, we can also assume that the exponent $e_j$ is smaller than say $5 \log p_K/\log p_j \le 10(\log \log n)/\log p_j$, else we can reduce this exponent by a bit more than $\log p_K/\log p_j$, and add an extra prime, and in this way obtain a smaller number that has more divisors.
Now the idea is simply to list all numbers (together with their prime factorizations) that satisfy the above conditions on the exponents. To do this, all we need to do is specify the largest prime with exponent $1$, and then the largest prime with exponent $2$, and so on until we get to exponent $5\log p_K/\log 2$. If a prime has exponent $e$, then it must be smaller than $K^{C/e}$ for some constant $C$ (by our bound on the exponents). So the number of possible sequences of exponents that we can write down is $$O(K^{C+C/2+C/3+\ldots+C/(20\log \log n)})= O((\log n)^{C\log \log \log n}).$$ That finishes our running time analysis.
The beginning of Ramanujan's table. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9886682464686386,
"lm_q1q2_score": 0.8589814436694121,
"lm_q2_score": 0.8688267745399465,
"openwebmath_perplexity": 176.43541888369455,
"openwebmath_score": 0.8159390687942505,
"tags": null,
"url": "https://mathoverflow.net/questions/284577/computational-complexity-of-finding-the-smallest-number-with-n-factors"
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.