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On Mar 23, 2023, at 1:24 PM, S.Y. Lee <syle...@gmail.com> wrote:
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Wolfram had recently announced the collaboration of chatGPT and wolfram alpha
ChatGPT Gets Its “Wolfram Superpowers”!—Stephen Wolfram Writings
They start to use chatgpt to generate the Wolfram code.
And it is likely to help the issues with correctness about math or science facts
because once it translates to the wolfram functions, and the wolfram function runs without error, the answer is correct.
However, the argument I'd give is that making a combination of code with
simplify, solve, integrate are not still informative because even though they do things logically correct,
they have problem that they can't inform people how to solve the problems in details.
So I'm thinking about an idea whether language models should really
generate a code that assembles the rules used in simplify, solve.
And then it can be more human readable (or can get more human readable logic from it)
if it is like compose(solve_trig, simplify_cos, ...)
(in some pseudo sympy code)
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Thanks Chris,
What I had meant to write was Sum(f(k), (k, 0, 10)).doit() ! My main issue was to use sum instead of Sum!
David
Find a 4x4 matrix which is a Jordan canonical form of the matrix A = matrix([[5, -1, 0, 0],[9, -1, 0, 0],[0, 0, 7, -2], [0, 0, 12, -3]]). assistant: To find the Jordan canonical form of a matrix, we need to find the eigenvalues and eigenvectors, and then use them to construct the Jordan blocks. First, we find the eigenvalues of A by solving for the determinant of (A - lambda*I), where I is the identity matrix: |5 - lambda -1 0 0 | |9 -1 - lambda 0 0 | |0 0 7 - lambda -2 | |0 0 12 -3 - lambda| = (5 - lambda)(-1 - lambda)(7 - lambda)(-3 - lambda) - 9*(-1)*(7 - lambda)*(12) = lambda^4 - 8*lambda^3 + 11*lambda^2 + 38*lambda - 360 = (lambda - 10)(lambda - 5)(lambda + 3)(lambda - 4) Therefore, the eigenvalues are: lambda1 = 10, lambda2 = 5, lambda3 = -3, lambda4 = 4. Next, we find the eigenvectors corresponding to each eigenvalue. For lambda1 = 10, we solve (A - 10*I)x = 0: | -5 -1 0 0 | | 9 -11 0 0 | | 0 0 -3 -2 | | 0 0 12 -13 | By performing elementary row operations to reduce this matrix, we obtain: | 1 0 0 1 | | 0 1 0 3/2 | | 0 0 1 2 | | 0 0 0 0 | So the eigenvector corresponding to lambda1 is x1 = [1, -3/2, -2, 1]. Similarly, for lambda2 = 5, we solve (A - 5*I)x = 0: | 0 -1 0 0 | | 9 -6 0 0 | | 0 0 2 -2 | | 0 0 12 -8 | By performing elementary row operations to reduce this matrix, we obtain: | 1 -3 0 0 | | 0 0 1 -1 | | 0 0 0 0 | | 0 0 0 0 | So the eigenvector corresponding to lambda2 is x2 = [3, 1, 0, 0]. For lambda3 = -3, we solve (A + 3*I)x = 0: | 8 -1 0 0 | | 9 -2 0 0 | | 0 0 4 -2 | | 0 0 12 -6 | By performing elementary row operations to reduce this matrix, we obtain: | 1 -1/2 0 0 | | 0 0 1 -1/2 | | 0 0 0 0 | | 0 0 0 0 | So the eigenvector corresponding to lambda3 is x3 = [1/2, 1, 1/2, 0]. Finally, for lambda4 = 4, we solve (A - 4*I)x = 0: | 1 -1 0 0 | | 9 -5 0 0 | | 0 0 3 -2 | | 0 0 12 -7 | By performing elementary row operations to reduce this matrix, we obtain: | 1 -1 0 0 | | 0 0 1 -2/3 | | 0 0 0 0 | | 0 0 0 0 | So the eigenvector corresponding to lambda4 is x4 = [1, 1/3, 0, 0]. We can now construct the Jordan canonical form of the matrix A by arranging these eigenvectors in blocks, as follows: J = [10 1 0 0 ] [0 10 0 0 ] [0 0 -3 1 ] [0 0 0 4 ] This is the desired 4x4 Jordan canonical form of A.
Example 2:
Find an invertible 3x3 real matrix P which puts the matrix A = matrix([[3,4,0],[-1, -3, -2],[1, 2, 1]]) into rational form. assistant: To put a matrix in rational form, we want to reduce it to row echelon form and then perform row operations to get all leading coefficients to be 1 and all other entries in the same columns to be 0. Here are the steps to do it for matrix A: Step 1: Choose the first element in the first row to be the pivot and eliminate all entries in the column below it. To do this, subtract the first row multiplied by (-1/3) from the second row, and add the first row to the third row. [3 4 0] [0 1 2] [4 6 1] Step 2: Choose the second element in the second row to be the pivot and eliminate all entries in the column below it. To do this, subtract the second row multiplied by 2 from the third row. [3 4 0] [0 1 2] [0 -2 -7] Step 3: Choose the third element in the third row to be the pivot and eliminate all entries in the column above it. To do this, add the second row multiplied by 2 to the third row. [3 4 0] [0 1 2] [0 0 -3] Now we have the row echelon form of matrix A. To get it in rational form, we can perform further row operations to get all leading coefficients to be 1. We can achieve this by multiplying the third row by -1/3. [3 4 0] [0 1 2] [0 0 1] Now we have the rational form of matrix A. To find the invertible matrix P that puts A into this form, we need to keep track of the row operations we performed. First, we subtracted the first row multiplied by (-1/3) from the second row. This is equivalent to multiplying by the matrix: [1 0 0] [-1/3 1 0] [0 0 1] Next, we added the first row to the third row. This is equivalent to multiplying by the matrix: [1 0 0] [0 1 0] [1 0 1] Then, we subtracted the second row multiplied by 2 from the third row. This is equivalent to multiplying by the matrix: [1 0 0] [0 1 0] [0 -2 1] Finally, we multiplied the third row by -1/3. This is equivalent to multiplying by the matrix: [1 0 0] [0 1 0] [0 0 -1/3] To find P, we multiply these matrices in order: P = [1 0 0] @ [-1/3 1 0] @ [1 0 0] @ [0 1 0] @ [0 -2 1] @ [1 0 0] @ [0 1 0] @ [0 0 -1/3] Simplifying this expression gives: P = matrix([[-1/3, -8/3, -2/3],[1/3, 5/3, 2/3],[-4/3, -2/3, 1/3]]) We can check that matrix A can be transformed to its rational form by multiplying it by P and verifying that the result is indeed the rational form: A_rational = P.inv() @ A @ P Output: A_rational = matrix([[1, 0, 0],[0, 1, 0],[0, 0, 1]])
Elimboto
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The norm of a vector is defined as the length or magnitude of the vector. The norm of a vector v = (v1, v2, ..., vn) is given by the formula:
||v|| = sqrt(v1^2 + v2^2 + ... + vn^2)
So, for the vector (1/2, 1, 1), the norm is:
|| (1/2, 1, 1) || = sqrt( (1/2)^2 + 1^2 + 1^2 ) = sqrt( 1/4 + 1 + 1 ) = sqrt( 6.25 ) = 2.5
Therefore, the norm of the vector (1/2, 1, 1) is 2.5.
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