# Jiapeng

Last seen: plus d'un an il y a Actif depuis 2022

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I was asked to use qr shift to calculate the eigenvalues of a matrix where the matrix with tolerance 10^(-16). Is qr shift a sta...

plus d'un an il y a | 1 réponse | 0

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Compute the roots of a general nth-degree polynomial
A = [ ]; a = compan(A); e = eig(a); I am trying to find the roots of a general polynomial of degree n using the eigenvalue...

plus d'un an il y a | 1 réponse | 1

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How to get the numerical rank of the matrix in matlab?
Suppose there exists a 3*3 matrix A. I know I can use rank(A) to get the rank of A. However, if there exists ε>0, how should I u...

plus d'un an il y a | 2 réponses | 0

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Picking a particular column to norm in matrix
A = [1.02 0.95 0.77 0.67 0.56 0.30 0.16 0.01]; b = [0.39 0.32 0.22 0.18 0.15 0.12 0.13 0.15]; da = [1.02+(1.2e-3) 0.95+(6.4e-...

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The results display problem
My code ishere. How about displaying every x and r from 1 to 4? A= [1.02 0.95 0.77 0.67 0.56 0.30 0.16 0.01]; b = [0.39 0.32 ...

plus d'un an il y a | 1 réponse | 0

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solve Ax=b by singular value decomposition
After I've got U,S,V by [U,S,V] = svd(A), how do I get x? Is it x = (V*(S^(-1)*U')*b?

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Singular value decomposition method for solving least squares problems
For Ax=y, how to show in matlab? In particular, how to show the inverse of the sum of the matrix.

plus d'un an il y a | 2 réponses | 0

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The problem of constants in linear least squares
The equation of the model is ay^2+bxy+cx+dy+e=x^2. Since e is a constant not associated with any variable, how should we get th...

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A question for orth
A = [1,-1,2;-1,1,-2;2,-2,4] If I enter B = orth(A) Then, B = -0.4082 0.4082 -0.8165 However, range of A shou...

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Formatting of result values.
For example, x=0.0012345678. How can I get x to be in the format 0.123457*10^(-2). There are six significant digits and if more...

plus d'un an il y a | 2 réponses | 0

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How to avoid rounding error
If I want to get the value of sqrt(10001),and it should to be 100.00499... However, MATLAB shows 100.005. How should I write t...

plus d'un an il y a | 1 réponse | 0

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