0 个投票

For any given number x, what's the easiest way to generate the following square matrix without using any loop:
0 x^(-0.5) x^(-1.0) x^(-1.5) x^(-2.0)
x^(-0.5) x^(-1.0) x^(-1.5) x^(-2.0) x^(-2.5)
x^(-1.0) x^(-1.5) x^(-2.0) x^(-2.5) x^(-3.0)
x^(-1.5) x^(-2.0) x^(-2.5) x^(-3.0) x^(-3.5)
x^(-2.0) x^(-2.5) x^(-3.0) x^(-3.5) x^(-4.0)
Here, in this example, I set the size of the matrix to be 5, but it has to be generated for any integer n.
Notice the matrix does look like the Vandermonde matrix, hence the idea of using repmat and cumprod commands..
Thanks in advance !

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KSSV
KSSV 2022-3-3
Ran in:

0 个投票

Ran in:
p = -(0:0.5:4) ; % power values
n = 5 ; % n value
ind1 = bsxfun(@plus, (1 : n), (0 : numel(p) - n).'); % make moving window indices
p = p(ind1) % power values
p = 5×5
0 -0.5000 -1.0000 -1.5000 -2.0000 -0.5000 -1.0000 -1.5000 -2.0000 -2.5000 -1.0000 -1.5000 -2.0000 -2.5000 -3.0000 -1.5000 -2.0000 -2.5000 -3.0000 -3.5000 -2.0000 -2.5000 -3.0000 -3.5000 -4.0000
x = rand ; % your x
V = x.^p % what you wanted
V = 5×5
1.0e+03 * 0.0010 0.0028 0.0077 0.0214 0.0595 0.0028 0.0077 0.0214 0.0595 0.1654 0.0077 0.0214 0.0595 0.1654 0.4594 0.0214 0.0595 0.1654 0.4594 1.2760 0.0595 0.1654 0.4594 1.2760 3.5443

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John D'Errico
John D'Errico 2022-3-3

1 个投票

The easiest way? Probably this line: (in R2016b or later)
x.^(((0:-1:1-n)' + (0:-1:1-n))/2)
If you want to use two lines of code, then it looks simpler yet.
N = (0:-1:1-n)/2;
x.^(N' + N)
Easrlier releases than R2016b would use bsxfun.

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