Vandermonde-like matrix

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 !

Answer 1

[KSSV]

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

[Mokrane Mahdi]

Thank you !

Answer 2

[John D'Errico]

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.

[Mokrane Mahdi]

Oh, this is way more easy ! Thanks !