Calculate this matrix in a more general and shorter way

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Rowan Humphreys 2021-11-25

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评论： Rowan Humphreys 2021-11-30

Is there a way of generating P in a way that doesn't involve writing sin/cos/sin/cos repeatedly? Perhaps like how you can generate an array by writing 'x = 1:0.1:10'?

the size of 'x' is 100by1 and 'w' is some constant.

P=[ones(size(x)),cos(w*x),sin(w*x),cos(2*w*x),sin(2*w*x),cos(3*w*x),sin(3*w*x),cos(4*w*x),sin(4*w*x),cos(5*w*x),sin(5*w*x)]

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Answer 1

Stephen23 2021-11-25

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x = rand(100,1);
w = pi;
m = (1:5).*w.*x;
m = [ones(100,1),reshape([cos(m);sin(m)],100,[])];

for comparison:

P = [ones(size(x)),cos(w*x),sin(w*x),cos(2*w*x),sin(2*w*x),cos(3*w*x),sin(3*w*x),cos(4*w*x),sin(4*w*x),cos(5*w*x),sin(5*w*x)];
isequal(P,m)
ans = logical
   1

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Rowan Humphreys 2021-11-30

Thank you! This worked instantly :)

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Answer 2

John D'Errico 2021-11-25

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x is 100x1. Now, suppose we want to generate multiple terms of the form sin(k*w*x), so k will be a vector. w is a fixed constant. Do you know how to do that?

x = (0:2:10)'; % arbitrary, your x is what you have. I've made x small so you can see what happens
w = 1; % again, my choice.
k = 1:4;

Ok, so what does this do in MATLAB?

x*w*k
ans = 6×4
   0     0     0
   4     6     8
   8    12    16
  12    18    24
  16    24    32
  20    30    40

This is just effectively a multiplication table. But do you see that it forms what you need? You can now compute sin(x*w*k), and cos(x*w*k).

In your case, x is a column vector, of size 100x1. Your code might look like:

x = ???? % whatever x is
w = ???? % again
n = 10; % how many terms?
P = [ones(size(x)),sin(x*w*k),cos(x*w*k)];

Now, the only problem is, all the sin terms are bundled together, then all the cos terms. If that bothers you, then reshuffle those columns. Something like this should work.

P(:,[2:2:end,3:2:end]) = P(:,[2:n+1,n+2:end]);

Now the sin and cos terms alternate.