# Expression representation in an Iteration

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Vezamafa Nzima on 6 Aug 2020
Commented: Vezamafa Nzima on 6 Aug 2020
I'd love to know how i can express the expression in a loop. n is the iteration number. I know it may seem trivial but i've been cracking my skull over it. My approach, which I figure is wrong anyways, was to index the second term with n. But the iteration number is greater than the the matrix elements and i get an error. Here is my attempt at the issue.
w=[randn(2,1)+ij*randn(2,1)];
W_k=w*ctranspose(w);
for n=1:N;
w=[randn(2,1)+ij*randn(2,1)];
W_k1=w*ctranspose(w);
t=W_k - W_k1;
end
Is this doing what the expression requires?

Walter Roberson on 6 Aug 2020
W_k = cell(N,1);
w=[randn(2,1)+ij*randn(2,1)];
W_k{1} = w*ctranspose(w);
for n=1:N;
w = randn(2,1)+ij*randn(2,1);
W_k{n+1} = w*ctranspose(w);
t = W_k{n} - W_k1{n+1};
%do something with t
end

#### 1 Comment

Vezamafa Nzima on 6 Aug 2020
Thank you very much. This exactly what i was looking for. Makes sense even just going through the code.

Sudheer Bhimireddy on 6 Aug 2020
Something like this ?
w_k(nx) = 0;
w_k_n(n_iter) = 0;
for i=1:nx
w_k_n(1) = w_k(i);
for j=1:n_iter
your_ans(i) = w_k(i) - w_k_n(j);
end
end

Vezamafa Nzima on 6 Aug 2020
Unfortunately it's not working. W_k and W_k_n should be matrices of the same size. The way i figure it also is that the iteration should not be limited by the size of the matrix. That is, i'd love the iteration to go on as many times as i choose.
Sudheer Bhimireddy on 6 Aug 2020
If both matrices are of same size and you want to iterate as many times as you want, then you can do one of the below two things:
1) Dynamically increase the size of both matrices as you iterate and fill the w_k with repetetive values. This way you will end up with redundant data.
OR
2) Over-write the w_k_n matrix once the iteration limit has crossed the size of the matrix. This way you might loose some data once you start overwriting.

R2019a

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