How to store all the T values from this nested loop?

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Tze Ting Ong
Tze Ting Ong 2018-3-3
评论： Tze Ting Ong 2018-3-5
T1=80; a=5; b=4;
for n=1:5
for x=1:5
y=1:4;
T(x,y)=(4*T1/pi)*(sin((2*n-1)*(pi*x/a))/(2*n-1)*sinh((2*n-1)*(pi*y)/a)/sinh((2*n-1)*(pi*b)/a));
end
end
  1 个评论
John D'Errico
John D'Errico 2018-3-3
Do you want to SUM over n?
The nested loop that you wrote will not even come close to a sum in that case.

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John D'Errico
John D'Errico 2018-3-3
编辑：John D'Errico 2018-3-3
Assuming that you wanted to sum over the values of n (which is the only thing that seems to make sense here)
In an older release of MATLAB, I might have written this as
T1=80; a=5; b=4;
[x,y] = ndgrid(1:5,1:4);
T = zeros(size(x));
for n = 1:5
T = T + (4*T1/pi)*(sin((2*n-1)*(pi*x/a))./(2*n-1).*sinh((2*n-1)*(pi*y)/a)/sinh((2*n-1)*(pi*b)/a));
end
T
T =
6.6567 16.505 36.12 71.671
10.524 25.047 47.86 74.706
10.524 25.047 47.86 74.706
6.6567 16.505 36.12 71.671
1.4076e-15 3.597e-15 9.1745e-15 6.2371e-14
Without looking at your code, I was surprised at the effective zeros on the last line, but then I looked slightly more carefully, and I see they arise from the term sin(2*n-1)*(x/a)). When x=5, we get zero.
In a newer (thus current) release, I might have written it differently.
T1=80; a=5; b=4;
x = (1:5).';
y = 1:4;
T = zeros(size(x));
for n = 1:5
T = T + (4*T1/pi)*(sin((2*n-1)*(pi*x/a))./(2*n-1).*sinh((2*n-1)*(pi*y)/a)/sinh((2*n-1)*(pi*b)/a));
end
T is the same as before, but it used implicit expansion of the vectors, thus allowing me to avoid the ndgrid call.
If I wanted to avoid the explicit loop completely, this is also quite easy.
T1=80; a=5; b=4;
x = (1:5).';
y = 1:4;
n = reshape(1:5,[1 1 5]);
T = (4*T1/pi)*(sin((2*n-1).*(pi*x/a))./(2*n-1).*sinh((2*n-1).*(pi*y)/a)./sinh((2*n-1).*(pi*b)/a));
T = sum(T,3);
Note here my careful use of the .* and ./ operators where needed. So T was originally created as a 5x4x5 array. Then I summed over the third dimension. I could have written that in one line, thus
T = sum((4*T1/pi)*(sin((2*n-1).*(pi*x/a))./(2*n-1).*sinh((2*n-1).*(pi*y)/a)./sinh((2*n-1).*(pi*b)/a)),3);
but that would have been less clear what I did.
In any event, again we see the same sum for T.
T
T =
6.6567 16.505 36.12 71.671
10.524 25.047 47.86 74.706
10.524 25.047 47.86 74.706
6.6567 16.505 36.12 71.671
1.4076e-15 3.597e-15 9.1745e-15 6.2371e-14
Be careful when writing expressions as I have done this, by expanding things to three dimensions. Had n been VERY large and/or x and y been very large vectors, then large intermediate arrays could be generated. So if x,y,n all were in the thousands of elements, we could easily be generating 3-d arrays with billions of elements, requiring many gigabytes of memory to store.

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