Hi
I am trying to do the following calululation in matlab:
But I do not how how to do the two sums in the front. Our data (d) is a 67x120 array.
Any suggestions would be appreciated :)

2 件のコメント

darova
darova 2020 年 3 月 2 日
Luna
Luna 2020 年 3 月 2 日
編集済み: Luna 2020 年 3 月 2 日
What have you done so far as coding? And what is d?

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Guillaume
Guillaume 2020 年 3 月 2 日
編集済み: Guillaume 2020 年 3 月 2 日

1 投票

s = (1:size(d, 1)).';
result = sum(sum(d .* cos(s))) / sum(cos(s));
Loops not needed, they're just a waste of time.

3 件のコメント

Luna
Luna 2020 年 3 月 2 日
編集済み: Luna 2020 年 3 月 2 日
My understanding of /Sigma[cos(s)] part, it should be divided into sum of cos(s) from 1 to what s is until that iteration. For example if s is 20 while j = 3;
d(20,3)*cos(20) / ( cos(1)+cos(2)+...+cos(20) )
so your result gives a vector (1x67). I assume that result must be a scalar.
Guillaume
Guillaume 2020 年 3 月 2 日
Yes, I made a mistake, there was a sum missing in the denominator. I assumed that the could be taken out of the double sum, i.e. it's over the whole range. It's not clear and we didn't get an answer to Darova's question. So, my code implement:
If it's a partial sum as you think, then:
s = (1:size(d, 1)).';
result = sum(sum(d .* cos(s) ./ cumsum(cos(s))));
which implements
Luna
Luna 2020 年 3 月 3 日
You are right.

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その他の回答 (1 件)

Luna
Luna 2020 年 3 月 2 日

0 投票

Maybe something like that would help:
d = rand(67,120);
sum_of_cos_s = 0;
result = 0;
for s = 1:67
for j = 1:120
sum_of_cos_s = sum_of_cos_s+cos(s);
result = result + (d(s,j)*cos(s))/sum_of_cos_s;
end
end

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