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Sum of matrix with variable dimension

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Favier
Favier 2016 年 10 月 29 日
コメント済み: Favier 2016 年 10 月 30 日
Hello, I have two matrices produced by meshgrid and a simple scalar:
r = 1:10;
N = 1:10;
[r,N] = meshgrid(r,N);
N0 = 4;
I want, for each value of N, to get the sum from N0 to N of r^n. However this Matlab commands is not working :
sum(r.^(N0:N));
For example for r(5,5) = 5 and N(5,5) = 5, the result would be 5^4 + 5^5 = 3750. What I want is to get the matrix resultant of this operation over each row and column of r and N.
Thank you for your help!
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Image Analyst
Image Analyst 2016 年 10 月 29 日
Why is the N value raised to the 5th power while the r value is raised to the 4th power? What does it mean to have an exponent that is a vector, like [4,5]?
Favier
Favier 2016 年 10 月 29 日
Thank you for your interest. Sorry I should have detailed my example a bit more. When I write 5^4 + 5^5 = 3750, it should be read as r(5,5)^N0 + r(5,5)^N(5,5) = 3750.
If N0 = 4 and N(5,5) = 7, then I want to calculate r(5,5)^N0 + r(5,5)^(N0+1) + r(5,5)^(N0+2) + r(5,5)^(N(5,5))
The exponent is a vector to simplify the representation. What I want is to calculate the sum of r^n with n from N0 to N for ALL my r and ALL my N. The larger the N, the more terms I have in the sum. That is the purpose of the vector exponent.

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Guillaume
Guillaume 2016 年 10 月 29 日
I'm not sure what should happen when N(y,x) < N0, should the result be 0?
If I understand correctly, this will do:
arrayfun(@(rr, nn) sum(rr .^(N0:nn)), r, N)
Note that the above works for N(y, x) < N0 because the sum of an empty vector is 0 instead of empty.
  1 件のコメント
Favier
Favier 2016 年 10 月 30 日
The result should indeed be zero when N < N0. I like your solution a lot.
Thank you very much!

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