Sum of series to infinity

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KB
KB 2015 年 9 月 15 日
回答済み: Walter Roberson 2015 年 9 月 15 日
I am trying to run the following code:
t = [0 60 120 180...3600];
l = length(t);
for i = 1:l
syms n;
sum = symsum(exp(-C*t(i)*pi^2*(2*n+1)^2)/(2*n+1)^2, n, 0, Inf);
y(i) = sum;
end
yp = 1-8*y/pi^2
But I am not getting any output where I am trying to solve 'C' value from a non linear fitting, could someone help me how to deal with the infinity sum ? I know that at t = 0, the value will be essentially would be pi^2/8 which eventually makes the yp value zero at t =0 which is important to catch, however it does not depend on the 'C' value.

採用された回答

Walter Roberson
Walter Roberson 2015 年 9 月 15 日
My tests indicate that there is no closed form solution for the infinite sum for t(i) values non-zero (unless C is 0). You will need to do numeric evaluations at each proposed C value.

その他の回答 (1 件)

Kirby Fears
Kirby Fears 2015 年 9 月 15 日
I'm not a symbolic toolbox user, but I think you need to set C as a symbolic variable to make a valid symbolic expression for symsum's first argument. Try using
syms n C;

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