Sum of series to infinity
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KB
le 15 Sep 2015
Réponse apportée : Walter Roberson
le 15 Sep 2015
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.
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Walter Roberson
le 15 Sep 2015
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.
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Kirby Fears
le 15 Sep 2015
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;
If that doesn't work, try reading this documentation page: http://www.mathworks.com/help/symbolic/create-symbolic-numbers-variables-and-expressions.html
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