Sum of series to infinity

4 vues (au cours des 30 derniers jours)
KB
KB 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.

Réponse acceptée

Walter Roberson
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.

Plus de réponses (1)

Kirby Fears
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;

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