Plotting an infinite series with multiple variables.

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Hi, recently I solved PDE of a 1D wave equation and came to the solution of:
my question while vague,also comes from a place of genuine interest. How exactly would I plot this? I tried splitting it into various components and then summing those components into one function but it just plotted lines for me which leads me here. Any and all help is appreciated; I'm rather novice at this program so detailed explaninations are very welcomed in reference to why you chose particular path way to achieving something.
quick chicken scratch code:
%variables
x=(0:.2:1);
t=(0:.2:1);
n=(0:.4:2);
A = 0.02.*(-1.*(((2.*(n.^2)-5).*sin(n.*pi))/((n.^4)-5.*(n.^2)+4)));
B = 0.02.*(sin(n.*pi.*x).^2);
C = 0.02.*(cos(n.*pi.*t));
u = A.*B.*C;
fplot(u);
  1 Comment
David Hill
David Hill on 18 Feb 2020
You could try something like:
[x,t]=meshgrid(0:.2:1,0:.2:1);
u=zeros(size(x));
for n=3:100
u=u-.02*(2*n^2-5)*sin(n*pi)/(n^4-5*n^2+4)*sin(n*pi*x).*cos(n*pi*t);
end
surf(x,t,u);
Your equation does not make much sense to me, since for n=1 and n=2 you are dividing by zero. How can n=0:.4:2 based on your equation?

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Accepted Answer

Raunak Gupta
Raunak Gupta on 21 Feb 2020
Hi,
David’s comment above provides a solution but has some missing terms. You may follow the bottom code for doing the same. Also as mentioned in comment n=1 and n=2, the value of u will be undefined so the summation should start at n=3. For the range of x and t you can change them in meshgrid below as per required.
% x and t ranges
xRange = 0:0.02:1;
tRange = 0:0.02:1;
[x,t] = meshgrid(xRange,tRange);
u = zeros(size(x));
for n=3:100
u = u - (0.02*(2*n^2-5)*sin(n*pi)/(n^4-5*n^2+4))*(sin(n*pi*x)^2).*cos(n*pi*t);
end
surf(x,t,u);
  2 Comments
Tristen Hernandez
Tristen Hernandez on 24 Feb 2020
This visualized it perfectly thank you so much! I appreciate the help and also thanks for poiinting out the error in my original equation,that was giving a lot of headache and I just overlooked it.
Rik
Rik on 26 Jan 2022
Comment posted as flag by Gabriel Zev Betancourt Pinto:
It was really helpful Thank you so much!

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