# How can I code a two variable function?

3 views (last 30 days)
Eduardo Fornazieri on 10 Jul 2021
Edited: ANKUR KUMAR on 11 Jul 2021
Hi, could you code this function from ?
I know I can't do it to infinity, so I choose the sum from 0 to 100. Please, help.
.
##### 0 CommentsShowHide -1 older comments

Sign in to comment.

### Accepted Answer

ANKUR KUMAR on 11 Jul 2021
Edited: ANKUR KUMAR on 11 Jul 2021
You can simply use for loop to iterate x and y. Refer to the below code just for your reference. You might need to make modification as per your need.
This might be the good starting point for you to write code independently.
Anyway, I am getting some of the NaN values.
func = @(n,x,y) (((-1)^n)/((2*n + 1)^3))* sech(((2*n + 1)*pi)/2)*cosh(((2*n + 1)*pi*x)/2)*cos(((2*n + 1)*pi*y)/2);
n_upper_limit=100;
for x=1:5
for y=1:5
u(x,y) = 1- y^2 - ((32)/(pi^3))*(sum(arrayfun(@(n) func(n,x,y), 1:n_upper_limit),'omitnan'));
end
end
disp(u)
1.0e+130 * -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 1.6033 0.0000 -1.6033 -0.0000 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
% For plotting u
imagesc(u)
title('Values of u')
xlabel('x values')
ylabel('y values')
xticks([1:5])
yticks([1:5])
colorbar
set(gca,'Ydir','normal')
##### 0 CommentsShowHide -1 older comments

Sign in to comment.

### Categories

Find more on Creating and Concatenating Matrices in Help Center and File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!