decreasing three loops to two loops with sum

I have this code and want to eliminate the inner most loop and use sum

How can I do that?

thetao=1;
alfa=1;
l=1;

theta2(1:7,1:21)=0;

T=[0.01, 0.1, 0.2, 0.5, 1, 2, 5];
x=0:0.05:l;
n=0:50;
for t=1:7
    for j=1:length(x)
        for k=1:length(n)
           
           theta1(t,j)=(sin((2*n(k)+1)*pi*x(j)/(2*l))*exp(-(2*n(k)+1).^2*pi^2*alfa*T(t)/(4*l^2))/((2*n(k)+1)*pi));
           theta2(t,j)=theta2(t,j)+theta1(t,j);
           theta(t,j)=thetao-4*thetao*theta2(t,j);
            
       end
    end
end

Answer 1

[Adam Danz]

----------- Updated answer ------------------------------

Vectorize the k-loop and use sum() to define theta2.

thetao=1;
alfa=1;
l=1;
theta2(1:7,1:21)=0;
T=[0.01, 0.1, 0.2, 0.5, 1, 2, 5];
x=0:0.05:l;
n=0:50;

for t=1:7
    for j=1:length(x)
        theta1 = sin((2*n+1)*pi*x(j)/(2*l)).* exp(-(2*n+1).^ 2*pi^2*alfa*T(t)/(4*l^2))./((2*n+1)*pi);
        theta2 = sum(theta1);
        theta(t,j)=thetao-4*thetao*theta2;
    end
end

To demonstrate that this produces the same values at the original version, run the block and above and then two two blocks below.

% Save the theta values above and then clear all variables 
theta_version2 = theta; 
clearvars -except theta_version2

% Original version
thetao=1;
alfa=1;
l=1;

theta2(1:7,1:21)=0;

T=[0.01, 0.1, 0.2, 0.5, 1, 2, 5];
x=0:0.05:l;
n=0:50;
for t=1:7
    for j=1:length(x)
        for k=1:length(n)
           
           theta1(t,j)=(sin((2*n(k)+1)*pi*x(j)/(2*l))*exp(-(2*n(k)+1).^2*pi^2*alfa*T(t)/(4*l^2))/((2*n(k)+1)*pi));
           theta2(t,j)=theta2(t,j)+theta1(t,j);
           theta(t,j)=thetao-4*thetao*theta2(t,j);
            
       end
    end
end

Now compare the results

max(abs(theta - theta_version2),[],'all')

% ans =
%    4.4409e-16

The max difference in theta between the original and new versions is 4.4409e-16. This is due to the precision in floating point calculations explained [here].

[Roya Rajabi]

They Dont give the the same result

[Adam Danz]

See my updated answer.

[Roya Rajabi]

Thanks a million it works