how to store each value of T0 in a matrix of order (l,240)?

for t=b:dt:240
    for x=a:dx:l
        for n=d:dn:5
            cn=(2/n*pi)*(((40/n*pi)*(sin(n*pi)))-(40*cos(n*pi)));
            T0=cn*(exp(-(n^2)*(pi^2)*t/(l^2)))*(sin(n*pi*x/l));
        end
    end
end

[Roger Stafford]

Your request implies that the triple nested for-loops will iterate L*240 times. That is certainly not evident and would depend very much on the values of b, a, d, dt, dx, dn, and L. Please tell us what these seven values are.

Answer 1

[KSSV]

t=b:dt:240 ;
x=a:dx:l ;
n=d:dn:5 ;
T0 = zeros(length(t),length(x),length(n)) ;
for i = 1:length(t)
    for j = 1:length(x)
        for k = length(n)
            cn=(2/n(k)*pi)*(((40/n(k)*pi)*(sin(n(k)*pi)))-(40*cos(n(k)*pi)));
            T0(i,j,k)=cn*(exp(-(n(k)^2)*(pi^2)*t(i)/(l^2)))*(sin(n(k)*pi*x(j)/l));
        end
    end
end