Evaluating a matrix at different time steps

Dear all, I have a matrix which is a transition matrix. I need to calculate the time evolution of the associated probability distribution. The matrix is raised to 't', the time variable and I want to evaluate the matrix at different time steps. Is it possible in Matlab? The code to generate the square matrix is given below which is working well for a given value if t.

if true
  N=3;
w=0.2;
t=1;
G=zeros(N);
for j=1:N         %rows
 for n=1:N      %columns
    for k=1:N %summation
     G(j,n)=G(j,n)+1/N*(((1-w)+w*cos((2*pi*k)/N)).^t*cos((2*pi*(j-n)*k)/N));
    end 
 end
end
disp(G)
end

Answer 1

[Star Strider]

Add a separate loop for ‘t’, and a third dimension to your ‘G’ matrix:

N=3;
w=0.2;
tv = linspace(0, 10, 10);                       % Define Time Vector
G=zeros(N,N,length(tv));                        % Preallocate Here
for t = 1:length(tv)                            % ‘Time’ Loop
    for j=1:N         %rows
        for n=1:N      %columns
            for k=1:N %summation
                G(j,n,t)=G(j,n,t)+1/N*(((1-w)+w*cos((2*pi*k)/N)).^tv(t)*cos((2*pi*(j-n)*k)/N));
            end
        end
    end
    % disp(G)
end

Define the vector of times you want to evaluate your matrix in ‘tv’. The rest of the code will then work. Each ‘page’ (third dimension elements) of ‘G’ will be ‘G’ at the corresponding times.

Also, move the preallocation step to be before the loop! Otherwise, it resets all previous values of ‘G’ to zero.

[Vipin Padinjarath]

That's great! It's working. Thank you very much.:-)

[Vipin Padinjarath]

But I have a doubt. When I set the value of t and evaluate the code (my old code) the values of the elements of the matrix are different from the corresponding matrix this new code generates. Why is that so? I am sorry, I am quite new to matlab.

[Star Strider]

My pleasure!

I see the problem. I forgot to include the ‘t’ dimension on the right-hand side reference to ‘G’. This should work correctly:

G(j,n,t)=G(j,n,t)+1/N*(((1-w)+w*cos((2*pi*k)/N)).^tv(t)*cos((2*pi*(j-n)*k)/N));
               ↑ ← ADDED ‘t’ HERE

(My apologies for the error.)

[Star Strider]

My pleasure!

[Vipin Padinjarath]

It works perfectly now!! Cheers!Thanks again!!

[Star Strider]

As always, my pleasure!

Answer 2

[Andrei Bobrov]

R2016b and later

N = 3;
w = 0.2;

k = 1:N;
k = reshape(k,1,1,1,[]);

jj = (1:N)';
n = 1:N;

t = (0:10);
t = reshape(t,1,1,[]);

g = ((1-w+w*cos(2*pi*k/N)).^t.*cos(2*pi*(jj-n).*k/N))/N;
G = sum(g,4);

R2016a and earlier

N = 3;
w = 0.2;

k = 1:N;
k = reshape(k,1,1,1,[]);

jj = (1:N)';
n = 1:N;

t = (0:10);
t = reshape(t,1,1,[]);

g0 = cos(bsxfun(@times,bsxfun(@minus,jj,n),2*pi*k)/N);
g1 = bsxfun(@power,1-w+w*cos(2*pi*k/N),t);
g = bsxfun(@times,g0,g1)/N;
G = sum(g,4);

[Vipin Padinjarath]

Thank you Andrei Bobrov. But for my standards, this is too much. I don't understand much of it. But will learn for sure. Thanks again.