Reputation: 200
I have a simple function below (I omitted the allocations, etc. for brevity) that I have been tryig to plot against it's x value for specific values of N and T but I keep getting a dimensions error. I think that when I try to plot this I am defining an array for x and then plotting Psum(N', x, T') for certain N' and T' against these x, however MATLAB doesn't seem to like this. Can someone give me some direction please.
function U = Psum(N, X, T)
for m = 1:N
A(1,m) = (1/(m*pi))*sin(m*pi*X)*T*exp(-(m^2)*(pi^2)*T);
% array terms of partial sum
end
M = -sum(A); % evaluate Nth partial sum
U = T*(1-X) + M; % output U(X,T) = T(1-X) + V(X,T)
end
I'm getting a similar error when I try to plot the following, I think there must be something wrong with my general approach
syms x;
f = @(x)((x/(100*pi))*(exp(-(100^2)*(pi^2)*x)));
x = 0:0.1:10000;
plot(x,f(x),'r')
title('PartialSum convergence');
xlabel('T');
ylabel('a_n');
the error I get here reads:
Error using *
Inner matrix dimensions must agree.
Upvotes: 0
Views: 838
Reputation: 32930
Here's the analysis of why you're getting a dimension mismatch error. From this line:
A(1,m) = (1/(m*pi))*sin(m*pi*X)*T*exp(-(m^2)*(pi^2)*T)
The element A(1, m)
is supposed to be a scalar value in a two-dimensional matrix. Now let's see what are the dimensions of each of the multiplicands:
(1/(m*pi))
is a scalar (that is, a 1×1 matrix).sin(m*pi*X)
has the same dimensions as X
. Let's assume its dimensions are q×n.exp(-(m^2)*(pi^2)*T)
has the same dimensions as T
, and is multiplied by T
.
Therefore T
must be a square matrix, so let's assume its dimensions are p×p.What we get is a q×n matrix multiplied by a square p×p matrix, and the result must be a scalar (that is, 1×1 matrix). This forces q=1 and n=p.
Now let's look at this line:
U = T*(1-X) + M
We are forced to conclude that p=1, otherwise T
cannot be multiplied by X
from the right.
This means that your code forces T
and X
to be scalar! No wonder you're getting a error :)
The remedy is simple: revise the computation in Psum
so that it can produce correct results for both a scalar X
and a vector X
. A possible fix would be adding another loop to iterate over all values of X
:
function U = Psum(N, X, T)
U = zeros(size(X));
for k = 1:numel(X) %// Iterate over all values of X
for m = 1:N
A(1,m) = (1/(m*pi))*sin(m*pi*X(k))*T*exp(-(m^2)*(pi^2)*T);
%// array terms of partial sum
end
M = -sum(A); % evaluate Nth partial sum
U(k) = T*(1-X(k)) + M; % output U(X,T) = T(1-X) + V(X,T)
end
end
The output of this function has the same dimensions as X
.
By the way, did you verify that Psum
produces that correct result for scalar inputs?
Upvotes: 2
Reputation: 1447
I don't fully understand what you are trying to accomplish, but just an observation for you: if your input X is a vector, line 3 can not be computed correctly
A(1,m) = (1/(m*pi))*sin(m*pi*X)*T*exp(-(m^2)*(pi^2)*T);
because the right hand side of the equation give you a vector, but the right hand side A(1,m)
is one element, not vector. so you have dimension mismatch.
Hope this helps!
Upvotes: 1