function handle and array operators - Matlab

Question

I keep getting a matrix dimension error, although i pay attention to the array operators from the below code. As I am not willing to strictly perform matrix operations could someone explain the origin of the dimension / size issue?

Thanks

Error using .* Matrix dimensions must agree

% integrand behaviour
f=@(k) k .* exp(-k);
x=0.5;
al=1;
bet=0.89;
theta0=(1/al) .*atan(bet .*tan(pi*al/2));
c=exp(-(pi .* x) ./2 * bet);
c2=1 /(2.*abs(bet));
v1=@(theta) 2/pi .* ((pi/2 + bet.*theta)/cos(theta)) .* exp(1/bet .* (pi/2 +
bet.*theta)...
.* tan(theta));
g=@(theta) c .* v1(theta);
y=@(theta) f(g(theta)) ;     % values of integrand
a=[-pi/2:0.2:pi/2]';
Y=y(a);
plot(a,Y)

Tamás Szabó · Accepted Answer · 2015-05-24 09:33:34Z

1

The source of your problem is here:

(pi/2 + bet.*theta)/cos(theta)

You use the / operator, where I think that you mean to use the ./ operator.

answered May 24, 2015 at 9:33

Tamás Szabó

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3 Comments

owner Over a year ago

it does not matter that much. I just realize that the source of the matter is the column vector 'a'. Actually the code works for any scalar (e.g a = pi/6, etc.) thus a solution is to make it work on vectors as far as i am concerned.

Tamás Szabó Over a year ago

It does matter. Indeed, your a is a 16 element column vector. The line I highlighted will turn this into a 16x16 matrix, which is obviously wrong. Remember, that / is a matrix operator, while ./ works element wise.

owner Over a year ago

Many thanks for your prompt feedback, as I definitely missed ./ cos(theta). You're right. Cheers

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