Matlab assumes a matrix is 1x2 instead of 1x9?

I have the following code: b should end up as a 1x9 matrix of values solved for in the equation.

D0=D(1,:); D1=D(2,:); D2=D(3,:); D3=D(4,:); D4=D(5,:); D5=D(6,:); D6=D(7,:); D7=D(8,:); D8=D(9,:);

Data=zeros(15,9);

Input=Data104cis;

   b0=[1,.1,.1,.1,.1,.1,.1,.1,.1];

for jj=1:15;

      y=Input(jj);

       fun = @(b) ((b(1).*D0) +(b(2).*D1)+ (b(3)*D2)+(b(4).*D3)+(b(5).*D4)+ (b(6).*D5) +(b(7).*D6)+(b(8)*D7)+(b(9)*D8));       

OLS =@(b) sum(y - fun(b)).^2;

fminsearch(OLS,b0);

Data(jj,:)=b;

b0=b;

end

instead, b somehow ends up as a 1x2 matrix and then the code crashes with a dimension mismatch error. I have no idea why it would assume b to be 1x2 when i have 9 b values shown in fun.

Answer 1

[Star Strider]

You’re not asking fminsearch to return anything!

Change:

fminsearch(OLS,b0);

to:

b = fminsearch(OLS,b0);

and see if that solves your problem.

Answer 2

[Matt J]

Your posted code doesn't define b anywhere, so if it is 1x2, it is because you assigned it that somewhere prior to the code you've shown. Presumably, you meant to assign b to the output of fminsearch,

 b=fminsearch(OLS,b0);

Incidentally, fun() could be defined much more simply and efficiently as

fun=@(b) b(:).'*D;

[John D'Errico]

But why would anyone use fminsearch to compute a 9 (NINE!) parameter OLS estimator? I've heard of using a Mack truck to take a pea to Boston, but this is different. Here Darya is using a human powered tricycle to pull a loaded boxcar with a full load of iron ore to Boston.

Time to learn how to use backslash for Darya!

[Matt J]

Yes. It is also suspicious that y is a scalar,

 y=Input(jj);

The solutions Data(jj,:) should all be identical, differing only by scale factors of Input(jj).

[Darya]

I hit clear and sure enough that fixed it. thanks for the clean up tip too!