my matrix is not square, although i think it is?

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Alex
Alex 2011-8-8
I have set the following least squares method up quickly. I know i am using a nonlinear technique to solve a linear problem, however i just wanted to get it working, before i choose a linear method but aparently my matrices are not square:
i = [1 2 3 4 5];
ti = [2 5 7 11 14];
Estimates3=fminsearch(@functionnumbers,1,options,T0,theta_f);
and the function code
function fnumbers = functionnumbers(params3,i,ti)
k = params3;
v0 = 1 + (k/2)*ti(1)^2/ti(1)
fnumbers = sum(ti.^2*(i/ti-(v0-(k/2)*ti))^2);
cheers
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Alex
Alex 2011-8-8
Sorry i posted that line incorrectly should have been
Estimates3=fminsearch(@functionnumbers,1,options,i,ti);
in terms of deleting old values, where are you referring to?

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采纳的回答

Walter Roberson
Walter Roberson 2011-8-8
fminsearch() accepts at most 3 inputs, and the function passed to it must accept a single input. See http://www.mathworks.com/help/techdoc/ref/fminsearch.html
For information on how to do this properly, please read this
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Alex
Alex 2011-8-9
So i'm not sure what the problem is in terms of "too many variables" but I have used this function before and passed it variables after the options before and it worked fine, at the moment when i run it i get the functionnumbers to display i and ti to make sure it recevies them and it does.
i have adjust the code. V0 does not need a matrix operator i dont think however fnumbers needs too:
v0 = (1 + (k/2)*ti(1)^2/ti(1))
fnumbers = sum(ti.^2*(i/ti-(v0-(k/2)*ti)).^2)
the error changes now from matrix must be square to inner matrix dimensions must agree. I tried transposing ti incase the problem was a pre multiplacation issue (matrix multiplication) but its not the case.
Walter Roberson
Walter Roberson 2011-8-9
Are you truly wanting to do a _matrix division_ between i and ti ?? If not then you should be using ./ instead of /
Are you truly wanting to do a matrix multiplication between ti.^2 and the rest of the expression? If not, then you should be using .* instead of *

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更多回答（1 个）

Alex
Alex 2011-8-9
So I have got it to run, I'm not sure why but breaking the equation into more equations seemed to work:
[code]function fnumbers = functionnumbers(k,i,ti)
v0 = (1 + (k/2)*ti(1)^2/ti(1)); Fitted_Curve = (v0*ti-(k/2)*ti.^2); Error_Vector = i - Fitted_Curve; fnumbers = sum(Error_Vector.^2);[/code]
calling it from
i = [1 2 3 4 5 6]; ti = [2.503 5.159 7.899 10.883 13.814 17.081];
Estimates3=fminsearch(@functionnumbers,0.004,options,i,ti)
however this isnt the most complicated problem, is there a better way to solve this because the value of k i am expecting for this data is about 0.03 but im getting 0.1 the equations non-linear because of the square, however is fminsearch overkill for this?
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Walter Roberson
Walter Roberson 2011-8-9
In the original equation you had "i" divided by "ti", but in the new equations that division appears to have vanished. It does not appear to me that your old and new equations are equivalent, but it could be that I missed something.
Alex
Alex 2011-8-10
Actually I missed something, Walter's answer was right. Thanks

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