lsqnonlin with jacobian problem
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TE=2:5:5*20;
S(1,:)=104*exp(-TE/10);
options=optimset('Algorithm','levenberg-marquardt','Display','off','Jacobian ','on','Tolfun',1e-6 );
P0=[59 30];
P=lsqnonlin(@test,P0,[],[],options,S,TE)
function [F,J]=test(P,S,TE)
Ft=P(1)*exp(-TE/P(2));
F=S-Ft;
if nargout >1
J(:,1)=exp(-TE/P(2));
J(:,2)=P(1)*TE.*exp(-TE/P(2))/(P(2)^2);
end
1 Comment
Walter Roberson
on 19 Jul 2012
Are you encountering an error message? If so, what message and where?
Answers (5)
Walter Roberson
on 19 Jul 2012
lsqnonlin() accepts a maximum of 5 parameters. See http://www.mathworks.com/help/toolbox/optim/ug/lsqnonlin.html and http://www.mathworks.com/help/toolbox/optim/ug/brhkghv-7.html
0 Comments
Mus Bohr
on 19 Jul 2012
1 Comment
Walter Roberson
on 19 Jul 2012
Edited: Walter Roberson
on 19 Jul 2012
P=lsqnonlin(@test,P0,[],[],options,S,TE)
- @test
- P0
- []
- []
- options
- S
- TE
That is 7 parameters. lsqnonlin() does not accept anything after "options".
Mus Bohr
on 19 Jul 2012
2 Comments
Walter Roberson
on 19 Jul 2012
parameters after "options" has no defined result, and so is subject to change at any time, without notice. We repeatedly get Questions here from people who have attempted to pass extra parameters in a similar manner only to have the function fail because of it. Is there a point in relying on accidental behavior when a simple and well-documented adjustment is available? http://www.mathworks.com/help/toolbox/optim/ug/brhkghv-7.html
Star Strider
on 19 Jul 2012
Swapping S and Ft so that F = Ft - S will likely solve your problem. In the objective function you gave it, the lsqnonlin function uses the Jacobian of F in its calculation, not the Jacobian of Ft, and while they may look the same, the derivatives of F = S - Ft will be the negative of the ones you posted, while the derivatives of F = Ft - S will have the same signs as those you posted.
This is likely the reason that with the ‘Jacobian’ option ‘off’, your function converged.
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