lsqnonlin with jacobian problem

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

[Walter Roberson]

Are you encountering an error message? If so, what message and where?

Answer 1

[Walter Roberson]

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

Answer 2

[Mus Bohr]

Thank you for your response.

My example contains only two variables (parameters). The result of the fitting is exact when I turn Jacobian 'off'. Otherwise, the result is completely erroneous.

Best,

MB

[Walter Roberson]

P=lsqnonlin(@test,P0,[],[],options,S,TE)

1. @test
2. P0
3. []
4. []
5. options
6. S
7. TE

That is 7 parameters. lsqnonlin() does not accept anything after "options".

Answer 3

[Mus Bohr]

Thank you for this precision, but lsqnonlin function accept parameters after "options".

Indeed, I found the solution to my problem. Actually, my scripts is correct, but I should to swap Ft and S (in 'test' function). In other words, F=Ft-S instead of F=S-Ft.

Anyway, thank you so much for you time.

Warm regards,

MB.

[Walter Roberson]

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

[Mus Bohr]

I take note. Again thank you.

Best,

MB

Answer 4

[Mus Bohr]

or provide Jacobian with the negative sign.

Answer 5

[Star Strider]

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.

[Mus Bohr]

Thank you for these precisions.

Best,

MB