Problem using lsqnonlin with nlparci toolbox

I have a problem when calculating the confidence intervals using lsqnonlin and nlparci toolbox.

I wanted to calculate the CI using the Jacobian, however, I am getting the following error: Error using nlparci (line 93) The size of J does not match the sizes of BETA and RESID. I am fitting 6 parameters.

Here is the syntax of the fitting function and of the calculation of the CI.

 [par_out, residual,jacobian] = lsqnonlin(@eq_2D3D1TR_set, MAT1,zeros(1,i),MAT2, options, tau_range, g_in_range,MAT3,FIX);
    
 ci = nlparci(par_out, residual, 'jacobian', jacobian);

@eq_2D3D1TR_set is the fitting method

MAT1 are the starting points

zeros(1,i) and MAT2 are the lower and upper bounds

options contain different options of the fitting: options=optimset('Display', 'final','LargeScale', 'Off', 'DiffMaxChange', 0.1, 'DiffMinChange', 0.00001,'TolFun', 1e-100, 'TolX', 0.001, 'MaxFunEvals', 5000, 'MaxIter', 1000);

tau_range, g_in_range are the input data with the selected range

FIX contains information about which of the parameters are fixed during the fitting (the GUI provides this information)

Currently I am fitting with all of the 6 parameters non-fixed.

Truly, when I have a look at the computed Jacobian and residual, then the Jacobian has only 1 column (instead of 6 since I have 6 parameters) and the residual contains only 1 number, what seems also not OK.

Answer 1

[Alan Weiss]

You can give the outputs of lsqnonlin any names you like, but calling the third output "jacobian" doesn't mean that it returns the Jacobian of your function:

   [par_out, residual,jacobian] = lsqnonlin(@...

The lsqnonlin function reference page shows that the function has this signature:

[x,resnorm,residual,exitflag,output,lambda,jacobian] = lsqnonlin(___)

In other words, the Jacobian is the seventh output, not the third.

Alan Weiss

MATLAB mathematical toolbox documentation

[Ed]

Thank you Allen very much, you just made my day!