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nlinfit with modelfun as an integral

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Armantas on 31 Jul 2014
Commented: Star Strider on 3 Aug 2014
Dear all,
I have discrete data A(x,y), which I want to fit by a specified function y=f(x). My function f(x) has the following restriction: df(x)/dlog(x) is equal to a sum of two Gaussians. df(x)/dlog(x) being the derivative of f(x) with respect to the argument log(x). Given this restriction f(x) can be expressed as an integral. This gives 6 fit parameters, namely: the heights of the Gaussians, their means and their standard deviations. I have gone this far:
par=[1e-9,1e-5,1,10000,1,1]; %initial Gaussian parameter guess
Integrand = @(x) ((par(3)/(par(5)*(2*pi)^0.5))*exp((-(log10(x)-log10(par(1))).^2)/(2*par(5)^2))+(par(4)/(par(6)*(2*pi)^0.5))*exp((-(log10(x)-log10(par(2))).^2)/(2*par(6)^2)))./(x*log(10)); %two gaussians
Integral = @(x) integral(Integrand,0,x);
I however get the following error:
Error using nlinfit (line 142)
Error evaluating model function '@(x)integral(Integrand,0,x)'.
Caused by:
Error using @(x)integral(Integrand,0,x)
Too many input arguments.
How can I fix this? Thank you

Accepted Answer

Star Strider
Star Strider on 1 Aug 2014
Edited: Star Strider on 1 Aug 2014
There are a few problems I can see. There could be more, but this should get you started:
Add ‘par’ to the Integrand function arguments:
Integrand = @(par,x) ((par(3)/(par(5) ...
The Integral function then becomes:
Integral = @(par,x) integral(@(x) Integrand(par,x),0,x);
I suggest that you name your initial parameter estimates ‘par0’ (or something other than ‘par’) to avoid confusion:
par0 = [1e-9,1e-5,1,10000,1,1]; %initial Gaussian parameter guess;
est_par = nlinfit(A(:,1),A(:,2),Integral,par0)
The ‘est_par’ assignment are the parameters estimated by nlinfit.
I can’t test your code (so no guarantees), but these changes should at least allow it to run.
Star Strider
Star Strider on 3 Aug 2014
My pleasure!
Yours is the most unusual curve-fitting design (with the integral) that I’ve thus far encountered, so I learned much from it. It is also interesting that The Statistics Toolbox function nlinfit could deal with the NaN values on its own.

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