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Re: FindFit and complex function?


Post working code to get better answers, but Re or Chop might help.

Bobby

On Fri, 24 Aug 2007 01:04:37 -0500, Michael Ignatov  
<mignatov at chemistry.ohio-state.edu> wrote:

> Hello everyone,
>
> I have ran into a problem of fitting my experimental data points to a
> function in mathematica.
> The function comes from a solution of five algebraic equations with 5
> variables. Very straight forward. Unfortunately cubic roots are present
> in the solution so when the function is evaluated at fixed parameters it
> always return a small imaginary part with the real part. The real parts
> are quite meaningful and imaginary parts are extremely small. For all
> practical purposes I can just drop the imaginary part. When I build a
> model function, which is just a linear combination of 5 solutions, I can
> evaluate it with fixed parameters and see that it describes my
> experimental points rather well. But FindFit fails to perform a search
> always stopping at the starting value. Any ideas on how to either drop
> imaginary part of the function completely or make FindFit work with a
> function that returns complex numbers?
>
> Thanks a lot for your time,
> Michael
>
>



-- 
DrMajorBob at bigfoot.com


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