MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: FindFit and complex function?

  • To: mathgroup at
  • Subject: [mg80557] Re: FindFit and complex function?
  • From: ben <benjamin.friedrich at>
  • Date: Sun, 26 Aug 2007 03:01:11 -0400 (EDT)
  • References: <falt1e$glh$>

Dear Michael,

have you tried Chop to chop off tiny imaginary parts ?

Can you use a constrained fit, to ensure that for all
parameter choices the fitting function, say f, takes real values?

Real[f] + If[ Chop[ Im[ f ] ] != 0, Infinity, 0]
instead of f ?


On 24 Aug., 08:15, Michael Ignatov <migna... at>
> 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

  • Prev by Date: Re: Solving Nonlinear Equations
  • Next by Date: Re: JLink
  • Previous by thread: Re: FindFit and complex function?
  • Next by thread: Re: FindFit and complex function?