Re: FindFit and complex function?
- To: mathgroup at smc.vnet.net
- Subject: [mg80582] Re: [mg80531] FindFit and complex function?
- From: DrMajorBob <drmajorbob at bigfoot.com>
- Date: Sun, 26 Aug 2007 03:14:09 -0400 (EDT)
- References: <1269514.1187937530861.JavaMail.root@m35>
- Reply-to: drmajorbob at bigfoot.com
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