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

FindFit and complex function?

• To: mathgroup at smc.vnet.net
• Subject: [mg80531] FindFit and complex function?
• From: Michael Ignatov <mignatov at chemistry.ohio-state.edu>
• Date: Fri, 24 Aug 2007 02:04:37 -0400 (EDT)

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

• Follow-Ups:
  • Re: FindFit and complex function?
    • From: Carl Woll <carlw@wolfram.com>