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