Re: FindFit and complex function?

• To: mathgroup at smc.vnet.net
• Subject: [mg80575] Re: FindFit and complex function?
• From: Bill Rowe <readnewsciv at sbcglobal.net>
• Date: Sun, 26 Aug 2007 03:10:32 -0400 (EDT)

```On 8/24/07 at 2:04 AM, mignatov at chemistry.ohio-state.edu (Michael
Ignatov) wrote:

>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?

It isn't entirely clear to me what expression you are trying to
fit your data to. It sounds like you may be using a polynomial.
If so, that tends to be numerically unstable. Try replacing the
polynomial with a linear combination of Chebyshev functions.
This will be much more stable and may eliminate your problems
with spurious imaginary parts.
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