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MathGroup Archive 2007

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Re: FindFit and complex function?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80538] Re: FindFit and complex function?
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Sun, 26 Aug 2007 02:51:16 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <falt1e$glh$1@smc.vnet.net>

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?

The easiest way to discard very small imaginary parts is to use *Chop*. 
For instance,

In[1]:= Exp[N[Range[4] Pi I]]

Out[1]= {-1. + 1.22465*10^-16 \[ImaginaryI],
  1.- 2.44929*10^-16 \[ImaginaryI], -1. + 3.67394*10^-16 \[ImaginaryI],
   1.- 4.89859*10^-16 \[ImaginaryI]}

In[2]:= Chop[%]

Out[2]= {-1., 1., -1., 1.}

-- 
Jean-Marc



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