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MathGroup Archive 2007

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Re: FindFit and complex function?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80579] Re: [mg80531] FindFit and complex function?
  • From: Michael Ignatov <mignatov at chemistry.ohio-state.edu>
  • Date: Sun, 26 Aug 2007 03:12:36 -0400 (EDT)
  • References: <200708240604.CAA16804@smc.vnet.net> <46CED255.1090804@wolfram.com>

Thank you all for all of the suggestions. Using
SetOptions[Roots, Cubics -> False, Quartics -> False];
before Solve worked like a charm. FindFit works with produced model no 
problem.

All best,
Michael

Carl Woll wrote:
> Michael Ignatov 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
>>  
>>
> Assuming that you are using Solve to solve your equations, you can set 
> an option to return Root objects instead of radicals. Root objects can 
> be numericalized more reliably than radicals. For example:
>
> In[136]:= Solve[x^3 - 3 x - 1 == 0, x] // N
>
> Out[136]= {{x ->
>   1.87939+ 0. \[ImaginaryI]}, {x -> -1.53209 +
>    1.11022*10^-16 \[ImaginaryI]}, {x -> -0.347296 +
>    1.11022*10^-16 \[ImaginaryI]}}
>
> Now, set options for Roots and try again:
>
> In[137]:= SetOptions[Roots, Cubics -> False, Quartics -> False];
>
> In[138]:= Solve[x^3 - 3 x - 1 == 0, x] // N
>
> Out[138]= {{x -> -1.53209}, {x -> -0.347296}, {x -> 1.87939}}
>
> This time there are no spurious imaginary parts.
>
> Carl Woll
> Wolfram Research


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