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Re: FindFit Mathematica 7

  • To: mathgroup at smc.vnet.net
  • Subject: [mg113989] Re: FindFit Mathematica 7
  • From: Curtis Osterhoudt <cfo at lanl.gov>
  • Date: Sat, 20 Nov 2010 06:11:30 -0500 (EST)

Once 1 is substituted into  a x + b for b, you're simply fitting the function a*x+1 to the data points. Any "extra" parameters which don't appear in the expression-to-fit (such "b", "yo", and "73skidoo" in  ax + 1) appear in the output as "b->0.". 

You should probably let Mathematica treat b as one of the free parameters (remove the substitution b-> 1) and it'll give you what you want. 




On Friday, November 19, 2010 04:58:16 Derivator wrote:
> 
> Thank you. Unfortunately, still
> 
> FindFit[{{0., 1.}, {1., -1}}, a x + b /. b -> 1, {a, b}, x]
> 
> yields
> 
> {a -> -2., b -> 0.}
> 
> instead of
> 
> {a -> -2., b -> 1.}
> 
> Regards,
> 
> L.L.
> 
> 


-- 
==================================
Curtis Osterhoudt
cfo at remove_this.lanl.and_this.gov
==================================


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