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Re: Using Fit to interpolate data

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  • Subject: [mg127454] Re: Using Fit to interpolate data
  • From: Bill Rowe <readnews at>
  • Date: Thu, 26 Jul 2012 03:33:25 -0400 (EDT)
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On 7/25/12 at 2:33 AM, carlsonkw at (Kris Carlson) wrote:

>On Tue, Jul 24, 2012 at 10:14 AM, jf.alcover <jf.alcover at>

>>Why not use FindFit and an Exp ?

>>fbddFit = Exp[-a x + b] /. FindFit[ fiberDataDensitiesFeierabend,
>>Exp[-a*x + b], {a, b}, x]

>Aha. That is helpful. I had tried using a decay function with Fit
>and couldn't get the syntax, but now I see it requires FindFit.
>Merci. The power law seems to work better, tho.

>In[277]:= fbddFit = Fit[fiberDataDensitiesFeierabend, {x^-13}, x]

>Out[277]= 2.4929*10^12/x^13

If your criteria for judging the fit is a plot then it is almost
always the case a power law fit will seem to fit non-linear data
better. And if the only thing you need to be able to do is find
a way to interpolate your data for a new prediction this is
likely to be OK.

But if the parameters are supposed to have some physical or
biological meaning (corresponding to some meaningful model) then
the high exponent is a very strong sign this isn't a good model
and will quite likely have poor prediction power in terms of any extrapolation.

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