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Re: Using Fit to interpolate data

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  • Subject: [mg127466] Re: Using Fit to interpolate data
  • From: Bill Rowe <readnews at>
  • Date: Fri, 27 Jul 2012 04:56:55 -0400 (EDT)
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On 7/26/12 at 3:34 AM, kjm at (Kevin J. McCann) wrote:

>One last go at it:
>fiberDataDensitiesFeierabend = {{16, 0}, {10.7, 0.11}, {10.4,
>0.19}, {9.77, 0.41}, {8.29, 3.05}, {7.14, 19.86}};
>model = A Exp[-a x + b]; params =
>FindFit[fiberDataDensitiesFeierabend, model, {A, a, b}, x]

Your choice for a model is a poor choice. Mathematically, there
is no difference between the following

a Exp[b x]
Exp[a x + b]
c Exp[a x + b]

But the last choice is numerically unstable since there are an
infinite number of choices for the parameters b and c that give
the exact same least squares fit to the data.

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