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Re: Exponential fit question.

  • To: mathgroup at smc.vnet.net
  • Subject: [mg28001] Re: Exponential fit question.
  • From: Martin Kraus <Martin.Kraus at informatik.uni-stuttgart.de>
  • Date: Wed, 28 Mar 2001 02:40:29 -0500 (EST)
  • Organization: Institut fuer Informatik, Universitaet Stuttgart
  • References: <99pdg6$lft@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

joe wrote:
> 
> hello.
> 
> I was wondering if someone could help me with the following problem.
> 
> I am trying to perform an exponential fit to the following data
> {{x,y}}
> 
> data
> ={{50,22},{64,62},{78,122},{93,269},{107,414},{122,507},{136,597}}
> 
> Fit[data,Exp[x],x]
> 
> what I get is
> 
> 1.94272422061017735^-63 *E^x Which is not correct.
> 
> With Excel I get 7.5*E^0.0034x which is correct.
> 
> How can I do this with Mathematica ?
> 
> Thanks.
> -Joseph.

Hi Joseph,

Fit calculates only coefficients for the functions you specify,
i.e., the result will be of the form a*Exp[x] 
while you need something like a*Exp[b*x].

In order to do so you could apply Log to all data values and
fit these logarithmic values to a linear function (a + b*x)
and then form Exp[a + b*x] which is a*Exp[b*x].

data = {{50, 22}, {64, 62}, {78, 122}, {93, 269}, {107, 414}, {122, 
      507}, {136, 597}}

data2 = N[Map[{#[[1]], Log[#[[2]]]} &, data]]

Simplify[Exp[Fit[data2, {1, x}, x]]]

This returns 5.22289*Exp[0.0379875*x] which is pretty good.
(I think the 0.0034 above should be 0.034.)
You may worry that Fit now minimized a quadratic deviation of
the logarithmic values; however, in practice this is often
desirable because the error of measurements often grows 
linearly with the data values.

Hope this helps

Martin Kraus


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