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