Re: Help fitting Exponential curves
- To: mathgroup at smc.vnet.net
- Subject: [mg29029] Re: Help fitting Exponential curves
- From: "Joseph C. Slater" <joseph.slater at wright.edu>
- Date: Fri, 25 May 2001 01:48:05 -0400 (EDT)
- Organization: Wright State University
- References: <9efk8o$1da@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <9efk8o$1da at smc.vnet.net>, Todd <tcs3a at virginia.edu> wrote:
> Hello,
> I am looking to fit some exponential curves to a form of
> P=Po*e^(-K*t) where K and Po are values I need. The P and t I have.
> I'm working with mathematica 4.1 and I can read the data and plot it fine
> but it doesn't want to give me reasonable fits.
> So I guess my question is how would you go about getting the Fitting to
> work on fitting a Pre-Exponential and an exponential variable. If I type
>
> PeakFit[x_] = Fit[Peak, Exp[-x], x]
>
> it only gives me a Pre-exponential, however if I type.
>
> PeakFit[x_] = Fit[Log[Peak], {-x,Log[x]}, x]
>
> It just doesn't like that. Could Someone Please point me in the right
> direction.
>
>
> Thank you,
>
> Todd
>
>
If you take the natural log of the data, you are then fitting to the
functions
ln(P [your data]) = ln Po + ln (e^-K*t)
which is
ln(P [your data]) = ln Po +-K*t
which can be written as
A+B*t where A is ln(Po) and B is -K. Now you can do a simply polynomial
curve fit to obtain A and B, from which P and K can be derived.
I don't recall the commands for all of this in Mathematica, but it can't
be too hard from here.
JS