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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


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