AW: Mathematica question
- To: mathgroup at smc.vnet.net
- Subject: [mg8139] AW: [mg8111] Mathematica question
- From: Buttgereit <Buttgereit at compuserve.com>
- Date: Fri, 15 Aug 1997 23:41:50 -0400
- Sender: owner-wri-mathgroup at wolfram.com
Hi Julia,
if you want to fit a exponential model - which is nonlinear - you have t=
o
use NonlinearFit from the Statistics packages:
<<Statistics`NonlinearFit`
will load this package.
You will get your fit using:
NonlinearFit[ data, model, vars, params ]
Say, you have the model: =
y=3D k exp( a t - dt) =
and your data are a list of tuples of one variable and the response
data=3D{ {t1, y1},{t2,y2}, ...,{tn,yn}}; =
then you put in
NonlinearFit[ data, k Exp[a t - dt],
t,{{k,startingvalue},{a,startingvalue},{dt,startingvalue}}]
If this does not work, try better starting values.
In Version 2.2.x you will get as result a list of replacement rules, say
{k->1.2, a->0.1, dt->.002}
You can insert this result in your model by
k Exp[a t - dt]/.{k->1.2, a->0.1, dt->.002}
to get
1.2 Exp[0.1 t - 0.002]
in this example. Version 3 will immediatly put in the computed values for=
your parameters.
If you have access to V.3 have a look at the help browser. You will find
more examples and basics concerning checking the quality of the fit there=
=2E
Good luck,
Peter
----------
Von: INTERNET:MEINERSH at CV3.CHEM.PURDUE.EDU[:]
Gesendet: Dienstag, 12. August 1997 08:34
An: INTERNET:mathgroup at smc.vnet.net
Betreff: [mg8111] Mathematica question
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From: MEINERSH at CV3.CHEM.PURDUE.EDU
To: mathgroup at smc.vnet.net
Subject: [mg8139] [mg8111] Mathematica question
Hello,
I am a beginner Mathematica user and am having a bit of trouble with a
curve fit. I'm sorry to bother you, but when I try to get into the
archives our computer doesn't seem to like it.
I am trying to fit a fluorescence lifetime decay that is at least a
double exponential. I am really confused when following the curve fit
section of the manuel and have tried a few things but with no success. I=
would appreciate if you could give me a few hints or insight to curve
fitting.
Thank you for your time,
Julia Meinershagen