       # Re: Extracting parameters from NonlinearFit

```Hi Jeff

suppose You have
myfit= 3.2*Exp[-0.5*t]

than
mfitpara= myfit /. a_*Exp[b_*t] :> {a,b}

will give You the desired parameters

Hope that helps
Jens

> Recently I began writing a program to do a monte carlo simulation of an
> NMR experiment. The whole routine is attached (I hope that works.) The
> problem is this:
> I perform a NonlinearFit utilizing an exponential function of the form
> Exp[-a*x]. The nonlinear fit gives me back an equation, something like
> 1*E(-a*x). I want to extract the a parameter out of the equation and do
> two things with it. Because I am performing this simulation several
> (thousand) times, I want to know the standard deviation of the a
> parameters. Second, I would like to get an average of the terms
> (without having to use the last two lines of my code). If anyone out
> there could help me out, I would greatly appreciate it; I'm really
> stuck.
>
> (PS, if you try to run my code, you may want to change the number of
> iterations to 10 or so. 10000 iterations takes a couple of hours.)
>
> Jeff Wank
> Dept of Chemistry & Biochemistry
> University of Colorado at Boulder
> (303) 492-8085
>
>
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> (*
> This routine takes a set of times, t, and fits a curve of the form \
> y=init*exp(-bx)
> where b is the rate constant (lamda) *)\
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>     \n (*\ initial\ value\ for\ lamda, \ the\ rate\ constant\ *) \n
>     lamda = 49.95; \n\n (*\ number\ of\ iterations\ *) \nn = 1; \n\n
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>     values = \((a[#1, t]&)\)/@Table[i, {i, 1, n}]; \n
>     newlist = \((Transpose[{t, #1}]&)\)/@values; \n
>     \n (*fitting\ the\ data*) \n\n\t
>     lamda1\  = \
>       \((NonlinearFit[newlist[\([#]\)], init*Exp[\(-a\)*x], {x}, {a,
> c}]&)\)/@
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```

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