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Re: FindFit

  • To: mathgroup at smc.vnet.net
  • Subject: [mg105824] Re: [mg105797] FindFit
  • From: Thomas Dowling <thomasgdowling at gmail.com>
  • Date: Sun, 20 Dec 2009 06:56:18 -0500 (EST)
  • References: <200912191126.GAA24631@smc.vnet.net>

Hello,

I think there  a few ways of approaching this problem.

1. The following are the data

data = {{40, 0.0624}, {50, 42.2 .276}, {58, 127.718}, {60,
    216.608}, {70, 2040.088}};

2. Use ListPlot to 'take a look':

dataplot =
 ListPlot[data, PlotMarkers -> {Style["\[FilledSquare]", Red], 10},
  AxesOrigin -> {0, 0}]

3. The data may be fitted to the exponential function as follows (note that
initial values have been chosen)

Clear[a, k];

model = ao E^(k t);

soln = FindFit[data, model, { {ao, 4}, {k, 0.05}}, t]

4. A fitted equation may now be generated

fittedeqn = ao E^ (k t) /. soln

5. 'Take a look' at the fitted plot

plotfitted =
 Plot[fittedeqn, {t, 0, Max[data[[All, 1]]] + 15},
  PlotStyle -> {Green}, AxesOrigin -> {0, 0}]

5. 'Show both plots superimposed:

Show[dataplot, plotfitted]

6.( Do you need to show a data point for time zero?)

Tom Dowling



On Sat, Dec 19, 2009 at 11:26 AM, jj <yohan2 at spray.se> wrote:

> Can anybody help me?
> I want to try to show my model (function) and my data in the same
> graph so I can see that my conclusions are correct.
> data= { {40,0.0624}, {50,42.2.276}, {58,127.718}, {60,216.608},
> {70,2040.088},
> I used FindFit for Exponential as my model to plot:
> t200= {Exp200}
>
> t200= {Exp300}
>
> model=aExp[kt];
>
> fit=FindFit[data,model,{a,k},t]
>
> Best regads jj
>
>


  • References:
    • FindFit
      • From: jj <yohan2@spray.se>
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