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Re: NonlinearFit for specific data

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  • Subject: [mg117844] Re: NonlinearFit for specific data
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Fri, 1 Apr 2011 02:35:33 -0500 (EST)

Here's a reasonable procedure:

data2 = {{0, 773.59}, {30, 2351.12}, {60, 4205.8}, {90,
     4222.69}, {120, 4222.69}, {150, 704.328}, {180, 1388.28}, {210,
     1388.28}, {240, 1821.23}, {270, 1821.23}, {300, 754.109}, {330,
     773.59}, {360, 773.59}};
{min, max} = Through[{Min, Max}@data2[[All, 1]]];
data3 = data2 /. {a_, b_} :> {a, Log@b};
quadratic = a + b x + c x^2;
fit1 = FindFit[data3, quadratic, {a, b, c}, x];
model = Exp@quadratic;
initial = fit1 /. Rule -> List
fit2 = FindFit[data2, model, initial, x]

{{a, 7.45656}, {b, 0.00453122}, {c, -0.0000207059}}

{a -> 6.97238, b -> 0.0332337, c -> -0.000192787}

maximum =
   Max[First@Maximize[model /. fit2, x], Max@data2[[All, -1]]];
Show[Plot[{model /. fit1, model /. fit2}, {x, min, max},
   PlotRange -> {0, maximum}], ListPlot@data2]

The plot suggests (IMHO) that such a model is a poor fit to this data, no  
matter WHAT parameters are chosen.

Bobby

On Thu, 31 Mar 2011 04:01:55 -0500, Alexandre Souza <costamfix at gmail.com>  
wrote:

> Dear Group,
> I have the following data :
> data2 = {  {0, 773.59},{30, 2351.12},{60, 4205.8}, {90, 4222.69},{120,
> 4222.69},{150, 704.328},
> {180, 1388.28},{210, 1388.28},{240, 1821.23},{270, 1821.23},{300,
> 754.109},{330, 773.59},{360, 773.59}};
> I want the following curve fit :
>  NonlinearFit[data2, Exp[ a + b x + c x^2], x, {a, b, c}]
> Unfortunatelly I get some "FindFit::sszero
> FindMinimum::sszero" message, and the given result is ackward.
> I also tried FindFit and NonlinearRegress without success.
> Any comment would be very welcome.
> Alex
>


-- 
DrMajorBob at yahoo.com


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