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From: South Light <southlight25@gmail.com>
Date: Thu, 14 Aug 2014 10:56:26 -0300
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 <CAKUdAB57SLGpa29vHcUas3-muVP5j+g7knNGmXQrH=mcsOt=nA@mail.gmail.com>
Subject: Re: [math] Curve fitting ...
To: Commons Users List <user@commons.apache.org>
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Hi Ted,

Thanks a lot for your suggestion but I need to add it using java.

Thanks
=E2=80=8Bagain=E2=80=8B
.=E2=80=8B


2014-08-14 2:22 GMT-03:00 Ted Dunning <ted.dunning@gmail.com>:

> Have you considered using an interactive system like R, Matlab or Octave?
>
> You might be happier.
>
> Or even have you considered goal search in Excel?
>
>
>
>
> On Wed, Aug 13, 2014 at 6:08 PM, South Light <southlight25@gmail.com>
> wrote:
>
> > Hi,
> >
> > May be someone can help me with this problem.
> >
> > Given the follow function: y =3D 10 ^ ((x + 82) / (-10 * A))
> >
> > I would like to found the A value witch curve fit better for a set of x=
,y
> > values, usually the set is about 20 to 25 x,y values.
> >
> > I use the CurveFitter class and the ParametricUnivariateFunction
> >
> >
> > ParametricUnivariateFunction function =3D new
>  ParametricUnivariateFunction()
> > {
> >
> > =E2=80=8B  =E2=80=8B
> > @Override
> > =E2=80=8B  =E2=80=8B
> > public double[] gradient(double x, double[] params) {
> > =E2=80=8B
> > (????? comment)
> > =E2=80=8B  =E2=80=8B
> > }
> >
> > =E2=80=8B  =E2=80=8B
> > @Override
> > =E2=80=8B  =E2=80=8B
> > public double value(double x, double[] params) {
> >
> > =E2=80=8B    =E2=80=8B
> > double a =3D params[0];
> >
> > =E2=80=8B    =E2=80=8B
> > return Math.pow(10, ((x + 82) /
> > =E2=80=8B(=E2=80=8B
> > -10 *
> > =E2=80=8Ba
> > =E2=80=8B)=E2=80=8B
> > ));
> >
> > =E2=80=8B  =E2=80=8B
> > }
> >
> > };
> >
> > LevenbergMarquardtOptimizer optimizer =3D new
> LevenbergMarquardtOptimizer();
> >
> > CurveFitter<ParametricUnivariateFunction> fitter =3D new
> > CurveFitter<ParametricUnivariateFunction>(optimizer);
> >
> > double[] x =3D {
> > -82
> > ,
> > =E2=80=8B-85
> > ,
> > =E2=80=8B-89
> >  };
> >
> > double[] y =3D {
> > =E2=80=8B1
> > ,
> > =E2=80=8B1.4
> > ,
> > =E2=80=8B2
> >  };
> >
> > for (int i =3D 0; i < x.length; i++)
> > =E2=80=8B  =E2=80=8B
> > fitter.addObservedPoint(x[i], y[i]);
> >
> > double[] result =3D fitter.fit(function, new double[] { 1, 10 });
> >
> >
> >
> > =E2=80=8BA. =E2=80=8BIs this the best way to solve the problem or there=
's another better
> > way?
> >
> > B. What do we need to write on the gradient area (????? comment) ?=E2=
=80=8B
> >
> > Any help will be more then welcome.
> >
> > Many thanks !!
> >
>

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