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Subject: Re: [math] Curve fitting ...
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From: Thomas Vandahl <tv@apache.org>
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Message-Id: <69315F3F-2346-4B44-B74D-C61E00325F1C@apache.org>
Date: Thu, 14 Aug 2014 18:47:24 +0200
To: Commons Users List <user@commons.apache.org>
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I take it, you ask for help for doing your homework?

Bye, Thomas=20

> Am 14.08.2014 um 15:56 schrieb South Light <southlight25@gmail.com>:
>=20
> Hi Ted,
>=20
> Thanks a lot for your suggestion but I need to add it using java.
>=20
> Thanks
> =E2=80=8Bagain=E2=80=8B
> .=E2=80=8B
>=20
>=20
>=20
> 2014-08-14 2:22 GMT-03:00 Ted Dunning <ted.dunning@gmail.com>:
>=20
>> Have you considered using an interactive system like R, Matlab or Octave?=

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

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