MATHEMATICA: Cubic Spline Interpolation
Rajesh Pendekanti (ERC ndd)
Hi!
I am looking for the Cubic Spline Interpolation in Mathematica.
ie given array {{x1,y1},{x2,y2},....,{xn,yn}}
with x1<x2< .... <xn
I need to be able to obtain the value of 'y' at
any 'x' ( x1<x<xn ).
Is there a pre-defined Mathematica function call for this??
Also is there a ftp / mosaic site with useful Mathematica
programs????
Thanks
Rajesh
Louis Gascoigne
>ie given array {{x1,y1},{x2,y2},....,{xn,yn}}
>with x1<x2< .... <xn
>I need to be able to obtain the value of 'y' at
>any 'x' ( x1<x<xn ).
>Is there a pre-defined Mathematica function call for this??
Why don't you just use a polynomial fit and then get a function in terms
of y?
Robert Knapp
In article <eapu294....@taurus.oac.uci.edu>,
Louis Gascoigne <eap...@taurus.oac.uci.edu> wrote:
> Why don't you just use a polynomial fit and then get a function in terms
> of y?
The appropriateness of this depends on circumstances--in amny cases, using an
excessively high order polynomial can lead to highly oscillatory
approximations.
Sinan Karasu <si...@u.washington.edu> wrote:
> Why not use NumericalAnalysis`SplineFit.m
> (or something like that...)
>
> Sinan
>
Actually it is NumericalMath`SplineFit`
An alternative to using cubic splines is to use the Mathematica function
Interpolation. This command constructs an InterpolatingFunction which is a
LOCAL approximation (default cubic) based on data at the nearest points.
If you data contains error, this may be better than a spline which is not
simply a local approximation. The advantage of the spline is that it is
continuously differentiable. (The result of Interpolation can be made
continuously differentiable by specifying derivatives with the data--in
which case it uses Hermite Interpolation.)
Here is an example:
In[7]:=
data = Table[ {x, N[Sin[x]]},{x, 0, 10}];
In[8]:=
fun = Interpolation[ data ];
In[9]:=
fun[1.5]
Out[9]=
0.975987
In[10]:=
Plot[ fun[x], {x, 0, 10}];
or the spline:
In[18]:= Needs["NumericalMath`SplineFit`"]
In[19]:= sp = SplineFit[data,Cubic]
Out[19]= SplineFunction[Cubic,{0., 10.},<>]
In[20]:= sp[1.5]
Out[20]= {1.5, 0.994193}
Robert Knapp
Wolfram Research, Inc.