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Re: derivative of cubic spline

  • To: mathgroup at smc.vnet.net
  • Subject: [mg66562] Re: derivative of cubic spline
  • From: Peter Pein <petsie at dordos.net>
  • Date: Sat, 20 May 2006 04:47:08 -0400 (EDT)
  • References: <e4ju6v$d56$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

derivSpline = ReplacePart[spline, Map[Append[Rest[#]*{1, 2, 3}, 0] &, spline[[4]], {2}], 4]

works in version 5.1


Jaccard Florian schrieb:
> Dear All,
> 
> I would like to know how to derivate a spline function constructed by
> Mathematica.
> 
> The question has already been posted in :
>  First derivative of interpolated spline
> 
> But I wasn't able to find an answer in mathgroup.
> 
> Example :
> 
> 
> data={{0, 1}, {1, 2.3}, {2, 2.5}, {3, 1.2}, {4, 0.47}, 
>   {5, 0.38}, {6, 0.76}}
> 
> << "Numericalmath`SplineFit`"
> 
> s = SplineFit[data, Cubic];
> 
> You can see that the spline is good :
> 
> ParametricPlot[s[x], {x, 0, 6}, PlotStyle -> Blue, 
>    Epilog -> {{Blue, Text[HoldForm[y = s(x)], 
>        {2.5, g1[2.5]}, Background -> White]}, 
>      ({PointSize[0.03], Red, Point[#1]} & ) /@ data}, 
>    PlotRange -> {-0.5, 3}, TextStyle -> 
>     {FontFamily -> Times}];
> 
> But D[s[t],t] or s'[t] or so one doesn't help...
> So I'm not able to see that the spline is better than Interpolation (in
> the sense of a continuous derivative)!
> 
> Please help!
> It's the first time I found something that is easier in MathCad than in
> Mathematica...
> 
> Regards
>  
> F.Jaccard
> florian.jaccard at he-arc.ch
>  
> 


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