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Splines without the Splines package

Klaus Schiefermayr, University
Posted 2 months ago

Hi there.
I want to compute Splines for a data set. Up to now I did this with the Splines package and the corresponding command SplineFit[...]. But it should also work with the command Interpolation[..], but I was not able to manage... Any idea?
POSTED BY: Klaus Schiefermayr
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J. M.
J. M.
Posted 2 months ago
POSTED BY: J. M.

Thanks a lot!
POSTED BY: Klaus Schiefermayr

Maybe this will help:

Cubic spline conditions in general leave two degrees of freedom. The spline conditions themselves require the function to pass through the data and to have continuous derivatives up to order two at the interior data points. There are many ways to complete a spline by adding two more conditions. It appears that Interpolation[] makes a "natural" spline, and the "Splines`" package makes a "not-a-knot" spline.
POSTED BY: Michael Rogers

Thanks anyway!!
POSTED BY: Klaus Schiefermayr

You are right, the two interpolations are different:

sp1 = SplineFit[data, Cubic];
sp3 = Interpolation[data, Method -> "Spline", InterpolationOrder -> 3];
plot2 = ParametricPlot[sp1[t], {t, 0, 4}];
plot3 = ParametricPlot[{t, sp3[t]}, {t, -2, 2}, PlotStyle -> Red];
Show[plot2, plot3]

The documentation on SplineFit states that the second derivative of the spline at the endpoints is set to 0. Interpolation does not do that, apparently. Sorry I can't help you more on this.
POSTED BY: Gianluca Gorni

That's clear. But the point is that with the command Interpolation[...] one obtains the function value 4.203... whereas the command SplineFit[...] gives the function value 4.127..., both at the point x=0.5
POSTED BY: Klaus Schiefermayr

SplineFit gives a SplineFunction object, which spits out couples {x,y}, while Interpolation gives y values only. Try this:

sp2 = Interpolation[data, Method -> "Spline"];
plot3 = ParametricPlot[{t, sp2[t]}, {t, -2, 2}];
Show[plot1, plot3]
POSTED BY: Gianluca Gorni
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