# Construct an extrapolation table with optimal rate of convergence for cubic spline approximation

Let $S$ be a cubic spline interpolant that approximates a function $f$ on the given nodes $x_{0},x_{1},...,x_{n}$ with the boundary conditions: $S''(x_{0})=0$ and $S'(x_{n})=f'(x_{n})$. Use $S$ to estimate $f(0.1234567)$ where $f(x)=xe^{x}$ and the nodes are $n+1$ uniformly distributed points on $[-1;1]$ for $n=20, 40, 80, 160, 320$. Construct an extrapolation table with optimal rates of convergence using these estimates.

Actually I have no difficulty computing the approximation to $f(0.1234567)$ using various choices of nodes:

0.139678933961527   0.139678959983576   0.139679035306608   0.139679035488050   0.139679035532356


What puzzles me is how to construct the extrapolation table with optimal rates of convergence. I'm supposed to use Richardson's extrapolation to generate such a table but then don't know how the truncation error in the the cubic spline approximation would look like. Any hints/suggestions to derive a predictable form of the error involved in the approximation would be much appreciated, thanks!

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The error in cubic spline approximation is typically proportional to $\delta^4$, where $\delta$ is the mesh size (the distance between nodes). So, take the values you have calculated, and measure their errors from the true value of $f(0.1234567)$. Make a table that has these errors and your mesh sizes (which will be $\delta = 2/20, 2/40, \ldots, 2/320$). Try to identify a relationship of the form $error \sim \delta^4$.
Thanks for the reply. I've tried plotting $ln(E)$ vs. $ln(h)$ where $E$ is the absolute error and $h$ is the mesh size and noticed it has a slope of ~ $2.00$ so I guess this cubic spline interpolant has error $O(h^{2})$, is that right? (Btw, here is the graph: i.imgur.com/JaVcHbW.png) –  drawar Mar 24 '13 at 12:44