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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 $error \sim \delta^4$.

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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: – drawar Mar 24 '13 at 12:44
Your graph certainly suggests that the error is order 2. But the theory says it should be order 4. For example, see deBoor's book "A Practical Guide to Splines" or the paper by C. A. Hall, Journal of Approximation Theory, 1968, pp. 209-218. – bubba Mar 26 '13 at 4:09

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