Richardson Extrapolation

Let T_n be the trapezium rule approximation with 2ⁿ subintervals to an integral I. then with $h_n=\frac{b-a}{2^n}$, we have

say, where we shall assume that $k=\frac{b-a}{12}f''(\eta )$ remains constant as we alter n. Suppose that with $h_{n-1}=\frac{b-a}{2^{n-1}}=2h_n$,

Eliminating k from equations () and (), we find

$$I=\frac{4T_n-T_{n-1}}{3}.$$

If our assumption about the constancy of k had been true, we should have the exact value of the integral, but even if it is only approximate, we should have a better answer.

We investigate the nature of the extrapolated answer a little further. Let $I=\int _a^bf(x)dx$. then

$$\begin{array} {lcl}T_n & = & \frac{h_n}{2}\left\{ f(a)+f(b) ... ...b) \right\} +2h_n\sum _{k=1}^{2^{n-1}-1}f(a+2kh_n),\end{array}$$

so

$$\begin{array} {lcl}\frac{4T_n-T_{n-1}}{3} & = & \frac{h_n}{3... ...a+2kh_n)+4f(a+(2k+1)h_n)+f(a+(2k+2)h_n) \right\} , \end{array}$$

which is just the composite Simpson rule.