# Richardson extrapolation Simpson's rule

"Assume that S(h) is equivalent to the (composite) Simpson's rule where h is the size of the step. Correct use of Richardson's extrapolation gives the formula: $R(h) = \frac {16S(h)-S(a)} {b}$. What's a and b?"

I'm completely lost here. Can't find any good resources online so don't really know where to start or how to look at the question. Some hints/tips would be greatly appreciated.

• It's taken from an exam in numerical methods at a Swedish university. – gbgult May 14 '18 at 17:49

## 1 Answer

Given a method $A(h) = \alpha_0 + \alpha_1 h^k +\mathcal{O}(h^{k+1})$ that approximates $\alpha_0$, the idea is to combine $A$ at varying step sizes to eliminate the lowest power of $h$, say $h^k$.

Since the error of Simpson's rule is $\mathcal{O}(h^4)$, we have $S(h) = \alpha_0 + \alpha_1 h^4 + \mathcal{O}(h^5)$ and \begin{align*} 16 S(h) &= 16 \alpha_0 + 16 \alpha_1 h^4 + \mathcal{O}(h^5) \\ S(2 h) &= \alpha_0 + 16 \alpha_1 h^4 + \mathcal{O}(h^5) \end{align*} which implies $$16 S(h) - S(2h) = 15\alpha_0 + \mathcal{O}(h^5).$$