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Richardson extrapolation

How to combine two approximations with steps $h$ and $h/2$ to cancel the dominant error term and raise the order, with the formulas for all terms and for even powers.

The idea: cancel the dominant error

If we know the error form of an approximation $N_1(h)$ of the exact value $M$ , we can combine $N_1(h)$ and $N_1(h/2)$ to remove the first error term. Repeating raises the order each time.

M=N_1(h)+k_1 h+k_2 h^2+k_3 h^3+\cdots\quad(\mathcal{O}(h))

General case and even powers

When the error has all powers of h, each step gains one order. If by symmetry only even powers appear (as in the central formula), each step gains two orders and the weights change.

DerivationDerivation: Richardson raises the orderView as its own page →

Start from the all-powers error and also evaluate with step $h/2$ :
$\begin{aligned} M&=N_1(h)+k_1 h+k_2 h^2+\cdots\\ M&=N_1\!\left(\tfrac{h}{2}\right)+k_1\tfrac{h}{2}+k_2\tfrac{h^2}{4}+\cdots \end{aligned}$
Compute $2\cdot(\text{second})-(\text{first})$ . The $k_1h$ term cancels and an $\mathcal{O}(h^2)$ error remains:
$M=N_1\!\left(\tfrac{h}{2}\right)+\left[N_1\!\left(\tfrac{h}{2}\right)-N_1(h)\right]-k_2\tfrac{h^2}{2}-\cdots$
Call the known part $N_2(h)$ : it is an order-2 approximation. Repeating with weights $\tfrac13$ , $\tfrac{1}{15}$ , … raises the order further.
$N_2(h)=N_1\!\left(\tfrac{h}{2}\right)+\left[N_1\!\left(\tfrac{h}{2}\right)-N_1(h)\right]$