Recipe

Richardson-extrapolate a difference approximation

Use when: you have an approximation $A(h)$ with a known, pure leading error order $h^p$ (for the central-difference formula, $p=2$ ), computed at a step size $h$ and at $h/2$

Goal: a new approximation, accurate to $O(h^{p+1})$ , with the shared leading error term of the two inputs canceled exactly

Taught in Richardson extrapolation

Quick recipe

Confirm the error order $p$
Compute the approximation at $h$ and at $h/2$
Combine as $\dfrac{2^pA(h/2)-A(h)}{2^p-1}$
Check that the improvement is dramatic, not incremental

1
Identify the power $p$ in the approximation's own error expansion. This is established when the formula itself is derived — Section 3 found $p=2$ for the central-difference formula.

Example

Section 3 proved the central-difference formula's leading error term is proportional to $h^2$ , not $h$ . So $p=2$ for $D(h)$ .
- a one-sided (forward- or backward-difference) approximation, $p=1$ — use $2^1=2$ in place of $4$ in Step 3.
- the central-difference approximation, $p=2$ — use $2^2=4$ , as the rest of this recipe does.
Why: the entire combination in Step 3 is built to cancel one specific power of $h$ . Using the wrong $p$ cancels nothing at all. It can make the result worse than either input.
2
Evaluate $A(h)$ and $A(h/2)$ with the same formula, the same $x$ , changing only the step size.

Example

For $f(x)=\sin x$ at $x=1$ : $D(0.1) = \dfrac{\sin(1.1)-\sin(0.9)}{0.2} \approx 0.539402$ ; $D(0.05) = \dfrac{\sin(1.05)-\sin(0.95)}{0.1} \approx 0.540077$ .

Why: two data points along the same known error curve are exactly what's needed. They isolate and cancel that curve's leading term. One point alone can't distinguish the true value from the error riding along with it.
3
For the central-difference formula ( $p=2$ ), this is $\dfrac{4A(h/2)-A(h)}{3}$ .

Example

$\frac{4(0.540077)-0.539402}{3} = \frac{2.160309-0.539402}{3} \approx 0.540302.$

Why: write $A(h)=\text{exact}+Ch^2$ and $A(h/2)=\text{exact}+\frac{Ch^2}{4}$ . The combination $4A(h/2)-A(h) = 3\cdot\text{exact}$ exactly. The $Ch^2$ terms cancel by construction. That leaves three copies of the true value to divide back out.
4
Compare the extrapolated value's own error against the two inputs' errors. It should drop by several orders of magnitude — not just by the same factor the inputs already differed by.

Example

$D(0.05)$ 's own error was about $2.25\times10^{-4}$ ; the extrapolated value's error is about $1.13\times10^{-7}$ — roughly $2{,}000$ times smaller, far past the factor of $4$ that halving $h$ alone bought between $D(0.1)$ and $D(0.05)$ .

Why: an extrapolation that only matches the improvement already visible between $A(h)$ and $A(h/2)$ hasn't canceled anything. A genuine order of the error series vanishing exactly should outperform naive halving by a wide margin.

Common mistakes

Using the combination weight for the wrong order: doubling instead of quadrupling (or the reverse) when the wrong $p$ is assumed. This cancels nothing, and can land further from the truth than either input.
Extrapolating two approximations that don't share one known error-order structure — say, one computed with the central formula and one with the forward formula. Step 3's algebra assumes both inputs lie on the same error curve.
Choosing $h$ so small before extrapolating that round-off (Section 4) has already contaminated both $A(h)$ and $A(h/2)$ . Richardson extrapolation cancels truncation error exactly. But it has no way to detect that the numbers it's combining are already mostly noise. So it confidently extrapolates that noise too.
Treating the extrapolated result as though it still carried the original formula's coarser error bound. It should carry the new, higher-order bound Step 3 actually earned instead.