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Richardson extrapolation cancels the second-order term with a fourth-order remainder

Statement

Let $I \subseteq R$ be open, $x \in I$ , and $h > 0$ with $[x - h, x + h] \subset I$ . Suppose that for every $0 < t \leq h$ the central difference has the actual uniform expansion

$D_{t}^{0} f (x) = f^{'} (x) + c_{2} t^{2} + r_{t}, ∣ r_{t} ∣ \leq M t^{4},$

where $c_{2}$ and $M \geq 0$ do not depend on $t$ . Then the Richardson estimate of Richardson extrapolation for a central difference with a fourth-order remainder satisfies

$∣ R_{h} f (x) - f^{'} (x) ∣ \leq \frac{5 M h^{4}}{12} .$

In particular, under A bounded fifth derivative gives the central-difference expansion required by Richardson extrapolation the bound holds with $c_{2} = f^{'''} (x) / 6$ and $M = M_{5} / 120$ , giving $∣ R_{h} f (x) - f^{'} (x) ∣ \leq M_{5} h^{4} / 288$ . If each of the four function values at $x \pm h / 2$ and $x \pm h$ is evaluated as $\hat{f} (y) = f (y) + e_{y}$ with $∣ e_{y} ∣ \leq δ$ , define ${\hat{R}}_{h} f (x)$ by replacing those four values in $R_{h} f (x)$ by their evaluations. Then

$∣ {\hat{R}}_{h} f (x) - f^{'} (x) ∣ \leq \frac{5 M h^{4}}{12} + \frac{3 δ}{h} .$

Facts & Assumptions

Given: The uniform expansion and evaluation-error bounds above.

[F1]

Richardson extrapolation for a central difference with a fourth-order remainder: $R_{h} f (x) = (4 D_{h / 2}^{0} f (x) - D_{h}^{0} f (x)) / 3$ .

[F2]

A bounded fifth derivative gives the central-difference expansion required by Richardson extrapolation: if $∣ f^{(5)} ∣ \leq M_{5}$ on the stencil interval, then $D_{h}^{0} f (x) = f^{'} (x) + (f^{'''} (x) / 6) h^{2} + r_{h}$ with $∣ r_{h} ∣ \leq M_{5} h^{4} / 120$ .

[F3]

Forward and central difference stencils with bounded evaluation errors: each function value error has absolute value at most $δ$ .

[F4]

Absolute value in an ordered field and Basic properties of the absolute value: absolute values are nonnegative and invariant under sign change.

[F5]

The triangle inequality: $∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ .

Proof

Given: The hypotheses listed above.

1.1givenF1algebra

Write $r_{h} = D_{h}^{0} f (x) - f^{'} (x) - c_{2} h^{2}$ and $r_{h / 2} = D_{h / 2}^{0} f (x) - f^{'} (x) - c_{2} h^{2} / 4$ . Substitution into [F1] cancels both copies of $f^{'} (x)$ correctly and gives $R_{h} f (x) - f^{'} (x) = (4 r_{h / 2} - r_{h}) / 3$ , since $4 (c_{2} h^{2} / 4) - c_{2} h^{2} = 0$ .

2.1givenF4F5step 1.1algebra

By [F4]–[F5], $∣ R_{h} f (x) - f^{'} (x) ∣ \leq (4 ∣ r_{h / 2} ∣ + ∣ r_{h} ∣) / 3 \leq (4 M (h / 2)^{4} + M h^{4}) / 3 = 5 M h^{4} / 12$ .

3.1givenF2step 2.1algebra

In the $C^{5}$ case, [F2] supplies the expansion at every $0 < t \leq h$ with the same $c_{2} = f^{'''} (x) / 6$ and $M = M_{5} / 120$ . Substituting in step 2.1 gives $5 (M_{5} / 120) h^{4} / 12 = M_{5} h^{4} / 288$ .

4.1givenF1F3F4F5step 2.1algebra∎

By the definition of ${\hat{R}}_{h}$ in the Statement, [F1] and the stencil formulas in [F3], the inexact perturbation is ${\hat{R}}_{h} f (x) - R_{h} f (x) = \frac{4}{3 h} (e_{+ h / 2} - e_{- h / 2}) - \frac{1}{6 h} (e_{+ h} - e_{- h})$ . Each difference has magnitude at most $2 δ$ by [F4]–[F5]. The total perturbation is therefore at most $8 δ / (3 h) + δ / (3 h) = 3 δ / h$ . Combining with step 2.1 proves the inexact bound.

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Sources

Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, finite differences (standard reference, not scraped)
Andreas Buttenschoen, Introduction to Scientific Computing, Numerical Differentiation (standard reference, not scraped)