LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A bounded fifth derivative gives the central-difference expansion required by Richardson extrapolation

Statement

Let $I \subseteq R$ be open, let $x \in I$ , and let $h > 0$ with $[x - h, x + h] \subset I$ . Suppose $f \in C^{5}$ on a neighborhood of this interval and $∣ f^{(5)} (y) ∣ \leq M_{5}$ throughout it. Then for some real $r_{h}$ ,

$D_{h}^{0} f (x) = f^{'} (x) + c_{2} h^{2} + r_{h}, c_{2} : = \frac{f^{'''} (x)}{6}, ∣ r_{h} ∣ \leq \frac{M_{5} h^{4}}{120} .$

The same coefficient $c_{2}$ works for every admissible $h$ at the fixed point $x$ .

Facts & Assumptions

Given: The domain, smoothness and fifth-derivative bound in the statement.

[F1]

Forward and central difference stencils with bounded evaluation errors: $D_{h}^{0} f (x) = (f (x + h) - f (x - h)) / (2 h)$ for $h > 0$ with both stencil points in the domain.

[F2]

The Lagrange and Cauchy forms of Taylor's remainder and Taylor polynomials and their remainders: for a $C^{5}$ function on the interval, the remainder after the degree- $4$ Taylor polynomial at $x$ is $f^{(5)} (ξ) (y - x)^{5} / 5!$ at an intermediate point $ξ$ .

[F3]

Absolute value in an ordered field and Basic properties of the absolute value: absolute values are nonnegative and $∣ - r ∣ = ∣ r ∣$ .

[F4]

The triangle inequality: $∣ r + s ∣ \leq ∣ r ∣ + ∣ s ∣$ .

Proof

Given: The hypotheses listed above.

1.1givenF2algebra

Apply [F2] at $x$ to $x + h$ and $x - h$ . The degree-four Taylor polynomials subtract to $2 h f^{'} (x) + h^{3} f^{'''} (x) / 3$ , because the constant, quadratic and fourth-degree terms agree at the two endpoints. Each remainder has magnitude at most $M_{5} h^{5} / 120$ .

2.1givenF1F3F4step 1.1algebra∎

By [F3]–[F4], the difference of the two remainders has magnitude at most $2 M_{5} h^{5} / 120$ . Divide the subtracted expansion by $2 h$ and use [F1]: the remainder in $D_{h}^{0} f (x)$ is bounded by $M_{5} h^{4} / 120$ , and the cubic term becomes $f^{'''} (x) h^{2} / 6$ .

Depends on

Used by

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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)