CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-01

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The Lagrange and Cauchy forms of Taylor's remainder

Statement

Under the hypotheses of Taylor's Schlömilch–Roche remainder formula, there are points $ξ_{L}, ξ_{C}$ between $a$ and $x$ such that $R_{n, a} f (x) = \frac{f^{(n + 1)} (ξ_{L})}{ι ((n + 1)!)} (x - a)^{n + 1}$ and $R_{n, a} f (x) = \frac{f^{(n + 1)} (ξ_{C})}{ι (n!)} (x - ξ_{C})^{n} (x - a) .$

Facts & Assumptions

Given: The hypotheses of the Schlömilch-Roche theorem.

[L1]

For each natural $1 \leq p \leq n + 1$ , the Schlömilch-Roche theorem gives a point $ξ$ strictly between $a$ and $x$ such that $R_{n, a} f (x) = \frac{f^{(n + 1)} (ξ)}{ι (p) ι (n!)} (x - ξ)^{n + 1 - p} (x - a)^{p} .$ (Taylor's Schlömilch–Roche remainder formula).

[L2]

Factorials and natural powers obey The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ and Integer powers $a^{m}$ , while the canonical embedding preserves products and positive naturals are nonzero (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

Proof

technique · direct

1.1

Set $p = n + 1$ in [L1]. Then $(x - ξ)^{0} = 1$ and $ι (n + 1) ι (n!) = ι ((n + 1)!)$ , giving the Lagrange form.

L1L2algebra

1.2

Set $p = 1$ . Since $ι (1) = 1$ , the formula becomes the Cauchy form.

L1L2algebra

2.1

These are the asserted special cases.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

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Sources

MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 lecture notes (standard reference, not scraped)
MathWorld, Schlömilch's remainder (standard reference, not scraped)
J. Lebl, Basic Analysis I, Taylor's theorem and related calculus (standard reference, not scraped)