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

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Taylor's Schlömilch–Roche remainder formula

Statement

Let $n \in N$ , let $a < x$ , and suppose $f$ has derivatives through order $n + 1$ on $[a, x]$ , with the usual endpoint continuity. For every natural $p$ with $1 \leq p \leq n + 1$ , some $ξ \in (a, x)$ satisfies $R_{n, a} f (x) = \frac{f^{(n + 1)} (ξ)}{ι (p) ι (n!)} (x - ξ)^{n + 1 - p} (x - a)^{p} .$ The reflected formula holds when $x < a$ .

Facts & Assumptions

Given: $f, a, x, n, p$ as stated.

[L1]

The Taylor polynomial and remainder are those of Taylor polynomials and their remainders, with coefficient identities from Taylor polynomials match the prescribed derivatives at the centre.

[L2]

The Cauchy mean-value quotient form is Cauchy's mean-value theorem in quotient form when the denominator derivative is nonzero.

[L3]

Finite-sum differentiation is licensed by Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, and Laws of finite sums and finite products.

[L4]

If $p \geq 1$ , then the canonical real $ι (p)$ is positive and nonzero (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

Proof

technique · direct

1.1L1L3algebra

Define $Φ (t) : = f (x) - \sum_{j = 0}^{n} f^{(j)} (t) (x - t)^{j} / ι (j!)$ and $Ψ (t) : = (x - t)^{p}$ . Telescoping after differentiating the sum gives $Φ^{'} (t) = - f^{(n + 1)} (t) (x - t)^{n} / ι (n!)$ , while $Ψ^{'} (t) = - ι (p) (x - t)^{p - 1}$ .

1.2givenL1L4algebra

We have $Φ (a) = R_{n, a} f (x)$ , $Φ (x) = 0$ , $Ψ (a) = (x - a)^{p}$ , and $Ψ (x) = 0$ . Also $Ψ^{'} \neq 0$ on $(a, x)$ , because $p \geq 1$ , $ι (p) > 0$ , and $x - t > 0$ .

2.1step 1.1step 1.2L2

Apply [L2] to $Φ, Ψ$ . For some $ξ \in (a, x)$ , $Φ (a) / Ψ (a) = Φ^{'} (ξ) / Ψ^{'} (ξ) = f^{(n + 1)} (ξ) (x - ξ)^{n + 1 - p} / (ι (p) ι (n!))$ .

3.1step 2.1algebra∎

Multiply by $(x - a)^{p}$ . If $x < a$ , interchange the interval endpoints; the same algebraic identity results.

Taylor's Schlömilch–Roche remainder formula

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources