TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Finite Euler–Maclaurin formula for the composite trapezoid

Statement

Let $a < b$ , let $N \geq 1$ and $p \geq 1$ be integers, put $h = (b - a) / N$ and $x_{i} = a + i h$ , and let $f \in C^{2 p} ([a, b])$ in the sense of Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ . Write $B_{m} (x)$ for Bernoulli polynomials by a normalized coefficient recursion and $B_{m} : = B_{m} (0)$ . For $u_{i} (x) : = (x - x_{i}) / h$ on $[x_{i}, x_{i + 1}]$ , the composite trapezoidal value $T_{N} (f)$ of Composite trapezoidal rule satisfies $T_{N} (f) = \int_{a}^{b} f (x) d x + \sum_{r = 1}^{p} \frac{B_{2 r} h^{2 r}}{(2 r)!} (f^{(2 r - 1)} (b) - f^{(2 r - 1)} (a)) - \frac{h^{2 p}}{(2 p)!} \sum_{i = 0}^{N - 1} \int_{x_{i}}^{x_{i + 1}} B_{2 p} (u_{i} (x)) f^{(2 p)} (x) d x .$ In particular, with $M_{2 p} : = \max_{0 \leq u \leq 1} ∣ B_{2 p} (u) ∣$ , the absolute value of the final remainder is at most $(b - a) M_{2 p} h^{2 p} ∥ f^{(2 p)} ∥_{\infty} / (2 p)!$ .

Facts & Assumptions

Given: A nondegenerate interval $[a, b]$ , integers $N, p \geq 1$ , the uniform mesh, and $f \in C^{2 p} ([a, b])$ .

[A1]

The parameters satisfy $a < b$ , $N \geq 1$ , $p \geq 1$ , $h = (b - a) / N > 0$ , and $f$ has continuous derivatives through order $2 p$ on $[a, b]$ .

[F1]

Composite trapezoidal rule: $T_{N} (f) = h (f (a) / 2 + \sum_{i = 1}^{N - 1} f (x_{i}) + f (b) / 2)$ .

[F2]

Bernoulli polynomials by a normalized coefficient recursion and Derivative, endpoint, parity, and boundedness identities for Bernoulli polynomials: $B_{1} (u) = u - \frac{1}{2}$ , $B_{m}^{'} = m B_{m - 1}$ , $B_{m} (1) = B_{m} (0)$ for $m \geq 2$ , and $B_{m} (0) = 0$ for odd $m \geq 3$ .

[F3]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : $C^{2 p} ([a, b])$ means the derivatives through order $2 p$ are continuous, with one-sided endpoint derivatives.

[F4]

The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , 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 $g (c) \neq 0$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , and 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: the derivatives and antiderivatives of the polynomial factors below obey the usual power, product, sum, and chain rules on each panel, so FTC applied to product derivatives gives the integration-by-parts identities.

[F5]

Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , and A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: the continuous panel integrands are Riemann-integrable, integrals are linear, and adjacent panel integrals add.

[F6]

The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ : $(k)!$ is the positive product of the first $k$ positive integer real images.

[F7]

For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent: a continuous real function on the compact interval $[0, 1]$ attains finite maximum and minimum values.

Proof

1.1givenA1F1F2F5F6algebra

On a panel $[x_{i}, x_{i + 1}]$ , put $u = (x - x_{i}) / h$ . Integration by parts using $B_{1} (u) = u - \frac{1}{2}$ gives $h \int_{x_{i}}^{x_{i + 1}} B_{1} (u) f^{'} (x) d x = \frac{h}{2} (f (x_{i}) + f (x_{i + 1})) - \int_{x_{i}}^{x_{i + 1}} f (x) d x .$ Summing this identity over the $N$ panels and using [F1] and [F5] yields $T_{N} (f) - \int_{a}^{b} f = J_{1}$ , where $J_{k} : = \frac{h^{k}}{k!} \sum_{i = 0}^{N - 1} \int_{x_{i}}^{x_{i + 1}} B_{k} (u_{i} (x)) f^{(k)} (x) d x$ for $1 \leq k \leq 2 p$ .

1.2givenA1F2F3F4F5algebra

For $1 \leq k < 2 p$ , differentiate $B_{k + 1} (u_{i} (x))$ to get $(k + 1) B_{k} (u_{i} (x)) / h$ . Integrating by parts on each panel therefore gives $J_{k} = \frac{h^{k + 1}}{(k + 1)!} \sum_{i = 0}^{N - 1} [B_{k + 1} (u_{i} (x)) f^{(k)} (x)]_{x_{i}}^{x_{i + 1}} - J_{k + 1} .$ Because $B_{k + 1} (1) = B_{k + 1} (0)$ by [F2], the interior endpoint terms telescope, leaving $\frac{h^{k + 1} B_{k + 1}}{(k + 1)!} (f^{(k)} (b) - f^{(k)} (a))$ as the boundary term.

2.1givenstep 1.2F2algebra

Write $b_{q} : = \frac{B_{q} h^{q}}{q!} (f^{(q - 1)} (b) - f^{(q - 1)} (a))$ for $q \geq 2$ . If $k + 1$ is odd and at least $3$ , [F2] makes that boundary term zero, so step 1.2 gives $J_{k} = - J_{k + 1}$ . If $k + 1 = 2 r$ is even, step 1.2 gives $J_{2 r - 1} = b_{2 r} - J_{2 r}$ . For each $r < p$ , the odd-index recurrence at $k = 2 r$ gives $J_{2 r} = - J_{2 r + 1}$ ; hence $J_{2 r - 1} = b_{2 r} + J_{2 r + 1}$ . Finite induction on $r = 1, \dots, p - 1$ (The principle of mathematical induction) iterates these identities; followed by the final even-index recurrence, it gives $J_{1} = \sum_{r = 1}^{p} \frac{B_{2 r} h^{2 r}}{(2 r)!} (f^{(2 r - 1)} (b) - f^{(2 r - 1)} (a)) - J_{2 p} .$

3.1givenA1F2F5F6F7step 1.1step 2.1algebra∎

The polynomial $B_{2 p}$ is continuous by [F2], so [F7] ensures $M_{2 p} = \max_{[0, 1]} ∣ B_{2 p} ∣$ exists and is finite. By the definition of $J_{2 p}$ and the bound $∣ B_{2 p} (u) ∣ \leq M_{2 p}$ on each unit cell, $∣ J_{2 p} ∣ \leq h^{2 p} M_{2 p} ∥ f^{(2 p)} ∥_{\infty} \sum_{i} (x_{i + 1} - x_{i}) / (2 p)! = (b - a) M_{2 p} h^{2 p} ∥ f^{(2 p)} ∥_{\infty} / (2 p)!$ . Substituting step 2.1 into step 1.1 proves the stated formula and remainder bound.

Depends on

Used by

Even-power expansion of trapezoid levels with a uniform remainder Corollary

Dependency tree · two levels

83 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

L. Ridgway Scott, Numerical Analysis, 2nd ed., §13.3 (standard reference, not scraped)
NIST Digital Library of Mathematical Functions, §3.5(iii), Romberg Integration (standard reference, not scraped)