LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge 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.

Composite midpoint enclosure for the scaled modified Bessel integral

Statement

Let $n \geq 0$ be an integer, $x \geq 0$ , $m \geq 1$ , $h = π / m$ , and $f (θ) = e^{- x (1 - \cos θ)} \cos (n θ), Q_{m} = h \sum_{j = 0}^{m - 1} f ((j + 1 / 2) h) .$ Then $∣ e^{- x} I_{n} (x) - \frac{Q_{m}}{π} ∣ \leq \frac{h^{2}}{24} ((x + n)^{2} + x) .$ The bound certifies the exact midpoint integral. Floating-point summation is certified only if its operations are enclosed separately.

Facts & Assumptions

Given: Integer $n \geq 0$ , real $x \geq 0$ , and integer $m \geq 1$ .

[F1]

Integral representations and a range bound for integer Bessel functions: $I_{n} (x) = \frac{1}{π} \int_{0}^{π} e^{x \cos θ} \cos (n θ) d θ, 0 \leq e^{- x} I_{n} (x) \leq 1.$

[F2]

Taylor's Lagrange remainder follows from repeated Rolle steps: for a $C^{2}$ function, the first-order Taylor remainder at $c$ is $f^{''} (ξ) u^{2} / 2$ for some point between $c$ and $c + u$ .

[L9]

The exponential function is smooth and $(\exp)^{'} = \exp$ , Sum, product, quotient, and chain rules from difference quotients, The derivatives of sine and cosine are cosine and minus sine: the displayed real functions are smooth on their stated domains, and their derivatives follow by the product, power and chain rules; iterate these rules for the finite higher derivatives used below.

Proof

1.1givenL9algebra

Put $g (θ) = e^{- x (1 - \cos θ)}$ . Since $1 - \cos θ \geq 0$ , we have $0 < g \leq 1$ on $[0, π]$ . Differentiation gives $g^{'} = - x \sin θ g$ and $g^{''} = (- x \cos θ + x^{2} \sin^{2} θ) g$ . Thus $∣ g^{'} ∣ \leq x$ and $∣ g^{''} ∣ \leq x + x^{2}$ . For $f = g \cos (n θ)$ , $∣ f^{''} ∣ \leq ∣ g^{''} ∣ + 2 n ∣ g^{'} ∣ + n^{2} ∣ g ∣ \leq x + x^{2} + 2 n x + n^{2} = (x + n)^{2} + x .$

2.1step 1.1F2algebra

On each panel centered at $c$ , [F2] bounds $∣ f (c + u) - f (c) - u f^{'} (c) ∣$ by $M u^{2} / 2$ , where $M = (x + n)^{2} + x$ by step 1.1. Integrating over $- h / 2 \leq u \leq h / 2$ cancels the odd linear term and bounds that panel's error by $M h^{3} / 24$ .

3.1F1step 2.1algebra∎

Summing the $m$ panel errors gives $∣ \int_{0}^{π} f - Q_{m} ∣ \leq m M h^{3} / 24 = π M h^{2} / 24$ . By [F1], $e^{- x} I_{n} (x) = π^{- 1} \int_{0}^{π} f (θ) d θ$ ; dividing by $π$ proves the claim.

Depends on

Used by

Dependency tree · two levels

29 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

DLMF, §10.32(ii), integral representations for modified Bessel functions (standard reference, not scraped)
N. M. Temme, “Numerical aspects of special functions,” Acta Numerica 17 (2008), §§2 and 4 (standard reference, not scraped)