TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-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.

Error bound after endpoint power regularization

Statement

Assume Countable Choice and the inputs of Finite endpoint power substitution followed by Gauss–Jacobi quadrature. Let $F_{L}, F_{R}$ be its transformed functions and define $C_{n} = \frac{1}{(2 n)!_{R}} \int_{0}^{1} {\tilde{p}}_{n} (s)^{2} d s,$ where ${\tilde{p}}_{n}$ is the monic degree- $n$ polynomial orthogonal for Lebesgue measure on $[0, 1]$ . Then the improper integral $I = \int_{a}^{b} f (x) d x$ exists, the algorithm's output is $\hat{I} = {\tilde{Q}}_{n} (F_{L}) + {\tilde{Q}}_{n} (F_{R})$ , and $∣ I - \hat{I} ∣ \leq C_{n} (∥ F_{L}^{(2 n)} ∥_{\infty} + ∥ F_{R}^{(2 n)} ∥_{\infty}) .$ Here ${\tilde{Q}}_{n} (H) = \frac{1}{2} \sum_{j = 1}^{n} w_{j} H ((1 + z_{j}) / 2)$ is the affine image on $[0, 1]$ of the $n$ -node $α = β = 0$ rule on $[- 1, 1]$ .

Facts & Assumptions

Given: Countable Choice, $a < b$ , $n \geq 1$ , $α, β > - 1$ , nonnegative integers $m_{a}, m_{b}$ , $h \in C^{2 n} ([a, b])$ , and positive integers $q_{a}, q_{b}$ with $q_{a} (α + 1) - 1 > 2 n$ and $q_{b} (β + 1) - 1 > 2 n$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): this axiom is used only to form the Lebesgue-density Gaussian rule.

[F1]

Finite endpoint power substitution followed by Gauss–Jacobi quadrature: $x_{L} (t) = a + (c - a) t^{q_{a}}$ and $x_{R} (t) = b - (b - c) t^{q_{b}}$ , and $F_{L, R} = f (x_{L, R}) ∣ x_{L, R}^{'} ∣$ on $(0, 1]$ , extended by zero at $0$ ; the output is the sum of the two shifted $n$ -node Gauss–Jacobi values.

[F2]

A sufficiently high power regularizes finite logarithmic endpoint factors: if $δ > r$ , then $t^{δ} P (\log t)$ , extended by zero at $0$ for any polynomial $P$ , is $C^{r}$ on $[0, 1]$ with its first $r$ derivatives zero at $0$ .

[F3]

Gauss–Jacobi exactness and the derivative remainder: for the weight $1$ on $[- 1, 1]$ , the $n$ -node Gaussian rule has error constant $(2 n)!^{- 1} \int_{- 1}^{1} p_{n}^{2}$ multiplying the supremum norm of the $2 n$ -th derivative.

[F4]

Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ : on a compact interval, differentiable substitution with integrable derivative gives the corresponding oriented change-of-variable identity for continuous integrands.

[F5]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : $C^{2 n}$ means continuous derivatives through order $2 n$ , including one-sided endpoint derivatives.

[F6]

The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ and Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$ : $(2 n)!_{R}$ is the positive real image of $\prod_{j = 1}^{2 n} j$ .

[F7]

Continuity and derivatives of positive-base real powers, 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 The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: away from the singular endpoint, the positive real powers, logarithms, and their products with $h$ have continuous derivatives through order $2 n$ .

[F8]

The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents and Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: positive-base powers obey product and iterated-power laws, while $\log (u v) = \log u + \log v$ . For a positive integer $q$ , iterating the logarithm product law gives $\log (t^{q}) = q \log t$ .

Proof

1.1A1F1F2F5F7F8givenalgebra

Put $c = (a + b) / 2$ . On the left half, $x_{L} (t) - a = (c - a) t^{q_{a}}$ and $b - x_{L} (t) = (b - a) (1 - t^{q_{a}} / 2) \geq (b - a) / 2$ . The endpoint logarithm is $\log ((c - a) / (b - a)) + q_{a} \log t$ , so its $m_{a}$ -th power is a polynomial in $\log t$ ; the opposite power and logarithm and $h (x_{L} (t))$ have $C^{2 n}$ derivatives because their arguments remain positive and bounded away from zero, by [F7]. Hence $F_{L} (t) = t^{δ_{L}} P_{L} (\log t) ϕ_{L} (t)$ times a constant, where $δ_{L} = q_{a} (α + 1) - 1 > 2 n$ , $\deg P_{L} \leq m_{a}$ , and $ϕ_{L} \in C^{2 n} ([0, 1])$ . By [F2] and the product rule, $F_{L}$ extends to $C^{2 n}$ at zero with the assigned value.

1.2F1F2F5F7F8givenalgebra

On the right half, $b - x_{R} (t) = (b - c) t^{q_{b}}$ and $x_{R} (t) - a = (b - a) (1 - t^{q_{b}} / 2) \geq (b - a) / 2$ . Its singular logarithm is $\log ((b - c) / (b - a)) + q_{b} \log t$ , while the opposite factors and $h (x_{R} (t))$ have $C^{2 n}$ derivatives by [F7]. The same factorization has $δ_{R} = q_{b} (β + 1) - 1 > 2 n$ and a polynomial in $\log t$ of degree at most $m_{b}$ ; [F2] therefore gives $F_{R} \in C^{2 n} ([0, 1])$ .

1.3F3F6givenalgebra

For $H \in C^{2 n} ([0, 1])$ , apply [F3] to $H ((1 + t) / 2)$ on $[- 1, 1]$ and multiply the integral and quadrature values by $1 / 2$ . The chain rule contributes $2^{- 2 n}$ to the $2 n$ -th derivative, so the shifted rule's error constant is $2^{- 2 n - 1} (2 n)!_{R}^{- 1} \int_{- 1}^{1} p_{n} (t)^{2} d t = (2 n)!_{R}^{- 1} \int_{0}^{1} {\tilde{p}}_{n} (s)^{2} d s = C_{n}$ ; the affine node polynomial is ${\tilde{p}}_{n} (s) = 2^{- n} p_{n} (2 s - 1)$ . Consequently its error is at most $C_{n} ∥ H^{(2 n)} ∥_{\infty}$ .

2.1F1F4step 1.1step 1.2

For every $0 < ε < 1$ , apply [F4] on $[ε, 1]$ . Since $x_{L}$ is increasing, $\int_{ε}^{1} F_{L} (t) d t = \int_{x_{L} (ε)}^{c} f (x) d x$ . Since $x_{R}$ is decreasing and $F_{R} (t) = - f (x_{R} (t)) x_{R}^{'} (t)$ , the oriented substitution gives $\int_{ε}^{1} F_{R} (t) d t = \int_{c}^{x_{R} (ε)} f (x) d x$ . Letting $ε ↓ 0$ , the endpoints tend to $a$ and $b$ , while $F_{L}, F_{R}$ are continuous on $[0, 1]$ by steps 1.1–1.2. Thus both improper half-integrals converge and their sum is $I = \int_{0}^{1} F_{L} + \int_{0}^{1} F_{R}$ .

3.1F1step 1.1step 1.2step 2.1step 1.3algebra∎

Apply step 1.3 to $F_{L}$ and $F_{R}$ , whose required smoothness was proved in steps 1.1–1.2. Since $I$ is their integral sum by step 2.1 and $\hat{I}$ is their quadrature sum by [F1], the triangle inequality gives the stated error bound.

Depends on

Used by

Gauss–Jacobi regularization for x to the minus one half times log x Example

Cited to discharge well-definedness by Finite endpoint power substitution followed by Gauss–Jacobi quadrature.

Dependency tree · two levels

93 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

J. Lebl, Basic Analysis I, §5.4.1–5.4.2, The logarithm and the exponential (standard reference, not scraped)
Christopher J. Bishop, The Riemann Mapping Theorem, Chapter 3 §3, Gauss-Jacobi quadrature lecture notes (standard reference, not scraped)