ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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.

A narrow peak reaches the adaptive Simpson depth cap

Example

Let $f (x) = 1 / (1 + 10000 (x - 1 / 2)^{2})$ on $[0, 1]$ . With tolerance $τ = 1 / 1000$ , depth cap $D = 0$ , and panel-call budget $P = 1$ , Finite-depth adaptive Simpson with an indicator returns the refined estimate with cap-reached status.

Facts & Assumptions

Given: The real-valued rational function $f$ on $[0, 1]$ , exact arithmetic, and the adaptive call with $D = 0$ , $P = 1$ , $τ = 1 / 1000$ .

[F1]

Finite-depth adaptive Simpson with an indicator: if the local indicator exceeds its tolerance and the depth cap is reached, the call returns its refined estimate with cap-reached status.

[F2]

Embedded Simpson refinement indicator: the local indicator is $∣ S_{1} - S_{0} ∣ / 15$ .

[F3]

Composite Simpson rule on an even uniform grid: on an even uniform grid, the composite Simpson value is the weighted sum of the sampled endpoint, midpoint and quarter-grid values.

Verification

1.1givenF2F3algebra

The endpoint values are $f (0) = f (1) = 1 / 2501$ , the midpoint value is $f (1 / 2) = 1$ , and the quarter-point values are $f (1 / 4) = f (3 / 4) = 1 / 626$ . The composite rule [F3] gives $S_{0} = 2 / 3 + 1 / 7503$ and $S_{1} = (1 / 12) (2 + 8 / 626 + 2 / 2501)$ .

2.1givenF2step 1.1algebra

Since $8 / 626 = 4 / 313 < 1 / 75$ and $2 / 2501 < 1 / 1000$ , $S_{1} < 6043 / 36000$ , while $S_{0} > 2 / 3$ . Therefore $S_{0} - S_{1} > 17957 / 36000$ , and the indicator exceeds $17957 / 540000 > 1 / 1000 = τ$ .

3.1givenF1step 2.1algebra∎

The indicator rejects the root, and $D = 0$ prevents subdivision. By [F1], the procedure returns the refined estimate with cap-reached status; it makes no tolerance-success claim.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, §5.7 (standard reference, not scraped)
Art B. Owen, Monte Carlo Theory, Methods and Examples, Chapter 7, §7.2 (standard reference, not scraped)