ExampleConstruction: AI-generatedVerification: 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.

A nonaligned one-dimensional micro window has resonance error

Example

Let $a (y) = 1$ on $[0, 1 / 2)$ and $a (y) = 2$ on $[1 / 2, 1)$ , extended with period one. At phase $θ = 0$ and nonaligned ratio $r = δ / ϵ = 3 / 2$ , the exact sampled harmonic tensor is $A_{r} = 6 / 5$ , whereas $A^{*} = 4 / 3$ . The discrepancy is $2 / 15 > 0$ even though the continuum coefficient is periodic.

Facts & Assumptions

Given: The two-phase periodic scalar coefficient, phase zero, and sample ratio $r = 3 / 2$ .

[F1]

One-dimensional nonaligned micro sampling resonance bound: the sampled tensor is the reciprocal of the average reciprocal coefficient and obeys $∣ A_{r} - A^{*} ∣ \leq 2 Λ^{2} / (λ r)$ .

[F2]

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: the affine pullback $y \mapsto 3 y / 2$ preserves measurability and null sets, so the sampled reciprocal integral is well defined on the stated coefficient class.

Proof

1.1F1F2algebra

By [F2], the sample pullback and its reciprocal are measurable on $[0, 1]$ . Over $0 \leq t \leq 3 / 2$ , the reciprocal coefficient equals $1$ for lengths $1 / 2$ , $1 / 2$ , and $1 / 2$ on the successive intervals $[0, 1 / 2)$ , $[1 / 2, 1)$ , and $[1, 3 / 2)$ , except that its value on the middle interval is $1 / 2$ . Thus its integral is $1 / 2 + 1 / 4 + 1 / 2 = 5 / 4$ ; dividing by $r = 3 / 2$ gives $\int_{0}^{1} a (3 y / 2)^{- 1} d y = 5 / 6$ , so $A_{r} = 6 / 5$ . The full-period reciprocal mean is $(1 + 1 / 2) / 2 = 3 / 4$ , hence $A^{*} = 4 / 3$ .

2.1F1step 1.1algebra∎

Therefore $∣ A_{r} - A^{*} ∣ = ∣ 6 / 5 - 4 / 3 ∣ = 2 / 15$ . Here $λ = 1$ , $Λ = 2$ , and $1 / r = 2 / 3$ , so the bound in [F1] is $2 \cdot 4 \cdot (2 / 3) = 16 / 3$ ; indeed $2 / 15 \leq 16 / 3$ . The example verifies both nonzero resonance and the stated conservative estimate.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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.