Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated
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 quadratic remainder at L2 radius n^(-1/4)

Example

Let Ψ(f)=∥f∥22 at f0=1. The derivative is nonzero, but the quadratic remainder is not negligible at perturbations of size n−1/4: for any unit vector v∈L2 and integers n≥1, hn=n−1/4v has n-scaled remainder equal to 1. This sequence does not satisfy the stronger radius premise in Nonlinear-functional BvM limit under fast L2 contraction.

Facts & Assumptions

Given: Ψ(f)=⟨f,f⟩L2, f0=1, and ∥v∥2=1.

[F1]

Nonlinear-functional BvM limit under fast L2 contraction: its Taylor remainder must be uniformly quadratic near f0, and its posterior L2 radius must be o(n−1/2) for the stated BvM conclusion.

Verification

Given: The functional and unit vector in the Example.

1.1givenalgebra

Expanding the inner product gives Ψ(f0+h)=∥f0∥22+2⟨f0,h⟩L2+∥h∥22. Thus the derivative is DΨf0[h]=2⟨1,h⟩L2, which is nonzero, and the remainder equals ∥h∥22.

2.1F1givenstep 1.1algebra∎

For integers n≥1 and hn=n−1/4v, ∥hn∥2=n−1/4 and n∥hn∥22=1. Hence the scaled quadratic remainder does not tend to zero along this sequence, so the linear approximation needed in [F1] is not justified at this radius.

Depends on

Used by

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Dependency tree · two levels

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