Alphabeta Math
TheoremStatement: AI-adaptedProof: 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.

Exponential coefficient decay gives a uniform partial-wave truncation bound

Statement

Assume Countable Choice and Dependent Choice. Let bℓ∈Hℓ for every integer ℓ≥0, where Hℓ is defined in Spherical harmonics and partial-wave conventions in three dimensions. Suppose constants C≥0 and ρ>1 satisfy ∥bℓ∥L2(S2)≤Cρ−ℓ for all ℓ. Then b(ω):=∑ℓ=0∞bℓ(ω) converges uniformly on S2, and for every integer L≥0, sup⁡ω∈S2∣b(ω)−∑ℓ=0Lbℓ(ω)∣≤C2L+34π ρ−(L+1)(1−ρ−1)2. This is a conditional angular-coefficient bound; it does not assert exponential decay for general scattering data.

Facts & Assumptions

Given: The harmonic spaces, constants C≥0, ρ>1, and coefficient bounds above.

[F1]

Spherical harmonics form a complete basis with the addition formula: for g∈Hℓ, ∥g∥∞≤(2ℓ+1)/(4π)∥g∥2.

Proof

Given: The data in the Statement.

1.1F1givenalgebra

By [F1], ∥bℓ∥∞≤C(2ℓ+1)/(4π)ρ−ℓ. The scalar series of these bounds converges because ρ>1, so the partial sums of ∑bℓ are uniformly Cauchy and converge uniformly to a function b.

2.1givenstep 1.1algebra∎

For ℓ=L+1+n with n≥0, 2ℓ+1≤(n+1)2L+3: squaring reduces this to 2L+3+2n≤(2L+3)(n+1)2, whose difference is nonnegative. Therefore the tail in step 1.1 is at most C(2L+3)/(4π)ρ−(L+1)∑n=0∞(n+1)ρ−n. For t=ρ−1∈(0,1), multiplying the geometric series by itself or differentiating its partial sums gives ∑n≥0(n+1)tn=(1−t)−2. This yields the claimed bound.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

38 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