Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-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 separated Cauchy block has a geometric rank bound

Statement

Let 0≤a<b, c≥0, and choose m,n≥1 points xi∈[0,a] and yj∈[b,b+c]. Put q=a/b<1 and define the Cauchy block Kij=1/(yj−xi). For any integer r≥1, the matrix K~ with entries K~ij=∑ℓ=0r−1xiℓyj−(ℓ+1) has rank at most r and satisfies ∣Kij−K~ij∣≤qr/(b−a) and ∥K−K~∥F≤mn qr/(b−a). If b−a≥ηmax⁡(a,c) and the H-matrix rank cap satisfies k≥r, the sampled clusters are admissible and K~ is a permitted rank-at-most-k far-block representation.

Facts & Assumptions

Given: The real intervals, their sample points, and the integer truncation order r.

[F1]

Finite cluster trees and geometric admissibility: admissibility requires positive distance and dist⁡(τ,σ)≥ηmax⁡(diam⁡(τ),diam⁡(σ)).

[F2]

Finite tensor arrays and the Frobenius inner product: the matrix Frobenius norm is the square root of the sum of squared entry moduli, and the rank is the dimension of the column image.

[F3]

H-matrix block representation: an admissible H-matrix block is stored as UV∗ with rank at most the declared cap.

Proof

1.1givenalgebra

For z=x/y one has (1−z)∑ℓ=0r−1zℓ=1−zr. Since y−x>0, division by y(1−x/y) gives the exact identity 1y−x=∑ℓ=0r−1xℓyℓ+1+xryr(y−x).

2.1F2step 1.1algebra

Because 0≤xi/yj≤a/b=q and yj−xi≥b−a, step 1.1 gives ∣Kij−K~ij∣=xir/(yjr(yj−xi))≤qr/(b−a). Each summand of K~ is the outer product of (xiℓ)i=1m and (yj−(ℓ+1))j=1n, so its column image lies in the span of at most r vectors and its rank is at most r.

3.1F1F2F3step 2.1algebra∎

Summing the squared entrywise bound from step 2.1 over the mn entries and using [F2] gives ∥K−K~∥F≤mn qr/(b−a). The sample clusters have distance at least b−a and diameters at most a and c; since b−a>0, the additional inequality in the Statement implies admissibility by [F1]. The rank bound in step 2.1 and k≥r then meet the far-block representation in [F3].

Depends on

Used by

Dependency tree · two levels

4 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