Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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.

Inside the common radius the product of two power-series sums is represented by the Cauchy product of their coefficients

Statement

Suppose f(x)=n0an(xc)nf(x)=\sum_{n\ge0}a_n(x-c)^n and g(x)=n0bn(xc)ng(x)=\sum_{n\ge0}b_n(x-c)^n have radii Rf,RgR_f,R_g. For xc<min(Rf,Rg)|x-c|<\min(R_f,R_g),

f(x)g(x)=n=0(k=0nakbnk)(xc)n,f(x)g(x)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_kb_{n-k}\right)(x-c)^n,

and the displayed product series converges absolutely.

Facts & Assumptions

Proof

technique · direct
1.1

Apply [L2] to the numerical series with terms ak(xc)ka_k(x-c)^k and bj(xc)jb_j(x-c)^j, whose absolute convergence is [L1].

L1L2
2.1

Its nnth Cauchy-product term is k=0nakbnk(xc)n\sum_{k=0}^{n}a_kb_{n-k}(x-c)^n, which gives the formula and absolute convergence.

step 1.1algebra

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 64 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources