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.

A convergent real power series with nonzero constant term has a convergent reciprocal power series on a smaller neighbourhood

Statement

Let f(x)=∑n≥0an(x−c)nf(x)=\sum_{n\ge0}a_n(x-c)^n have positive radius and a0≠0a_0\ne0. Then on some neighbourhood of cc, 1/f1/f is represented by a convergent real power series about cc.

Facts & Assumptions

Given: The convergent power series ff with f(c)=a0≠0f(c)=a_0\ne0.

[L1]

For ∣z∣<1|z|<1, ∑m≥0(−z)m=1/(1+z)\sum_{m\ge0}(-z)^m=1/(1+z) (For ∣r∣<1|r| < 1, ∑k≥0rk=1/(1−r)\sum_{k \ge 0} r^k = 1/(1-r), and for ∣r∣≥1|r| \ge 1 the series diverges).

[L2]

A power series converges absolutely inside its radius, and its sum is continuous there (A real power series converges absolutely inside its radius and diverges outside it, while either behaviour may occur at an endpoint, The sum of a real power series is continuous at every point strictly inside its interval of convergence).

[L3]

Finite powers of a power series are represented by repeated Cauchy products (Inside the common radius the product of two power-series sums is represented by the Cauchy product of their coefficients).

[L4]

Absolutely convergent double series may be regrouped without changing their sum (Fubini for double series: if ∑i∑j∣aij∣\sum_i \sum_j |a_{ij}| converges then both iterated sums and the sum along every bijection N→N×N\mathbb{N} \to \mathbb{N} \times \mathbb{N} converge to one and the same value).

Proof

technique · constructive
1.1

Write f=a0+hf=a_0+h, where h(c)=0h(c)=0. By absolute convergence, choose r>0r>0 inside the radius so small that Br:=∑n≥1∣an∣rn<∣a0∣B_r:=\sum_{n\ge1}|a_n|r^n<|a_0|.

constructL2choose
2.1

By [L1], 1/f(x)=a0−1∑m≥0(−h(x)/a0)m1/f(x)=a_0^{-1}\sum_{m\ge0}(-h(x)/a_0)^m for ∣x−c∣≤r|x-c|\le r. Expand each power by [L3].

step 1.1L1L3
3.1

The total absolute sum of the expanded terms is bounded by ∣a0∣−1∑m(Br/∣a0∣)m<∞|a_0|^{-1}\sum_m(B_r/|a_0|)^m<\infty. By [L4], regrouping by powers of x−cx-c gives a convergent reciprocal power series on the neighbourhood.

step 1.1step 2.1L1L4discharge-construct∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 137 results over 28 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