LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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 complex power series with nonzero constant term has a convergent reciprocal power series locally

Statement

If $f (z) = \sum_{n \geq 0} c_{n} (z - a)^{n}$ converges near $a$ and $c_{0} \neq 0$ , then $1 / f$ is represented by a convergent power series on some neighbourhood of $a$ . Its coefficients $d_{n}$ satisfy $d_{0} = c_{0}^{- 1}$ and $d_{n} = - c_{0}^{- 1} \sum_{k = 1}^{n} c_{k} d_{n - k}$ for $n \geq 1$ .

Facts & Assumptions

Given: A convergent complex power series $f$ with $f (a) = c_{0} \neq 0$ .

[L1]

A composition of convergent complex power series has a local power-series expansion when the inner series has zero constant term (A composition of convergent complex power series has a convergent local power-series expansion when the inner sum maps the centre to the outer centre).

[L2]

A complex power-series sum is holomorphic throughout its open disc of convergence (Inside its disc of convergence a complex power series is holomorphic and may be differentiated term by term).

[L3]

Complex differentiability at a point implies continuity there (Complex differentiability at a point implies continuity there).

[L4]

Products of convergent complex power series have the finite Cauchy-convolution coefficients on their common disc (Products of convergent complex power series are represented by their Cauchy-product coefficients on the common disc).

[L5]

Two power-series representations about the same centre that agree on a neighbourhood have equal coefficients (A complex power-series representation about a fixed centre has unique coefficients).

Proof

technique · direct

1.1

Put $H (z) = 1 - f (z) / c_{0}$ , so $H (a) = 0$ . By [L2] and [L3], $f$ is continuous at $a$ ; since $∣ H (z) - H (a) ∣ = ∣ f (z) - f (a) ∣ / ∣ c_{0} ∣$ and $c_{0} \neq 0$ , $H$ is continuous there. Hence on a sufficiently small disc one has $∣ H (z) ∣ < 1$ .

L2L3choosealgebra

2.1

The finite identity $(1 - H) \sum_{m = 0}^{N} H^{m} = 1 - H^{N + 1}$ and $∣ H ∣ < 1$ give $(1 - H)^{- 1} = \sum_{m \geq 0} H^{m}$ . By [L1], this composition has a local power series.

step 1.1L1algebra

3.1

Hence $1 / f = c_{0}^{- 1} (1 - H)^{- 1}$ has a local power series. Multiply it by $f$ using [L4] and compare with the constant series $1$ by [L5]; the constant coefficient gives $d_{0} = c_{0}^{- 1}$ , while for $n \geq 1$ the nonempty sum over $1 \leq k \leq n$ gives the displayed recursion. The nonzero constant term licenses division. If $f$ is constant, the same recursion gives $d_{n} = 0$ for every $n \geq 1$ .

step 2.1L4L5algebra∎

Depends on

Used by

Complex analytic functions are closed under finite linear combinations, products, quotients with nonzero denominator, and composition Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 49 results over 13 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

Power-series supplementary notes, Colby College (standard reference, not scraped)