TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 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 complex power-series sum re-expands about every interior point, at least to the distance from that point to the original boundary

Statement

If $f (z) = \sum_{n \geq 0} c_{n} (z - a)^{n}$ has radius $R$ and $∣ b - a ∣ < R$ , then $f (b + h) = \sum_{k \geq 0} \frac{f^{(k)} (b)}{k!} h^{k} whenever ∣ h ∣ < R - ∣ b - a ∣ .$ The displayed bound is a guaranteed radius, not necessarily the exact radius of the new series.

Facts & Assumptions

Given: A complex power-series sum $f$ and an interior point $b$ .

[L1]

Under $∣ b - a ∣ + ∣ h ∣ < R$ , the binomial double series is absolutely convergent and may be regrouped (The binomial double series for re-expanding a complex power series is absolutely convergent and may be regrouped).

[L2]

The coefficient of a representation about $b$ is $f^{(k)} (b) / k!$ (The coefficients of a complex power series are its derivatives at the centre divided by the corresponding factorials).

[L3]

Every derivative of a power-series sum is obtained termwise (A complex power-series sum has complex derivatives of every order, obtained by repeated termwise differentiation).

Proof

technique · direct

1.1L1algebra

Put $z = b + h$ . The binomial expansion and [L1] give $f (b + h) = \sum_{k \geq 0} d_{k} h^{k}$ , where $d_{k} = \sum_{n \geq k} (\binom{n}{k}) c_{n} (b - a)^{n - k}$ .

2.1step 1.1L3

By [L3], evaluating the $k$ th derivative at $b$ gives $f^{(k)} (b) = k! d_{k}$ .

3.1step 1.1step 2.1L2∎

By [L2], $d_{k} = f^{(k)} (b) / k!$ , proving the stated expansion for $∣ h ∣ < R - ∣ b - a ∣$ . The bound remains valid when $R = + \infty$ .

Depends on

Used by

Dependency tree · two levels

10 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

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