CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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The coefficients of a complex power series are its derivatives at the centre divided by the corresponding factorials

Statement

If $f (z) = \sum_{n \geq 0} c_{n} (z - a)^{n}$ near $a$ , then for every $n \in N$ , $c_{n} = \frac{f^{(n)} (a)}{n!} .$

Facts & Assumptions

Given: A complex power-series representation of $f$ about $a$ .

[L1]

The $n$ th derivative is obtained by repeated termwise differentiation with falling-factorial coefficients (A complex power-series sum has complex derivatives of every order, obtained by repeated termwise differentiation).

[L2]

Complex natural powers are defined by the recursion $z^{0} = 1$ and $z^{n + 1} = z^{n} z$ for $n \in N$ ; in particular $0^{0} = 1$ (Integer powers in the complex field).

[L3]

The factorial $n!$ is nonzero (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

Proof

technique · direct

1.1

Evaluate [L1] at $z = a$ , so every remaining power is $0^{m}$ with $m = k - n \geq 0$ . From the recursion of [L2], $0^{m + 1} = 0^{m} \cdot 0 = 0$ for every $m \in N$ , so $0^{m} = 0$ for every positive $m$ , while $0^{0} = 1$ by the base clause of [L2]. Hence every term with a positive remaining power vanishes and the term indexed by $n$ is $n! c_{n}$ .

L1L2algebra

2.1

Thus $f^{(n)} (a) = n! c_{n}$ ; divide by the nonzero factorial from [L3]. For $n = 0$ , this reads $c_{0} = f (a)$ .

step 1.1L3algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 55 results over 16 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

L. Ahlfors, Complex Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)