CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck 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 power-series sum is infinitely differentiable inside its radius and satisfies $a_{n} = f^{(n)} (c) / ι (n!)$ at its centre

Statement

Let $f (x) = \sum_{n \geq 0} a_{n} (x - c)^{n}$ have positive or infinite radius $R$ . Define $f^{(0)} : = f$ and $f^{(m + 1)} : = (f^{(m)})^{'}$ . Then every derivative exists on $∣ x - c ∣ < R$ , and for each $m \in N$ ,

$f^{(m)} (x) = \sum_{j = 0}^{\infty} ι ⁣ ((m + j)^{\underline{m}}) a_{m + j} (x - c)^{j} .$

In particular,

$f^{(m)} (c) = ι (m!) a_{m}, a_{m} = \frac{f^{(m)} (c)}{ι (m!)} .$

Facts & Assumptions

Given: A power-series sum $f$ of radius $R > 0$ and the recursively defined derivatives $f^{(m)}$ .

[L1]

A power series may be differentiated term by term throughout its open radius, and its first derived series has the same radius (Inside its radius a real power series may be differentiated term by term, and the differentiated series has the same radius).

[L2]

Falling factorials satisfy $n^{\underline{0}} = 1$ and $n^{\underline{k + 1}} = n^{\underline{k}} (n - k)$ for all natural $n, k$ , with truncated difference $n - k$ , and $n^{\underline{n}} = n! \neq 0$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ); the canonical embedding into $R$ is multiplicative and injective (Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$ ).

[L3]

The induction principle on $N$ (The principle of mathematical induction).

Proof

technique · induction

1.1

For $m = 0$ , the formula reads $f (x) = \sum_{j \geq 0} ι (j^{\underline{0}}) a_{j} (x - c)^{j} = \sum_{j \geq 0} a_{j} (x - c)^{j}$ , so the claim holds.

basegivenL2

1.2

Fix $m$ and assume the displayed formula for $f^{(m)}$ , with radius $R$ .

2.1

By [L1], differentiate the series in step 1.2 term by term. Reindexing $j + 1$ as $j$ gives $f^{(m + 1)} (x) = \sum_{j \geq 0} ι (j + 1) ι ((m + j + 1)^{\underline{m}}) a_{m + j + 1} (x - c)^{j}$ .

step 1.2L1

3.1

The falling-factorial recursion with $n = m + j + 1$ and $k = m$ gives $(m + j + 1)^{\underline{m + 1}} = (m + j + 1)^{\underline{m}} (j + 1)$ . Since $ι$ preserves products, step 2.1 is precisely the asserted formula with $m + 1$ in place of $m$ .

step 2.1L2algebra

4.1

By [L3], the derivative formula holds for every $m$ . At each induction step [L1] also preserves the radius $R$ , so every derived series has radius $R$ .

step 1.1step 3.1L1L3discharge-induction

5.1

At $x = c$ , every term with $j \geq 1$ vanishes and the $j = 0$ term is $ι (m^{\underline{m}}) a_{m} = ι (m!) a_{m}$ ; since $m! \neq 0$ , division gives the coefficient formula.

step 4.1L2discharge-induction∎

Depends on

Used by

Dependency tree · two levels

37 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, Encyclopedia of Mathematics (standard reference, not scraped)
E. Randles, Supplementary Notes for Real Analysis (standard reference, not scraped)

A power-series sum is infinitely differentiable inside its radius and satisfies an=f(n)(c)/ι(n!)an​=f(n)(c)/ι(n!) at its centre