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TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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A power-series sum may be re-expanded about every interior point, with coefficients given by its derivatives there

Statement

Suppose f(x)=∑n≥0an(x−c)nf(x)=\sum_{n\ge0}a_n(x-c)^n has radius RR, and let dd satisfy ∣d−c∣<R|d-c|<R. Then for every real xx with

∣d−c∣+∣x−d∣<R|d-c|+|x-d|<R

one has

f(x)=∑k=0∞f(k)(d)ι(k!)(x−d)k.f(x)=\sum_{k=0}^{\infty}\frac{f^{(k)}(d)}{\iota(k!)}(x-d)^k.

Thus the sum may be re-expanded about every interior point.

Facts & Assumptions

Given: The series for ff and the interior point dd.

[L1]

The binomial double series is absolutely convergent when ∣d−c∣+∣x−d∣<R|d-c|+|x-d|<R and may be regrouped by powers of x−dx-d (The binomial double series used to re-expand a power series at an interior point is absolutely convergent and may be regrouped).

[L2]

Repeated termwise differentiation gives f(k)(d)=∑n≥kι(nk‾)an(d−c)n−kf^{(k)}(d)=\sum_{n\ge k}\iota(n^{\underline k})a_n(d-c)^{n-k} (A power-series sum is infinitely differentiable inside its radius and satisfies an=f(n)(c)/ι(n!)a_n=f^{(n)}(c)/\iota(n!) at its centre).

[L3]

(nk)k!=nk‾\binom nk k!=n^{\underline k} for k≤nk\le n ((nk) k! (n−k)!=n!\binom{n}{k}\,k!\,(n-k)! = n! for k≤nk \le n; hence (nk) k!=nk‾\binom{n}{k}\,k! = n^{\underline{k}}, the quotient n!/(k!(n−k)!)n!/(k!(n-k)!) is a natural number, and (nk)=(nn−k)\binom{n}{k} = \binom{n}{n-k}).

Proof

technique · direct
1.1

Fix xx satisfying the stated inequality and set h:=x−dh:=x-d. By [L1], f(x)=∑k≥0bkhkf(x)=\sum_{k\ge0}b_kh^k, where bk:=∑n≥kι ⁣(nk)an(d−c)n−kb_k:=\sum_{n\ge k}\iota\!\binom nk a_n(d-c)^{n-k}.

givenL1
2.1

By [L2] and [L3], ι(k!)bk=f(k)(d)\iota(k!)b_k=f^{(k)}(d) for every kk.

step 1.1L2L3algebra
3.1

Substituting the coefficient identity from step 2.1 into the series in step 1.1 proves the formula.

step 1.1step 2.1∎

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