Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-07-31
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  • 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.
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A composition of convergent real power series has a convergent power-series expansion wherever the inner series maps a neighbourhood into the outer disk of convergence

Statement

Let F(y)=∑m≥0bm(y−e)m have positive radius S, and let G(x)−e=∑n≥1cn(x−d)n converge near d. If some r>0 satisfies

Br:=∑n≥1∣cn∣rn<S,

then F∘G is represented for ∣x−d∣<r by a convergent power series about d, obtained by expanding and regrouping ∑mbm(G(x)−e)m.

Facts & Assumptions

Given: The outer and inner series and r from the statement.

[L1]

The Cauchy product of two absolutely convergent series converges absolutely, and its absolute sum is at most the product of the two absolute sums (If ∑ak and ∑bk both converge absolutely then their Cauchy product converges absolutely, with sum AB).

Proof

technique · constructive
1.1

For each m, repeatedly use [L1] to expand (G(x)−e)m in powers of x−d; take the zeroth power to be 1. For ∣x−d∣≤r, the inner numerical series is absolutely convergent with absolute sum at most Br, so the expanded mth power has absolute term sum at most Brm.

givenconstructL1
2.1

Consequently, for ∣x−d∣≤r, the sum of absolute values of all expanded terms with outer degree m is at most ∣bm∣Brm. The series of these bounds converges because Br<S and [L2] applies.

step 1.1L2algebra
3.1

By [L3], regroup the absolutely convergent expansion by total powers of x−d. The resulting power series converges on ∣x−d∣<r and sums to ∑mbm(G(x)−e)m=F(G(x)).

step 1.1step 2.1L3discharge-construct∎

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Sources