ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck 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 rational function with nonvanishing denominator is locally represented by geometric-series expansions

Statement

The rational function $r (x) = 1 / (2 - x)$ is real analytic on $R ∖ {2}$ . At every $c \neq 2$ it has the local expansion

$\frac{1}{2 - x} = \sum_{n = 0}^{\infty} \frac{(x - c)^{n}}{(2 - c)^{n + 1}} (∣ x - c ∣ < ∣ 2 - c ∣) .$

Facts & Assumptions

Given: A centre $c \neq 2$ .

[L1]

The geometric series converges for arguments of absolute value below $1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L2]

A nonvanishing analytic denominator has a local analytic reciprocal (A convergent real power series with nonzero constant term has a convergent reciprocal power series on a smaller neighbourhood, Real-analytic functions are closed under sums, products and compositions, and under quotients where the denominator is nonzero).

Verification

technique · direct

1.1

Factor $2 - x = (2 - c) (1 - (x - c) / (2 - c))$ and apply [L1]. The resulting series is exactly the displayed one and converges when $∣ x - c ∣ < ∣ 2 - c ∣$ .

givenL1algebra

2.1

Since every $c \neq 2$ admits this positive-radius local representation, $r$ is real analytic on its domain, in agreement with [L2].

step 1.1L2∎

Depends on

Used by

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Sources

Analytic function, Encyclopedia of Mathematics (standard reference, not scraped)
Power series, Encyclopedia of Mathematics (standard reference, not scraped)
Northwestern Math 320-2 lecture notes (standard reference, not scraped)