ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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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.
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The compositional inverse of $x / (1 - x)$ is $x / (1 + x)$

Example

Over every commutative ring,

$f (x) = \frac{x}{1 - x} = x + x^{2} + x^{3} + \dots$

has compositional inverse

$g (x) = \frac{x}{1 + x} = x - x^{2} + x^{3} - x^{4} + \dots .$

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

If $g$ and $h$ both have zero constant coefficient then $(f \circ g) \circ h = f \circ (g \circ h)$ ; also $f \circ x = f$ and $x \circ f = f$ (Substitution by a zero-constant series is a ring homomorphism, and composition is associative when both inner series have zero constant coefficient).

[F2]

A formal power series is a unit exactly when its constant coefficient is a unit (A formal power series is a unit exactly when its constant coefficient is a unit).

[F3]

Two formal series are equal if and only if all their coefficients are equal (Coefficient extraction is $R$ -linear, separates formal series, shifts under multiplication by $x^{k}$ , and converts products to finite convolution).

[F4]

For a commutative ring $R$ and $f \in x R ⟦ x ⟧$ , there is a unique $g \in x R ⟦ x ⟧$ with $f \circ g = x = g \circ f$ exactly when $[x] f$ is a unit (A zero-constant formal series has a compositional inverse exactly when its linear coefficient is a unit).

Verification

technique · simplify both admissible compositions

1.1

Both $f$ and $g$ have zero constant coefficient and unit linear coefficient. Formal substitution and ring algebra give $f \circ g = g / (1 - g) = x$ because $1 - g = (1 + x)^{- 1}$ , and $g \circ f = f / (1 + f) = x$ because $1 + f = (1 - x)^{- 1}$ .

givenF1F2

2.1

Thus $g$ is a two-sided compositional inverse of $f$ , and uniqueness gives the claim. Multiplying $\sum_{n \geq 0} x^{n}$ by $1 - x$ , and $\sum_{n \geq 0} (- 1)^{n} x^{n}$ by $1 + x$ , gives constant coefficient $1$ and every later coefficient $0$ ; extensionality and inverse uniqueness give the two displayed expansions. Equivalently, the inverse equation $(1 + x) g = x$ yields $[x] g = 1$ and the alternating recursion $[x^{n}] g = - [x^{n - 1}] g$ for $n \geq 2$ . These calculations also hold in the zero ring.

step 1.1givenF2F3F4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 24 results over 13 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

Benjamin Sambale, An Invitation to Formal Power Series (standard reference, not scraped)

The compositional inverse of x/(1−x)x/(1−x) is x/(1+x)x/(1+x)