TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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$2 x C (x) = 1 - (1 - 4 x)^{1 / 2}$ , where $(1 - 4 x)^{1 / 2}$ is the unique square root with constant coefficient $1$

Statement

In $Q ⟦ x ⟧$ , with $C$ the Catalan generating function (The Catalan generating function $C (x) = \sum_{n \geq 0} C_{n} x^{n}$ in $Q ⟦ x ⟧$ ) and $(1 - 4 x)^{1 / 2}$ the formal binomial power of Formal exponential, logarithm, and binomial powers over a commutative $Q$ -algebra,

$1 - 2 x C = (1 - 4 x)^{1 / 2}, equivalently 2 x C = 1 - (1 - 4 x)^{1 / 2} .$

The series $(1 - 4 x)^{1 / 2}$ is the unique element of $1 + x Q ⟦ x ⟧$ whose square is $1 - 4 x$ (Every $1 + u$ with $u \in x R ⟦ x ⟧$ has a unique $k$ th root with constant coefficient $1$ in a commutative $Q$ -algebra), and the content of the theorem is that $1 - 2 x C$ is that element. No square root is chosen, no branch is selected and no substitution for $x$ is made.

Facts & Assumptions

Given: the Catalan generating function $C \in Q ⟦ x ⟧$ .

[F1]

$C = 1 + x C^{2}$ in $Q ⟦ x ⟧$ ( $C (x) = 1 + x C (x)^{2}$ ).

[F2]

For every $n \geq 0$ , $[x^{n}] C = C_{n}$ , and $Q ⟦ x ⟧$ is a commutative $Q$ -algebra (The Catalan generating function $C (x) = \sum_{n \geq 0} C_{n} x^{n}$ in $Q ⟦ x ⟧$ ).

[L1]

For a commutative $Q$ -algebra $R$ , $u \in x R ⟦ x ⟧$ and $k \geq 1$ , there is a unique $v \in 1 + x R ⟦ x ⟧$ with $v^{k} = 1 + u$ , namely $v = (1 + u)^{1 / k}$ (Every $1 + u$ with $u \in x R ⟦ x ⟧$ has a unique $k$ th root with constant coefficient $1$ in a commutative $Q$ -algebra).

[L2]

$[x^{n}] (f + g) = [x^{n}] f + [x^{n}] g$ , $[x^{n}] (r f) = r [x^{n}] f$ , and $[x^{n}] (x^{k} f) = [x^{n - k}] f$ for $k \leq n$ and $0$ for $k > n$ (Coefficient extraction is $R$ -linear, separates formal series, shifts under multiplication by $x^{k}$ , and converts products to finite convolution).

[L3]

The coefficientwise sum and Cauchy product make $Q ⟦ x ⟧$ a commutative ring (Cauchy multiplication makes $R ⟦ x ⟧$ a commutative ring containing $R [x]$ as the finitely supported subring).

[L4]

For $u \in x R ⟦ x ⟧$ and $c \in R$ the formal binomial power is $(1 + u)^{c} : = \exp (c \log (1 + u))$ (Formal exponential, logarithm, and binomial powers over a commutative $Q$ -algebra).

Proof

technique · direct

1.1

Expanding in the commutative ring $Q ⟦ x ⟧$ gives $(1 - 2 x C)^{2} = 1 - 4 x C + 4 x^{2} C^{2} = 1 - 4 x (C - x C^{2})$ , and [F1] says $C - x C^{2} = 1$ , so $(1 - 2 x C)^{2} = 1 - 4 x$ .

F1L3

1.2

The series $1 - 2 x C$ lies in $1 + x Q ⟦ x ⟧$ : its coefficient at the index $0$ is $1 - 0 = 1$ by [L2], since $[x^{0}] (x C) = 0$ .

F2L2

2.1

The series $- 4 x$ lies in $x Q ⟦ x ⟧$ , so [L1] with $k = 2$ supplies exactly one element of $1 + x Q ⟦ x ⟧$ whose square is $1 - 4 x$ , namely $(1 - 4 x)^{1 / 2}$ as defined in [L4]. By steps 1.1 and 1.2 the series $1 - 2 x C$ is such an element, so it is that one: $1 - 2 x C = (1 - 4 x)^{1 / 2}$ , and adding $2 x C - (1 - 4 x)^{1 / 2}$ to both sides gives $2 x C = 1 - (1 - 4 x)^{1 / 2}$ .

L1L3L4step 1.1step 1.2∎

Remarks

The root is identified, not chosen. Both primary sources for this page solve the quadratic by the quadratic formula and then pick the branch by letting $x$ tend to $0$ . That is an analytic argument about a function, and there is no function here: $x$ is an indeterminate and no value is substituted for it. The uniqueness clause of Every $1 + u$ with $u \in x R ⟦ x ⟧$ has a unique $k$ th root with constant coefficient $1$ in a commutative $Q$ -algebra replaces the branch choice with an identification, and it is the only step of this page where the sources use an argument the library may not.
Why the identity is stated with the factor $2 x$ left in place. The series $2 x$ is not a unit of $Q ⟦ x ⟧$ , since its coefficient at $0$ is $0$ , so $C$ cannot be obtained by dividing. Every coefficient statement below is derived from the cleared identity by extracting a coefficient, which is legitimate at every index.

Depends on

Used by

Dependency tree · two levels

20 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

A. Postnikov (notes by A. Lin), MIT 18.212 Algebraic Combinatorics, Spring 2019, Proposition 7 (standard reference, not scraped)
D. Guichard, An Introduction to Combinatorics and Graph Theory, §3.5 (standard reference, not scraped)

2x C(x)=1−(1−4x)1/22xC(x)=1−(1−4x)1/2, where (1−4x)1/2(1−4x)1/2 is the unique square root with constant coefficient 11