Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13
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Lagrange inversion gives the Catalan coefficients of the inverse of x−x2

Example

The compositional inverse w of x−x2 in Q⟦x⟧ is

w=x+x2+2x3+5x4+14x5+42x6+⋯ ,

and for n≥1,

[xn]w=1n(2n−2n−1).

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

If K contains Q, ϕ∈K⟦u⟧ has nonzero constant term, w∈xK⟦x⟧ is the unique solution of w=xϕ(w), H∈K⟦u⟧, and n≥1, then [xn]H(w)=1n[un−1]H′(u)ϕ(u)n (Lagrange–Bürmann inversion extracts coefficients of a compositional inverse and of functions of it).

[F2]

In a commutative Q-algebra, for u∈xR⟦x⟧ and c∈R, formal binomial powers satisfy (1+u)c=∑n≥0c(c−1)⋯(c−n+1)un/n! (Formal exp⁡ and log⁡ are inverse homomorphisms and formal binomial powers obey the expected addition laws).

[F3]

For a commutative ring R and f∈xR⟦x⟧, there is a unique g∈xR⟦x⟧ with f∘g=x=g∘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 · put the inverse equation in Lagrange form
1.1

The equation x=w−w2 is equivalent to w=x(1−w)−1. Lagrange inversion with ϕ(u)=(1−u)−1 and H(u)=u gives [xn]w=1n[un−1](1−u)−n.

givenF1
2.1

Apply the generalized-binomial formula with exponent −n and argument −u. The coefficient of un−1 is (−1)n−1(−n)(−n−1)⋯(−2n+2)/(n−1)!=(2n−2n−1). At n=1,…,6 this yields 1,1,2,5,14,42.

step 1.1givenalgebraF2
3.1

Direct substitution of these coefficients gives w−w2≡x(modx7), and uniqueness of the compositional inverse confirms the displayed initial segment.

step 2.1givenF3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources