Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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Lagrange inversion gives the Catalan coefficients of the inverse of xx2

Example

The compositional inverse w of xx2 in Qx is

w=x+x2+2x3+5x4+14x5+42x6+,

and for n1,

[xn]w=1n(2n2n1).

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

If K contains Q, ϕKu has nonzero constant term, wxKx is the unique solution of w=xϕ(w), HKu, and n1, then [xn]H(w)=1n[un1]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 uxRx and cR, formal binomial powers satisfy (1+u)c=n0c(c1)(cn+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 fxRx, there is a unique gxRx with fg=x=gf 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=ww2 is equivalent to w=x(1w)1. Lagrange inversion with ϕ(u)=(1u)1 and H(u)=u gives [xn]w=1n[un1](1u)n.

givenF1
2.1

Apply the generalized-binomial formula with exponent n and argument u. The coefficient of un1 is (1)n1(n)(n1)(2n+2)/(n1)!=(2n2n1). At n=1,,6 this yields 1,1,2,5,14,42.

step 1.1givenalgebraF2
3.1

Direct substitution of these coefficients gives ww2x(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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 63 results over 20 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