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Lagrange–Bürmann inversion extracts coefficients of a compositional inverse and of functions of it

Statement

Let $K$ be a field containing $Q$ , let $ϕ \in K ⟦ u ⟧$ have $ϕ (0) \neq 0$ , and let $w \in x K ⟦ x ⟧$ be the unique solution of

$w = x ϕ (w) .$

Then for $H \in K ⟦ u ⟧$ and $n \geq 1$ ,

$[x^{n}] H (w) = \frac{1}{n} [u^{n - 1}] H^{'} (u) ϕ (u)^{n} .$

In particular, for $1 \leq k \leq n$ ,

$[x^{n}] w^{k} = \frac{k}{n} [u^{n - k}] ϕ (u)^{n},$

and this coefficient is $0$ when $k > n$ .

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

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).

[F2]

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).

[F3]

Formal residues satisfy integration by parts, and Laurent substitution with nonzero linear term satisfies ${res}_{x} ((F \circ g) D g) = {res}_{x} (F)$ over a field containing $Q$ (Formal residues satisfy integration by parts, logarithmic differentiation, and change of variables).

Proof

technique · formal residues

1.1

Put $ψ (u) = u / ϕ (u)$ . Its linear coefficient is $ϕ (0)^{- 1} \neq 0$ , so it has a unique compositional inverse $w$ ; the inverse identity $ψ (w) = x$ is exactly $w = x ϕ (w)$ .

givenF1F2

1.2

Coefficient extraction is residue extraction: $[x^{n}] H (w) = {res}_{x} (H (w) x^{- n - 1})$ . Change variables $x = ψ (u)$ to obtain ${res}_{u} (H (u) ψ (u)^{- n - 1} ψ^{'} (u))$ .

givenF3

2.1

Since $D (ψ^{- n}) = - n ψ^{- n - 1} ψ^{'}$ , integration by parts transforms step 1.2 into $\frac{1}{n} {res}_{u} (H^{'} (u) ψ (u)^{- n})$ . Substituting $ψ (u) = u / ϕ (u)$ gives $\frac{1}{n} {res}_{u} (H^{'} (u) u^{- n} ϕ (u)^{n}) = \frac{1}{n} [u^{n - 1}] H^{'} (u) ϕ (u)^{n}$ .

step 1.2givenF3

3.1

Taking $H (u) = u^{k}$ gives the second formula. If $k > n$ , the requested exponent $n - k$ is negative while $ϕ (u)^{n}$ is a power series, so the coefficient is $0$ ; the same vanishing also follows from ${ord}_{x} (w^{k}) = k > n$ .

step 2.1given

4.1

Steps 1.1-3.1 prove existence, uniqueness, the general Lagrange–Bürmann formula, and both ranges of the power specialization.

step 1.1step 2.1step 3.1∎

Depends on

Used by

Lagrange inversion gives the Catalan coefficients of the inverse of x-x² Example

Dependency tree · two levels

16 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

Benjamin Sambale, An Invitation to Formal Power Series (standard reference, not scraped)
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics (standard reference, not scraped)