The power series for the inverse function of y(1 − y)^{t}

Noam D. Elkies

Fix t (usually an integer, but may be real or even complex); For small x let y(x) be the small solution of y(1 − y)^{t} = x. Let a_{n} be the coefficient of the power series expansion

y = ∑_{n=1}^{∞} a_{n}x^{n}.

For t = 2, the a_{n} are Catalan numbers; larger integers t yield generalized Catalan numbers. The formula

a_{n} = \frac{1}{n}\binom{(t + 1)n − 2}{n − 1}  (1)

is usually proved using the residue theorem. The following derivation uses only elementary calculus, and as an added bonus gives the power series expansion of y^{β} for all β.

Differentiate both sides of y(1 − y)^{t} = x to find

1 = (1 − \frac{ty}{1 − y} )(1 − y)^{t}y′ = \frac{1 − (t + 1)y}{1 − y} \frac{x}{y} y′.

Multiply by (1 − y)y^{β} to obtain

y^{β} − y^{1+β} = x(y^{β−1} − (t + 1)y^{β})y′ = x\frac{d}{dx} (\frac{y^{β}}{β} − (t + 1)\frac{y^{β+1}}{β + 1})  (2)

(for β ≠ 0, −1). Now recall that the operator x d/dx takes any power series ∑_{n} a_{n}x^{n} to ∑_{n} na_{n}x^{n}. Thus if for any c we expand y^{c} in a power series

y^{c} = ∑_{m=0}^{∞} a_{m+c}(c)x^{m+c}

then our identity (2) amounts to

a_{m+β}(β) − a_{m+β}(1 + β) = (m + β) (\frac{a_{m+β}(β)}{β} − (t + 1)\frac{a_{m+β}(1 + β)}{1 + β}) ,

which simplifies to the recursion

a_{m+β}(β) = \frac{β}{1 + β} \frac{t(m + β) + m − 1}{m} a_{m+β}(1 + β).

Applying this m times, we find

a_{m+β}(β) = \frac{β}{m + β} \binom{t(m + β) + m − 1}{m} a_{m+β}(m + β).

But a_{m+β}(m + β) is the leading coefficient of y^{m+β}, and thus equals 1. We conclude that

a_{m+β}(β) = \frac{β}{m + β} \binom{t(m + β) + m − 1}{m} .

In particular, taking β = 1 and m = n − 1 we recover the formula (1) for the x^{n} coefficient a_{n} = a_{n}(1) of y.