The power series for the inverse function of ye^{−y}

Noam D. Elkies

For small x let y(x) be the small solution of ye^{−y} = x. Abel showed that this function has the curious power series

y = ∑_{n=1}^{∞} n^{n−1} \frac{x^{n}}{n!} .

One standard application of this is the enumeration of trees on n labeled vertices (there are n^{n−2}). Abel’s power series 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 ye^{−y} = x to find

1 = (1 − y)e^{−y}y′ = (1 − y)\frac{x}{y}y′.

Multiply by y^{β} to obtain

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

(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 (1) amounts to

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

which simplifies to the recursion

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

Applying this m times, we find

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

In particular, taking β = 1 and m = n − 1 we find that the x^{n} coefficient of y is n^{n−1}/n!, as claimed.