Yet another proof of the series in x for the powers y^{β} of the solution of y(1 − y)^{t} = x

Noam D. Elkies

Proposition. Fix t, and for each β ∉ {0, −1, −2, −3, …} let S_{t,β}(x) be the power series

S_{t,β}(x) := ∑_{j=0}^{∞} \frac{β}{j + β} \binom{t(j + β) + j − 1}{j} x^{j+β}.

Then S_{t,β}(x) = y^{β}, where y = S_{t,1}(x) is the unique solution of y(1 − y)^{t} = x in (Q(t))[[x]].

Proof : It is enough to check that S_{t,β}(y(1 − y)^{t}) = y^{β}. We have

S_{t,β}(y(1 − y)^{t}) = ∑_{j=0}^{∞} \frac{β}{j + β} \binom{t(j + β) + j − 1}{j} y^{j+β}(1 − y)^{t(j+β)}

= ∑_{j=0}^{∞} \frac{β}{j + β} \binom{t(j + β) + j − 1}{j} y^{j+β} ∑_{k=0}^{∞}(−1)^{k}\binom{t(j + β)}{k}y^{t}

= y^{β} + β ∑_{m=1}^{∞} y^{m+β} ∑_{j=0}^{m} \frac{(−1)^{m−j}}{j + β} \binom{t(j + β) + j − 1}{j}\binom{t(j + β)}{m − j} ,

where we took m = j + k in the last transformation. Using the factorial formula for combinatorial coefficients we rewrite this as

S_{t,β}(y(1 − y)^{t}) = y^{β} + βt ∑_{m=1}^{∞} \frac{y^{m+β}}{m!} ∑_{j=0}^{m}(−1)^{m−j}\binom{m}{j} \frac{(t(j + β) + j − 1)!}{(t(j + β) + j − m)!} .

But the inner sum is the m-th finite difference of

\frac{(t(j + β) + j − 1)!}{(t(j + β) + j − m)!} = ∏_{i=1}^{m−1}(t(j + β) + j − i),

which as a function of j is a polynomial of degree m−1. Hence its m-th finite difference vanishes. We conclude that S_{t,β}(y(1 − y)^{t}) = y^{β}, as desired. ♢

Remark: This approach was suggested by R. Stanley’s proof of the power series for the solution of ye^{−y} = x (see the bottom of page 28 of Enumerative Combinatorics, Vol. 2). That proof can be recovered from ours by setting β = 1 and taking a scaled limit as k → ∞.