Inversion of a power series without a linear term

Question

Could someone explain me how to invert $$ z = y e^{-y} = e^{-1} - \frac{1}{2e}(y - 1)^2 + \frac{1}{3e}(y - 1)^3 - \frac{1}{8e}(y - 1)^4 + \cdots $$ around $y=1, z=e^{-1}$, so that $y$ is expressed as a series of $(1 - ez)$ ? This is a part of example VI.8 in "Analytic combinatorics" by Flajolet and Sedgewick. They skips the details of the inversion process. Actually I am not so familiar with manipulating series expansions. If someone could give some details of the method, I'll greatly appreciate for it.

Answer 1

Well, this is like the Lagrange inversion theorem (also see Wikipedia) for the equation $z=f(y)$, but with $f'(y_0)$ vanishing. You compute the expansion for $y$ by writing down the general form of its power series with unknown coefficients: $$ y = 1 + \alpha (1-ez)^{1/2} + \beta(1-ez) + O((1-ez)^{3/2}). $$ Then you substitute: $$z = \frac{1}{e} - \frac1{2e}\left(\alpha (1-ez)^{1/2} + \beta(1-ez) + \cdots\right)^2 + \frac1{3e}\left(\alpha(1-ez)^{1/2}+\cdots\right)^{3/2} + \cdots, $$ and equate coefficients on the left and right: $$ \frac1e-\frac{1-ez}{e} = \frac1e - \frac{\alpha^2}{2e}(1-ez) + \left(-\frac{\alpha\beta}{e} + \frac{\alpha^3}{3e}\right)(1-ez)^{3/2} + \cdots. $$ Solving gives $$ \alpha = \sqrt2, \qquad \beta = \frac23, $$ so $$ y = 1 + \sqrt{2} (1-ez)^{1/2} + \frac23(1-ez) + O((1-ez)^{3/2}). $$

You can find more terms by doing the same with longer power series. I don't know if there is a closed form for the series coefficients.