# Inversion of a power series without a linear term

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.

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They have simply written the Taylor series of the function around $y=1$ –  Ali Jul 12 '13 at 6:35

## 1 Answer

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.

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How can I sure that the powers should be 0, 1/2, 1, 3/2, ... to write the general form with unknown coefficients? –  kld Jul 12 '13 at 7:16
Well, if the question were instead about $z=a+(y-1)^2$, you would clearly need to write $y=1+(z-a)^{1/2}$. So you can either do this by trial and error, or by noting that since the first term should be $\propto (z-a)^{1/2}$, the other terms must be powers of it, so $\propto (z-a)^{n/2}$. In other words, $y$ is an analytic function of $(z-a)^{1/2}$ rather than $(z-a)$. Flajolet and Sedgewick prove this at some point, I believe, in Theorem VI.6. –  Kirill Jul 12 '13 at 12:36
The function $z:=ye^{-y}$ is an algebroidal function of $y$. This implies that the inverse function also is an algebroidal function of $z$ and the one is represented as a Puiseux series . See the article by A. Khovanskii to this end. –  user64494 Jul 12 '13 at 16:43