Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
They have simply written the Taylor series of the function around $y=1$ – Ali Jul 12 '13 at 6:35
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.