# What is the name of $\frac{e^z-1}{z}$ and how to invert it?

I came across this complex function in my work $f(z)=\frac{e^z-1}{z}$. Is there a reference to $f(z)$? What is its name in the literature? More importantly, is the function inversible? If so, what is $f^{-1}(z)$? Thanks.

-
I don't know about the name, but certainly it can't be globally invertible. For example, it assumes the value 0 infinitely often. The only invertible global holomorphic functions are the polynomials of degree 1. – Angelo May 2 '12 at 10:08
– Neil Strickland May 2 '12 at 10:31
Thanks, if I write the function as a series instead, i.e. $f(z) = \sum_{k=0}^\infty \frac{z^k}{(k+1)!}$, is it invertible? – Minh-Tri Pham May 2 '12 at 11:19
It is strictly increasing on the reals, so it is invertible there. – Gerald Edgar May 2 '12 at 12:51
The Wikipedia article gives a series for 1/f(z), but the question was about $f^{-1}(z)$. – Michael Renardy May 2 '12 at 15:00

Let $y=(e^z-1)/z$ and $x=-1/y$. Then $xe^x=(x-z)e^{x-z}$. Hence $$x-z=W(xe^x).$$ Here W is an appropriately chosen branch of the Lambert function (ProductLog[-1,.] in Mathematica).

-
Interesting. Of course $x=W(x e^x)$ for some other branch of $W$. – Gerald Edgar May 2 '12 at 15:25
...so I guess this means: solutions to $y=f(z)$ are $$z=-\frac{1 + W_k \left(-\frac{\operatorname{e} ^{-1/y}}{y}\right) y}{y}$$ where $W_k$ are the branches of the Lambert W function. – Gerald Edgar May 2 '12 at 15:37
We need to rule out the principal branch of the Lambert function, which would simply give z=0. – Michael Renardy May 2 '12 at 15:51
Thanks. This is precisely the answer I have been looking for. – Minh-Tri Pham May 2 '12 at 16:32

As for the name, according to wikipedia the Todd genus is given by:

$$\mathrm{Td}(z)=\frac{z}{1-e^{-z}}.$$

So, $f(z)=1/\mathrm{Td}(-z)$.

-
Interesting, but does not answer the question. Why the upvotes??? – András Bátkai May 2 '12 at 19:55
@András Bátkai: probably because it's a near-answer ["$1/{\rm Td}(-z)$" feels closer to a named function than "$(e^z-1)/z$"] and makes a possibly unexpected connection with research-level mathematics. – Noam D. Elkies May 3 '12 at 0:56
@András Bátkai: Let's not be competitive - if someone posts a useful answer/remark I'm glad to upvote it. – Qfwfq May 3 '12 at 7:50
There was no competitiveness. I was just surprised that the accepted answer got (at the time) less upvote then this one. Which is, of course, informative. – András Bátkai May 3 '12 at 8:20

the coefficients of $f^{-1}(z)$ is just bernoulli numbers

$$\frac{z}{e^z-1}=\sum_{n=0}^{\infty}B_n\frac{z^n}{n!}$$

and we have Bernoulli numbers of higher order $\alpha$

$$\big(\frac{z}{e^z-1}\big)^\alpha=\sum_{n=0}^{\infty}B_n^{(\alpha)}\frac{z^n}{n!}$$

where $\alpha=-1$ in the case $f(z)$ and Todd genus can be computed in terms of Bernoulli numbers

-

A related thread (about relations between two real branches of lambert function):

Equation between the two branches of the lambert w function

A (perhaps more complete?) solution involving both the real branches could be written as:

$z=W_0(-\frac{e^\frac{-1}{f}}{f})-W_{-1}(-\frac{e^\frac{-1}{f}}{f})=-\frac{1}{f}-W_{-1}(-\frac{e^\frac{-1}{f}}{f})$ for $f >1$

$z=W_{-1}(-\frac{e^\frac{-1}{f}}{f})-W_{0}(-\frac{e^\frac{-1}{f}}{f})=-\frac{1}{f}-W_{0}(-\frac{e^\frac{-1}{f}}{f})$ for $f < 1$

-
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. – Stefan Kohl Feb 12 '15 at 21:47
I have edited my post adding details and providing an answer to the question. – giorgiomugnaini Feb 14 '15 at 9:29