I came across this complex function in my work $f(z)=\frac{e^z1}{z}$. 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.

$\begingroup$ 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. $\endgroup$ – Angelo May 2 '12 at 10:08

4$\begingroup$ en.wikipedia.org/wiki/Bernoulli_number#Generating_function is relevant. $\endgroup$ – Neil Strickland May 2 '12 at 10:31

$\begingroup$ 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? $\endgroup$ – MinhTri Pham May 2 '12 at 11:19

$\begingroup$ It is strictly increasing on the reals, so it is invertible there. $\endgroup$ – Gerald Edgar May 2 '12 at 12:51

3$\begingroup$ The Wikipedia article gives a series for 1/f(z), but the question was about $f^{1}(z)$. $\endgroup$ – Michael Renardy May 2 '12 at 15:00
Let $y=(e^z1)/z$ and $x=1/y$. Then $xe^x=(xz)e^{xz}$. Hence $$xz=W(xe^x).$$ Here W is an appropriately chosen branch of the Lambert function (ProductLog[1,.] in Mathematica).

$\begingroup$ Interesting. Of course $x=W(x e^x)$ for some other branch of $W$. $\endgroup$ – Gerald Edgar May 2 '12 at 15:25

1$\begingroup$ ...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. $\endgroup$ – Gerald Edgar May 2 '12 at 15:37

$\begingroup$ We need to rule out the principal branch of the Lambert function, which would simply give z=0. $\endgroup$ – Michael Renardy May 2 '12 at 15:51

$\begingroup$ Thanks. This is precisely the answer I have been looking for. $\endgroup$ – MinhTri Pham May 2 '12 at 16:32
As for the name, according to wikipedia the Todd genus is given by:
$$\mathrm{Td}(z)=\frac{z}{1e^{z}}.$$
So, $f(z)=1/\mathrm{Td}(z)$.

3$\begingroup$ Interesting, but does not answer the question. Why the upvotes??? $\endgroup$ – András Bátkai May 2 '12 at 19:55

4$\begingroup$ @András Bátkai: probably because it's a nearanswer ["$1/{\rm Td}(z)$" feels closer to a named function than "$(e^z1)/z$"] and makes a possibly unexpected connection with researchlevel mathematics. $\endgroup$ – Noam D. Elkies May 3 '12 at 0:56

1$\begingroup$ @András Bátkai: Let's not be competitive  if someone posts a useful answer/remark I'm glad to upvote it. $\endgroup$ – Qfwfq May 3 '12 at 7:50

3$\begingroup$ 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. $\endgroup$ – András Bátkai May 3 '12 at 8:20
the coefficients of $f^{1}(z)$ is just bernoulli numbers
$$\frac{z}{e^z1}=\sum_{n=0}^{\infty}B_n\frac{z^n}{n!}$$
and we have Bernoulli numbers of higher order $\alpha$
$$\big(\frac{z}{e^z1}\big)^\alpha=\sum_{n=0}^{\infty}B_n^{(\alpha)}\frac{z^n}{n!}$$
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 $

1$\begingroup$ 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. $\endgroup$ – Stefan Kohl Feb 12 '15 at 21:47

$\begingroup$ I have edited my post adding details and providing an answer to the question. $\endgroup$ – giorgiomugnaini Feb 14 '15 at 9:29