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.
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$\begingroup$ 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. $\endgroup$– AngeloMay 2, 2012 at 10:08
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4$\begingroup$ en.wikipedia.org/wiki/Bernoulli_number#Generating_function is relevant. $\endgroup$– Neil StricklandMay 2, 2012 at 10:31
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$\begingroup$ 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? $\endgroup$– Minh-Tri PhamMay 2, 2012 at 11:19
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$\begingroup$ It is strictly increasing on the reals, so it is invertible there. $\endgroup$– Gerald EdgarMay 2, 2012 at 12:51
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3$\begingroup$ The Wikipedia article gives a series for 1/f(z), but the question was about $f^{-1}(z)$. $\endgroup$– Michael RenardyMay 2, 2012 at 15:00
4 Answers
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).
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$\begingroup$ Interesting. Of course $x=W(x e^x)$ for some other branch of $W$. $\endgroup$ May 2, 2012 at 15:25
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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. $\endgroup$ May 2, 2012 at 15:37
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$\begingroup$ We need to rule out the principal branch of the Lambert function, which would simply give z=0. $\endgroup$ May 2, 2012 at 15:51
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$\begingroup$ Thanks. This is precisely the answer I have been looking for. $\endgroup$ May 2, 2012 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, $f(z)=1/\mathrm{Td}(-z)$.
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3$\begingroup$ Interesting, but does not answer the question. Why the upvotes??? $\endgroup$ May 2, 2012 at 19:55
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4$\begingroup$ @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. $\endgroup$ May 3, 2012 at 0:56
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1$\begingroup$ @András Bátkai: Let's not be competitive - if someone posts a useful answer/remark I'm glad to upvote it. $\endgroup$– QfwfqMay 3, 2012 at 7:50
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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. $\endgroup$ May 3, 2012 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 $
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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, 2015 at 21:47
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$\begingroup$ I have edited my post adding details and providing an answer to the question. $\endgroup$ Feb 14, 2015 at 9:29