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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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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
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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
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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
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2 Answers 2

up vote 11 down vote accepted

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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Interesting. Of course $x=W(x e^x)$ for some other branch of $W$. –  Gerald Edgar May 2 '12 at 15:25
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...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
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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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Interesting, but does not answer the question. Why the upvotes??? –  András Bátkai May 2 '12 at 19:55
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@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
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@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
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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
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