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I feel like finding the inverse of $y=xe^x$ should have an easy answer but can't find it.

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up vote 14 down vote accepted, see the section Applications.

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Thanks, it makes me feel better that the answer is actually more complex than I thought. – G.P. Burdell Nov 14 '10 at 17:49

Inversion of $xe^{-x}$ is a problem with combinatorial structure, from which one can also invert $xe^x$ and derive most of the properties at the Wikipedia page. At least for theoretical understanding it is easier to treat the inverse of $x\exp(-x)$ as the basic object and then translate it into "Lambert W" terms.

Inverse of $f(x) = x \exp(-x)$ is the generating function for rooted trees $\Sigma n^{n-1} X^n/n!$. That the series represents the inverse is equivalent to a simple combinatorial fact ("removing the root and its edges from a tree, gives a collection of rooted trees"). The inverse of $xe^x$ is then $W(x)= -f^{-1}(-x)$, which is the same series with minus signs. The radius of convergence $1/e$ follows from Stirling's approximation to $n!$.

The coefficients of $W(x)^r$ given in the Wikipedia article are, up to some signs, Abel polynomials. This also follows from the combinatorial interpretation: $(f^{-1}(x))^r$ is the generating function whose $x^n$ coefficient is the number maps from a set of size $n$ to a set of $r$ elements with a rooted tree on each fiber of the map.

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Asymptotically, $x \approx \log y - \log\log y$.

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This is an interesting answer without using lamber-W...But how accurate is this? Can i have a plot b/w the real solution and your approx solution? Thanks! – NeilRoy Jun 13 '15 at 6:35
Yes, you can create the plot on your own and share it here. – Yuval Filmus Jun 13 '15 at 11:49

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