Starting with the two-branched Lambert W function (from Wikipedia):
Suppose we just flip it like this:
Is there a single power series for this $y=W^{-1}(x)$?
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Lambert W function, $y=W(x)$ is a solution for $y \mathrm{e}^y = x$. Hence $W^{-1}(y) = y \mathrm{e}^y$. |
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