Inverse function for $f(x)=x/(1-e^{-x})$

inverse functionlambert-w

I'm looking for the inverse function for $f(x)=x/(1-e^{-x})$, over the domain $x>0$.

Wolfram says that the answer is $f^{-1}(x)=x+W(-xe^{-x})$, for $x>1$, where $W(x)$ is the Lambert W function. Graphically this looks correct, but I'm struggling to verify it for myself on paper. I would greatly appreciate it if someone could provide the working for this. Thanks in advance!

Best Answer

Your formulation uses $x$ for both for the variable in $f(x)$ and for the different variable in $f^{-1}(x)$, which may cause confusion. Let's instead describe this as solving $y =\frac{x}{1-e^{-x}}$ for $x$.

The Lambert W function provides the solution(s) to $we^w =z$ with $w=W_n(z)$ which suggests here to get your desired solution you should be considering $z=-ye^{-y}$ and $w = x-y$. We now need to manipulate the original equation into a suitable form:

  • $y =\dfrac{x}{1-e^{-x}}$
  • $x = y -ye^{-x}$
  • $x -y = -ye^{-x}$
  • $(x -y)e^{x-y} = -ye^{-y}$

which we can now solve using the Lambert W function as

  • $x-y = W_n(-ye^{-y})$
  • $x=y+W_n(-ye^{-y})$

and this is the desired solution as given by Wolfram Alpha.

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