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

Question

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!

Answer 1

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.

Answer 2

We use one of the both defining equations of the inverse function:

$$\frac{x}{1-e^{-x}}=y$$

1.) Elementary functions

$$x-y+ye^{-x}=0$$

We see, this equation is a polynomial equation of more than one algebraically independent monomials ($x,e^{-x}$) and with no univariate factor. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (operations) we can read from the equation. See also Joseph Fels Ritt 1925.

$$xe^x-ye^x+y=0$$

We see, for algebraic $y$, this equation is an irreducible algebraic equation of both $x$ and $e^x$ simultaneously. Algebraic equations of this kind cannot have solutions except $0$ that are elementary numbers. See Ferng-Ching Lin 1983 and Timothy Chow 1999.
The elementary function on the left-hand side of zeroing equations of this kind cannot have partial inverses that are elementary functions therefore.

2.) Lambert W

$$(y-x)e^x=y$$

We see, both the factor and the exponent of the exponential term are linear functions of the solution variable ($x$). Such equations can be solved in terms of Lambert W - as shown by the other answer.