Inverse of $x(e^{-x} - 1)$ for $x \geq 0$?

Question

Does anyone know if the inverse of $$f(x) = x(e^{-x} - 1)$$ on $\{ x \geq 0 \}$ is expressible in terms of known special functions? The natural function to consider is the Lambert $W$-function, but I haven't been able to see a relationship.

Context: During some research, I needed to compute the solution to $x(ap^{-x} - 1) = C$, where $a > 1$, $p \in (0,1)$, and $C > 0$. I believe I could do this if I had an expression for $f^{-1} (x)$.

Answer 1

Another (simpler) solution.

Let $p^{-x}=e^y$

$$x \left(a \,p^{-x}-1\right)=C \implies y(a e^y-1)=K \quad \text{where} \quad K=-C \log(p) ~>~0 $$

So, the solution of the equation is lower bounded by the solution of $$a y e^y=K \implies y_0=W\left(\frac{K}{a}\right)$$ One iteration of Newton method gives $$y_1=y_0-\frac {y_0(a e^{y_0}-1)-K}{a e^{y_0} (y_0+1)-1 }=\frac{a e^{y_0} y_0^2+K}{a e^{y_0} (y_0+1)-1}$$

Trying for $a=2$, $p=\frac 13$, $C=123456$, we have $y_0=8.93461$ and $y_1=8.93467$ which is the solution for this number of figures. Then $x=8.13268$.

Answer 2

Consider the function $$f(x)=x \left(a \,p^{-x}-1\right)-C$$ for which $$f'(x)=x p^{-2 x} \left(p^x-a\right) \left(a x \log (p)-a+p^x\right)$$ the first derivative cancels at $$x_1=0\qquad\qquad x_2=\frac{\log (a)}{\log (p)}\qquad\qquad x_3=\frac{1-W\left(\frac{e}{a}\right)}{\log (p)}$$we have $$f(x_1)=-C\qquad\qquad f(x_2)=-C\qquad\qquad f(x_3)=-C+\frac{\left(W\left(\frac{e}{a}\right)-1\right)^2}{W\left(\frac{e}{a}\right) \log (p)}$$ $$f''(x_1)=-2 a \log (p)\quad f''(x_2)=(\log (a)-2) \log (p)\quad f''(x_3)=-\left(1+\frac{1}{W\left(\frac{e}{a}\right)}\right) \log (p)$$ So, since $p \in (0,1)$, $x_1$ and $x_3$ correspond to minimum points. Remains the problem of $x_2$; since it is negative, then no problem.

Now, you need numerical methods.

What I would suggest is to solve is $$g(x)=\log\Big[x \left(a \,p^{-x}-1\right)\Big]-\log(C)$$

Trying for $a=2$, $p=\frac 13$, $C=123456$, Newton iterates would be

$$\left( \begin{array}{cc} n & x_n \\ 0 & 1.00000 \\ 1 & 5.36270 \\ 2 & 8.05263 \\ 3 & 8.13264 \\ 4 & 8.13268 \end{array} \right)$$

In odger to generate some estimate, I expanded $g(x)$ as a series around $x=1$ and later used series reversion to get

$$x \sim 1+t-\frac{(a-p)^2+a p \log ^2(p)}{2 (a-p) (a \log (p)-a+p)}t^3+O(t^3)$$ where $$t=-\frac{(a-p) \left(\log (C)-\log \left(\frac{a}{p}-1\right)\right)}{a \log (p)-a+p}$$

For the worked example, this gives $$t=\frac{5 \log \left(\frac{123456}{5}\right)}{5+\log (729)}$$

and, using the expansion to $O(t^2)$ gives $x_0=5.36270$ while using the expansion to $O(t^3)$ gives $x_0=10.6567$. I suppose that it is coincidence that the average of these two estimes gives $8.00969$ which is quite close to the solution.

Notice that $$g(1)=\log \left(\frac{a-p}{Cp}\right) \qquad \text{and} \qquad g''(1)=-\frac{a p \log ^2(p)}{(a-p)^2}-1~~<~~0$$ So, by Darboux theorem, if $g(1)<0$ (that is to say $C > \frac {a-p}p$) starting with $x_0=1+t$ ensures that the solution will be reached without any overshoot.