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I am trying to invert (or to estimate the inverse of) $$y=\frac{1-e^{-x}}{x}$$ for $y\in(0,1)$. The function 'looks' monotonically decreasing between $x=0$ and $x=\infty$, but I have not been able to show this.

Plot of y=(1-Exp[-x])/x against x

I have been able to compute the inverse function numerically, but I am wondering if there is an analytical solution or approximation that would help speed things up.

Mathematica tells me that the inverse is $$x=\frac{1+y\cdot\text{ProductLog}[-e^{-1/y}/y]}{y}$$ where $\text{ProductLog}[z]$ is the solution to $z=we^w$. I have tried re-arranging the latter expression but I cannot arrive at the original function. Plotting the latter function on $y\in(0,1)$, it looks plausible, but I don't want to use this formula without understanding where it comes from.

Plot of x against y

Can anyone show me how to invert the original function or help me estimate the inverse to some degree of precision?

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    $\begingroup$ I don't get. You already know the inverse function depends on the Lambert function, so you just have to check Lambert function asymptotics on the related Wikipedia page. Or you may solve $y=\frac{1-e^{-x}}{x}$ through Newton's method with starting point $x_0=-2\log(y)$. $\endgroup$ – Jack D'Aurizio Sep 7 '15 at 17:21
  • $\begingroup$ Thanks, I did know know about the Lambert function. I will investigate it. $\endgroup$ – JS1204 Sep 7 '15 at 17:24
  • $\begingroup$ OK using the asymptotics of the Lambert function I think I can approximate Mathematica's formula for the inverse. However, I would still like to know how to arrive at Mathematica's answer for the inverse function. $\endgroup$ – JS1204 Sep 7 '15 at 17:36
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Derivation to obtain the Lambert $W$ function : \begin{align} y&=\frac{1-e^{-x}}x\\ x&=\frac{1-e^{-x}}y\\ x\,e^x&=\frac{e^{x}-1}y\\ \left(x-\frac 1y\right)\,e^x&=-\frac 1y\\ \left(x-\frac 1y\right)\,e^{\large{x-\frac 1y}}&=-\frac {\large{e^{-\frac 1y}}}y\\ x-\frac 1y&=W\left(-\frac {\large{e^{-\frac 1y}}}y\right)\\ \end{align} and the wished formula : $\quad\boxed{\displaystyle x=\frac 1y+W\left(-\frac {\large{e^{-\frac 1y}}}y\right)}$

At this point (as indicated by robjohn) we may use the fact that $\;y\in(0,1)\;$ and observe that the parameter of the Lambert-$W$ function $\,-\dfrac 1y\;e^{-\large{\frac 1y}}\;$ will belong to $\;\left(-\dfrac 1e,\;0\right)$.

The implications are :

  • for any $\,y\in(0,1)\;$ we have two real solutions from the two branches of the Lambert-$W$ function (see the picture in the Wikipedia link and the discussion about the image of $W$ under or above $-1$ corresponding to the parameter $-\dfrac 1e$) :$$x_1=\frac 1y+W\left(-\frac {\large{e^{-\frac 1y}}}y\right),\;x_2=\frac 1y+W_{-1}\left(-\frac {\large{e^{-\frac 1y}}}y\right)$$
  • the parameter of $W$ may be written as $\;u\,e^u\,$ for $u=-\dfrac 1y\;$ but $\;W_{-1}(u\,e^u)=u\;$ in the second case so that $\;x_2=\dfrac 1y-\dfrac 1y\;$ with the solutions becoming simply : $$x_1=\frac 1y+W\left(-\frac {\large{e^{-\frac 1y}}}y\right),\ x_2=0$$ ($x_2=0$ is rather a solution of $\:x\;y=1-e^{-x}\;$ than of the initial equation)
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    $\begingroup$ Since the argument is negative, it should be noted that there are two branches of $\mathrm{W}$. This is important since one value of $\mathrm{W}\left(-\frac1ye^{-\frac1y}\right)$ is $-\frac1y$ which gives $x=0$. $\endgroup$ – robjohn Sep 7 '15 at 22:53
  • $\begingroup$ Thanks @robjohn: (I didn't notice the bounds on $y$) I'll edit this again. Cheers, $\endgroup$ – Raymond Manzoni Sep 7 '15 at 22:58
  • $\begingroup$ Thanks! How do you get from the 4th line to the 5th line in your derivation? $\endgroup$ – JS1204 Sep 7 '15 at 22:58
  • $\begingroup$ @js86: It is a multiplication by $e^{-1/y}$ and the previous line is a subtraction of $e^x/y$. $\endgroup$ – Raymond Manzoni Sep 7 '15 at 23:00
  • $\begingroup$ of course, thanks! $\endgroup$ – JS1204 Sep 7 '15 at 23:02

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