# Solve equation $\tfrac 1x (e^x-1) = \alpha$

I have the equation $\tfrac 1x (e^x-1) = \alpha$ for an positive $\alpha \in \mathbb{R}^+$ which I want to solve for $x\in \mathbb R$ (most of all I am interested in the solution $x > 0$ for $\alpha > 1$). How can I do this?

My attempt

I defined $\phi(x) = \tfrac 1x (e^x-1)$ which can be continuously extended to $x=0$ with $\phi(0)=1$ ($\phi$ is the difference quotient $\frac{e^x-e^0}{x-0}$ of the exponential function). Therefore it is an entire function. Its Taylor series is

$$\phi(x) = \frac 1x (e^x-1) = \frac 1x (1+x+\frac{x^2}{2!}+\frac{x^3}{3!} + \ldots -1) = \sum_{n=0}^\infty \frac{x^n}{(n+1)!}$$

Now I can calculate the power series of the inverse function $\phi^{-1}$ with the methods of Lagrange inversion theorem or the Faà di Bruno's formula. Is there a better approach?

Diagram of $\phi(x)=\begin{cases} \tfrac 1x (e^x-1) & ;x\ne 0 \\ 1 & ;x=0\end{cases}$:

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Do you have to solve it on $\Bbb R$ or on $\Bbb C$? Anywhere, it is equivalent to $e^x - 1 = \alpha x$ and on $\Bbb R$ by the analysis of the graph and using convexity you have no solutions if $\alpha<1$, one solution $x = 0$ when $\alpha = 1$ and two solutions for $\alpha>1$ - but I am not sure whether the latter can be found explicitly. Maybe, with some special functions. –  Ilya Jan 5 '13 at 14:53
I am interested in the second (real) solution of $e^x-1=\alpha x$ for $\alpha > 1$ –  tampis Jan 5 '13 at 15:04
Here is a related problem. –  Mhenni Benghorbal Jan 5 '13 at 15:43

The solution can be expressed in terms of Lambert's W (the inverse function of $x \mapsto xe^x$, implemented as ProductLog in Mathematica). The equation $$\frac{e^x - 1}{x} = \alpha$$ is equivalent to $$(-\alpha^{-1} - x) e^{-\alpha^{-1} - x} = - e^{-\alpha^{-1}}\cdot \alpha^{-1}$$ and therefore $$- \alpha^{-1} - x = W(-e^{-\alpha^{-1} }\alpha^{-1})$$ or $$x = - \frac{1}{\alpha} - W\left( -\frac{e^{-\alpha^{-1}}}{\alpha} \right) \, .$$

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$$x = -\frac 1\alpha -W\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)$$

$\alpha > 0$ implies $-\tfrac 1\alpha \in \mathbb{R}^{-}$ and thus $-\tfrac 1\alpha e^{-\tfrac 1\alpha} \in \left[-\tfrac 1e,0\right)$ (The function $z\mapsto ze^z$ maps $\mathbb{R}^-$ to $\left[-\tfrac 1e,0\right)$) The Lambert $W$ function has two branches $W_0$ and $W_{-1}$ on the interval $\left[-\tfrac 1e,0\right)$:

So we have the two solutions

$$x_1 = -\frac 1\alpha -W_0\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)$$ $$x_2 = -\frac 1\alpha -W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)$$

One of $W_0\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)$ and $W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)$ will always be $-\tfrac 1\alpha$ as $W$ is the inverse of the function $z \mapsto ze^z$. This solution of $W$ would give $x=0$ which must be canceled out for $\alpha \ne 1$ as $\phi(x)=1$ just for $x=0$.

Case $\alpha=1$: For $\alpha=1$ is $-\tfrac 1\alpha e^{-\tfrac 1\alpha}=-\tfrac 1e$ and thus $W_0\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)=W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)=-1$. This gives $\phi^{-1}(1)=0$ as expected.

Case $\alpha > 1$: $\alpha > 1 \Rightarrow 0 < \tfrac 1 \alpha < 1 \Rightarrow -1 < -\tfrac 1 \alpha < 0$.

Because $W_0(y) \ge -1$ it must be $W_0\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)=-\tfrac 1\alpha$ and so

$$\phi^{-1}(\alpha) = -\frac 1\alpha -W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)\text{ for } \alpha > 1$$

Case $\alpha < 1$: $0 < \alpha < 1 \Rightarrow \tfrac 1 \alpha > 1 \Rightarrow -\tfrac 1\alpha < -1$

Because $W_{-1}(y) \le -1$ we have $W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)=-\tfrac 1\alpha$ and thus

$$\phi^{-1}(\alpha) = -\frac 1\alpha -W_{0}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)\text{ for } \alpha < 1$$

Solution

The solution is

$$\phi^{-1}(\alpha) = \begin{cases} -\frac 1\alpha -W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right) & ; \alpha > 1 \\ 0 & ; \alpha = 1 \\-\frac 1\alpha -W_{0}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right) & ; \alpha < 1 \end{cases}$$

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