I'm trying to understand a demonstration on a paper and since my math knowledge is limited i can't understand how do i deduce the value of $\,\gamma$ given a value for $k$ and $n$.
The equation is the following: $\gamma = n\frac{k+\textit{W}(-ke^{-k})}{k}$, where $\textit{W}(x)$ is the largest solution of $x = We^W$
Lets say for instance that k = 2, what would be the value of $W(-2e^{-2})$?
$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
For any $k>0$ there are two real values, $\Wp(-k\e^{-k})$ and $\Wm(-k\e^{-k})$, where $\Wp$ is the principal branch and $\Wm$ is the other real branch of the Lambert $\W$ function, and the following relation holds:
\begin{align} \Wm(-k\e^{-k})&\le -1 \le \Wp(-k\e^{-k}) <0 . \end{align}
As for this specific value, for $k=2$,
\begin{align}
\Wm(-2\e^{-2})&=-2
,\\
\Wp(-2\e^{-2})&=-2\,\varrho
\approx -0.40637574
,
\end{align}
where $\varrho\approx 0.20318787\ $ is so-called the rumor constant.
The maximum of the two values for $k>0$ is always $\Wp(-k\e^{-k})$, and for $k\in(0,1)$ it is simple, $\Wp(-k\e^{-k})=-k$, but for $k>1$ we must have $\Wm(-k\e^{-k})=-k<-1$, so the maximum is still $\Wp(-k\e^{-k})\in(-1,0)$.
$\endgroup$
