Taking $\alpha$ as given, I am trying to know under which restrictions on $n$, the following holds: $$ 0 \leq n+\alpha W\left(-e^{\frac{-n}{\alpha}} \right) \leq 1.$$ i.e I want to solve for $n$ the following two inequalities: a) $0 \leq n+\alpha W\left(-e^{\frac{-n}{\alpha}} \right)$, and b) $n+\alpha W\left(-e^{\frac{-n}{\alpha}} \right) \leq 1$. In addition, I know $\alpha >0$, and that both $n$ and $\alpha$ are reals.
The Lambert W function being defined only over the domain $[-1/e, + \infty)$, it must be in addition that $n \geq \alpha$. Which then ensures that a) holds.
However I struggle to find any solution for the inequality b).
Hence, does anyone is able and kind enough to solve the following for $n$ ? $$ n+\alpha W\left(-e^{\frac{-n}{\alpha}} \right) \leq 1 $$
With $n\geq \alpha >0$ and both $n$ and $\alpha$ in the Reals.
Thank you,
Let $f(n) = n + \alpha W(-\mathrm{e}^{-n/\alpha}) - 1$.
We have, for all $n > \alpha$, $$f'(n) = \frac{1}{1 + W(-\mathrm{e}^{-n/\alpha})} > 0.$$
Also, we have \begin{align*} &n + \alpha W(-\mathrm{e}^{-n/\alpha}) - 1 = 0\\ \Longleftrightarrow & \, W(-\mathrm{e}^{-n/\alpha}) = \frac{1 - n}{\alpha}\\ \Longleftrightarrow &\, -\mathrm{e}^{-n/\alpha} = \frac{1 - n}{\alpha}\mathrm{e}^{\frac{1 - n}{\alpha}}\\ \Longleftrightarrow &\, n = 1 + \alpha \mathrm{e}^{-1/\alpha}. \end{align*}
Thus, the solution to $n + \alpha W(-\mathrm{e}^{-n/\alpha}) \le 1$ is $n \le 1 + \alpha \mathrm{e}^{-1/\alpha}$.