I am trying to characterize the set i
\begin{align}
A_{a,k}=\{ x \ge 0: x^k \exp(-x) > a \}
\end{align}
in terms of inequaliti on $x$ where $a$ and $k$ are some give positive numbers. That is I am trying to find
\begin{align}
A_{a,k}=\{ x \ge 0: g_1(a) \le x \le g_2(a) \}.
\end{align}
The function $x^k \exp(-x)$ has maximum at $x=\sqrt{k}$. So, we have to assume that $a \le k^{k/2}e^{-k}$ for the set to be non-empty.
To characterize $g_1$ and $g_2$ we would need to solve
\begin{align}
x^k \exp(-x) = a
\end{align}
This should have to solution if $a$ is strictly less than $k^{k/2}e^{-k}$. However, I cannot find these.
WolframAlpha gives me a solution in terms of $W$ function
\begin{align}
x= -k W \left(-\frac{a^{1/k}}{k} \right)
\end{align}
However, it doesn't give me the second solution.
Finally my question is: What are $g_1$ and $g_2$. ?
I think
\begin{align}
g_1 = -k W_0 \left(-\frac{a^{1/k}}{k} \right)
\end{align}
where $W_0 $ is the zero branch
Best Answer
Expression \begin{align} x&= -k \operatorname{W} \left(-\frac{a^{1/k}}{k} \right) \end{align}
means two real solutions,
\begin{align} x_{0}&= -k \operatorname{W_0} \left(-\frac{a^{1/k}}{k} \right) ,\\ x_{-1}&= -k \operatorname{W_{-1}} \left(-\frac{a^{1/k}}{k} \right) \\ \text{for }\quad -\tfrac1k{a^{1/k}}\in(-\tfrac1{\mathrm{e}},0) , \end{align}
where $\operatorname{W_0},\operatorname{W_{-1}}$ are so-called two real branches of the Lambert W function.
In this case $x_0<x_{-1}$, since for the argument $z:\ -\tfrac1{\mathrm{e}}<z<0$,
\begin{align} \operatorname{W_{-1}(z)}&<-1<\operatorname{W_{0}(z)}<0 . \end{align}