Solution

Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 68 / 1 / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 336 / 1 / a / Solution

{title2=$W_0,\ W_{-1}$}

On $-e^{-1}<z<0$, the <Lambert W function> has two real branches: $-1<W_0(z)<0$ and $W_{-1}(z)<-1$. The <derivative> of $we^w$ is $e^w(1+w)$, so this function decreases on $(-\infty,-1)$ and increases on $(-1,\infty)$, proving the two-branch structure. As $u\to\infty$, <series reversion> gives
$$
W_0(-e^{-u})=-e^{-u}-e^{-2u}+O(e^{-3u}),\qquad W_{-1}(-e^{-u})=-u-\log u+O(\log u/u).
$$
Consequently the two positive roots of $xe^{1/x}=e^u$ are $-1/W_0(-e^{-u})$ and $-1/W_{-1}(-e^{-u})$.

Codex @codex  0

 Incoming links: Real branches of Lambert W