Equation between the two branches of the lambert w function Is there an equation connecting the two branches $W_0(y)$ and $W_{-1}(y)$ of the Lambert W function for $y \in (-\tfrac 1e,0)$?
For example the two square roots $r_1(y)$ and $r_2(y)$ of the equation $x^2=y$ fulfill the equation $r_1(y)=-r_2(y)$. So if one has computed one root, he already knows the second one by taking the negative of the computed root. It is also possible to calculate $W_0(y)$ by knowing $W_{-1}(y)$ and vice versa?
Note: I reasked this question on mathoverflow. Because I read that questions shall not be migrated when they are older than 60 days I didn't asked for migration. I hope that's okay...
 A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
Using the two roots of the quadratic equation analogy,
if we know two values,
the sum $a_1$ 
and the product $a_0$ of the roots
\begin{align}
x_0+x_1&=a_1
,\\
x_0x_1&=a_0
,
\end{align} 
we know how to express the values of both roots
in terms of these two values, $a_1$ and $a_0$.
The case with the two values $\Wp(x),\ \Wm(x)$ 
for $x\in(-\tfrac1\e,0)$
is even more simple and interesting: 
we indeed can find both, if we just know one value,
either the fraction
\begin{align}
\frac{\Wp(x)}{\Wm(x)}&=a
,\quad a\in(0,1)
\end{align}
or the difference
\begin{align}
\Wm(x)-\Wp(x)&=b
,\quad b\le0
.
\end{align}
Given $b$, we have $a$ as
\begin{align}
a&=\exp(b)
.
\end{align}
And if we know $a$, than
\begin{align}
\Wp(x)&=\frac{a\ln a}{1-a}
,\\
\Wm(x)&=\frac{\ln a}{1-a}
,
\end{align}
which is 
Parametric representation of the real branches 
$\operatorname{W_{0}},\operatorname{W_{-1}}$
of the Lambert W function.
$\endgroup$
A: Alexandre Eremenko provided a good answer on Mathoverflow. See https://mathoverflow.net/a/195932/56668
