Equation between the two branches of the lambert w function My question: 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 asked the question two years ago on math.stackexchange.com. Unfortunately I didn't get an answer or comment there. That's why I decided to reask the question here and I hope that's okay.
I read that questions shall not be migrated when they are older than 60 days. That's why I did not ask for migration on MSE...
 A: You should specify what do you mean by "equation". If you mean algebraic equation,
then evidently there is none. Because an algebraic equation has finitely many solutions,
and if there is an algebraic equation $F(W_0,W_{-1})=0$, analytic continuation
will give you infinitely many solutions. This follows from the description of the
Riemann surface of $W^{-1}$ in  Corless, R. M.; Gonnet, G. H.; Hare, D. E. G.; Jeffrey, D. J.; Knuth, D. E. On the Lambert W function. Adv. Comput. Math. 5 (1996), no. 4, 329–359.
More precisely, let $W_0(x)>W_{-1}(x)$ on $(-1/e,0)$. All branches except $W_0$ share a logarithmic branch point at $0$. Start from some point on this interval,
say $x_0=-1/(2e)$ then describe a circle of radius $1/(2e)$ around zero. As you continue
analytically around this circle, $W_0$ does not change, while $W_{-1}$ becomes $W_{-2}$, $W_{-3}$ etc.
EDIT. In your comment you say "I only can give you an intuitive explanation of what I mean by "equation between two branches". It seems that you mean an equation $F(x,y)=0$ where $F$ is an elementary function. To this the answer is $W_0\exp(W_0)=W_1\exp(W_1)$.
I also recommend the book
MR3154823 Khovanskii, Askold Galois theory, coverings, and Riemann surfaces. Translated from the 2006 Russian original. Springer, Heidelberg, 2013.
A: The two branches are related in a trascendental way. Their difference can be used to solve other trascendental equations.
If $y(x)=\frac{x}{1-e^x}$
then :
$x=W_0(ye^y)-W_{-1}(ye^y)=y-W_{-1}(ye^y)$ for  $-1< x <0 $
$x=W_{-1}(ye^y)-W_{0}(ye^y)=y-W_{0}(ye^y)$ for  $x < -1 $
See
http://www.apmaths.uwo.ca/~djeffrey/Offprints/SYNASC2014.pdf
