Confusion about two branches of Lambert $W$ function

Question

I am confused about the usage of the two branches $W_{0}$ and $W_{-1}$ of Lambert $W$ function. Suppose $\epsilon > 0$ and we have an equation for the form: $$-xe^{-x} = -e^{-1-\epsilon}$$ Note that $-e^{-1-\epsilon} > -e^{-1}$. Does this mean that the above equation has solutions: $$-x = W_{0}(-e^{-1-\epsilon}) \quad \mbox{and} \quad -x = W_{-1}(e^{-1-\epsilon})$$ or am I doing something wrong?

Answer 1

You have a sign error in the second of your solutions.

$$-xe^{-x}=-e^{-1-\epsilon}$$

$$-x=W_k(-e^{-1-\epsilon})\ \ \ \ \ (\forall k\in\mathbb{Z})$$

If you use the real branches $W_0$ and $W_{-1}$, you get the two solutions

$$-x=W_0(-e^{-1-\epsilon}) \quad \mbox{and} \quad -x=W_{-1}(\color{red}-e^{-1-\epsilon}).$$

Answer 2

When in doubt, think of the graphs...

$y = x e^x$

---

$y = W(x)$