Solution of $W_0(x)-W_{-1}(x)=1$

Using the equation (1)(second page of the paper)

If $d=W_0(z)-W_{-1}(z)$, $$\frac{d}{e^{-d}-1}\text{exp}\left(\frac{d}{e^{-d}-1}\right)=z$$

Here, $d=1$, and your result follows:

$$z=-\frac{e}{e-1}\text{exp}\left(\frac{e-1+1}{1-e}\right)=-\frac1{e-1} \text{exp}\left(\frac{1}{1-e}\right) $$

Proof of the equation:

Note that $\text{(strange W)}_{mn}=W_m-W_n$(I spent a while to figure this out)


Note that the inverse of $W$ is a rather simple function. Then we're looking for a value $x$ such that $$x e^x = (x+1)e^{x+1}$$ and this directly gives $$x = \frac{e}{1-e}.$$ So for $W$ we get the argument $$ \frac{e}{1-e}\exp\left(\frac{e}{1-e}\right) = \frac1{1-e}\exp\left(\frac{1}{1-e}\right).$$

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Lambert W

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