If we solve the equation $x\cdot e^x=c$ with $-\frac{1}{e}< c<0$ , we get two solutions, namely $productlog(0,c)$ and $productlog(-1,c)$ , sometimes written as $W_0(c)$ and $W_{-1}(c)$. $W_{-1}(c)$ gives the solution with $x<-1$
If $c$ tends to $0$, what is the asymptotic formula for $W_{-1}(c)$ ?
I couldn't get an answer with Wolfram Alpha.
Motivation : If $c$ is negative, but very near $0$, I would like to have a good start-value for the Newton-method for the solution smaller than $-1$
For $z=0$ the all branches $W_k$ have a similar expansion, see 1 formula 4.22: $$W_k(z) = \ln z + 2\pi i k - \ln( \ln z + 2\pi i k) + \cdots.$$
In 1 there is section 5 on numerical computation where they suggest Halley's method.
For pure real computation see D. Veberic, Having Fun with Lambert W(x) Function, http://arxiv.org/abs/1003.1628:
$$W_{−1}(x) = \ln(−x)−\ln(−(\ln(−x)− \ln(−(\ln(− x ) − ... )))) $$