# Series expansion for ProductLog[-1,x]

Is there a way to get Mathematica to expand Series[ProductLog[-1, x], {x, -Exp[-1], 1}]? I'm on Mathematica 12, and it just returns Out[] = ProductLog[-1, x].

I can expand the main branch Series[ProductLog[x], {x, -Exp[-1], 1}]=-1+Sqrt[2 E] Sqrt[x+1/E]-2/3 E (x+1/E)+O[x+1/E]^(3/2). The -1 branch should have the same expansion, except without the alternating signs. (See e.g. https://arxiv.org/pdf/1003.1628.pdf )

I wonder if Mathematica simply doesn't know that this simple expansion exists, or if I'm doing something wrong?

(I can also write Series[x Exp[x], {x, -1, 3}] // InverseSeries, but that gives a weird result full of System'SeriesDump'z$561645's.) • Try: Series[ProductLog[-1, x], {x, -Exp[-1], 5}, Assumptions -> x > -1] ? Jul 16 '19 at 17:31 • @MariuszIwaniuk Awesome! Thanks! Is there a similar trick to get PadeApproximant to work? It also works on the principal branch but not on$W_{-1}\$. Jul 16 '19 at 17:59

As Mariusz says, you can give Series an assumption:

Series[ProductLog[-1, x], {x, -Exp[-1], 1}, Assumptions -> x > -1] //TeXForm


$$-1-\sqrt{2 e} \sqrt{x+\frac{1}{e}}-\frac{2}{3} e \left(x+\frac{1}{e}\right)+O\left(\left(x+\frac{1}{e}\right)^{3/2}\right)$$

For PadeApproximant, you can instead use Assuming:

Assuming[
x > -1,

$$\frac{\frac{301}{540} e \left(x+\frac{1}{e}\right)-\frac{14}{45} \sqrt{2 e} \sqrt{x+\frac{1}{e}}-1}{\frac{83}{540} e \left(x+\frac{1}{e}\right)-\frac{31}{45} \sqrt{2 e} \sqrt{x+\frac{1}{e}}+1}$$
• Thank you! A last one: How do I get Series[ProductLog[-1, x],{x,0,1}] to produce reasonable results (without imaginary numbers in it)? any Assumptions or Assuming statements I've tried don't appear to help. Jul 16 '19 at 21:15