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.)

  • 4
    $\begingroup$ Try: Series[ProductLog[-1, x], {x, -Exp[-1], 5}, Assumptions -> x > -1] ? $\endgroup$ Jul 16, 2019 at 17:31
  • $\begingroup$ @MariuszIwaniuk Awesome! Thanks! Is there a similar trick to get PadeApproximant to work? It also works on the principal branch but not on $W_{-1}$. $\endgroup$ Jul 16, 2019 at 17:59

1 Answer 1


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:

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

$\frac{\frac{301}{540} e \left(x+\frac{1}{e}\right)-\frac{14}{45} \sqrt{2 e} \sqrt{x+\frac{1}{e}}-1}{\frac{83}{540} e \left(x+\frac{1}{e}\right)-\frac{31}{45} \sqrt{2 e} \sqrt{x+\frac{1}{e}}+1}$

  • $\begingroup$ 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. $\endgroup$ Jul 16, 2019 at 21:15

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