Power series expansion involving Lambert-W function

Need to power series expand this term:

$-\frac{e}{2z} W\left(\frac{-2z}{e^2} \right) \left( \log \left( - \frac{e}{2z} W \left( \frac{-2z}{e^2}\right) \right) -1 \right)$

I tried expanding this with Wolfram but it doesn't give me Higher Order Terms. Can anyone with Maple/Mathematica help out with this? I'm working on doing this by hand but it's gonna be tedious from the looks of it. Any suggestions welcome!!

Use the MATHEMATICA commands

f = -E/(2 z) ProductLog[-2 z/E^2] (Log [-E/(2 z) ProductLog[-2 z/E^2]] - 1) Series[f, {z, 1, 10}]//N

and you will obtain

$$f(z)=-0.880194-0.22445 (z-1.)-0.153651 (z-1.)^2-0.164248 (z-1.)^3-0.224172 (z-1.)^4-0.355448 (z-1.)^5-0.622729 (z-1.)^6-1.17077 (z-1.)^7-2.31929 (z-1.)^8-4.78304 (z-1.)^9-10.1829 (z-1.)^{10}+O\left((z-1.)^{11}\right)$$

or if you prefer the lenghty form (only four therms)

$$f(z) = \frac{1}{2} e W\left(-\frac{2}{e^2}\right) \left(W\left(-\frac{2}{e^2}\right)+2\right)-\frac{1}{2} \left(e W\left(-\frac{2}{e^2}\right)^2\right) (z-1)+\frac{e W\left(-\frac{2}{e^2}\right)^3 (z-1)^2}{2 W\left(-\frac{2}{e^2}\right)+2}-\frac{\left(e W\left(-\frac{2}{e^2}\right)^4 \left(3 W\left(-\frac{2}{e^2}\right)+4\right)\right) (z-1)^3}{6 \left(W\left(-\frac{2}{e^2}\right)+1\right)^3}+\frac{e W\left(-\frac{2}{e^2}\right)^5 \left(2 W\left(-\frac{2}{e^2}\right) \left(6 W\left(-\frac{2}{e^2}\right)+17\right)+25\right) (z-1)^4}{24 \left(W\left(-\frac{2}{e^2}\right)+1\right)^5}+O\left((z-1)^5\right)$$

etc.

You can also use the fact

$$f(z) = y(z)(\ln y(z) -1)$$

with

$$y(z) = -\frac{e W\left(-\frac{2 z}{e^2}\right)}{2 z}$$

and

$$y_0 = y(1) = -\frac{1}{2} e W\left(-\frac{2}{e^2}\right)$$

and then find the expansion for $y$. This can be done with MATHEMATICA or with bare hand.

• thanks so much! this expansion is about z=0, I also need to expand it about z=1. Could you help out with that? – anikfaisal Jul 26 '18 at 20:07
• @anikfaisal OK. Now the expansion near $z=1$ To avoid length formulations, use the command //N – Cesareo Jul 26 '18 at 20:15
• this is so very helpful!! thanks again – anikfaisal Jul 26 '18 at 20:21