I have read a few other posts with similar question but all of them are for a positive value of x, and/or require you to use a computer to compute the result.
I was wondering if it is possible to manually calculate/approximate the Lambert W function for an x which has two real solutions thus meaning $-1/e < x < 0$.
A simpler way to formulate my question would be to ask, how to manually calculate the numerical values of:
$W_0 (x)$ and $W_-1 (x)$ where $-1/e < x < 0 $
(I would prefer not to use newton's method)
Best Answer
This is just a partial answer.
Since I faced this specific problem a few years ago for $W_0(x)$, what I found as "best" are Padé approximants.
The simplest ones are
For $\color{red}{-\frac 1e \leq x \leq -\frac 1{2e}}$ $$W_0(x) \approx \frac{-1+\frac{14\sqrt{2}}{45} \sqrt{e x+1}+\frac{301}{540} (e x+1)}{1+\frac{31\sqrt{2}}{45} \sqrt{e x+1}+\frac{83}{540} (e x+1)}\tag 1$$ For $\color{red}{-\frac 1{2e} \leq x \leq 0}$
$$W_0(x) \approx \frac{x+\frac{4 }{3}x^2}{1+\frac{7 }{3}x+\frac{5 }{6}x^2}\tag 2$$
For sure, for more accuracy, I built similar expressions with more terms.
Edit
For the other branch, you can use
$$W_{-1}(x)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(-2+L_2)}{2L_1^2}+\frac{L_2(6-9L_2+2L_2^2)}{6L_1^3}+\frac{L_2(-12+36L_2-22L_2^2+3L_2^3)}{12L_1^4}+\frac{L_2(60-300L_2+350L_2^2-125 L_2^3+12 L_2^4)}{60L_1^5}+\cdots$$ where $L_1=\log(-x)$ and $L_2=\log(-L1)$
Update
Calling $f(x)$ and $g(x)$ the expressions given in $(1)$ and $(2)$, let us consider the error function $$\Phi(a)=\int_{-\frac 1 e}^a \left(W(x)-f(x)\right)^2+\int^0_a \left(W(x)-g(x)\right)^2$$ and get the following values $$\left( \begin{array}{cc} a & 10 ^{10} \,\Phi(a) \\ -0.350 & 472433 \\ -0.325 & 72500 \\ -0.300 & 13455 \\ -0.275 & 2646 \\ -0.250 & 523 \\ -0.225 & 109 \\ \color{red}{ -0.200} &\color{red}{ 44} \\ -0.175 & 57 \\ -0.150 & 100 \\ -0.125 & 168 \\ -0.100 & 269 \\ -0.075 & 411 \\ -0.050 & 605 \\ -0.025 & 863 \\ 0.000 & 1197 \end{array} \right)$$