# $W_{-1} (x)$ series expansion?

By the Lagrange Inversion Theorem, one can derive the series expansion for the principal branch of $$W_0(x)$$: $$W_0(x)= \sum_{n=1}^{\infty} \frac{(-n)^{n-1}x^n}{n!}, \, |x| \leq \frac1e$$

For $$x \in \mathbb{R}$$.

Is there also a series expansion for $$W_{-1}(x)$$? If so, how can it be derived?

• Here is a double series expansion for any branch of the W lambert function May 5, 2022 at 20:21

## 2 Answers

Lagrange reversion:

$$\text W_{-1}(z)=\ln(z)-2i\pi-\ln(\ln(z)-2i\pi)-\sum_{n=1}^\infty\sum_{m=1}^n\frac{S_n^{(m-n+1)}\ln^m(\ln(z)-2i\pi)}{(2\pi i-\ln(z))^nm!}\tag1$$

from The Generalized Lambert function from Wolfram Functions and the Stirling numbers of the first kind $$S_n^{(m)}$$. If both sums go to $$\infty$$, then they are interchangeable. The derivation uses Lagrange reversion:

$$We^W=z\implies W=\ln(z)-2\pi i -\ln(W)\implies W_{-1}(z)=\ln(z)-2\pi i+\sum_{n=1}^\infty\frac{(-1)^n}{n!}\left.\frac{d^{n-1}\ln^n(a)}{da^{n-1}}\right|_{a=\ln(z)-2\pi i}$$

Differentiating gives a pattern with factorial power $$u^{(v)}$$:

$$\frac{d^{n-1}}{dx^{n-1}}\ln^n(x)=x^{1-n}\sum_{m=1}^{n-1}n^{(m)}S_{n-1}^{(m)}\ln^{n-m}(x)$$

The $$n=1$$ term does not fit this formula, so remove $$\ln(\ln(z)-2\pi i)$$ and simplify:

$$W_{-1}(z)=\ln(z)-2\pi i-\ln(\ln(z)-2\pi i)+\sum_{n=2}^\infty\sum_{m=1}^{n-1}\frac{S_{n-1}^{(m)}(-1)^n\ln^{n-m}(\ln(z)-2\pi i)}{(\ln(z)-2\pi i)^{n-1}(n-m)!}$$

which works. Then, some index shift gives $$(1)$$.

Inverse gamma regularized:

$$-Q^{-1}(2,-ez)-1\mathop=^{-\frac1e\le z<0}\text W_{-1}(z)=…-\frac{43}{135}(ex+1)^2-\frac{11}{72}(2ex+2)^\frac32-\frac23(ex+1)-\sqrt{2ex+2}-1$$

From the series expansion of Inverse Gamma Regularized. Please correct me and give me feedback.

The series coefficients have a recurrence relation mentioned in quantile mechanics

• Using this integral, we can get an integral representation for lambert w, numerically checked. Generalizing, it would work for any branch and give another one other than this known one Feb 5 at 3:25

Let $$y=xe^x$$; so $$x=W_{-1}(y)$$, where $$y\to 0^-$$ as $$x\to-\infty$$. Note that $$y=0$$ is a logarithmic branch point for $$W_{-1}(y)$$. The series for $$x$$ in terms of $$y$$ presumably begins like this: $$x = \log(-y) - \log(-\log(-y)) + \dots\qquad\text{as } y \to -\infty$$

Derivation ... I prefer asymptotics for $$X \to +\infty$$, so write $$X=-x$$ and $$Y = -1/y$$. So $$X \to +\infty$$ and $$Y\to + \infty$$, with $$e^X = XY,$$ $$X = \log X + \log Y \tag0$$ Now $$\log X= o(X)$$ so from $$(0)$$ we get $$X = \log Y + o(\log Y) \tag1$$ That is the first approximation.
Next, compute $$\log X = \log(\log Y + \log X) = \log\left((\log Y)\left(1+\frac{\log X}{\log Y}\right)\right) = \log\log Y + \log\left(1+\frac{\log X}{\log Y}\right)$$ Thus $$X = \log Y + \log\log Y + \log\left(1+\frac{\log X}{\log Y}\right)$$ Here, $$\log\log Y \to +\infty$$, but $$\log X/\log Y \to 0$$ so $$\log\left(1+\frac{\log X}{\log Y}\right) \to 0$$, so $$X = \log Y + \log\log Y + o(1) \tag2$$ That is the second approximation. We could continue this to get as many terms as we want.
The next term is not $$\log\log\log Y$$. I think the next one is $$X = \log Y + \log\log Y + \frac{\log\log Y}{\log Y} + \dots \tag3$$

Substitute in $$(2)$$ for $$x, y$$ to get $$W_{-1}(y) = x = \log(-y) - \log(-\log(-y)) + o(1) \qquad\text{as } y \to -\infty$$

• Presumably? I don't see the derivation May 5, 2022 at 20:41