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

Question

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?

Answer 1

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

Answer 2

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 $$ W_{-1}(y) = x = \log(-y) - \log(-\log(-y)) + o(1) \qquad\text{as } y \to -\infty $$