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The Lambert $W$ function has the property that

$$ W'(x) = \frac{W(x)}{x[1+W(x)]}, $$

and using this one can show that its Taylor expansion about $x=a$ has the form

$$ W(x) = W(a) + \sum_{n=1}^{\infty} \frac{P_n[W(a)]}{n!\,[1+W(a)]^{2n-1}} \left(\frac{W(a)}{a}\right)^n (x-a)^n, $$

where $(P_n(x))_{n\in\mathbb{N}}$ is a sequence of polynomials defined recursively by $P_1(x) = 1$ and

$$ \tag{$*$} P_{n+1}(x) = \Bigl[1-(n+1)(3+x)\Bigr] P_n(x) + (1+x) P_n'(x). $$

Calculating the first 12 polynomials in the sequence I see that the constant term $P_n(0)$ is nonzero. I would like to know that $P_n(0) \neq 0$ for all $n$, but I don't see how to show it.

It appears that the recurrence $(*)$ can be solved "explicitly" as a mess of sums and products, but this seems like trading one problem for a bigger one.

Antonio Vargas
  • 24,993

1 Answers1

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Drive the point $a$ to zero so that $W(a)$ is becoming small: $W(0) = 0$. Then, since $\frac{W(a)}{a} = \exp(-W(a))$ , $\lim_{a \to 0} \frac{W(a)}{a} = 1$. Now, consider Taylor expansion about $x=0$: $$ W(x) = \sum_{n=1}^\infty \frac{x^n}{n!} P_n(0) $$ Comparing with the well known series expansion for the $W(x)$ around the origin we find: $$ P_n(0) = (-n)^{n-1} $$

Sasha
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