Asymptotic evaluation of a quantity

Can we say that the following quantity (a recursion of logarithms):

$W_{-1}(x)=\ln \cfrac{-x}{-\ln \cfrac{-x}{-\ln \cfrac{-x}{...}}}$

is $\Theta(\ln x)$? i.e., asimptotically upper and lower bounded?

• Are these minus signs of any use ? – Yves Daoust May 30 '16 at 10:30
• Yes, the sign - distinguishes this negative branch from the positive branch (in the Lambert function). – Dingo13 May 30 '16 at 14:39
• Don't they simplify ? – Yves Daoust May 30 '16 at 15:51

Your actual question seems to be about the asymptotics of $W_{-1}(x)$. I'm not sure how the expression for $W_{-1}(x)$ in your question is relevant.

Complete asymptotic expansions for the branches of $W$ were obtained in Corless et. al. "On the Lambert W Function" [PDF Link] and Jeffrey et. al. "Unwinding the branches of the Lambert W function" [HTML link].

In particular, taking only the largest term in the double series in equation (4.19) in the first paper,

\begin{align*} W_{-1}(z) &= \log(-x) - \log (-\log(-x)) + \Theta\!\left(\frac{\log(-\log(-x))}{\log(-x)}\right) \end{align*}

as $x \to 0^-$ with $x \in \mathbb R$.

• Thanks very much for your answer. Is the last term in the right member a $\theta$ or a small $o$ ? In fact, I need to show that $W_{-1}(-\log^{-c}n)$ is in $\Omega(\log(\log^{-c}n))$, where $c$ is a positive constant. Can we use directly your formula? – Dingo13 Jun 1 '16 at 16:48
• @Dingo13, It's $\Theta$, just as I've written it. See this link for the definition. Using just the first term of the expression I gave, $$W_{-1}(-\log^{-c} n) \sim \log \log^{-c} n \qquad \text{as } n \to \infty,$$ so certainly $W_{-1}(-\log^{-c} n) = \Omega(\log \log^{-c} n)$ as $n \to \infty$. – Antonio Vargas Jun 1 '16 at 19:29

At convergence, $$y=\ln\left(\frac{x}{y}\right),$$ or

$$x=ye^y$$ and $$y=W(x).$$

• Thanks for you answer. Can you give more information ? – Dingo13 May 30 '16 at 14:40
• @Dingo13: about what ? This is straightforward. – Yves Daoust May 30 '16 at 15:18
• in fact I don't really understand. You've written a recursive equation $y=\ln(x/y)$. The upper and lower bounds should not depend on $y$? doesn't it? What is the progression in the reasoning? – Dingo13 May 30 '16 at 15:47
• @Dingo13: I haven't written any bound, just solved the recurrence. – Yves Daoust May 30 '16 at 15:51
• You say that $W(x)=W_{-1}(x)$ ? I'm looking for a rough lower and upper bound for $W_{-1}(x)$, using $\Theta$ notation. – Dingo13 May 30 '16 at 16:03