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How does $W$ behave near $+\infty$ compared to $\log$? In particular, I'm interested in the asymptotic expansion of

$\frac{W(x)}{\ln(x)}$

near $\infty$ (but along the positive real line, if that matters). Clearly $W(x)\sim\ln(x)$ as $x$ increases, and the next term looks hyperbolic.

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    I don't know. It is the exponentiated recurrence for numerators of Dirichlet series for logarithms.2013-10-21

3 Answers 3

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According to Theorem 2.7 of this paper, for every $x \geq e$, $ \log x - \log \log x + \frac{1}{2}\frac{{\log \log x}}{{\log x}} \le W(x) \le \log x - \log \log x + \frac{e}{{e - 1}}\frac{{\log \log x}}{{\log x}} $ (with equality only for $x=e$). Note that $\frac{e}{{e - 1}} \approx 1.582$.

EDIT (cf. Juan's answer). According to Wolfram MathWorld, an asymptotic formula which yields reasonably accurate results for sufficiently large $x$ is $ W(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 - 125L_2^3 + 12L_2^4 )}}{{60L_1^5 }} + O\bigg[\bigg(\frac{{L_2 }}{{L_1 }}\bigg)^6 \bigg], $ where $ L_1 = \log x $ $ L_2 = \log \log x. $ Useful references can be found in that link.

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    Good, the world makes sense again. :) Thanks for the answer.2011-03-16
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One of the original papers by Corless et al. on the Lambert function gives the following series expansion at $\infty$ for the (principal branch of the) Lambert function:

$W_0(z)=\ln\;z-\ln\ln\;z+\sum_{n=1}^\infty\left(\frac{-1}{\ln\;z}\right)^n \sum_{m=1}^n (-1)^m \left[n\atop{n-m+1}\right]\frac{(\ln\ln\;z)^m}{m!}$

where $\left[n\atop m\right]$ is a Stirling cycle number.

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    +1 for referring to the Stirling number used as the 'Stirling cycle' number. Whoever originally named them '.. of the first kind' and '...of the 2nd kind' cost the world precious brain energy in trying to remember which was which.2011-03-16
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This Mathematica program takes a while to evaluate:

Clear[x, a, nn, b, z] nn = 200; z = 100 a = Series[Exp[-x], {x, N[Log[z], 500], nn}]; b = Normal[InverseSeries[Series[x/a, {x, 0, nn}]]]; x = z; N[b, 20] N[LambertW[x], 30] 

It appears to give LambertW(100) = 3.3856301402900501849... but it uses 200 terms of the power series of Exp[-x] expanded at Log[100] to get there.