12

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.

J. M. ain't a mathematician
  • 75,051
Charles
  • 32,122
  • charles, it goes to zero as $x\rightarrow \pm \infty$ its slower than log. –  Mar 16 '11 at 05:59
  • @Approximist: I don't think so. W(10^10)/ln(10^10) = 0.86..., W(10^100)/ln(10^100) = 0.97..., and so forth. – Charles Mar 16 '11 at 06:03
  • http://www.wolframalpha.com/input/?i=lambert+w+function%2F+ln+x Go to the end of the page –  Mar 16 '11 at 06:05
  • 3
    @Approximist: Don't believe everything you read on W|A! In this case the software misunderstood your meaning. Try http://www.wolframalpha.com/input/?i=limit+ProductLog[x]%2FLog[x]+as+x+goes+to+Infinity instead. – Charles Mar 16 '11 at 06:08
  • Thanks for pointing that out! I think $1$ is a highly probable limit for large x. lambertw(10^100); log(10^100) 224.843; 230.258 (this is clear if we just take first two terms in the expansion of W(z)=ln z - ln ln z) –  Mar 16 '11 at 06:18
  • This appears to work only for integers at infinity: http://math.stackexchange.com/questions/531276/lambertwk-k-by-tetration-for-natural-numbers – Mats Granvik Oct 18 '13 at 20:59
  • @MatsGranvik: How so? – Charles Oct 20 '13 at 18:44
  • I don't know. It is the exponentiated recurrence for numerators of Dirichlet series for logarithms. – Mats Granvik Oct 21 '13 at 10:45

3 Answers3

11

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.

Shai Covo
  • 24,077
  • The Mathematica result seems to contradict the Corless-Jeffrey-Knuth result that Juan Joder quotes below. (The error term O(1/x) would instead be O((log log x/log x)^3) by that formula.) Thoughts? – Charles Mar 16 '11 at 13:14
  • I deleted the paragraph referring to that result. – Shai Covo Mar 16 '11 at 14:20
  • Good, the world makes sense again. :) Thanks for the answer. – Charles Mar 16 '11 at 14:28
8

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.

Juan Joder
  • 81
  • 3
    +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. – Mitch Mar 16 '11 at 15:41
0

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.

Mats Granvik
  • 7,396