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Some questions about the Lambert W function

  1. Oct 23, 2011 #1
    I was wondering, what would be the solution for [tex]\lim_{x \rightarrow -\infty} W(x) [/tex] where W(x) is the Lambert-W function.

    Wolfram|Alpha gives the result [tex]\infty[/tex] but the graph certainly does not imply this (In fact,[tex]\lim_{x\rightarrow \infty} x\,\exp(x)=\infty [/tex]) I only graphed it it between -10 and 10 and behind -1.5 or so, the graph goes straight down so you can't even see the function beyond -1.75 or so. Does the function make a sharp turn somewhere else behind? Is the function even continuous?


    [tex]\lim_{\omega \rightarrow -\infty} \omega \,\exp(\omega)=0[/tex] and [tex]\lim_{\omega\rightarrow 0} \omega\,\exp(\omega)=0[/tex]. So does this imply that [tex] W(0)=0 \mbox{ and } W(-\infty)=0 [/tex].

    And one final question, Wolfram|Alpha gives a long complicated result for the inverse function of x^x involving the Lambert W function. How does one use the Lambert W function to do so?

    Last edited: Oct 23, 2011
  2. jcsd
  3. Oct 23, 2011 #2


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    Homework Helper

    For the real branches of the Lambert Function, [itex]W_{0}(x)[/itex] or [itex]W_{-1}(x)[/itex], the limit doesn't exist. The plot of x Exp[x] never dips down to -Infinity, so there is no limit. In fact, these two branches are only defined on the range [itex]x \geq -1/e[/itex]. The complex branches may have a limit as x tends to -Infinity. That may be the result that wolframalpha is giving you.

    As for x = y^y, write this as x = exp[y ln y] and then take the log of both sides. Then note that y = e^(ln y). (There are some other subtleties regarding branches of the complex logarithm, but this is the basic result).
  4. Oct 24, 2011 #3
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