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For $x>e$, express $x/W(x)$ with respect to $\ln(x)$ and $\ln \ln(x)$, where $W(\cdot)$ is the Lambert-W function.

In Wikipedia, we can find the expression of $W(x)$ with respect to $\ln(x)$ and $\ln \ln(x)$. I am not sure how they got that. I need to do this for $x/W(x)$.

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  • $\begingroup$ Google does not give any related result to the word ‘polylog expression’. $\endgroup$
    – Szeto
    Commented Jun 28, 2018 at 7:33
  • $\begingroup$ Where does this task come from? $\endgroup$
    – gammatester
    Commented Jun 28, 2018 at 7:42

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Di you mean "productlog"? The word "polylog" may be an abbreviated form of polylogarithm. In the Wolfram Language $\,\texttt{ProductLog[x]}\,$ is used for $\,W(x).\,$ We have $$\, W(x) = x - x^2 + 3x^3/2! - 16x^4/3! + 125x^5/4! + O(x^5) \,$$ and $$\, x/W(x) = 1 + x - x^2/2 + 2x^3/3 - 9x^4/8 + 32x^5/15 + O(x^6). \,$$ These power series have a limited radius of convergence. The Wikipedia series using $\, L_1:=\ln(x)\,$ and $\, L_2:=\ln\ln(x) \,$ are asymptotic but you can just substitute that series into $\, x/W(x). \,$ Explicity, $$\, x/W(x) = x\left(1/L_1 + L_2/L_1^2 + L_2(L_2-1)/L_1^3 + L_2(2 -5L_2 +L_2^2)/(2L_1^4) + O(1/L_1^5)\right). $$

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  • $\begingroup$ Would you have an idea for the expansion of $\sqrt{\frac x{W(x)}}$ for large values of $x$ ? Thanks in advance. $\endgroup$
    – Claude Leibovici
    Commented May 17, 2021 at 5:43
  • $\begingroup$ @ClaudeLeibovici I would try square rooting my $\,x/W(x)\,$ expansion to get $\,\sqrt{x/L_1}(1 + L_2/(2L_1) + L_2(3L_2-4)/(8L_1^2)+\dots).$ $\endgroup$
    – Somos
    Commented May 17, 2021 at 11:40
  • $\begingroup$ Thanks ! I never feel comfortable with log's raised to fractional powers. Cheers :-) $\endgroup$
    – Claude Leibovici
    Commented May 17, 2021 at 12:58

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