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I'm looking for a series for $W_0(x)$ for x $\in [\frac{-1}{e},\infty [$ but every time i found only for $x\in [\frac{-1}{e}, \frac{1}{e}]$

and what about a series for $W_-1(x)$

if it is no series with x$\in [\frac{-1}{e},\infty [$ so how we can have an expression for those function ?

also what about $W_k(x)$ how they evaluate $W_k(x)$ or what is the expression of $W_k(x)$

i mean in the (expression) : series... any thing lead to evaluate any x

please help

thanks for all

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What kind of series? The Taylor series for $W_0(x)$ at $x=0$ has radius of convergence $1/e$, so it cannot represent anything when $x>1/e$.

There is a transseries you could use, like this $$ W_0(x) = \log(x) - \log(\log(x)) + \frac{\log(\log(x))}{\log(x)} + \cdots $$ as $x \to \infty$.

(comput in Maple: http://www.mapleprimes.com/questions/120720-Asymptotics-Of-Lambert-W )

added: More terms $$ \ln \left( x \right) -\ln \left( \ln \left( x \right) \right) +{ \frac {\ln \left( \ln \left( x \right) \right) }{\ln \left( x \right) }}+{\frac {-\ln \left( \ln \left( x \right) \right) +1/2\, \left( \ln \left( \ln \left( x \right) \right) \right) ^{2}}{ \left( \ln \left( x \right) \right) ^{2}}}+{\frac {\ln \left( \ln \left( x \right) \right) -3/2\, \left( \ln \left( \ln \left( x \right) \right) \right) ^{2}+1/3\, \left( \ln \left( \ln \left( x \right) \right) \right) ^{3}}{ \left( \ln \left( x \right) \right) ^{3}}}+ \left( -\ln \left( \ln \left( x \right) \right) +3 \, \left( \ln \left( \ln \left( x \right) \right) \right) ^{2}-{ \frac {11}{6}}\, \left( \ln \left( \ln \left( x \right) \right) \right) ^{3}+1/4\, \left( \ln \left( \ln \left( x \right) \right) \right) ^{4} \right) \left( \ln \left( x \right) \right) ^{-4}+ \left( \ln \left( \ln \left( x \right) \right) -5\, \left( \ln \left( \ln \left( x \right) \right) \right) ^{2}+{\frac {35}{6}}\, \left( \ln \left( \ln \left( x \right) \right) \right) ^{3}-{ \frac {25}{12}}\, \left( \ln \left( \ln \left( x \right) \right) \right) ^{4}+1/5\, \left( \ln \left( \ln \left( x \right) \right) \right) ^{5} \right) \left( \ln \left( x \right) \right) ^{-5}+O \left( {\frac { \left( \ln \left( \ln \left( x \right) \right) \right) ^{6}}{ \left( \ln \left( x \right) \right) ^{6}}} \right) $$

I put $x=11$ in there and got $1.80705$, but according to Maple, the true value is $W(11) = 1.80650\dots$.

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  • $\begingroup$ for example how i can find $W(6)$ or $W_2(5)$ ..... etc. $\endgroup$ – mhd.math May 21 '13 at 18:29
  • $\begingroup$ that sereis for x>1/e ? $\endgroup$ – mhd.math May 21 '13 at 18:33
  • $\begingroup$ I'd use the Corless et al reference in another answer: math.stackexchange.com/a/398435/442 $\endgroup$ – GEdgar May 21 '13 at 18:33
  • $\begingroup$ I have it and i explored the page and saw your series for $x\rightarrow \infty $ .. for example is this mean I can evaluate $W(11)$ by that series ? $\endgroup$ – mhd.math May 21 '13 at 18:40
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    $\begingroup$ Maybe more practical: solve $x^x=11$ numerically for $x$ using Newton's method. $\endgroup$ – GEdgar May 21 '13 at 18:53

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