best approximation to the LambertW(x) or exp(LambertW(x)) what is the best available approximation ( say up to 10 digits ) for LambertW(x) or exp(LambertW(x)) for x > 2000
 A: Your question is not really clear about what you mean by 'approximation.'  The Lambert W(x) function is implemented in various software packages, as ProductLog[x] in Mathematica, for example.  And Mathematica can compute numerical values for specific x out to as many digits as you like:
N[ProductLog[2000],25]=5.836731494908178747954545
You may instead be asking about the asymptotic expansion of W(x) as x goes to $\infty$.  Here the Mathematica input
FullSimplify[Normal[Series[ProductLog[x], {x, Infinity, 0}]], 
 Assumptions -> x > E]
returns
$\log (x)-\log (\log (x))+\log (\log (x))/\log (x)+(\log (\log (x))-2) \log (\log
   (x))/(2 \log ^2(x))$.
So one may say that
$W(x) = \log (x)-\log (\log (x))+\log (\log (x))/\log (x)+O\left(1/\log(x)\right)$.
(If you're unfamiliar with asymptotic analysis and the Big Oh notation, you might start with
http://en.wikipedia.org/wiki/Big_O_notation  )
A: You want to approximate the solution $w:=W(x)$ of the equation $we^w =x$, respectively, the solution $u:=\exp(W(x))$ of of the equation $u\log(u)=x$, in dependence on the parameter $x\ge 2000$. In general, solutions of such equations are easily approximated by means of iterative methods. You may use the Newton method (check the linked article for the quadratic bounds on the approximation). Anyway, the Newton Method is very fast once you are conveniently close to the solution: I guess that the best thing here is to start with a more rough but more stable method, and pass to the Newton iteration as soon as the error became sufficiently small. Anyway, for these particular equations,  of course, you may find ready the study on various approximations in all details: check e.g. the wiki article on the Lambert W function and the external links therein.
A: If you want a good approximation on a given interval, you can sometimes do significantly better than asymptotics using Chebyshev/Pade approximations or the Remez algorithm.
A: As Pietro says, $u=\exp(W(x))$ satisfies $u=x/\ln u$.  This is a contraction mapping for large enough $u$, so just start with any old approximation, like $u=x/\ln x$, and do $u:=x/\ln u$ until it converges. For $x>1000$ it doesn't take more than about a dozen iterations to get 10 digits.
