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How to find asymptotic of function s=s(n): $s^s = n$.

Please help me to solve this problem.

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Sorry, what is your function again? – Berci Sep 19 '12 at 17:55
s^s = n, where s is a function of n. – noname_0 Sep 19 '12 at 17:56
I think yes. How did you get this? – noname_0 Sep 19 '12 at 18:04

If $s\log s=\log n$, then $s\lt\log n$ for every $n\gt e^e$, because then $\log n\log\log n\gt\log n$. Likewise, for every $a\lt1$, $(\log n)^a\log(\log n)^a=a(\log n)^a\log\log n\ll\log n$, hence $s\gt(\log n)^a$ for every $n$ large enough. This proves that $$ \log s\sim\log\log n. $$ One can go further, plugging back into the relation $s\log s=n$ any available expansion of $s$ one has, to get a more precise one. This yields, for example, $$ s=\frac{\log n}{\log\log n}\,\left(1+\frac{\log\log\log n}{\log\log n}\,(1+o(1))\right). $$

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$a(\log n)^a\log\log n\ll\log n$ How did you get this? It does not seem right – noname_0 Sep 19 '12 at 21:04
Of course this holds. Could you prove that $\log\log n\ll(\log n)^b$ for every $b\gt0$? – Did Sep 19 '12 at 21:15

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