How can I find asymptotic for $\chi(n)$, if $\chi^{\chi^\chi} = n$. Is here self-qualification estimation? I tried to take the logarithm of both sides, but to nothing has come.
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2You took the logarithm only once? Try doing it twice. – Harald Hanche-Olsen Dec 08 '14 at 17:27
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I have got $xln(x) + ln(ln(x)) = ln(ln(n))$, and what should i do next? $xln(x) = O(ln(ln(n))) ? $ – neznaika94 Dec 08 '14 at 17:35
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So if I have $xln(x) = O(ln(ln(n)))$, can I say that $lnx + ln(ln(x)) = O(ln(ln(ln(n))))$? – neznaika94 Dec 08 '14 at 18:03
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@AntonioVargas Only yesterday, eh? Duplicates tend to flock together, it seems. – Harald Hanche-Olsen Dec 08 '14 at 18:32
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1@HaraldHanche-Olsen,, when this happens I always wonder if the askers are taking the same class. – Antonio Vargas Dec 08 '14 at 18:36