What is a good asymptotic approximation of $\left(\log n\right)^{\log n}$? It certainly grows faster than $e^{\log n}=n$. But how much faster?
If we let $x=(\log n)^{\log n}$, then we have $\log x=\log n\log\log n$, which doesn't help that much.
Approximation of $\left(\log n\right)^{\log n}$
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algebra-precalculus
logarithms
1 Answers
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Well, $(\log(n))^{\log(n)}=e^{\log(n) \log(\log(n))}=n^{\log(\log(n))}$. Now $\log(\log(n))$ diverges, albeit slowly, so what you have is superpolynomial. On the other hand this grows a lot slower than an exponential, even slower than $e^{n^c}$ for arbitrarily small positive $c$. I don't know a better way to characterize the asymptotic growth rate than that; you'd have to tell me what other superpolynomial+subexponential function you want to compare to.
However, for "moderately large" $n$, you can think of $\log(\log(n))$ as not being all that big. For example, $\log(\log(10^{100}))<5.5$.