2
$\begingroup$

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.

$\endgroup$
0
4
$\begingroup$

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, $\log(\log(10^{100}))<5.5$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.