# how to tell which function asymptotically grows faster than other?

But functions in my case seems to be complicated to me..

Which of the following functions asymptotically grows the fastest as $n$ goes to infinity?

$(\log \log(n))!$

$(\log\log(n))^{\log(n)}$

$(\log\log(n))^{\log\log\log(n)}$

$(\log(n))^{\log\log(n)}$

$2^{\sqrt{\log\log(n)}}$

for example take the first and second function that are

$f(n) = (\log\log(n))!$

$g(n) = (\log\log(n))^{\log(n)}$

now calculating

$$\lim_{n \to \infty} \frac{f(n)}{g(n)}$$

$f(n)$ it's derivative can't be calculated.

Then how can I tell which function asymptotically grows fastest ?

The answer given in the book is $(\log\log(n))^{\log(n)}$

• Welcome to MSE. Your formatting will look nicer if you write \log instead of log. (Same goes for \sin, \max, etc.) Aug 31 '18 at 9:43
• I suppose the first is $f(n)=\log\bigl(\log(n!)\bigr)$. Aug 31 '18 at 10:19

Take logarithms. To compare the second with the first, for example, ee have $$\log((\log\log n)^{\log n}) = \log n \cdot\log\log\log n \tag{1}$$ On the other hand, we know that $\log(x!) \sim x\log x,$ so that $$\log(\log \log n)!) \sim \log\log n \cdot \log\log\log n\tag{2}$$ Comparing $(1)$ and $(2)$, we see that $(1)$ grows much faster.
• What is $\log(x!)$ if $x$ is not a positive integer? Aug 31 '18 at 11:09