[Solved]: $\log^*(n)$ runtime analysis

Problem Detail: 

So I know that $\log^*$ means iterated logarithm, so $\log^*(3)$ = $(\log\log\log\log...)$ until $n \leq 1$.

I'm trying to solve the following:

is

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

little $o$, little $\omega$, or $\Theta$ of

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

In terms of the interior functions, $\log^*(2^{2^n})$ is much bigger than $\log^*(n)$, but squaring the $\log^*(n)$ is throwing me off.

I know that $\log(n)^2$ is $O(n)$, but I don't think that property holds for the iterative logarithm.

I tried applying the master method, but I'm having trouble with the properties of a $\log^*(n)$ function. I tried setting n to be max (i.e. $n = 5$), but this didn't really simplify the problem.

Does anyone have any tips as to how I should approach this?

Asked By : gfppaste

Answered By : Zach Langley

Recall that for $k > 1$, by definition we have $\log^*k = \log^*(\log{k}) + 1$.

By applying the definition twice, we see that $\log^*(2^{2^n}) = \log^*n + 2$. Now we can compare $\log^*n + 2$ and $(\log^*n)^2$.

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Question Source : http://cs.stackexchange.com/questions/1339