Asymptotic Complexity of Logarithms and Powers

Question

So, clearly, log(n) is O(n). But, what about (log(n))^2? What about sqrt(n) or log(n)--what bounds what?

There\'s a family of comparisons like this:
n^a versu

Answer 1

log(n)^a is always O(n^b), for any positive constants a, b.

Are you looking for a proof? All such problems can be reduced to seeing that log(n) is O(n), by the following trick:

log(n)^a = O(n^b) is equivalent to: log(n) = O(n^{b/a}), since raising to the 1/a power is an increasing function. This is equivalent to log(m^{a/b}) = O(m), by setting m = n^{b/a}. This is equivalent to log(m) = O(m), since log(m^{a/b}) = (a/b)*log(m).

You can prove that log(n) = O(n) by induction, focusing on the case where n is a power of 2.