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:

nᵃ   (vs.)   (log(n))ᵇ

I run into these comparisons a lot, and I've never come up with a good way to solve them. Hints for tactics for solving the general case?

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[EDIT: I'm not talking about the computational complexity of calculating the values of these functions. I'm talking about the functions themselves. E.g., f(n) = n is an upper bound on g(n) = log(n) because f(n) ≤ c g(n) for c = 1 and n₀ > 0.]

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.

Answer 2

log n -- O(log n)
sqrt n -- O(sqrt n)
n^2 -- O(n^2)
(log n)^2 -- O((log n)^2)

n^a versus (log(n))^b

You need either bases or powers the same. So use your math to change n^a to log(n)^(whatever it gets to get this base) or (whatever it gets to get this power)^b. There is no general case

Answer 3

I run into these comparisons a lot (...) Hints for tactics for solving the general case?

As you as about general case and that you following a lot into such questions. Here is what I recommend :

Use limit definition of BigO notation, once you know:

f(n) = O(g(n)) iff limit (n approaches +inf) f(n)/g(n) exists and is not +inf

You can use Computer Algebra System, for example opensource Maxima, here is in Maxima documentation about limits .

For more detailed info and example - check out THIS answer