Big-Oh ascending order of some functions f(n)

Question

Put these function in the ascending order of their growth in terms of n: 2^((log n)^0.5), 2^n, 2^(n/2), n^(4/3), n(log n)^3, n^(log n), 2^(n^2), n!

Answer 1

The way you solve this problem is by seeing which function approaches infinity fastest as n approaches infinity. Suppose you have some functions:

f(n) = n!
g(n) = n
h(n) = 3^n

You can compare any two functions by evaluating their ratio as n approaches infinity. For example:

Lim      f(n) /
n->∞         / g(n)

If the result is greater than 1, then f(n) is asymptotically greater than g(n). If the result is 1, then they are asymptotically the same. If the result is less than one, then f(n) is asymptotically less than g(n).

You may find that some are easy to solve, like f(n) and g(n), and you can simplify the expression to get a definite value at the limit.

    f(n) /       = n! /    = n (n-1)! /   = (n-1)!
        / g(n)       / n             / n

The limit as n goes to infinity of this expression is infinity, which means that f(n) is asymptotically greater than g(n).

Other expressions are not as simple. If evaluating the limit gives you an indeterminite form, you need to use L'Hôpital's rule.

Lim      g(n) /        =  Lim    n /     =  ∞ /
n->∞         / h(n)       n->∞    / 3^n      / ∞

According to L'Hôpital's rule, we can evaluate a limit by substituting g'(n) and h'(n).

Lim      g'(n) /         =  Lim    1 /           =  1 /
n->∞          / h'(n)       n->∞    / 3^n ln(3)      / ∞

This limit is clearly less than 1, so we can say that g(n) is asymptotically less than h(n).

Answer 2

Partial answer: 2^((log n)^0.5), 2^(n/2), 2^n, n!, 2^(n^2).

Need to put these as well: n^(4/3), n(log n)^3, n^(log n)