Big-O problem, need help

Question

f(n) = max(n^2, n^1.5 log^16 n)

f(n) should be O(n^2),Omega(n^2), O(n^1.5 log^16 n)， or Omega(n^1.5 log^16 n)?

Can anyone help me with it and explain why?

Answer 1

For n, f(n1)=n^2 f(n2)=n^3/2 f(n3)=log^16n so, f(n3)=16nlog(base2) therefore, we can draw the conclusion: 16nlog(base2)>n^2>n^3/2

Now, as we know omegan(n) gives lower bound for the function, so, for any function O(f(n))>Omega(f(n)) so, the two function with omega bound is not going to give you the desired result.

And, for very large values of n n^2>n^3/2>16nlog(base2) so, as the generalization, f(n) can be denoted as O(n^1.5log^16n)