What would be the log(O(n * log(n)))?

Question

I am currently learning about big O notation in my algorithms class and I stumbled upon this particular problem that perplexed me. Can we represent O(n lg(n)) as c * n* lg n if we know the formal definition holds? Meaning, if f(n) <= c n lg n, and the definition holds true for some constants, then O(n lg n) can be represented as c* n* lg n? And if my assumption is true, then we could do:

= lg(O(n lg n))

= lg(c* n* lg n)

= lg(c) + lg(n) +lg(lg(n))

If lg(n) is the highest order term in this case, then would this simplify to be O(lg(n))? Since all the lower order terms would eventually be overlapped by the highest order term?

Answer 1

Yes, you are totally right and your math is correct.

Answer 2

Your question is a little hard to follow. I think you are asking:

Suppose we have a function f that grows asymptotically at or less than a rate of O(n lg n) . Does the function lg ∘ f grow asymptotically at or less than a rate of O(lg n) ?

Yes, it does.

UPDATE: Commenter Paul Hankin points out a counterexample. Perhaps this is correct?

Suppose we have a function f that grows asymptotically at exactly a rate of O(n lg n) . Does the function lg ∘ f grow asymptotically at a rate of O(lg n) ?

I think the answer to that is yes.

Answer 3

You're trying to show that lg(O(n lg n)) = O(lg n) . This is somewhat non-standard notation, but it means that for all f in O(n lg n) , there's a g in O(lg n) such that log(f) = g . This is explained on wikipedia here: https://en.wikipedia.org/wiki/Big_O_notation#Multiple_usages

The statement isn't quite true as written. For example, f = 2^(-n) is in O(n lg n) , but lg(f) = -n which isn't in O(lg n) . But if we limit ourselves to f 's which are greater than 1 , then it's true.

If f = O(n lg n) , then there's a c such that for all sufficiently large n , f(n) < cn lg n . Then lg(f(n)) < lg(c) + lg(n) + lg(lg(n)) = O(lg n) . Since lg(f(n)) > 0 , we have have that lg(f(n)) = O(lg n) . This is essentially your proof from the question with a little additional care taken about definitions and making sure that lg(f(n)) isn't large and negative.