Big Oh Notation - formal definition

Question

I\'m reading a textbook right now for my Java III class. We\'re reading about Big-Oh and I\'m a little confused by its formal definition.

Formal Definition: \"A func

Answer 1

The whole thing about picking numbers is just this: To make it easier. Because you're allowed to pick any numbers you like for N and c, the author just picks something, where it's most easy to see. And that's what you can also do (when writing an exam etc).

So while it would often be possible to use a smaller N, the reasoning would become a little bit harder (often requiring some previous knowledge about analysis - we've all learnt years before, that x doesn't grow as fast as x^2... But do you want to write down the analysis proof?)

Keep it simple, is the message :-) It's just a bit strange to get used to this at first.

Answer 2

Probably the reason that they said 50n<=50n^2 for n>=50 is that if n is less than 1, than n^2 < n. Of course, if n is a positive integer, then yes 50n<=50n^2. In this case, it seems that n is assumed to be a positive integer, although the formal definition they give doesn't state that explicitly.

I can see why saying 50n<=50n^2 for n>=50 may seem a little silly. But it's still true. The book doesn't say 50n<=50n^2 holds ONLY for n>=50; that would be false.

As an analogy, if I say "all of my siblings speak English", that would be true, even though there are a lot of people who speak English who are not my siblings.

Regarding the proof, we might split it into different statements.

 (1): 4n^2 + 50n - 10 <= 4n^2 + 50n  (for all n)
 (2): 4n^2 + 50n <= 4n^2 + 50n^2 (for all n>=50)  
 (3): 4n^2 + 50n^2 = 54 n^2 (for all n, including all n>=50)
 (4): Therefore, 4n^2 + 50n - 10 <= 54n^2 for all n>=50
 (5): Therefore, for f(n)=4n^2 + 50n - 10, g(n)=n^2, N=50, and c=54, 
           the statement  f(n) <= c g(n) for all n >= N is true
 (6): Therefore, by definition 4n^2 + 50n - 10=O(n^2). 

It should be clear that each of these statements is true, either on its own (1,2,3), or as a result of the previous statements.

Answer 3

Formal definition:

• f(n) = O(g(n)) means there exist c > 0 and n₀ such that for any n >= n₀ f(n) <= c*g(n)
• f(n) = o(g(n)) means for any c > 0 there exist n₀ such that for any n >= n₀ f(n) <= c*g(n)

As you can note there are slightly different :)

Answer 4

50n <= 50n^2 for n >= 50

because if n is 50, then 50n is the same as n^2, because 50*50 equals 50^2.

Substituting n^2 for 50n we get

n^2 <= 50n^2 for n >= 50

which is obvious.

Answer 5

Keep in mind that you're looking for "an upper bound on f(n) when n is sufficiently large." Thus, if you can show that f(n) is less than or equal to some cg(n) for values of n greater than N, this means cg(n) is an upper bound for f(n) and f(n)'s complexity is therefore O(g(n)).

The examples given are intended to show that the given function f(n) can never grow beyond c*g(n) for any n > N. By manipulating an initial upper bound so it can be expressed more simply (if 4n^2 + 50n is an upper bound on f(n) then so is 4n^2 + 50n^2, which is equal to 54n^2, which becomes your 54*g(n) where c = 54 and g(n) = n^2), the authors can show that f(n) is bounded by c*g(n), which has complexity O(g(n)) and therefore so does f(n).