algorithm - 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 function f(n) is of order at most g(n) - that is, f(n) = O(g(n)) - if a positive real number c and positive integer N exist such that f(n) <= c g(n) for all n >= N. That is, c g(n) is an upper bound on f(n) when n is sufficiently large."

Ok, that makes sense. But hold on, keep reading...the book gave me this example:

"In segment 9.14, we said that an algorithm that uses 5n + 3 operations is O(n). We now can show that 5n + 3 = O(n) by using the formal definition of Big Oh.

When n >= 3, 5n + 3 <= 5n + n = 6n. Thus, if we let f(n) = 5n + 3, g(n) = n, c = 6, N = 3, we have shown that f(n) <= 6 g(n) for n >= 3, or 5n + 3 = O(n). That is, if an algorithm requires time directly proportional to 5n + 3, it is O(n)."

Ok, this kind of makes sense to me. They're saying that if n = 3 or greater, 5n + 3 takes less time than if n was less than 3 - thus 5n + n = 6n - right? Makes sense, since if n was 2, 5n + 3 = 13 while 6n = 12 but when n is 3 or greater 5n + 3 will always be less than or equal to 6n.

Here's where I get confused. They give me another example:

Example 2: "Let's show that 4n^2 + 50n - 10 = O(n^2). It is easy to see that: 4n^2 + 50n - 10 <= 4n^2 + 50n for any n. Since 50n <= 50n^2 for n

= 50, 4n^2 + 50n - 10 <= 4n^2 + 50n^2 = 54n^2 for n >= 50. Thus, with c = 54 and N = 50, we have shown that 4n^2 + 50n - 10 = O(n^2)."

This statement doesn't make sense: 50n <= 50n^2 for n >= 50.

Isn't any n going to make the 50n less than 50n^2? Not just greater than or equal to 50? Why did they even mention that 50n <= 50n^2? What does that have to do with the problem?

Also, 4n^2 + 50n - 10 <= 4n^2 + 50n^2 = 54n^2 for n >= 50 is going to be true no matter what n is.

And how in the world does picking numbers show that f(n) = O(g(n))?

Answer 1

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).