The big O notationm, what does it mean?

[Amok]

Homework Statement

Do you know what O(x) means? This O was put after a Taylor polynomial, and I don't get what it means. I've read the definition, and I don't really get it, what does it actually tell me?

For example, what does writing exp(x) = 1 + x + x^2/4 + O(x^3) , where Rn(x) is O(x^3) (Rn(x) is the remainder) tell me?

 

[Office_Shredder]

I feel a compelling need to point out it's x²/2 actually. Regardless.

If you have f(x) = o(p(x)) for some f and p, it means that as x goes to zero, f goes to zero faster than p, i.e. f/p -> 0 as x goes to zero. This is meant to indicate that near zero something is so small, it's going to vanish when you take a limit to zero.

O is the opposite. f(x) = O(p(x)) means that as x goes to infinity, f is bounded above by p, i.e. there is some constant C such that f(x) <= C*p(x) for all large x.

So in your Taylor series example, I don't agree with what's written (assuming you meant little o as you stated in your second sentence) as the remainder is

$$x^3/6 + ...$$ and when you divide through by x³ you get $$1/6 + x*(stuff that's at least constant)$$ which doesn't go to zero. It IS o(x² as when you divide through by x² you get

$$x/6 + x^2/24 + ... = x*(stuff)$$ and that goes to zero as x goes to zero

 

[Amok]

I feel a compelling need to point out it's x²/2 actually. Regardless.

If you have f(x) = o(p(x)) for some f and p, it means that as x goes to zero, f goes to zero faster than p, i.e. f/p -> 0 as x goes to zero. This is meant to indicate that near zero something is so small, it's going to vanish when you take a limit to zero.

O is the opposite. f(x) = O(p(x)) means that as x goes to infinity, f is bounded above by p, i.e. there is some constant C such that f(x) <= C*p(x) for all large x.

So in your Taylor series example, I don't agree with what's written (assuming you meant little o as you stated in your second sentence) as the remainder is

$$x^3/6 + ...$$ and when you divide through by x³ you get $$1/6 + x*(stuff that's at least constant)$$ which doesn't go to zero. It IS o(x² as when you divide through by x² you get

$$x/6 + x^2/24 + ... = x*(stuff)$$ and that goes to zero as x goes to zero

Yes, it's 2! = 2 :D

Anyway, it's a big O, not a little o, I don't know why I wrote little o, I was tired I guess. Sorry. I mean big O.