Big O Notation: Better Approximation than o

[diligence]

While discussing Taylor's theorem, my professor pointed out that for n=2, Taylor's Theorem says:

$f(x) = f(x_{0}) + f'(x_{0})(x - x_{0}) + O(|x - x_{0}|^{2})$

He then emphasized that $O(|x - x_{0}|^{2})$ is a much better approximation than $o(|x - x_{0}|)$.

But how is $O(|x - x_{0}|^{2})$ a better approximation than $o(|x - x_{0}|)$?
(I'm assuming he means as x goes to x_0)

I know in this situation (as x goes to x_o) if something is little o, it means it goes to zero faster than whatever its being compared to goes to zero. And if something is Big O of the same thing squared, then it's bounded as the thing it's being compared to, squared, goes to zero. And I understand that if something goes to zero, then that same thing squared goes to zero much faster, but I can't see exactly why we can conclude that Big O of something squared is better than little o of the same quantity not squared. For instance, if something is little o when compared to a quantity that goes to zero, how do you know it's not also little o to that quantity squared? In that case, certainly little o is better than Big O.

I get the basic concept of Big O/little o, but I guess I'm still prone to confusion during application.

 

[LCKurtz]

I agree with you. Perhaps your professor mis-spoke or you misunderstood what he said. Or he is just mistaken. It happens.

 

[diligence]

ok thanks. i'll talk to him about it. I could have very well written the wrong thing during the firestorm that is notetaking in that class.

 

[Citan Uzuki]

Your professor is correct. Consider the function |x-x₀|^3/2. This is o(|x-x₀|), but not O(|x-x₀|²).