How do these big-oh equalities hold?

Question

Chong and Zak, in their book An Introduction to Optimization, 4th ed, write

... we write ... $f(x) = O(g(x))$ to mean that the quotient $\frac{||f(x)||}{|g(x)|} $ is bounded near $0$; that is, there exist numbers $K>0$ and $\delta > 0$ such that if $||x||<\delta$, $x \in \Omega$, then $\frac{||f(x)||}{|g(x)|} \leq K$.

then give following example:

$\begin{bmatrix} x^3 \\ 2x^2+3x^4 \end{bmatrix} = O(x^2)$

Neither the definition nor the example make sense to me.

The definition does not make sense to me, because the quotient should be bounded in the infinity I think, not near $0$?

Can you explain?

Answer 1

Whenever you write $f(x)=O(g(x))$ it needs to be understood what limit you are taking. Often this is obvious from the context, but sometimes you need to be explicit. By far the most common assumption is that you mean $f(x)=O(g(x))$ as $x\to\infty$, but the definition given is for $f(x)=O(g(x))$ as $x\to0$.

This vector (if that's what's intended by the square brackets) is $O(x^2)$ as $x\to0$ because both its components are: if $|x|<1$ then $\frac{|2x^2+3x^4|}{|x^2|}=2+3x^2<5$.

However, $2x^2+3x^4$ is not $O(x^2)$ as $x\to\infty$.

Answer 2

If $f(x)=\begin{bmatrix} x^3 \\ 2x^2+3x^4 \end{bmatrix} $, then an easy computation gives

$\frac{||f(x)||}{|g(x)|}=\sqrt{4+13x^2+9x^4}$, hence

$\frac{||f(x)||}{|g(x)|} \le \sqrt{4+13+9} \le 6$ for $|x| <1$

This shows that $\begin{bmatrix} x^3 \\ 2x^2+3x^4 \end{bmatrix} = O(x^2)$