# How do these big-oh equalities hold?

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?

• How do you define a norm of a vector with function components? Or is it a quotient? If it is a quotient use \frac – Yanko Oct 26 '18 at 10:30
• @Yanko. I do not see a problem there. Square root of sum of squares of elements. It would be a function itself. – an apprentice Oct 26 '18 at 10:35

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 $$x\to\infty$$.

• Accepted for the explanation in the first paragraph. Second paragraph may be wrong though. – an apprentice Oct 29 '18 at 9:35

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