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I'm using the following definition for Big-O:

$f(x)$ is $O(g(x))$, when $x \rightarrow a$, if
$\exists \space M > 0$ and $\delta > 0$ s.t.
$|f(x)| \leq M|g(x)|$ when $|x-a| < \delta$

Now for example for the function $f(x)=xsin(x)$ it's easy to show that:

Let $M = 1$.
Then $\forall x \in \mathbb{R}$

$$|f(x)| = |x||sin(x)|≤|x|\cdot 1=M|x|$$

i.e. $f(x)$ is $O(x)$, when $x \rightarrow 0$.

But what did/do I need the $\delta$ and the inequality $|x-0|=|x|< \delta$ for?

mavavilj
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  • Without the restriction to an interval around $a$, $f=O(g)$ would be impossible if $f(x)\ne0=g(x)$ for some $x$, even located "far away" from $a$. Recall that $O$ is a local comparison relation. – Did Sep 10 '16 at 19:14
  • @Did But is it not a global property in the case of my function? Since the inequalities hold $\forall x \in \mathbb{R}$. – mavavilj Sep 11 '16 at 07:22
  • So is the $\delta$ similar to $M$ that it can be fixed, rather than like in $\epsilon - \delta$ -proofs where you sometimes fix it with the $\epsilon$ so that rather than being a single fixed value, it can take many values? But in the case of my function I think I can pick any $\delta$. – mavavilj Sep 11 '16 at 07:24
  • Yeah, very much like showing the constant function continuous, your $\delta $ is not so important, since $\sin (x)$ is bounded. – Andres Mejia Sep 11 '16 at 09:21
  • @AndresMejia But I wonder what would be an example where you need the $\delta$ and the $|x-a|< \delta$ inequality? – mavavilj Sep 11 '16 at 09:23
  • The function I gave meets the requirement that $\delta <1$ since the function will not be $O (x) $ for $x>1$. – Andres Mejia Sep 11 '16 at 09:24
  • "But I wonder what would be an example where you need the δ and the |x−a|<δ inequality?" There is an example of exactly that in my first comment. Try $f(x)=1$, $g(x)=\cos x$, $a=0$. – Did Sep 11 '16 at 09:32
  • "But is it not a global property in the case of my function?" For $f(x)=x\sin x$ and $g(x)=x$, indeed $|f(x)|\leqslant|g(x)|$ holds for every $x$. This implies that $f=O(g)$ at $0$ but is not a requirement for that. – Did Sep 11 '16 at 09:34

1 Answers1

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Let $f: \mathbb R \to \mathbb R$ be defined by

$f(x)= \begin{cases} 5x & x \in [-1,1] \\ x^3 &\textrm{for}\,, |x|>1\\ \end{cases}$

Then $f=O(x)$ at $0$,(or as $x \to 0$), but not outside of this interval, the important thing, is that it is $O(x)$ when you are sufficiently close to $0$. The definition allows you to say something about local properties for functions such as this one, even if it acts oddly far away from the point in question.

Andres Mejia
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  • But is it not the case that in the case of my function, any $a$ works since the inequalities hold for $\forall x \in \mathbb{R}$? I.e. it's not a local, but a global property. – mavavilj Sep 11 '16 at 07:20