Big-O Interpretation

Solution 1:

Note: We usually abuse the notation and write $f(x) = \mathcal{O}(g(x))$ instead of $f(x) \in \mathcal{O}(g(x))$. However, $g(x) \not = \mathcal{O}(f(x))$ in general, because $g(x) \notin \mathcal{O}(f(x))$ in general.

If $f(x) \in \mathcal{O}(g(x))$ then for large $x$, $f(x)$ has the same rate of growth as $g(x)$ or $f(x)$ has a smaller rate of growth than $g(x)$. When we write

$$\sin(x) = x + \mathcal{O}(x^3) \text{ as $x \rightarrow 0$}$$

we mean $\sin(x)$ is equal to $x$ plus some quantity that is "Big Oh of $x^3$." The last quantity is not stated exactly, but "Big Oh" tells us that the absolute value of the last quantity is no more than a positive constant times $x^3$. We can even write many familiar results from calculus such as $\sin(x) \leq 1$ and $n! \sim {(2\pi)^{1/2}}{n^{1/2}}{n^n}{e^{-n}}$ (Stirling's approximation) using the Big Oh:

$$\sin(x) = \mathcal{O}(1) \text{ as $x \rightarrow 0$}$$ $$n! = \mathcal{O}({n^{1/2}}{n^n}{e^{-n}}) \text{ as $x \rightarrow +\infty$}.$$

Another familiar result

$$\mathop{\lim}\limits_{x \to 0}\frac{{\sin(x)}}{x} = 1$$

can be written as

$$\sin(x) = \mathcal{O}(x) \text{ as $x \rightarrow 0$}$$

but since the limit is $1$, we can actually write $\sin(x) \sim x \text{ as $x \rightarrow 0$}$.

You can think of "Big Oh" as $$\mathop{\lim \sup}\limits_{x \to \infty} \left|\frac{{f(x)}}{{g(x)}}\right| = K \in \mathbb{R^+} \Rightarrow f(x) \in \mathcal{O}(g(x)).$$

Technically:

$$\mathop{\lim \sup}\limits_{x \to \infty} \left|\frac{{f(x)}}{{g(x)}}\right| = K \in \mathbb{R^+} \Rightarrow f(x) \in \mathcal{O}(g(x)) \wedge f(x) \in \Theta(g(x)) \wedge f(x) \in \Omega(g(x))$$

where $\Theta$ and $\Omega$ are related asymptotic notations.

See Concrete Mathematics by Graham, Knuth, and Patashnik for a good introduction to asymptotics.

Solution 2:

On of the main reasons to use big-Oh/asymptotic/Landau notation is to understand how some complicated function works by expanding it in easier functions, taking the argument to $\infty$ (or wherever you want) and looking at the largest terms on RHS. In you case the idea is that as $x \to 0$, $\sin x =O(x)$ by expanding in Maclaurin series. Another interesting example I'll show here is Harmonic sum:

$$ H(n)=\sum_{k=1}^{n} \frac{1}{k} = O( \log n) $$