# The big $O$ versus little $o$ notation.

I'm familiar to both the $O$ and little $o$ notation. I know they are of great use when studying limits, asymptotics and series. However, I'm not sure when to use one over another, particularly when solving problems. For example, you can put that, for $x \to 0$

$$\sin x = x + o(x)$$

since

$$\frac{\sin x}{x}-1 \to 0$$

I've also recently read about a great definition for the derivative and strong derivative (Knuth), respectively, for $\epsilon \to 0$:

$$f(x+\epsilon)=f(x)+\epsilon f'(x)+o(\epsilon)$$

which clearly means that

$$\frac{f(x+\epsilon)-f(x)}{\epsilon}- f'(x) \to 0$$

and the same for

$$f(x+\epsilon)=f(x)+\epsilon f'(x)+o(\epsilon ^2)$$

Now consider the following

$$\lim_{x \rightarrow 0}{\frac{\int_{0}^{x} \frac{\sin(t)}{t}-\tan(x)}{2x(1-\cos(x))}}$$

I know that

$$2x\left( {1 - \cos x} \right) = {x^3} + o\left( {{x^3}} \right)$$

$$\tan x = x + \frac{x^3}{3} + o\left( x^3 \right)$$

$$\int\limits_0^x {\frac{{\sin t}}{t}dt = x -\frac{x^3}{3\cdot 3!} +o\left( x^3 \right)}$$

So this gives

$$\frac{{x - \frac{{{x^3}}}{{3 \cdot 3!}} + o\left( {{x^3}} \right) - x - \frac{{{x^3}}}{3} - o\left( {{x^3}} \right)}}{{{x^3} + o\left( {{x^3}} \right)}} =$$

$$\frac{{\left( { - \frac{1}{3} - \frac{1}{{18}}} \right){x^3} + o\left( {{x^3}} \right)}}{{{x^3} + o\left( {{x^3}} \right)}}$$

$$\frac{{\left( { - \frac{7}{{18}}} \right){x^3} + o\left( {{x^3}} \right)}}{{{x^3} + o\left( {{x^3}} \right)}} = - \frac{7}{{18}}$$

So how could I use the big $O$ to solve the same problem? And why should I choose big $O$ over little $o$, when $o$ seems more accurate than $O$?

NOTE: When I talk about the big $O$ I'm considering it for any $x \to a$, not only for asymptotic behaviour $x \to \infty$. For example, for $x \to 0$

$$\sin x = O(x)$$

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Well, actually $O(x)$ says less than $o(x)$ and it is nice in a sense, to use minimal tools to reach the aim. On the other hand, I don't feel like using $O$ if $o$ suits perfectly. –  savick01 Feb 25 '12 at 1:55
@savick01 That's kind of my preocupation: why use $O$ when $o$ seems to give a stronger result? –  Pedro Tamaroff Feb 25 '12 at 1:58
It's not a question of "less" or "more". $\sin x = x + o(x)$ says less than $\sin x = x + O(x^2)$, which says less than $\sin x = x + o(x^2)$ and so on. –  GEdgar Feb 25 '12 at 3:14
It gives a stronger result, but raughly speaking we need more knowledge (assumptions) to get a stronger result. Everything depends on the particular application. Anyway, I meet $o$ much much more frequently (if it is mathematics not related to IT, of course). –  savick01 Feb 25 '12 at 9:26
That frequency may have some reason - we use $o,O$ during examining asymptotic behavior of some $g$. In that case we usually want to find a function $f$ such that $g/f \to 1$. If we have $g=\tilde{g}+o(\tilde{g})$ then $\lim g/f = \lim \tilde{g}/f$ and big $O$ would give us just nothing. We may know that $g=\tilde{g} + O(h)$ but then we are interested if $h=o(\tilde{g})$. In IT we may want to know $g/f < M$ or $\limsup g/f < M$ for some $M$, and then we can replace all the $o$ with $O$ and still get the answer. That's what I meant by "$o$ is stronger", but GEdgar is obviously right. –  savick01 Feb 25 '12 at 9:44

Let's look at $\sin x$ as $x\to0$. You can write $\sin x=O(x)$, or $\sin x=x+o(x^2)$, or $\sin x=x+O(x^3)$, or $\sin x=x-(1/3)x^3+o(x^4)$, or .... You use whichever one works out nicest in whatever problem it is that you are solving.