Order estimates question and big O notation

Question

How can I show that $y(x) = 1 - \cos(x)$ is $\mathcal{O}(x^2)$ for $|x| <<1$ ?

Additionally, with the $|x| << 1$ is there a precise definition? I tried to google it but nothing conclusive returned. I'm assuming it just means $x \to 0$. Is there any reason for the $|x| <<$ term to use the number $1$? Or is it just a matter of convetion.

Would you say that $\cos(x) \leq x$ in this case, or use a strict inequality as $x \to 0$ and $x \neq 0$

Is it sufficient to do the following:

$\lim_{x \to 0} \frac{1-cos(x)}{x^2} = \lim_{x \to 0} \frac{1}{x^2} - \frac{1}{x} \neq 0$

Answer 1

To show that $f(x)$ is of $ \mathcal O(g(x))$ for $x\rightarrow 0$, one has to show that, for a constant, $|C|<\infty$, the following holds: $$\lim_{x\rightarrow 0}\frac{}{}\frac{f(x)}{g(x)}=C.$$

In words: $f(x)$ tends to somewhere in the Order of $g(x)$ when $x$ tends to the same number or even $\pm \infty$. This is "big Oh" if the quotient above tends to a constant, and it is "little oh" if the quotient tends to $0$. In our case $$\lim_{x\rightarrow0}\frac{1-\cos(x)}{x^2}=\lim_{x\rightarrow0}\frac{\sin(x)}{2x}=\frac{1}{2}.$$ (Where the L'Hospital rule was used.)

Answer 2

How can I show that $y(x) = 1 - \cos(x)$ is $\mathcal{O}(x^2)$ for $|x| <<1$ ?

Hint. Recall the Maclaurin series, which says that for any function $f \in \mathcal{C}^2(-a,a)$ you have, as $x \to 0$, $$ f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+\int_0^x (x-t)^{2}f^{(2)}(t)dt. $$ Then applying it to $x \longmapsto \cos x$ you readily obtain $$ \cos (x)=\cos (0)-\sin (0)\:x-\cos''(0)\frac{x^2}{2}-\int_0^x (x-t)^{2}\cos(t)dt $$ (alternately you may just integrate the preceding integral twice by parts) and since $$ \left|\int_0^x (x-t)^{2}\cos(t)dt\right|\leq \int_0^x (x-t)^{2}|\cos(t)|dt\leq \int_0^x (x-t)^{2}dt=\frac{x^3}{3} $$ you get, as $x \to 0$, $$ \cos (x)=1-\frac{x^2}{2}+\mathcal{O}(x^3)=1+\mathcal{O}(x^2). $$