How to prove a Big-O bound

Question

In "Numerical Analysis: Mathematics of Scientific Computing" (3rd ed), chapter 1.2, the author says that $sin x = х - \frac{x^3}{6} + \mathcal{O}(х^5) \quad (х \rightarrow 0)$. This means that on a neighborhood of $0$, there exists a constant $C$ such that $\left|sin x - х + \frac{x^3}{6}\right| \leq C\left|х^5\right|$ and that it can be verified by using Taylor's theorem up to $n=4$.

Doing that, I got to $\frac{\left|x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} +R_4(x)- x+ \frac{x^3}{6}\right|}{\left|х^5\right|} =\left|\frac{1}{5!} - \frac{x^2}{7!} + \frac{x^4}{9!} \right| + \frac{\left|R_4(x)\right|}{\left|х^5\right|}$

And I got stuck there. I know that since I'm in a neighborhood of $0$, $\frac{x^2}{7!} + \frac{x^4}{9!}$ go to $0$, and $\frac{1}{5!}$ is just a constant, but I don't know what to do with the $\frac{\left|R_4(x)\right|}{\left|х^5\right|}$ factor.

How can I prove that it is bounded? how can I find the $C$ that I need?

And to that effect, the first problem of that section is to find for the smallest integer $k$ such that $arctan(x) = x+ \mathcal{O}(х^k)$ as $(х \rightarrow 0)$

I suppose that I need to do something similar, that is, to calculate $\frac{\left|x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9}- ... +R_n(x)- x\right|}{\left|х^k\right|}$. I assume that I should use $n = 3$, but I don't know how to prove that.

Am I on the right path?

Answer 1

I think you overcomplicated it for yourself and they meant $n=4$ to include the $0$ terms.

We know by Taylor's theorem that there exists an $R(x)$ such that $$ \sin(x)=x-\frac{x^3}{3!}+R_5(x)x^5 $$ with $\lim_{x\rightarrow 0} R_5(x)=0$.

By definition of the limit, if we choose $\varepsilon=1$, there must exist $\delta$ such that if $|x|<\delta$ then $|R(x)| < \varepsilon=1$.

Bringing this together, we can choose $C=1$ and say that for all values of $x$ with $|x|<\delta$, we know that $$ |R(x)x^5|<C|x^5| $$ which proves our claim that $$ \sin(x) = x-\frac{x^3}{3!}+\mathcal{O}(x^5) $$

Answer 2

The power series of the sine is an alternating series. If the size of the terms is monotonically falling, then the partial sums are alternatingly upper and lower bounds of the limit. This is one corollary to the Leibniz test.

The series remainder to $\sin x-x+\frac{x^3}6$ is $$ \frac{x^5}{5!}-\frac{x^7}{7!}\pm...+(-1)^k\frac{x^{2k+1}}{(2k+1)!}+... $$ Its terms are falling if $x^2<6\cdot 7$. Starting with the lowest partial series, the remainder value lies between $0$ and $\frac{x^5}{120}$. So $$C=\frac1{120}$$ is one possibility.