The question: Let $f(x) = \sin(x)$. The taylor polynomial is the following: $P_{n}(x) = \sum_{n =1}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}$. Find the smallest value of n such that $|P_{n}(x) - f(x)| < 10^{-4}$, for all $x \in [0,1]$. Use $P_{n}(0.2)$ to approximate $f(0.2)$ and compare to the actual value.
Attempt:
Well, $|P_{n}(x) - f(x)| = |R_n(x)|$, which $R_{n}(x)=$ $f^{n+1}(φ(x))\frac{x^{n+1}}{(n+1)!}$. But $R_{n}(x) = (-1)^{n} \frac{x^{2n}}{(2n)!}$. So $|R_{n}(x)| = |\frac{x^{2n}x^{n+1}}{(2n)! (n+1)!} |.$
This is where I am stumped. Could I use the the fact that $f^{n+1}(x) = \cos(x)$ and apply it to the remainder formula?
I have no idea how to start with the second question.
Can someone give me some hints on how to solve this problem? Thank you very much for the help!
Note that for $|x|\le 1$, we have (SEE THIS)
$$\begin{align} \left|\sin(x)-\sum_{n=0}^N \frac{(-1)^nx^{2n+1}}{(2n+1)!}\right|&=\left|\sum_{n=N+1}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\right|\\\\ &\le \frac{|x|^{2N+3}}{(2N+3)!}\\\\ &\le \frac{1}{(2N+3)!} \end{align}$$
We find that the smallest number $N$ for which $(2N+3)!\ge 10^4$ is $N=3$. Therefore, for $|x|\le 1$
$$\left|\sin(x)-\left(x-\frac{x^3}6+\frac{x^5}{120}-\frac{x^7}{5040}\right)\right|\le 10^{-4}$$