TITLE: Maclaurin series accuracy
QUESTION [1 upvotes]: Find an $n_1$ such that the $n_1$th-order Taylor polynomial for $\sin(x)$ about $x=0$ gives an approximation of $\sin(x)$ with an error of less than $5\cdot 10^{-10}$, for all $x$ between $0$ and $\dfrac{1}{2}\pi$. Your $n_1$ doesn't have to be the smallest possible $n_1$, but show it satisfies the requirements, without using a calculator. You can use $ 2< \pi < 4$.  

So I know the following things. The maclaurin series for $\sin(x)$. 
$$ \sin(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - ... $$
And the error in the approximation:
$$ E_n(x) = \dfrac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1}$$ where $s$ is some number between $a$ and $x$. I'm not sure, but I believe we have $a=0$. 
How do I tackle this problem? Should you set the error equal to $5\cdot10^{-10}$ and solve for $n$? Can we just choose $\pi = 3$ or $\pi = 3.5$ or whatever (I don't really understand what is meant by that)? What should we fill in for $s$ and $x$, and how can we show it satisfies the requirement for all possible $x$ in $[0, \dfrac{1}{2} \pi]$.

REPLY [0 votes]: Hint:
$\displaystyle|E_n(x)|=\frac{|f^{n+1}(s)|}{(n+1)!}|(x-a)^{n+1}|$ where $|f^{n+1}(s)|\le1$ since
$f^{n+1}(s)=\pm\sin s$ or $f^{n+1}(s)=\pm\cos s$, so
$\displaystyle|E_n(x)|\le\frac{|x-0|^{n+1}}{(n+1)!}\le\frac{(\frac{\pi}{2})^{n+1}}{(n+1)!}\le\frac{2^{n+1}}{(n+1)!}$ since $\pi<4$.
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