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Show that $|\sin(x) - x + \frac{1}{6} x^3| < .08$ for $ |x| < \frac{1}{2}\pi$.

How large do you have to take $k$ so that the $k$-th order Taylor polynomial of $\sin(x)$ about $a = 0$ approximates $sin(x)$ to within $.01$ for $|x| < \frac{1}{2}\pi$.

I'm having trouble solving both of these questions. Here's what I've been thinking:

So the first thing I've observed is that $x + \frac{1}{6}x^3%$ is the 4th-order Tp of $\sin(x)$. Next, I know that $\sin(x)$ is bounded by $[-1, 1]$ in the given interval of $x$.

So I thought I could argue that the remainder defined by $r_n,a(x)$:

$r_n,a(x) \leq \frac{M}{(k+1)!} < .01$, for $M = 1$ (because bounded above by 1).

then it is a matter of determining the value of $k$, for which I believe $k = 100$.

But I am not sure how to answer the first question - how do I show that it less than 0.08. Both questions I am quite unsure about, so help is appreciated.

1 Answers 1

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Taylor expansion of the sine function is $$\sin( x)= \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$$ So, let us suppose that you limit to $m$ terms. Write $$\sin( x)= \sum^{m}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}+\sum^{\infty}_{n=m+1} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$$ Since it is an alternating series, the remainder is smaller than the first neglected term $$|R_m|=\frac{x^{2m+3}}{(2m+3)!}$$ and you want that $|R_m|<\epsilon$. Considering the range $|x| < \frac{\pi}{2}$, then, for any $x$ $$|R_m|<\frac{\pi^{2m+3}}{2^{2m+3}(2m+3)!}$$ So, for a given $\epsilon$, $m$ should be determined solving $$\frac{\pi^{2m+3}}{2^{2m+3}(2m+3)!}=\epsilon$$

Your case is simpler since you are given $m=3$ which makes $$|R_3|<\frac{\pi ^5}{3840}\approx 0.0797$$ Similarly, you would have $$|R_4|<\frac{\pi ^7}{645120}\approx 0.0047$$ $$|R_5|<\frac{\pi ^9}{185794560}\approx 0.0002$$

So, for your problem, you just need one extra term, that is to say $$\sin(x)\approx x-\frac{x^3}{6}+\frac{x^5}{120}$$