Show that $|\sin x - x + \frac{1}{6}x^3| < 0.08$ for $|x| \le \frac{\pi}2$. 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 $0.01$ for $|x|\le \frac{\pi}2$?
My question has to do with how to interpret the $|x| \le\frac{\pi}2$ part. The Taylor polynomial can be about any number, in other words it doesn't have to be around $0$ like in this question, so that would mean the interval $|x| \le\frac{ \pi}2$ may not correspond to where the Taylor polynomial is. Or is it implying a distance of $\frac{\pi}2$ away from wherever my Taylor polynomial is approximating around?
The Taylor series is specified as being expanded about the point $x=0$ (called the center of the series. By asking to you approximate $\sin x$ with its third order Taylor polynomial (about $x=0$) on the interval $|x|\le \pi/2$, you are being told the interval on which to apply Taylor's Remainder Theorem.
Suppose $f(x)$ has Taylor series centered at $x=a$ given by $$\sum_{n=0}^\infty c_n(x-a)^n, \quad\text{where }c_n={f^{(n)}(a)\over n!},$$ and let $$ R_N(x):=f(x)-\sum_{n=0}^N c_n(x-a)^n $$ denote the remainder or error between $f(x)$ and the first $N$ terms of its Taylor series. Then we have:
So you can see how the interval $I$ (along with the center, $a$) is involved in a discussion of error analysis.