I'm tasked with determining the values of $x$ for which $f(x)=\cos{x}\approx 1-\frac{x^2}{2!}+\frac{x^4}{4!}$ has an error no greater than 0.001.
Using the error for Taylor polynomials $E = |R_n(x)| = \frac{(x-c)^{n+1}}{(n+1)!}\max|f^{n+1}(x)|$
I've determined that $$E = \frac{x^5}{5!}\le0.001$$
since $f^{(5)}(z)=-\sin{z}$ is bounded by -1 and 1, and no specific interval is given in the problem motivation.
Therefore, values of $x$ for which the 4th-order Maclaurin polynomial approximate $\sin{x}$ with an error no greater than 0.001 should be $$(-0.001\times5!)^{1/5}\le x \le (0.001\times5!)^{1/5}$$ or $$-0.65438939 \le x \le 0.65438939$$ This is not the answer given in the solution manual. I'm clearly making a mistake, and I don't believe it's in my arithmetic. What bad assumption (or incorrect application of the error formula) am I using here?
As it turns out, when $x$ is sufficiently small, $\sin(x)$ is close enough to $x$ such that $$\frac{f^{(5)}(z)x^{5}}{5!} \approx \frac{x^6}{5!}$$
which gives an interval of $$-0.936410984 \le x \le0.936410984$$
This is pretty close to the answer, though for some reason the book takes the error up to the sixth order.