# Determining the values for which a Maclaurin Polynomial approximates $f(x)=\cos(x)$ within an error of 0.001

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?

• Maybe they took into account that $|\sin x|\le |x|$, so it would be the $6$-th root? Apr 21 '19 at 23:46
• Interesting. Could you elaborate? Apr 21 '19 at 23:47
• We'd have $|R_n(x)| =\Bigl|\frac{x^5}{120}\Bigr||\sin c|$ for some $c$ between $0$ and $x$, so $|R_n(x)| \le\Bigl|\frac{x^5}{120}\Bigr||x|=\Bigl|\frac{x^6}{120}\Bigr|$, and ultimately $|x|^6\le 0.6$, whence $|x| \le\dots$ Apr 21 '19 at 23:55

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 $$-0.936410984 \le x \le0.936410984$$
• Oh! yes. The Maclaurin expansion for the cosine has no term of degree $5$, so the expansion at order $4$ is ipso facto the expansion at order $5$, and the error term can be calculated accordingly. Apr 22 '19 at 0:13
• Okay, so then here's a gut-check: The Maclaurin expansion for sine has no even-ordered terms. So if I had to calculate the error of the $k$th order approximation of sine, then I would use $k$+2 in my error formula? Apr 22 '19 at 0:16
• If $k$ is odd, of course. Apr 22 '19 at 9:13