Let
$$
f(x) = \sin(x), \quad a = \frac{\pi}{6}, \quad n = 4, \quad 0 \leq x \leq \frac{\pi}{3}
$$
Find a fourth degree ($n=4$) Taylor polynomial for $f$.
$$
T_{4}(x) = \frac{1}{2}+\frac{\sqrt{3}}{2}\left(x – \frac{\pi}{6}\right)-\frac{1}{4}\left(x – \frac{\pi}{6}\right)^2-\frac{\sqrt{3}}{12}\left(x – \frac{\pi}{6}\right)^3+\frac{1}{48}\left(x – \frac{\pi}{6}\right)^4
$$
Now, I need to estimate the accuracy of the approximation $f(x) \approx T_{n}(x)$ using Taylor's Inequality using the given interval.
However, I'm a bit loss on how to proceed.
Best Answer
The remainder is $R_n=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some $c$ between $x$ and $a$. In this case, the derivatives of $f(x)$ are all $\pm\sin x$ or $\pm \cos x$, so $|f^{(n+1)}(c)|\leq 1$. Thus $|R_n|\leq \frac{|x-\frac{\pi}{6}|^{n+1}}{(n+1)!}$. You can plug in $n=4$ to finish the problem.
Edit: The preceding argument was for all of $\mathbb R$, but the problem is for the specific interval $[0,\pi/3]$, and the OP wants an absolute numerical bound. Restricting to $[0,\pi/3]$ does not let us improve our bound on the fifth derivative $f^{(5)}(x)=\cos(x)$ since $\cos(x)$ achieves the value $1$ on that interval. So $$R_4\leq \frac{1}{5!}|x-\pi/6|^5.$$ To get an absolute numerical bound, notice that $|x-\pi/6|\leq \pi/6$ on the given interval.