# Finding the Accuracy of a Taylor Polynomial for the Approximation $f(x) \approx T_{n}(x)$

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.

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.
• I'm not too clear in regard to the endpoints of the given interval. That is, do the endpoints of the given interval need to be checked to get a particular value for $\left|R_{4}(x)\right|$? – JCM Dec 5 '14 at 7:44
• @JCM Oh, I see. If you want an absolute numerical bound, then yes, $x=\pi/3$ and $x=0$ will give you the most the error could be, because $|(x-\pi/6)^5|$ is maximized in those cases. It depends if you want a bound that depends on $x$ or not. As for your second statement about the fifth derivative, you don't get to choose $c$. What if $c$ is close to $0$? – Cheerful Parsnip Dec 5 '14 at 11:08