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.

  • $\begingroup$ @JCM, this is exactly correct answer. $\endgroup$ – erfan soheil Dec 4 '14 at 12:25
  • $\begingroup$ 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 $\left|R_{4}(x)\right|$? $\endgroup$ – JCM Dec 5 '14 at 7:44
  • $\begingroup$ @JCM: I'm not sure what you mean about the endpoints. You can just use the formula as is. In this particular case, the bound on the 5th derivative holds for the entire real line, so the formula holds in general. $\endgroup$ – Cheerful Parsnip Dec 5 '14 at 10:43
  • $\begingroup$ @JCM: note that I am using the "mean value" form of the remainder. $\endgroup$ – Cheerful Parsnip Dec 5 '14 at 10:44
  • $\begingroup$ @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 $0$? $\endgroup$ – Cheerful Parsnip Dec 5 '14 at 11:08

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