0
$\begingroup$

I need to find $k\in N$ such that $$ \dfrac{k}{10\ ^ 5} \le \arctan(0.1) \lt \dfrac{k +1}{10\ ^ 5} $$

I tried using Lagrange Remainder formula to find that k but with no luck.

If I am using Leibniz estimate for the reminder than I know that $$ |\arctan(0.1) - S_k|\le a_{k+1}$$ where $a_k = \dfrac{(0.1)^{2k+1}}{2k+1}$

is there an easy way to get that $k$ that I need ?

Thanks for helping.

$\endgroup$
  • $\begingroup$ Use the error estimate for alternating series. Lagrange form of the remainder is too much work. $\endgroup$ – André Nicolas Jun 23 '16 at 6:44
  • $\begingroup$ @AndréNicolas is that what you meant ? if so , how can i continue from here ? $\endgroup$ – user335501 Jun 23 '16 at 7:17
  • $\begingroup$ You are double-using $k$. If we drop the term $(0.1)^5/5$ our error has absolute value $\lt (0.1)^5/5$. So our estimate is $0.1-(0.001/3)$. Express this as $k/10^5$ using your calculator or by hand. Will be back in maybe $7$ hours to check what you have done. $\endgroup$ – André Nicolas Jun 23 '16 at 8:15
2
$\begingroup$

$$\arctan(x) = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)}x^{2n+1} $$ for any $x\in(-1,1)$, so: $$ \frac{1}{10}-\frac{1}{3000}<\arctan\left(\frac{1}{10}\right) < \frac{1}{10}-\frac{1}{3000}+\frac{1}{500000}$$ and we may take $k$ as: $$ k = \left\lfloor \frac{100}{3}\cdot 299\right\rfloor = \color{red}{996}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy