4
$\begingroup$

How do you prove that for $0<x\leq1$, it is true that

$$x-\displaystyle\frac{x^3}{3}<\arctan x<x-\displaystyle\frac{x^3}{6}$$

$\endgroup$
  • $\begingroup$ You could look at the derivatives to find out. All these functions vanish at $0$, so inequalities between the derivatives lift to the corresponding inequalities for the functions. $\endgroup$ – Daniel Fischer Dec 23 '14 at 11:41
  • 1
    $\begingroup$ The first inequality can be derived from the series expansion of $\arctan{x}$. $\endgroup$ – gst Dec 23 '14 at 11:42
  • $\begingroup$ I have edited your question to improve the title and wording, please check that I did not change the meaning of your question. $\endgroup$ – Alice Ryhl Dec 23 '14 at 11:47
  • $\begingroup$ I really do not get the right hand side of this inequality by using shortened taylor expansions $\endgroup$ – Raio Dec 23 '14 at 11:48
  • 1
    $\begingroup$ On the other hand you can show each inequality seperatly, e.g. let $f_1(x)=\arctan{x}-\left(x-\frac{x^3}{3}\right)$ and then show that $f_1(0) = 0$ and show that the derivative of $f_1$ is strictly increasing. The same goes for the second case with an appropriate function $f_2$. $\endgroup$ – gst Dec 23 '14 at 11:49
4
$\begingroup$

The first inequality is trivial, since $\arctan x = x - \frac{x^3}{3}+\frac{\xi^5}{5}$ for some $\xi \in (0,1)$. For the second, just remark that $$ \frac{d}{dx} \left( \arctan x - x + \frac{x^3}{6} \right) = \frac{x^2}{2}+\frac{1}{x^2+1}-1, $$ and this quantity is negative for $x \in (0,1)$. Hence $$ \arctan x - x + \frac{x^3}{6} < 0 $$ for $x \in (0,1]$.

$\endgroup$
  • 1
    $\begingroup$ Strictly you still have to show that at some point in your interval the inequality is actually valid. $\endgroup$ – gst Dec 23 '14 at 11:57
  • $\begingroup$ Just take the limit as $x \to 0+$. $\endgroup$ – Siminore Dec 23 '14 at 14:19
  • $\begingroup$ Yeah sure I just thought it should be mentioned ;-) Because otherwise it's incomplete. $\endgroup$ – gst Dec 23 '14 at 14:21

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.