Inequality with arctangent between shortened taylor expansion and equation similar to the taylor expansion

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

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

• 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. – Daniel Fischer Dec 23 '14 at 11:41
• The first inequality can be derived from the series expansion of $\arctan{x}$. – gst Dec 23 '14 at 11:42
• I have edited your question to improve the title and wording, please check that I did not change the meaning of your question. – Alice Ryhl Dec 23 '14 at 11:47
• I really do not get the right hand side of this inequality by using shortened taylor expansions – Raio Dec 23 '14 at 11:48
• 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 $f_2$. – gst Dec 23 '14 at 11:49

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 $x \in (0,1]$.
• Just take the limit as $x \to 0+$. – Siminore Dec 23 '14 at 14:19