# How do I prove $x - \frac{x^3}3 < \arctan x < x$?

Prove that $$x - \frac{x^3}3 < \arctan x < x$$ for every $x>0$?

I tried taking the limit of $x-x^3/3$ and $\arctan(x)$ as $x$ approaches $0$, but I get $0$ which makes sense since they're both $0$ at $x=0$

I'm not sure what else to do algebraically. Would appreciate some help.

Note that $$1-t^2 \le\frac{1}{1+t^2} \le 1$$ for a real $t$ (the inequalities hold with equality only when $t=0$). Therefore, $$\int_0^{x}(1-t^2)\, dt \le \int_0^{x}\frac{1}{1+t^2}\, dt \le \int_0^{x}1\, dt$$ for $x \ge 0$. That is, $$x-\frac{x^3}{3} \le \arctan(x) \le x.$$

The inequalities hold with equality only when $x=0$.

As suggested in the comments, for $x > 0$, we have $$\int_{x/2}^{x}(1-t^2)\, dt < \int_{x/2}^{x}\frac{1}{1+t^2}\, dt < \int_{x/2}^{x} 1\, dt.$$ Therefore, $$\int_{0}^{x}(1-t^2)\, dt < \int_{0}^{x}\frac{1}{1+t^2}\, dt < \int_{0}^{x} 1\, dt$$ for $x > 0$.

• It's easy to extend this to strict inequalities: for $0<x/2<t<x$, we have $\frac{1}{1+t^2}-(1-t^2)>0$ and $1-\frac{1}{1+t^2}>0$. Therefore $\int_{x/2}^x(1-t^2)\,dt<\int_{x/2}^x\frac{1}{1+t^2}\,dt<\int_{x/2}^x 1\,dt$. Splitting the integrals as $\int_0^x=\int_0^{x/2}+\int_{x/2}^x$ gives the claim. Commented Jul 20, 2018 at 20:48
• @robjohn: the inequalities are strict because if $f$ is continuous and non-negative on $[a, b]$ and $f(c) >0$ for some $c\in[a, b]$ then $\int_{a} ^{b} f(x) \, dx>0$. So continuous functions need to have strict inequality at just one point of the interval to ensure the strict inequality of their integrals. Commented Jul 21, 2018 at 2:48
• @robjohn : Yeah, I hope OP includes something similar to my comment in the answer. Commented Jul 21, 2018 at 3:10

Hint:

$$\frac{d}{dx}\arctan x=(1+x^2)^{-1}$$

$$(1+x^k)^n=\sum_{r=0}^\infty\dfrac{n(n-1)\cdots(n-r+1)}{r!}x^{kr}$$

• Hmm.... Enlighten me on where in a solution one might use that second equality? Commented Jul 20, 2018 at 3:09
• Shouldn't the $x^r$ be $x^{kr}$?
– robjohn
Commented Jul 20, 2018 at 3:54

# Proof

(1) Let $$f(x)=\arctan x,~~~x>0.$$

By Lagrange's mean value theorem，we obtain $$f(x)-f(0)=f'(\xi_1)(x-0),$$where $$0<\xi_1 Thus, $$\arctan x=\frac{1}{1+\xi_1^2}\cdot x

(2) Let $$g(x)=\arctan x-x+\frac{x^3}{3},~~~x>0.$$

By Lagrange's mean value theorem，we obtain $$g(x)-g(0)=g'(\xi_2)(x-0),$$where $$0<\xi_2 Thus, $$\arctan x-x+\frac{x^3}{3}=\frac{\xi_2^4}{\xi_2^2+1}\cdot x>0,$$namely,$$\arctan x>x-\frac{x^3}{3}.$$

Combining (1) and (2), we are done.