# Approximations to inverse tangent function

## Abstract

In this paper, we present a sharp Shafer-type inequality for the inverse tangent function. Based on the PadÃ© approximation method, we give approximations to the inverse tangent function. Based on the obtained result, we establish new bounds for arctanx.

## 1 Introduction

In 1966, Shafer [1] posed, as a problem, the following inequality:

$$\frac{3x}{1+2\sqrt{1+x^{2}}}< \arctan x, \quad x>0.$$
(1.1)

Three proofs of it were later given in [2]. Shaferâ€™s inequality (1.1) was sharpened and generalized by Qi et al. in [3]. AÂ survey and expository of some old and new inequalities associated with trigonometric functions can be found in [4]. Chen et al. [5] presented a new method to sharpen bounds of both sincx and arcsinx functions, and the inequalities in exponential form as well.

For each $$a>0$$, Chen and Cheung [6] determined the largest number b and the smallest number c such that the inequalities

$$\frac{b x}{1+a\sqrt{1+x^{2}}}\leq \arctan x\leq \frac{c x}{1+a\sqrt{1+x ^{2}}}$$
(1.2)

are valid for all $$x\geq 0$$. More precisely, these author proved that the largest number b and the smallest number c required by inequality (1.2) are

\begin{aligned} & \text{when } 0< a\leq \frac{\pi }{2},\quad b =\frac{\pi }{2}a,\qquad c=1+a; \\ & \text{when } \frac{\pi }{2}< a\leq \frac{2}{\pi -2},\quad b = \frac{4(a^{2}-1)}{a^{2}},\qquad c=1+a; \\ & \text{when } \frac{2}{\pi -2}< a< 2,\quad b =\frac{4(a^{2}-1)}{a^{2}},\qquad c= \frac{\pi }{2}a; \\ & \text{when } 2\leq a< \infty,\quad b =1+a, \qquad c=\frac{\pi }{2}a. \end{aligned}

In 1974, Shafer [7] indicated several elementary quadratic approximations of selected functions without proof. Subsequently, Shafer [8] established these results as analytic inequalities. For example, Shafer [8] proved that, for $$x>0$$,

$$\frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x.$$
(1.3)

The inequality (1.3) can also be found in [9]. The inequality (1.3) is an improvement of the inequality (1.1).

Zhu [10] developed (1.3) to produce a symmetric double inequality. More precisely, the author proved that, for $$x>0$$,

$$\frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x< \frac{8x}{3+\sqrt{25+\frac{256}{ \pi^{2}}x^{2}}},$$
(1.4)

where the constants $$80/3$$ and $$256/\pi^{2}$$ are the best possible.

### Remark 1.1

For $$x>0$$, the following symmetric double inequality holds:

$$\frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x< \frac{\frac{2 \sqrt{15}\pi }{3}x}{3+\sqrt{25+\frac{80}{3}x^{2}}},$$
(1.5)

where the constants 8 and $$\frac{2\sqrt{15}\pi }{3}$$ are the best possible. We here point out that, for $$x>0$$, the upper bound in (1.4) is better than the upper bound in (1.5).

Based on the following power series expansion:

$$\arctan x \biggl( 3+\sqrt{25+\frac{80}{3}x^{2}} \biggr) =8x+ \frac{32}{4725}x^{7}-\frac{64}{4725}x^{9}+ \frac{25\text{,}376}{1\text{,}299\text{,}375}x^{11}- \cdots,$$

Sun and Chen [11] presented a new upper bound and proved that, for $$x>0$$,

$$\arctan x< \frac{8x+\frac{32}{4725}x^{7}}{3+\sqrt{25+\frac{80}{3}x ^{2}}}.$$
(1.6)

Moreover, these authors pointed out that, for $$0< x< x_{0}=1.4243\ldots$$â€‰, the upper bound in (1.6) is better than the upper bound in (1.4). In fact, we have the following approximation formulas near the origin:

\begin{aligned}& \arctan x-\frac{8x}{3+\sqrt{25+\frac{256}{\pi^{2}}x^{2}}}=O\bigl(x^{3}\bigr), \\& \arctan x-\frac{3x}{1+2\sqrt{1+x^{2}}}=O\bigl(x^{5}\bigr), \\& \arctan x-\frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}=O\bigl(x^{7}\bigr), \end{aligned}

and

$$\arctan x-\frac{8x+\frac{32}{4725}x^{7}}{3+\sqrt{25+\frac{80}{3}x ^{2}}}=O\bigl(x^{9}\bigr).$$

Nishizawa [12] proved that, for $$x>0$$,

$$\frac{\pi^{2}x}{4+\sqrt{(\pi^{2}-4)^{2}+(2\pi x)^{2}}}< \arctan x< \frac{ \pi^{2}x}{4+\sqrt{32+(2\pi x)^{2}}},$$
(1.7)

where the constants $$(\pi^{2}-4)^{2}$$ and 32 are the best possible.

Using the Maple software, we derive the following asymptotic formulas in the Appendix:

\begin{aligned}& \frac{\arctan x}{x}=\frac{\pi^{2}}{4+\sqrt{32+(2\pi x)^{2}}}-\frac{12- \pi^{2}}{3\pi^{2} x^{4}}+O \biggl( \frac{1}{x^{5}} \biggr), \end{aligned}
(1.8)
\begin{aligned}& \frac{\arctan x}{x} =\frac{3\pi^{2}}{24-\pi^{2}+\sqrt{432-24\pi^{2}+ \pi^{4}-12\pi (12-\pi^{2})x+(6\pi x)^{2}}} \\& \hphantom{\frac{\arctan x}{x} =} {} +\frac{\pi^{4}-72}{18\pi^{3}x^{5}}+O \biggl( \frac{1}{x^{6}} \biggr), \end{aligned}
(1.9)

and

\begin{aligned} x \biggl( \frac{\pi }{2}-\arctan x \biggr) = \frac{x^{2}+\frac{4}{15}}{x ^{2}+\frac{3}{5}}+O \biggl( \frac{1}{x^{6}} \biggr) \end{aligned}
(1.10)

as $$x\to \infty$$.

In this paper, motivated by (1.9), we establish a symmetric double inequality for arctanx. Based on the PadÃ© approximation method, we develop the approximation formula (1.10) to produce a general result. More precisely, we determine the coefficients $$a_{j}$$ and $$b_{j}$$ ($$1 \leq j\leq k$$) such that

$$x \biggl( \frac{\pi }{2}-\arctan x \biggr) =\frac{x^{2k}+a_{1}x^{2(k-1)}+ \cdots +a_{k}}{x^{2k}+b_{1}x^{2(k-1)}+\cdots +b_{k}}+ O \biggl( \frac{1}{x ^{4k+2}} \biggr), \quad x\to \infty,$$

where $$k\geq 1$$ is any given integer. Based on the obtained result, we establish new bounds for arctanx.

Some computations in this paper were performed using Maple software.

## 2 Lemma

It is well known that

\begin{aligned} \sum_{k=0}^{2n+1}(-1)^{k} \frac{x^{2k+1}}{(2k+1)!}< \sin x< \sum_{k=0} ^{2n}(-1)^{k}\frac{x^{2k+1}}{(2k+1)!} \end{aligned}
(2.1)

and

\begin{aligned} \sum_{k=0}^{2n+1}(-1)^{k} \frac{x^{2k}}{(2k)!}< \cos x< \sum_{k=0}^{2n}(-1)^{k} \frac{x ^{2k}}{(2k)!} \end{aligned}
(2.2)

for $$x>0$$ and $$n\in \mathbb{N}_{0}:=\mathbb{N}\cup \{0\}$$, where $$\mathbb{N}$$ denotes the set of positive integers.

The following lemma will be used in our present investigation.

### Lemma 2.1

For $$0< u<\pi /2$$,

\begin{aligned} &\cos u\sin^{2}u>u^{2}-\frac{5}{6}u^{4}+ \frac{91}{360}u^{6}- \frac{41}{1008}u^{8} \end{aligned}
(2.3)

and

\begin{aligned} \sin^{3} u>u^{3}-\frac{1}{2}u^{5}+ \frac{13}{120}u^{7}-\frac{41}{3024}u ^{9}. \end{aligned}
(2.4)

### Proof

We find that

\begin{aligned} \cos u\sin^{2}u =&\frac{1}{4} \bigl(\cos u-\cos (3u) \bigr) \\ =&u^{2}- \frac{5}{6}u^{4}+\frac{91}{360}u^{6}- \frac{41}{1008}u^{8}+\sum_{n=5} ^{\infty }(-1)^{n-1}w_{n}(u) \end{aligned}
(2.5)

and

\begin{aligned} \sin^{3} u =&\frac{1}{4} \bigl(3\sin u-\sin (3u) \bigr) \\ =&u^{3}-\frac{1}{2}u ^{5}+\frac{13}{120}u^{7}- \frac{41}{3024}u^{9}+\sum_{n=5}^{\infty }(-1)^{n-1}W _{n}(u), \end{aligned}
(2.6)

where

\begin{aligned} w_{n}(u)=\frac{9^{n}-1}{(2n)!}u^{2n} \quad \text{and}\quad W_{n}(u)=\frac{3(9^{n}-1)}{4 \cdot (2n+1)!}u^{2n+1}. \end{aligned}

Elementary calculations reveal that, for $$0< u<\pi /2$$ and $$n\geq 5$$,

\begin{aligned} \frac{w_{n+1}(u)}{w_{n}(u)} &= \frac{u^{2}(9^{n+1}-1)}{2(2n+1)(n+1)(9^{n}-1)}< \frac{(\pi /2)^{2}(9^{n+1}-1)}{2(2n+1)(n+1)(9^{n}-1)} \\ &< \frac{3\cdot 9^{n+1}}{2(2n+1)(n+1)(9^{n}-1)}= \frac{27}{2(2n+1)(n+1)} \biggl\{ 1+\frac{1}{9^{n}-1} \biggr\} \\ &\leq \frac{27}{2(2n+1)(n+1)} \biggl\{ 1+\frac{1}{9^{5}-1} \biggr\} = \frac{1\text{,}594\text{,}323}{118\text{,}096(2n+1)(n+1)}< 1 \end{aligned}

and

\begin{aligned} \frac{W_{n+1}(u)}{W_{n}(u)} &= \frac{u^{2}(9^{n+1}-1)}{2(2n+3)(n+1)(9^{n}-1)}< \frac{w_{n+1}(u)}{w _{n}(u)}< 1. \end{aligned}

Therefore, for fixed $$u\in (0,\pi /2)$$, the sequences $$n\longmapsto w _{n}(u)$$ and $$n\longmapsto W_{n}(u)$$ are both strictly decreasing for $$n\geq 5$$. From (2.5) and (2.6), we obtain the desired results (2.3) and (2.4).â€ƒâ–¡

The proof of Theorem 3.1 makes use of the inequalities (2.1)â€“(2.4).

## 3 Sharp Shafer-type inequality

Equation (1.9) motivated us to establish a symmetric double inequality for arctanx.

### Theorem 3.1

For $$x>0$$, we have

\begin{aligned} &\frac{3\pi^{2}x}{24-\pi^{2}+\sqrt{\alpha -12\pi (12-\pi^{2})x+36\pi ^{2}x^{2}}} \\ &\quad < \arctan x \\ &\quad < \frac{3\pi^{2}x}{24-\pi^{2}+\sqrt{\beta -12\pi (12-\pi^{2})x+36\pi ^{2}x^{2}}}, \end{aligned}
(3.1)

with the best possible constants

\begin{aligned} &\alpha =432-24\pi^{2}+ \pi^{4}=292.538\ldots \quad \textit{and} \\ &\beta =576-192\pi^{2}+ 16\pi^{4}=239.581\ldots. \end{aligned}
(3.2)

### Proof

The inequality (3.1) can be written for $$x>0$$ as

\begin{aligned} \beta < \biggl( \frac{3\pi^{2} x^{2}}{\arctan x}-\bigl(24-\pi^{2} \bigr) \biggr) ^{2}+12 \pi \bigl(12-\pi^{2}\bigr)x-36 \pi^{2} x^{2}< \alpha. \end{aligned}
(3.3)

By the elementary change of variable $$t = \arctan x$$ ($$x>0$$), (3.3) becomes

\begin{aligned} \beta < \vartheta (t)< \alpha,\quad 0< t< \frac{\pi }{2}, \end{aligned}
(3.4)

where

\begin{aligned} \vartheta (t)={}& \biggl( \frac{3\pi^{2}\tan^{2} t}{t}-\bigl(24-\pi^{2}\bigr) \biggr) ^{2} \\ &{}+12\pi \bigl(12-\pi^{2}\bigr)\tan t-36 \pi^{2}\tan^{2}t. \end{aligned}

Elementary calculations reveal that

\begin{aligned} &\lim_{t\to 0^{+}}\vartheta (t)=576-192\pi^{2}+ 16 \pi^{4} \quad \text{and} \\ & \lim_{t\to \pi /2^{-}}\vartheta (t)=432-24\pi^{2}+\pi ^{4}. \end{aligned}

In order to prove (3.4), it suffices to show that $$\vartheta (t)$$ is strictly increasing for $$0 < t < \pi /2$$.

Differentiation yields

\begin{aligned} t^{3}\cos^{3}t\vartheta '(t) =&\bigl(24\pi t- \pi^{3} t\bigr)\sin t\cos^{2}t+\bigl(3\pi ^{3} t-12 \pi t^{3}\bigr)\sin t \\ &{} - \bigl(3\pi^{3}+\bigl(24\pi -\pi^{3}\bigr) t^{2}-\bigl(24-2\pi^{2}\bigr) t^{3} \bigr)\cos t+3 \pi^{3}\cos^{3} t \\ =:&\lambda (t). \end{aligned}

We now consider two cases to prove $$\lambda (t)>0$$ for $$0< t<\pi /2$$.

Case 1: $$0< t\leq 0.6$$.

Using (2.1) and (2.2), we have, for $$0< t\leq 0.6$$,

\begin{aligned} \lambda (t) =& \biggl( 6\pi -\frac{1}{4}\pi^{3} \biggr) t\sin (3t)+ \frac{3}{4}\pi^{3}\cos (3t)+ \biggl\{ \biggl( 6\pi + \frac{11}{4}\pi^{3} \biggr) t-12 \pi t^{3} \biggr\} \sin t \\ &{} - \biggl( \frac{3}{4}\pi^{3}-\bigl(\pi^{3}-24 \pi\bigr)t^{2}-\bigl(24-2\pi^{2}\bigr)t ^{3} \biggr) \cos t \\ >& \biggl( 6\pi -\frac{1}{4}\pi^{3} \biggr) t \biggl( 3t- \frac{9}{2}t^{3}+ \frac{81}{40}t^{5}- \frac{243}{560}t^{7} \biggr) \\ &{}+\frac{3}{4}\pi^{3} \biggl( 1-\frac{9}{2}t^{2}+\frac{27}{8}t^{4}- \frac{81}{80}t^{6} \biggr) \\ &{} + \biggl\{ \biggl( 6\pi +\frac{11}{4}\pi^{3} \biggr) t-12 \pi t^{3} \biggr\} \biggl( t-\frac{1}{6}t^{3} \biggr) \\ &{} - \biggl( \frac{3}{4}\pi^{3}-\bigl(\pi^{3}-24\pi \bigr)t^{2}-\bigl(24-2\pi^{2}\bigr)t ^{3} \biggr) \biggl( 1-\frac{1}{2}t^{2}+\frac{1}{24}t^{4} \biggr) \\ =&t^{3} \biggl\{ 24-2\pi^{2}- \biggl( 28\pi - \frac{8}{3}\pi^{3} \biggr) t- \bigl( 12-\pi^{2} \bigr) t^{2} \biggr\} \\ &{} +t^{6} \biggl\{ \frac{263}{20}\pi -\frac{235}{192} \pi^{3}+ \biggl( 1- \frac{1}{12}\pi^{2} \biggr) t- \biggl( \frac{729}{280}\pi - \frac{243}{2240}\pi^{3} \biggr) t^{2} \biggr\} . \end{aligned}

Each function in curly braces is positive for $$t\in (0,0.6]$$. Thus, $$\lambda (t)>0$$ for $$t\in (0,0.6]$$.

Case 2: $$0.6< t<\pi /2$$.

We now prove $$\lambda (t)>0$$ for $$0.6< t<\pi /2$$. Replacing t by $$\frac{\pi }{2}-u$$ leads to an equivalent inequality:

\begin{aligned} \mu (u)>0,\quad 0< u< \frac{\pi }{2}-0.6, \end{aligned}

where

\begin{aligned} \mu (u) ={}&\bigl(24\pi -\pi^{3} \bigr) \biggl( \frac{\pi }{2}-u \biggr) \cos u\sin^{2}u+ \biggl\{ 3\pi^{3} \biggl( \frac{\pi }{2}-u \biggr) -12\pi \biggl( \frac{ \pi }{2}-u \biggr) ^{3} \biggr\} \cos u \\ &{} - \biggl\{ 3\pi^{3}+\bigl(24\pi -\pi^{3}\bigr) \biggl( \frac{\pi }{2}-u \biggr) ^{2}-\bigl(24-2\pi^{2}\bigr) \biggl( \frac{\pi }{2}-u \biggr) ^{3} \biggr\} \sin u+3 \pi^{3}\sin^{3} u. \end{aligned}

Using (2.1)â€“(2.4), we have, for $$0< u<\frac{\pi }{2}-0.6$$,

\begin{aligned} \mu (u) >&\bigl(24\pi -\pi^{3} \bigr) \biggl( \frac{\pi }{2}-u \biggr) \biggl( u^{2}- \frac{5}{6}u^{4}+ \frac{91}{360}u^{6}-\frac{41}{1008}u^{8} \biggr) \\ &{} + \biggl\{ 3\pi^{3} \biggl( \frac{\pi }{2}-u \biggr) -12\pi \biggl( \frac{ \pi }{2}-u \biggr) ^{3} \biggr\} \biggl( 1- \frac{1}{2}u^{2}+\frac{1}{24}u ^{4}- \frac{1}{720}u^{6} \biggr) \\ &{} - \biggl\{ 3\pi^{3}+\bigl(24\pi -\pi^{3}\bigr) \biggl( \frac{\pi }{2}-u \biggr) ^{2}-\bigl(24-2\pi^{2}\bigr) \biggl( \frac{\pi }{2}-u \biggr) ^{3} \biggr\} \biggl( u- \frac{1}{6}u^{3}+\frac{1}{120}u^{5} \biggr) \\ &{} +3\pi^{3} \biggl( u^{3}-\frac{1}{2}u^{5}+ \frac{13}{120}u^{7}- \frac{41}{3024}u^{9} \biggr) \\ =&u^{4} \biggl\{ \frac{1}{3}\pi^{4}-24+ \biggl( 12\pi -\frac{9}{5}\pi^{3} \biggr) u+ \biggl( 2\pi^{2}- \frac{11}{90}\pi^{4}+4 \biggr) u^{2} \\ &{}+ \biggl( - \frac{82}{15}\pi +\frac{199}{360}\pi^{3} \biggr) u^{3} \\ &{} + \biggl( -\frac{1}{5}-\frac{25}{56}\pi^{2}+ \frac{41}{2016}\pi^{4} \biggr) u ^{4}+ \biggl( \frac{403}{420}\pi -\frac{41}{504}\pi^{3} \biggr) u^{5} \biggr\} >0. \end{aligned}

We then obtain $$\lambda (t)>0$$ and $$\vartheta '(t)>0$$ for all $$0< t<\pi /2$$. Hence, $$\vartheta (t)$$ is strictly increasing for $$0 < t < \pi /2$$. The proof is complete.â€ƒâ–¡

From (1.7) and (3.1), we obtain the following approximation formulas:

$$\frac{\arctan n}{n} \approx \frac{\pi^{2}}{4+\sqrt{32+(2\pi n)^{2}}}=:a_{n}$$
(3.5)

and

\begin{aligned} \frac{\arctan n}{n}\approx \frac{3\pi^{2}}{24-\pi^{2}+\sqrt{432-24 \pi^{2}+\pi^{4}-12\pi (12-\pi^{2})n+(6\pi n)^{2}}}=:b_{n}, \end{aligned}
(3.6)

as $$n\to \infty$$.

The following numerical computations (see TableÂ 1) would show that, for $$n\in \mathbb{N}$$, Eq.Â (3.6) is sharper than Eq.Â (3.5).

In fact, we have, as $$n\to \infty$$,

$$\frac{\arctan n}{n} = a_{n}+O \biggl( \frac{1}{n^{4}} \biggr) \quad \text{and}\quad \frac{\arctan n}{n}= b_{n}+O \biggl( \frac{1}{n^{5}} \biggr).$$

## 4 Approximations to arctanx

For later use, we introduce the PadÃ© approximant (see [13â€“16]). Let f be a formal power series,

\begin{aligned} f(t)=c_{0}+c_{1}t+c_{2}t^{2}+ \cdots. \end{aligned}
(4.1)

The PadÃ© approximation of order $$(p, q)$$ of the function f is the rational function, denoted by

\begin{aligned}{} [p/q]_{f}(t)=\frac{\sum_{j=0}^{p}a_{j}t^{j}}{1+\sum_{j=1}^{q}b_{j}t ^{j}}, \end{aligned}
(4.2)

where $$p\geq 0$$ and $$q\geq 1$$ are two given integers, the coefficients $$a_{j}$$ and $$b_{j}$$ are given by (see [13â€“15])

\begin{aligned} \textstyle\begin{cases} a_{0}=c_{0}, \\ a_{1}=c_{0}b_{1}+c_{1}, \\ a_{2}=c_{0}b_{2}+c_{1}b_{1}+c_{2}, \\ \vdots \\ a_{p} = c_{0}b_{p}+\cdots + c_{p-1}b_{1} + c_{p}, \\ 0 = c_{p+1} + c_{p}b_{1} + \cdots + c_{p-q+1}b_{q}, \\ \vdots & \\ 0 = c_{p+q} + c_{p+q-1}b_{1} + \cdots + c_{p}b_{q}, \end{cases}\displaystyle \end{aligned}
(4.3)

and the following holds:

\begin{aligned}{} [p/q]_{f}(t)- f (t) = O\bigl(t^{p+q+1} \bigr). \end{aligned}
(4.4)

Thus, the first $$p + q + 1$$ coefficients of the series expansion of $$[p/q]_{f}$$ are identical to those of f.

From the expansion (see [17, p. 81])

\begin{aligned} \arctan x=\frac{\pi }{2}+\sum_{j=1}^{\infty } \frac{(-1)^{j}}{(2j-1)x ^{2j-1}},\quad \vert x\vert >1, \end{aligned}

we obtain

\begin{aligned} x \biggl( \frac{\pi }{2}-\arctan x \biggr) =\sum _{j=0}^{\infty }\frac{c _{j}}{x^{2j}}=1-\frac{1}{3x^{2}}+ \frac{1}{5x^{4}}-\frac{1}{7x^{6}}+ \cdots, \end{aligned}
(4.5)

where

\begin{aligned} c_{j}=\frac{(-1)^{j}}{2j+1}\quad \text{for } j\geq 0. \end{aligned}
(4.6)

Let

\begin{aligned} f(t)=\sum_{j=0}^{\infty } \frac{c_{j}}{t^{j}}, \end{aligned}
(4.7)

with the coefficients $$c_{j}$$ given in (4.6). Then we have

\begin{aligned} f\bigl(x^{2}\bigr)=\sum_{j=0}^{\infty } \frac{c_{j}}{x^{2j}}=x \biggl( \frac{\pi }{2}- \arctan x \biggr). \end{aligned}
(4.8)

In what follows, the function f is given in (4.7).

Based on the PadÃ© approximation method, we now give a derivation of Eq.Â (1.10). To this end, we consider

\begin{aligned}{} [1/1]_{f}(t)=\frac{\sum_{j=0}^{1}a_{j}t^{-j}}{1+\sum_{j=1}^{1}b_{j}t ^{-j}}. \end{aligned}

Noting that

\begin{aligned} c_{0}=1, \qquad c_{1}=-\frac{1}{3}, \qquad c_{2}=\frac{1}{5}, \end{aligned}
(4.9)

holds, we have, by (4.3),

\begin{aligned} \textstyle\begin{cases} a_{0}=1, \\ a_{1}=b_{1}-\frac{1}{3}, \\ 0 =\frac{1}{5} -\frac{1}{3}b_{1}, \end{cases}\displaystyle \end{aligned}

that is,

\begin{aligned} a_{0}=1,\qquad a_{1}=\frac{4}{15}, \qquad b_{1}=\frac{3}{5}. \end{aligned}

We thus obtain

$$[1/1]_{f}(t)= \frac{1+\frac{4}{15t}}{1+\frac{3}{5t}}= \frac{15t+4}{3(5t+3)},$$
(4.10)

and we have, by (4.4),

$$f(t)=\frac{15t+4}{3(5t+3)}+O \biggl( \frac{1}{t^{3}} \biggr), \quad t\to \infty.$$
(4.11)

Replacing t by $$x^{2}$$ in (4.11) yields (1.10).

From the PadÃ© approximation method and the expansion (4.7), we now present a general result.

### Theorem 4.1

The PadÃ© approximation of order $$(p, q)$$ of the function $$f(t)=\sum_{j=0}^{\infty }\frac{c_{j}}{t^{j}}$$ (at the point $$t=\infty$$) is the following rational function:

\begin{aligned}{} [p/q]_{f}(t)&=\frac{1+\sum_{j=1}^{p}a_{j}t^{-j}}{1+\sum_{j=1}^{q}b_{j}t ^{-j}} \\ &=t^{q-p} \biggl( \frac{t^{p}+a_{1}t^{p-1}+\cdots +a_{p}}{t^{q}+b _{1}t^{q-1}+\cdots +b_{q}} \biggr), \end{aligned}
(4.12)

where $$p\geq 1$$ and $$q\geq 1$$ are any given integers, the coefficients $$a_{j}$$ and $$b_{j}$$ are given by

\begin{aligned} \textstyle\begin{cases} a_{1}=b_{1}+c_{1}, \\ a_{2}=b_{2}+c_{1}b_{1}+c_{2}, \\ \vdots \\ a_{p} = b_{p}+\cdots + c_{p-1}b_{1} + c_{p}, \\ 0 = c_{p+1} + c_{p}b_{1} + \cdots + c_{p-q+1}b_{q}, \\ \vdots & \\ 0 = c_{p+q} + c_{p+q-1}b_{1} + \cdots + c_{p}b_{q}, \end{cases}\displaystyle \end{aligned}
(4.13)

and $$c_{j}$$ is given in (4.6), and the following holds:

\begin{aligned} f (t) -[p/q]_{f}(t) = O \biggl( \frac{1}{t^{p+q+1}} \biggr),\quad t\to \infty. \end{aligned}
(4.14)

In particular, replacing t by $$x^{2}$$ in (4.14) yields

\begin{aligned} &x \biggl( \frac{\pi }{2}-\arctan x \biggr) \\ &\quad =x^{2(q-p)} \biggl( \frac{x^{2p}+a _{1}x^{2(p-1)}+\cdots +a_{p}}{x^{2q}+b_{1}x^{2(q-1)}+\cdots +b_{q}} \biggr) + O \biggl( \frac{1}{x^{2(p+q+1)}} \biggr), \quad x\to \infty, \end{aligned}
(4.15)

with the coefficients $$a_{j}$$ and $$b_{j}$$ given by (4.13).

Setting $$(p, q)=(k, k)$$ in (4.15), we obtain the following corollary.

### Corollary 4.1

As $$x\to \infty$$,

\begin{aligned} x \biggl( \frac{\pi }{2}-\arctan x \biggr) = \frac{x^{2k}+a_{1}x^{2(k-1)}+ \cdots +a_{k}}{x^{2k}+b_{1}x^{2(k-1)}+\cdots +b_{k}}+ O \biggl( \frac{1}{x ^{4k+2}} \biggr), \end{aligned}
(4.16)

where $$k\geq 1$$ is any given integer, the coefficients $$a_{j}$$ and $$b_{j}$$ ($$1\leq j\leq k$$) are given by

\begin{aligned} \textstyle\begin{cases} a_{1}=b_{1}+c_{1}, \\ a_{2}=b_{2}+c_{1}b_{1}+c_{2}, \\ \vdots \\ a_{k} = b_{k}+\cdots + c_{k-1}b_{1} + c_{k}, \\ 0 = c_{k+1} + c_{k}b_{1} + \cdots + c_{1}b_{k}, \\ \vdots & \\ 0 = c_{2k} + c_{2k-1}b_{1} + \cdots + c_{k}b_{k}, \end{cases}\displaystyle \end{aligned}
(4.17)

and $$c_{j}$$ is given in (4.6).

Setting $$k=2$$ in (4.16) yields, as $$x\to \infty$$,

\begin{aligned} x \biggl( \frac{\pi }{2}-\arctan x \biggr) = \frac{945x^{4}+735x^{2}+64}{15(63x ^{4}+70x^{2}+15)}+O \biggl( \frac{1}{x^{10}} \biggr), \end{aligned}
(4.18)

which gives

\begin{aligned} \arctan x=\frac{\pi }{2}-\frac{945x^{4}+735x^{2}+64}{15x(63x^{4}+70x ^{2}+15)}+ O \biggl( \frac{1}{x^{11}} \biggr). \end{aligned}

Using the Maple software, we find, as $$x\to \infty$$,

\begin{aligned} \arctan x={}&\frac{\pi }{2}-\frac{945x^{4}+735x^{2}+64}{15x(63x^{4}+70x ^{2}+15)}+ \frac{64}{43\text{,}659x^{11}} \\ &{}-\frac{1856}{464\text{,}373x^{13}}+ O \biggl( \frac{1}{x ^{15}} \biggr). \end{aligned}
(4.19)

Equation (4.19) motivated us to establish new bounds for arctanx.

### Theorem 4.2

For $$x>0$$, we have

\begin{aligned} &\frac{\pi }{2}-\frac{945x^{4}+735x^{2}+64}{15x(63x^{4}+70x^{2}+15)}+\frac{64}{43\text{,}659x ^{11}}- \frac{1856}{464\text{,}373x^{13}} \\ &\quad < \arctan x< \frac{\pi }{2}-\frac{945x^{4}+735x^{2}+64}{15x(63x^{4}+70x ^{2}+15)}+\frac{64}{43\text{,}659x^{11}}. \end{aligned}
(4.20)

### Proof

For $$x>0$$, let

\begin{aligned} I(x)=\arctan x- \biggl( \frac{\pi }{2}-\frac{945x^{4}+735x^{2}+64}{15x(63x ^{4}+70x^{2}+15)}+ \frac{64}{43\text{,}659x^{11}}-\frac{1856}{464\text{,}373x^{13}} \biggr) \end{aligned}

and

\begin{aligned} J(x)=\arctan x- \biggl( \frac{\pi }{2}-\frac{945x^{4}+735x^{2}+64}{15x(63x ^{4}+70x^{2}+15)}+ \frac{64}{43\text{,}659x^{11}} \biggr). \end{aligned}

Differentiation yields

\begin{aligned} I'(x)=-\frac{64(230\text{,}391x^{8}+372\text{,}680x^{6}+236\text{,}885x^{4}+65\text{,}400x^{2}+6525)}{35\text{,}721x ^{14}(1+x^{2})(63x^{4}+70x^{2}+15)^{2}}< 0 \end{aligned}

and

\begin{aligned} J'(x)=\frac{64(12\text{,}789x^{8}+15\text{,}610x^{6}+8890x^{4}+2325x^{2}+225)}{3969x ^{12}(1+x^{2})(63x^{4}+70x^{2}+15)^{2}}>0. \end{aligned}

Hence, $$I(x)$$ is strictly decreasing and $$J(x)$$ is strictly increasing for $$x>0$$, and we have

\begin{aligned} I(x)>\lim_{t\to \infty }I(t)=0 \quad \text{and}\quad J(x)< \lim _{t\to \infty }J(t)=0 \quad \text{for } x>0. \end{aligned}

The proof is complete.â€ƒâ–¡

### Remark 4.1

We point out that, for $$x>1.0213\ldots$$â€‰, the lower bound in (4.20) is better than the one in (1.7). For $$x>0.854439\ldots$$â€‰, the upper bound in (4.20) is better than the one in (1.7). For $$x>0.947273\ldots$$â€‰, the lower bound in (4.20) is better than the one in (3.1). For $$x>0.792793\ldots$$â€‰, the upper bound in (4.20) is better than the one in (3.1).

## 5 Conclusions

In this paper, we establish a symmetric double inequality for arctanx (Theorem 3.1). We determine the coefficients $$a_{j}$$ and $$b_{j}$$ ($$1\leq j\leq k$$) such that

$$x \biggl( \frac{\pi }{2}-\arctan x \biggr) =\frac{x^{2k}+a_{1}x^{2(k-1)}+ \cdots +a_{k}}{x^{2k}+b_{1}x^{2(k-1)}+\cdots +b_{k}}+ O \biggl( \frac{1}{x ^{4k+2}} \biggr), \quad x\to \infty,$$

where $$k\geq 1$$ is any given integer (see Corollary 4.1). Based on the obtained result, we establish new bounds for arctanx (Theorem 4.2).

## References

1. Shafer, R.E.: Problem E1867. Am. Math. Mon. 73, 309 (1966)

2. Shafer, R.E.: Problems E1867 (solution). Am. Math. Mon. 74, 726â€“727 (1967)

3. Qi, F., Zhang, S.Q., Guo, B.N.: Sharpening and generalizations of Shaferâ€™s inequality for the arc tangent function. J.Â Inequal. Appl. 2009, Article ID 930294 (2009). https://doi.org/10.1155/2009/930294

4. Qi, F., Niu, D.W., Guo, B.N.: Refinements, generalizations, and applications of Jordanâ€™s inequality and related problems. J.Â Inequal. Appl. 2009, Article ID 271923 (2009). https://doi.org/10.1155/2009/271923

5. Chen, X.D., Shi, J., Wang, Y., Pan, X.: A new method for sharpening the bounds of several special functions. Results Math. 72, 695â€“702 (2017). https://doi.org/10.1007/s00025-017-0700-x

6. Chen, C.P., Cheung, W.S., Wang, W.: On Shafer and Carlson inequalities. J.Â Inequal. Appl. 2011, Article ID 840206 (2011). https://doi.org/10.1155/2011/840206

7. Shafer, R.E.: On quadratic approximation. SIAM J. Numer. Anal. 11, 447â€“460 (1974). https://doi.org/10.1137/0711037

8. Shafer, R.E.: Analytic inequalities obtained by quadratic approximation. Publ. Elektroteh. Fak. Univ. Beogr., Ser. Mat. Fiz. 57(598), 96â€“97 (1977)

9. Shafer, R.E.: On quadratic approximation, II. Publ. Elektroteh. Fak. Univ. Beogr., Ser. Mat. Fiz. 602(633), 163â€“170 (1978)

10. Zhu, L.: On a quadratic estimate of Shafer. J.Â Math. Inequal. 2, 571â€“574 (2008). http://files.ele-math.com/articles/jmi-02-51.pdf

11. Sun, J.-L., Chen, C.-P.: Shafer-type inequalities for inverse trigonometric functions and Gauss lemniscate functions. J.Â Inequal. Appl. 2016, 212 (2016). https://doi.org/10.1186/s13660-016-1157-2

12. Nishizawa, Y.: Refined quadratic estimations of Shaferâ€™s inequality. J.Â Inequal. Appl. 2017, 40 (2017). https://doi.org/10.1186/s13660-017-1312-4

13. Bercu, G.: PadÃ© approximant related to remarkable inequalities involving trigonometric functions. J.Â Inequal. Appl. 2016, 99 (2016). https://doi.org/10.1186/s13660-016-1044-x

14. Bercu, G.: The natural approach of trigonometric inequalitiesâ€”PadÃ© approximant. J.Â Math. Inequal. 11, 181â€“191 (2017). https://doi.org/10.7153/jmi-11-18

15. Bercu, G., Wu, S.: Refinements of certain hyperbolic inequalities via the PadÃ© approximation method. J.Â Nonlinear Sci. Appl. 9, 5011â€“5020 (2016). https://doi.org/10.22436/jnsa.009.07.05

17. Abramowitz, M., Stegun, I.A. (eds.): Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing edn. Applied Mathematics Series, vol.Â 55. National Bureau of Standards, Washington (1972)

## Acknowledgements

We thank the editor and referees for their careful reading and valuable suggestions to make the article more easily readable.

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### Contributions

Both authors contributed equally to this work. Both authors read and approved the final manuscript.

### Corresponding author

Correspondence to Chao-Ping Chen.

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## Appendix: â€‰A derivation of (1.8), (1.9), and (1.10)

### Appendix: â€‰A derivation of (1.8), (1.9), and (1.10)

Define the function $$F(x)$$ by

$$F(x)=\frac{\arctan x}{x}-\frac{1}{a+\sqrt{b+c x^{2}}}.$$

We are interested in finding the values of the parameters a, b, and c such that $$F(x)$$ converges as fast as possible to zero, as $$x\to \infty$$. This provides the best approximations of the form

\begin{aligned} \frac{\arctan x}{x}\approx \frac{1}{a+\sqrt{b+c x^{2}}},\quad x\to \infty. \end{aligned}

Using the Maple software, we find, as $$x\to \infty$$,

\begin{aligned} F(x) ={}&\frac{\pi \sqrt{c}-2}{2\sqrt{c}x}+\frac{a-c}{cx^{2}}+\frac{b-2a ^{2}}{2c^{3/2}x^{3}} \\ &{}+ \frac{3a^{3}+c^{2}-3ab}{3c^{2} x^{4}}+O \biggl( \frac{1}{x ^{5}} \biggr). \end{aligned}

The three parameters a, b, and c, which produce the fastest convergence of the function $$F(x)$$, are given by

$$\textstyle\begin{cases} \pi \sqrt{c}-2=0, \\ a-c=0, \\ b-2a^{2}=0, \end{cases}$$

namely, if

\begin{aligned} a=\frac{4}{\pi^{2}},\qquad b=\frac{32}{\pi^{4}}, \qquad c= \frac{4}{\pi ^{2}}. \end{aligned}

We then obtain, as $$x\to \infty$$,

\begin{aligned} \frac{\arctan x}{x} &=\frac{1}{\frac{4}{\pi^{2}}+\sqrt{\frac{32}{\pi ^{4}}+\frac{4}{\pi^{2}} x^{2}}}-\frac{12-\pi^{2}}{3\pi^{2} x^{4}}+O \biggl( \frac{1}{x^{5}} \biggr) \\ &=\frac{\pi^{2}}{4+\sqrt{32+(2\pi x)^{2}}}-\frac{12-\pi^{2}}{3\pi ^{2} x^{4}}+O \biggl( \frac{1}{x^{5}} \biggr). \end{aligned}

Define the function $$G(x)$$ by

$$G(x)=\frac{\arctan x}{x}-\frac{1}{p+\sqrt{q+r x+s x^{2}}}.$$

Using the Maple software, we find, as $$x\to \infty$$,

\begin{aligned} G(x) =&\frac{\pi \sqrt{s}-2}{2\sqrt{s}x}+\frac{r-2s^{3/2}+2p \sqrt{s}}{2s^{3/2}x^{2}}+\frac{4qs-3r^{2}-8sp^{2}-8\sqrt{s}pr}{8s ^{5/2}x^{3}} \\ &{} +\frac{-48s^{3/2}pq+48\sqrt{s}pr^{2}-36rqs+15r^{3}+48s^{3/2}p ^{3}+72sp^{2}r+16s^{7/2}}{48s^{7/2}x^{4}} \\ &{} +\bigl(120sqr^{2}-128\sqrt{s}pr^{3}-128s^{2}p^{4}-240sp ^{2}r^{2}+256rs^{3/2}pq+192s^{2}p^{2}q \\ &{}-256s^{3/2}p^{3}r-35r^{4}-48q ^{2}s^{2}\bigr)/\bigl(128s^{9/2}x^{5} \bigr) \\ &{} +O \biggl( \frac{1}{x^{6}} \biggr). \end{aligned}

For

\begin{aligned} p=\frac{24-\pi^{2}}{3\pi^{2}},\qquad q=\frac{432-24\pi^{2}+\pi^{4}}{9 \pi^{4}}, \qquad r=- \frac{4(12-\pi^{2})}{3\pi^{3}},\qquad s=\frac{4}{\pi ^{2}}, \end{aligned}

we obtain, as $$x\to \infty$$,

\begin{aligned} \frac{\arctan x}{x} =&\frac{3\pi^{2}}{24-\pi^{2}+\sqrt{432-24\pi^{2}+ \pi^{4}-12\pi (12-\pi^{2})x+(6\pi x)^{2}}} \\ &{} +\frac{\pi^{4}-72}{18\pi^{3}x^{5}}+O \biggl( \frac{1}{x^{6}} \biggr). \end{aligned}

Define the function $$H(x)$$ by

$$H(x)=x \biggl( \frac{\pi }{2}-\arctan x \biggr) -\frac{x^{2}+a_{1}x+a_{2}}{x ^{2}+b_{1}x+b_{2}}.$$

Using the Maple software, we find, as $$x\to \infty$$,

\begin{aligned} H(x) =&\frac{b_{1}-a_{1}}{x}-\frac{3a_{2}-3b_{2}-3a_{1}b_{1}+3b_{1} ^{2}+1}{3x^{2}}+\frac{a_{1}b_{2}-2b_{1}b_{2}+a_{2}b_{1}-a_{1}b_{1} ^{2}+b_{1}^{3}}{x^{3}} \\ &{} -\frac{-1-5a_{2}b_{2}+5b_{2}^{2}+10a_{1}b_{1}b_{2}-15b_{1} ^{2}b_{2}+5a_{2}b_{1}^{2}-5a_{1}b_{1}^{3}+5b_{1}^{4}}{5x^{4}} \\ &{} +\frac{-a_{1}b_{2}^{2}+3b_{1}b_{2}^{2}-2a_{2}b_{1}b_{2}+3a _{1}b_{1}^{2}b_{2}-4b_{1}^{3}b_{2}+a_{2}b_{1}^{3}-a_{1}b_{1}^{4}+b _{1}^{5}}{x^{5}}+O \biggl( \frac{1}{x^{6}} \biggr). \end{aligned}

For

\begin{aligned} a_{1}=0, \qquad b_{1}=0, \qquad a_{2}= \frac{4}{15}, \qquad b_{2}= \frac{3}{5}, \end{aligned}

we obtain, as $$x\to \infty$$,

\begin{aligned} x \biggl( \frac{\pi }{2}-\arctan x \biggr) =\frac{x^{2}+\frac{4}{15}}{x ^{2}+\frac{3}{5}}+O \biggl( \frac{1}{x^{6}} \biggr). \end{aligned}

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Qiao, QX., Chen, CP. Approximations to inverse tangent function. J Inequal Appl 2018, 141 (2018). https://doi.org/10.1186/s13660-018-1734-7