1 Introduction
In 1966, Shafer [1] posed, as a problem, the following inequality:
Three proofs of it were later given in [2]. Shafer’s inequality (1.1) was sharpened and generalized by Qi et al. in [3]. A 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.
$$ \frac{3x}{1+2\sqrt{1+x^{2}}}< \arctan x, \quad x>0. $$
(1.1)
For each \(a>0\), Chen and Cheung [6] determined the largest number b and the smallest number c such that the inequalities
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
$$ \frac{b x}{1+a\sqrt{1+x^{2}}}\leq \arctan x\leq \frac{c x}{1+a\sqrt{1+x ^{2}}} $$
(1.2)
$$\begin{aligned} & \text{when } 0< a\leq \frac{\pi }{2},\quad b =\frac{\pi }{2}a,\qquad c=1+a; \\ & \text{when } \frac{\pi }{2}< a\leq \frac{2}{\pi -2},\quad b = \frac{4(a^{2}-1)}{a^{2}},\qquad c=1+a; \\ & \text{when } \frac{2}{\pi -2}< a< 2,\quad b =\frac{4(a^{2}-1)}{a^{2}},\qquad c= \frac{\pi }{2}a; \\ & \text{when } 2\leq a< \infty,\quad b =1+a, \qquad c=\frac{\pi }{2}a. \end{aligned}$$
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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 \(x>0\),
The inequality (1.3) can also be found in [9]. The inequality (1.3) is an improvement of the inequality (1.1).
$$ \frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x. $$
(1.3)
Zhu [10] developed (1.3) to produce a symmetric double inequality. More precisely, the author proved that, for \(x>0\),
where the constants \(80/3\) and \(256/\pi^{2}\) are the best possible.
$$ \frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x< \frac{8x}{3+\sqrt{25+\frac{256}{ \pi^{2}}x^{2}}}, $$
(1.4)
Remark 1.1
For \(x>0\), the following symmetric double inequality holds:
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).
$$ \frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x< \frac{\frac{2 \sqrt{15}\pi }{3}x}{3+\sqrt{25+\frac{80}{3}x^{2}}}, $$
(1.5)
Based on the following power series expansion:
Sun and Chen [11] presented a new upper bound and proved that, for \(x>0\),
Moreover, these authors pointed out that, for \(0< x< x_{0}=1.4243\ldots \) , 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:
and
$$\arctan x \biggl( 3+\sqrt{25+\frac{80}{3}x^{2}} \biggr) =8x+ \frac{32}{4725}x^{7}-\frac{64}{4725}x^{9}+ \frac{25\text{,}376}{1\text{,}299\text{,}375}x^{11}- \cdots, $$
$$ \arctan x< \frac{8x+\frac{32}{4725}x^{7}}{3+\sqrt{25+\frac{80}{3}x ^{2}}}. $$
(1.6)
$$\begin{aligned}& \arctan x-\frac{8x}{3+\sqrt{25+\frac{256}{\pi^{2}}x^{2}}}=O\bigl(x^{3}\bigr), \\& \arctan x-\frac{3x}{1+2\sqrt{1+x^{2}}}=O\bigl(x^{5}\bigr), \\& \arctan x-\frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}=O\bigl(x^{7}\bigr), \end{aligned}$$
$$\arctan x-\frac{8x+\frac{32}{4725}x^{7}}{3+\sqrt{25+\frac{80}{3}x ^{2}}}=O\bigl(x^{9}\bigr). $$
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Nishizawa [12] proved that, for \(x>0\),
where the constants \((\pi^{2}-4)^{2}\) and 32 are the best possible.
$$ \frac{\pi^{2}x}{4+\sqrt{(\pi^{2}-4)^{2}+(2\pi x)^{2}}}< \arctan x< \frac{ \pi^{2}x}{4+\sqrt{32+(2\pi x)^{2}}}, $$
(1.7)
Using the Maple software, we derive the following asymptotic formulas in the Appendix:
and
as \(x\to \infty \).
$$\begin{aligned}& \frac{\arctan x}{x}=\frac{\pi^{2}}{4+\sqrt{32+(2\pi x)^{2}}}-\frac{12- \pi^{2}}{3\pi^{2} x^{4}}+O \biggl( \frac{1}{x^{5}} \biggr), \end{aligned}$$
(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)
$$\begin{aligned} x \biggl( \frac{\pi }{2}-\arctan x \biggr) = \frac{x^{2}+\frac{4}{15}}{x ^{2}+\frac{3}{5}}+O \biggl( \frac{1}{x^{6}} \biggr) \end{aligned}$$
(1.10)
In this paper, motivated by (1.9), we establish a symmetric double inequality for arctanx. Based on the Padé 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
where \(k\geq 1\) is any given integer. Based on the obtained result, we establish new bounds for arctanx.
$$x \biggl( \frac{\pi }{2}-\arctan x \biggr) =\frac{x^{2k}+a_{1}x^{2(k-1)}+ \cdots +a_{k}}{x^{2k}+b_{1}x^{2(k-1)}+\cdots +b_{k}}+ O \biggl( \frac{1}{x ^{4k+2}} \biggr), \quad x\to \infty, $$
Some computations in this paper were performed using Maple software.
2 Lemma
It is well known that
and
for \(x>0\) and \(n\in \mathbb{N}_{0}:=\mathbb{N}\cup \{0\}\), where \(\mathbb{N}\) denotes the set of positive integers.
$$\begin{aligned} \sum_{k=0}^{2n+1}(-1)^{k} \frac{x^{2k+1}}{(2k+1)!}< \sin x< \sum_{k=0} ^{2n}(-1)^{k}\frac{x^{2k+1}}{(2k+1)!} \end{aligned}$$
(2.1)
$$\begin{aligned} \sum_{k=0}^{2n+1}(-1)^{k} \frac{x^{2k}}{(2k)!}< \cos x< \sum_{k=0}^{2n}(-1)^{k} \frac{x ^{2k}}{(2k)!} \end{aligned}$$
(2.2)
The following lemma will be used in our present investigation.
Lemma 2.1
For
\(0< u<\pi /2\),
and
$$\begin{aligned} &\cos u\sin^{2}u>u^{2}-\frac{5}{6}u^{4}+ \frac{91}{360}u^{6}- \frac{41}{1008}u^{8} \end{aligned}$$
(2.3)
$$\begin{aligned} \sin^{3} u>u^{3}-\frac{1}{2}u^{5}+ \frac{13}{120}u^{7}-\frac{41}{3024}u ^{9}. \end{aligned}$$
(2.4)
Proof
We find that
and
where
$$\begin{aligned} \cos u\sin^{2}u =&\frac{1}{4} \bigl(\cos u-\cos (3u) \bigr) \\ =&u^{2}- \frac{5}{6}u^{4}+\frac{91}{360}u^{6}- \frac{41}{1008}u^{8}+\sum_{n=5} ^{\infty }(-1)^{n-1}w_{n}(u) \end{aligned}$$
(2.5)
$$\begin{aligned} \sin^{3} u =&\frac{1}{4} \bigl(3\sin u-\sin (3u) \bigr) \\ =&u^{3}-\frac{1}{2}u ^{5}+\frac{13}{120}u^{7}- \frac{41}{3024}u^{9}+\sum_{n=5}^{\infty }(-1)^{n-1}W _{n}(u), \end{aligned}$$
(2.6)
$$\begin{aligned} w_{n}(u)=\frac{9^{n}-1}{(2n)!}u^{2n} \quad \text{and}\quad W_{n}(u)=\frac{3(9^{n}-1)}{4 \cdot (2n+1)!}u^{2n+1}. \end{aligned}$$
Elementary calculations reveal that, for \(0< u<\pi /2\) and \(n\geq 5\),
and
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). □
$$\begin{aligned} \frac{w_{n+1}(u)}{w_{n}(u)} &= \frac{u^{2}(9^{n+1}-1)}{2(2n+1)(n+1)(9^{n}-1)}< \frac{(\pi /2)^{2}(9^{n+1}-1)}{2(2n+1)(n+1)(9^{n}-1)} \\ &< \frac{3\cdot 9^{n+1}}{2(2n+1)(n+1)(9^{n}-1)}= \frac{27}{2(2n+1)(n+1)} \biggl\{ 1+\frac{1}{9^{n}-1} \biggr\} \\ &\leq \frac{27}{2(2n+1)(n+1)} \biggl\{ 1+\frac{1}{9^{5}-1} \biggr\} = \frac{1\text{,}594\text{,}323}{118\text{,}096(2n+1)(n+1)}< 1 \end{aligned}$$
$$\begin{aligned} \frac{W_{n+1}(u)}{W_{n}(u)} &= \frac{u^{2}(9^{n+1}-1)}{2(2n+3)(n+1)(9^{n}-1)}< \frac{w_{n+1}(u)}{w _{n}(u)}< 1. \end{aligned}$$
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
with the best possible constants
$$\begin{aligned} &\frac{3\pi^{2}x}{24-\pi^{2}+\sqrt{\alpha -12\pi (12-\pi^{2})x+36\pi ^{2}x^{2}}} \\ &\quad < \arctan x \\ &\quad < \frac{3\pi^{2}x}{24-\pi^{2}+\sqrt{\beta -12\pi (12-\pi^{2})x+36\pi ^{2}x^{2}}}, \end{aligned}$$
(3.1)
$$ \begin{aligned} &\alpha =432-24\pi^{2}+ \pi^{4}=292.538\ldots \quad \textit{and} \\ &\beta =576-192\pi^{2}+ 16\pi^{4}=239.581\ldots. \end{aligned} $$
(3.2)
Proof
The inequality (3.1) can be written for \(x>0\) as
By the elementary change of variable \(t = \arctan x\) (\(x>0\)), (3.3) becomes
where
Elementary calculations reveal that
In order to prove (3.4), it suffices to show that \(\vartheta (t)\) is strictly increasing for \(0 < t < \pi /2\).
$$\begin{aligned} \beta < \biggl( \frac{3\pi^{2} x^{2}}{\arctan x}-\bigl(24-\pi^{2} \bigr) \biggr) ^{2}+12 \pi \bigl(12-\pi^{2}\bigr)x-36 \pi^{2} x^{2}< \alpha. \end{aligned}$$
(3.3)
$$\begin{aligned} \beta < \vartheta (t)< \alpha,\quad 0< t< \frac{\pi }{2}, \end{aligned}$$
(3.4)
$$\begin{aligned} \vartheta (t)={}& \biggl( \frac{3\pi^{2}\tan^{2} t}{t}-\bigl(24-\pi^{2}\bigr) \biggr) ^{2} \\ &{}+12\pi \bigl(12-\pi^{2}\bigr)\tan t-36 \pi^{2}\tan^{2}t. \end{aligned}$$
$$\begin{aligned} &\lim_{t\to 0^{+}}\vartheta (t)=576-192\pi^{2}+ 16 \pi^{4} \quad \text{and} \\ & \lim_{t\to \pi /2^{-}}\vartheta (t)=432-24\pi^{2}+\pi ^{4}. \end{aligned}$$
Differentiation yields
We now consider two cases to prove \(\lambda (t)>0\) for \(0< t<\pi /2\).
$$\begin{aligned} t^{3}\cos^{3}t\vartheta '(t) =&\bigl(24\pi t- \pi^{3} t\bigr)\sin t\cos^{2}t+\bigl(3\pi ^{3} t-12 \pi t^{3}\bigr)\sin t \\ &{} - \bigl(3\pi^{3}+\bigl(24\pi -\pi^{3}\bigr) t^{2}-\bigl(24-2\pi^{2}\bigr) t^{3} \bigr)\cos t+3 \pi^{3}\cos^{3} t \\ =:&\lambda (t). \end{aligned}$$
Case 1: \(0< t\leq 0.6\).
Using (2.1) and (2.2), we have, for \(0< t\leq 0.6\),
Each function in curly braces is positive for \(t\in (0,0.6]\). Thus, \(\lambda (t)>0\) for \(t\in (0,0.6]\).
$$\begin{aligned} \lambda (t) =& \biggl( 6\pi -\frac{1}{4}\pi^{3} \biggr) t\sin (3t)+ \frac{3}{4}\pi^{3}\cos (3t)+ \biggl\{ \biggl( 6\pi + \frac{11}{4}\pi^{3} \biggr) t-12 \pi t^{3} \biggr\} \sin t \\ &{} - \biggl( \frac{3}{4}\pi^{3}-\bigl(\pi^{3}-24 \pi\bigr)t^{2}-\bigl(24-2\pi^{2}\bigr)t ^{3} \biggr) \cos t \\ >& \biggl( 6\pi -\frac{1}{4}\pi^{3} \biggr) t \biggl( 3t- \frac{9}{2}t^{3}+ \frac{81}{40}t^{5}- \frac{243}{560}t^{7} \biggr) \\ &{}+\frac{3}{4}\pi^{3} \biggl( 1-\frac{9}{2}t^{2}+\frac{27}{8}t^{4}- \frac{81}{80}t^{6} \biggr) \\ &{} + \biggl\{ \biggl( 6\pi +\frac{11}{4}\pi^{3} \biggr) t-12 \pi t^{3} \biggr\} \biggl( t-\frac{1}{6}t^{3} \biggr) \\ &{} - \biggl( \frac{3}{4}\pi^{3}-\bigl(\pi^{3}-24\pi \bigr)t^{2}-\bigl(24-2\pi^{2}\bigr)t ^{3} \biggr) \biggl( 1-\frac{1}{2}t^{2}+\frac{1}{24}t^{4} \biggr) \\ =&t^{3} \biggl\{ 24-2\pi^{2}- \biggl( 28\pi - \frac{8}{3}\pi^{3} \biggr) t- \bigl( 12-\pi^{2} \bigr) t^{2} \biggr\} \\ &{} +t^{6} \biggl\{ \frac{263}{20}\pi -\frac{235}{192} \pi^{3}+ \biggl( 1- \frac{1}{12}\pi^{2} \biggr) t- \biggl( \frac{729}{280}\pi - \frac{243}{2240}\pi^{3} \biggr) t^{2} \biggr\} . \end{aligned}$$
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:
where
$$\begin{aligned} \mu (u)>0,\quad 0< u< \frac{\pi }{2}-0.6, \end{aligned}$$
$$\begin{aligned} \mu (u) ={}&\bigl(24\pi -\pi^{3} \bigr) \biggl( \frac{\pi }{2}-u \biggr) \cos u\sin^{2}u+ \biggl\{ 3\pi^{3} \biggl( \frac{\pi }{2}-u \biggr) -12\pi \biggl( \frac{ \pi }{2}-u \biggr) ^{3} \biggr\} \cos u \\ &{} - \biggl\{ 3\pi^{3}+\bigl(24\pi -\pi^{3}\bigr) \biggl( \frac{\pi }{2}-u \biggr) ^{2}-\bigl(24-2\pi^{2}\bigr) \biggl( \frac{\pi }{2}-u \biggr) ^{3} \biggr\} \sin u+3 \pi^{3}\sin^{3} u. \end{aligned}$$
Using (2.1)–(2.4), we have, for \(0< u<\frac{\pi }{2}-0.6\),
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. □
$$\begin{aligned} \mu (u) >&\bigl(24\pi -\pi^{3} \bigr) \biggl( \frac{\pi }{2}-u \biggr) \biggl( u^{2}- \frac{5}{6}u^{4}+ \frac{91}{360}u^{6}-\frac{41}{1008}u^{8} \biggr) \\ &{} + \biggl\{ 3\pi^{3} \biggl( \frac{\pi }{2}-u \biggr) -12\pi \biggl( \frac{ \pi }{2}-u \biggr) ^{3} \biggr\} \biggl( 1- \frac{1}{2}u^{2}+\frac{1}{24}u ^{4}- \frac{1}{720}u^{6} \biggr) \\ &{} - \biggl\{ 3\pi^{3}+\bigl(24\pi -\pi^{3}\bigr) \biggl( \frac{\pi }{2}-u \biggr) ^{2}-\bigl(24-2\pi^{2}\bigr) \biggl( \frac{\pi }{2}-u \biggr) ^{3} \biggr\} \biggl( u- \frac{1}{6}u^{3}+\frac{1}{120}u^{5} \biggr) \\ &{} +3\pi^{3} \biggl( u^{3}-\frac{1}{2}u^{5}+ \frac{13}{120}u^{7}- \frac{41}{3024}u^{9} \biggr) \\ =&u^{4} \biggl\{ \frac{1}{3}\pi^{4}-24+ \biggl( 12\pi -\frac{9}{5}\pi^{3} \biggr) u+ \biggl( 2\pi^{2}- \frac{11}{90}\pi^{4}+4 \biggr) u^{2} \\ &{}+ \biggl( - \frac{82}{15}\pi +\frac{199}{360}\pi^{3} \biggr) u^{3} \\ &{} + \biggl( -\frac{1}{5}-\frac{25}{56}\pi^{2}+ \frac{41}{2016}\pi^{4} \biggr) u ^{4}+ \biggl( \frac{403}{420}\pi -\frac{41}{504}\pi^{3} \biggr) u^{5} \biggr\} >0. \end{aligned}$$
From (1.7) and (3.1), we obtain the following approximation formulas:
and
as \(n\to \infty \).
$$ \frac{\arctan n}{n} \approx \frac{\pi^{2}}{4+\sqrt{32+(2\pi n)^{2}}}=:a_{n} $$
(3.5)
$$\begin{aligned} \frac{\arctan n}{n}\approx \frac{3\pi^{2}}{24-\pi^{2}+\sqrt{432-24 \pi^{2}+\pi^{4}-12\pi (12-\pi^{2})n+(6\pi n)^{2}}}=:b_{n}, \end{aligned}$$
(3.6)
The following numerical computations (see Table 1) would show that, for \(n\in \mathbb{N}\), Eq. (3.6) is sharper than Eq. (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 Padé approximant (see [13‐16]). Let f be a formal power series,
The Padé approximation of order \((p, q)\) of the function f is the rational function, denoted by
where \(p\geq 0\) and \(q\geq 1\) are two given integers, the coefficients \(a_{j}\) and \(b_{j}\) are given by (see [13‐15])
and the following holds:
Thus, the first \(p + q + 1\) coefficients of the series expansion of \([p/q]_{f}\) are identical to those of f.
$$\begin{aligned} f(t)=c_{0}+c_{1}t+c_{2}t^{2}+ \cdots. \end{aligned}$$
(4.1)
$$\begin{aligned}{} [p/q]_{f}(t)=\frac{\sum_{j=0}^{p}a_{j}t^{j}}{1+\sum_{j=1}^{q}b_{j}t ^{j}}, \end{aligned}$$
(4.2)
$$\begin{aligned} \textstyle\begin{cases} a_{0}=c_{0}, \\ a_{1}=c_{0}b_{1}+c_{1}, \\ a_{2}=c_{0}b_{2}+c_{1}b_{1}+c_{2}, \\ \vdots \\ a_{p} = c_{0}b_{p}+\cdots + c_{p-1}b_{1} + c_{p}, \\ 0 = c_{p+1} + c_{p}b_{1} + \cdots + c_{p-q+1}b_{q}, \\ \vdots & \\ 0 = c_{p+q} + c_{p+q-1}b_{1} + \cdots + c_{p}b_{q}, \end{cases}\displaystyle \end{aligned}$$
(4.3)
$$\begin{aligned}{} [p/q]_{f}(t)- f (t) = O\bigl(t^{p+q+1} \bigr). \end{aligned}$$
(4.4)
From the expansion (see [17, p. 81])
we obtain
where
$$\begin{aligned} \arctan x=\frac{\pi }{2}+\sum_{j=1}^{\infty } \frac{(-1)^{j}}{(2j-1)x ^{2j-1}},\quad \vert x\vert >1, \end{aligned}$$
$$\begin{aligned} x \biggl( \frac{\pi }{2}-\arctan x \biggr) =\sum _{j=0}^{\infty }\frac{c _{j}}{x^{2j}}=1-\frac{1}{3x^{2}}+ \frac{1}{5x^{4}}-\frac{1}{7x^{6}}+ \cdots, \end{aligned}$$
(4.5)
$$\begin{aligned} c_{j}=\frac{(-1)^{j}}{2j+1}\quad \text{for } j\geq 0. \end{aligned}$$
(4.6)
Let
with the coefficients \(c_{j}\) given in (4.6). Then we have
In what follows, the function f is given in (4.7).
$$\begin{aligned} f(t)=\sum_{j=0}^{\infty } \frac{c_{j}}{t^{j}}, \end{aligned}$$
(4.7)
$$\begin{aligned} f\bigl(x^{2}\bigr)=\sum_{j=0}^{\infty } \frac{c_{j}}{x^{2j}}=x \biggl( \frac{\pi }{2}- \arctan x \biggr). \end{aligned}$$
(4.8)
Based on the Padé approximation method, we now give a derivation of Eq. (1.10). To this end, we consider
Noting that
holds, we have, by (4.3),
that is,
We thus obtain
and we have, by (4.4),
Replacing t by \(x^{2}\) in (4.11) yields (1.10).
$$\begin{aligned}{} [1/1]_{f}(t)=\frac{\sum_{j=0}^{1}a_{j}t^{-j}}{1+\sum_{j=1}^{1}b_{j}t ^{-j}}. \end{aligned}$$
$$\begin{aligned} c_{0}=1, \qquad c_{1}=-\frac{1}{3}, \qquad c_{2}=\frac{1}{5}, \end{aligned}$$
(4.9)
$$\begin{aligned} \textstyle\begin{cases} a_{0}=1, \\ a_{1}=b_{1}-\frac{1}{3}, \\ 0 =\frac{1}{5} -\frac{1}{3}b_{1}, \end{cases}\displaystyle \end{aligned}$$
$$\begin{aligned} a_{0}=1,\qquad a_{1}=\frac{4}{15}, \qquad b_{1}=\frac{3}{5}. \end{aligned}$$
$$ [1/1]_{f}(t)= \frac{1+\frac{4}{15t}}{1+\frac{3}{5t}}= \frac{15t+4}{3(5t+3)}, $$
(4.10)
$$ f(t)=\frac{15t+4}{3(5t+3)}+O \biggl( \frac{1}{t^{3}} \biggr), \quad t\to \infty. $$
(4.11)
From the Padé approximation method and the expansion (4.7), we now present a general result.
Theorem 4.1
The Padé 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:
where
\(p\geq 1\)
and
\(q\geq 1\)
are any given integers, the coefficients
\(a_{j}\)
and
\(b_{j}\)
are given by
and
\(c_{j}\)
is given in (4.6), and the following holds:
In particular, replacing
t
by
\(x^{2}\)
in (4.14) yields
with the coefficients
\(a_{j}\)
and
\(b_{j}\)
given by (4.13).
$$\begin{aligned}{} [p/q]_{f}(t)&=\frac{1+\sum_{j=1}^{p}a_{j}t^{-j}}{1+\sum_{j=1}^{q}b_{j}t ^{-j}} \\ &=t^{q-p} \biggl( \frac{t^{p}+a_{1}t^{p-1}+\cdots +a_{p}}{t^{q}+b _{1}t^{q-1}+\cdots +b_{q}} \biggr), \end{aligned}$$
(4.12)
$$\begin{aligned} \textstyle\begin{cases} a_{1}=b_{1}+c_{1}, \\ a_{2}=b_{2}+c_{1}b_{1}+c_{2}, \\ \vdots \\ a_{p} = b_{p}+\cdots + c_{p-1}b_{1} + c_{p}, \\ 0 = c_{p+1} + c_{p}b_{1} + \cdots + c_{p-q+1}b_{q}, \\ \vdots & \\ 0 = c_{p+q} + c_{p+q-1}b_{1} + \cdots + c_{p}b_{q}, \end{cases}\displaystyle \end{aligned}$$
(4.13)
$$\begin{aligned} f (t) -[p/q]_{f}(t) = O \biggl( \frac{1}{t^{p+q+1}} \biggr),\quad t\to \infty. \end{aligned}$$
(4.14)
$$\begin{aligned} &x \biggl( \frac{\pi }{2}-\arctan x \biggr) \\ &\quad =x^{2(q-p)} \biggl( \frac{x^{2p}+a _{1}x^{2(p-1)}+\cdots +a_{p}}{x^{2q}+b_{1}x^{2(q-1)}+\cdots +b_{q}} \biggr) + O \biggl( \frac{1}{x^{2(p+q+1)}} \biggr), \quad x\to \infty, \end{aligned}$$
(4.15)
Setting \((p, q)=(k, k)\) in (4.15), we obtain the following corollary.
Corollary 4.1
As
\(x\to \infty \),
where
\(k\geq 1\)
is any given integer, the coefficients
\(a_{j}\)
and
\(b_{j}\) (\(1\leq j\leq k\)) are given by
and
\(c_{j}\)
is given in (4.6).
$$\begin{aligned} x \biggl( \frac{\pi }{2}-\arctan x \biggr) = \frac{x^{2k}+a_{1}x^{2(k-1)}+ \cdots +a_{k}}{x^{2k}+b_{1}x^{2(k-1)}+\cdots +b_{k}}+ O \biggl( \frac{1}{x ^{4k+2}} \biggr), \end{aligned}$$
(4.16)
$$\begin{aligned} \textstyle\begin{cases} a_{1}=b_{1}+c_{1}, \\ a_{2}=b_{2}+c_{1}b_{1}+c_{2}, \\ \vdots \\ a_{k} = b_{k}+\cdots + c_{k-1}b_{1} + c_{k}, \\ 0 = c_{k+1} + c_{k}b_{1} + \cdots + c_{1}b_{k}, \\ \vdots & \\ 0 = c_{2k} + c_{2k-1}b_{1} + \cdots + c_{k}b_{k}, \end{cases}\displaystyle \end{aligned}$$
(4.17)
Setting \(k=2\) in (4.16) yields, as \(x\to \infty \),
which gives
$$\begin{aligned} x \biggl( \frac{\pi }{2}-\arctan x \biggr) = \frac{945x^{4}+735x^{2}+64}{15(63x ^{4}+70x^{2}+15)}+O \biggl( \frac{1}{x^{10}} \biggr), \end{aligned}$$
(4.18)
$$\begin{aligned} \arctan x=\frac{\pi }{2}-\frac{945x^{4}+735x^{2}+64}{15x(63x^{4}+70x ^{2}+15)}+ O \biggl( \frac{1}{x^{11}} \biggr). \end{aligned}$$
Using the Maple software, we find, as \(x\to \infty \),
Equation (4.19) motivated us to establish new bounds for arctanx.
$$\begin{aligned} \arctan x={}&\frac{\pi }{2}-\frac{945x^{4}+735x^{2}+64}{15x(63x^{4}+70x ^{2}+15)}+ \frac{64}{43\text{,}659x^{11}} \\ &{}-\frac{1856}{464\text{,}373x^{13}}+ O \biggl( \frac{1}{x ^{15}} \biggr). \end{aligned}$$
(4.19)
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
and
Differentiation yields
and
Hence, \(I(x)\) is strictly decreasing and \(J(x)\) is strictly increasing for \(x>0\), and we have
The proof is complete. □
$$\begin{aligned} I(x)=\arctan x- \biggl( \frac{\pi }{2}-\frac{945x^{4}+735x^{2}+64}{15x(63x ^{4}+70x^{2}+15)}+ \frac{64}{43\text{,}659x^{11}}-\frac{1856}{464\text{,}373x^{13}} \biggr) \end{aligned}$$
$$\begin{aligned} J(x)=\arctan x- \biggl( \frac{\pi }{2}-\frac{945x^{4}+735x^{2}+64}{15x(63x ^{4}+70x^{2}+15)}+ \frac{64}{43\text{,}659x^{11}} \biggr). \end{aligned}$$
$$\begin{aligned} I'(x)=-\frac{64(230\text{,}391x^{8}+372\text{,}680x^{6}+236\text{,}885x^{4}+65\text{,}400x^{2}+6525)}{35\text{,}721x ^{14}(1+x^{2})(63x^{4}+70x^{2}+15)^{2}}< 0 \end{aligned}$$
$$\begin{aligned} J'(x)=\frac{64(12\text{,}789x^{8}+15\text{,}610x^{6}+8890x^{4}+2325x^{2}+225)}{3969x ^{12}(1+x^{2})(63x^{4}+70x^{2}+15)^{2}}>0. \end{aligned}$$
$$\begin{aligned} I(x)>\lim_{t\to \infty }I(t)=0 \quad \text{and}\quad J(x)< \lim _{t\to \infty }J(t)=0 \quad \text{for } x>0. \end{aligned}$$
Remark 4.1
We point out that, for \(x>1.0213\ldots \) , the lower bound in (4.20) is better than the one in (1.7). For \(x>0.854439\ldots \) , the upper bound in (4.20) is better than the one in (1.7). For \(x>0.947273\ldots \) , the lower bound in (4.20) is better than the one in (3.1). For \(x>0.792793\ldots \) , 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
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).
$$x \biggl( \frac{\pi }{2}-\arctan x \biggr) =\frac{x^{2k}+a_{1}x^{2(k-1)}+ \cdots +a_{k}}{x^{2k}+b_{1}x^{2(k-1)}+\cdots +b_{k}}+ O \biggl( \frac{1}{x ^{4k+2}} \biggr), \quad x\to \infty, $$
Acknowledgements
We thank the editor and referees for their careful reading and valuable suggestions to make the article more easily readable.
Competing interests
The authors declare that they have no competing interests.
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