Open Access

Shafer-type inequalities for inverse trigonometric functions and Gauss lemniscate functions

Journal of Inequalities and Applications20162016:212

https://doi.org/10.1186/s13660-016-1157-2

Received: 1 July 2016

Accepted: 31 August 2016

Published: 7 September 2016

Abstract

In this paper, we present Shafer-type inequalities for inverse trigonometric functions and Gauss lemniscate functions.

Keywords

inverse trigonometric functions lemniscate function inequalities

MSC

26D07

1 Introduction

Shafer [1] indicated several elementary quadratic approximations of selected functions without proof. Subsequently, Shafer [2] established these results as analytic inequalities. For example, Shafer [2] proved that for \(x>0\),
$$ \frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x. $$
(1.1)
The inequality (1.1) can also be found in [3]. Also in [2], Shafer proved that for \(0< x<1\),
$$ \frac{8x}{3\sqrt{1-x^{2}}+\sqrt{25+\frac{5}{3}x^{2}}}< \arcsin x. $$
(1.2)
Zhu [4] proved that the function
$$ F(x)=\frac{ (\frac{8x}{\arctan x}-3 )^{2}-25}{x^{2}} $$
is strictly decreasing for \(x>0\), and
$$ \lim_{x\to0^{+}}F(x)=\frac{80}{3}\quad \text{and} \quad \lim _{x\to \infty}F(x)=\frac{256}{\pi^{2}}. $$
From this one derives the following double inequality:
$$ \frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x< \frac{8x}{3+\sqrt {25+\frac{256}{\pi^{2}}x^{2}}}, \quad x>0. $$
(1.3)
The constants \(80/3\) and \(256/\pi^{2}\) are the best possible. In [4], (1.3) is called Shafer-type inequality.
Using the Maple software, we find that
$$ \arctan x \biggl(3+\sqrt{25+\frac{80}{3}x^{2}} \biggr)=8x+ \frac {32}{4\text{,}725}x^{7}-\frac{64}{4\text{,}725}x^{9}+ \frac{25\text{,}376}{1\text{,}299\text{,}375}x^{11}-\cdots. $$
This fact motivated us to present a new upper bound for arctanx, which is the first aim of the present paper.

Theorem 1.1

For \(x>0\),
$$ \arctan x< \frac{8x+\frac{32}{4\text{,}725}x^{7}}{3+\sqrt{25+\frac{80}{3}x^{2}}}. $$
(1.4)

The second aim of the present paper is to develop (1.2) to produce a symmetric double inequality.

Theorem 1.2

For \(0< x<1\), we have
$$ \frac{8x}{3\sqrt{1-x^{2}}+\sqrt{25+ax^{2}}}< \arcsin x< \frac{8x}{3\sqrt {1-x^{2}}+\sqrt{25+bx^{2}}} $$
(1.5)
with the best possible constants
$$ a=\frac{5}{3}=1.666666\ldots \quad \textit{and}\quad b= \frac{256-25\pi ^{2}}{\pi^{2}}=0.938223\ldots. $$
(1.6)

Recently, some famous inequalities for trigonometric and inverse trigonometric functions have been improved (see, for example, [58]).

The lemniscate, also called the lemniscate of Bernoulli, is the locus of points \((x, y)\) in the plane satisfying the equation \((x^{2} + y^{2})^{2} = x^{2} + y^{2}\). In polar coordinates \((r, \theta)\), the equation becomes \(r^{2} = \cos(2\theta)\) and its arc length is given by the function
$$ \operatorname {arcsl}x= \int_{0}^{x}\frac{1}{\sqrt{1-t^{4}}}\,\mathrm{d}t,\quad |x|\leq1, $$
(1.7)
where arcslx is called the arc lemniscate sine function studied by Gauss in 1797-1798. Another lemniscate function investigated by Gauss is the hyperbolic arc lemniscate sine function, defined as
$$ \operatorname {arcslh}x= \int_{0}^{x}\frac{1}{\sqrt{1+t^{4}}}\,\mathrm{d}t,\quad x \in \mathbb{R}. $$
(1.8)
Functions (1.7) and (1.8) can be found (see [9], Chapter 1, [10], p.259 and [1119]).
Another pair of lemniscate functions, the arc lemniscate tangent arctl and the hyperbolic arc lemniscate tangent arctlh, have been introduced in [12], (3.1)-(3.2). Therein it has been proven that
$$ \operatorname {arctl}x=\operatorname {arcsl}\biggl(\frac{x}{\sqrt[4]{1+x^{4}}} \biggr), \quad x\in \mathbb{R} $$
(1.9)
and
$$ \operatorname {arctlh}x=\operatorname {arcslh}\biggl(\frac{x}{\sqrt[4]{1-x^{4}}} \biggr),\quad |x|< 1 $$
(1.10)
(see [12], Proposition 3.1).

In analogy with (1.1), we here establish Shafer-type inequalities for the lemniscate functions, which is the last aim of the present paper.

Theorem 1.3

For \(0< x<1\),
$$ \frac{10x}{5+\sqrt{25-10x^{4}}}< \operatorname {arcsl}x $$
(1.11)
and
$$ \frac{10x}{5+\sqrt{25-15x^{4}}}< \operatorname {arctlh}x. $$
(1.12)

Theorem 1.4

For \(x>0\),
$$ \frac{95x}{80+\sqrt{225+285x^{4}}}< \operatorname {arcslh}x. $$
(1.13)

We present the following conjecture.

Conjecture 1.1

For \(x>0\),
$$ \operatorname {arcslh}x< \frac{95x+\frac{931}{2\text{,}925}x^{13}}{80+\sqrt{225+285x^{4}}} $$
(1.14)
and
$$ \frac{1\text{,}210x}{940+9\sqrt{900+1\text{,}210x^{4}}}< \operatorname {arctl}x< \frac{1\text{,}210x+\frac {2\text{,}078\text{,}417}{280\text{,}800}x^{13}}{940+9\sqrt{900+1\text{,}210x^{4}}}. $$
(1.15)

2 Lemmas

The following lemmas have been proved in [17].

Lemma 2.1

For \(|x|<1\),
$$ \operatorname {arcsl}x =\sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi }(4n+1)\cdot n!}x^{4n+1}=x+\frac{1}{10} x^{5}+ \frac{1}{24} x^{9}+\cdots. $$
(2.1)

Lemma 2.2

For \(0< x<1\),
$$ \operatorname {arctlh}x=\sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{3}{4})}{\Gamma(\frac {3}{4})\cdot (4n+1)\cdot n!}x^{4n+1}=x+\frac{3}{20}x^{5}+ \frac{7}{96}x^{9}+\cdots. $$
(2.2)

3 Proofs of Theorems 1.1 to 1.4

Proof of Theorem 1.1

The inequality (1.11) is obtained by considering the function \(f(x)\) defined by
$$ f(x)=\arctan x-\frac{8x+\frac{32}{4\text{,}725}x^{7}}{3+\sqrt{25+\frac {80}{3}x^{2}}}, \quad x>0. $$
Differentiation yields
$$ f'(x)=-\frac{2g(x)}{(1+x^{2})(9+\sqrt{225+240x^{2}})^{2}\sqrt{225+240x^{2}}}, $$
where
$$\begin{aligned} g(x) =&\bigl(-7\text{,}875-2\text{,}100x^{2}+112x^{6}+112x^{8} \bigr)\sqrt{225+240x^{2}} \\ &{} +118\text{,}125+94\text{,}500x^{2}+2\text{,}800x^{6}+5 \text{,}360x^{8}+2\text{,}560x^{10}. \end{aligned}$$
We now show that
$$ g(x)>0, \quad x>0. $$
(3.1)
By an elementary change of variable
$$ t=\sqrt{225+240x^{2}}, \quad t>15, $$
the inequality (3.1) is equivalent to
$$\begin{aligned}& \biggl[-7\text{,}875-2\text{,}100 \biggl(\frac{t^{2}-225}{240} \biggr)+112 \biggl( \frac {t^{2}-225}{240} \biggr)^{3}+112 \biggl(\frac{t^{2}-225}{240} \biggr)^{4} \biggr]t \\& \qquad {}+118\text{,}125+94\text{,}500 \biggl(\frac{t^{2}-225}{240} \biggr)+2 \text{,}800 \biggl(\frac{t^{2}-225}{240} \biggr)^{3}+5\text{,}360 \biggl( \frac{t^{2}-225}{240} \biggr)^{4} \\& \qquad {}+2\text{,}560 \biggl(\frac{t^{2}-225}{240} \biggr)^{5} \\& \quad =\frac {(2t^{6}+141t^{5}+4\text{,}515t^{4}+93\text{,}690t^{3}+1\text{,}562\text{,}400t^{2}+24\text{,}053\text{,}625t+362\text{,}626\text{,}875)(t-15)^{4}}{622\text{,}080\text{,}000} \\& \quad >0\quad \text{for } t>15, \end{aligned}$$
which is true. Hence, we have
$$ g(x)>0\quad \text{and} \quad f'(x)< 0\quad \text{for } x>0. $$
So, \(f(x)\) is strictly decreasing for \(x>0\), and we have
$$ f(x)< f(0)=0, \quad x>0. $$
The proof is complete. □

Remark 3.1

Let \(x_{0}=1.4243\ldots\) . Then we have
$$ \frac{8x+\frac{32}{4\text{,}725}x^{7}}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \frac {8x}{3+\sqrt{25+\frac{256}{\pi^{2}}x^{2}}},\quad 0< x< x_{0}. $$
This shows that for \(0< x< x_{0}\), the upper bound in (1.11) is better than the upper bound in (1.3). In fact, for \(x\to0\), we have
$$\begin{aligned}& \arctan x-\frac{8x}{3+\sqrt{25+\frac{256}{\pi^{2}}x^{2}}}=O\bigl(x^{3}\bigr), \\& \arctan x-\frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}=O\bigl(x^{7}\bigr), \end{aligned}$$
and
$$ \arctan x-\frac{8x+\frac{32}{4\text{,}725}x^{7}}{3+\sqrt{25+\frac{80}{3}x^{2}}}=O\bigl(x^{9}\bigr). $$

Proof of Theorem 1.2

The double inequality (1.11) can be written for \(0< x<1\) as
$$ b< F(x)< a, $$
where
$$ F(x)=\frac{ (\frac{8x}{\arcsin x}-3\sqrt{1-x^{2}} )^{2}-25}{x^{2}}, \quad 0< x< 1. $$
By an elementary change of variable,
$$ x=\sin t,\quad 0< t< \frac{\pi}{2}, $$
we have
$$ G(t)=F(\sin t)=\frac{ (\frac{8\sin t}{t}-3\cos t )^{2}-25}{\sin^{2}t},\quad 0< t< \frac{\pi}{2}. $$
We now prove that \(F(x)\) is strictly decreasing for \(0< x<1\). It suffices to show that \(G(t)\) is strictly decreasing for \(0< t<\pi/2\). Differentiation yields
$$\begin{aligned} \begin{aligned}[b] -\frac{t^{3}\sin^{3}t}{16}G'(t)={}&8\sin^{3} t-3t^{2}\sin t-\bigl(2t^{3}+3t\bigr)\cos t+3t \cos^{3}t \\ ={}&8 \biggl(\frac{3\sin t-\sin(3t)}{4} \biggr)-3t^{2}\sin t- \bigl(2t^{3}+3t\bigr)\cos t \\ &{}+3t \biggl(\frac{\cos(3t)+3\cos t}{4} \biggr) \\ ={}&\bigl(6-3t^{2}\bigr)\sin t-2\sin(3t)+\frac{3}{4}t\cos(3t)- \biggl(\frac {3}{4}t+2t^{3} \biggr)\cos t \\ ={}&\frac{16}{945}t^{9}-\frac{16}{4\text{,}725}t^{11}+\sum _{n=6}^{\infty }(-1)^{n}u_{n}(t), \end{aligned} \end{aligned}$$
(3.2)
where
$$ u_{n}(t)=\frac{21+2n+48n^{2}+64n^{3}+(6n-21)\cdot9^{n}}{4\cdot(2n+1)!}t^{2n+1}. $$
Elementary calculations reveal that for \(0< t<\pi/2\) and \(n\geq6\),
$$\begin{aligned} \frac{u_{n+1}(t)}{u_{n}(t)}&=\frac{t^{2} (135+290n+240n^{2}+64n^{3}+(54n-135)\cdot9^{n} )}{2(n+1)(2n+3) (21+2n+48n^{2}+64n^{3}+(6n-21)\cdot9^{n} )} \\ &< \frac{(\pi/2)^{2}}{n+1}\frac{135+290n+240n^{2}+64n^{3}+(54n-135)\cdot 9^{n}}{2(2n+3) (21+2n+48n^{2}+64n^{3}+(6n-21)\cdot9^{n} )} \\ &< \frac{135+290n+240n^{2}+64n^{3}+(54n-135)\cdot9^{n}}{2(2n+3) (21+2n+48n^{2}+64n^{3}+(6n-21)\cdot9^{n} )} \end{aligned}$$
and
$$\begin{aligned}& 2(2n+3) \bigl(21+2n+48n^{2}+64n^{3}+(6n-21) \cdot9^{n} \bigr) \\& \qquad {}- \bigl(135+290n+240n^{2}+64n^{3}+(54n-135) \cdot9^{n} \bigr) \\& \quad =\bigl(24n^{2}-102n+9\bigr)9^{n}+256n^{4}+512n^{3}+56n^{2}-194n-9>0. \end{aligned}$$
We then obtain, for \(0< t<\pi/2\) and \(n\geq6\),
$$ \frac{u_{n+1}(t)}{u_{n}(t)}< 1. $$
Hence, for every \(t\in(0,\pi/2)\), the sequence \(n\longmapsto u_{n}(t)\) is strictly decreasing for \(n\geq6\). We then obtain from (3.2)
$$ -\frac{t^{3}\sin^{3}t}{16}G'(t)>t^{9} \biggl( \frac{16}{945}-\frac {16}{4\text{,}725}t^{2} \biggr)>0,\quad 0< t< \frac{\pi}{2}, $$
which implies \(G'(t)<0\) for \(0< t<\pi/2\). Hence, \(G(t)\) is strictly decreasing for \(0< t<\pi/2\), and \(F(x)\) is strictly decreasing for \(0< x<1\). So, we have
$$ \frac{256-25\pi^{2}}{\pi^{2}}=\lim_{t\to1}F(t)< F(x)=\frac{ (\frac{8x}{\arcsin x}-3\sqrt{1-x^{2}} )^{2}-25}{x^{2}}< \lim _{t\to 0}F(t)=\frac{5}{3} $$
for all \(x\in(0, 1)\), with the constants \(5/3\) and \((256-25\pi^{2})/\pi^{2}\) being best possible. The proof is complete. □

Proof of Theorem 1.3

By (2.1), we find that for \(0< x<1\),
$$\begin{aligned} \bigl(25-10x^{4}\bigr)- \biggl(\frac{10x}{\operatorname {arcsl}x}-5 \biggr)^{2}&>\bigl(25-10x^{4}\bigr)- \biggl( \frac{10x}{x+\frac{1}{10}x^{5}+\frac{1}{24}x^{9}}-5 \biggr)^{2} \\ &=\frac{10x^{8}(3\text{,}120-1\text{,}344x^{4}-120x^{8}-25x^{12})}{(120+12x^{4}+5x^{8})^{2}}. \end{aligned}$$
Noting that
$$ 3\text{,}120-1\text{,}344t-120t^{2}-25t^{3}>0\quad \text{for } 0< t< 1, $$
we obtain, for \(0< x<1\),
$$ \bigl(25-10x^{4}\bigr)- \biggl(\frac{10x}{\operatorname {arcsl}x}-5 \biggr)^{2}>0, $$
which implies (1.11).
By (2.2), we find that for \(0< x<1\),
$$\begin{aligned} \bigl(25-15x^{4}\bigr)- \biggl(\frac{10x}{\operatorname {arctlh}x}-5 \biggr)^{2}&>\bigl(25-15x^{4}\bigr)- \biggl( \frac{10x}{x+\frac{1}{10}x^{5}+\frac{1}{24}x^{9}}-5 \biggr)^{2} \\ & =\frac{15x^{8}A(x)}{(24\text{,}960+3\text{,}744x^{4}+1\text{,}820x^{8}+1\text{,}155x^{12})^{2}}, \end{aligned}$$
(3.3)
where
$$\begin{aligned} A(x) =&\bigl(115\text{,}947\text{,}520-71\text{,}285\text{,}760x^{8}-4 \text{,}204\text{,}200x^{16}\bigr) \\ &{} +x^{4}\bigl(87\text{,}320\text{,}064-11\text{,}961 \text{,}040x^{8}-1\text{,}334\text{,}025x^{16}\bigr). \end{aligned}$$
Noting that for \(0< t<1\),
$$ 115\text{,}947\text{,}520-71\text{,}285\text{,}760t-4\text{,}204 \text{,}200t^{2}>0 $$
and
$$ 87\text{,}320\text{,}064-11 \text{,}961\text{,}040t-1\text{,}334\text{,}025t^{2}>0, $$
we obtain \(A(x)>0\) for \(0< x<1\). From (3.3), we obtain (1.12). The proof is complete. □

Proof of Theorem 1.4

The inequality (1.13) is obtained by considering the function \(h(x)\) defined by
$$ h(x)=\operatorname {arcslh}x-\frac{95x}{80+\sqrt{225+285x^{4}}}, \quad x>0. $$
Differentiation yields
$$ h'(x)=\frac{1}{\sqrt{1+x^{4}}}-\frac{475(16\sqrt {225+285x^{4}}+45-57x^{4})}{(80+\sqrt{225+285x^{4}})^{2}\sqrt{225+285x^{4}}}. $$
By an elementary change of variable
$$ t=\sqrt{225+285x^{4}}, \quad x>0 \qquad \biggl(\text{or } x=\sqrt [4]{\frac{t^{2}-225}{285}}, t>15 \biggr), $$
(3.4)
we have
$$\begin{aligned}& \frac{1}{\sqrt{1+x^{4}}}-\frac{475(16\sqrt {225+285x^{4}}+45-57x^{4})}{(80+\sqrt{225+285x^{4}})^{2}\sqrt{225+285x^{4}}} \\& \quad =\frac{285}{\sqrt{17\text{,}100+285t^{2}}}+\frac {95(t^{2}-80t-450)}{t(80+t)^{2}}=\frac{95I(t)}{t(80+t)^{2}}, \end{aligned}$$
where
$$ I(t)=\frac{19\text{,}200t+480t^{2}+3t^{3}}{\sqrt {17\text{,}100+285t^{2}}}+t^{2}-80t-450, \quad t>15. $$
We now prove that
$$ h'(x)>0, \quad x>0. $$
It suffices to show that
$$ I(t)>0, \quad t>15. $$
Differentiation yields
$$\begin{aligned}& I'(t)=\frac{6(192\text{,}000+9\text{,}600t+90t^{2}+80t^{3}+t^{4})}{(60+t^{2})\sqrt {17\text{,}100+285t^{2}}}+2t-80, \\& I''(t)=\frac{6(576\text{,}000-565\text{,}200t-4\text{,}800t^{2}+150t^{3}+t^{5})}{(60+t^{2})^{2}\sqrt {17\text{,}100+285t^{2}}}+2, \end{aligned}$$
and
$$ I'''(t)=\frac{10\text{,}800(-18\text{,}840-1\text{,}920t+1\text{,}271t^{2}+8t^{3})}{(60+t^{2})^{3}\sqrt {17\text{,}100+285t^{2}}}>0 \quad \text{for } t>15. $$
Thus, we have, for \(t>15\),
$$ I''(t)>I''(15)=0\quad \Longrightarrow\quad I'(t)>I'(15)=0\quad \Longrightarrow\quad I(t)>I(15)=0. $$
Hence, \(h'(x)>0\) holds for \(x>0\), and we have
$$ h(x)>h(0)=0, \quad x>0. $$
The proof is complete. □

Declarations

Acknowledgements

The authors thank the referees for helpful comments.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematics and Informatics, Henan Polytechnic University

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© Sun and Chen 2016