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

Abstract

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

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, [5–8]).

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 [11–19]).

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. □