Sharp Arcsine-Type Bounds and Analytic Approximations for the Gauss Lemniscate Sine Function

Abstract

This paper develops two novel arcsine-type analytic approximation formulas for $\mathrm{arcsl}\left(x\right)$, the Gauss lemniscate sine function, each equipped with a monotonic and bounded remainder term of order $x^{29}$ as $x\to 0$, which demonstrates the high accuracy of the obtained formulas near the origin. Based on these approximations, we establish several new sharp inequalities valid on the interval $(0,1)$. Numerical evidence confirms that the resulting bounds provide improved accuracy compared with existing estimates in the literature, particularly in a neighborhood of the origin.

1. Introduction

The lemniscate is defined as the set of points in the plane such that the product of their distances to two fixed points, called the foci, remains constant. Let the foci be located at $(\pm \alpha ,0)$ and let the product of the distances be ${\beta }^2$. Then, the Cartesian equation of the lemniscate is

$${(x^2+y^2)}^2={\beta }^4-{\alpha }^4+2{\alpha }^2(x^2-y^2).$$

(1)

For $\beta =\alpha =1/\sqrt{2}$, we obtain the Bernoulli lemniscate curve

$${(x^2+y^2)}^2=x^2-y^2,$$

(2)

which, when expressed in polar coordinates $(r,\theta )$, becomes

$$r^2=cos2\theta .$$

The total arc length $\mathbf{L}$ of the lemniscate curve (2) can be expressed as

$$\mathbf{L}=4{\int }_0^1{\displaystyle \frac{dt}{\sqrt{1-t^4}}}={\displaystyle \frac{4\sqrt{\pi }\Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}}\approx 5.24412,$$

(3)

where $\Gamma \left(x\right)$ is the ordinary Euler Gamma function.

More generally, the arc length from the origin to a point on the curve can be represented by the integral

$$\mathrm{arcsl}\left(x\right)={\int }_0^x{\displaystyle \frac{dt}{\sqrt{1-t^4}}},\left|x\right|\le 1,$$

(4)

which defines the arc lemniscate sine function. This function was systematically studied by Gauss during the years 1797–1798. In a similar manner, Gauss also introduced the hyperbolic arc lemniscate sine function, given by

$$\mathrm{arcslh}\left(x\right)={\int }_0^x{\displaystyle \frac{dt}{\sqrt{1+t^4}}},x\in \mathbb{R}.$$

(5)

Further properties and discussions of the functions defined in (4) and (5) may be found in the literature; see, for example, [1,2,3,4,5,6,7,8].

Another set of lemniscate functions consists of the arc lemniscate tangent $\mathrm{arctl}$ and the hyperbolic arc lemniscate tangent $\mathrm{arctlh}$, which were introduced in [4]. It was shown therein that

$$\mathrm{arctl}\left(x\right)=\mathrm{arcsl}\left({\displaystyle \frac{x}{\sqrt[4]{1+x^4}}}\right),x\in \mathbb{R},$$

(6)

and

$$\mathrm{arctlh}\left(x\right)=\mathrm{arcslh}\left({\displaystyle \frac{x}{\sqrt[4]{1-x^4}}}\right),\left|x\right|<1.$$

(7)

A variety of inequalities have been derived to obtain upper and lower bounds for the function $\mathrm{arcsl}\left(x\right)$. In 2012, Neuman [9] presented the following bounds

$${\displaystyle \frac{x\sqrt{5}}{\sqrt{2\sqrt{1-x^4}+3}}}<\mathrm{arcsl}\left(x\right)<{\displaystyle \frac{x}{\sqrt[10]{1-x^4}}},0<\left|x\right|<1.$$

(8)

In 2016, Sun and Chen [10] deduced the following inequality

$${\displaystyle \frac{10x}{\sqrt{25-10x^4}+5}}<\mathrm{arcsl}\left(x\right),0<x<1.$$

(9)

Also, in 2016, Liu and Chen [11] deduced the following inequality

$${\displaystyle \frac{x\left(23623x^8-214500x^4+265200\right)}{15\left(2445x^8-16068x^4+17680\right)}}<\mathrm{arcsl}\left(x\right),0<x<1.$$

(10)

In 2017, Mahmoud and Agarwal [12] proved the following double inequality for $0<x<1$

$${\displaystyle \frac{5x^{12}+18x^8+360x^4-240\sqrt{1-x^4}+240}{480x^3}}<\mathrm{arcsl}\left(x\right)<{\displaystyle \frac{x\left(\sqrt{1-x^4}\left(5x^8+24x^4+360\right)+30\right)}{390\sqrt{1-x^4}}}$$

(11)

and they established new inequality for the function $\mathrm{arcsl}\left(x\right)$ expressed in terms of the Lerch zeta function

$$\Phi (x;s;a)={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{x^k}{{(k+a)}^s}},\left|x\right|<1.$$

They showed that, for $0<x<1$,

$${\displaystyle \frac{1}{8}}x\Phi \left(x^4;{\displaystyle \frac{3}{2}};{\displaystyle \frac{1}{4}}\right)<\mathrm{arcsl}\left(x\right)<{\displaystyle \frac{1}{4}}x\Phi \left(x^4;{\displaystyle \frac{3}{2}};{\displaystyle \frac{1}{4}}\right).$$

In 2020, Alzer and Kwong [13] demonstrated that the upper constant ${\displaystyle \frac{1}{4}}$ can be improved by replacing it with the sharp value ${\displaystyle \frac{\mathbf{L}}{4\Phi \left(1;\frac{3}{2};\frac{1}{4}\right)}}\approx 0.12836.$

In 2020, Wei, He, and Wang [14] proved the double inequality

$${\displaystyle \frac{\mathbf{L}x}{(\mathbf{L}-1)\sqrt[4]{1-x^4}+1}}<\mathrm{arcsl}\left(x\right)<{\displaystyle \frac{\mathbf{L}x}{(\mathbf{L}-1)\sqrt{1-x^4}+1}},0<\left|x\right|<1.$$

(12)

In 2021, Qian and Wang [15] presented the double inequality

$${\displaystyle \frac{x}{\sqrt[4]{1-\frac{2x^4}{5}}}}<\mathrm{arcsl}\left(x\right)<{\displaystyle \frac{x}{\sqrt[4]{1-\left(1-{\mathbf{L}}^{-4}\right)x^4}}},0<x<1.$$

(13)

In 2024, Zhao and Wang [16] deduced the double inequality

$${\displaystyle \frac{x}{{\left(\frac{1}{2}{\left(1-x^4\right)}^{8/15}+\frac{1}{2}\right)}^{3/8}}}<\mathrm{arcsl}\left(x\right)<{\displaystyle \frac{x}{{\left(\frac{29}{65}{\left(1-x^4\right)}^{13/27}+\frac{36}{65}\right)}^{27/58}}},0<\left|x\right|<1.$$

(14)

In 2025, Zhang, Shi, Xu and Qian [17] proved the following double inequality for $0<\left|x\right|<1$

$${\displaystyle \frac{5\mathbf{L}x}{2(4\mathbf{L}-5)\sqrt[4]{1-x^4}+(5-3\mathbf{L})\sqrt{1-x^4}+5}}<\mathrm{arcsl}\left(x\right)<{\displaystyle \frac{75x}{16\sqrt[4]{1-x^4}+7\sqrt{1-x^4}+52}}.$$

(15)

These results show the continuing interest in obtaining sharper analytic approximations and simpler computable bounds for $\mathrm{arcsl}\left(x\right)$.

In this paper, we present two new arcsine-type analytic approximations of the function $\mathrm{arcsl}\left(x\right)$ with monotonic, sharply bounded remainders. Unlike the previously known bounds, the proposed approximations possess monotonic remainders of order $x^{29}$ near the origin. Additionally, we derive new sharp inequalities for the function $\mathrm{arcsl}\left(x\right)$. Numerical computations indicate that our bounds improve upon the previously mentioned inequalities in (8)–(15), particularly near zero.

In this work, Mathematica (version 10) was employed to carry out all necessary symbolic and numerical computations.

2. Main Results

Consider the power series

$$f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }}{u}_n{x}^n,g\left(x\right)={\displaystyle \sum _{n=0}^{\infty }}{v}_n{x}^n,$$

having a common radius of convergence $R>0$, and assume that $v_n>0$ for all $n\ge 0$. The following result provides an effective criterion for determining the monotonic behavior of the quotient $f\left(x\right)/g\left(x\right)$ on the interval $(0,R)$. According to the Biernacki–Krzyż theorem [18], if the sequence ${\left\{{\displaystyle \frac{u_n}{v_n}}\right\}}_{n\ge 0}$ is increasing (decreasing), then the function $x⟼{\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}$ is increasing (decreasing) on $(0,R)$. Moreover, if the sequence $\left\{{\displaystyle \frac{u_n}{v_n}}\right\}$ is strictly monotone, then the quotient $f\left(x\right)/g\left(x\right)$ is strictly monotone on $(0,R)$.

The Taylor series expansion of the function ${\sin }^{-1}\left(x^2\right)$ is given by

$${\displaystyle \frac{{\sin }^{-1}\left(x^2\right)}{x}}={\displaystyle \sum _{n=0}^{\infty }}{b}_n{x}^{4n+1}={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{4^{-n}\left(2n\right)!}{(2n+1){(n!)}^2}}{x}^{4n+1},x\in (-1,1)$$

(16)

and the function $\mathrm{arcsl}\left(x\right)$ can be expressed in terms of the hypergeometric function as

$$\mathrm{arcsl}\left(x\right)=x_2{F}_1\left({\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{2}};{\displaystyle \frac{5}{4}};x^4\right)={\displaystyle \sum _{n=0}^{\infty }}{a}_n{x}^{4n+1}={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{{\left(\frac{1}{4}\right)}_n{\left(\frac{1}{2}\right)}_n}{{\left(\frac{5}{4}\right)}_{n}n!}}{x}^{4n+1},x\in [-1,1].$$

(17)

Now, the coefficients $b_n,a_n>0$ for $n\ge 0$ and the sequence $c_n={\displaystyle \frac{a_n}{b_n}}$ satisfies that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{c_{n+1}}{c_n}}& & ={\displaystyle \frac{a_{n+1}}{a_n}}·{\displaystyle \frac{b_n}{b_{n+1}}}={\displaystyle \frac{\left(\frac{1}{4}+n\right)\left(\frac{1}{2}+n\right)}{\left(\frac{5}{4}+n\right)(n+1)}}·{\displaystyle \frac{4\left(2n\right)!}{(2n+2)!}}·{\displaystyle \frac{2n+3}{2n+1}}·{\displaystyle \frac{{((n+1)!)}^2}{{(n!)}^2}}\hfill \\ & & ={\displaystyle \frac{(4n+1)(2n+1)}{2(4n+5)(n+1)}}·{\displaystyle \frac{4}{(2n+2)(2n+1)}}·{\displaystyle \frac{2n+3}{2n+1}}·{(n+1)}^2\hfill \\ & & ={\displaystyle \frac{8n^2+14n+3}{8n^2+14n+5}}<1,n\ge 0.\hfill \end{array}$$

Then $c_n$ is a strictly decreasing sequence, and using Biernacki–Krzyż theorem, we have that the function ${\displaystyle \frac{\mathrm{arcsl}\left(x\right)}{\frac{{\sin }^{-1}\left(x^2\right)}{x}}}$ is strictly decreasing for $0<x<1$. Also,

$$\underset{x\to 0}{lim}{\displaystyle \frac{\mathrm{arcsl}\left(x\right)}{\frac{{\sin }^{-1}\left(x^2\right)}{x}}}=\underset{x\to 0}{lim}\left\{{\displaystyle \frac{x+\frac{x^5}{10}+O\left(x^7\right)}{x+\frac{x^5}{6}+O\left(x^7\right)}}\right\}=1$$

and

$$\underset{x\to 1}{lim}{\displaystyle \frac{\mathrm{arcsl}\left(x\right)}{\frac{{\sin }^{-1}\left(x^2\right)}{x}}}={\displaystyle \frac{\mathbf{L}/4}{\pi /2}}={\displaystyle \frac{2\Gamma \left(\frac{5}{4}\right)}{\sqrt{\pi }\Gamma \left(\frac{3}{4}\right)}}\approx 0.834627.$$

Therefore, we get our first result:

Theorem 1.

The lemniscate sine function $\mathit{arcsl}\left(x\right)$ admits the following approximation formula:

$$\mathit{arcsl}\left(x\right)={\displaystyle \frac{{\mathit{sin}}^{-1}\left(x^2\right)}{x}}{\mu }_1\left(x\right),0<x<1,$$

(18)

where ${\mu }_1\left(x\right)$ is a continuous, decreasing, and bounded function satisfying

$${\displaystyle \frac{\mathbf{L}}{2\pi }}<{\mu }_1\left(x\right)<1,0<x<1$$

with sharp bounds.

As a consequence, we obtain

$${\displaystyle \frac{\mathbf{L}}{2\pi }}{\displaystyle \frac{{\mathit{sin}}^{-1}\left(x^2\right)}{x}}<\mathit{arcsl}\left(x\right)<{\displaystyle \frac{{\mathit{sin}}^{-1}\left(x^2\right)}{x}},0<x<1.$$

(19)

To improve the approximations of the function $\mathrm{arcsl}\left(x\right)$ in inequality (19), we will use the power series expansions in the relations (16) and (17). Now, consider

$$\mathrm{arcsl}\left(x\right)-{\displaystyle \frac{{sin}^{-1}\left(x^2\right)}{x}}\left({\displaystyle \sum _{n=0}^5}{q}_n{x}^{4n}\right)={\displaystyle \sum _{n=0}^5}{p}_n{x}^{4n+1}+O\left(x^{25}\right),x\to 0$$

(20)

where

$$\begin{array}{ccc}& & p_0=1-q_0\hfill \\ & & p_1=-{\displaystyle \frac{q_0}{6}}-q_1+{\displaystyle \frac{1}{10}}\hfill \\ & & p_2=-{\displaystyle \frac{3q_0}{40}}-q_2-{\displaystyle \frac{q_1}{6}}+{\displaystyle \frac{1}{24}}\hfill \\ & & p_3=-{\displaystyle \frac{5q_0}{112}}-q_3-{\displaystyle \frac{q_2}{6}}-{\displaystyle \frac{3q_1}{40}}+{\displaystyle \frac{5}{208}}\hfill \\ & & p_4=-{\displaystyle \frac{35q_0}{1152}}-q_4-{\displaystyle \frac{q_3}{6}}-{\displaystyle \frac{3q_2}{40}}-{\displaystyle \frac{5q_1}{112}}+{\displaystyle \frac{35}{2176}}\hfill \\ & & p_5=-{\displaystyle \frac{63q_0}{2816}}-q_5-{\displaystyle \frac{q_4}{6}}-{\displaystyle \frac{3q_3}{40}}-{\displaystyle \frac{5q_2}{112}}-{\displaystyle \frac{35q_1}{1152}}+{\displaystyle \frac{3}{256}}.\hfill \end{array}$$

The solution of the system $p_i=0$; $i=0,1,2,3,4,5$ is given by

$$q_0=1,q_1=-{\displaystyle \frac{1}{15}},q_2=-{\displaystyle \frac{1}{45}},q_3=-{\displaystyle \frac{731}{61425}},q_4=-{\displaystyle \frac{3433}{447525}},q_5=-{\displaystyle \frac{403531}{73841625}}.$$

Also, consider

$$\mathrm{arcsl}\left(x\right)-{\displaystyle \frac{(1-x^4){sin}^{-1}\left(x^2\right)}{x}}\left({\displaystyle \sum _{n=0}^6}{\delta }_n{x}^{4n}\right)={\displaystyle \sum _{n=0}^6}{\lambda }_n{x}^{4n+1}+O\left(x^{29}\right),x\to 0$$

(21)

where

$$\begin{array}{ccc}& & {\lambda }_0=1-{\delta }_0\hfill \\ & & {\lambda }_1={\displaystyle \frac{5{\delta }_0}{6}}-{\delta }_1+{\displaystyle \frac{1}{10}}\hfill \\ & & {\lambda }_2={\displaystyle \frac{11{\delta }_0}{120}}+{\displaystyle \frac{5{\delta }_1}{6}}-{\delta }_2+{\displaystyle \frac{1}{24}}\hfill \\ & & {\lambda }_3={\displaystyle \frac{17{\delta }_0}{560}}+{\displaystyle \frac{11{\delta }_1}{120}}+{\displaystyle \frac{5{\delta }_2}{6}}-{\delta }_3+{\displaystyle \frac{5}{208}}\hfill \\ & & {\lambda }_4={\displaystyle \frac{115{\delta }_0}{8064}}+{\displaystyle \frac{17{\delta }_1}{560}}+{\displaystyle \frac{11{\delta }_2}{120}}+{\displaystyle \frac{5{\delta }_3}{6}}-{\delta }_4+{\displaystyle \frac{35}{2176}}\hfill \\ & & {\lambda }_5={\displaystyle \frac{203{\delta }_0}{25344}}+{\displaystyle \frac{115{\delta }_1}{8064}}+{\displaystyle \frac{17{\delta }_2}{560}}+{\displaystyle \frac{11{\delta }_3}{120}}+{\displaystyle \frac{5{\delta }_4}{6}}-{\delta }_5+{\displaystyle \frac{3}{256}}\hfill \\ & & {\lambda }_6={\displaystyle \frac{735{\delta }_0}{146432}}+{\displaystyle \frac{203{\delta }_1}{25344}}+{\displaystyle \frac{115{\delta }_2}{8064}}+{\displaystyle \frac{17{\delta }_3}{560}}+{\displaystyle \frac{11{\delta }_4}{120}}+{\displaystyle \frac{5{\delta }_5}{6}}-{\delta }_6+{\displaystyle \frac{231}{25600}}.\hfill \end{array}$$

The solution of the system ${\lambda }_i=0$; $i=0,1,2,3,4,5,6$ is given by

$${\delta }_0=1,{\delta }_1={\displaystyle \frac{14}{15}},{\delta }_2={\displaystyle \frac{41}{45}},{\delta }_3={\displaystyle \frac{55234}{61425}},{\delta }_4={\displaystyle \frac{2792903}{3132675}},{\delta }_5={\displaystyle \frac{458004278}{516891375}},{\delta }_6={\displaystyle \frac{9573093983}{10854718875}}.$$

Therefore, we get our second result:

Theorem 2.

The function $\mathit{arcsl}\left(x\right)$ has the following approximations

$$\mathit{arcsl}\left(x\right)=\left(1-{\displaystyle \frac{x^4}{15}}-{\displaystyle \frac{x^8}{45}}-{\displaystyle \frac{731x^{12}}{61425}}-{\displaystyle \frac{3433x^{16}}{447525}}-{\displaystyle \frac{403531x^{20}}{73841625}}\right){\displaystyle \frac{{sin}^{-1}\left(x^2\right)}{x}}-O\left(x^{29}\right),x\to 0$$

(22)

and

$$\begin{array}{ccc}\hfill \mathit{arcsl}\left(x\right)=& & \left(1+{\displaystyle \frac{14x^4}{15}}+{\displaystyle \frac{41x^8}{45}}+{\displaystyle \frac{55234x^{12}}{61425}}+{\displaystyle \frac{2792903x^{16}}{3132675}}+{\displaystyle \frac{458004278x^{20}}{516891375}}\right.\hfill \\ & & \left.+{\displaystyle \frac{9573093983x^{24}}{10854718875}}\right){\displaystyle \frac{\left(1-x^4\right){sin}^{-1}\left(x^2\right)}{x}}+O\left(x^{29}\right),x\to 0.\hfill \end{array}$$

(23)

Now, we use our new approximations of the function $\mathrm{arcsl}\left(x\right)$ given in (22) and (23) to obtain our third result, which provides improvements to the bounds in (19).

Theorem 3.

The following double inequality holds

$$\begin{array}{ccc}\hfill & & {\displaystyle \frac{\left(\frac{9573093983x^{24}}{10854718875}+\frac{458004278x^{20}}{516891375}+\frac{2792903x^{16}}{3132675}+\frac{55234x^{12}}{61425}+\frac{41x^8}{45}+\frac{14x^4}{15}+1\right)\left(1-x^4\right){sin}^{-1}\left(x^2\right)}{x}}\hfill \\ & & <x\mathrm{arcsl}\left(x\right)<\hfill \\ & & {\displaystyle \frac{\left(-\frac{403531x^{20}}{73841625}-\frac{3433x^{16}}{447525}-\frac{731x^{12}}{61425}-\frac{x^8}{45}-\frac{x^4}{15}+1\right){sin}^{-1}\left(x^2\right)}{x}},0<x<1.\hfill \end{array}$$

(24)

Proof. 

Use the power series expansions in the relations (16) and (17) to get

$$\begin{array}{ccc}& & \mathrm{arcsl}\left(x\right)-{\displaystyle \frac{\left(-\frac{403531x^{20}}{73841625}-\frac{3433x^{16}}{447525}-\frac{731x^{12}}{61425}-\frac{x^8}{45}-\frac{x^4}{15}+1\right){sin}^{-1}\left(x^2\right)}{x}}\hfill \\ & & =-{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{{\sigma }_n{x}^{4n+2}/516891375\sqrt{\pi }}{(2n+3)(2n+5)(2n+7)(2n+9)(2n+11)(2n+13)(4n+25)n!}}<0,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill {\sigma }_n=& & 2\left(2n\right(4n\left(2n\right(n(798234362n+15270577053)+113088741647)+807454120911)\hfill \\ & & +5533310071525)+7238708173125)\Gamma \left(n+3/2\right)>0,n\ge 0.\hfill \end{array}$$

Also,

$$\begin{array}{ccc}& & \mathrm{arcsl}\left(x\right)-\left({\displaystyle \frac{9573093983x^{24}}{10854718875}}+{\displaystyle \frac{458004278x^{20}}{516891375}}+{\displaystyle \frac{2792903x^{16}}{3132675}}\right.\hfill \\ & & \left.+{\displaystyle \frac{55234x^{12}}{61425}}+{\displaystyle \frac{41x^8}{45}}+{\displaystyle \frac{14x^4}{15}}+1\right){\displaystyle \frac{{sin}^{-1}\left(x^2\right)}{x{\left(1-x^4\right)}^{-1}}}={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{{\varrho }_n{x}^{4n+29}}{{\tau }_n}}>0,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill {\varrho }_n& & =2\left(n\right(2n\left(2n\right(2n\left(2n\right(2n\left(6n\right(7236479250n+345249486277)+39034505054705)\hfill \\ & & +779783587908595)+9140558461400331)+64699705852706285)\hfill \\ & & +270282325628780011)+607882342084483281)\hfill \\ & & +280323693181026495)\Gamma \left(n+1/2\right)>0,n\ge 0\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill {\tau }_n=& & 10854718875\sqrt{\pi }(2n+1)(2n+3)(2n+5)(2n+7)(2n+9)(2n+11)(2n+13)\hfill \\ & & (2n+15)(4n+29)n!>0,n\ge 0.\hfill \end{array}$$

□

Our fourth result will be the following approximation formula:

Theorem 4.

The lemniscate sine function $\mathit{arcsl}\left(x\right)$ admits the following approximation formula:

$$\mathit{arcsl}\left(x\right)={\displaystyle \frac{\left(1-\frac{x^4}{15}-\frac{x^8}{45}-\frac{731x^{12}}{61425}-\frac{3433x^{16}}{447525}-\frac{403531x^{20}}{73841625}\right){sin}^{-1}\left(x^2\right)}{x}}{\mu }_2\left(x\right),0<x<1,$$

(25)

where ${\mu }_2\left(x\right)$ is a continuous, decreasing, and bounded function satisfying

$${\displaystyle \frac{516891375\mathbf{L}}{916008556\pi }}<{\mu }_2\left(x\right)<1,0<x<1$$

with sharp bounds.

Proof. 

For $0<x<1$, the function

$$\begin{array}{ccc}\hfill f_1\left(x\right)& & ={\displaystyle \frac{\left(-\frac{403531x^{20}}{73841625}-\frac{3433x^{16}}{447525}-\frac{731x^{12}}{61425}-\frac{x^8}{45}-\frac{x^4}{15}+1\right){sin}^{-1}\left(x^2\right)}{x}}\hfill \\ & & =x+{\displaystyle \frac{x^5}{10}}+{\displaystyle \frac{x^9}{24}}+{\displaystyle \frac{5x^{13}}{208}}+{\displaystyle \frac{35x^{17}}{2176}}+{\displaystyle \frac{3x^{21}}{256}}+{\displaystyle \sum _{n=0}^{\infty }}{\alpha }_n{x}^{4n+25},\hfill \end{array}$$

where ${\alpha }_n={\displaystyle \frac{\Gamma (n+3/2)d_n}{\sqrt{\pi }(n+6)!h_n}}$ with

$$\begin{array}{ccc}\hfill d_n& & =468996380672n^{10}+16252256212480n^9+248678083388160n^8\hfill \\ & & +2209979295043200n^7+12616411124582976n^6+48279034131336960n^5\hfill \\ & & +125226333285927040n^4+217035933885468200n^3+240104235128274252n^2\hfill \\ & & +152799203308683510n+42386065121131875\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill h_n& & =1058593536000n^6+25406244864000n^5+244799755200000n^4\hfill \\ & & +1206796631040000n^3+3192916590864000n^2+4269837027456000n\hfill \\ & & +2235203710740000.\hfill \end{array}$$

Also,

$$\mathrm{arcsl}\left(x\right)=x+{\displaystyle \frac{x^5}{10}}+{\displaystyle \frac{x^9}{24}}+{\displaystyle \frac{5x^{13}}{208}}+{\displaystyle \frac{35x^{17}}{2176}}+{\displaystyle \frac{3x^{21}}{256}}+{\displaystyle \sum _{n=0}^{\infty }}{a}_{n+6}{x}^{4n+25}.$$

Using the relation

$${\displaystyle \frac{{\alpha }_{n+1}}{{\alpha }_n}}-{\displaystyle \frac{a_{n+7}}{a_{n+6}}}={\displaystyle \frac{{\eta }_n}{{\varsigma }_n}}>0,n=0,1,2,\dots$$

where

$$\begin{array}{ccc}\hfill {\eta }_n& & =2238929649664n^{11}+100771031791616n^{10}+2039418896102400n^9\hfill \\ & & +24493089662192640n^8+193920067483848192n^7+1062553286436033408n^6\hfill \\ & & +4110697475799291200n^5+11225987253486117760n^4+21203144097471274944n^3\hfill \\ & & +26371424482193246976n^2+19432340612605497600n+6424364864685081600\hfill \end{array}$$

and

$$\begin{array}{ccc}& & {\varsigma }_n=3751971045376n^{13}+211623419936768n^{12}+5408721684104192n^{11}\hfill \\ & & +82872009932968960n^{10}+848538869653651968n^9+6125496463694555904n^8\hfill \\ & & +32041043986645724096n^7+122822503376806119040n^6+344605151580572701456n^5\hfill \\ & & +697997028024012424928n^4+990572009074684486512n^3+931172366668921949700n^2\hfill \\ & & +518722402192688582325n+129065568293846559375.\hfill \end{array}$$

Then ${\beta }_n={\displaystyle \frac{a_{n+6}}{{\alpha }_n}}$ is a decreasing sequence, and using Biernacki–Krzyż theorem, we have that the function ${\displaystyle \frac{\mathrm{arcsl}\left(x\right)}{f_1\left(x\right)}}$ is decreasing for $0<x<1$. Also,

$$\underset{x\to 0}{lim}{\displaystyle \frac{\mathrm{arcsl}\left(x\right)}{f_1\left(x\right)}}=\underset{x\to 0}{lim}\left\{{\displaystyle \frac{x+\frac{x^5}{10}+\frac{x^9}{24}+O\left(x^{12}\right)}{x+\frac{x^5}{10}+\frac{x^9}{24}+O\left(x^{12}\right)}}\right\}=1$$

and

$$\underset{x\to 1}{lim}{\displaystyle \frac{\mathrm{arcsl}\left(x\right)}{f_1\left(x\right)}}={\displaystyle \frac{\mathbf{L}/4}{\frac{229002139\pi }{516891375}}}={\displaystyle \frac{516891375\Gamma \left(\frac{5}{4}\right)}{229002139\sqrt{\pi }\Gamma \left(\frac{3}{4}\right)}}\approx 0.941938.$$

□

3. Numerical Comparisons

Consider the following functions for $0<x<1$, which represent the upper bounds in inequalities (8)–(15):

$U_1\left(x\right)={\displaystyle \frac{x}{\sqrt[10]{1-x^4}}}$

$U_2\left(x\right)={\displaystyle \frac{x\left(\sqrt{1-x^4}\left(5x^8+24x^4+360\right)+30\right)}{390\sqrt{1-x^4}}}$

$U_3\left(x\right)=\mathrm{arcsl}\left(1\right)/\Phi \left(1;{\displaystyle \frac{3}{2}};{\displaystyle \frac{1}{4}}\right)x\Phi \left(x^4;{\displaystyle \frac{3}{2}};{\displaystyle \frac{1}{4}}\right)$

$U_4\left(x\right)={\displaystyle \frac{\mathbf{L}x}{(\mathbf{L}-1)\sqrt{1-x^4}+1}}$

$U_5\left(x\right)={\displaystyle \frac{x}{\sqrt[4]{1-\left(1-{\mathbf{L}}^{-4}\right)x^4}}}$

$U_6\left(x\right)={\displaystyle \frac{x}{{\left(\frac{29}{65}{\left(1-x^4\right)}^{13/27}+\frac{36}{65}\right)}^{27/58}}}$

$U_7\left(x\right)={\displaystyle \frac{75x}{16\sqrt[4]{1-x^4}+7\sqrt{1-x^4}+52}}$

and our new upper bound

$U_8\left(x\right)=\left.-{\displaystyle \frac{403531x^{20}}{73841625}}-{\displaystyle \frac{3433x^{16}}{447525}}-{\displaystyle \frac{731x^{12}}{61425}}-{\displaystyle \frac{x^8}{45}}-{\displaystyle \frac{x^4}{15}}+1\right){\displaystyle \frac{{sin}^{-1}\left(x^2\right)}{x}}$.

Using Mathematica software, we can see that

$$\begin{array}{ccc}& & U_6\left(x\right)<U_1\left(x\right),0<x<0.999977\dots \hfill \\ & & U_6\left(x\right)<U_2\left(x\right),0<x<1\hfill \\ & & U_6\left(x\right)<U_3\left(x\right),0<x<0.998996\dots \hfill \\ & & U_6\left(x\right)<U_4\left(x\right),0<x<0.999597\dots \hfill \\ & & U_6\left(x\right)<U_5\left(x\right),0<x<1\hfill \\ & & U_6\left(x\right)<U_7\left(x\right),0<x<1\hfill \\ \hfill \mathrm{and}& & \\ & & U_8\left(x\right)<U_6\left(x\right),0<x<0.663569\dots .\hfill \end{array}$$

Hence, our new upper bound $U_8\left(x\right)$ has an advantage over the bounds $U_1\left(x\right)-U_7\left(x\right)$ in the domain $0<x<0.663569\dots$ For illustration, we present the graph of the difference $U_8\left(x\right)-U_6\left(x\right)$, which clearly exhibits the intervals where one bound is superior to the other (see Figure 1).

Consider the following functions for $0<x<1$, which represent the lower bounds in inequalities (8)–(15):

$L_1\left(x\right)={\displaystyle \frac{x\sqrt{5}}{\sqrt{2\sqrt{1-x^4}+3}}}$

$L_2\left(x\right)={\displaystyle \frac{10x}{\sqrt{25-10x^4}+5}}$

$L_3\left(x\right)={\displaystyle \frac{x\left(23623x^8-214500x^4+265200\right)}{15\left(2445x^8-16068x^4+17680\right)}}$

$L_4\left(x\right)={\displaystyle \frac{5x^{12}+18x^8+360x^4-240\sqrt{1-x^4}+240}{480x^3}}$

$L_5\left(x\right)={\displaystyle \frac{1}{8}}x\Phi \left(x^4;{\displaystyle \frac{3}{2}};{\displaystyle \frac{1}{4}}\right)$

$L_6\left(x\right)={\displaystyle \frac{\mathbf{L}x}{(\mathbf{L}-1)\sqrt[4]{1-x^4}+1}}$

$L_7\left(x\right)={\displaystyle \frac{x}{\sqrt[4]{1-\frac{2x^4}{5}}}}$

$L_8\left(x\right)={\displaystyle \frac{x}{{\left(\frac{1}{2}{\left(1-x^4\right)}^{8/15}+\frac{1}{2}\right)}^{3/8}}}$

$L_9\left(x\right)={\displaystyle \frac{5\mathbf{L}x}{2(4\mathbf{L}-5)\sqrt[4]{1-x^4}+(5-3\mathbf{L})\sqrt{1-x^4}+5}}$

and our new lower bound

$L_{10}\left(x\right)={\displaystyle \frac{\left(\frac{9573093983x^{24}}{10854718875}+\frac{458004278x^{20}}{516891375}+\frac{2792903x^{16}}{3132675}+\frac{55234x^{12}}{61425}+\frac{41x^8}{45}+\frac{14x^4}{15}+1\right)\left(\left(1-x^4\right){sin}^{-1}\left(x^2\right)\right)}{x}}$.

Using Mathematica software, we can see that

$$\begin{array}{ccc}& & L_1\left(x\right)<L_3\left(x\right),0<x<0.960799\dots \hfill \\ & & L_2\left(x\right)<L_3\left(x\right),0<x<1\hfill \\ & & L_4\left(x\right)<L_3\left(x\right),0<x<0.968077\dots \hfill \\ & & L_5\left(x\right)<L_3\left(x\right),0<x<0.99025\dots \hfill \\ & & L_6\left(x\right)<L_3\left(x\right),0<x<0.99914\dots \hfill \\ & & L_7\left(x\right)<L_3\left(x\right),0<x<1\hfill \\ & & L_8\left(x\right)<L_3\left(x\right),0<x<0.850394\dots \hfill \\ & & L_9\left(x\right)<L_3\left(x\right),0<x<0.98046\dots \hfill \\ \hfill \mathrm{and}& & \\ & & L_3\left(x\right)<L_{10}\left(x\right),0<x<0.387457\dots .\hfill \end{array}$$

Hence, our new lower bound $L_{10}\left(x\right)$ has an advantage over the bounds $L_1\left(x\right)-L_9\left(x\right)$ in the domain $0<x<0.387457\dots$ For illustration, we present the graph of the difference $L_3\left(x\right)-L_{10}\left(x\right)$, which clearly exhibits the intervals where one bound is superior to the other (see Figure 2).

4. Discussion

We have studied several properties of the Gauss lemniscate sine function. Particularly, we established two new arcsine-type analytic approximation Formulas (18) and (25), each with a monotonic and bounded remainder of order $x^{29}$ near the origin, allowing precise control of the approximation error. These formulas lead to the sharp double inequality (24), which holds on the interval $(0,1)$. Numerical results show that the proposed bounds offer improved accuracy compared with existing estimates, especially near the origin. Furthermore, the obtained approximations may be useful in numerical computation and computer-based evaluations involving the Gauss lemniscate sine function, where accurate approximations are often required in the treatment of real applied problems. Overall, these findings provide practical and theoretical tools for further studies related to special functions, analytic inequalities, and applications involving the Gauss lemniscate sine function. As a direction for future research, it would be of interest to generalize these approximation methods to other classes of special functions and to develop higher-order expansions that may further enhance the precision of analytical and numerical results.