New Bounds for the Modified Bessel Function of the First Kind and Toader-Qi Mean

Abstract

Let $I_p\left(x\right)$ be the modified Bessel function of the first kind of order p. The upper and lower bounds in the form of simple rational functions about $cosht$ and $(sinht)/t$ for the function $I_0\left(x\right)$ are obtained. The corresponding inequalities for the Toader-Qi mean do not match those in the existing literature.

1. Introduction

We know that the modified Bessel function of the first kind of order p is denoted as $I_p(x)$ (([1] p. 77), ([2] p. 374)) and has important applications in many fields of natural science. The function $I_p(x)$ denotes a class of solutions of the second-order differential equation ([2] p. 374)

$$x^2{y}^{″}\left(x\right)+x{y}^{\prime }\left(x\right)-\left(x^2+p^2\right)y\left(x\right)=0,$$

(1)

which can be written as the infinite series

$$I_p\left(x\right)={\left(\frac{x}{2}\right)}^p\sum _{n=0}^{\infty }\frac{{\left(x/2\right)}^{2n}}{n!\Gamma \left(p+n+1\right)},x\in \mathbb{R},p\in \mathbb{R}\backslash \{-1,-2,\dots \},$$

(2)

where $\Gamma \left(x\right)$ is the gamma function ([2] p. 255). Moreover, the normalized modified Bessel function defined by ${\mathcal{I}}_p:\mathbb{R}\to [1,+\infty )$ as ${\mathcal{I}}_p(x)=2^p\Gamma (p+1)x^{-p}{I}_p(x)$ was considered in [3]. Then, we have

$${\mathcal{I}}_{p+1/2}(x)=2^{p+1/2}\Gamma (p+3/2)x^{-(p+1/2)}{I}_{p+1/2}(x).$$

(3)

Particularly, for $p=0$, $p=1$ in the above, the function ${\mathcal{I}}_{p+1/2}(x)$ reduces to some elementary functions, as follows ([1] p. 54):

$${\mathcal{I}}_{1/2}(x)=\sqrt{\frac{\pi }{2x}}{I}_{1/2}(x)=\frac{sinhx}{x},{\mathcal{I}}_{3/2}(x)=\frac{3}{x}\sqrt{\frac{\pi }{2x}}{I}_{3/2}(x)=-3\left(\frac{sinhx}{x^3}-\frac{coshx}{x^2}\right).$$

(4)

The properties and various inequalities of this special function ${\mathcal{I}}_p(x)$ are widely discussed in References [3,4,5,6,7,8,9,10,11].

Throughout this article, we assume that a and b are two different positive real numbers. This paper focuses on

$$I_0\left(x\right)=\sum _{n=0}^{\infty }\frac{{\left(x/2\right)}^{2n}}{n!\Gamma \left(n+1\right)}=\sum _{n=0}^{\infty }\frac{x^{2n}}{2^{2n}{\left(n!\right)}^2},$$

(5)

which is the modified Bessel function of the first kind of order 0, which is related to the Toader-Qi mean (see [12,13]) defined by

$$TQ\left(a,b\right)=\frac{2}{\pi }{\int }_0^{\pi /2}{a}^{{cos}^2\theta }{b}^{{sin}^2\theta }d\theta =\sqrt{ab}{I}_0\left(ln\sqrt{\frac{a}{b}}\right).$$

(6)

Let

$$\begin{array}{cc}A\equiv A\left(a,b\right)={\displaystyle \frac{a+b}{2}},\hfill & G\equiv G\left(a,b\right)=\sqrt{ab},\hfill \\ L\equiv L\left(a,b\right)={\displaystyle \frac{a-b}{lna-lnb}},\hfill & I\equiv I\left(a,b\right)=e^{-1}{\left({\displaystyle \frac{b^b}{a^a}}\right)}^{1/\left(b-a\right)}\hfill \end{array}$$

(7)

and

$$A_{\nu }\equiv A_{\nu }\left(a,b\right)={\left(\frac{a^{\nu }+b^{\nu }}{2}\right)}^{1/\nu }\mathrm{if}\nu \ne 0$$

be the arithmetic mean, geometric mean, logarithmic mean, exponential mean, and power mean of order $\nu$, respectively. Then, an inequality chain (see [14,15,16])

$$G<L<A_{1/3}<\frac{A+2G}{3}<A_{1/2}<\frac{2A+G}{3}<A_{2/3}<I<A_{ln2}<A_1$$

(8)

holds. Alzer [17] also provided the following result:

$$\sqrt{AG}<\sqrt{LI}<\frac{L+I}{2}<\frac{A+G}{2}.$$

(9)

From now on, let us focus on the Toader-Qi mean of $I_0\left(x\right)$. In 2017, Qi, Shi, Liu, and Yang [13] proved that the inequalities

$$L<TQ<\frac{A+G}{2}<\frac{2A+G}{3}<I$$

(10)

hold Yang and Chu ([18] Theorem 3.4) established the following results:

$$L^{3/4}{A}^{1/4}<TQ<\sqrt{L{A}_{2/3}}<\frac{3}{4}L+\frac{1}{4}A,$$

(11)

$$\sqrt{\frac{2AL}{\pi }}<TQ<\sqrt{AL},$$

(12)

$$\sqrt{\left(\frac{2}{\pi }A+1-\frac{2}{\pi }\right)L}<TQ<\sqrt{\left({\lambda }_{0}A+1-{\lambda }_0\right)L},$$

(13)

where ${\lambda }_0=0.6766\dots$ Yang, Chu, and Song [19] and Yang, Tian, and Zhu ([20] Remark 4) obtained

$$\sqrt{\frac{e}{\pi }}\sqrt{LI}<\frac{2^{4/3}}{\sqrt{\pi }}\sqrt{L{A}_{2/3}}<TQ<\sqrt{L{A}_{2/3}}<\sqrt{LI}.$$

(14)

Let $b>a>0$ and $t=ln\sqrt{a/b}$. Then, those means mentioned above can be represented in terms of hyperbolic functions:

$$\begin{array}{ccc}\hfill \frac{A}{G}& =& cosht,\frac{L}{G}=\frac{sinht}{t},\frac{I}{G}=exp\left(\frac{t}{tanht}-1\right),\hfill \\ \hfill \frac{TQ}{G}& =& I_0\left(t\right),\frac{A_p}{G}={cosh}^{1/p}\left(pt\right)\mathrm{for}p\ne 0.\hfill \end{array}$$

Correspondingly, the inequalities (10)–(14) mentioned above are equivalent to

$$\frac{sinht}{t}<I_0\left(t\right)<\frac{cosht+1}{2}<\frac{2cosht+1}{3}<exp\left(\frac{t}{tanht}-1\right),$$

(15)

$${\left(\frac{sinht}{t}\right)}^{3/4}{\left(cosht\right)}^{1/4}<I_0\left(t\right)<\sqrt{\frac{sinht}{t}{cosh}^{3/2}\left(\frac{2t}{3}\right)}<\frac{3}{4}\frac{sinht}{t}+\frac{1}{4}cosht$$

(16)

$$\sqrt{\frac{sinh\left(2t\right)}{\pi t}}<I_0\left(t\right)<\sqrt{\frac{sinh\left(2t\right)}{2t}},$$

(17)

$$\sqrt{\left(\frac{2}{\pi }cosht+1-\frac{2}{\pi }\right)\frac{sinht}{t}}<I_0\left(t\right)<\sqrt{\left({\lambda }_{0}cosht+1-{\lambda }_0\right)\frac{sinht}{t}},$$

(18)

and

$$\begin{array}{ccc}& & \sqrt{\frac{e}{\pi }}\sqrt{\frac{sinht}{t}exp\left(\frac{t}{tanht}-1\right)}\hfill \\ & <& \frac{2^{3/4}}{\sqrt{\pi }}\sqrt{\frac{sinht}{t}{cosh}^{3/2}\left(\frac{2t}{3}\right)}\hfill \\ & <& I_0\left(t\right)<\sqrt{\frac{sinht}{t}{cosh}^{3/2}\left(\frac{2t}{3}\right)}<\sqrt{\frac{sinht}{t}exp\left(\frac{t}{tanht}-1\right)}\hfill \end{array}$$

(19)

hold for $t>0$.

In this paper, we first consider finding the bound of $TQ\left(a,b\right)$ in the following form:

$$\frac{x_1{A}^2+y_{1}AL+z_1{G}^2}{A+w_{1}G},$$

which is equivalent to searching for a bound for $TQ\left(a,b\right)/G=I_0\left(t\right)$ in the form of

$$\begin{array}{ccc}\hfill \frac{x_1{A}^2+y_{1}AL+z_1{G}^2}{\left(A+w_{1}G\right)G}& =& \frac{x_1{A}^2+y_{1}AL+z_1{G}^2}{AG+w_1{G}^2}\hfill \\ & =& \frac{x_1{\left(A/G\right)}^2+y_1\left(A/G\right)\left(L/G\right)+z_1}{A/G+w_1}\hfill \\ & =& \frac{x_1{\left(cosht\right)}^2+y_1\left(cosht\right)\left(\left(sinht\right)/t\right)+z_1}{cosht+w_1}.\hfill \end{array}$$

In this way, we can consider the power series expansion

$$\begin{array}{ccc}& & \left(cosht+w_1\right)I_0\left(t\right)-\left[x_1{\left(cosht\right)}^2+y_1\left(cosht\right)\left(\left(sinht\right)/t\right)+z_1\right]\hfill \\ & =& \left(-x_1-y_1-z_1+w_1+1\right)-t^2\left(x_1+\frac{2}{3}{y}_1-\frac{1}{4}{w}_1-\frac{3}{4}\right)\hfill \\ & & -t^4\left(\frac{1}{3}{x}_1+\frac{2}{15}{y}_1-\frac{1}{64}{w}_1-\frac{35}{192}\right)-t^6\left(\frac{2}{45}{x}_1+\frac{4}{315}{y}_1-\frac{1}{2304}{w}_1-\frac{77}{3840}\right)\hfill \\ & & -t^8\left(\frac{1}{315}{x}_1+\frac{2}{2835}{y}_1-\frac{1}{\mathrm{147,456}}{w}_1-\frac{143}{\mathrm{114,688}}\right)\hfill \\ & & -t^{10}\left(\frac{2}{\mathrm{14,175}}{x}_1+\frac{4}{\mathrm{155,925}}{y}_1-\frac{1}{\mathrm{14,745,600}}{w}_1-\frac{\mathrm{46,189}}{\mathrm{928,972,800}}\right)+O\left(t^{12}\right).\hfill \end{array}$$

When letting

$$\left\{\begin{array}{c}-x_1-y_1-z_1+w_1+1=0\\ x_1+\frac{2}{3}{y}_1-\frac{1}{4}{w}_1-\frac{3}{4}=0\\ \frac{1}{3}{x}_1+\frac{2}{15}{y}_1-\frac{1}{64}{w}_1-\frac{35}{192}=0\\ \frac{2}{45}{x}_1+\frac{4}{315}{y}_1-\frac{1}{2304}{w}_1-\frac{77}{3840}=0\end{array}\right.$$

we can obtain that $x_1$ = 2263/11,456, $y_1$ = 20,475/22,912, $z_1$ = 1879/22,912 and $w_1=31/179$ and find that

$$\begin{array}{ccc}& & \left(cosht+\frac{31}{179}\right)I_0\left(t\right)-\left[\frac{2263}{\mathrm{11,456}}{\left(cosht\right)}^2+\frac{\mathrm{20,475}}{\mathrm{22,912}}\frac{sinhtcosht}{t}+\frac{1879}{\mathrm{22,912}}\right]\hfill \\ & =& -\frac{209}{\mathrm{21,995,520}}{t}^8-\frac{\mathrm{16,217}}{\mathrm{15,242,895,360}}{t}^{10}+O\left(t^{12}\right),\hfill \end{array}$$

which motivates us to prove the inequality

$$\begin{array}{ccc}\hfill I_0\left(t\right)& <& \frac{\frac{2263}{\mathrm{11,456}}{\left(cosht\right)}^2+\frac{\mathrm{20,475}}{\mathrm{22,912}}\frac{sinhtcosht}{t}+\frac{1879}{\mathrm{22,912}}}{cosht+\frac{31}{179}}\hfill \\ & =& \frac{8284t+\mathrm{20,475}sinh2t+4526tcosh2t}{256t\left(179cosht+31\right)},t>0.\hfill \end{array}$$

Similarly, we can find the bound for $TQ\left(a,b\right)$ in the form of

$$\frac{x_2{L}^2+y_{2}LG+z_2{G}^2}{A+w_{2}G}$$

and obtain the exact bound by the approximation method used just now. The main conclusions of this paper are as follows.

Theorem 1.

The inequality

$$\begin{array}{ccc}\hfill I_0\left(t\right)& <& \frac{\frac{2263}{\mathrm{11,456}}{\left(cosht\right)}^2+\frac{\mathrm{20,475}}{\mathrm{22,912}}\frac{sinhtcosht}{t}+\frac{1879}{\mathrm{22,912}}}{cosht+\frac{31}{179}}\hfill \\ & =& \frac{8284t+\mathrm{20,475}sinh2t+4526tcosh2t}{256t\left(179cosht+31\right)}\hfill \end{array}$$

(20)

holds for $t>0$, or equivalently,

$$\begin{array}{ccc}\hfill TQ\left(a,b\right)& <& \frac{\frac{2263}{\mathrm{11,456}}{A}^2+\frac{\mathrm{20,475}}{\mathrm{22,912}}AL+\frac{1879}{\mathrm{22,912}}{G}^2}{A+\frac{31}{179}G}\hfill \\ & =& \frac{\mathrm{20,475}AL+4526A^2+1879G^2}{128\left(179A+31G\right)}\hfill \end{array}$$

(21)

holds for $a,b>0$ with $a\ne b$.

Theorem 2.

The inequality

$$\begin{array}{ccc}\hfill I_0\left(t\right)& >& \frac{\frac{4659}{568}{\left(\frac{sinht}{t}\right)}^2+\frac{7941}{284}\frac{sinht}{t}-\frac{4877}{568}}{cosht+\frac{1887}{71}}\hfill \\ & =& \frac{4659cosh2t+\mathrm{31,764}tsinht-9754t^2-4659}{16t^2\left(71cosht+1887\right)}\hfill \end{array}$$

(22)

holds for $t>0$, or equivalently,

$$\begin{array}{ccc}\hfill TQ\left(a,b\right)& >& \frac{\frac{4659}{568}{L}^2+\frac{7941}{284}LG-\frac{4877}{568}{G}^2}{A+\frac{1887}{71}G}\hfill \\ & =& \frac{4659L^2+\mathrm{15,882}GL-4877G^2}{8\left(71A+1887G\right)}\hfill \end{array}$$

(23)

holds for $a,b>0$ with $a\ne b$.

2. Lemmas

The precise power series representation of $\left(cosht\right)I_0\left(t\right)$ is required to establish two theorems in this article. We thank Prof. Dr. H. W. Gould for compiling an important result on the finite sum of combinatorial numbers in his monograph [21], which we consider as the substance of Lemma 1.

Lemma 1

([21] (3.175)). Let $\left(\genfrac{}{}{0pt}{}{n}{k}\right)$ be the number of different ways to choose k elements from a given set with n distinct elements, that is

$$\left(\genfrac{}{}{0pt}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}.$$

Then, for all $x\in \mathbb{R}$,

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{x}{2k}\right)\left(\genfrac{}{}{0pt}{}{x-2k}{n-k}\right)2^{2k}=\left(\genfrac{}{}{0pt}{}{2x}{2n}\right).$$

(24)

Lemma 2.

We have

$$I_0\left(t\right)cosht=\sum _{n=0}^{\infty }\frac{\left(4n-1\right)!!}{{\left[\left(2n\right)!\right]}^2}{t}^{2n}=:\sum _{n=0}^{\infty }{\kappa }_n{t}^{2n},$$

(25)

where

$${\kappa }_n=\frac{\left(4n-1\right)!!}{{\left[\left(2n\right)!\right]}^2}.$$

(26)

Proof. 

Using the Cauchy product formula, we have

$$\begin{array}{ccc}\hfill I_0\left(t\right)cosht& =& \left(\sum _{n=0}^{\infty }\frac{1}{2^{2n}{\left(n!\right)}^2}{t}^{2n}\right)\left(\sum _{n=0}^{\infty }\frac{1}{\left(2n\right)!}{t}^{2n}\right)\hfill \\ & =& \sum _{n=0}^{\infty }\sum _{k=0}^n\frac{1}{2^{2k}{\left(k!\right)}^2}\frac{1}{\left(2n-2k\right)!}{t}^{2n}\hfill \\ & =& :\sum _{n=0}^{\infty }{\kappa }_n{t}^{2n},\hfill \end{array}$$

where

$${\kappa }_n=\sum _{k=0}^n\frac{1}{2^{2k}{\left(k!\right)}^2}\frac{1}{\left(2n-2k\right)!}.$$

Since

$$\begin{array}{ccc}\hfill \left(\genfrac{}{}{0pt}{}{2n}{2k}\right)& =& \frac{\left(2n\right)!}{\left(2k\right)!(2n-2k)!},\hfill \\ \hfill \left(2k\right)!& =& {\left(k!\right)}^2\left(\genfrac{}{}{0pt}{}{2k}{k}\right),\hfill \end{array}$$

we have

$$\begin{array}{ccc}\hfill {\kappa }_n& =& \sum _{k=0}^n\frac{1}{2^{2k}{\left(k!\right)}^2\left(2n-2k\right)!}=\frac{1}{\left(2n\right)!}\sum _{k=0}^n\frac{\left(2k\right)!}{2^{2k}{\left(k!\right)}^2}\left(\genfrac{}{}{0pt}{}{2n}{2k}\right)\hfill \\ & =& \frac{1}{\left(2n\right)!}\sum _{k=0}^n\frac{{\left(k!\right)}^2\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}{2^{2k}{\left(k!\right)}^2}\left(\genfrac{}{}{0pt}{}{2n}{2k}\right)=\frac{1}{\left(2n\right)!}\sum _{k=0}^n\frac{1}{2^{2k}}\left(\genfrac{}{}{0pt}{}{2n}{2k}\right)\left(\genfrac{}{}{0pt}{}{2k}{k}\right).\hfill \end{array}$$

Letting $x=2n$ in (24) gives

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{2n}{2k}\right)\left(\genfrac{}{}{0pt}{}{2n-2k}{n-k}\right)2^{2k}=\left(\genfrac{}{}{0pt}{}{4n}{2n}\right).$$

On the one hand, let $n-k=l.$ Then

$$\begin{array}{ccc}\hfill \sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{2n}{2k}\right)\left(\genfrac{}{}{0pt}{}{2n-2k}{n-k}\right)2^{2k}& =& \sum _{l=0}^n\left(\genfrac{}{}{0pt}{}{2n}{2n-2l}\right)\left(\genfrac{}{}{0pt}{}{2l}{l}\right)2^{2(n-l)}\hfill \\ & =& 2^{2n}\sum _{l=0}^n\frac{1}{2^{2l}}\left(\genfrac{}{}{0pt}{}{2n}{2n-2l}\right)\left(\genfrac{}{}{0pt}{}{2l}{l}\right)\hfill \\ & =& 2^{2n}\sum _{k=0}^n\frac{1}{2^{2k}}\left(\genfrac{}{}{0pt}{}{2n}{2n-2k}\right)\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\hfill \\ & =& 2^{2n}\sum _{k=0}^n\frac{1}{2^{2k}}\left(\genfrac{}{}{0pt}{}{2n}{2k}\right)\left(\genfrac{}{}{0pt}{}{2k}{k}\right).\hfill \end{array}$$

On the other hand, we have

$$\left(\genfrac{}{}{0pt}{}{4n}{2n}\right)=\frac{(4n)!}{{\left[(2n)!\right]}^2}.$$

Therefore,

$$2^{2n}\sum _{k=0}^n\frac{1}{2^{2k}}\left(\genfrac{}{}{0pt}{}{2n}{2k}\right)\left(\genfrac{}{}{0pt}{}{2k}{k}\right)=\frac{(4n)!}{{\left[(2n)!\right]}^2}⟺\sum _{k=0}^n\frac{1}{2^{2k}}\left(\genfrac{}{}{0pt}{}{2n}{2k}\right)\left(\genfrac{}{}{0pt}{}{2k}{k}\right)=\frac{(4n)!}{2^{2n}{\left[(2n)!\right]}^2}$$

and

$$\begin{array}{ccc}\hfill {\kappa }_n& =& \frac{1}{\left(2n\right)!}\frac{(4n)!}{2^{2n}{\left[(2n)!\right]}^2}=\frac{1}{\left(2n\right)!}\frac{(4n)(4n-1)(4n-2)(4n-3)\cdots 4·3·2·1}{2^{2n}{\left[(2n)!\right]}^2}\hfill \\ & =& \frac{1}{\left(2n\right)!}\frac{(4n)(4n-2)\cdots 4·2(4n-1)!!}{2^{2n}{\left[(2n)!\right]}^2}=\frac{1}{\left(2n\right)!}\frac{2^{2n}(2n)!(4n-1)!!}{2^{2n}{\left[(2n)!\right]}^2}\hfill \\ & =& \frac{(4n-1)!!}{{\left[(2n)!\right]}^2}.\hfill \end{array}$$

 □

3. Proofs of Theorems 1 and 2

3.1. Proof of Theorem 1

The inequality to be proved is

$$\begin{array}{ccc}\hfill f_1(t)& =& :8284+\mathrm{20,475}\frac{1}{t}sinh2t+4526cosh2t\hfill \\ & >& 256\left(179cosht+31\right)I_0\left(t\right)=:g_1(t),t>0.\hfill \end{array}$$

(27)

Using the power series expansion of hyperbolic sine and hyperbolic cosine functions, we have

$$\begin{array}{ccc}\hfill f_1(t)& =& 8284+\mathrm{20,475}\sum _{n=0}^{\infty }\frac{2^{2n+1}}{\left(2n+1\right)!}{t}^{2n}+4526\sum _{n=0}^{\infty }\frac{2^{2n}}{\left(2n\right)!}{t}^{2n}\hfill \\ & =& 8284+\sum _{n=0}^{\infty }\left[\frac{\mathrm{20,475}·2^{2n+1}}{\left(2n+1\right)!}+\frac{4526·2^{2n}}{\left(2n\right)!}\right]t^{2n}\hfill \\ & =& \mathrm{53,760}+\mathrm{36,352}{t}^2+\frac{\mathrm{25,432}}{3}{t}^4+\frac{\mathrm{41,504}}{45}{t}^6\hfill \\ & & +\sum _{n=4}^{\infty }\left[\frac{\mathrm{20,475}·2^{2n+1}}{\left(2n+1\right)!}+\frac{4526·2^{2n}}{\left(2n\right)!}\right]t^{2n}.\hfill \end{array}$$

Then, using Formulas (5) and (25), we obtain

$$\begin{array}{ccc}\hfill g_1(t)& =& \mathrm{45,824}{I}_0\left(t\right)cosht+7936I_0\left(t\right)\hfill \\ & =& \mathrm{45,824}\sum _{n=0}^{\infty }{\kappa }_n{t}^{2n}+7936\sum _{n=0}^{\infty }\frac{1}{2^{2n}{\left(n!\right)}^2}{t}^{2n}\hfill \\ & =& \sum _{n=0}^{\infty }\left[\mathrm{45,824}{\kappa }_n+\frac{7936}{2^{2n}{\left(n!\right)}^2}\right]t^{2n}\hfill \\ & =& \mathrm{53,760}+\mathrm{36,352}{t}^2+\frac{\mathrm{25,432}}{3}{t}^4+\frac{\mathrm{41,504}}{45}{t}^6\hfill \\ & & +\sum _{n=4}^{\infty }\left[\mathrm{45,824}{\kappa }_n+\frac{7936}{2^{2n}{\left(n!\right)}^2}\right]t^{2n}.\hfill \end{array}$$

The inequality (27) is proved when the following inequality holds:

$$\mathrm{45,824}{\kappa }_n+\frac{7936}{2^{2n}{\left(n!\right)}^2}<\frac{\mathrm{20,475}·2^{2n+1}}{\left(2n+1\right)!}+\frac{4526·2^{2n}}{\left(2n\right)!},n\ge 4$$

$$⟺\mathrm{45,824}\frac{\left(4n-1\right)!!}{{\left(\left(2n\right)!\right)}^2}+\frac{7936}{2^{2n}{\left(n!\right)}^2}<\frac{\mathrm{20,475}·2^{2n+1}}{\left(2n+1\right)!}+\frac{4526·2^{2n}}{\left(2n\right)!},n\ge 4$$

$$⟺\mathrm{45,824}\frac{\left(4n-1\right)!!}{{\left(\left(2n\right)!\right)}^2}<\frac{\mathrm{20,475}·2^{2n+1}}{\left(2n+1\right)!}+\frac{4526·2^{2n}}{\left(2n\right)!}-\frac{7936}{2^{2n}{\left(n!\right)}^2},n\ge 4$$

(28)

Next, we prove (28) by mathematical induction. It is not difficult to verify that the above Formula (28) is true for $n=4$. Suppose (28) holds for m, that is

$$\mathrm{45,824}\frac{\left(4m-1\right)!!}{{\left(\left(2m\right)!\right)}^2}<\frac{\mathrm{20,475}·2^{2m+1}}{\left(2m+1\right)!}+\frac{4526·2^{2m}}{\left(2m\right)!}-\frac{7936}{2^{2m}{\left(m!\right)}^2},m\ge 4.$$

(29)

By (29) we have

$$\begin{array}{ccc}& & \mathrm{45,824}\frac{\left(4m+3\right)!!}{{\left[\left(2m+2\right)!\right]}^2}\hfill \\ & =& \frac{\left(4m+3\right)\left(4m+1\right)}{{\left(2m+2\right)}^2{\left(2m+1\right)}^2}\left[\mathrm{45,824}\frac{\left(4m-1\right)!!}{{\left(\left(2m\right)!\right)}^2}\right]\hfill \\ & <& \frac{\left(4m+3\right)\left(4m+1\right)}{{\left(2m+2\right)}^2{\left(2m+1\right)}^2}\left[\frac{\mathrm{20,475}·2^{2m+1}}{\left(2m+1\right)!}+\frac{4526·2^{2m}}{\left(2m\right)!}-\frac{7936}{2^{2m}{\left(m!\right)}^2}\right].\hfill \end{array}$$

The inequality (28) is proved when proving

$$\begin{array}{ccc}& & \frac{\left(4m+3\right)\left(4m+1\right)}{{\left(2m+2\right)}^2{\left(2m+1\right)}^2}\left[\frac{\mathrm{20,475}·2^{2m+1}}{\left(2m+1\right)!}+\frac{4526·2^{2m}}{\left(2m\right)!}-\frac{7936}{2^{2m}{\left(m!\right)}^2}\right]\hfill \\ & <& \frac{\mathrm{20,475}·2^{2m+3}}{\left(2m+3\right)!}+\frac{4526·2^{2m+2}}{\left(2m+2\right)!}-\frac{7936}{2^{2m+2}{\left(\left(m+1\right)!\right)}^2},\hfill \end{array}$$

that is

$$\begin{array}{ccc}& & \frac{\left(4m+3\right)\left(4m+1\right)}{{\left(2m+2\right)}^2{\left(2m+1\right)}^2}\frac{\mathrm{20,475}·2^{2m+1}}{\left(2m+1\right)!}+\frac{\left(4m+3\right)\left(4m+1\right)}{{\left(2m+2\right)}^2{\left(2m+1\right)}^2}\frac{4526·2^{2m}}{\left(2m\right)!}\hfill \\ & & -\left(\frac{\mathrm{20,475}·2^{2m+3}}{\left(2m+3\right)!}+\frac{4526·2^{2m+2}}{\left(2m+2\right)!}\right)\hfill \\ & <& \frac{\left(4m+3\right)\left(4m+1\right)}{{\left(2m+2\right)}^2{\left(2m+1\right)}^2}\frac{7936}{2^{2m}{\left(m!\right)}^2}-\frac{7936}{2^{2m+2}{\left(\left(m+1\right)!\right)}^2}\hfill \end{array}$$

or

$$-2\times 2^{2m}\frac{-\mathrm{70,909}m-\mathrm{68,754}{m}^2+\mathrm{36,208}{m}^3+6735}{\left(2m+3\right)!\left(m+1\right){\left(2m+1\right)}^2}<\frac{3968\left(6m+6m^2+1\right)}{\left(m+1\right){!}^2{2}^{2m}{\left(2m+1\right)}^2}.$$

Since 36,208$m^3$ − 68,754$m^2$ − 70,909$m+6735>0$ for $m\ge 4$, the above inequality is obviously true.

The proof of the Theorem 1 has been completed.

3.2. Proof of Theorem 2

The conclusion to be proved is equivalent to

$$\begin{array}{ccc}\hfill f_2(t)& =& :\left(1136t^{2}cosht+\mathrm{30,192}{t}^2\right)I_0\left(t\right)\hfill \\ & >& 4659cosh2t+\mathrm{31,764}tsinht-9754t^2-4659=:g_2(t),t>0.\hfill \end{array}$$

(30)

Using Formulas (5) and (25), we obtain

$$\begin{array}{ccc}\hfill f_2(t)& =& 1136t^2{I}_0\left(t\right)cosht+\mathrm{30,192}{t}^2{I}_0\left(t\right)\hfill \\ & =& 1136\sum _{n=0}^{\infty }{\kappa }_n{t}^{2n+2}+\mathrm{30,192}\sum _{n=0}^{\infty }\frac{1}{2^{2n}{\left(n!\right)}^2}{t}^{2n+2}\hfill \\ & =& \sum _{n=0}^{\infty }\left[1136{\kappa }_n+\frac{\mathrm{30,192}}{2^{2n}{\left(n!\right)}^2}\right]t^{2n+2}\hfill \\ & =& \mathrm{31,328}{t}^2+8400t^4+\frac{4073}{6}{t}^6+\frac{2153}{60}{t}^8+\sum _{n=4}^{\infty }\left[1136{\kappa }_n+\frac{\mathrm{30,192}}{2^{2n}{\left(n!\right)}^2}\right]t^{2n+2}.\hfill \end{array}$$

Using the power series expansion of hyperbolic sine and hyperbolic cosine functions, we have

$$\begin{array}{ccc}\hfill g_2(t)& =& 4659\sum _{n=0}^{\infty }\frac{2^{2n}}{\left(2n\right)!}{t}^{2n}+\mathrm{31,764}\sum _{n=0}^{\infty }\frac{1}{\left(2n+1\right)!}{t}^{2n+2}-9754t^2-4659\hfill \\ & =& 4659\sum _{n=-1}^{\infty }\frac{2^{2n+2}}{\left(2n+2\right)!}{t}^{2n+2}+\mathrm{31,764}\sum _{n=0}^{\infty }\frac{1}{\left(2n+1\right)!}{t}^{2n+2}-9754t^2-4659\hfill \\ & =& \mathrm{31,328}{t}^2+8400t^4+\frac{4073}{6}{t}^6+\frac{2153}{60}{t}^8\hfill \\ & & +4659\sum _{n=4}^{\infty }\frac{2^{2n+2}}{\left(2n+2\right)!}{t}^{2n+2}+\mathrm{31,764}\sum _{n=4}^{\infty }\frac{1}{\left(2n+1\right)!}{t}^{2n+2}\hfill \\ & =& \mathrm{31,328}{t}^2+8400t^4+\frac{4073}{6}{t}^6+\frac{2153}{60}{t}^8\hfill \\ & & +\sum _{n=4}^{\infty }\left[\frac{4659·2^{2n+2}}{\left(2n+2\right)!}+\frac{\mathrm{31,764}}{\left(2n+1\right)!}\right]t^{2n+2}.\hfill \end{array}$$

The inequality (27) is proved when the following inequality

$$1136{\kappa }_n+\frac{\mathrm{30,192}}{2^{2n}{\left(n!\right)}^2}>\frac{4659·2^{2n+2}}{\left(2n+2\right)!}+\frac{\mathrm{31,764}}{\left(2n+1\right)!}$$

holds for $n\ge 4,$ that is,

$$\frac{1136\left(4n-1\right)!!}{{\left(\left(2n\right)!\right)}^2}+\frac{\mathrm{30,192}}{2^{2n}{\left(n!\right)}^2}>\frac{4659·2^{2n+2}}{\left(2n+2\right)!}+\frac{\mathrm{31,764}}{\left(2n+1\right)!},n\ge 4$$

or

$$\frac{1136\left(4n-1\right)!!}{{\left(\left(2n\right)!\right)}^2}>\frac{4659}{\left(2n+2\right)!}{2}^{2n+2}+\frac{\mathrm{31,764}}{\left(2n+1\right)!}-\frac{\mathrm{30,192}}{2^{2n}{\left(n!\right)}^2},n\ge 4.$$

(31)

It can be verified that the above formula holds when $n=4$. Suppose that Formula (31) holds for m, i.e.,

$$\frac{1136\left(4m-1\right)!!}{{\left(\left(2m\right)!\right)}^2}>\frac{4659}{\left(2m+2\right)!}{2}^{2m+2}+\frac{\mathrm{31,764}}{\left(2m+1\right)!}-\frac{\mathrm{30,192}}{2^{2m}{\left(m!\right)}^2},m\ge 4.$$

(32)

Next, we prove that Formula (31) is also true for $m+1$. By (32), we have

$$\begin{array}{ccc}& & \frac{1136\left(4m+3\right)!!}{{\left(\left(2m+2\right)!\right)}^2}\hfill \\ & =& \frac{\left(4m+3\right)\left(4m+1\right)}{{\left(\left(2m+2\right)\left(2m+1\right)\right)}^2}\frac{1136\left(4m-1\right)!!}{{\left(\left(2m\right)!\right)}^2}\hfill \\ & >& \frac{\left(4m+3\right)\left(4m+1\right)}{{\left(\left(2m+2\right)\left(2m+1\right)\right)}^2}\left[\frac{4659}{\left(2m+2\right)!}{2}^{2m+2}+\frac{\mathrm{31,764}}{\left(2m+1\right)!}-\frac{\mathrm{30,192}}{2^{2m}{\left(m!\right)}^2}\right].\hfill \end{array}$$

In this way, we complete the proof of (31) when proving

$$\begin{array}{ccc}& & \frac{\left(4m+3\right)\left(4m+1\right)}{{\left(\left(2m+2\right)\left(2m+1\right)\right)}^2}\left[\frac{4659}{\left(2m+2\right)!}{2}^{2m+2}+\frac{\mathrm{31,764}}{\left(2m+1\right)!}-\frac{\mathrm{30,192}}{2^{2m}{\left(m!\right)}^2}\right]\hfill \\ & >& \frac{4659}{\left(2m+4\right)!}{2}^{2m+4}+\frac{\mathrm{31,764}}{\left(2m+3\right)!}-\frac{\mathrm{30,192}}{2^{2m+2}{\left(\left(m+1\right)!\right)}^2}\hfill \end{array}$$

$$\begin{array}{ccc}& ⟺& \frac{\left(4m+3\right)\left(4m+1\right)}{{\left(\left(2m+2\right)\left(2m+1\right)\right)}^2}\frac{4659}{\left(2m+2\right)!}{2}^{2m+2}+\frac{\left(4m+3\right)\left(4m+1\right)}{{\left(\left(2m+2\right)\left(2m+1\right)\right)}^2}\frac{\mathrm{31,764}}{\left(2m+1\right)!}\hfill \\ & & -\frac{\left(4m+3\right)\left(4m+1\right)}{{\left(\left(2m+2\right)\left(2m+1\right)\right)}^2}\frac{\mathrm{30,192}}{2^{2m}{\left(m!\right)}^2}\hfill \\ & >& \frac{4659}{\left(2m+4\right)!}{2}^{2m+4}+\frac{\mathrm{31,764}}{\left(2m+3\right)!}-\frac{\mathrm{30,192}}{2^{2m+2}{\left(\left(m+1\right)!\right)}^2}\hfill \end{array}$$

$$\begin{array}{ccc}& ⟺& \frac{\left(4m+3\right)\left(4m+1\right)}{{\left(\left(2m+2\right)\left(2m+1\right)\right)}^2}\frac{4659}{\left(2m+2\right)!}{2}^{2m+2}+\frac{\left(4m+3\right)\left(4m+1\right)}{{\left(\left(2m+2\right)\left(2m+1\right)\right)}^2}\frac{\mathrm{31,764}}{\left(2m+1\right)!}\hfill \\ & & -\left(\frac{4659}{\left(2m+4\right)!}{2}^{2m+4}+\frac{\mathrm{31,764}}{\left(2m+3\right)!}\right)\hfill \\ & >& \frac{\left(4m+3\right)\left(4m+1\right)}{{\left(\left(2m+2\right)\left(2m+1\right)\right)}^2}\frac{\mathrm{30,192}}{2^{2m}{\left(m!\right)}^2}-\frac{\mathrm{30,192}}{2^{2m+2}{\left(\left(m+1\right)!\right)}^2}\hfill \end{array}$$

$$\begin{array}{ccc}& ⟺& \frac{6\left[\begin{array}{c}\left(\mathrm{74,544}{m}^3+\mathrm{170,830}{m}^2+\mathrm{107,157}m+\mathrm{15,530}\right)2^{2m}\\ +\mathrm{127,056}{m}^5+\mathrm{719,984}{m}^4+\mathrm{1,503,496}{m}^3\\ +\mathrm{1,413,498}{m}^2+\mathrm{577,046}m+\mathrm{74,116}\end{array}\right]}{\left(2m+4\right)!{\left(m+1\right)}^2{\left(2m+1\right)}^2}\hfill \\ & >& \frac{\mathrm{15,096}\left(6m+6m^2+1\right)}{{\left[\left(m+1\right)!\right]}^2{2}^{2m}{\left(2m+1\right)}^2}\hfill \end{array}$$

$$\begin{array}{ccc}& ⟺& \frac{\left[\begin{array}{c}\left(\mathrm{74,544}{m}^3+\mathrm{170,830}{m}^2+\mathrm{107,157}m+\mathrm{15,530}\right)2^{2m}\\ +\mathrm{127,056}{m}^5+\mathrm{719,984}{m}^4+\mathrm{1,503,496}{m}^3\\ +\mathrm{1,413,498}{m}^2+\mathrm{577,046}m+\mathrm{74,116}\end{array}\right]}{\left(2m+4\right)!}\hfill \\ & >& \frac{2516\left(6m+6m^2+1\right)}{{\left(m!\right)}^2{2}^{2m}},\hfill \end{array}$$

that is,

$$\frac{{\left(m!\right)}^2{2}^{2m}}{\left(2m+4\right)!}>\frac{2516\left(6m+6m^2+1\right)}{\left[\begin{array}{c}\left(\mathrm{74,544}{m}^3+\mathrm{170,830}{m}^2+\mathrm{107,157}m+\mathrm{15,530}\right)2^{2m}\\ +\mathrm{127,056}{m}^5+\mathrm{719,984}{m}^4+\mathrm{1,503,496}{m}^3\\ +\mathrm{1,413,498}{m}^2+\mathrm{577,046}m+\mathrm{74,116}\end{array}\right]}$$

(33)

holds for $m\ge 4$.

Then, we use mathematical induction to prove (33). It is not difficult to verify that the above Formula (33) is true for $m=4$. Suppose that (33) holds for $m=n$, i.e.,

$$\begin{array}{ccc}\hfill f(n)& =& :\frac{{\left(n!\right)}^2{2}^{2n}}{\left(2n+4\right)!}\hfill \\ & >& \frac{2516\left(6n+6n^2+1\right)}{\left[\begin{array}{c}\left(\mathrm{74,544}{n}^3+\mathrm{170,830}{n}^2+\mathrm{107,157}n+\mathrm{15,530}\right)2^{2n}\\ +\mathrm{127,056}{n}^5+\mathrm{719,984}{n}^4+\mathrm{1,503,496}{n}^3\\ +\mathrm{1,413,498}{n}^2+\mathrm{577,046}n+\mathrm{74,116}\end{array}\right]}=:g(n),n\ge 4.\hfill \end{array}$$

(34)

By (34), we have

$$\begin{array}{ccc}\hfill f(n+1)& =& \frac{{\left(\left(n+1\right)!\right)}^2{2}^{2n+2}}{\left(2n+6\right)!}=\frac{4{\left(n+1\right)}^2}{\left(2n+6\right)\left(2n+5\right)}\frac{n{!}^2{2}^{2n}}{\left(2n+4\right)!}\hfill \\ & >& \frac{4{\left(n+1\right)}^2}{\left(2n+6\right)\left(2n+5\right)}\frac{2516\left(6n+6n^2+1\right)}{\left[\begin{array}{c}\left(\mathrm{74,544}{n}^3+\mathrm{170,830}{n}^2+\mathrm{107,157}n+\mathrm{15,530}\right)2^{2n}\\ +\mathrm{127,056}{n}^5+\mathrm{719,984}{n}^4+\mathrm{1,503,496}{n}^3\\ +\mathrm{1,413,498}{n}^2+\mathrm{577,046}n+\mathrm{74,116}\end{array}\right]},\hfill \end{array}$$

so the proof of (33) is complete when we can prove that

$$\begin{array}{ccc}\hfill \frac{A}{B}& =& :\frac{2516·4{\left(n+1\right)}^2\left(6n+6n^2+1\right)}{\left(2n+6\right)\left(2n+5\right)\left[\begin{array}{c}\left(\mathrm{74,544}{n}^3+\mathrm{170,830}{n}^2+\mathrm{107,157}n+\mathrm{15,530}\right)2^{2n}\\ +\mathrm{127,056}{n}^5+\mathrm{719,984}{n}^4+\mathrm{1,503,496}{n}^3\\ +\mathrm{1,413,498}{n}^2+\mathrm{577,046}n+\mathrm{74,116}\end{array}\right]}\hfill \\ & >& g(n+1)\hfill \\ & =& \frac{2516\left(6\left(n+1\right)+6{\left(n+1\right)}^2+1\right)}{\left[\begin{array}{c}\left(\mathrm{74,544}{\left(n+1\right)}^3+\mathrm{170,830}{\left(n+1\right)}^2+\mathrm{107,157}\left(n+1\right)+\mathrm{15,530}\right)2^{2n+2}\\ +\mathrm{127,056}{\left(n+1\right)}^5+\mathrm{719,984}{\left(n+1\right)}^4+\mathrm{1,503,496}{\left(n+1\right)}^3\\ +\mathrm{1,413,498}{\left(n+1\right)}^2+\mathrm{577,046}\left(n+1\right)+\mathrm{74,116}\end{array}\right]}\hfill \\ & =& :\frac{C}{D}.\hfill \end{array}$$

In fact,

$$\begin{array}{ccc}& & (AD-BC)/2516\hfill \\ & =& 9318\times 2^{2n}\left(\begin{array}{c}349n+4227n^2+\mathrm{13,036}{n}^3+\mathrm{17,962}{n}^4\\ +\mathrm{12,512}{n}^5+4296n^6+576n^7-18\end{array}\right)\hfill \\ & & -10588\left(n+3\right)\left(n+2\right)\left(n+1\right)\left(\begin{array}{c}1238n+2702n^2+2496n^3\\ +1008n^4+144n^5+177\end{array}\right)\hfill \\ & >& 0\hfill \end{array}$$

(35)

holds for all $n\ge 4.$ The above formula is obviously equivalent to

$$2^{2n}>\frac{\left[\begin{array}{c}5294\left(n+1\right)\left(n+2\right)\left(n+3\right)\\ \left(1238n+2702n^2+2496n^3+1008n^4+144n^5+177\right)\end{array}\right]}{4659\left[\begin{array}{c}349n+4227n^2+\mathrm{13,036}{n}^3+\mathrm{17,962}{n}^4\\ +\mathrm{12,512}{n}^5+4296n^6+576n^7-18\end{array}\right]},n\ge 4.$$

(36)

We use mathematical induction for the third time in the process of proving Theorem 2. The Formula (36) is obviously true for $n=4$. We assume that (36) holds for $n=m$, that is,

$$2^{2m}>\frac{\left[\begin{array}{c}5294\left(m+1\right)\left(m+2\right)\left(m+3\right)\\ \left(1238m+2702m^2+2496m^3+1008m^4+144m^5+177\right)\end{array}\right]}{4659\left[\begin{array}{c}349m+4227m^2+\mathrm{13,036}{m}^3+\mathrm{17,962}{m}^4\\ +\mathrm{12,512}{m}^5+4296m^6+576m^7-18\end{array}\right]}.$$

(37)

It is proved below that (36) holds for $n=m+1$. By (37), we have

$$2^{2m+2}=4·2^{2m}>4·\frac{\left[\begin{array}{c}5294\left(m+1\right)\left(m+2\right)\left(m+3\right)\\ \left(1238m+2702m^2+2496m^3+1008m^4+144m^5+177\right)\end{array}\right]}{4659\left[\begin{array}{c}349m+4227m^2+\mathrm{13,036}{m}^3+\mathrm{17,962}{m}^4\\ +\mathrm{12,512}{m}^5+4296m^6+576m^7-18\end{array}\right]},$$

so the proof of (36) is complete when proving

$$\begin{array}{ccc}& & 4·\frac{5294}{4659}\frac{\left[\begin{array}{c}\left(m+1\right)\left(m+2\right)\left(m+3\right)\\ \left(1238m+2702m^2+2496m^3+1008m^4+144m^5+177\right)\end{array}\right]}{\left[\begin{array}{c}349m+4227m^2+\mathrm{13,036}{m}^3+\mathrm{17,962}{m}^4\\ +\mathrm{12,512}{m}^5+4296m^6+576m^7-18\end{array}\right]}\hfill \\ & >& \frac{5294}{4659}\frac{\left[\begin{array}{c}\left(m+2\right)\left(m+3\right)\left(m+4\right)\\ \left(\begin{array}{c}1238\left(m+1\right)+2702{\left(m+1\right)}^2+2496{\left(m+1\right)}^3\\ +1008{\left(m+1\right)}^4+144{\left(m+1\right)}^5+177\end{array}\right)\end{array}\right]}{\left[\begin{array}{c}349\left(m+1\right)+4227{\left(m+1\right)}^2+\mathrm{13,036}{\left(m+1\right)}^3+\mathrm{17,962}{\left(m+1\right)}^4\\ +\mathrm{12,512}{\left(m+1\right)}^5+4296{\left(m+1\right)}^6+576{\left(m+1\right)}^7-18\end{array}\right]}\hfill \end{array}$$

$$\begin{array}{ccc}& ⟺& \frac{a}{b}=:\frac{4\left(m+1\right)\left[\begin{array}{c}1238m+2702m^2+2496m^3\\ +1008m^4+144m^5+177\end{array}\right]}{\left[\begin{array}{c}349m+4227m^2+\mathrm{13,036}{m}^3+\mathrm{17,962}{m}^4\\ +\mathrm{12,512}{m}^5+4296m^6+576m^7-18\end{array}\right]}\hfill \\ & >& \frac{\left(m+4\right)\left[\begin{array}{c}1238\left(m+1\right)+2702{\left(m+1\right)}^2+2496{\left(m+1\right)}^3\\ +1008{\left(m+1\right)}^4+144{\left(m+1\right)}^5+177\end{array}\right]}{\left[\begin{array}{c}349\left(m+1\right)+4227{\left(m+1\right)}^2+\mathrm{13,036}{\left(m+1\right)}^3+\mathrm{17,962}{\left(m+1\right)}^4\\ +\mathrm{12,512}{\left(m+1\right)}^5+4296{\left(m+1\right)}^6+576{\left(m+1\right)}^7-18\end{array}\right]}\hfill \\ & =& :\frac{c}{d}.\hfill \end{array}$$

We calculated that

$$\begin{array}{ccc}\hfill ad-bc& =& \mathrm{248,832}{m}^{13}+\mathrm{5,505,408}{m}^{12}+\mathrm{55,199,232}{m}^{11}+\mathrm{332,101,728}{m}^{10}\hfill \\ & & +\mathrm{1,337,449,104}{m}^9+\mathrm{3,805,389,696}{m}^8+\mathrm{7,862,263,692}{m}^7\hfill \\ & & +\mathrm{11,923,195,508}{m}^6+\mathrm{13,236,235,612}{m}^5+\mathrm{10,578,592,104}{m}^4\hfill \\ & & +\mathrm{5,876,921,975}{m}^3+\mathrm{2,125,982,239}{m}^2+\mathrm{440,485,650}m+\mathrm{38,040,600}\hfill \\ & >& 0.\hfill \end{array}$$

4. Comparisons of New and Old Results

Through observations, in the literature, the more accurate upper and lower bounds of $I_0\left(t\right)$ are the second inequality in (11)

$$I_0\left(t\right)<\sqrt{\frac{sinht}{t}{cosh}^{3/2}\left(\frac{2t}{3}\right)}$$

(38)

and the left-hand side of the inequality in (11)

$${\left(\frac{sinht}{t}\right)}^{3/4}{\left(cosht\right)}^{1/4}<I_0\left(t\right).$$

(39)

Next, we compare them with the results of this paper.

(i) The numerical results show that inequality (20) is sharper than the one (38) on (0, 5.2904) while that inequality (38) is sharper than the one (20) on (5.2904, ∞). In other words, the two inequalities (21) and the second inequality in (11) cannot be compared.

(ii) The numerical results show that inequality (22) is sharper than the one (39) on (0, 4.916) while that inequality (39) is sharper than the one (22) on (4.916, ∞). In other words, the two inequalities (23) and the left-hand side of (11) cannot be compared.

5. Conclusions

Let $I_p\left(x\right)$ be the modified Bessel function of the first kind of order p. The double inequality

$$\frac{4659cosh2t+\mathrm{31,764}tsinht-9754t^2-4659}{16t^2\left(71cosht+1887\right)}<I_0\left(t\right)<\frac{8284t+\mathrm{20,475}sinh2t+4526tcosh2t}{256t\left(179cosht+31\right)}$$

was proved to hold for $t>0$. The corresponding inequalities for a Toader-Qi mean are not in accordance with the conclusions in previous literature.