New Lower Bound for the Generalized Elliptic Integral of the First Kind

Abstract

In this paper, we obtain a new simple rational approximation for ${\mathcal{K}}_a\left(r\right)$: the inequality $2{\mathcal{K}}_a\left(r\right)/\pi$$>g_2\left(r^{\prime }\right)/g_1\left(r^{\prime }\right)$ holds for all $r\in (0,1)$, where ${\mathcal{K}}_a\left(r\right)$ is the generalized elliptic integral of the first kind, $r^{\prime }=\sqrt{1-r^2}$, $g_1\left(r^{\prime }\right)$ and $g_2\left(r^{\prime }\right)$ are specific primary and quadratic polynomials about $r^{\prime }$, respectively. In particular, when a is taken as 1/2, 1/3, 1/4, 1/5 and 1/6 respectively, we can obtain some new specific lower bounds of the corresponding functions.

1. Introduction

For $r\in (0,1),$$a\in (0,1),$ and $r^{\prime }=\sqrt{1-r^2},$ the generalized elliptic integral of the first kind (see [1], Section 5.5) is defined by

$$\left\{\begin{array}{c}{\mathcal{K}}_a\equiv {\mathcal{K}}_a\left(r\right)=\frac{\pi }{2}F\left(a,1-a;1;r^2\right),\\ {\mathcal{K}}_a\left(0^{+}\right)=\pi /2,{\mathcal{K}}_a\left(1^{-}\right)=\infty ,\end{array}\right.$$

which can be expressed as a power series

$${\mathcal{K}}_a\left(r\right)=\frac{\pi }{2}\sum _{n=0}^{\infty }\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}{r}^{2n},$$

where ${\left(a\right)}_0=1$ for $a\ne 0$, ${\left(a\right)}_n\equiv a(a+1)(a+2)\cdots (a+n-1)=\mathsf{\Gamma }\left(a+n\right)/\mathsf{\Gamma }\left(a\right)$ is the shifted factorial function (or Pochhammer symbol) and $\mathsf{\Gamma }\left(x\right)={\int }_0^{\infty }{t}^{x-1}{e}^{-t}dt$$\left(x>0\right)$ is the gamma function (see [1,2,3]). In 2000, this special function was rediscovered by [4], in which the relationships between the modular function and the generalized elliptic integrals were revealed. In this way, the role of the generalized elliptic integrals in the field of geometric function theory is prominent. Of course, we know that the generalized elliptic integral of the first kind ${\mathcal{K}}_a\left(r\right)$ cannot be expressed by elementary function, and the simple and accurate estimation for ${\mathcal{K}}_a\left(r\right)$ becomes more and more urgent. In the particular case $a=1/2,$ the function ${\mathcal{K}}_a\left(r\right)$ reduces to $\mathcal{K}\left(r\right),$ the well-known complete elliptic integral of the first kind:

$$\mathcal{K}\left(r\right)={\int }_0^{\pi /2}\frac{dt}{\sqrt{1-r^2{sin}^2\left(t\right)}},$$

which is the particular case of the Gaussian hypergeometric function can be a special power series [5,6,7,8,9,10,11]:

$$\mathcal{K}\left(r\right)=\frac{\pi }{2}F\left(\frac{1}{2},\frac{1}{2};1;r^2\right)=\frac{\pi }{2}\sum _{n=0}^{\infty }\frac{{\left(\frac{1}{2}\right)}_n^2}{{(n!)}^2}{r}^{2n}=\frac{\pi }{2}\sum _{n=0}^{\infty }{\left[\frac{(2n-1)!!}{\left(2n\right)!!}\right]}^2{r}^{2n}.$$

(1)

Many researchers have obtained some results about this special function ${\mathcal{K}}_a\left(r\right)$ (see [12,13,14,15,16]). In [15], the upper and lower bounds for ${\mathcal{K}}_a\left(r\right)$ were shown, and a double inequality was proved as follows:

$$1+\alpha {\left(r^{\prime }\right)}^2<\frac{{\mathcal{K}}_a\left(r\right)}{sin\left(\pi a\right)log\left(e^{R\left(a\right)/2}/r^{\prime }\right)}<1+\beta {\left(r^{\prime }\right)}^2$$

holds for all $a\in (0,1/2]$ and $r\in (0,1)$ if and only if $\alpha \le \pi /\left[R\right(a)sin(\pi a\left)\right]-1$ and $\beta \ge a\left(1-a\right)$, where $R\left(x\right)$ is the Ramanujan constant function (see [17]). When $a=1/2$, the above inequality becomes the following estimation of the function $\mathcal{K}\left(r\right)$:

$$A\left(r^{\prime }\right)=:\left[1+\left(\frac{\pi }{4log2}-1\right){\left(r^{\prime }\right)}^2\right]log(4/r^{\prime })<\mathcal{K}\left(r\right)<\left[1+\frac{1}{4}{\left(r^{\prime }\right)}^2\right]log(4/r^{\prime }).$$

Recently, in [18], the author has shown a new concise bound for ${\mathcal{K}}_a\left(r\right)$ by a rational function of the arugment $r^{\prime }$ and obtained an inequality as follows.

Proposition 1.

Let $a\in (0,1),$$r\in (0,1),$ and $r^{\prime }=\sqrt{1-r^2}$. Then

$$\frac{2}{\pi }{\mathcal{K}}_a\left(r\right)>\frac{\left(1-3a+3a^2\right)r^{\prime }+\left(1+3a-3a^2\right)}{\left(1+a-a^2\right)r^{\prime }+\left(1-a+a^2\right)}=:{\mathcal{G}}_{1,1}\left(r^{\prime }\right).$$

(2)

The aim of this paper is to provide a better lower bound for ${\mathcal{K}}_a\left(r\right)$ and to obtain the following result.

Theorem 1.

Let $a\in (0,1),$$r\in (0,1),$$r^{\prime }=\sqrt{1-r^2},$ and

$$\begin{array}{ccc}\hfill u_1\left(a\right)& =& 2a-4a^2+4a^3-2a^4+3,\hfill \\ \hfill v_1\left(a\right)& =& 7a-5a^2-4a^3+2a^4+6,\hfill \\ \hfill u_2\left(a\right)& =& 17a+3a^2-35a^3+5a^4+15a^5-5a^6+3,\hfill \\ \hfill v_2\left(a\right)& =& -5a-19a^2+38a^3+6a^4-30a^5+10a^6+6,\hfill \\ \hfill w_2\left(a\right)& =& 3a-7a^2+3a^3+11a^4-15a^5+5a^6.\hfill \end{array}$$

Then

$$\frac{2}{\pi }{\mathcal{K}}_a\left(r\right)>\frac{u_2\left(a\right)+v_2\left(a\right)r^{\prime }-w_2\left(a\right){\left(r^{\prime }\right)}^2}{u_1\left(a\right)+v_1\left(a\right)r^{\prime }}=:{\mathcal{G}}_{2,1}\left(r^{\prime }\right).$$

(3)

It is not hard to find out that

$${\mathcal{G}}_{2,1}\left(r^{\prime }\right)-{\mathcal{G}}_{1,1}\left(r^{\prime }\right)=\frac{a\left(1-a\right)\left(1+a-a^2\right)\left(3-4a-a^2+10a^3-5a^4\right)}{\left[u_1\left(a\right)+v_1\left(a\right)r^{\prime }\right]\left[\left(1+a-a^2\right)r^{\prime }+\left(1-a+a^2\right)\right]}{\left(1-r^{\prime }\right)}^3>0$$

due to

$$\begin{array}{ccc}\hfill 1+a-a^2& =& 1+a\left(1-a\right)>0,\hfill \\ \hfill 3-4a-a^2+10a^3-5a^4& =& \frac{27}{16}+\frac{1}{16}\left[20a\left(1-a\right)+21\right]{\left(2a-1\right)}^2>0.\hfill \end{array}$$

On the other hand, we compute to obtain that

$$\begin{array}{ccc}\hfill \frac{2}{\pi }{\mathcal{K}}_a\left(r\right)-{\mathcal{G}}_{2,1}\left(r^{\prime }\right)& =& \frac{1}{5184}\frac{a\left(1-a\right)\left(2-a\right)\left(1+a\right)\left(9+3a-20a^2+27a^3+4a^4-21a^5+7a^6\right)}{1+a-a^2}{r}^8\hfill \\ & & +O\left(r^{10}\right),\hfill \\ \hfill \frac{2}{\pi }{\mathcal{K}}_a\left(r\right)-{\mathcal{G}}_{1,1}\left(r^{\prime }\right)& =& \frac{1}{288}a\left(1-a\right)\left(-10a^4+20a^3-2a^2-8a+6\right)r^6+O\left(r^8\right).\hfill \end{array}$$

The above two aspects illustrate the fact that ${\mathcal{G}}_{2,1}\left(r^{\prime }\right)$ is sharper than ${\mathcal{G}}_{1,1}\left(r^{\prime }\right)$ as the lower bound of $2{\mathcal{K}}_a\left(r\right)/\pi$, and the approximation accuracy of ${\mathcal{G}}_{2,1}\left(r^{\prime }\right)$ and $2{\mathcal{K}}_a\left(r\right)/\pi$ is higher than that of ${\mathcal{G}}_{1,1}\left(r^{\prime }\right)$ and $2{\mathcal{K}}_a\left(r\right)/\pi$. Letting $a=1/2,1/3,1/4,1/5$ and $1/6$ in the above theorem, respectively, we immediately draw the following corollaries.

Corollary 1.

Let $r\in (0,1)$ and $r^{\prime }=\sqrt{1-r^2}$. Then

$${\mathcal{K}}_{1/2}\left(r\right)\equiv \mathcal{K}\left(r\right)>\frac{\pi }{2}\frac{22r^{\prime }-3{\left(r^{\prime }\right)}^2+61}{8\left(7r^{\prime }+3\right)}=:B\left(r^{\prime }\right).$$

(4)

Corollary 2.

Let $r\in (0,1)$ and $r^{\prime }=\sqrt{1-r^2}$. Then

$${\mathcal{K}}_{1/3}\left(r\right)>\frac{\pi }{2}\frac{2620r^{\prime }-302{\left(r^{\prime }\right)}^2+5701}{9\left(620r^{\prime }+271\right)}.$$

(5)

Corollary 3.

Let $r\in (0,1)$ and $r^{\prime }=\sqrt{1-r^2}$. Then

$${\mathcal{K}}_{1/4}\left(r\right)>\frac{\pi }{2}\frac{1890r^{\prime }-177{\left(r^{\prime }\right)}^2+3151}{32\left(105r^{\prime }+47\right)}.$$

(6)

Corollary 4.

Let $r\in (0,1)$ and $r^{\prime }=\sqrt{1-r^2}$. Then

$${\mathcal{K}}_{1/5}\left(r\right)>\frac{\pi }{2}\frac{1578r^{\prime }-124{\left(r^{\prime }\right)}^2+2171}{5\left(498r^{\prime }+227\right)}.$$

(7)

Corollary 5.

Let $r\in (0,1)$ and $r^{\prime }=\sqrt{1-r^2}$. Then

$${\mathcal{K}}_{1/6}\left(r\right)>\frac{\pi }{2}\frac{224686r^{\prime }-15215{\left(r^{\prime }\right)}^2+268753}{72\left(4543r^{\prime }+2099\right)}.$$

(8)

It must be pointed out that the lower bound for the function ${\mathcal{K}}_a\left(r\right)$ given in this paper is concise and accurate. In order to express this advantage, we especially list the following two absolute error tables, one is the error table between the function ${\mathcal{K}}_{1/2}\left(r\right)\equiv \mathcal{K}\left(r\right)$ and the lower bound $B\left(r^{\prime }\right)$ given in this paper, and the other is the error table between the lower bound $B\left(r^{\prime }\right)$ given in this paper and the one $A\left(r^{\prime }\right)$ given in [17]. The latter shows the fact that the lower bound $B\left(r^{\prime }\right)$ for $\mathcal{K}\left(r\right)$ is better than the one $A\left(r^{\prime }\right)$ on large interval $\left(0,0.896\right)$ while the opposite is true on the small interval $\left(0.897,1\right)$.

r	$0.2$	$0.4$	$0.6$	$0.8$	0.9
$\mathcal{K}\left(r\right)-B\left(r^{\prime }\right)$	$9.264\times {10}^{-8}$	$1.0829\times {10}^{-6}$	$4.6501\times {10}^{-5}$	$1.2589\times {10}^{-3}$	$7.2937\times {10}^{-3}$

r	$0.2$	$0.4$	$0.6$	0.896	0.897	0.9
$B\left(r^{\prime }\right)-A\left(r^{\prime }\right)$	$4.3276\times {10}^{-4}$	$1.8001\times {10}^{-3}$	$4.1813\times {10}^{-3}$	$2.2286\times {10}^{-5}$	$-1.5716\times {10}^{-4}$	$-7.2626\times {10}^{-4}$

This paper does not find a concise upper bound for ${\mathcal{K}}_a\left(r\right)$, which is the lack of this paper and the direction of our future research.

2. Lemmas

In order to prove our main results, we need following lemmas.

Lemma 1.

Let $a\in \left(0,1\right)$. Then

$$\begin{array}{ccc}\hfill p_1\left(a\right)& =& 320a^{16}-2560a^{15}-12358a^{14}+131306a^{13}+47912a^{12}-2110930a^{11}+2431366a^{10}\hfill \\ & & +10627888a^9-24415064a^8+4758790a^7+30676747a^6-34613219a^5\hfill \\ & & +6595332a^4+14044275a^3-6691680a^2-1468125a+850500,\hfill \\ \hfill p_2\left(a\right)& =& 160a^8-640a^7+1181a^6-1303a^5+926a^4-427a^3-962a^2+1065a+630,\hfill \\ \hfill p_3\left(a\right)& =& \left(5a-a^2-8a^3+4a^4+3\right)\left(3a-20a^2+27a^3+4a^4-21a^5+7a^6+9\right),\hfill \\ \hfill p_4\left(a\right)& =& -64a^{12}+384a^{11}+91a^{10}-3975a^9+6627a^8+1566a^7-10967a^6+6501a^5\hfill \\ & & +2387a^4-3720a^3-99a^2+1269a+405,\hfill \\ \hfill p_5\left(a\right)& =& 24a^{14}-168a^{13}-374a^{12}+4428a^{11}-3226a^{10}-28464a^9+56916a^8-1308a^7\hfill \\ & & -80278a^6+69948a^5-287a^4-32610a^3+8595a^2+6804a+405,\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill q_0\left(a\right)& =& 11750400a^{18}-105753600a^{17}-414880920a^{16}+5716128960a^{15}-22946263740a^{14}\hfill \\ & & +52201803780a^{13}-50419694970a^{12}-84859350840a^{11}+345782467890a^{10}\hfill \\ & & -473759668260a^9+207934633170a^8+326470029120a^7-499680890010a^6\hfill \\ & & +73459525860a^5+239504763330a^4-128601169560a^3-45687166290a^2\hfill \\ & & +55393735680a+18631140480,\hfill \\ \hfill q_1\left(a\right)& =& 7355520a^{18}-66199680a^{17}-460485396a^{16}+5184409248a^{15}-20094499782a^{14}\hfill \\ & & +44682495354a^{13}-38298994641a^{12}-95681011932a^{11}+351060807177a^{10}\hfill \\ & & -484747863318a^9+228437739261a^8+326235974616a^7-529181029413a^6\hfill \\ & & +96006877038a^5+247228800669a^4-149141453148a^3-45535993317a^2\hfill \\ & & +64363071744a+21076451424,\hfill \\ \hfill q_2\left(a\right)& =& 1820640a^{18}-16385760a^{17}-202237032a^{16}+1989306816a^{15}-7439602170a^{14}\hfill \\ & & +15964408950a^{13}-10768123071a^{12}-47586780516a^{11}+156419035635a^{10}\hfill \\ & & -215734933230a^9+106769231949a^8+142465947240a^7-242127573681a^6\hfill \\ & & +51761145432a^5+110150008335a^4-74410069662a^3-19469546451a^2\hfill \\ & & +32234346576a+10304154018,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill q_3\left(a\right)& =& 222640a^{18}-2003760a^{17}-46593592a^{16}+418167296a^{15}-1501328128a^{14}\hfill \\ & & +3032404256a^{13}-1037284638a^{12}-13580494252a^{11}+39972088993a^{10}\hfill \\ & & -54555595847a^9+27792888997a^8+35530299826a^7-62535737692a^6\hfill \\ & & +15321758146a^5+27649917354a^4-20919117537a^3-4649493582a^2\hfill \\ & & +9109901520a+2848107501,\hfill \\ \hfill q_4\left(a\right)& =& 13440a^{18}-120960a^{17}-6150472a^{16}+51945536a^{15}-176449004a^{14}+316499988a^{13}\hfill \\ & & +118174802a^{12}-2424383888a^{11}+6410969255a^{10}-8576271881a^9+4410075650a^8\hfill \\ & & +5538352566a^7-9972614216a^6+2752273964a^5+4272836978a^4-3632097435a^3\hfill \\ & & -674968491a^2+1591914168a+487415637,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill q_5\left(a\right)& =& 320a^{18}-2880a^{17}-467856a^{16}+3808128a^{15}-11760928a^{14}+15455776a^{13}\hfill \\ & & +41621888a^{12}-276537952a^{11}+661130866a^{10}-858496122a^9+436434583a^8\hfill \\ & & +552928736a^7-1005339808a^6+309195430a^5+415612463a^4-399454548a^3\hfill \\ & & -60507162a^2+176379066a+52940871,\hfill \\ \hfill q_6\left(a\right)& =& -19040a^{16}+152320a^{15}-373208a^{14}-53144a^{13}+4544644a^{12}-19646432a^{11}\hfill \\ & & +42827192a^{10}-53461732a^9+26236688a^8+34556840a^7-62537458a^6\hfill \\ & & +21302234a^5+24791703a^4-27209088a^3-3225393a^2+12113874a+3566997,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill q_7\left(a\right)& =& -320a^{16}+2560a^{15}-1024a^{14}-37632a^{13}+233248a^{12}-793792a^{11}+1593832a^{10}\hfill \\ & & -1894696a^9+871444a^8+1237344a^7-2192896a^6+825160a^5+826276a^4\hfill \\ & & -1050528a^3-90936a^2+471960a+136404,\hfill \\ \hfill q_8\left(a\right)& =& 160a^{14}-1120a^{13}+4752a^{12}-13952a^{11}+26096a^{10}-29280a^9+12128a^8+19456a^7\hfill \\ & & -33144a^6+13784a^5+11716a^4-17616a^3-972a^2+7992a+2268\hfill \end{array}$$

are all positive.

Proof. 

Due to the similarity of methods, we only take the positive definite proofs of two functions $q_0\left(a\right)$ and $q_6\left(a\right)$, which are the most representative of all functions, as the examples for the proofs of other functions.

Let ${\left(a-1/2\right)}^2=t$. Then $0\le t<1/4$ and

$$\begin{array}{ccc}\hfill q_0\left(a\right)& =& \frac{254072971474335}{8192}-\frac{34837458958305}{1024}{\left(a-\frac{1}{2}\right)}^2-\frac{27049849283385}{256}{\left(a-\frac{1}{2}\right)}^4\hfill \\ & & +\frac{12290616417015}{64}{\left(a-\frac{1}{2}\right)}^6-\frac{473696824275}{8}{\left(a-\frac{1}{2}\right)}^8\hfill \\ & & -\frac{207114049035}{4}{\left(a-\frac{1}{2}\right)}^{10}+27734907465{\left(a-\frac{1}{2}\right)}^{12}+736483860{\left(a-\frac{1}{2}\right)}^{14}\hfill \\ & & -864333720{\left(a-\frac{1}{2}\right)}^{16}+11750400{\left(a-\frac{1}{2}\right)}^{18}\hfill \end{array}$$

$$\begin{array}{ccc}& =& \frac{254072971474335}{8192}-\frac{34837458958305}{1024}t-\frac{27049849283385}{256}{t}^2+\frac{12290616417015}{64}{t}^3\hfill \\ & & -\frac{473696824275}{8}{t}^4-\frac{207114049035}{4}{t}^5+27734907465t^6+736483860t^7\hfill \\ & & -864333720t^8+11750400t^9\hfill \end{array}$$

$$\begin{array}{ccc}& =& 18631140480-55393735680\left(t-\frac{1}{4}\right)+9706569390{\left(t-\frac{1}{4}\right)}^2\hfill \\ & & +109188030780{\left(t-\frac{1}{4}\right)}^3-97765898400{\left(t-\frac{1}{4}\right)}^4-9960024600{\left(t-\frac{1}{4}\right)}^5\hfill \\ & & +27526592610{\left(t-\frac{1}{4}\right)}^6-965745180{\left(t-\frac{1}{4}\right)}^7-837895320{\left(t-\frac{1}{4}\right)}^8\hfill \\ & & +11750400{\left(t-\frac{1}{4}\right)}^9\hfill \end{array}$$

$$\begin{array}{ccc}& =& 18631140480+\left[-55393735680\left(t-\frac{1}{4}\right)+109188030780{\left(t-\frac{1}{4}\right)}^3\right]\hfill \\ & & +\left[9706569390{\left(t-\frac{1}{4}\right)}^2-97765898400{\left(t-\frac{1}{4}\right)}^4\right]\hfill \\ & & +\left[-9960024600{\left(t-\frac{1}{4}\right)}^5+27526592610{\left(t-\frac{1}{4}\right)}^6\right]\hfill \\ & & +\left[-965745180{\left(t-\frac{1}{4}\right)}^7-837895320{\left(t-\frac{1}{4}\right)}^8+11750400{\left(t-\frac{1}{4}\right)}^9\right]\hfill \end{array}$$

$$\begin{array}{ccc}& =& 18631140480+\frac{557685}{16}\left(1-4t\right)\left(-783152t^2+391576t+348365\right)\hfill \\ & & +\frac{61965}{4}\left(-394440t^2+197220t+14509\right){\left(1-4t\right)}^2+\frac{2295}{2048}\left(14676839-23988316t\right){\left(1-4t\right)}^5\hfill \\ & & +\frac{2295}{8192}\left(-2560t^2+183828t+164605\right){\left(1-4t\right)}^7\hfill \\ & >& 0,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill q_6\left(a\right)& =& \frac{7122778335}{1024}-\frac{7183362609}{512}{\left(a-\frac{1}{2}\right)}^2-\frac{343651995}{128}{\left(a-\frac{1}{2}\right)}^4\hfill \\ & & +\frac{673429279}{32}{\left(a-\frac{1}{2}\right)}^6-\frac{80731943}{8}{\left(a-\frac{1}{2}\right)}^8-\frac{7156763}{2}{\left(a-\frac{1}{2}\right)}^{10}\hfill \\ & & +2206126{\left(a-\frac{1}{2}\right)}^{12}+197992{\left(a-\frac{1}{2}\right)}^{14}-19040{\left(a-\frac{1}{2}\right)}^{16}\hfill \end{array}$$

$$\begin{array}{ccc}& =& \frac{7122778335}{1024}-\frac{7183362609}{512}t-\frac{343651995}{128}{t}^2+\frac{673429279}{32}{t}^3-\frac{80731943}{8}{t}^4\hfill \\ & & -\frac{7156763}{2}{t}^5+2206126t^6+197992t^7-19040t^8\hfill \\ & =& :\frac{1}{1024}g\left(t\right),\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill g\left(t\right)& =& -19496960t^8+202743808t^7+2259073024t^6-3664262656t^5-10333688704t^4\hfill \\ & & +21549736928t^3-2749215960t^2-14366725218t+7122778335\hfill \\ & =& 3652604928-12404606976\left(t-\frac{1}{4}\right)+9101804544{\left(t-\frac{1}{4}\right)}^2+9658497024{\left(t-\frac{1}{4}\right)}^3\hfill \\ & & -12690591744{\left(t-\frac{1}{4}\right)}^4-26611712{\left(t-\frac{1}{4}\right)}^5+2579755008{\left(t-\frac{1}{4}\right)}^6\hfill \\ & & +163749888{\left(t-\frac{1}{4}\right)}^7-19496960{\left(t-\frac{1}{4}\right)}^8\hfill \end{array}$$

$$\begin{array}{ccc}& =& 3652604928+12404606976\left(\frac{1}{4}-t\right)+9101804544{\left(\frac{1}{4}-t\right)}^2-9658497024{\left(\frac{1}{4}-t\right)}^3\hfill \\ & & -12690591744{\left(\frac{1}{4}-t\right)}^4+26611712{\left(\frac{1}{4}-t\right)}^5+2579755008{\left(\frac{1}{4}-t\right)}^6\hfill \\ & & -163749888{\left(\frac{1}{4}-t\right)}^7-19496960{\left(\frac{1}{4}-t\right)}^8\hfill \end{array}$$

$$\begin{array}{ccc}& =& 3652604928+\left[12404606976\left(\frac{1}{4}-t\right)-9658497024{\left(\frac{1}{4}-t\right)}^3\right]\hfill \\ & & +\left[9101804544{\left(\frac{1}{4}-t\right)}^2-12690591744{\left(\frac{1}{4}-t\right)}^4\right]\hfill \\ & & +\left[26611712{\left(\frac{1}{4}-t\right)}^5-163749888{\left(\frac{1}{4}-t\right)}^7\right]\hfill \\ & & +\left[2579755008{\left(\frac{1}{4}-t\right)}^6-19496960{\left(\frac{1}{4}-t\right)}^8\right]\hfill \end{array}$$

$$\begin{array}{ccc}& =& 3652604928+10368\left(\frac{1}{4}-t\right)\left(465784t-931568t^2+1138209\right)\hfill \\ & & +768{\left(\frac{1}{4}-t\right)}^2\left(8262104t-16524208t^2+10818545\right)\hfill \\ & & +512{\left(\frac{1}{4}-t\right)}^5\left(159912t-319824t^2+31987\right)\hfill \\ & & +2048{\left(\frac{1}{4}-t\right)}^6\left(4760t-9520t^2+1259051\right)\hfill \\ & >& 0.\hfill \end{array}$$

□

Lemma 2.

Let $a\in \left(0,1\right),$$n\ge 7,$ and

$$\begin{array}{ccc}\hfill P\left(n\right)& =& \left(n-a\right)\left(a+n-1\right)\left[\begin{array}{c}\left(-135a^4+270a^3+81a^2-216a-81\right)\\ +\left(10a^6-30a^5+97a^4-144a^3-71a^2+138a+63\right)n\end{array}\right],\hfill \\ \hfill Q\left(n\right)& =& \left(2n-1\right)\left(2n-3\right)\left[\begin{array}{c}\left(-36a^8+144a^7-135a^6-99a^5+135a^4+63a^3-45a^2-27a\right)\\ +\left(-36a^6+108a^5-27a^4-126a^3+9a^2+72a+27\right)n\\ +\left(4a^8-16a^7+32a^6-40a^5+20a^4+8a^3-20a^2+12a+9\right)n^2\end{array}\right].\hfill \end{array}$$

Then

$$\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}>2a\left(a+1\right)\left(1-a\right)\left(2-a\right)\frac{P\left(n\right)}{Q\left(n\right)}\frac{{\left(\frac{1}{2}\right)}_n}{n!}$$

(9)

holds.

Proof. 

By using mathematical induction we can prove $\left(9\right)$. When $n=7$, if the difference between the left and right sides of the above formula is $r\left(a\right)$, then

$$r\left(a\right)=\frac{1}{10080}a\left(1-a\right)\left(2-a\right)\left(a+6\right)\left(7-a\right)\left(a+1\right)\frac{p_1\left(a\right)}{p_2\left(a\right)},$$

where $p_1\left(a\right)$ and $p_2\left(a\right)$ are defined as Lemma 1. By Lemma 1 we have $p_1\left(a\right)>0$ and $p_2\left(a\right)>0,$ so $r\left(a\right)>0$ which implies $\left(9\right)$ holds for $n=7$. Assuming that $\left(9\right)$ holds for $n=m\ge 8$, that is,

$$\frac{{\left(a\right)}_m{\left(1-a\right)}_m}{m!}>\frac{2a\left(a+1\right)\left(1-a\right)\left(2-a\right)P\left(m\right)}{Q\left(m\right)}{\left(\frac{1}{2}\right)}_m.$$

(10)

Next, we prove that $\left(9\right)$ is valid for $n=m+1$. By $\left(10\right)$, we have

$$\begin{array}{ccc}\hfill \frac{{\left(a\right)}_{m+1}{\left(1-a\right)}_{m+1}}{\left(m+1\right)!}& =& \frac{\left(a+m\right)\left(1-a+m\right)}{\left(m+1\right)}\frac{{\left(a\right)}_m{\left(1-a\right)}_m}{m!}\hfill \\ & >& \frac{\left(a+m\right)\left(1-a+m\right)}{\left(m+1\right)}\frac{2a\left(1-a\right)\left(2-a\right)\left(a+1\right)P\left(m\right)}{Q\left(m\right)}{\left(\frac{1}{2}\right)}_m,\hfill \end{array}$$

in order to complete the proof of $\left(9\right)$, it suffices to show that

$$\begin{array}{ccc}& & \frac{\left(a+m\right)\left(1-a+m\right)}{\left(m+1\right)}\frac{2a\left(1-a\right)\left(2-a\right)\left(a+1\right)P\left(m\right)}{Q\left(m\right)}{\left(\frac{1}{2}\right)}_m\hfill \\ & >& \frac{2a\left(a+1\right)\left(1-a\right)\left(2-a\right)P(m+1)}{Q(m+1)}{\left(\frac{1}{2}\right)}_{m+1},\hfill \end{array}$$

that is

$$\begin{array}{ccc}\hfill \frac{\left(a+m\right)\left(1-a+m\right)}{\left(m+1\right)}\frac{P\left(m\right)}{Q\left(m\right)}{\left(\frac{1}{2}\right)}_m& >& \frac{P(m+1)}{Q(m+1)}{\left(\frac{1}{2}\right)}_{m+1}\hfill \\ & =& \frac{P(m+1)}{Q(m+1)}{\left(\frac{1}{2}\right)}_m\left(\frac{1}{2}+m\right)\hfill \\ & ⟺& \\ \hfill \frac{\left(a+m\right)\left(1-a+m\right)}{\left(m+1\right)}\frac{P\left(m\right)}{Q\left(m\right)}& >& \frac{P(m+1)}{Q(m+1)}\frac{\left(2m+1\right)}{2}\hfill \\ & ⟺& \\ \hfill 2\left(a+m\right)\left(1-a+m\right)P\left(m\right)Q(m+1)& >& \left(m+1\right)\left(2m+1\right)P(m+1)Q\left(m\right).\hfill \end{array}$$

In fact,

$$2\left(a+m\right)\left(1-a+m\right)P\left(m\right)Q(m+1)-\left(m+1\right)\left(2m+1\right)P(m+1)Q\left(m\right)=\sum _{k=0}^8{q}_k\left(a\right){\left(m-8\right)}^k>0$$

for $m\ge 8$ because the coefficients of the power square of $\left(m-8\right)$ are positive by Lemma 1. □

Lemma 3

([19,20]). Let ${\left\{a_k\right\}}_{k=0}^{\infty }$ be a nonnegative real sequence with $a_m>0$ and ${\sum }_{k=m+1}^{\infty }{a}_k>0$, and

$$S\left(t\right)=-\sum _{k=0}^m{a}_k{t}^k+\sum _{k=m+1}^{\infty }{a}_k{t}^k$$

be a convergent power series on the interval $(0,r)$$(r>0)$. Then the following statements are true:

$\left(1\right)$ If $S\left(r^{-}\right)\le 0$, then $S\left(t\right)<0$ for all $t\in (0,r)$;

$\left(2\right)$ If $S\left(r^{-}\right)>0$, then there exists $t_0\in (0,r)$ such that $S\left(t\right)<0$ for $t\in (0,t_0)$ and $S\left(t\right)>0$ for $t\in (t_0,r)$.

3. Proof of Main Result

By

$$\frac{2}{\pi }{\mathcal{K}}_a\left(r\right)=\sum _{n=0}^{\infty }\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}{r}^{2n},{\left(1-r^2\right)}^{\rho }=\sum _{n=0}^{\infty }\frac{{\left(-\rho \right)}_n}{n!}{r}^{2n}$$

we have

$$\begin{array}{ccc}\hfill \mathcal{F}\left(r\right)& =& :\left[u_1\left(a\right)-v_1\left(a\right)r^{\prime }\right]\left[u_1\left(a\right)+v_1\left(a\right)r^{\prime }\right]\frac{2}{\pi }{\mathcal{K}}_a\left(r\right)\hfill \\ & & -\left[u_1\left(a\right)-v_1\left(a\right)r^{\prime }\right]\left[u_2\left(a\right)+v_2\left(a\right)r^{\prime }-w_2\left(a\right){\left(r^{\prime }\right)}^2\right]\hfill \\ & =& \left[\begin{array}{c}\left(-27-72a-9a^2+126a^3+27a^4-108a^5+36a^6\right)\\ +\left(36+84a-11a^2-118a^3-7a^4+68a^5-4a^6-16a^7+4a^8\right)r^2\end{array}\right]\frac{2}{\pi }{\mathcal{K}}_a\left(r\right)\hfill \\ & & +18a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-5a+5a^2-3\right)\left(-a+a^2-1\right)\sqrt{1-r^2}\hfill \\ & & +a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-2a+2a^2-3\right)\left(4a+a^2-10a^3+5a^4-3\right)r^2\sqrt{1-r^2}\hfill \\ & & -\left(90a^8-360a^7+252a^6+504a^5-513a^4-234a^3+225a^2+36a-27\right)\hfill \\ & & -r^2\left(\begin{array}{c}10a^{10}-50a^9-20a^8+380a^7-403a^6-331a^5\\ +508a^4+79a^3-194a^2+21a+36\end{array}\right)\hfill \end{array}$$

$$\begin{array}{ccc}& =& \left[\begin{array}{c}\left(-27-72a-9a^2+126a^3+27a^4-108a^5+36a^6\right)\\ +\left(36+84a-11a^2-118a^3-7a^4+68a^5-4a^6-16a^7+4a^8\right)r^2\end{array}\right]\sum _{n=0}^{\infty }\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}{r}^{2n}\hfill \\ & & +18a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-5a+5a^2-3\right)\left(-a+a^2-1\right)\sum _{n=0}^{\infty }\frac{{\left(-\frac{1}{2}\right)}_n}{n!}{r}^{2n}\hfill \\ & & +a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-2a+2a^2-3\right)\left(4a+a^2-10a^3+5a^4-3\right)r^2\sum _{n=0}^{\infty }\frac{{\left(-\frac{1}{2}\right)}_n}{n!}{r}^{2n}\hfill \\ & & -\left(90a^8-360a^7+252a^6+504a^5-513a^4-234a^3+225a^2+36a-27\right)\hfill \\ & & -r^2\left(10a^{10}-50a^9-20a^8+380a^7-403a^6-331a^5+508a^4+79a^3-194a^2+21a+36\right)\hfill \end{array}$$

$$\begin{array}{ccc}& =& \left(-27-72a-9a^2+126a^3+27a^4-108a^5+36a^6\right)\sum _{n=0}^{\infty }\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}{r}^{2n}\hfill \\ & & +\left(36+84a-11a^2-118a^3-7a^4+68a^5-4a^6-16a^7+4a^8\right)\sum _{n=0}^{\infty }\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}{r}^{2n+2}\hfill \\ & & +18a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-5a+5a^2-3\right)\left(-a+a^2-1\right)\sum _{n=0}^{\infty }\frac{{\left(-\frac{1}{2}\right)}_n}{n!}{r}^{2n}\hfill \\ & & +a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-2a+2a^2-3\right)\left(4a+a^2-10a^3+5a^4-3\right)\sum _{n=0}^{\infty }\frac{{\left(-\frac{1}{2}\right)}_n}{n!}{r}^{2n+2}\hfill \\ & & -\left(90a^8-360a^7+252a^6+504a^5-513a^4-234a^3+225a^2+36a-27\right)\hfill \\ & & -r^2\left(10a^{10}-50a^9-20a^8+380a^7-403a^6-331a^5+508a^4+79a^3-194a^2+21a+36\right)\hfill \end{array}$$

$$\begin{array}{ccc}& =& \left(-27-72a-9a^2+126a^3+27a^4-108a^5+36a^6\right)\sum _{n=4}^{\infty }\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}{r}^{2n}\hfill \\ & & +\left(36+84a-11a^2-118a^3-7a^4+68a^5-4a^6-16a^7+4a^8\right)\sum _{n=3}^{\infty }\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}{r}^{2n+2}\hfill \\ & & +18a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-5a+5a^2-3\right)\left(-a+a^2-1\right)\sum _{n=4}^{\infty }\frac{{\left(-\frac{1}{2}\right)}_n}{n!}{r}^{2n}\hfill \\ & & +a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-2a+2a^2-3\right)\left(4a+a^2-10a^3+5a^4-3\right)\sum _{n=3}^{\infty }\frac{{\left(-\frac{1}{2}\right)}_n}{n!}{r}^{2n+2}\hfill \end{array}$$

$$\begin{array}{ccc}& =& \left(-27-72a-9a^2+126a^3+27a^4-108a^5+36a^6\right)\sum _{n=4}^{\infty }\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}{r}^{2n}\hfill \\ & & +\left(36+84a-11a^2-118a^3-7a^4+68a^5-4a^6-16a^7+4a^8\right)\sum _{n=4}^{\infty }\frac{{\left(a\right)}_{n-1}{\left(1-a\right)}_{n-1}}{{(\left(n-1\right)!)}^2}{r}^{2n}\hfill \\ & & +18a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-5a+5a^2-3\right)\left(-a+a^2-1\right)\sum _{n=4}^{\infty }\frac{{\left(-\frac{1}{2}\right)}_n}{n!}{r}^{2n}\hfill \\ & & +a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-2a+2a^2-3\right)\left(4a+a^2-10a^3+5a^4-3\right)\sum _{n=4}^{\infty }\frac{{\left(-\frac{1}{2}\right)}_{n-1}}{\left(n-1\right)!}{r}^{2n}\hfill \\ & =& :\sum _{n=4}^{\infty }{c}_n{r}^{2n},\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill c_n& =& \left(-27-72a-9a^2+126a^3+27a^4-108a^5+36a^6\right)\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}\hfill \\ & & +\left(36+84a-11a^2-118a^3-7a^4+68a^5-4a^6-16a^7+4a^8\right)\frac{{\left(a\right)}_{n-1}{\left(1-a\right)}_{n-1}}{{(\left(n-1\right)!)}^2}\hfill \\ & & +18a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-5a+5a^2-3\right)\left(-a+a^2-1\right)\frac{{\left(-\frac{1}{2}\right)}_n}{n!}\hfill \\ & & +a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-2a+2a^2-3\right)\left(4a+a^2-10a^3+5a^4-3\right)\frac{{\left(-\frac{1}{2}\right)}_{n-1}}{\left(n-1\right)!}.\hfill \end{array}$$

Since

$${\left(-\frac{1}{2}\right)}_n=\left(-\frac{1}{2}\right){\left(\frac{1}{2}\right)}_{n-1}$$

we have

$$\begin{array}{ccc}\hfill c_n& =& \left(-27-72a-9a^2+126a^3+27a^4-108a^5+36a^6\right)\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}\hfill \\ & & +\left(36+84a-11a^2-118a^3-7a^4+68a^5-4a^6-16a^7+4a^8\right)\hfill \\ & & \times \frac{{\left(a\right)}_n{\left(1-a\right)}_n}{\left(a+n-1\right)\left(1-a+n-1\right){(\left(n-1\right)!)}^2}\hfill \\ & & +\left(-\frac{1}{2}\right)18a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-5a+5a^2-3\right)\left(-a+a^2-1\right)\frac{{\left(\frac{1}{2}\right)}_{n-1}}{n!}\hfill \\ & & +\left(-\frac{1}{2}\right)a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-2a+2a^2-3\right)\left(4a+a^2-10a^3+5a^4-3\right)\frac{{\left(\frac{1}{2}\right)}_{n-2}}{\left(n-1\right)!}.\hfill \end{array}$$

Considering

$${\left(\frac{1}{2}\right)}_{n-1}=\frac{{\left(\frac{1}{2}\right)}_n}{\frac{1}{2}+n-1}$$

we calculate to obtain

$$\begin{array}{ccc}\hfill c_n& =& \left(-27-72a-9a^2+126a^3+27a^4-108a^5+36a^6\right)\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}\hfill \\ & & +\left(36+84a-11a^2-118a^3-7a^4+68a^5-4a^6-16a^7+4a^8\right)\hfill \\ & & \times \frac{n^2}{\left(a+n-1\right)\left(1-a+n-1\right)}\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}\hfill \\ & & +\left(-\frac{1}{2}\right)18a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-5a+5a^2-3\right)\left(-a+a^2-1\right)\frac{1}{\left(\frac{1}{2}+n-1\right)}\frac{{\left(\frac{1}{2}\right)}_n}{n!}\hfill \\ & & +\left(-\frac{1}{2}\right)a\left(a-1\right)\left(a-2\right)\left(a+1\right)\left(-2a+2a^2-3\right)\left(4a+a^2-10a^3+5a^4-3\right)\hfill \\ & & \times \frac{n}{\left(\frac{1}{2}+n-1\right)\left(\frac{1}{2}+n-2\right)}\frac{{\left(\frac{1}{2}\right)}_n}{n!}\hfill \end{array}$$

$$\begin{array}{ccc}& =& \frac{\left[\begin{array}{c}\left(-36a^8+144a^7-135a^6-99a^5+135a^4+63a^3-45a^2-27a\right)\\ +\left(-36a^6+108a^5-27a^4-126a^3+9a^2+72a+27\right)n\\ +\left(4a^8-16a^7+32a^6-40a^5+20a^4+8a^3-20a^2+12a+9\right)n^2\end{array}\right]}{\left(n-a\right)\left(a+n-1\right)}\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}\hfill \\ & & -\frac{2a\left(1-a\right)\left(2-a\right)\left(a+1\right)\left[\begin{array}{c}\left(-135a^4+270a^3+81a^2-216a-81\right)\\ +\left(10a^6-30a^5+97a^4-144a^3-71a^2+138a+63\right)n\end{array}\right]}{\left(2n-1\right)\left(2n-3\right)}\frac{{\left(\frac{1}{2}\right)}_n}{n!}\hfill \end{array}$$

$$=\frac{Q\left(n\right)}{\left(2n-1\right)\left(2n-3\right)\left(n-a\right)\left(a+n-1\right)}\left[\frac{{\left(a\right)}_n{\left(1-a\right)}_n}{{(n!)}^2}-2a\left(1-a\right)\left(2-a\right)\left(a+1\right)\frac{P\left(n\right)}{Q\left(n\right)}\frac{{\left(\frac{1}{2}\right)}_n}{n!}\right],$$

where $P\left(n\right)$ and $Q\left(n\right)$ are shown in Lemma 2. We can calculate directly

$$\begin{array}{ccc}\hfill c_4& =& -\frac{1}{576}a\left(1-a\right)\left(2-a\right)\left(a+1\right)p_3\left(a\right),\hfill \\ \hfill c_5& =& -\frac{1}{14400}a\left(1-a\right)\left(2-a\right)\left(a+1\right)p_4\left(a\right),\hfill \\ \hfill c_6& =& -\frac{1}{115200}a\left(1-a\right)\left(2-a\right)\left(a+1\right)p_5\left(a\right).\hfill \end{array}$$

By Lemma 1 and Lemma 2 we know that $c_n<0$ for $4\le n\le 6$ and $c_n>0$ for all $n\ge 7$. Since

$$\mathcal{F}\left(1^{-}\right)=u_1^2\left(a\right)\frac{2}{\pi }{\mathcal{K}}_a\left(1^{-}\right)-u_1\left(a\right)u_2\left(a\right)=\left[\frac{2}{\pi }{\mathcal{K}}_a\left(1^{-}\right)-\frac{u_2\left(a\right)}{u_1\left(a\right)}\right]u_1^2\left(a\right)=\infty ,$$

by Lemma 3, we obtain that there exists $r_0\in (0,1)$ such that $\mathcal{F}\left(r\right)<0$ for $r\in (0,r_0)$ and $\mathcal{F}\left(r\right)>0$ for $r\in (r_0,1).$ At the same time, since

$$\mathcal{F}\left(r\right)=\left[u_1\left(a\right)-v_1\left(a\right)r^{\prime }\right]\left\{\left[u_1\left(a\right)+v_1\left(a\right)r^{\prime }\right]\frac{2}{\pi }{\mathcal{K}}_a\left(r\right)-\left[u_2\left(a\right)+v_2\left(a\right)r^{\prime }-w_2\left(a\right){\left(r^{\prime }\right)}^2\right]\right\},$$

and the function

$$\begin{array}{ccc}\hfill \mathcal{S}\left(r\right)& =& :u_1\left(a\right)-v_1\left(a\right)r^{\prime }\hfill \\ & =& \frac{\left(u_1\left(a\right)-v_1\left(a\right)\sqrt{1-r^2}\right)\left(u_1\left(a\right)+v_1\left(a\right)\sqrt{1-r^2}\right)}{u_1\left(a\right)+v_1\left(a\right)\sqrt{1-r^2}}\hfill \\ & =& \frac{u_1^2\left(a\right)-v_1^2\left(a\right)\left(1-r^2\right)}{u_1\left(a\right)+v_1\left(a\right)\sqrt{1-r^2}}=\frac{v_1^2\left(a\right)r^2-\left[v_1^2\left(a\right)-u_1^2\left(a\right)\right]}{u_1\left(a\right)+v_1\left(a\right)\sqrt{1-r^2}}\hfill \\ & =& \frac{v_1^2\left(a\right)}{u_1\left(a\right)+v_1\left(a\right)\sqrt{1-r^2}}\left\{r^2-\frac{v_1^2\left(a\right)-u_1^2\left(a\right)}{v_1^2\left(a\right)}\right\}\hfill \\ & =& \frac{v_1^2\left(a\right)}{u_1\left(a\right)+v_1\left(a\right)\sqrt{1-r^2}}\left[r-\frac{\sqrt{v_1^2\left(a\right)-u_1^2\left(a\right)}}{v_1\left(a\right)}\right]\left[r+\frac{\sqrt{v_1^2\left(a\right)-u_1^2\left(a\right)}}{v_1\left(a\right)}\right]\hfill \\ & =& \frac{v_1^2\left(a\right)}{u_1\left(a\right)+v_1\left(a\right)\sqrt{1-r^2}}\left[r-\frac{3\sqrt{\left(5a-a^2-8a^3+4a^4+3\right)\left(a-a^2+1\right)}}{\left(2-a\right)\left(a+1\right)\left(-2a^2+2a+3\right)}\right]\hfill \\ & & \times \left[r+\frac{3\sqrt{\left(5a-a^2-8a^3+4a^4+3\right)\left(a-a^2+1\right)}}{\left(2-a\right)\left(a+1\right)\left(-2a^2+2a+3\right)}\right]\hfill \end{array}$$

has a unique zero on $\left(0,1\right)$, which leads to

$$r_0=\frac{3\sqrt{\left(5a-a^2-8a^3+4a^4+3\right)\left(a-a^2+1\right)}}{\left(2-a\right)\left(a+1\right)\left(3+2a-2a^2\right)}.$$

The fact that $r_0\in \left(0,1\right)$ is true, because

$$r_0>0⟺3+2a-2a^2=3+2a\left(1-a\right)>0$$

and

$$\begin{array}{ccc}\hfill r_0& <& 1⟺9\left(5a-a^2-8a^3+4a^4+3\right)\left(a-a^2+1\right)<{\left[\left(2-a\right)\left(a+1\right)\left(3+2a-2a^2\right)\right]}^2\hfill \\ & ⟺& {\left[\left(2-a\right)\left(a+1\right)\left(3+2a-2a^2\right)\right]}^2-9\left(5a-a^2-8a^3+4a^4+3\right)\left(a-a^2+1\right)\hfill \\ & =& {\left(-3-2a+4a^2-4a^3+2a^4\right)}^2>0\hfill \end{array}$$

hold for all $a\in \left(0,1\right)$. Clearly, $\mathcal{S}\left(r\right)<0$ for $r\in (0,r_0)$ and $\mathcal{S}\left(r\right)>0$ for $r\in (r_0,1)$. In a word, the two functions $\mathcal{F}\left(r\right)$ and $\mathcal{S}\left(r\right)$ have the same sign on both sides of the point $r_0$, so we have

$$\frac{2}{\pi }{\mathcal{K}}_a\left(r\right)-{\mathcal{G}}_{2,1}\left(r^{\prime }\right)=\frac{\mathcal{F}\left(r\right)}{\mathcal{S}\left(r\right)\left[u_1\left(a\right)+v_1\left(a\right)r^{\prime }\right]}>0$$

holds for all $r\in \left(0,1\right)$ with $r\ne r_0$. Because these two functions ${\mathcal{K}}_a\left(r\right)$ and ${\mathcal{G}}_{2,1}\left(r^{\prime }\right)$ are continuous at the point $r_0$, the inequality $\left(2\right)$ still holds for $r=r_0$.

The proof of Theorem 1 is complete.

4. Conclusions

In this paper, we obtain a new simple rational approximation for ${\mathcal{K}}_a\left(r\right)$:

$$\frac{2}{\pi }{\mathcal{K}}_a\left(r\right)>\frac{u_2\left(a\right)+v_2\left(a\right)r^{\prime }-w_2\left(a\right){\left(r^{\prime }\right)}^2}{u_1\left(a\right)+v_1\left(a\right)r^{\prime }}=:{\mathcal{G}}_{2,1}\left(r^{\prime }\right),$$

which holds for all $r\in (0,1),$ where ${\mathcal{K}}_a\left(r\right)$ is the generalized elliptic integral of the first kind, $r^{\prime }=\sqrt{1-r^2},$ and

$$\begin{array}{ccc}\hfill u_1\left(a\right)& =& 2a-4a^2+4a^3-2a^4+3,\hfill \\ \hfill v_1\left(a\right)& =& 7a-5a^2-4a^3+2a^4+6,\hfill \\ \hfill u_2\left(a\right)& =& 17a+3a^2-35a^3+5a^4+15a^5-5a^6+3,\hfill \\ \hfill v_2\left(a\right)& =& -5a-19a^2+38a^3+6a^4-30a^5+10a^6+6,\hfill \\ \hfill w_2\left(a\right)& =& 3a-7a^2+3a^3+11a^4-15a^5+5a^6.\hfill \end{array}$$

In this way, when a takes any number in $(0,1/2]$, we will obtain a more accurate estimate of the corresponding function.