On some refinements for inequalities involving zero-balanced hypergeometric function

1.   Introduction

Let $a, b, c\in\mathbb{R}$ with $c\neq0, -1, -2, \cdots$. Then the Gaussian hypergeometric function $F(a, b;c; r)$ ^[1,2,3] is defined by

where $(a)_0 = 1$, $(a)_n = a(a+1)(a+2)\cdots(a+n-1) = \Gamma(a+n)/\Gamma(a)$ for $n\in\mathbb{N} = \{1, 2, \cdots\}$ is the Pochhammer symbol and $\Gamma(x) = \int_0^\infty t^{x-1}e^{-t}dt$ is the classical Euler gamma function ^[4,5]. If $c = a+b$, then $F(a, b;a+b; r)$ is said to be zero-balanced. In particular, the complete elliptic integral $\mathcal{K}(r)$ ^[6,7] and generalized complete elliptic integral $\mathcal{K}_a(r)$ $(r\in(0, 1), a\in(0, 1/2])$ ^[8] of the first kind are the special cases of the Gaussian hypergeometric function $F(a, b;c; r)$. Indeed, $\mathcal{K}(r)$ and $\mathcal{K}_a(r)$ can be expressed by

and

In 2016, Takeuchi ^[9] introduced the complete $p$-elliptic integral $K_p(r)$ of the first kind in terms of the Gaussian hypergeometric functions as follows

where $\pi_p$ is given by

and

is the Beta function.

Recently, the Gaussian hypergeometric function and its special cases have attracted the attention of many researchers ^{[10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25]} due to they have wide applications in pure and applied mathematics ^{[26,27,28,29,30,31,32,33,34,35,36,37,38,39,40]}.

Anderson et al. ^[41] proved that the double inequality

holds for all $r\in(0, 1)$.

Motivated by inequality (1.3), many researchers provided its improvements, variants, refinements and generalizations. For example, Alzer and Richards ^[42] proved that the double inequality

holds for all $a\in(0, 1/2]$ and $r\in(0, 1)$ with the best possible factors $\sigma(a) = a(1-a)$ and $\tau_a = 0$.

Recently, Yin et al. ^[43] generalized inequality (1.4) to the case of complete $p$-elliptic integral of the first kind $\mathcal{K}_p(r)$ and a sharp improvement of (1.4) was presented by Zhao et al. in ^[8]. Ismail [42, p. 1669] asked whether the inequality (1.4) can be extended to the zero-balanced hypergeometric function. Inspired by this question, Richards ^[44] proved the following result which is to answer the question by Ismail from another point of view.

Theorem A. (See ^[44]) Let $a, b > 0$. Then the double inequality

holds for all $r\in(0, 1)$ with the best possible exponents $\lambda(a, b) = ab/(a+b)$ and $\mu(a, b) = 0$.

The main purpose of this paper is to generalize the inequality (1.4) to the case of zero-balanced hypergeometric function and also make some refinements of (1.4) and Theorem A under certain restriction of $a, b$, which gives an affirmative answer to the question by Ismail. Our main result is the following Theorem 1.1.

Throughout this paper, we mainly focus on the parameters satisfying $a+b\geq6ab/5$ for $a, b > 0$. By the symmetry, our parameters might be only consider as $0 < a < b$ and $a+b\geq6ab/5$, which is equivalent to

For convenience, we denote $\sigma(a, b)$ and $\tau(a, b)$ by $\sigma$ and $\tau$ simply if no risk for confusion, where

Theorem 1.1. Let $a, b > 0$ with $a+b\geq6ab/5$. Then the double inequality

holds for all $r\in(0, 1)$ if and only if $\alpha\leq\alpha_0$ and $\beta\geq\beta_0$, where

Remark 1.2. For later use, we need discuss about the sign of $\alpha_0$ and $\alpha_0+\tau$ for $a, b > 0$ with $a+b\geq6ab/5$.

● Let $\widehat{\alpha}(a, b) = a (2 + a) + (1 - a) (2 + a) b + (1 - a) b^2$. Then it is easy to see that $\widehat{\alpha}(a, b) > 0$ for $0 < a\leq1$ and $b > 0$. Moreover, it follows from (1.5) that

for $1 < a\leq5/3$ and $a < b\leq 5a/(6a-5)$. This yields $\alpha_0 < 0$.

● By calculations, we obtain

From the above expression, we clearly see that $\alpha_0+\tau > 0$ for $0 < a\leq1/2$ (or $1\leq a\leq5/3$) and $b > 0$, and $\alpha_0+\tau < 0$ for $1/2 < a\leq5/6$ and sufficiently large $b > 0$.

As mentioned in (1.2), if $b = 1-a$, then $F(a, b;a+b; r^2)/F(a, b;a+b; r)$ reduces to the ratio of generalized complete elliptic integral of the first kind $\mathcal{K}_a(r)/\mathcal{K}_a(\sqrt{r})$ and $a(1-a)\leq1/4$ for $0 < a < 1$. That is to say Theorem 1.1 in ^[8] can be derived from our Theorem 1.1 as a corollary.

Corollary 1.3. Let $a\in(0, 1/2]$. Then the double inequality

holds for all $r\in(0, 1)$, where $\sigma(a) = \sigma(a, 1-a), \tau(a) = \tau(a, 1-a)$ and $\lambda(a) = \alpha_0(a, 1-a)$, $\mu(a) = \beta_0(a, 1-a)$ are defined in the literature ^[8].

Remark 1.4. Corollary 1.3 gives an affirmative answer to the question by Ismail. Moreover, the bounds for inequality (1.7) are better than (1.4). Indeed, from $\lambda(a) < 0$, $\sigma(a) = \tau(a)+\mu(a)$ and $\tau(a)+\lambda(a) = a(1-a)\big[(1- 2 a) (6 + 13 a + 33 a^2) + 2 a^3 (25 + 4 a)\big]/36 > 0$ for $a\in(0, 1/2]$, we obtain

and

for $r\in(0, 1)$ and $a\in(0, 1/2].$

Remark 1.5. Under the assumption of Theorem 1.1, the upper bound of (1.6) is better than that in Theorem A due to

from $\sigma = \tau+\beta_0$. On the other hand, in order to compare the lower bound of (1.6) and Theorem A, it suffices to take into account the sign of $f(r): = (1+\tau r^2+\alpha_0r^3)(1+r)^\sigma-(1+\sigma r)$.

Differentiation yields

where

It follows from

and $\alpha_0 < 0$ that $\hat{f}(r)$ can be regarded as the special polynomial defined in Lemma 2.1. We can verify but miss the details that $\hat{f}(1) < 0$ for $a, b > 0$ with $a+b\geq6ab/5$. This in conjunction with Lemma 2.1 implies that there exists $r^*\in(0, 1)$ such that $\hat{f}(r) > 0$ for $r\in(0, r^*)$ and $\hat{f}(r) < 0$ for $r\in(r^*, 1)$. Combining this with (1.8) and $f(0) = f'(0) = 0$, we conclude that $f(r) > 0$ for $r\in(0, r^*]$ and $f(r)$ is strictly concave on $(r^*, 1)$.

For $a, b > 0$ with $a+b\geq 6ab/5$, we have the following two conclusion:

● if $f(1)\geq0$, then $f(r) > \min\{f(r^*), f(1)\}\geq0$ for $r\in(r^*, 1)$. This yields the lower bound of (1.6) is better than that in Theorem A for $r\in(0, 1)$ and we refer to see the domain of $a, b$ illustrated in Figure 1;

● if $f(1) < 0$, then there exists $r'\in(r^*, 1)$ such that the lower bound of (1.6) is better than that in Theorem A for $r\in(0, r')$.

2.   Lemmas

In this section, we introduce some notations and present some technical lemmas, which will be used in proving our main result.

Let $a, b > 0$ and

Then we clearly see from (1.1) that $F(a, b;a+b; r)$ can be expressed simply as

It is easy to verify that $\Delta_n$ satisfies the recurrence relation

and also $\Delta_n$ is strictly decreasing for $n\geq0$ if $a+b\geq ab$.

The following lemma provides a simple criterion to determine the sign of a class of special polynomial.

Lemma 2.1. (See [45, Lemma 7) Let $n, m\in\mathbb{N}\cup\{0\}$ with $n > m$ and $P_n(t)$ be the polynomial of degree $n$ defined by

where $a_m, a_n > 0$ and $a_i\geq0$ for $0\leq i\leq n-1$ with $i\neq m$. Then there exist $t_0\in(0, \infty)$ such that $P_n(t_0) = 0$ and $P_n(t) > 0$ for $t\in(0, t_0)$ and $P_n(t) < 0$ for $t\in(t_0, \infty)$.

Lemma 2.2. (1) For $a, b > 0$, then $n\Delta_n(a, b)$ is strictly increasing for $n\geq1$. In particular,

for $m > n\geq1$.

(2) For $a, b > 0$ with $a+b\geq6ab/5$, then $\Delta_{n+1}/\Delta_n$ is strictly increasing for $n\geq1$.

(3) For $a, b > 0$ with $a+b\geq ab$, then $\Delta_{n+1}/\Delta_n > \Delta_{2n+2}/\Delta_{2n}$ holds for $n\geq0$.

Proof. (1) From the recurrence relation (2.2) of $\Delta_n$, we clearly see that

for $a, b > 0$ and $n\geq1$. This yields the monotonicity of $n\Delta_n(a, b)$ with respect to $n$ and inequality (2.3) follows directly from the monotonicity of $n\Delta_n(a, b)$.

(2) Taking the differentiation of (2.2) with respect to $n$ yields

where

● If $0 < a\leq 1$ and $b > 0$, then we clearly see that $\delta_1(n; a, b) > 0$ for $n\geq1$.

● If $1 < a\leq5/3$, then it follows from (1.5) that

for $n\geq1$.

Therefore, the proof is completed from (2.4) and $\delta_1(n; a, b) > 0$.

(3) From (2.2) we clearly see that

where

Combining this with $a+b\geq ab$ and (2.5) yields the desired result.

Lemma 2.3. Let $a, b > 0$ with $a+b\geq 6ab/5$ and $\rho_n(a, b) = (1+\sigma)\Delta_{n+2}-2(\alpha_0+\tau)\Delta_{2n+2}-2\Delta_{2n+4}$. Then $\rho_n(a, b) > 0$ for $n\geq2$.

Proof. Let $\hat{\rho}_n(a, b) = (1+\sigma)\Delta_{n+2}-2\Delta_{2n+4}$. Then Remark 1.2 makes us to know that the sign of $\alpha_0+\tau$ can not be determined. We divide into two cases to complete the proof by mathematical induction.

Case 1 $\alpha_0+\tau\leq0$. It suffices to show that $\hat{\rho}_n(a, b) > 0$ for $n\geq0$.

From the definition of $\Delta_n$, we compute that

where

It follows from $0 < a\leq1$ that $\eta(a, b) > 0$. For $a > 1$, we clearly see that $\eta(a, b) > 0$ from (1.5) and Lemma 2.1 together with

This in conjunction with (2.6) yields $\hat{\rho}_0(a, b) > 0$.

We assume that $\hat{\rho}_k(a, b) > 0$, namely, $(1+\sigma)\Delta_{k+2} > 2\Delta_{2k+4}$ for $k\geq0.$ By the induction hypothesis, it follows from Lemma 2.2(3) and $n = k+2$ that

This completes the proof of Case 1.

Case 2 $\alpha_0+\tau > 0$.

By simple calculations, $\rho_2(a, b)$ can be simplified as

where

Obviously, it follows that $\epsilon_0(a) > 0, \epsilon_1(a) > 0$ for $a > 0$ and $\epsilon_7(a)\geq0$ for $0 < a\leq1$ and $\epsilon_7(a) < 0$ for $1 < a\leq5/3.$ Moreover, $\epsilon_k(a)$ $(k = 2, 3, \cdots, 6)$ is the special polynomial defined in Lemma 2.1. It follows from Lemma 2.1 and

that $\epsilon_2(a) > 0, \epsilon_3(a) > 0$ for $0 < a\leq5/3$ and there exits $a_k\in(0, 5/3)$ $(k = 4, 5, 6)$ such that $\epsilon_k(a) > 0$ for $0 < a < a_k$ and $\epsilon_k(a) < 0$ for $a_k < a\leq5/3$. The roots $a_k$ can be computed numerically as follows $a_4 = 1.65774\cdots > a_5 = 1.41613\cdots > a_6 = 1.22785\cdots > 1$. So

● for $0 < a\leq1$, then it follows that $\sum_{k = 0}^7\epsilon_k(a)b^k > 0$;

● for $1 < a\leq5/3$, then we can consider the following intervals $(1, a_6],$ $(a_6, a_5],$ $(a_5, a_4]$, $(a_4, 5/3]$ and $\sum_{k = 0}^7\epsilon_k(a)b^k$ can be regarded as the special polynomial defined in Lemma 2.1 on each such interval, which yields $\sum_{k = 0}^7\epsilon_k(a)b^k > 0$ following from (1.5) and Lemma 2.1 together with

This in conjunction with (2.7) implies $\rho_2(a, b) > 0$.

Suppose that $\rho_k(a, b) > 0$ for $k\geq2$. We now prove that $\rho_{k+1}(a, b) > 0$.

The induction hypothesis enables us to know that

It follows from Lemma 2.2(2) and (3) with $n = k+2$ that

This in conjunction with (2.8) yields

Lemma 2.4. Let $a, b > 0$ with $a+b\geq6ab/5$ and

Then (i) $A_n > 0$; (ii) $A_n+B_n > 0$ for $n\geq0.$

Proof. (ⅰ) By Lemma 2.2(2) and Bernoulli's inequality we know that

for $n\geq2$. This in conjunction with (2.2) and the monotonicity of $\Delta_n$ with respect to $n$ gives

where $\delta_3(n; a, b) = \zeta_0(a, b)+\zeta_1(a, b)n+\zeta_2(a, b)n^2$ and $\zeta_0(a, b) = \sum_{k = 0}^5\varepsilon^0_k(a)b^k$, $\zeta_1(a, b) = \sum_{k = 0}^5\varepsilon^1_k(a)b^k$, $\zeta_2(a, b) = \sum_{k = 0}^4\varepsilon^2_k(a)b^k$, and

From the above expressions, we clearly see that $\zeta_2(a, b) > 0$ for $0 < a\leq5/3$ and $b > 0$. Moreover,

which gives $2\zeta_2(a, b)+\zeta_1(a, b) > 0$ for $0 < a\leq1$. By the same argument as in Case 2 of Lemma 2.3, $2\zeta_2(a, b)+\zeta_1(a, b)$ can be regarded as the special polynomial of $b$ defined in Lemma 2.1 for $1 < a\leq5/3$, which implies $2\zeta_2(a, b)+\zeta_1(a, b) > 0$ from Lemma 2.1 and

From the above discussion, we clearly see that for $n\geq2$,

Similarly, it follows from (2.10) that $\hat{\zeta}(a, b) > 0$ for $0 < a\leq1$ and $\hat{\zeta}(a, b) > 0$ for $1 < a\leq5/3$ from Lemma 2.1 and

This in conjunction with (2.9) and (2.10) implies that $A_n > 0$ for $n\geq2$.

It remains to verify the sign of $A_0$ and $A_1$. We only give the details of calculations for $A_0$ and similar for $A_1$.

By the definition of $\Delta_n$, $A_0$ can be rewritten as

where

The same argument as in (2.10) enables us to know that $\hat{\zeta}_0(a, b) > 0$ for $0 < a\leq1$ and $\hat{\zeta}_0(a, b) > 0$ for $1 < a\leq5/3$ from Lemma 2.1 and

(ⅱ) We first verify the sign of $A_0+B_0$ and $A_1+B_1$. Since the method is similar as above, we only give some necessary expression for $A_0+B_0$ and then similar for $A_1+B_1$.

By calculations, we obtain

where

We can compute easily that $\hat{\zeta}_1(a, b) > 0$ for $0 < a\leq1$ and $\hat{\zeta}_1(a, b) > 0$ for $1 < a\leq5/3$ from Lemma 2.1 and

For $n\geq2$, it follows from $\alpha_0 < 0$ and Lemma 2.3 together with the monotonicity of $\Delta_n$ with respect to $n$ that

Lemma 2.5. Let $a, b > 0$ with $a+b\geq ab$ and

Then (i) $C_n < 0$; (ii) $C_n+D_n < 0$ for $n\geq0.$

Proof. (ⅰ) By calculations, we obtain

The relation $\sigma = \beta_0+\tau$ allows us to rewrite $C_n$ as

where

From Lemma 2.2(1) it suffices to show $\hat{C}_n < 0$ for $n\geq1$.

For $n\geq1$, Lemma 2.2(1) and $\tau, \beta_0 > 0$ together with the monotonicity of $\Delta_n$ with respect to $n$ lead to

Simplifying $\hat{C}^*_n$ gives rise to

This in conjunction with (2.11) and (2.12) completes the proof of (ⅰ).

(ⅱ) For $n\geq0$, it follows from Lemma 2.2(1) and $\sigma = \tau+\beta_0$ together with the monotonicity of $\Delta_n$ with respect to $n$ that

This completes the proof.

3.   Proof of Theorem 1.1

Proof. Define

and

In order to prove the inequalities (1.6) is valid, it suffices to show $\varphi_{a, b}(r) > 0$ and $\phi_{a, b}(r) < 0$ for $r\in(0, 1)$.

From (2.1), we can rewrite $\varphi_{a, b}(r)$ and $\phi_{a, b}(r)$, in terms of power series, as

where $A_n, B_n$ and $C_n, D_n$ are defined in Lemma 2.4 and Lemma 2.5, respectively.

From (3.1) and (3.2), we only need to prove that $A_n+B_nr > 0$ and $C_n+D_nr < 0$ for $r\in(0, 1)$ and $n\geq0$.

Case 1 $A_n+B_nr > 0$.

● If $B_n\geq0$, then it follows from Lemma 2.4(i) that $A_n+B_nr\geq A_n > 0$ for $r\in(0, 1)$;

● If $B_n < 0$, then Lemma 2.4(ii) enables us to know that $A_n+B_nr > A_n+B_n > 0$ for $r\in(0, 1)$.

Case 2 $C_n+D_nr < 0$.

● If $D_n\leq0$, then it follows from Lemma 2.5(i) that $C_n+D_nr\leq C_n < 0$ for $r\in(0, 1)$;

● If $D_n > 0$, then from Lemma 2.5(ii) we clearly see that $C_n+D_nr < C_n+D_n < 0$ for $r\in(0, 1)$.

We are now in a position to prove $\alpha_0$ and $\beta_0$ are the best possible constants.

Let

Then we clearly see that

For $\alpha_0 < c < \beta_0$, then it follows from (3.4) that there exist sufficiently small $r_1, r_2\in(0, 1)$ such that $\Phi_{a, b}(r) < c$ for $r\in(0, r_1)$ and $\Phi_{a, b}(r) > c$ for $r\in(1-r_2, 1)$. In other words,

and

This completes the proof of Theorem 1.1.

Remark 3.1. In the proof of Theorem A ^[44], the parameters $a$ and $b$ are required to meet the condition that $a+b > ab$, although this is still more relaxed than our parameter's condition $a+b\geq6ab/5$ in Theorem 1.1.

It is natural to ask that can our parameter's condition be relaxed to $a+b > ab$? From Lemma 2.5, we clearly see that the upper bound in Theorem 1.1 still holds for $a+b\geq ab$ but the lower bound is not true. The reason is that $A_0(a, b) < 0$ for $3/2\leq a\leq3$ and $a+b = ab$. More precisely,

for $3/2\leq a\leq3$. By the continuity, it follows from (3.1) that there exists a sufficiently small $r_{a, b}\in(0, 1)$ such that $\varphi_{a, b}(r) < 0$ for $r\in(0, r_{a, b})$ when $a$ and $b$ lie in a very narrow strip near the curve segment $\{(a, b)|\ 3/2\leq a\leq3, a+b = ab\}$. We focus on the condition that $a+b\geq6ab/5$ just to make the calculation simpler. Of course, one can give some refinements for our parameter condition.

To this end, numerical experiment results allow us to pose the following conjecture.

Conjecture. Let $a, b > 0$ with $a+b\geq6ab/5$ and $\Phi_{a, b}(r)$ be defined as in (3.3). Then $\Phi_{a, b}(r)$ is strictly increasing from $(0, 1)$ onto $\big(\alpha_0, \beta_0\big)$.

4.   Discussions

This paper deals with the zero-balanced hypergeometric function ${_2F_1}(a, b;a+b; r)$. In this study, we present an elegant double inequality for ${_2F_1}(a, b;a+b; r^2)/{_2F_1}(a, b;a+b; r)$, which gives some refinements for some previously known results and also answers to the question by Ismail in the affirmative.

5.   Conclusions

In this paper, we have established a sharp double inequality involving the ratio of zero-balanced hypergeometric function ${_2F_1}(a, b;a+b; r^2)/{_2F_1}(a, b;a+b; r)$. More precisely, the double inequality

holds for all $r\in(0, 1)$ with the best possible constants $\alpha_0$ and $\beta_0$, where

Acknowledgments

The authors would like to thank the anonymous referees for their valuable comments and suggestions, which led to considerable improvement of the article.

This work was supported by the National Natural Science Foundation of China (Grant Nos. 11971142, 61673169, 11701176, 11871202) and the Natural Science Foundation of Zhejiang Province (Grant No. LY19A010012).

Conflict of interest

The authors declare that they have no competing interests.