Landen inequalities for hypergeometric functions

Theorem 1.2. For $a, b \in (0,1)$ , c = a + b and C = const. > 1, with $c \le 1$ , define the functions f and g on (0,1) by $$f(r) = (1 + \sqrt{r})F(a, b; c; r) - F(a, b; c; 4\sqrt{r}/(1 + \sqrt{r})^2),$$ $$g(r) = CF(a, b; c; r) - F(a, b; c; 4\sqrt{r}/(1 + \sqrt{r})^2),$$ respectively. Then we have: (1) For any $a,b \in (0,1)$ with $c \leq 1$ (c < 1, respectively), f is (strictly, respectively) increasing from (0,1) onto $(0,(R-\log 16)/B)$ , where B=B(a,b) and R=R(a,b) are defined by (1.4) and (1.6), respectively. In particular, for $a,b \in (0,1)$ with $c=a+b \leq 1$ , and for all $r \in (0,1)$ , (1.8) $$F(a,b;c;(2\sqrt{r}/(1+r))^2) \le (1+r)F(a,b;c;r^2)$$ $$\le F(a,b;c;(2\sqrt{r}/(1+r))^2) + [(R-\log 16)/B]$$ and (1.9) $$\frac{1+r}{2}F(a,b;c;1-r^2) \le F\left(a,b;c;\left(\frac{1-r}{1+r}\right)^2\right) \\ \le \frac{1+r}{2}\left[F(a,b;c;1-r^2) + \frac{1}{B}(R-\log 16)\right],$$ with equality in each instance if and only if a = b = 1/2. (2) If $C \geq 2$ , then g(r) > 0 for all $r \in (0,1)$ . Moreover, if $1 < C \leq 2$ , then g is strictly decreasing from (0,1) onto $(C_1, C-1)$ , where $$C_1 = \begin{cases} -\infty, & \text{if } 1 < C < 2, \\ (R - \log 16)/B, & \text{if } C = 2. \end{cases}$$ In particular, for all $r \in (0,1)$ , (1.10) $$F\left(a, b; c; \left(\frac{2\sqrt{r}}{1+r}\right)^{2}\right) - 1 < 2F(a, b; c; r) - 1$$ $$< F\left(a, b; c; \left(\frac{2\sqrt{r}}{1+r}\right)^{2}\right) < 2F(a, b; c; r) - \frac{1}{B}(R - \log 16).$$ (3) If 1 < C < 2, then $F(a,b;c;(2\sqrt{r}/(1+r))^2)$ and $CF(a,b;c;r^2)$ are not directly comparable for all $r \in (0,1)$ , that is, neither the inequality $$F(a, b; c; (2\sqrt{r}/(1+r))^2) \ge CF(a, b; c; r^2)$$ nor its reversed inequality holds for all $r \in (0,1)$ . From Theorem 1.2, we see that the first and second identities in (1.7) are the special case of (1.8) and (1.9), respectively, when a = b = 1/2. Parts (2) and (3) of Theorem 1.2 imply that in the Open Problem 1.1 the best value of C, for which $h(2\sqrt{r}/(1+r)) \leq Ch(r)$ , is 2. It should be noted that in (1.8) and (1.9), the hypergeometric functions have the same sets of parameters, in analogue with (1.7) where $\mathcal{K}(r)$ is this function. The so-called quadratic transformation formulas (see e.g. [AS, 15.3.15–15.3.32]) also yield formulas which give (1.7) as a particular case, but those formulas involve hypergeometric functions with two different sets of parameters. In [AVV1, p. 79], the following problem was put forward. (For the particular case a = b = 1/2 the answer is known to be affirmative.) OPEN PROBLEM 1.3. Is it true that the function $$W(r) \equiv F(a, b; a + b; r) + \frac{C}{r}\log(1 - r),$$ where C = 1/B(a, b), is monotone on (0, 1) for suitable a and b? In [AVV1, Theorem 6], Gauss' asymptotic formula (1.3) was refined by finding the lower and upper bounds for W(r), while in [ABRVV, Theorem 1.4] the function W(r) was shown to be monotone when $a, b \in (0, 1)$ or $a, b \in (1, \infty)$ . Our second result gives a full solution to the Open Problem 1.3.

Theorem 1.4. For $a, b \in (0, \infty)$ , c = a + b, let $a_1 = 1 - ab, a_2 = 2ab - a - b, a_3 = |a_1| + |a_2|$ , and define the function f on (0, 1) by $$f(r) = BF(a, b; c; r) + \frac{1}{r}\log(1 - r),$$ where B = B(a, b). Then we have: - (1) If $a_3 = 0$ , then $f(r) \equiv 0$ . - (2) If $a_3 \neq 0$ and $a_1 \geq \max\{0, a_2\}$ , then f is strictly increasing from (0, 1) onto (B 1, R), where R = R(a, b). - (3) If $a_3 \neq 0$ and $a_1 \leq \min\{0, a_2\}$ , then f is strictly decreasing from (0, 1) onto (R, B 1). - (4) In the other cases not stated in parts (1)–(3), that is, $a_2 < a_1 < 0$ , f is not always monotone on (0,1). Remark 1.1. It seems that there would be two possible cases not stated in parts (1)–(3) of Theorem 1.4: $$a_2 < a_1 < 0$$ if $a_2 < 0$ , and, $0 \le a_1 \le a_2$ if $a_2 \ge 0$ . However, one can show that there are no values of a and b such that $a_2 \ge a_1 \ge 0$ with $a_3 \ne 0$ . In fact, if $a_2 > a_1 = 0$ , then ab = 1 and $a_2 = 2 - (a+b) > 0$ so that $2 > a + b = a + (1/a) \ge 2$ , a contradiction. If $a_2 \ge a_1 > 0$ , then $$(1.11) a+b-3ab+1 \le 0 \text{ and } ab < 1.$$ Since if $a, b \le 1$ then $a_2 = a(b-1) + b(a-1) \le 0$ and since if $a, b \ge 1$ then $a_1 \le 0$ , we may assume that $0 < a < 1 \le b$ . Thus it follows from (1.11) that $a \ge (b+1)/(3b-1)$ and hence $$a_1 = 1 - ab \le 1 - \frac{b(b+1)}{3b-1} = -\frac{(b-1)^2}{3b-1} \le 0,$$ a contradiction again. It is not difficult for us to employ the Monotone l'Hôpital's rule [AVV2, Theorem 1.25] to prove that the function $$f(r) \equiv [\mathcal{K}(r) - \log(4/r')][(r'/r)^2 \log(1/r')]^{-1}$$ is strictly decreasing from (0,1) onto $(1/4,\pi-\log 16)$ , (see Corollary 2.1). Here and in the sequel, we let $r'=\sqrt{1-r^2}$ for $r\in(0,1)$ . Consider the function $$g(r) \equiv \frac{B(a,b)F(a,b;a+b;r) + \log(1-r) - R(a,b)}{[(1-r)/r]\log(1/(1-r))}.$$ Then for a = b = 1/2 we have $g(r^2) = f(r)$ . Therefore, it is natural to ask if the function g(r) is monotone on (0,1) for $a,b \in (0,\infty)$ . Our next result answers this question.

Theorem 1.5. For $a, b \in (0, \infty)$ , let $A_1 = A_1(a, b) = a + b + ab - 3$ , $A_2 = A_2(a, b) = a + b - 3ab + 1$ , $A = |A_1| + |A_2|$ , and define the function f on (0, 1) by $$f(r) = \frac{BF(a,b;a+b;r) + \log(1-r) - R}{[(1-r)/r]\log[1/(1-r)]},$$ where B = B(a,b) and R = R(a,b) are defined by (1.4) and (1.6), respectively. Then we have the following conclusions: - (1) If A = 0, then $f(r) \equiv 1$ . - (2) If $A \neq 0$ and $A_1 \leq \min\{0, A_2\}$ , then f is strictly decreasing from (0, 1) onto (ab, B R). In particular, with this condition, for all $r \in (0, 1)$ , (1.12) $$\begin{cases} ab\frac{1-r}{r}\log\frac{1}{1-r} < BF(a,b;a+b;r) + \log(1-r) - R \\ < (B-R)\frac{1-r}{r}\log\frac{1}{1-r}. \end{cases}$$ (3) If $A \neq 0$ and $A_1 \geq \max\{0, A_2\}$ , then f is strictly increasing from (0, 1) onto (B - R, ab). In particular, under this condition, for all $r \in (0, 1)$ , $$(1.13) \begin{cases} (B-R)\frac{1-r}{r}\log\frac{1}{1-r} < BF(a,b;a+b;r) + \log(1-r) - R \\ < ab\frac{1-r}{r}\log\frac{1}{1-r}. \end{cases}$$ (4) In the other cases not stated in parts (1)–(3), namely, $0 < A_1 < A_2$ , f is not always monotone on (0,1) for all a and b with $0 < A_1 < A_2$ . Theorem 1.5 gives two sided estimates for the O-term in Ramanujan's asymptotic formula (1.4).

Corollary 2.1. The function $$f(r) \equiv \frac{\mathcal{K}(r) - \log(4/r')}{(r'/r)^2 \log(1/r')}$$ is strictly decreasing from (0,1) onto $(1/4,\pi-\log 16)$ . We shall next compare Corollary 2.1 to some earlier bounds for the function $\mathcal{K}(r)$ . For this purpose we need the following proposition.

Proposition 2.2. The function $g(r) \equiv (\log(1/r'))/(r^2 \log(4/r'))$ is strictly increasing and convex from (0,1) onto $(1/\log 16,1)$ . In particular, for all $r \in (0,1)$ , $$1/\log 16 < g(r) < (1/\log 16) + [1 - (1/\log 16)]r.$$

Corollary 2.4. Let $D_1$ and $D_2$ be as in Remark 2.1, $D_4$ and $D_5$ as in Remark 2.2, B = B(a, b) and R = R(a, b). Then for all $r \in (0, 1)$ , (2.37) $$\max \left\{ ab \frac{1-r}{r} \log \frac{1}{1-r}, B-R-1 + \frac{1-r}{r} \log \frac{1}{1-r} \right\}$$ $$< BF(a,b;a+b;r) + \log(1-r) - R$$ $$< (B-R) \frac{1-r}{r} \log \frac{1}{1-r}$$ if $(a,b) \in D_1$ , and $$(2.38) \quad (B-R)\frac{1-r}{r}\log\frac{1}{1-r} < BF(a,b;a+b;r) + \log(1-r) - R$$ $$< \min\left\{ab\frac{1-r}{r}\log\frac{1}{1-r}, B-R-1 + \frac{1-r}{r}\log\frac{1}{1-r}\right\}$$ if $(a,b) \in D_5$ .

Theorem 2.5. For $a, b, c \in (0, \infty)$ with c < a + b, let d = a + b - c, and define the function f on (0,1) by $$f(r) = (1 - r)^d F(a, b; c; r).$$ Then (2.41) $$f'(r) = \frac{1}{c}(c-a)(c-b)(1-r)^{d-1}F(a,b;c+1;r),$$ and we have: - (1) If c = a or c = b, then $f(r) \equiv 1$ . - (2) If $c < \min\{a,b\}$ or $c > \max\{a,b\}$ , then f is strictly increasing from (0,1) onto (1,D), where D = D(a,b,c) is defined by (2.40). Moreover, if $ab \ge (d-1)(c+1)$ (resp. ab < (d-1)(c+1)), then f is convex (resp. concave) on (0,1). - (3) If $\min\{a,b\} < c < \max\{a,b\}$ , then f is strictly decreasing from (0,1) onto (D,1). Moreover, if $ab \ge (d-1)(c+1)$ (resp. ab < (d-1)(c+1)), then f is concave (resp. convex) on (0,1).