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On quotients and differences of hypergeometric functions
Journal of Inequalities and Applications volume 2011, Article number: 141 (2011)
Abstract
For Gaussian hypergeometric functions F(x) = F(a, b; c; x), a, b, c > 0, we consider the quotient Q F (x, y) = (F(x) + F(y))/F(z) and the difference D F (x, y) = F(x) + F(y) - F(z) for 0 < x, y < 1 with z = x + y - xy. We give best possible bounds for both expressions under various hypotheses about the parameter triple (a, b; c).
2010 Mathematics Subject Classification: 26D06; 33C05.
1. Introduction
Among special functions, the hypergeometric function has perhaps the widest range of applications. For instance, several well-known classes of special functions such as complete elliptic integrals, Legendre functions, Chebyshev and Jacobi polynomials, and some elementary functions, such as the logarithm, are particular cases of it, cf. [1]. In a recent article [2] the authors studied various extensions of the Bernoulli inequality for functions of logarithmic type. In particular, the zero-balanced hypergeometric function F(a, b; a + b; x), a, b > 0 occurs in these studies, because it has a logarithmic singularity at x = 1, see (2.8) below. We now continue the discussion of some of the questions for quotients and differences of hypergeometric functions that were left open in [2].
Motivated by the asymptotic behavior of the function F(x) = F(a, b; c; x) when x → 1-, see (2.8), we define for 0 < x, y < 1, a, b, c > 0
Our task in this article is to give tight bounds for these quotients and differences assuming various relationships between the parameters a, b, c.
For the general case, we can formulate the following theorem.
Theorem 1.2. For a, b, c > 0 and 0 < x, y < 1 let Q F be as in (1.1). Then,
The bounds in (1.3) are best possible as can be seen by taking [1, 15.1.8]
Then,
and the conclusion follows immediately. Similarly,
Theorem 1.4. For a, b > 0, c > a + b and 0 < x, y < 1, we have
with as the best possible constant.
Most intriguing is the zero-balanced case. For example,
Theorem 1.6. For a, b > 0 and 0 < x, y < 1 let D F be as in (1.1). Then,
with R = R(a, b) = -2γ - ψ(a) - ψ(b), B = B(a, b).
Both bounding constants are best possible.
In the sequel, we shall give a complete answer to an open question posed in [2].
2. Preliminary results
In this section, we recall some well-known properties of the Gaussian hypergeometric function F(a, b; c; x) and certain of its combinations with other functions, for further applications.
It is well known that hypergeometric functions are closely related to the classical gamma function Γ(x), the psi function ψ(x), and the beta function B(x, y). For Re x > 0, Re y > 0, these functions are defined by
respectively (cf. [1, Chap. 6]). It is well known that the gamma function satisfies the difference equation [1, 6.1.15]
and the reflection property [1, 6.1.17]
We shall also need the function
where γ is the Euler-Mascheroni constant given by
Given complex numbers a, b, and c with c ≠ 0, -1, -2,..., the Gaussian hypergeometric function is the analytic continuation to the slit plane ℂ\[1, ∞) of the series
Here (a, 0) = 1 for a ≠ 0, and (a, n) is the shifted factorial function or the Appell symbol
for n ∈ ℕ\{0}, where ℕ = {0, 1, 2,...}.
The hypergeometric function has the following simple differentiation formula ([1, 15.2.1])
An important tool for our study is the following classification of the behavior of the hypergeometric function near x = 1 in the three cases a + b < c, a + b = c, and a + b > c:
Some basic properties of this series may be found in standard handbooks, see for example [1]. For some rational triples (a, b, c), the functions F(a, b; c; x) can be expressed in terms of well-known elementary function. A particular case that is often used in this article is [1, 15.1.3]
It is clear that for a, b, c > 0 the function F(a, b; c; x) is a strictly increasing map from [0, 1) into [1, ∞) and that by (2.8) we see that it is onto [1, ∞) if a + b ≥ c. For a, b > 0 we see by (2.8) that xF(a, b; a + b; x) defines an increasing homeomorphism from [0, 1) onto [0, ∞).
Theorem 2.10. [3],[[4], Theorem 1.52] For a, b > 0, let B = B(a, b) be as in (2.1), and let R = R(a, b) be as in (2.4). Then the following are true.
-
(1)
The function is strictly increasing from (0, 1) onto (ab/(a + b), 1/B).
-
(2)
The function f2(x) ≡ BF (a, b; a + b; x) + log(1 - x) is strictly decreasing from (0, 1) onto (R, B).
-
(3)
The function f3(x) ≡ BF (a, b; a + b; x) + (1/x) log(1 - x) is increasing from (0, 1) onto (B - 1, R) if a, b ∈ (0, 1).
-
(4)
The function f3 is decreasing from (0, 1) onto (R, B - 1) if a, b ∈ (1, ∞).
-
(5)
The function
is decreasing from (0, 1) onto (1/B, 1) if a, b ∈ (0, 1).
-
(6)
If a, b > 1, then f4 is increasing from (0, 1) onto (1, 1/B).
-
(7)
If a = b = 1, then f4(x) = 1 for all x ∈ (0, 1).
We also need the following refinement of some parts of Theorem 2.10.
Lemma 2.11. [[5], Cor. 2.14] For a, b > 0, let B = B(a, b) be as in (2.1), and let R = R(a, b) be as in (2.4) and denote
(1) If a ∈ (0, ∞) and b ∈ (0, 1/a], then the function f is decreasing with range (1/B, 1);
(2) If a ∈ (1/2, ∞) and b ≥ a/(2a - 1), then f is increasing from (0, 1) to the range (1, 1/B).
(3) If a ∈ (0, ∞) and b ∈ (0, 1/a], then the function h defined by
is increasing from (0, 1) onto (B - 1, R).
(4) If a ∈ (1/3, ∞) and b ≥ (1 + a)/(3a - 1), then h is increasing from (0, 1) onto (R, B - 1).
For brevity, we write ℝ+ = (0, ∞).
Lemma 2.12. (Cf. [4, 1.24, 7.42(1)]) (1) If E(t)/t is an increasing function on ℝ+, then E is sub-additive, i.e., for each x, y > 0 we have that
(2) If E(t)/t decreases on ℝ+, then E is a super-additive function, that is
for x, y ∈ ℝ+.
3. Main results
By (2.8), the zero-balanced hypergeometric function F(a, b; a + b; x) has a logarithmic singularity at x = 1. We shall now demonstrate that its behavior is nearly logarithmic also in the sense that some basic identities of the logarithm yield functional inequalities for the zero-balanced function.
Next, writing the basic addition formula for the logarithm
in terms of the function g in (2.9), we have
Based on this observation and a few computer experiments, we posed in [2] the following question:
Question 3.1. Fix c, d > 0 and let g(x) = xF(c, d; c + d; x) for x ∈ (0, 1) and set
for x, y ∈ (0, 1).
-
(1)
For which values of c and d, this function is bounded from below and above?
-
(2)
Is it true that
-
a)
Q g (x, y) ≥ 1, if cd ≤ 1?
-
b)
Q g (x, y) ≤ 1, if c, d > 1?
-
c)
Are there counterparts of Theorem 1.6 for the function
We shall give a complete answer to this question in the sequel.
Note first that the quotient Q g is always bounded. Namely,
Theorem 3.2. For all c, d > 0 and all x, y ∈ (0, 1) we have that
A refinement of these bounds for some particular (c, d) pairs is given by the following two assertions.
Theorem 3.3. (1) For c, d > 0, cd ≤ 1 and x, y ∈ (0, 1) we have
(2) For c, d > 0, 1/c + 1/d ≤ 2 and x, y ∈ (0, 1) we have
Note that parts (1) and (3) of Lemma 2.11 imply that for c, d > 0, cd ≤ 1, (c, d) ≠ (1, 1) we have
Theorem 3.5. For cd ≤ 1 and x, y ∈ (0, 1) we have
We shall prove now the hypothesis from the second part of Question 3.1 under the condition 1/c + 1/d ≤ 2 in part (b) which, in particular, includes the case c > 1, d > 1.
Theorem 3.6. Fix c, d > 0 and let Q and g be as in Question 3.1.
-
(1)
If cd ≤ 1 then Q g (x, y) ≥ 1 for all x, y ∈ (0, 1).
-
(2)
If 1/c + 1/d ≤ 2, then Q g (x, y) ≤ 1 for all x, y ∈ (0, 1).
Counterparts of Theorem 1.6 for the difference D g are given in the next assertion.
Theorem 3.7. Fix c, d > 0 and let D be as in Question 3.1.
-
(1)
If cd ≤ 1, then
for all x, y ∈ (0, 1).
-
(2)
If 1/c + 1/d ≤ 2, then
for all x, y ∈ (0, 1).
Combining the results above, we obtain the following two-sided bounds for the quotient Q g .
Corollary 3.8. Fix c, d > 0 and let Q be as in Question 3.1.
-
(1)
If cd ≤ 1, then
for all x, y ∈ (0, 1).
-
(2)
If 1/c + 1/d ≤ 2, then
for all x, y ∈ (0, 1).
The assertions above represent a valuable tool for estimating quotients and differences of a hypergeometric function with different arguments. To illustrate this point, we give an example.
In [2], motivated by the relation with g as in (2.9), the authors asked the question about the bounds for the function S(t) defined by
where g(t):= tF(a, b; a + b; t), a, b > 0.
An answer follows instantly applying Corollary 3.8.
Theorem 3.9. Let, t ∈ (0, 1), with g(t):= tF (a, b; a + b; t), a, b > 0.
-
(1)
If ab ≤ 1, then
-
(2)
If 1/a + 1/b ≤ 2, then
4. Proofs
4.1. Proof of Theorem 1.2
The proof is based solely on the monotonicity property of the function F(x) = F(a, b; c; x). Namely, for x, y ∈ (0, 1), put z = x + y - xy, z ∈ (0, 1). Since
and F(u) is monotone increasing in u, we obtain
The left-hand bound is trivial. □
4.2. Proof of Theorem 1.4
The assertion of this theorem is a consequence of the previous one and (2.8). Indeed, from (1.3) we get
that is,
□
4.3. Proof of Theorem 1.6
Consider the function
where y, y ∈ (0, 1), is an independent parameter.
Since
We get (1 - x) s'(x)
Therefore, s(x) is monotone decreasing on (0, 1) and, consequently,
Also, by Theorem 2.10, part 2, we obtain
Since D F (0+, 0+) = 1 and D F (1-, 1-) = R/B, cited bounds are best possible. □
4.4. Proof of Theorem 3.2
Analogously to the proof of Theorem 1.2, we have
□
4.5. Proof of Theorem 3.3
Lemma 2.11 (1) yields
for u ∈ (0, 1), cd ≤ 1.
Therefore,
The lower bound is proved in the same way.
Applying part (2) of Lemma 2.11, we bound Q g similarly in the case 1/c + 1/d ≤ 2. □
Remark 4.1. From parts (1) and (3) of Lemma 2.11, we conclude that
for c, d > 0, cd ≤ 1, (c, d) ≠ (1, 1).
4.6. Proof of Theorem 3.4
Let us write B = B(c, d), R = R(c, d) and L = log(1/((1 - x)(1 - y))) > 0. By Lemma 2.11 (3) we have
Since x + y < 2(x + y - xy) we obtain by (4.2)
and
□
4.7. Proof of Theorem 3.6
By the first part of Lemma 2.11, f is monotone decreasing for cd ≤ 1.
Hence, for 0 < × < y < 1 we have
Putting 1 - x = e-u, 1 - y = e-v; u, v ∈ (0, ∞), we get that the inequality
holds whenever 0 < u < v < ∞.
This means that the function G(t)/t is monotone decreasing, where
By Lemma 2.12, it follows that G is super-additive, that is
which is equivalent to
and the proof of the first part of Theorem 3.6 is complete.
The proof of the second part is similar. Note that the condition c ∈ (1/2, ∞), d ≥ c/(2c - 1) of Lemma 2.11 is equivalent to the condition 1/c + 1/d ≤ 2 of Theorem 3.6. □
4.8. Proof of Theorem 3.7
-
(1)
The left-hand side of this inequality is a direct consequence of part (1) of Theorem 3.6.
Next, from Lemma 2.11, part (3) for u ∈ (0, 1), cd ≤ 1, we get
Hence, by the terminology from Theorem 3.2, we obtain
since Remark 4.1 yields R - B + 1 > 0 and B - 1 > 0.
-
(2)
To prove this part we shall use Lemma 2.11, part (4). Because d ≥ c/(2c - 1) ≥ (c + 1)/(3c - 1) and (1/2, ∞) ⊂ (1/3, ∞), we conclude that this assertion is valid under the condition 1/c + 1/d ≤ 2.
Therefore, for u ∈ (0, 1), 1/c + 1/d ≤ 2, (c, d) ≠ (1, 1), we get
and, as above,
since parts (1) and (4) of Lemma 2.11 give R - B + 1 < 0 and B - 1 < 0.
Because the right-hand inequality is a consequence of Theorem 3.6, part (2), the proof is complete. □
4.9. Proof of Theorem 3.9
Putting, we obtain z = x + y - xy = t. Therefore,
The rest is an application of Corollary 3.8. □
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Acknowledgements
The authors are indebted to the referee for his/her constructive comments. The research of Matti Vuorinen was supported by the Academy of Finland, Project 2600066611. This project also supported Slavko Simić' visit to Finland.
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Some of the main problems in this paper were motivated by earlier work and computer experiments of MV. SS found most proofs. All authors read and approved the final manuscript.
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Simić, S., Vuorinen, M. On quotients and differences of hypergeometric functions. J Inequal Appl 2011, 141 (2011). https://doi.org/10.1186/1029-242X-2011-141
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DOI: https://doi.org/10.1186/1029-242X-2011-141