Abstract
For zero-balanced Gaussian hypergeometric functions , , we determine maximal regions of plane where well-known Landen identities for the complete elliptic integral of the first kind turn on respective inequalities valid for each . Thereby an exhausting answer is given to the open problem from the work by Anderson et al., 1990.
1. Introduction
Among special functions, the hypergeometric function has perhaps the widest range of applications. For instance, several well-known classes of mathematical physics are particular or limiting cases of it. For real numbers , , and with , the Gaussian hypergeometric function is defined by for , where for , and for . For many rational triples the function (1.1) can be expressed in terms of elementary functions and long lists of such particular cases are given in [1].
It is clear that small changes of the parameters will have small influence on the value of . In this paper we will study to what extent some well-known properties of the complete elliptic integral of the first kind, can be extended to for close to . Recall that is called zero-balanced if . In the zero-balanced case, there is a logarithmic singularity at and Gauss proved the asymptotic formula as tends to 1, where is the classical beta function. Note that and , see [2, Chapter 6].
Ramanujan found a much sharper asymptotic formula as tends to 1 (see also [3].) Here and in the sequel, and is the Euler-Mascheroni constant. Ramanujan’s formula (1.6) is a particular case of another well-known formula given in [2, 15.3.10].
We shall use in the sequel the following assertion which is a mixture of Biernacki-Krzyz and related results on the ratio of formal power series [4, 5].
Lemma 1.1. Suppose that the power series and have the radius of convergence and for all . Denote also (1)If the sequence is monotone increasing then is also monotone increasing on .(2)If the sequence is monotone decreasing then is also monotone decreasing on .(3)If the sequence is monotone increasing (decreasing) for and monotone decreasing (increasing) for , then there exists such that is increasing (decreasing) on and decreasing (increasing) on .
Some of the most important properties of the elliptic integral are the Landen identities [6, p. 507]: where , . In [4, Page 79], the following problem was raised.
Open Problem 1. Find an analog of Landen’s transformation formulas in (1.9) for . In particular, if and , is it true that for some constant and all ?
Since for , must be greater than 1.
Some other forms of Landen inequalities can be found in [7, 8].
In [4, pp. 20-21] and [9, Theorem 1.4] Gauss’ asymptotic formula (1.4) was refined by finding the lower and upper bounds for when or . Our second result gives a full solution to Open Problem 1.
We wish to point out that in [10, Theorem 1.2(1)] it was claimed that for , , the function is increasing in . As pointed out by Baricz [11] the proof contains a gap and the correct proof will be given here.
We also found another area in plane where the function is monotone decreasing in .
2. Main Results
Our first result shows that Landen inequalities hold not only in the neighborhood of the point but also in some unbounded parts of plane.
Theorem 2.1. For all with one has that the inequality
holds for each . Also, for , , the reversed inequality
takes place for each .
In the remaining region neither of the above inequalities holds for each .
The disjoint regions in plane and , where Landen inequalities hold, are shown on the Figure 1.
The only common point of the graphs in Figure 1 is where equality sign holds.
Two-sided bounds for the ratio of target functions are also possible.
Theorem 2.2. For each and , one has
For the inequalities are reversed,
The bounds in both pairs of inequalities are sharp and equality is reached for .
Some numerical estimations of the constant in Open Problem 1 follows.
Corollary 2.3. Let be defined as in Open Problem 1. Then, for each and , one has
In the region one has
Two-sided bounds for the difference exist in a smaller region (see Figure 1), where and in .
Theorem 2.4. Let be the classical Beta function and be defined by (1.7).
For , , one has
If , , then
The second Landen identity has the following counterpart for hypergeometric functions. The resulting inequalities might be called Landen inequalities for zero-balanced hypergeometric functions.
Theorem 2.5. Let .
For and each , one has
If , then
For , one has
3. Proofs
Throughout this section we denote where are fixed positive parameters and with the regions , , defined as above.
The basic results, which makes possible all proofs in the sequel, are contained in the following.
Lemma 3.1. The function is monotone decreasing in on and monotone increasing on .
The function is monotone decreasing on and monotone increasing on .
It should be noted that a general result of this kind was given in [12, Theorem 2.31].
Proof. We shall use Lemma 1.1 in the proof.
Since , , applying the lemma one can see that the monotonicity of depends on the sign of
Since and
it follows that(1)if , that is, , then ; hence for and is monotone decreasing in ;(2)if , that is, then , that is, , and is monotone increasing in .
In the second case we have , and, proceeding analogously, we get(3)If , that is, , , let . Then and
Since , it follows that both , are nonpositive. Therefore because both constants cannot be zero simultaneously. By Lemma 1.1, we conclude that the function is monotone decreasing in .
(4)If , that is, , , then , hence . Also . Therefore and . As above, we conclude that , and is monotone increasing in this case.
Proof of Theorem 2.1. By the above lemma, for each we have on and on .
Putting , , we get on ,
that is, by Landen’s identity,
The second inequality is obtained analogously.
It is easily seen by (3.3) that in the remaining region the sequence decreases and then increases. By Lemma 1.1, part 3, this means that the function , for some , decreases in and increases in . Therefore, putting and , one concludes that neither of the given inequalities holds for each .
Proof of Theorem 2.2. Since is monotone decreasing on , applying Gauss formula, we obtain
Therefore,
by the Landen identity.
The inequality valid on can be proved similarly.
Proof of Theorem 2.4. Both assertions of this theorem are a consequence of the following.Lemma 3.2. The function
is monotone increasing in on and monotone decreasing on .Proof. Let . Then
Hence
We used here the well-known formula
On the other hand, differentiating the first Landen identity we get
Since is monotone decreasing on and , we get , that is,
This, together with (3.15), yields
By (3.14) again, we get
Hence,
The last expression is positive on because and, by Lemma 3.1, the function is monotone decreasing on .
Therefore we proved that the function is monotone increasing in on .
Remark 3.3. Due to the remark in Section 1, this proof gives an affirmative answer to the 12-years-old hypothesis risen in [10].
Since is increasing on , we get
Hence, proceeding as before, it follows that
since is monotone increasing on .
Therefore is monotone decreasing in on and the proof of Lemma 3.2 is done.
By Lemma 3.2 we obtain on and on .
Evidently, .
Applying Ramanujan formula (1.6), we get
The assertion of Theorem 2.4 follows.
Proof of Theorem 2.5. Changing variable , we obtain
Putting this in Theorems 2.2 and 2.4, we obtain the assertions of Theorem 2.5.
Remark 3.4. As the referee notes, the results from Theorems 2.1 and 2.2 can be generalized for . This is left to the readers.
Acknowledgments
The authors are indebted to the referee for his/her constructive comments. The research of M. Vuorinen was supported by the Academy of Finland (Project 2600066611).