# Landen Inequalities for Zero-Balanced Hypergeometric Functions

Authors: Slavko Simić; Matti Vuorinen
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

For zero-balanced Gaussian hypergeometric functionsF(a,b;a+b;x), a,b>0, we determine maximal regions of ab plane where well-known Landen identities for the complete elliptic integral of the first kind turn on respective inequalities valid for each x∈(0,1). Thereby an exhausting answer is given to the open problem from the work by Anderson et al., 1990.


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## Body

## 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 numbersa, b, and c with c≠0,-1,-2,…, the Gaussian hypergeometric function is defined by(1.1)F(a,b;c;x)≔2F1(a,b;c;x)=∑n=0∞(a,n)(b,n)(c,n)xnn!,
for x∈(-1,1), where(1.2)(a,n):=a(a+1)(a+2)⋯(a+n-1),
for n=1,2,…, and (a,0)=1 for a≠0. For many rational triples (a,b,c) 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 parametersa,b,c will have small influence on the value of F(a,b;c;x). In this paper we will study to what extent some well-known properties of the complete elliptic integral of the first kind,(1.3)K(x)≡π2F(12,12;1;x2)=∫0π/2(1-x2sin⁡2t)-1/2dt,x∈(0,1),
can be extended to F(a,b;a+b;x) for (a,b) close to (1/2,1/2). Recall that F(a,b;c;r) is called zero-balanced if c=a+b. In the zero-balanced case, there is a logarithmic singularity at r=1 and Gauss proved the asymptotic formula(1.4)F(a,b;a+b;r)~-1B(a,b)log⁡(1-r),
as r tends to 1, where(1.5)B(z,w)≡Γ(z)Γ(w)Γ(z+w),Rez>0,Rew>0
is the classical beta function. Note that Γ(1/2)=π and B(1/2,1/2)=π, see [2, Chapter 6].Ramanujan found a much sharper asymptotic formula(1.6)B(a,b)F(a,b;a+b;r)+log⁡(1-r)=R(a,b)+O((1-r)log⁡(1-r)),
as r tends to 1 (see also [3].) Here and in the sequel,(1.7)R(a,b)≡-Ψ(a)-Ψ(b)-2γ,R(12,12)=log⁡16,Ψ(z)≡ddz(log⁡Γ(z))=Γ′(z)Γ(z),Rez>0,
and γ 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 seriesf(x)=∑n≥0f̂nxn and g(x)=∑n≥0ĝnxn have the radius of convergence r>0 and ĝn>0 for all n∈{0,1,2,…}. Denote also
(1.8)h(x)=f(x)g(x)=∑n≥0ĥnxn.(1)
If the sequence{f̂n/ĝn}n≥0 is monotone increasing then h(x) is also monotone increasing on (0,r).(2)
If the sequence{f̂n/ĝn}n≥0 is monotone decreasing then h(x) is also monotone decreasing on (0,r).(3)
If the sequence{f̂n/ĝn} is monotone increasing (decreasing) for 0<n≤n0 and monotone decreasing (increasing) for n>n0, then there exists x0∈(0,r) such that h(x) is increasing (decreasing) on (0,x0) and decreasing (increasing) on (x0,r).Some of the most important properties of the elliptic integral𝒦(r) are the Landen identities [6, p. 507]:(1.9)K(2r1+r)=(1+r)K(r),K(1-r1+r)=1+r2K′(r),
where 𝒦′(r)=𝒦(1-r2), r∈(0,1). In [4, Page 79], the following problem was raised.Open Problem 1.
Find an analog of Landen’s transformation formulas in (1.9) for F(a,b;a+b;r). In particular, if k(r)=F(a,b;a+b;r2) and a,b∈(0,1), is it true that
(1.10)k(2r1+r)≤Ck(r)
for some constant C and all r∈(0,1)?Since2r/(1+r)>r for r∈(0,1), C 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(1.11)W(r)=B(a,b)F(a,b;a+b;r)+(1x)log⁡(1-r),
when a,b∈(0,1) or a,b∈(1,∞). 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 a,b∈(0,1), c=a+b≤1, the function(1.12)s(r)=(1+r)F(a,b;c;r)-F(a,b;c;4r(1+r)2)
is increasing in r∈(0,1). As pointed out by Baricz [11] the proof contains a gap and the correct proof will be given here.We also found another area inab plane where the function s(r) is monotone decreasing in r∈(0,1).


## 2. Main Results

Our first result shows that Landen inequalities hold not only in the neighborhood of the pointa=b=1/2 but also in some unbounded parts of ab plane.Theorem 2.1.
For alla,b>0 with ab≤1/4 one has that the inequality
(2.1)F(a,b;a+b;4r(1+r)2)≤(1+r)F(a,b;a+b;r2)
holds for each r∈(0,1). Also, for a,b>0, 1/a+1/b≤4, the reversed inequality
(2.2)F(a,b;a+b;4r(1+r)2)≥(1+r)F(a,b;a+b;r2),
takes place for each r∈(0,1).
In the remaining regiona,b>0⋀ab>1/4⋀1/a+1/b>4 neither of the above inequalities holds for each r∈(0,1).The disjoint regions inab plane D1={(a,b)∣a,b>0,ab≤1/4} and D2={(a,b)∣a,b>0,1/a+1/b≤4}, where Landen inequalities hold, are shown on the Figure 1.Figure 1
The domainsDj, j=1,2,3 visualized.The only common point of the graphs in Figure1 is (1/2,1/2) where equality sign holds.Two-sided bounds for the ratio of target functions are also possible.Theorem 2.2.
For eachr∈(0,1) and (a,b)∈D1, one has
(2.3)1≤(1+r)F(a,b;a+b;r2)F(a,b;a+b;4r/(1+r)2)≤B(a,b)π.
For(a,b)∈D2 the inequalities are reversed,(2.4)B(a,b)π≤(1+r)F(a,b;a+b;r2)F(a,b;a+b;4r/(1+r)2)≤1.
The bounds in both pairs of inequalities are sharp and equality is reached fora=b=1/2.Some numerical estimations of the constantC in Open Problem 1 follows.Corollary 2.3.
Letk(·) be defined as in Open Problem 1. Then, for each r∈(0,1) and (a,b)∈D1, one has
(2.5)πB(a,b)k(r)<k(2r1+r)<2k(r).
In the regionD2 one has(2.6)k(r)<k(2r1+r)<2πB(a,b)k(r).Two-sided bounds for the difference exist in a smaller regionD3⊂D1 (see Figure 1), where D3={(a,b)∣a,b>0,a+b≤1} and in D2.Theorem 2.4.
LetB=B(a,b) be the classical Beta function and R=R(a,b) be defined by (1.7).
Fora,b>0, a+b≤1, one has(2.7)0≤(1+r)F(a,b;a+b;r)-F(a,b;a+b;4r(1+r)2)≤R-log⁡16B.
Ifa,b>0, 1/a+1/b≤4, then(2.8)0≤F(a,b;a+b;4r(1+r)2)-(1+r)F(a,b;a+b;r)≤log⁡16-RB.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.
LetF(x)=F(a,b;a+b;x).
For(a,b)∈D1 and each x∈(0,1), one has(2.9)12<F(((1-x)/(1+x))2)(1+x)F(1-x2)<B(a,b)2π.
If(a,b)∈D3, then(2.10)(1+x)F(1-x2)≤2F((1-x1+x)2)≤(1+x)[F(1-x2)+R-log⁡16B].
For(a,b)∈D2, one has(2.11)B(a,b)2π<F(((1-x)/(1+x))2)(1+x)F(1-x2)<12,0≤(1+x)F(1-x2)-2F((1-x1+x)2)≤(1+x)(log⁡16-R)B.


## 3. Proofs

Throughout this section we denote(3.1)F(x)=F(a,b;a+b;x),G(x)=F(a,b;a+b+1;x),
where a,b,(a,b)≠(1/2,1/2) are fixed positive parameters and(3.2)F0(x)=F(12,12;1;x),G0(x)=F(12,12;2;x),
with the regions D1, D2, D3 defined as above.The basic results, which makes possible all proofs in the sequel, are contained in the following.Lemma 3.1.
(1) The function f(r)=F(r)/F0(r) is monotone decreasing in r∈(0,1) on D1 and monotone increasing on D2.
(2) The function g(r)=G(r)/G0(r) is monotone decreasing on D3 and monotone increasing on D2.It should be noted that a general result of this kind was given in [12, Theorem 2.31].Proof.
We shall use Lemma1.1 in the proof.
SinceF̂n=(a)n(b)n/(a+b)n(1)n, F0̂n=((1/2)n/(1)n)2, applying the lemma one can see that the monotonicity of {F̂n/F0̂n} depends on the sign of(3.3)Tn=T(a,b;n)=n(ab-14)+ab-a+b4=C1n+C2.
Since(a,b)≠(1/2,1/2) and(3.4)C2=abab+1/2C1-(a-b)24,
it follows that(1)
ifC1≤0, that is, (a,b)∈D1, then C2<0; hence Tn<0 for n=0,1,2,… and f(r) is monotone decreasing in r∈(0,1);(2)
ifC2≥0, that is, (a,b)∈D2 then C1>0, that is, Tn>0, n=0,1,2,… and f(r) is monotone increasing in r.
In the second case we haveĜn=(a)n(b)n/(a+b+1)n(1)n, G0̂n=((1/2)n/(1)n)2/(n+1) and, proceeding analogously, we get(3.5)Tn′=n(ab+a+b-54)+2ab-a+b4-14=C3n+C4.(3)
If(a,b)∈D3, that is, a,b>0, a+b≤1, let a+b=k>0. Then ab≤k2/4 and(3.6)C3≤k24+k-54=(k-1)(k+5)4;C4≤k22-k4-14=(k-1)(2k+1)4.
Since0<k≤1, it follows that both C3, C4 are nonpositive. Therefore Tn′<0,n=0,1,2,… because both constants cannot be zero simultaneously. By Lemma 1.1, we conclude that the function g(r) is monotone decreasing in r∈(0,1).
(4)
If(a,b)∈D2, that is, a,b>0, 1/a+1/b≤4, then 4ab≥a+b≥2ab, hence ab≥1/4. Also a+b≥2ab≥2·(1/2)=1. Therefore C3≥0 and C4=(ab-1/4)+(4ab-a-b)/4≥0. As above, we conclude that Tn′>0, n=0,1,2,… and g(r) is monotone increasing in this case.Proof of Theorem2.1.
By the above lemma, for each0<x<y<1 we have f(x)>f(y) on D1 and f(x)<f(y) on D2.
Puttingx=x(r)=r2, y=y(r)=4r/(1+r)2, we get on D1,(3.7)F(r2)F0(r2)>F(y)F0(y),
that is, by Landen’s identity,
(3.8)F(y)<F0(y)F0(r2)F(r2)=(1+r)F(r2).
The second inequality is obtained analogously.
It is easily seen by (3.3) that in the remaining region the sequence {F̂n/F0̂n} decreases and then increases. By Lemma 1.1, part 3, this means that the function f(r), for some r0∈(0,1), decreases in (0,r0) and increases in (r0,1). Therefore, putting 0<x(r)<y(r)<r0 and r0<x(r)<y(r)<1, one concludes that neither of the given inequalities holds for each r∈(0,1).Proof of Theorem2.2.
Sincef(r) is monotone decreasing on D1, applying Gauss formula, we obtain
(3.9)1=lim⁡r→0+F(r)F0(r)>F(r)F0(r)>lim⁡r→1-F(r)F0(r)=B(1/2,1/2)B(a,b)=πB(a,b).
Therefore,(3.10)F(y(r))F(x(r))<B(a,b)πF0(y(r))F0(x(r))=(1+r)B(a,b)π,
by the Landen identity.
The inequality valid onD2 can be proved similarly.Proof of Theorem2.4.
Both assertions of this theorem are a consequence of the following.Lemma 3.2.
The function(3.11)s(r)=(1+r)F(a,b;a+b;r)-F(a,b;a+b;4r(1+r)2)
is monotone increasing in r∈(0,1) on D3 and monotone decreasing on D2.Proof.
Letz=4r/(1+r)2. Then
(3.12)1-z=(1-r)2(1+r)2;dzdr=2(1-r)r(1+r)3.
Hence
(3.13)s1(r):=2r(1-r)s′(r)=(1-r)F(a,b;a+b;r)+2r(1-r)F′(a,b;a+b;r)-41+r(1-z)F′(a,b;a+b;z)=(1-r)F(a,b;a+b;r)+2aba+brF(a,b;a+b+1;r)-4ab(a+b)(1+r)F(a,b;a+b+1;z)=(1-r)F(r)+2aba+brG(r)-4ab(a+b)(1+r)G(z).
We used here the well-known formula
(3.14)(1-x)F′(a,b;a+b;x)=aba+bF(a,b;a+b+1;x).
On the other hand, differentiating the first Landen identity we get(3.15)11+rG0(z)=(1-r)F0(r)+12rG0(r).
Sinceg(r) is monotone decreasing on D3 and 0<r<z<1, we get g(r)>g(z), that is,(3.16)G(z)<G0(z)G0(r)G(r).
This, together with (3.15), yields(3.17)s1(r)>(1-r)F(r)+2aba+brG(r)-4ab(a+b)(1+r)G0(z)G0(r)G(r)=(1-r)F(r)+2aba+brG(r)-4ab(a+b)((1-r)F0(r)G0(r)+12r)G(r)=(1-r)(F(r)-4ab(a+b)F0(r)G0(r)G(r)).
By (3.14) again, we get(3.18)4ab(a+b)G(r)G0(r)=F′(r)F0′(r).
Hence,(3.19)2rs′(r)>F(r)-F′(r)F0′(r)F0(r)=F2(r)F0′(r)(F0(r)F(r))′.
The last expression is positive on D3 because D3⊂D1 and, by Lemma 3.1, the function f(r)=F(r)/F0(r) is monotone decreasing on D1.
Therefore we proved that the functions(r) is monotone increasing in r∈(0,1) on D3.
Remark 3.3.
Due to the remark in Section1, this proof gives an affirmative answer to the 12-years-old hypothesis risen in [10].
Sinceg(r) is increasing on D2, we get(3.20)G(z)>G0(z)G0(r)G(r).
Hence, proceeding as before, it follows that(3.21)2rs′(r)<F2(r)F0′(r)(F0(r)F(r))′<0,
since f(r)=F(r)/F0(r) is monotone increasing on D2.
Therefores(r) is monotone decreasing in r∈(0,1) on D2 and the proof of Lemma 3.2 is done.
By Lemma3.2 we obtain limr→0+s(r)<s(r)<limr→1-s(r) on D3 and limr→1-s(r)<s(r)<limr→0+s(r) on D2.
Evidently,limr→0+s(r)=0.
Applying Ramanujan formula (1.6), we get(3.22)lim⁡r→1-s(r)=lim⁡r→1-⁡(R-2log⁡(1-r)+log⁡(1-z)+o(1))B=lim⁡r→1-(R-2log⁡(1-r)(1+r)+2log⁡⁡((1-r)/(1+r))+o(1))B=(R-log⁡16)B.
The assertion of Theorem2.4 follows.Proof of Theorem2.5.
Changing variable(1-r)/(1+r)=x∈(0,1), we obtain
(3.23)r=1-x1+x;1+r=21+x;4r(1+r)2=1-x2.
Putting this in Theorems2.2 and 2.4, we obtain the assertions of Theorem 2.5.Remark 3.4.
As the referee notes, the results from Theorems2.1 and 2.2 can be generalized for F(a,b,c;r). This is left to the readers.



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