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Research article Special Issues

Monotonicity, convexity and inequalities involving zero-balanced Gaussian hypergeometric function

  • We generalize several monotonicity and convexity properties as well as sharp inequalities for the complete elliptic integrals to the zero-balanced Gaussian hypergeometric function F(a,b;a+b;x).

    Citation: Li Xu, Lu Chen, Ti-Ren Huang. Monotonicity, convexity and inequalities involving zero-balanced Gaussian hypergeometric function[J]. AIMS Mathematics, 2022, 7(7): 12471-12482. doi: 10.3934/math.2022692

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  • We generalize several monotonicity and convexity properties as well as sharp inequalities for the complete elliptic integrals to the zero-balanced Gaussian hypergeometric function F(a,b;a+b;x).



    For a,b,cR with c0,1,2,, the Gaussian hypergeometric function is defined by [1,3,4,25]

    F(a,b;c;x)=2F1(a,b;c;x)=n=0(a,n)(b,n)(c,n)xnn!,x(1,1), (1.1)

    where (a,n) denotes the shifted factorial function (a,n)a(a+1)(a+n1) for nN, and (a,0)=1 for a0. It is well known that F(a,b;c;x) is widely applied in geometric function theory, theory of mean values as well as in many other fields of mathematics and some other disciplines. Many elementary functions and special functions in mathematical physics are particular or limiting cases of the Gaussian hypergeometric function. F(a,b;c;x) is said to be zero-balanced if c=a+b. For the known properties of F(a,b;c;x), the readers are referred to [1,3,4,7,8,13,15,18,20,21,23,26].

    As the special cases of the Gaussian hypergeometric function, for r(0,1),a(0,1), the generalized elliptic integrals of the first and the second kinds are defined by [5,6,24].

    Ka(r)=π2F(a,1a;1;r2), (1.2)
    Ea(r)=π2F(a1,1a;1;r2), (1.3)
    Ka(0)=π2,Ka(1)=,Ea(0)=π2,Ea(1)=sin(πa)2(1a).

    Set Ka(r)=Ka(r),Ea(r)=Ea(r). Here and hereafter we always let r=1r2 for r[0,1]. Note that, when a=1/2, the functions Ka(r) and Ea(r) reduce to the classical complete elliptic integrals K(r) and E(r) of the first and second kind [3,4].

    K(r)=π2F(1/2,1/2;1;r2)forr(0,1),K(0)=π2,K(1)=, (1.4)
    E(r)=π2F(1/2,1/2;1;r2)forr(0,1),E(0)=π2,E(1)=1. (1.5)

    Set K(r)=K(r),E(r)=E(r).

    Let γ=limn(1+12+13++1nlogn)=0.577156649 be the Euler-Mascheroni constant. For a>0,b>0, the Gamma, Psi and Beta functions are defined respectively by [1,3,4]

    Γ(a)=0ta1etdt,ψ(a)=Γ(a)Γ(a),B(a,b)=Γ(a)Γ(b)Γ(a+b).

    The so-called Ramanujan R-function R(a) is defined by [11,13]

    R(a)=2γψ(a)ψ(1a),a(0,1), (1.6)

    which is the special case of the following function of two parameters a and b.

    R(a,b)=2γψ(a)ψ(b),fora,b(0,+),

    which is sometimes called the Ramanujan constant although it is a function of a and b ( [11]).

    For a(0,1), we set [5,12]

    B(a)=B(a,1a)=Γ(a)Γ(1a)=πsin(πa). (1.7)

    By the symmetry, we can sometimes assume that a(0,1/2] in (1.6) and (1.7).

    In the past few years, the complete elliptic integrals [5], the generalized elliptic integrals [11,12] and the p-elliptic integrals [14,27] have been studied by many authors. The recent interest is motivated by applications to geometric function theory.

    In 2000, Anderson, Qiu, Vamanamurthy, and Vuorinen obtained some results in [6] as follows.

    Theorem 1.1 (See [6]). Let a(0,1/2] be given, and let b=1a,c=(sin(πa))/b. Then the functions

    (1)

    ˆf1(r)r2Ka(r)/Ea(r)

    is decreasing from (0,1) onto (0,1).

    (2)

    ˆf2(r)[(π/2)2(rKa(r))2]/[Ea(r)r2Ka(r)]

    is increasing from (0,1) onto (π(a2+b2)/(2a),π2/(2c)).

    (3)

    ˆf3(r)(Ea(r)r2Ka(r))/r2

    is increasing and convex from (0,1) onto (πa/2,c/2).

    In order to estimate the Robin capacity, Anderson, Qiu, and Vamanamurthy in [5] dealt with the convexity properties of the functions (Er2K)/(r)2 and [(Er2K)/r2]/[(Er2K)/(r)2]. They proved the following theorem.

    Theorem 1.2 (See [5]). For r(0,1), there exists r0(0,1) such that the function

    F1(r)Er2Kr2,

    is concave on (0,r0) and convex on (r0,1).

    Theorem 1.3 (See [5]). For r(0,1), the function

    F2(r)[Er2K]/r2[Er2K]/(r)2

    is strictly increasing and convex from (0,1) onto (π/4,4/π). In particular, for r(0,1)

    π4<F2(r)<π4+(4ππ4)r.

    The corresponding properties of the additive counterpart (Er2K)/r2(Er2K)/(r)2 are obtained by Alzer and Richard in [2]. They proved the following theorem.

    Theorem 1.4 (See [2]). For r(0,1), the function

    F3(r)Er2Kr2Er2Kr2,

    is strictly increasing and convex from (0,1) onto (π/41,1π/4). Moreover, for all r(0,1), the double inequality

    π41+αr<F3(r)<π41+βr,

    holds for all r(0,1) with the best constants

    α=0,andβ=2π/2=0.42920.

    In 2019, Wang et al. [22] generalized Theorem 1.2 to the generalized elliptic integrals and obtained analogous properties. In 2017, Huang et al. [12] generalized Theorem 1.3 and Theorem 1.4 to the generalized elliptic integrals. So it is natural to ask that how to extend these results to zero-balanced Gaussian hypergeometric function F(a,b;a+b;x)? The purpose of this paper is to solve this question.

    For the purpose, we require some more properties of the zero-balanced Gaussian hypergeometric function, so we will give some lemmas in Section 2. In the last section, we will present our main results and their proofs.

    In this section, we give a definition and some lemmas needed in the proofs of our main results in Section 3. Firstly, we recall some formulas below. By [1,3,15], the hypergeometric function has the following simple differentiation formula

    ddxF(a,b;c;x)=abcF(a+1,b+1;c+1;x)

    and it is well known that

    F(a,b;c;x)=(1x)cabF(ca,cb;c;x), (2.1)
    F(a,b;c;1)=Γ(c)Γ(cab)Γ(ca)Γ(cb),c>a+b, (2.2)
    B(a,b)F(a,b;a+b;x)+log(1x)=R(a,b)+O((1x)log(1x)),x1. (2.3)

    Definition 2.1. A function f is said to be strictly completely monotonic on an interval IR if (1)nf(n)(x)>0 for all xI and n=0,1,2,3. If (1)nf(n)(x)0 for all xI and n=0,1,2,3, then f is called completely monotonic on I.

    Completely monotonic functions play a dominant role in areas such as numerical analysis [19], probability theory [10], special function theory [1] and physics [9].

    The following lemmas will be frequently applied in the sequel.

    Lemma 2.1. (See [15,Lemma 2.1]) Let <a<b<, f,g:[a,b]R becontinuous on [a,b] and differentiable on (a,b), andg(x)0 on (a,b). If f(x)/g(x) isincreasing (decreasing) on (a,b), then so are the functions

    f(x)f(a)g(x)g(a),f(x)f(b)g(x)g(b).

    If f(x)/g(x) is strictly monotone, then themonotonicity in the conclusion is also strict.

    Lemma 2.2. (See [16,Lemma 2.4]) Suppose that r(0,) is the common radius of convergence of the real power series A(x)=n=0anxn and B(x)=n=0bnxn with bn>0, and {an/bn} is a non-constant sequence. Let φ(x)=A(x)/B(x).

    (1) If there is an n0N such that the sequence {an/bn} is increasing (decreasing) for 0nn0, and decreasing (increasing) for nn0, then φ is increasing (decreasing) on (0,r) if and only if φ(r)0 (φ(r)0, respectively).

    (2) If there is an n0N such that the sequence {an/bn} is increasing (decreasing) for 0nn0, and decreasing (increasing) for nn0, and if φ(r)<0 (φ(r)>0), then there exists a number x0(0,r) such that φ is strictly increasing (decreasing) on (0,x0] and decreasing (increasing, respectively) on [x0,r).

    Lemma 2.3. (See [17,Lemma 1.1]) Suppose that the power series f(x)=n=0anxn and g(x)=n=0bnxn have the radius of convergence r>0 and that bn>0 for all n{0,1,2,}. Let h(x)=f(x)/g(x).If the sequence {an/bn} is (strictly) increasing (decreasing), then h(x) is also (strictly) increasing (decreasing) on (0,r).

    Lemma 2.4. For a>0,b>0,x(0,1) and a1, we define the function

    H(x)F(a1,b;a+b;x)(1x)F(a,b;a+b;x)x,

    then we get

    H(x)=aa+bF(a,b;a+b+1;x),

    and H is strictly increasing and convex from (0,1) onto (aa+b,1bB(a,b)).

    Proof. By the definition of the Gaussian hypergeometric function in (1.1), we have

    H(x)=F(a1,b;a+b;x)(1x)F(a,b;a+b;x)x=1x[n=0(a1,n)(b,n)(a+b,n)xnn!(1x)n=0(a,n)(b,n)(a+b,n)xnn!]=n=0(a,n)(b,n)(a+b,n+1)(n+1)!a(n+1)xn=aa+bn=0(a,n)(b,n)(a+b+1,n)n!xn=aa+bF(a,b;a+b+1;x).

    It is clear that H(0+)=aa+b. By (2.2), H(1)=1bB(a,b). The remainings conclusion is clear.

    Remark 2.5. Let a(0,1/2],b=1a,x=r2,r(0,1). From Lemma 2.4, we can obtain the properties of the function ˆf3(r) in Theorem 1.1 which is proved by Anderson et al. in [6].

    Theorem 3.1. For a>0,b>0,x(0,1) and a1, the function

    f1(x)(1x)F(a,b;a+b;x)F(a1,b;a+b;x)

    is strictly decreasing from (0,1) onto (0,1).

    Proof. We can write f1(x) as

    f1(x)=1F(a1,b;a+b;x)(1x)F(a,b;a+b;x)F(a1,b;a+b;x). (3.1)

    Clearly, f1(0+)=1 and by (2.3), f1(1)=0. By (1.1),

    F(a1,b;a+b;x)(1x)F(a,b;a+b;x)=n=0a1n+a1anxn[n=0anxnn=1an1xn]=n=1(nn+a1an+an1)xn=n=1an+a+b1an1xn, (3.2)

    where an=[(a,n)(b,n)]/[(a+b,n)n!]. By (3.2), F(a1,b;a+b;x)(1x)F(a,b;a+b;x) is positive and strictly increasing on (0,1). It is easy to see that F(a1,b;a+b;x) is positive for a>0 and is strictly decreasing (increasing) for a(0,1)(a(1,), respectively). Hence, for a(0,1), f1 is strictly decreasing on (0,1).

    Next, we will study the monotonicity of f1 for a(1,). By (1.1),

    f1(x)=(1x)F(a,b;a+b;x)F(a1,b;a+b;x)=n=0anxnn=1an1xnn=0a1n+a1anxn=n=0bnxnn=0cnxn,

    where b0=1,bn=anan1,n1 and cn=a1n+a1an. Since

    b0c0b1c1=1a1a0a1aa1=1ab(a+b)(a1)b=a(a1)b>0, (3.3)

    so by (3.3), b0c0>b1c1. For n1,

    bncn=anan1a1n+a1an=n+a1a1(1an1an)=n+a1a1[1(a+b+n1)n(a+n1)(b+n1)]=n+a1a1ababn+1(a+n1)(b+n1)=1a1[a(b1)b+n11], (3.4)
    bn+1cn+1bncn=a(b1)a1(1b+n1b+n1). (3.5)

    Therefore, for a>1,b>1, by (3.3) and (3.5), {bn/cn} is strictly decreasing for n=1,2,3,. By Lemma 2.3, the function f1 is strictly decreasing.

    Then, we study the monotonicity of g1 for a(1,),b(0,1].

    Differentiation of f1(x) gives

    f1(x)=F(a,b;a+b;x)F(a1,b;a+b;x)+aba+bF(a,b;a+b+1;x)F(a1,b;a+b;x)(a1)ba+b(1x)F(a,b;a+b;x)F(a,b+1;a+b+1;x)(F(a1,b;a+b;x))2.

    For a>1,b>0, x(0,1), we have

    (1x)F(a,b;a+b;x)F(a,b+1;a+b+1;x)(F(a1,b;a+b;x))2>0

    and

    limx1F(a,b;a+b;x)F(a1,b;a+b;x)=,limx1aba+bF(a,b;a+b+1;x)F(a1,b;a+b;x)=b.

    Hence f1(1)=. By (3.5), for b(0,1], {bn/cn} is strictly increasing for n=1,2,3,, and since {bn/cn} is strictly decreasing on 0n1, and f1(1)=<0. Hence by Lemma 2.2(1), f1 is strictly decreasing on (0,1).

    Theorem 3.2. For a>0,b>0,x(0,1) and a1, the function

    f2(x)1[1xF(a,b;a+b;x)]2F(a1,b;a+b;x)(1x)F(a,b;a+b;x)

    is strictly increasing on (0,1) if 0<a+b2ab<1.

    Proof. Let h1(x)=1[1xF(a,b;a+b;x)]2 and h2(x)=F(a1,b;a+b;x)(1x)F(a,b;a+b;x), since h1(0+)=h2(0+)=0, then

    h1(x)h2(x)=2[1xF(a,b;a+b;x)]×[121xF(a,b;a+b;x)+1xaba+b(1x)1F(a,b;a+b+1;x)](a1)ba+bF(a,b+1;a+b+1;x)+F(a,b;a+b;x)aba+bF(a,b;a+b+1;x)=F(a,b;a+b;x)n=0(12aba+b+n)anxnn=0a(n+1)a+b+nanxn=F(a,b;a+b;x)G(x) (3.6)

    where an=[(a,n)(b,n)]/[(a+b,n)n!] and

    G(x)=n=0(12aba+b+n)anxnn=0a(n+1)a+b+nanxn. (3.7)

    Let An=(12aba+b+n)an,Bn=a(n+1)a+b+nan, then, An/Bn=[a+b2ab+n]/a(n+1). Simple computations give

    An+1Bn+1AnBn=2abab+1a(n+1)(n+2).

    Since 0<a+b2ab<1, we can get G(x)>0 and An+1/Bn+1An/Bn0, hence {An/Bn} is increasing, then G(x) is increasing on (0,1) by Lemma 2.3. Therefore h1(x)/h2(x) is a product of two positive and increasing functions, the monotonicity of f2 follows from Lemma 2.1.

    Remark 3.3. Let a(0,1/2],b=1a,x=r2,r(0,1).

    (1) The properties of the function ˆf1(r) in Theorem 1.1 follow from Theorem 3.1.

    (2) Since a+b2ab=2(a12)2+12(0,1), we can obtain the properties of the function ˆf2(r) in Theorem 1.1 by Theorem 3.2.

    Theorem 3.4. For a>0,b>0,x(0,1) and a1, the function

    f3(x)F(a1,b;a+b;1x)xF(a,b;a+b;1x)1x

    is completely monotonic on (0,1). In particular, for a>0,b>0 and a1, f3 is decreasing and convex on (0,1).

    Proof. By Lemma 2.4, f3(x) can be written as

    f3(x)=aa+bF(a,b;a+b+1;1x),

    differentiation gives

    f(n)3(x)=(1)na(a,n)(b,n)(a+b)(a+b+1,n)F(a+n,b+n;a+b+n+1;1x), (3.8)

    for a>0,b>0, x(0,1), it is easy to obtain (1)nf(n)3(x)>0. By Definition 2.1, f3 is completely monotonic on (0,1). Let n=1 and n=2, so f3(x)<0,f3(x)>0, then the monotonicity and convexity of f3 follow.

    Remark 3.5. (1) Let a=b=1/2,x=r2,r(0,1). From Theorem 3.4, we can obtain the properties of the function (Er2K)/(r)2 which is proved by Anderson, Qiu, and Vamanamurthy in [5].

    (2) Let a(0,1/2],b=1a,x=r2,r(0,1). From Theorem 3.4, we can obtain the properties of the function (Ear2Ka)/(r)2 which is proved by Wang, Zhang, and Chu in [22].

    Theorem 3.6. Let a>0,b>0,x(0,1) and a1, we define the function

    f4(x)F(a1,b;a+b;x)(1x)F(a,b;a+b;x)xF(a1,b;a+b;1x)xF(a,b;a+b;1x)1x,

    then f(2n1)4 is decreasing(increasing) on (0,1/2] ([1/2,1),respectively) and f(2n)4 is strictly increasing from (0,1) onto (,+) for n=1,2,, nN.

    Proof. By Lemma 2.4, f4(x) can be written as

    f4(x)=aa+b[F(a,b;a+b+1;x)F(a,b;a+b+1;1x)]. (3.9)

    For n=1,2,, nN, differentiation of (3.9) gives

    f(2n)4(x)=aa+b(a,2n)(b,2n)(a+b+1,2n)×[F(a+2n,b+2n;a+b+2n+1;x)F(a+2n,b+2n;a+b+2n+1;1x)]=aa+b(a,2n)(b,2n)(a+b+1,2n)×n=0(a+2n,n)(b+2n,n)(a+b+2n+1,n)n![xn(1x)n], (3.10)
    f(2n1)4(x)=aa+b(a,2n1)(b,2n1)(a+b+1,2n1)×[F(a+2n1,b+2n1;a+b+2n;x)+F(a+2n1,b+2n1;a+b+2n;1x)]. (3.11)

    By (3.10) and (3.11), for a>0,b>0 and a1, f(2n1)4(x)>0 on (0,1), and f(2n)4(x)<0 on (0,1/2), f(2n)4(x)>0 on (1/2,1). Therefore, f(2n)4 is strictly increasing on (0,1), and f(2n1)4 is strictly decreasing on (0,1/2) and increasing on (1/2,1) for n=1,2,. Clearly, f(2n)4(0+)=,f(2n)4(1)=+.

    The following Corollary is obtained by Theorem 3.6.

    Corollary 3.7. Let a>0,b>0,x(0,1) and a1, the function f4 is strictly increasing from (0,1) onto (aa+b1b1B(a,b),1b1B(a,b)aa+b) and f4 is concave on (0,1/2] and convex on [1/2,1). Moreover, for all x(0,1), the double inequality

    aa+b1b1B(a,b)+α1x<f4(x)<aa+b1b1B(a,b)+β1x,

    holds for all x(0,1) with the best possible constants

    α1=0,andβ1=2bB(a,b)2aa+b.

    Proof. Let n=1 in (3.10),

    f4(x)=a2b(a+b)(a+b+1)×[F(a+1,b+1;a+b+2;x)+F(a+1,b+1;a+b+2;1x)], (3.12)

    for a>0,b>0,x(0,1) and a1, f4(x)>0, hence the monotonicity of f4 follows. By Theorem 3.6, since f(2n1)4 is decreasing(increasing) on (0,1/2]([1/2,1),respectively), hence f4(x)<0 on (0,1/2) and f4(x)>0 on (1/2,1), so the concavity and convexity of f4 are obtained. The limiting values are easy to know f4(0+)=aa+b1b1B(a,b),f4(1)=1b1B(a,b)aa+b by (2.2) and (3.9), and thereby the double inequality in Corollary 3.7 follows.

    Remark 3.8. (1) Let a=b=1/2,x=r2,r(0,1). From Theorem 3.6, we can obtain the properties of the function (Er2K)/r2(Er2K)/(r)2 which is proved by Alzer and Richard in [2].

    (2) Let a(0,1/2],b=1a,x=r2,r(0,1). From Theorem 3.6, we can obtain the properties of the function [Ear2Ka]/r2[Ear2Ka]/(r)2 which is proved by Huang, Tan, and Zhang in [12].

    Theorem 3.9. Let a>0,b>0,x(0,1) and a1, the function

    f5(x)F(a1,b;a+b;x)(1x)F(a,b;a+b;x)x×1xF(a1,b;a+b;1x)xF(a,b;a+b;1x),

    is strictly increasing from (0,1) onto (abB(a,b)a+b,a+babB(a,b)). Moreover, for all x(0,1), the double inequality

    aba+bB(a,b)<f5(x)<a+babB(a,b),

    holds for all x(0,1).

    Proof. By Lemma 2.4, f5(x) can be written as

    f5(x)=F(a,b;a+b+1;x)F(a,b;a+b+1;1x). (3.13)

    Since F(a,b;a+b+1;x) is positive and strictly increasing on (0,1) and F(a,b;a+b+1;1x) is positive and strictly decreasing on (0,1). Hence f5(x) is a product of two positive and increasing functions, then f5 is strictly increasing on (0,1). The limit values are easy to know f5(0+)=abB(a,b)/(a+b),f5(1)=(a+b)/[abB(a,b)] from (2.2) and (3.13), and thereby the double inequality in Theorem 3.9 follows.

    Remark 3.10. (1) Let a=b=1/2,x=r2,r(0,1). From Theorem 3.9, we can obtain the properties of the function [(Er2K)/r2]/[(Er2K)/(r)2] are obtained by Anderson, Qiu, and Vamanamurthy in [5].

    (2) Let a(0,1/2],b=1a,x=r2,r(0,1). From Theorem 3.9, we can obtain the properties of the function [(Ear2Ka)/r2]/[(Ear2Ka)/(r)2] which is proved by Huang, Tan, and Zhang in [12].

    Many properties of generalized elliptic integrals currently have been published in the literature. Results from zero-balanced Gaussian hypergeometric function F(a,b;a+b;x) are presented in this paper. Firstly we propose some primary properties of the zero-balanced Gaussian hypergeometric function which required in the proofs of main results. Then we generalized the monotonicity and convexity properties of elliptic integrals to the zero-balanced Gaussian hypergeometric function. We obtain some new properties and give sharp inequalities for the zero-balanced Gaussian hypergeometric function F(a;b;a+b;x). These studies will validly improve the theory of special functions and their applications in the natural sciences and engineering.

    The authors would like to express their sincere thanks to the editor and the anonymous reviewers for their helpful comments and suggestions.

    The authors declare that they have no competing interests.



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