JFS Journal of Function Spaces 2314-8888 2314-8896 Hindawi Publishing Corporation 10.1155/2016/3698463 3698463 Research Article Optimal Bounds for Gaussian Arithmetic-Geometric Mean with Applications to Complete Elliptic Integral Wang Hua 1,2 2 http://orcid.org/0000-0002-9206-3097 Qian Wei-Mao 3 http://orcid.org/0000-0002-0944-2134 Chu Yu-Ming 4 Stens Rudolf L. 1 Department of Mathematics and System Science National University of Defense Technology Changsha 410073 China nudt.edu.cn 2 Department of Mathematics Changsha University of Science and Technology Changsha 410014 China csust.edu.cn 3 School of Distance Education Huzhou Broadcast and TV University Huzhou 313000 China hutc.zj.cn 4 School of Mathematics and Computation Sciences Hunan City University Yiyang 413000 China hncu.net 2016 2072016 2016 08 05 2016 11 06 2016 23 06 2016 2072016 2016 Copyright © 2016 Hua Wang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We present the best possible parameters α1,β1,α2,β2R and α3,β3(1/2,1) such that the double inequalities Qα1(a,b)A1-α1(a,b)<AG[A(a,b),Q(a,b)]<Qβ1(a,b)A1-β1(a,b), α2Q(a,b)+(1-α2)A(a,b)<AG[A(a,b),Q(a,b)]<β2Q(a,b)+(1-β2)A(a,b), Q[α3a+(1-α3)b,α3b+(1-α3)a]<AG[A(a,b),Q(a,b)]<Q[β3a+(1-β3)b,β3b+(1-β3)a] hold for all a,b>0 with ab, where A(a,b), Q(a,b), and AG(a,b) are the arithmetic, quadratic, and Gauss arithmetic-geometric means of a and b, respectively. As applications, we find several new bounds for the complete elliptic integrals of the first and second kind.

Natural Science Foundation of China 11371125 61374086 11401191 Natural Science Foundation of Zhejiang Province LY13A010004 Natural Science Foundation of the Zhejiang Broadcast and TV University XKT-15G17
1. Introduction

Let r(0,1) and a,b>0. Then the elliptic elliptic integral of the first kind K(r) and second kind E(r), Gaussian arithmetic-geometric mean AG(a,b), arithmetic mean A(a,b), and quadratic mean Q(a,b) are, respectively, given by (1)Kr=0π/211-r2sin2tdt,Er=0π/21-r2sin2tdt,(2)AGa,b=π20π/2dt/a2cos2t+b2sin2t,Aa,b=a+b2,Qa,b=a2+b22.

The Gauss identity  shows that (3)AG1,r=π2Krfor all r(0,1), where and in what follows r=1-r2.

It is well known that the elliptic elliptic integrals K(r) and E(r) and the Gaussian arithmetic-geometric mean AG(a,b) have many applications in mathematics, physics, mechanics, and engineering . Recently, the bounds for the Gaussian arithmetic-geometric mean AG(a,b) have attracted the attention of many researchers.

The inequalities (4)1+r2AG1,r<AG1,r<π2log4/r,(5)La,b<AGa,b<L3/2a,bfor all r(0,1) and a,b>0 with ab can be found in the literature , where L(a,b)=(a-b)/(loga-logb) and Lp(a,b)=L1/pap,bp are, respectively, the logarithmic and pth generalized logarithmic means of a and b. The first inequality of (5) is due to Carlson and Vuorinen .

By using a variant of L’Hospital’s rule and representation theorems with elliptic integrals, Vamanamurthy and Vuorinen  proved, among other results, the inequalities (6)AGa,b<Aa,bLa,b,(7)La,b<AGa,b<π2La,b,(8)AGa,b<Ia,b<Aa,b,(9)AGa,b<Aa,b+Ga,b2,(10)Aa,b<AGa2,b2AGa,b<Qa,b,(11)AGa,b>L1/λaλ,bλfor all a,b>0 with ab and λ(0,1], where I(a,b)=(bb/aa)1/(b-a)/e is the identric mean of a and b.

By use of the homogeneity of the above means and a series representation of AG(a,b) due to Gauss, Sándor  obtained, among other results, new proofs for inequalities (7), (8) and a counterpart of inequality (9): (12)AGa,b>Aa,bGa,b,for all a,b>0 with ab, where G(a,b)=ab is the geometric mean of a and b. Inequalities (9) and (12) show that AG lies between the arithmetic and geometric means of A and G. In , Sándor provided new proofs for inequalities (6) and (8), (9), (10), and (12) by using only elementary methods for recurrent sequences and found much stronger forms of these results.

Neuman and Sándor  gave the comparison of the Gaussian arithmetic-geometric mean and the Schwab-Borchardt mean.

The upper bounds π/2log(4/r) for AG(1,r) in (4) were replaced by π(1-r2/9)/[2log(4/r)] due to Kühnau .

Qiu and Vamanamurthy  presented that 4π/[(9-r2)(2log2-logr)] and (9-r2)π/[18.192×(2log2-logr)] are, respectively, the lower and upper bounds for AG(1,r) with r(0,1). Alzer and Qiu  proved that λ=3/4 and μ=2/π are the best possible parameters such that the double inequality (13)1λ/La,b+1-λ/Aa,b<AGa,b<1μ/La,b+1-μ/Aa,bholds for all a,b>0 with ab.

Chu and Wang  proved that the double inequality (14)Spa,b<AGa,b<Sqa,bholds for all a,b>0 with ab if and only if p1/2 and q1, where Sp(a,b)=[ap-1+bp-1/(a+b)]1/(p-2)(p2) and S2(a,b)=aabb1/(a+b) is the pth Gini mean of a and b. In , Yang et al. proved that the inequalities(15)S7/4,-1/4a,b<AGa,b<A1/4a,bL3/4a,b,AGa,b<Sp,1a,bS1-p,1a,bhold for all p(1/2,1) and a,b>0 with ab, where Sp,q(a,b)=[q(ap-bp)/(p(aq-bq))]1/(p-q) is the Stolarsky mean  of a and b.

Let a,b>0 with ab and x[1/2,1]. Then it is not difficult to verify that the function f(x)=Q[xa+(1-x)b,xb+(1-x)a] is continuous and strictly increasing on the interval [1/2,1]. Note that(16)f12=Aa,b=minAa,b,Qa,b<AGAa,b,Qa,b,(17)AGAa,b,Qa,b<maxAa,b,Qa,b=Qa,b=f1.

Inequalities (16) give us the motivation to deal with the best possible parameters α1,β1,α2,β2R and α3,β3(1/2,1) such that the double inequalities (18)Qα1a,bA1-α1a,b<AGAa,b,Qa,b<Qβ1a,bA1-β1a,b,α2Qa,b+1-α2Aa,b<AGAa,b,Qa,b<β2Qa,b+1-β2Aa,b,Qα3a+1-α3b,α3b+1-α3a<AGAa,b,Qa,b<Qβ3a+1-β3b,β3b+1-β3ahold for all a,b>0 with ab.

2. Lemmas

In order to prove our main results we need several derivative formulas and particular values for K(r) and E(r), which we present in this section.

K ( r ) and E(r) satisfy the formulas (see )(19)dKrdr=Er-r2Krrr2,dErdr=Er-Krr,K0+=E0+=π2,K1-=,E1-=1,K22=Γ21/44π=1.85407467,E22=4Γ23/4+Γ21/48π=1.35064388,where Γ(x)=0tx-1e-tdt(x>0) is the classical Euler gamma function.

Lemma 1 (see [<xref ref-type="bibr" rid="B24">24</xref>, Theorem  1.25]).

Let -<a<b<, f,g:[a,b]R be continuous on [a,b] and differentiable on (a,b) and g(x)0 on (a,b). Then both functions (20)fx-fagx-ga,fx-fbgx-gbare increasing (decreasing) on (a,b) if f(x)/g(x) is increasing (decreasing) on (a,b). If f(x)/g(x) is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2 (see [<xref ref-type="bibr" rid="B24">24</xref>, Theorem  3.21(1), Theorem  3.21(7), and Exercises  3.43(32)]).

The following statements are true:

The function rE(r)-r2K(r)/r2 is strictly increasing from (0,1) onto (π/4,1).

The function rrλK(r) is strictly decreasing from (0,1) onto (0,π/2) if λ1/2.

The function r[K(r)-E(r)]/[r2K(r)] is strictly increasing from (0,1) onto (1/2,1).

Lemma 3.

Let p=1/2+2π2-2K2(2/2)/[4K(2/2)]=0.8299 and h(r) be defined by (21)hr=Er-r2Krr2r2K3r-16p1-pπ2.Then there exists r0(0,2/2) such that h(r)<0 for r(0,r0) and h(r)>0 for r(r0,2/2).

Proof.

From (21) we clearly see that h(r) can be rewritten as (22)hr=Er-r2Krr2r2/3Kr-3-16p1-pπ2.

It follows from Lemma 2(1) and (2) together with (22) that h(r) is strictly increasing on (0,2/2).

Numerical computations show that (23)h0+=2π2-3K22/2π2K22/2=-0.02612<0,h22-=4π2E2/2-2K32/2π2K32/2=0.03708>0.

Therefore, Lemma 3 follows easily from (23) and the monotonicity of h(r) on the interval (0,2/2).

3. Main Results Theorem 4.

The double inequality (24)Qα1a,bA1-α1a,b<AGAa,b,Qa,b<Qβ1a,bA1-β1a,bholds for all a,b>0 with ab if and only if α11/2 and β12[logπ-logK(2/2)]/log2-1=0.5215.

Proof.

Since A(a,b), Q(a,b), and AG(a,b) are symmetric and homogenous of degree 1, without loss of generality, we assume that a>b>0. Let r=(a-b)/2(a2+b2)(0,2/2). Then (2) and (3) lead to (25)AGAa,b,Qa,b=πAa,b2rKr,Qa,b=Aa,br,(26)logAGAa,b,Qa,b-logAa,blogQa,b-logAa,b=logKr+logr+log2-logπlogr.Let (27)f1r=logKr+logr+log2-logπ,f2r=logr,fr=f1rf2r.Then simple computations give (28)f10+=f20+=0,f1rf2r=Kr-Err2Kr.

It follows from Lemmas 1, 2(3) and (27) and (28) that (29)f0+=12and f(r) is strictly increasing on the interval (0,2/2).

Note that (30)f22-=2logπ-logK2/2log2-1.

Therefore, Theorem 4 follows easily from (26), (27), (29), and (30) and the monotonicity of f(r) on the interval (0,2/2).

Remark 5.

The left side inequality of Theorem 4 for α11/2 can be derived directly from the fact that AG(a,b)>G(a,b)=ab and Q(a,b)>A(a,b) for all a,b>0 with ab.

Theorem 6.

The double inequality (31)α2Qa,b+1-α2Aa,b<AGAa,b,Qa,b<β2Qa,b+1-β2Aa,bholds for all a,b>0 with ab if and only if α2[π-2K(2/2)]/[(2-2)K(2/2)]=0.4783  and β21/2.

Proof.

Without loss of generality, we assume that a>b>0. Let r=(a-b)/2(a2+b2)(0,2/2). Then it follows from (2) and (3) that (32)AGAa,b,Qa,b-Aa,bQa,b-Aa,b=π/2Kr-r1-r.Let (33)g1r=π2Kr-r,g2r=1-r,gr=g1rg2r.

Then simple computations lead to (34)g10+=g20+=0,g1rg2r=1-π2Er-r2Krr2r1/2Kr-2.

It follows from Lemmas 1, 2(1) and (2) together with (33) and (34) that (35)g0+=12and g(r) is strictly decreasing on the interval (0,2/2).

Note that (36)g22-=π-2K2/22-2K2/2.

Therefore, Theorem 6 follows easily from (32), (33), (35), and (36) and the monotonicity of g(r) on the interval (0,2/2).

Remark 7.

The right side inequality of Theorem 6 for β21/2 can be derived directly from the fact that AG(a,b)<A(a,b)=(a+b)/2 and Q(a,b)>A(a,b) for all a,b>0 with ab.

Theorem 8.

Let α3,β3(1/2,1). Then the double inequality (37)Qα3a+1-α3b,α3b+1-α3a<AGAa,b,Qa,b<Qβ3a+1-β3b,β3b+1-β3aholds for all a,b>0 with ab if and only if α31/2+2π2-2K2(2/2)/[4K(2/2)]=0.8299 and β31/2+2/4=0.8535.

Proof.

Without loss of generality, we assume that a>b>0. Let r=(a-b)/2(a2+b2)(0,2/2) and p(1/2,1). Then (2) and (3) lead to (38)Qpa+1-pb,pb+1-pa=1-4p1-pr2rAa,b,(39)Qpa+1-pb,pb+1-pa-AGAa,b,Qa,b=Aa,b1-4p1-pr2+π/2KrrHr,where(40)Hr=1-4p1-pr2-π24K2r,(41)H0+=0,(42)H22-=1-2p1-p-π24K22/2,(43)Hr=π2r2hr,where h(r) is defined by (21).

We divide the proof into four cases.

Case  1 (p=p0=1/2+2π2-2K2(2/2)/[4K(2/2)]). Then (42) becomes(44)H22-=0.

It follows from Lemma 3 and (43) that there exists r0(0,2/2) such that H(r) is strictly decreasing on (0,r0] and strictly increasing on [r0,2/2). Therefore, (45)Qp0a+1-p0b,p0b+1-p0a<AGAa,b,Qa,bfollows from (39), (41), and (44) together with the piecewise monotonicity of H(r) on the interval (0,2/2).

Case  2 (p=p0=1/2+2/4). Then we clearly see that (46)Qp0a+1-p0b,p0b+1-p0a=A2a,b+Q2a,b2=QAa,b,Qa,b>AAa,b,Qa,b>AGAa,b,Qa,b.

Case  3 (1/2+2π2-2K2(2/2)/[4K(2/2)]<p<1). Then (42) leads to (47)H22->0.

Equation (39) and inequality (47) imply that there exists small enough 0<δ1<2/2 such that (48)Qpa+1-pb,pb+1-pa>AGAa,b,Qa,bfor all a,b>0 with |a-b|/2(a2+b2)(2/2-δ1,2/2).

Case  4 (1/2<p<1/2+2/4). Then (40) leads to (49)Hr=4p-122-12r2+or2.Note that (50)4p-122-12r2<0.

Equations (39) and (49) together with inequality (50) imply that there exists small enough 0<δ2<2/2 such that (51)Qpa+1-pb,pb+1-pa<AGAa,b,Qa,bfor all a,b>0 with a-b/2(a2+b2)(0,δ2).

4. Applications

In this section, we use Theorems 4, 6, and 8 to present several bounds for the complete elliptic integrals K(r) and E(r).

From Theorems 4, 6, and 8 we get Theorem 9 immediately.

Theorem 9.

Let λ1=2[logπ-logK(2/2)]/log2-1=0.5215, λ2=[π-2K(2/2)]/[(2-2)K(2/2)]=0.4783, and λ3=2-π2/[2K2(2/2)]=0.5644. Then the double inequalities (52)π21-r21-λ1/2<Kr<π2λ2+1-λ21-r2,π21-1/2r2<Kr<π21-λ3r2hold for all r(0,2/2).

It follows from the inequality (53)π24<ErKr<π24rgiven in  that (54)π24Kr<Er<π24rKr.

Theorem 9 and (54) lead to the following.

Theorem 10.

Let λ1=2[logπ-logK(2/2)]/log2-1=0.5215, λ2=[π-2K(2/2)]/[(2-2)K(2/2)]=0.4783, and λ3=2-π2/[2K2(2/2)]=0.5644. Then the double inequalities (55)π2λ2+1-λ2r<Er<π2r1/2-λ1,π21-λ3r2<Er<π22-r22rhold for all r(0,2/2).

Competing Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The research was supported by the Natural Science Foundation of China under Grants 11371125, 61374086, and 11401191, the Natural Science Foundation of Zhejiang Province under Grant LY13A010004, and the Natural Science Foundation of the Zhejiang Broadcast and TV University under Grant XKT-15G17.

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