# Optimal Bounds for Gaussian Arithmetic-Geometric Mean with Applications to Complete Elliptic Integral

Authors: Hua Wang; Wei-Mao Qian; Yu-Ming Chu
Journal: Journal of Function Spaces
 (2016)

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## Abstract

We present the best possible parametersα1,β1,α2,β2∈R 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 a≠b, 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.


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## Body

## 1. Introduction

Letr∈(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=π2∫0π/2dt/a2cos2t+b2sin2t,Aa,b=a+b2,Qa,b=a2+b22.The Gauss identity [1–3] 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 integralsK(r) and E(r) and the Gaussian arithmetic-geometric mean AG(a,b) have many applications in mathematics, physics, mechanics, and engineering [4–9]. 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<π2log⁡4/r,(5)La,b<AGa,b<L3/2a,bfor all r∈(0,1) and a,b>0 with a≠b can be found in the literature [10–12], where L(a,b)=(a-b)/(log⁡a-log⁡b) 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 [13].By using a variant of L’Hospital’s rule and representation theorems with elliptic integrals, Vamanamurthy and Vuorinen [14] 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 a≠b 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 ofAG(a,b) due to Gauss, Sándor [15] 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 a≠b, 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 [16], 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 [17] gave the comparison of the Gaussian arithmetic-geometric mean and the Schwab-Borchardt mean.The upper boundsπ/2log⁡(4/r) for AG(1,r) in (4) were replaced by π(1-r2/9)/[2log⁡(4/r)] due to Kühnau [18].Qiu and Vamanamurthy [19] presented that 4π/[(9-r2)(2log⁡2-log⁡r)] and (9-r2)π/[18.192×(2log⁡2-log⁡r)] are, respectively, the lower and upper bounds for AG(1,r) with r∈(0,1). Alzer and Qiu [20] 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 a≠b.Chu and Wang [21] proved that the double inequality (14)Spa,b<AGa,b<Sqa,bholds for all a,b>0 with a≠b if and only if p≤1/2 and q≥1, where Sp(a,b)=[ap-1+bp-1/(a+b)]1/(p-2)(p≠2) and S2(a,b)=aabb1/(a+b) is the pth Gini mean of a and b. In [22], 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 a≠b, where Sp,q(a,b)=[q(ap-bp)/(p(aq-bq))]1/(p-q) is the Stolarsky mean [23] of a and b.Leta,b>0 with a≠b 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=min⁡Aa,b,Qa,b<AGAa,b,Qa,b,(17)AGAa,b,Qa,b<max⁡Aa,b,Qa,b=Qa,b=f1.Inequalities (16) give us the motivation to deal with the best possible parameters α1,β1,α2,β2∈R 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 a≠b.


## 2. Lemmas

In order to prove our main results we need several derivative formulas and particular values forK(r) and E(r), which we present in this section.K ( r ) and E(r) satisfy the formulas (see [24])(19)dKrdr=Er-r′2Krrr′2,dErdr=Er-Krr,K0+=E0+=π2,K1-=∞,E1-=1,K22=Γ21/44π=1.85407467…,E22=4Γ23/4+Γ21/48π=1.35064388…,where Γ(x)=∫0∞‍tx-1e-tdt(x>0) is the classical Euler gamma function.Lemma 1 (see [24, 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 [24, Theorem  3.21(1), Theorem  3.21(7), and Exercises  3.43(32)]).
The following statements are true:(1)
The functionr→E(r)-r′2K(r)/r2 is strictly increasing from (0,1) onto (π/4,1).(2)
The functionr→r′λK(r) is strictly decreasing from (0,1) onto (0,π/2) if λ≥1/2.(3)
The functionr→[K(r)-E(r)]/[r2K(r)] is strictly increasing from (0,1) onto (1/2,1).Lemma 3.
Letp=1/2+2π2-2K2(2/2)/[4K(2/2)]=0.8299… and h(r) be defined by (21)hr=Er-r′2Krr2r′2K3r-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-r′2Krr2r′2/3Kr-3-16p1-pπ2.
It follows from Lemma2(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, Lemma3 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 a≠b if and only if α1≤1/2 and β1≥2[log⁡π-log⁡K(2/2)]/log⁡2-1=0.5215….Proof.
SinceA(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,b2r′Kr,Qa,b=Aa,br′,(26)log⁡AGAa,b,Qa,b-log⁡Aa,blog⁡Qa,b-log⁡Aa,b=log⁡Kr+log⁡r′+log⁡2-log⁡πlog⁡r′.Let (27)f1r=log⁡Kr+log⁡r′+log⁡2-log⁡π,f2r=log⁡r′,fr=f1rf2r.Then simple computations give (28)f10+=f20+=0,f1′rf2′r=Kr-Err2Kr.
It follows from Lemmas1, 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-=2log⁡π-log⁡K2/2log⁡2-1.
Therefore, Theorem4 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 Theorem4 for α1≤1/2 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 a≠b.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 a≠b if and only if α2≤[π-2K(2/2)]/[(2-2)K(2/2)]=0.4783…  and β2≥1/2.Proof.
Without loss of generality, we assume thata>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-r′1-r′.Let (33)g1r=π2Kr-r′,g2r=1-r′,gr=g1rg2r.
Then simple computations lead to(34)g10+=g20+=0,g1′rg2′r=1-π2Er-r′2Krr2r′1/2Kr-2.
It follows from Lemmas1, 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, Theorem6 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 Theorem6 for β2≥1/2 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 a≠b.Theorem 8.
Letα3,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 a≠b if and only if α3≤1/2+2π2-2K2(2/2)/[4K(2/2)]=0.8299… and β3≥1/2+2/4=0.8535….Proof.
Without loss of generality, we assume thata>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-pr2r′Aa,b,(39)Qpa+1-pb,pb+1-pa-AGAa,b,Qa,b=Aa,b1-4p1-pr2+π/2Krr′Hr,where(40)Hr=1-4p1-pr2-π24K2r,(41)H0+=0,(42)H22-=1-2p1-p-π24K22/2,(43)H′r=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 Lemma3 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)Qp0∗a+1-p0∗b,p0∗b+1-p0∗a=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 Theorems4, 6, and 8 to present several bounds for the complete elliptic integrals K(r) and E(r).From Theorems4, 6, and 8 we get Theorem 9 immediately.Theorem 9.
Letλ1=2[log⁡π-log⁡K(2/2)]/log⁡2-1=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<π24r′given in [24] that (54)π24Kr<Er<π24r′Kr.Theorem9 and (54) lead to the following.Theorem 10.
Letλ1=2[log⁡π-log⁡K(2/2)]/log⁡2-1=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<π2r′1/2-λ1,π21-λ3r2<Er<π22-r22r′hold for all r∈(0,2/2).



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