Central JSM Mathematics and Statistics Review Article *Corresponding author Rohan J Dalpatadu, Department of Mathematical Sciences, University of Nevada, Las Vegas, USA The Arithmetic -Geometric- Submitted: 21 August 2014 Accepted: 22 August 2014 Harmonic Mean Published: 22 August 2014 Copyright Rohan J Dalpatadu* © 2014 Dalpatadu Department of Mathematical Sciences, University of Nevada, USA OPEN ACCESS Abstract The arithmetic-geometric mean of two positive numbers has been defined to be the common limit of the sequences {an} and {bn} defined recursively by 1 2 an+ 1= () a n b n , b n+ 1=() a n + b n / 2. This mean has applications to elliptic integrals. The basic idea of this paper is to generalize the above mean to one that uses three numbersby using the harmonic mean to define a third sequence as proposed by Dalpatadu and use elementary techniques to prove the existence of such a mean and obtain results on rates of convergence. INTRODUCTION sequences {an} and {bn} The arithmetic-geometric1 mean of two positive numbers a ≤ b, denoted by AGM(a, b), is defined to be the common limit of the 2 an+ 1= () a n b n , n=defined 0,1,2,...; recursively a0 = a, by [1] ()an+ b n b + = , n= 0,1,2,...; b= b. n 1 2 0 π/2 2 dθ It has also been shown that. [2] the reciprocal of this mean equals π ∫ a2 cos 2θ + b 2 sin 2 θ 0 π2 π We wish to generalize thed aboveφ d θ mean and explore the possibility of evaluating integrals such as the ellipsoidal surface integral 1 ⋅ ∫ ∫ a2 cos 2θ sin 2 φ + b 2 sin 2 θ sin 2 φ + c 2 cos 2 φ 2 φ=0 θ= 0 via three sequences {an n n In this paper, we will define the arithmetic-geometric-harmonic mean of three positive numbers a ≤ b ≤ c }, {b }, {c }. We shall establish the existence of such a mean and obtain a couple of results on the convergence of these sequences. A relationship between this mean and the above surface integral has not yet been established. We will show that a natural generalization suchTHE as ARITHMETIC the one suggested -GEOMETRIC-HARMONIC above is not the proper application MEAN of the arithmetic-geometric-harmonic mean. In this section, the arithmetic-geometric-harmonic mean is defined and its existence is proved. Furthermore, two important results onDefinition rates of convergence 1 are also established. We will first define the threen sequencesn n using the arithmetic, geometric, and harmonic means. Let 0 < a ≤ b ≤c.3 Define {a }, {b }, {c } recursively by a n+ 1 = , n= 0,1,2,...; a0 = a, 1 1 1 + + an b n c n 1 3 bn+ 1= () a n b n c n , n= 0,1,2,...; b0 = b, Cite this article: Dalpatadu RJ (2014) The Arithmetic -Geometric-Harmonic Mean. JSM Math Stat 1(1): 1002. Moraru (2014) Email: Central ()an+ b n + c n c + = , n= 0,1,2,...; c = c. n 1 3 0 TheProposition first of the 1: propositions 0< an ≤ b n ≤ deals c n , n with = 0,1,2,... the ordering of the three sequences. Proof: n n n for some non negative integer n. Then a c an+ 1 > 0 The result holds if n = 0. Suppose 0 < a ≤ b ≤ c n+1 ≤ bn+1 ≤ n+1, because harmonic mean ≤ geometric mean ≤ arithemetic mean of three positive number. Furthermore, . Hence result. TheProposition next proposition 2: an+ 1≥ deals a n , n with = 0,1,2,... the monotonicity of the sequence obtained via the harmonic mean. 3 3 Proof: a n+ 1 = ≥ = an . 1 1 1 1 1 1 + + + + an b n c n an a n a n Lemma 1: a+Le bt + 0 c < a ≤ b ≤ c. Then1 ac≤ b2 ⇔ ≤ ()a2 b 2 c 2 3 . 1 1 1 + + a b c a+ b + c 1 Proof: − ()a2 b 2 c 2 3 1 1 1 + + a b c 1 2 2 2 1 2 2 2 3 1 1 1 a bc+ ab c + abc 2 2 2 ()a b c 4 4 4 = − ()a b c 3 = ()a bc3 +() ab c3 +() abc3 − bc − ca − ab + + + + bc ca ab bc ca ab 1 1 ()a2 b 2 c 2 3 2 1 2 1 2 1 ()a2 b 2 c 2 3 1 2 2 1 1 2 = a3− (bc) 3 b 3 − (ca) 3 c 3 − (ab) 3 = (bc)3− a 3 c 3 − (ab) 3 (ca) 3− b 3 . bc+ ca + ab bc+ ca + ab 1 2 2 1 1 2 The result follows because 0 < a ≤ b ≤ c, implies that 2 3 (bc)3− a 3 ≥ 0 c3− (ab) 3 ≥ 0 ac≤ b if and only if ()ac3 ≤ b . with equality if and only if a = b = c, with equality if and only if a = b = c 2 2 WeProposition need the next3: proposition in order to establishac≤ the b ⇔monotonicity an c n ≤ b n , nof = the 0,1,2,... second sequence. Proof: Let 0 < a ≤ b ≤ c. Then a+ b + c 1 The2 definitionn− 1 nof − 1 the n −three 1 sequences2 2 gives 2 3 us: an c n− b n = − (an− 1 b n − 1 c n − 1 ) , n= 0 ,1 ,2,... 1 1 1 + + an− 1 b n − 1 c n − 1 Now the result follows easily whenThen we use≤ mathematical2 ⇔ ≤ induction = along with the above lemma. Proposition 4: ac b bn+ 1 b n , n 0,1,2,... 1 Let 0 < a ≤ 3b ≤ c. bn+ 1− b n =() a n b n c n − b n Proof: 1 1 2 3 3 3 = (bn ) (a n c n )− (b n ) . 0 < an n n ≤2 b ≤ c by Proposition2 1, ac≤ b ⇔ an c n ≤ b n , n = 0,1,2,... 1 2 2 3 3 by Proposition 3, acn n− b n ≤ 0 ⇔ (ac)n n − (b) n ≤ 0 . Hence result. The following proposition proves the monotonicity of the sequence defined as the arithmetic mean. JSM Math Stat 1(1): 1001 (2014) 2/4 Moraru (2014) Email: Central Proposition 5: cn+ 1≤ c n , n = 0,1,2,... an+ b n + c n cn+ c n + c n Proof: c + = ≤ = c . n 1 3 3 n Remark 1: {an} n} n} is a decreasing bounded sequence. Thus lim a n lim bn lim cn n→∞ is ann→∞ increasingn→∞ bounded sequence, {b is a monotonic bounded sequence, and {c Proposition 6: A =, B = C. , and exist. Let these limits be A, B, and C, respectively. Proof: a Byn+ Proposition b n + c n 1, AA +≤ BB +≤ CC. c + = ⇒C = ⋅ n 1 3 3 Therefore, A + B 2C ≥ 2B and A ≥ B. Since A ≤ B, A = B and it follows that DefinitionA = C. 2 n n n}. Let 0 < a ≤ b ≤ c. The Arithmetic-Geometric-Harmonic Mean of the numbers a, b, and c, denoted by AGHM (a,b,c), is the common limit of the sequences {a }, {b }, and {c n 2 WeProposition finally end 7: this cn −section a n ≤ by establishing (c − a), n = 1,2,... two results on rates of convergence. 3 Proof: cn− a n ≤ c n − a n− 1 an− 1+ b n − 1 + c n − 1 bn− 1+ c n − 1 − 2a n − 1 2cn− 1− 2a n − 1 = − a − =(Proposition 2) ≤ 3 n 1 3 3 2 (Proposition 1) =(c− − a − ). 3 n 1 n 1 RemarkMathematical 2: induction with the above inequality will give us the desired result. 2 2 Proposition 7(c indicatesn− 1− a n − 1 )that (cwe n− 1have− a n −fairly 1 ) rapid convergence. Proposition 8: cn− a n ≤ ≤ , n= 1,2,... 3a n− 1 3a an− 1+ b n − 1 + c n − 1 3an− 1 b n − 1 c n − 1 Proof: cn− a n = − 3 bn1n1− c −+ c n1n1 − a − + a n1n1 − b − a2 b+ a b2 + a 2 c + a c2 + b 2 c + b c2 − 6a b c = n1n1−− n1n1 −− n1n1 −− n1n1 −− n1n1 −− n1n1 −− n1n1n1− − − 3(an1n1− b −+ b n1n1 − c − + c n1n1 − a − ) a(b− c)b(c2 + − a)c(a2 + − b)2 = n1n1− − n1 − n1n1− − n1 − n1n1 − − n1 − 3(b c+ c a + a b ) n1n1− − n1n1 − − n1n1 − − (a+ b + c )(c − a )2 ≤ n-1 n1− n1 − n1 − n1 − 3(bn1n1− c −+ c n1n1 − a − + a n1n1 − b − ) (a+ b + c )(c − a )2 (c− a )2 (c− a )2 ≤ n-1 n1− n1 − n1 − n1 − = n− 1 n − 1 ≤ n− 1 n − 1 ⋅ 3(a c+ a a + a b ) 3a 3a n1n1− − n1n1 − − n1n1 − − n− 1 Remark 3: Proposition 8 shows us that we have quadratic convergence. Therefore, it is possible to approximate AGHM(a,b,c) fairly accurately in a few iterations. JSM Math Stat 1(1): 1001 (2014) 3/4 Moraru (2014) Email: Central Applications In this section, we show that the original conjecture proposed by [1] is not true and provide a different application (that does not π π directly involve a surface integral) using2 the existence and properties of the AGHM. 1 1 dφ d θ Conjecture: = . AGH(a,b,c) 2π2 1 ∫ ∫ a2 cos 2θ sin 2 φ + b 2 sin 2 θ sin 2 φ + c 2 cos 2 φ 2 φ=0 θ= 0 This conjecture was made in January 1989 by [1] and is not true because of the following: AGM(1,2) is approximately 1.4567550309. AGHM(1,1,2)π2 π is approximately 1.2632465593. π2 π 1 dφ d θ 1 dφ d θ = 2π2 1 2π2 1 ∫ ∫ 12 cos 2θ sin 2 φ + 1 2 sin 2 θ sin 2 φ + 2 2 cos 2 φ 2 ∫ ∫ 12 sin 2φ + 2 2 cos 2 φ 2 φ=0 θ= 0 φ=0 θ= 0 π π/2 1 dφ d θ 2 dφ d θ 1 1 = = = ≠ ⋅ π 1 π 1 AGM(1,2) AGHM(1,1,2) ∫ 12 sin 2φ + 2 2 cos 2 φ 2 ∫ 12 sin 2φ + 2 2 cos 2 φ 2 φ=0 φ=0 RECENTHowever, APPLICATIONS it is possible that the reciprocal of the AGHM is equal to some closed surface integral.