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Bounds of the Mertens Functions

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Bounds of the Mertens Functions Advances in Dynamical Systems and Applications. ISSN 0973-5321, Volume 16, Number 1, (2021) pp. 35-44 © Research India Publications https://dx.doi.org/10.37622/ADSA/16.1.2021.35-44 Bounds of the Mertens Functions Darrell Coxa, Sourangshu Ghoshb, Eldar Sultanowc aGrayson County College, Denison, TX 75020, USA bDepartment of Civil Engineering , Indian Institute of Technology Kharagpur, West Bengal, India cPotsdam University, 14482 Potsdam, Germany Abstract In this paper we derive new properties of Mertens function and discuss about a likely upper bound of the absolute value of the Mertens function √log(푥!) > |푀(푥)| when 푥 > 1. Using this likely bound we show that we have a sufficient condition to prove the Riemann Hypothesis. 1. INTRODUCTION We define the Mobius Function 휇(푘). Depending on the factorization of n into prime factors the function can take various values in {−1, 0, 1} 휇(푛) = 1 if n has even number of prime factors and it is also square- free(divisible by no perfect square other than 1) 휇(푛) = −1 if n has odd number of prime factors and it is also square-free 휇(푛) = 0 if n is divisible by a perfect square. 푛 Mertens function is defined as 푀(푛) = ∑푘=1 휇(푘) where 휇(푘) is the Mobius function. It can be restated as the difference in the number of square-free integers up to 푥 that have even number of prime factors and the number of square-free integers up to 푥 that have odd number of prime factors. The Mertens function rather grows very slowly since the Mobius function takes only the value 0, ±1 in both the positive and 36 Darrell Cox, Sourangshu Ghosh, Eldar Sultanow negative directions and keeps oscillating in a chaotic manner. Mertens after verifying all the numbers up to 10,000 conjectured that the absolute value of 푀(푥) is always bounded by √푥. This conjecture was later disproved by Odlyzko and te Riele1. This 2 conjecture is replaced by a weaker one by Stieltjes who conjectured that 푀(푥) = 1 푂(푥2). Littlewood3 proved that the Riemann hypothesis is equivalent to the statement 1 − −휖 that for every 휖 > 0 the function 푀(푥)푥 2 approaches zero as x → ∞. This proves 1 +휖 that the Riemann Hypothesis is equivalent to conjecture that 푀(푥) = 푂(푥2 ) which gives a rather very strong upper bound to the growth of 푀(푥). Although there exists no analytic formula, Titchmarsh4 showed that if the Riemann Hypothesis is true and if there exist no multiple non-trivial Riemann zeta function zeros, then there must exist a sequence 푇푘 which satisfies 푘 ≤ 푇푘 ≤ 푘 + 1 such that the following result holds: 휌 ∞ 푛+1 푥 (−1) 2휋 2푛 푀0(푥) = lim ∑ − 2 + ∑ ( ) 푘→∞ 휌ζ′(휌) (2푛)! 푛ζ(2n + 1) 푥 휌 푛=1 |훾|<푇푘 Where ζ (z) is the Riemann zeta function, and 휌 are all the all nontrivial zeros of the 1 Riemann zeta function and 푀 (푥) is defined as 푀 (푥) = 푀(푥) − 휇(푥) 푖푓 푥 ∈ 푍+, 0 0 2 푀(푥) Otherwise (Odlyzko and te Riele1) 2. NEW PROPERTIES OF MERTENS FUNCTION 5 푥 푥 Lehman proved that ∑푖=1 푀(⌊푥/푖⌋)= 1. In general, ∑푖=1 푀(⌊푥/(푖푛)⌋)= 1, 푛 = 1, 2, 3, . , 푥(since⌊⌊푥/푛⌋/푖⌋ = (⌊푥/(푖푛)⌋). Let 푅′ denote a square matrix where element (i, j) equals 1 if 푗 divides 푖 or 0 otherwise. (In a Redheffer matrix, element (푖, 푗) equals 1 if 푖 divides 푗 or if j = 1. Redheffer6 proved that the determinant of the matrix equals the Mertens Function 푀(푥).) Let 푇 denote the matrix obtained from 푅′ by element-by-element multiplication of the columns 푥 푥 푥 푥 by 푀 (⌊ ⌋) , 푀 (⌊ ⌋) , 푀 (⌊ ⌋) … 푀(⌊ ⌋ ). For example, the T matrix for x = 12 is 1 2 3 푥 −2 0 0 0 0 0 0 0 0 0 0 0 −1 −1 0 0 0 0 0 0 0 0 0 0 −1 0 −1 0 0 0 0 0 0 0 0 0 −1 −1 0 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 푇 = [ ] 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0 0 0 1 Bounds of the Mertens Functions 37 푥 Theorem (1) : ∑푖=1 푀(⌊푥/푖⌋) 푖 = 퐴(푥) 푥 Proof: Let us now take 퐴(푥) = ∑푖=1 φ(i) where φ is Euler’s totient function. Let 푈 denote the matrix obtained from 푇 by element-by-element multiplication of the columns by 휑(푗). The sum of the columns of 푈 then equals 퐴(푥). Now since 푖 = 푥 ∑푑|푖 φ(d) we can write ∑푖=1 푀(⌊푥/푖⌋) 푖 (the sum of the rows of U) equals 퐴(푥). By the Schwarz inequality, 퐴(푥)/√푥(푥 + 1)(2푥 + 1)/6 is a lower bound 푥 2 푥 7 of √∑푖=1 푀(⌊푥/푖⌋) . 퐴(푥) = ∑푖=1 φ(i) Is further simplified by Walfisz and Hardy and Wright8 as 1 푥 푥 3 퐴(푥) = ∑푥 φ(i) = ∑푥 휇(푘) ⌊ ⌋ (1 + ⌊ ⌋) = 푥2 + 푖=1 2 푘=1 푘 푘 휋2 푂(푛(log 푛)2/3(log log 푛)4/3) 3 2 퐴(푥) 푥 3√3 푥2 √∑푥 푀(⌊푥/푖⌋)2 > > 휋2 = 푖=1 푥(푥 + 1)(2푥 + 1) 푥(푥 + 1)(2푥 + 1) 휋 ( )( ) √ √ √푥 푥 + 1 푥 + 1/2 6 6 This can be further simplified to 3√3 1 √∑푥 푀(⌊푥/푖⌋)2 /푥 > 푖=1 휋 √(1+1/푥 )(1 + 1/2푥) Taking limit of infinity on both the sides, we get 3√3 lim √∑푥 푀(⌊푥/푖⌋)2 /푥 > 푥→∞ 푖=1 휋 27 This shows that ∑푥 푀(⌊푥/푖⌋)2 at large values of x is greater than 푥2. 푖=1 휋2 Let 훬(푖) denote the Mangoldt function (훬(푖) equals 푙표푔(푝) if 푖 = 푝푚 for some 9 푥 prime 푝 and some 푚 ≥ 1 or 0 otherwise). Mertens proved that ∑푖=1 푀(⌊푥/푖⌋)푙표푔 푖 = 휓(푥) where 휓(푥) denotes the second Chebyshev function (ψ(x) =∑푖≤푥 훬(푖) ). 푥 Theorem (2) : ∑푥 푀 (⌊ ⌋) log(푖) 휎 (푖)/2 = 푙표푔(푥!) 푖=1 푖 0 푥 Proof: Let 휎푥(푖) denote the sum of positive divisors function (휎푥(푖) = ∑푑|푖 푑 ). Replacing 휑(푗) with 푙표푔(푗) in the U matrix gives a similar result. 푎1 푎푘 Let 휆(푛) denote the Liouville function (휆(1) = 1 or if = 푝1 . 푝푘 , 휆(푛) = 푎1+..+푎푘 (−1) ). ∑푑|푛 휆(푑) Equals 1 if n is a perfect square or 0 otherwise. Let (푥) = ∑n≤x 휆(푑) . Let us also assume 퐻(푥) = ∑n≤x 휇(푛)log (푛) . 퐻(푥)/(푥 푙표푔(푥)) → 0 as 푥 → ∞ and lim 푀(푥)/푥 − 퐻(푥)/(푥 푙표푔(푥))) = 0. The statement lim 푀(푥)/ 푥→∞ 푥→∞ 푥 = 0 is equivalent to the prime number theorem. Also, 훬(푛) = − ∑d|n 휇(푑)log (푑) . (Apostol10). The generalization of the Euler’s totient function is Jordan totient function. Let it be denoted as 퐽푘(푥) which is defined as number of set of k positive integers which are all less than or equal to n that will form a co-prime set of (푘 + 1) positive integers 38 Darrell Cox, Sourangshu Ghosh, Eldar Sultanow 푥 푘 together with 푛. Let us define (푥) = ∑푖=1 퐽푘(푖) . It is known that ∑푑|푛 퐽푘(푑) = 푛 . Then we get the following theorem. 푥 Theorem (3): ∑푥 푀 (⌊ ⌋) 푖푘 = 퐵(푥) 푖=1 푖 퐵(푥) is expanded by McCarthy11 to be 푛푟+1 퐵(푥) = ∑푥 퐽 (푖) = + 푂(푛푟) 푖=1 푘 (푟+1)ζ(r+1) 푘+1 푥 푥 푘 푛 푘 We therefore get ∑ 푀 (⌊ ⌋) 푖 = 퐵(푥) = + 푂(푛 ) . 푖=1 푖 (푘+1)ζ(k+1) Likewise we can derive some other similar relationships using the 푇 matrix that are as listed below: 푥 Theorem (4): ∑푥 푀 (⌊ ⌋) 휎 (푖) = ∑푥 푖푘 for 푘 ∈ 푍+ 푖=1 푖 푘 푖=1 푥 Theorem (5): ∑푥 푀 (⌊ ⌋) where the summation is over those 푖 values that are 푖=1 푖 perfect squares equals 퐿(푥) 푥 Theorem (6): ∑푥 푀 (⌊ ⌋) 훬(푖) = −퐻(푥) 푖=1 푖 푥 6 Theorem (7): ∑푥 푀 (⌊ ⌋) 2휔(푛) = ∑푥 |휇(푖)| ~ 푥2 + 푂(√푛) = 푖=1 푖 푖=1 휋2 푁표. 표푓 푆푞푢푎푟푒 퐹푟푒푒 퐼푛푡푒푔푒푟푠 푥 Theorem (8): ∑푥 푀 (⌊ ⌋) 푑(푛2) = ∑푥 2휔(푖) where 푑(푥) is the sum of all the 푖=1 푖 푖=1 divisors of 푥 푥 Theorem (9): ∑푥 푀 (⌊ ⌋) 푑2(푛) = ∑푥 푑(푖2) 푖=1 푖 푖=1 푥 푖 휇2(푖) Theorem (10): ∑푥 푀 (⌊ ⌋) ( ) = ∑푥 푖=1 푖 φ(i) 푖=1 φ(i) Similarly many other relationships can be found between various arithmetic functions and the Mertens Functions. 3. A LIKELY UPPER BOUND OF |M(x)| The following conjecture is based on data collected for x ≤ 500, 000.
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