Evidence for the Riemann Hypothesis L´eo Ag´elas

∗

September 20, 2014

Abstract Riemann Hypothesis (that all non-trivial zeros of the zeta function have real part one-half) is arguably the most important unsolved problem in contemporary mathematics due to its deep relation to the fundamental building blocks of the integers, the primes. The proof of the Riemann hypothesis will immediately verify a slew of dependent theorems ([4], [22]). In this paper, we give another evidence in favour of Riemann Hypothesis using a probabilistic approach. Keywords Riemann hypothesis; complex function; analytic function.

1

Introduction

Riemann Hypothesis has defied proof so far, and very complicated and advanced abstract mathematics is often brought to bear on it. Does it need abstract mathematics, or just a flash of elementary inspiration. The Riemann zeta function, defined ([3] p. 807) by the series, ζ(s) =

∞ X 1 , s n n=1

s ∈ C,

(1)

is analytic in Re(s) > 1 (see [4]). The basic connection between complex analysis and prime numbers is the following beautiful identity of Euler (see [24],[4],[5],[22]): −1 Y 1 ζ(s) = , Re(s) > 1, (2) 1− s p p∈P

where P = {2, 3, 5, 7, 11, 13, ...} is the set of prime positive integers p. Equation (2) is used to shown that ([4]), ζ(s) 6= 0 in Re(s) > 1. (3) Riemann proved (see [23], [10]), that ζ(s) has an analytic continuation to the whole complex plane except for a simple pole at s = 1. Moreover, he showed that ζ(s) satisfies the functional equation (see [25],[10],[4]), πs (4) ζ(s) = 2s π s−1 sin( )Γ(1 − s)ζ(1 − s), 2 where Γ(s) is the complex gamma function. From (4), it can be deduced that, 1. ζ(s) is nonzero in Re(s) < 0, except for the real zeros {−2m}m∈N∗ , ∗ Department

of Mathematics, ([email protected])

IFPEN, 1 & 4 avenue Bois Pr´ eau,

1

92852 Rueil-Malmaison Cedex,

France

2. {−2m}m∈N∗ are the only real zeros of ζ(s), 3. ζ(s) possesses infinitely many non real zeros in the strip 0 ≤ Re(s) ≤ 1, (the so-called) critical strip for ζ(s) From (4), it also follows that ζ(s) has no zeros on Re(s) = 0, whence, ζ(s) possesses infinitely many non real zeros in 0 < Re(s) < 1.

(5)

It is interesting to mention that,

1 ζ(s) has infinitely many non real zero in Re(s) = (Hardy [7]), 2 1 1 ζ(s) has at least of its zeros on Re(s) = (Levinson [19]). 3 2 It also follows from (4) that if ζ(s) = 0 where s is non real then, {s, 1 − s, 1 − s} are also zeros of ζ. In 1859, B. Riemann formulated the following conjecture, Conjecture 1.1 (Riemann Hypothesis) . All non real zeros of ζ(s) lies exactly on Re(s) =

1 . 2

We know that the zeta function was introduced as an analytic tool for studying prime numbers and some of the most important applications of the zeta functions belong to prime number theory. It was shown (see [25] p. 45), independently in 1896 by Hadamard and de la Vall´ee-Poussin, that ζ(s) has no zeros on Re(s) = 1, which provided the first proof of the Prime Number Theorem : π(x) ∼

x log x

(x → +∞),

(6)

where π(x) = { number of primes p for which p ≤ x} (where x > 0). The key point in both of their proofs was that ζ(s) 6= 0 for any s = 1 + i t, i.e., along the line with σ = 1. Their proof comes with an explicit error estimate: they showed in fact (see for example Theorem 6.9 in [20]),   x √ , (7) π(x) = li(x) + O exp(c log x) uniformly for x ≥ 2. Here li(x) is the logarithmic integral, Z x dt li(x) = . log t 2 Furthermore, the error estimate in (7) might be improved, indeed there exists an equivalence between the error term in the Prime Number Theorem and the real part of the zeros of ζ(s) which states as follow,
π(x) = li(x) + O xθ+ǫ , (8)

is equivalent to ζ(s) 6= 0 for Re(s) > θ, 12 ≤ θ < 1 (see Theorem 12.3 [15]). Moreover, Von Koch proved that the Riemann hypothesis is equivalent to the ”best possible” bound for the error of the Prime Number Theorem (see [17]), namely Riemann hypothesis is equivalent to,
√ (9) π(x) = li(x) + O x log(x) . In this paper, we give an evidence in favour of Riemann Hypothesis using probabilistic arguments.

2

2

Riemann Hypothesis

Let us introduce the Mobius function µ   1 µ(n) = (−1)r  0

defined for all positive integers by if n = 1, if n is the product of r distinct primes, if n is divisible by the square of a prime.

(10)

The Mobius function has generating functions ∞ X µ(n) 1 = , s n ζ(s) n=1

for Re(s) > 1 (see [2] p. 229, see also [21] p.130, [9] p. 245-249, [18], [6]). It is known (see, for example, Titchmarsh [[25], p. 370]) that a necessary and sufficient condition for the truth of the Riemann hypothesis is that 1

M (x) = O(x 2 +ǫ ), for all ǫ > 0, where M (x) =

X

n≤x

(11)

µ(n) for x ≥ 1. 1

As it was mentionned in [8] and [12], the condition M (x) = 0(x 2 +ǫ ) would be true if the Mobius sequence {µ(n)} were a random sequence, taking the values −1, 0, and 1, with specified probabilities, those of −1 and 1 being equal. Our proof follows this line, one argument in favour of this probabilistic approach is that the Mobius sequence {µ(n)} satisfies what we call the strong law of large numbers in probability theory, namely (see for example [20], p. 182),

lim

n X

µ(k)

k=1

n→∞

= 0.

n

Let us precise our approach. If A is a set of positive integers, we denote by AN the number of its elements among the first N integers. If the limit lim

N →∞

AN = Prob(A), N

exists, it is called the probability of a positive integer to be in A. For any a ∈ N∗ , we introduce the set of positive integers divisible by a, Ea = {n ∈ N∗ , a|n}. Let us 1 show that Prob(Ea ) = . Indeed, for any N ∈ N∗ , let EN = {n ∈ N∗ , a|n and n ≤ N }, we have a   N N N and therefore, we get card(EN ) = − 1 < card(EN ) ≤ which yields to, a a a 1 card(EN ) 1 1 − ≤ ≤ . a N N a card(EN ) 1 1 = which give us Prob(Ea ) = , this reveals that the probability N →∞ N a a 1 that a positive integer to be divisible by a is . a For p and q two distinct primes, let us show that the events of being divisible by p and q are independent. For this, we consider the set Ep,q = {n ∈ N∗ , p|n and q|n}. We notice that for any n ∈ Ep,q , we have p|n, q|n and since p, q are two distinct primes, then (p, q) = 1, therefore we infer that pq|n which implies that Then, we deduce that lim

3

n ∈ Epq , this means that Ep,q ⊂ Epq . Moreover, for any n ∈ Epq , we have pq|n which implies p|n and q|n and thus n ∈ Ep,q , hence we deduce that Epq ⊂ Ep,q . Therefore, we get Ep,q = Epq , Then we deduce that, Prob(Ep,q )

= Prob(Epq ) 1 = pq = Prob(Ep )Prob(Eq ),

which reveals that the events of being divisible by p and q are independent. This holds, of course, for any number of distinct primes. For any integer n > 1, if the distinct prime factors of n are p1 , . . . , pr , and if pi occurs as a factor ai times, then thanks to the fundamental theorem of arithmetic (see for example Theorem 1.10 in [2]), we get the unique representation of n as products of powers of its prime divisors, namely, n=

r Y

pai i .

(12)

i=1

Then, by introducing αp (n) the power with which the prime p appears in the (unique) representation of n as a product of powers of its prime divisors, we get, Y n= pαp (n) . (13) p

The product

Y

is understood as the product over the infinite set of prime numbers ranked in ascending

p

order, notice also that αp (n) can be equal to zero. A positive integer n is a square-free number if and only if in the prime factorization of n, (13), no prime number occurs more than once, another way of stating the same is that µ(n) 6= 0. Then, by introducing the function n 7→ ρp (n) defined on N∗ as follows,  1, p|n ρp (n) = (14) 0, p6 | n, we infer that if n is a square-free number, the unique representation of n is, Y n= pρp (n) .

(15)

p

We define ν(n) ≡

X

ρp (n), the number of prime divisors. Since the events of being divisible by any

p

number of distinct primes are independent, then thanks to (15), we deduce that the functions ρp ∈ {0, 1} are independent, this suggests that, in some sense, the distribution of values of ν(n) may be given by a ν(n) − log log n √ normal law. Indeed, the Erd˜ os-Kac Theorem states that the probability distribution of log log n is the standard normal distribution N (0, 1) (see Theorem  7.21 in [20], [11],[16]), more precisely, for any  ν(n) − log log n √ ≤ b , we have, a < b, by setting Ea,b = n ∈ N∗ ; a ≤ log log n Z b 2 1 Prob (Ea,b ) = √ e−y /2 dy. 2π a   ν(n) − log log n ≤ gn where gn is a sequence approaching infinity: Moreover, by setting Eg = n ∈ N∗ ; √ log log n lim gn = ∞, we have (see [16], [11]), n→∞

Prob (Eg ) = 1. 4

This reveals that for larger value of n that ν(n) behaves like a Gaussian random variable and is approximatively equal to log log n, we notice that for any n square-free number that µ(n) = (−1)ν(n) , then among the square free numbers, successive evaluations of µ(n) = ±1 may be independent and µ(n) behaves like a random variable, and therefore among the square-free numbers each evaluation of µ(n) may behave like an party of coin tossing. ( ) Y βp From (15), we notice that the set of square-free numbers, F , can be written as F = p , βp ∈ {0, 1} . p

Then, the set F is in bijection with the set of infinite binary sequences {0, 1}∞ . This bijection reveals again the independence between square-free numbers. The set {0, 1} can be seen as a coin tossing where 0 and 1 means respectively Tails and Heads. Thus, a square-free number can be seen as an endless party of coin tossing. Therefore, the event ”a square-free number have an even number of prime factors” matches with the event ”we get an even number of Heads after an endless party of coin tossing”. Let us calculate the probability that a square-free number may have an even number of prime factors. Let N > 1, the probability, pN , to get an even number of Heads after a party of N coin tosses is given by X  N  2k k∈N,2k≤N pN = . (16) 2N

Furthermore, we have (see [1], p. 236), X

k∈N,2k≤N



N 2k



= 2N −1 .

(17)

1 . Therefore, for any N > 1, the probability to get an even number 2 1 of Heads after a party of N coin tosses is . Then, we infer that the probability to get an even number 2 1 of Heads after an endless party of coin tossing is which implies that the probability that a square-free 2 1 number may have an even number of prime factors is . Therefore, we deduce that the probability that a 2 1 square-free number may have an odd number of prime factors is also . For any n square-free number, n 2 has an even number of prime factors if and only if µ(n) = 1 and n has an odd number of prime factors if and only if µ(n) = −1. Therefore, we deduce that for n square-free number, the probability that µ(n) = 1 1 1 is equal to and the probability that µ(n) = −1 is also equal to . 2 2 Thanks to the Kolmogorov’s law of iterated logarithm which holds also for p−random sequences (see [26]) and states that for any random walk, {Sn } with the increment of zero mean and finite variance, σ 2 , the following holds almost surely (see [13]): Then, from (16), we deduce pN =

Sn lim sup p = 1, n→∞ 2σ 2 n log log n

Sn = −1. lim inf p n→∞ 2σ 2 n log log n

(18)

(19)

Therefore, if we can apply Kolmogorov’s law of iterated logarithm to {Mn } which behaves like a random walk, with the increment of zero mean and variance of one, we will obtain, M (n) lim sup √ = 1, 2n log log n n→∞ 5

(20)

M (n) lim inf √ = −1. n→∞ 2n log log n

(21)

We observe for all x ≥ 1, M (x) = M ([x]) and from (20), (21), we will obtain, M (x) = 1, 2x log log x

(22)

M (x) = −1. lim inf √ x→∞ 2x log log x

(23)

lim sup √ x→∞

Then, thanks to (22) and (23), we infer that (11) holds and therefore Riemann Hypothesis will be proved.

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