On Robin’s Inequality and Riemann Hypothesis Dr. Jamal Salah Department of Basic Sciences College of Applied Sciences A’Sharqiya University Ibra, A’Sharqiya, Oman damous73@yahoo.com

Contents Prime Number Theory (PNT). Euler’s Approach to PNT. Riemann’s Hypothesis (RH). Robin’s Inequality (RI). Observations. References.

Prime Number Theory Infinitude of primes has been proved first by Euclid. Therefore, the matter arises is how the primes are distributed amongst the positive integers i.e. how many primes are less than a given real number. Let π(x) = { # p; p prime and p ≤ x} (The prime numbers counting function). Bonus 1: the largest known prime number is: 257,885,161 − 1, which is Mersenne prime with 17,425,170 digits it’s part of Great Internet Mersenne Prime Search (GIMPS). There is a reward of $ 150000 for finding a prime with more than 100 million digits!! Remark: Mersenne Prime is of the form :

Prime Number Theory (Cont’d) Gauss stated the following asymptotic estimation : Gauss later refined his result to: The above estimation was called later PNT. The main objective of mathematicians is to minimize the estimation error. Chebyshev obtained the true order of by proving that

Euler’s approach to PNT Euler proved the infinitude of primes by factorizing the real harmonic series as follows: With over all prime number and

Riemann’s approach to PNT Inspired by Euler’s series representation, Riemann replaced the real valued variable x by a complex variable s and introduced the following Zeta function: It is obvious that Zeta doesn’t have roots in the domain of convergence , therefore it is not possible to factorize the left-hand side of the series. Bearing in mind that complex valued functions have a unique extension over the whole complex field, Riemann used the Hankel Contour integral together with Gamma function in order to extend Zeta over the whole complex field, with no singularity and with a pole at s = 1.

Riemann’s Approach to PNT (Cont’d) The extension produced the following functional equation: The immediate result is the zeros provided automatically from the term This zeros are the set of all negative even integers: -2, -4, -6,… Unfortunately, these zeros do not provide any factorization of Zeta function, consequently we call them simple or trivial zeros.

Riemann’s approach to PNT (Cont’d) In order to omit the impact of trivial zeros, Riemann refined his functional equation by introducing the following function: Which is an Entire function and satisfies:

Riemann Approach to PNT (Cont’d) The first non trivial zeros were calculating by Riemann, he manipulated the previous functional equation by setting , and observing the change of the sign of The first non trivial zeros are then he posed his celebrate hypothesis: Riemann Hypothesis: For more details see (Bombieri 2010) and (Sabbagh 2002)

Riemann’s Approach to PNT (Cont’d) Riemann Hypothesis provides the best approximation to the prime-counting function: So far 10,000,000,000,000 zeros have been checked, and they all satisfy RH! The ZetaGrid project has used a large network of computers to calculate the non trivial zeros. But unfortunately, till now none of this network of ideas offers a plausible strategy to prove RH. Check: www.zetagrid.net Bonus 2: Prove RH & win $ 1,000,000 from the Clay Mathematics Institute.

Riemann’s Approach to PNT (Cont’d)

Some values of Zeta

Mystery of Zeta: by Euler We can extend Zeta function by multiplication gradually in order to be entire. For the instance, lets consider this 1 step extension: Now,

Robin’s Inequality

Robin’s Inequality (Cont’d) In 1984 Robin proved that: * Later Robin proved that the following inequality is Equivalent to RH See: (Choie et.al 2007), (Lagarias 2002) and (Caveney et.al 2011)

Observations So far, the set of numbers that do not satisfy RI is: A= {1,2,3,4,5,6,8,9,10,12,16,18,20,24,30,36,48,60,72,84,120,180,240,360,720, 840,2520,5040} There are 28 numbers not satisfying RI. 7 is the 1st number satisfy RI, while 7! = 5040 is the last not to satisfy. For every n less than or equal 7, n! doesn’t satisfy RI.

Observations: Statistics If we let N be the set of all numbers less than or equal 5040. let p be the probability that a number satisfies RI, then: p= 5012/5040=0.99444…, q= 1-0.994= 0.0555..6 Therefore we will have the following Binomial distribution: How can we show that any other sample must have p = 1???

Observations: factorial Recall: In a letter correspondence, l'Hopital asked Leibniz: ‘’ What if the order of the derivative is ½ ‘’? To which Leibniz replied in a prophetical way, ‘’ Thus it follows that will be equal to an apparent paradox, from which one day useful consequences will be drawn.“ This letter of Leibniz was dated 30th September, 1695. So 30th September is considered as the birthday of fractional calculus. The question is what’s the concept of fractional Factorial?

Observation: Quadratic Equations.

Observations: Calculus.

References Choie., Y.-J, lichiardopol, N., Moree., P., Sole., P.: On Robin’s criterion for the Riemann Hypothesis. J. Theor. Nombres Bordeaux 19, (2007) 351-366. Lagarias, J. C: An elementary problem equivalent to the Riemann Hypothesis. Amer. Math. Monthly 109, (2002) 534-543. G. Caveney, J. L. Nicolas, J. Sondow, Robin’s Theorem, Primes, and a new elementary reformulation of the Riemann Hypothesis, arXiv: 1110.5078 v1 [math.NT] 23 Oct (2011). E. Bombieri, Problems of the Millennium: The Riemann Hypothesis, Clay Mathematics institution, 30 Oct (2010). http://www.claymath.org/millennium/RH/riemann.pdf. K. Sabbagh, The Riemann Hypothesis: The Greatest Unsolved Problem in Mathematics, Farrar, Straus, and Giroux, (2002).