The Riemann Hypothesis
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1 The Riemann Hypothesis Matilde N. Laĺın GAME Seminar, Special Series, History of Mathematics University of Alberta March 5, 2008 Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, 2008 / 28
2 Introduction Fundamental Theorem of Arithmetic prime numbers bricks of Z Theorem (Euclid, 300BC) There are many primes. Assume finitely many p,..., p n. N = p... p n + new prime dividing N Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
3 Introduction Fundamental Theorem of Arithmetic prime numbers bricks of Z Theorem (Euclid, 300BC) There are many primes. Assume finitely many p,..., p n. N = p... p n + new prime dividing N Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
4 Euler and the zeta function Mengoli 644: ( the Basel problem ) Find Euler 735: infinite product sin x x n= n 2 = π2 6. = x 2 3! + x 4 5! x 6 sin x x Compare coefficients of x 2. = ( x 2 π 2 n 2 7! +... ). Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
5 Euler and the zeta function Mengoli 644: ( the Basel problem ) Find Euler 735: infinite product sin x x n= n 2 = π2 6. = x 2 3! + x 4 5! x 6 sin x x Compare coefficients of x 2. = ( x 2 π 2 n 2 7! +... ). Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
6 Theorem (Euler (737), Euler s product) ζ(s) = n s = ( ) p s s R > (True for Re(s) > ) p prime q n= ( p s ) = p prime p s = + p s + p 2s +..., ( ) ( p s = + 2 s + ) ( 2 2s q s + ) q 2s +... = n prime divisors q n s. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
7 Theorem (Euler (737), Euler s product) ζ(s) = n s = ( ) p s s R > (True for Re(s) > ) p prime q n= ( p s ) = p prime p s = + p s + p 2s +..., ( ) ( p s = + 2 s + ) ( 2 2s q s + ) q 2s +... = n prime divisors q n s. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
8 Another proof for primes! ln n= n = p prime = p prime < p prime ln( p ) = + p + p = p prime p prime p prime p prime ( p + 2p 2 + ) 3p 3 + ( p p + ) 4p p ( p 2 + p + p ) 2 + p prime. p(p ) Euler concluded that p prime p = ln ln. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
9 The Prime Number Theorem p<x p p should have density ln s ln(ln x) = ln x dt t = x e ds s ln s. π(x) = {p x p prime} Gauss 792: Conjectured Li(x) = x 2 dt ln t Theorem (Prime Number Theorem) π(x) Li(x) Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
10 Legendre 796: π(x) x ln x Another version of Prime Number Theorem Li(x) = x ln x + x (ln x) 2 + π(x) x ln x. 2x (ln x) N!x ( ) x (ln x) N + O (ln x) N, Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
11 x π(x) π(x) x x ln x Li(x) π(x) π(x) , , ,498 6, ,579 44, ,76, , ,847,534 2,592,592, ,052,5 20,758,029 3, ,844,570,422,669 89,604,962,452,052, ,220,89,602,560,98,840 49,347,93,044,659,70 222,744, ln 0 = Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
12 Chebyschev 848, 850: Study π(x) by analytic methods, in connection to ζ(s). π(x) if lim x x ln(x) exists, it is < π(x) x ln(x) <. θ(x) = p x ln p, ψ(x) = p n x ln p. Prime number theorem is equivalent to θ(x) x, ψ(x) x. Bertrand s postulate 845: p prime n < p < 2n for any integer n 2. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
13 The work of Riemann Riemann 859: Ueber die Anzahl der Primzahlen unter einer gegebenen Grösse ζ(s) as a complex function!!! Gamma function: Γ(s) = 0 x s e x dx meromorphic, simple poles at s =, 2,..., Γ(n + ) = n! n Z >0 Γ(s) n s = x s e nx dx, 0 ζ(s) = e x Γ(s) 0 e x x s dx Re(s) >. extension to C with a pole at s =. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
14 Theorem (functional identity) ( π s s ) 2 Γ ζ(s) = π s 2 π s 2 Γ ( s 2 = Γ ( ) s 2 π s 2 n = s ) ζ(s) = 0 0 x s 2 ω(x)dx + 2 Γ ( s 2 x s 2 e n2 πx dx, x s 2 ( n= x s 2 ω ) ζ( s). e n2 πx ) dx ( ) dx x Where ω(x) = n= e n2 πx satisfies a functional equation ( π s s ) 2 Γ ζ(s) = 2 s(s ) + (x s 2 + x s+ 2 )ω(x)dx. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, 2008 / 28
15 Theorem (functional identity) ( π s s ) 2 Γ ζ(s) = π s 2 π s 2 Γ ( s 2 = Γ ( ) s 2 π s 2 n = s ) ζ(s) = 0 0 x s 2 ω(x)dx + 2 Γ ( s 2 x s 2 e n2 πx dx, x s 2 ( n= x s 2 ω ) ζ( s). e n2 πx ) dx ( ) dx x Where ω(x) = n= e n2 πx satisfies a functional equation ( π s s ) 2 Γ ζ(s) = 2 s(s ) + (x s 2 + x s+ 2 )ω(x)dx. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, 2008 / 28
16 Euler product no zeros with Re >. Functional equation no zeros with Re < 0 except trivial zeros: s = 2, 4,.... All complex zeros are in 0 Re(s). Theorem (conj by Riemann, proved by von Mangold 905) Let N(t) = {σ + it 0 < σ <, 0 < t < T, ζ(σ + it) = 0}. Then. N(T ) = T 2π ln T 2π T 2π + O(ln T ). Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
17 ln ζ(s) = p Differentiating, n= np ns = Π(x) = π(x) + 2 π (x 2 ζ (s) ζ(s) = After Mellin transform Theorem (von Mangoldt 905) ψ(x) = x 0 0 x s dπ(x) = s x s dψ(x) = s ρ, ζ(ρ)=0, Im(ρ) 0 ) + ( ) 3 π x Π(x)x s dx, ψ(x)x s dx. x ρ ρ ζ (0) ζ(0) 2 ln ( x 2) Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
18 ψ(x) = x ρ, ζ(ρ)=0, Im(ρ) 0 x ρ ρ ζ (0) ζ(0) 2 ln ( x 2) ζ (0) ζ(0) = ln(2π) corresponds to the pole in s = 0 2 ln ( x 2) = n= x 2n 2n corresponds to the trivial zeros The sum can not be absolutely convergent (left side is not continuous). Infinitely many ρ. x ρ = x Re ρ (we had Re ρ ) Re ρ < ψ x Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
19 Riemann Hypothesis The nontrivial zeros of ζ(s) have real part 2 Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
20 After Riemann Hadamard and de la Vallée Poussin in 896 : Prime Number Theorem π(x) = Li(x) + O (xe ) a ln x for a > 0. von Koch 90: Riemann hypothesis equivalent to π(x) = Li(x) + O( x ln x). Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
21 After Riemann Hadamard and de la Vallée Poussin in 896 : Prime Number Theorem π(x) = Li(x) + O (xe ) a ln x for a > 0. von Koch 90: Riemann hypothesis equivalent to π(x) = Li(x) + O( x ln x). Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
22 Hilbert 900: ICM in Paris, includes RH in a list of 23 unsolved problems. Early 900s: Littlewood gets RH as thesis problem(!!!) he can not solve it but he stills does many contributions and gets his PhD. (Littlewood is my Greatgreatgreatgreatgrandsupervisor) Littlewood 94: π(n) < Li(n) fails for infinitely many n Hardy and Littlewood 922: RH implies Golbach s weak conjecture Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
23 Hilbert 900: ICM in Paris, includes RH in a list of 23 unsolved problems. Early 900s: Littlewood gets RH as thesis problem(!!!) he can not solve it but he stills does many contributions and gets his PhD. (Littlewood is my Greatgreatgreatgreatgrandsupervisor) Littlewood 94: π(n) < Li(n) fails for infinitely many n Hardy and Littlewood 922: RH implies Golbach s weak conjecture Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
24 Hilbert 900: ICM in Paris, includes RH in a list of 23 unsolved problems. Early 900s: Littlewood gets RH as thesis problem(!!!) he can not solve it but he stills does many contributions and gets his PhD. (Littlewood is my Greatgreatgreatgreatgrandsupervisor) Littlewood 94: π(n) < Li(n) fails for infinitely many n Hardy and Littlewood 922: RH implies Golbach s weak conjecture Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
25 Hilbert 900: ICM in Paris, includes RH in a list of 23 unsolved problems. Early 900s: Littlewood gets RH as thesis problem(!!!) he can not solve it but he stills does many contributions and gets his PhD. (Littlewood is my Greatgreatgreatgreatgrandsupervisor) Littlewood 94: π(n) < Li(n) fails for infinitely many n Hardy and Littlewood 922: RH implies Golbach s weak conjecture Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
26 Lindelöf hypothesis Lindelöf 890: For every ɛ > 0, ζ ( ) 2 + it = O (t ɛ ), as t. (consequence of the Riemann Hypothesis). Hardy and Littlewood: exponent 4 + ɛ. Weyl: exponent 6 + ɛ. The original question remains unanswered. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
27 Lindelöf hypothesis Lindelöf 890: For every ɛ > 0, ζ ( ) 2 + it = O (t ɛ ), as t. (consequence of the Riemann Hypothesis). Hardy and Littlewood: exponent 4 + ɛ. Weyl: exponent 6 + ɛ. The original question remains unanswered. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
28 Mertens conjecture M(x) = n x µ(n) Möbius function: Mertens 897: if n = µ(n) = ( ) k if n = p... p k 0 if p 2 n M(x) x Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
29 Littlewood 92: implies the Riemann Hypothesis. ζ(s) = n= µ(n) n s = p ( p s ). Converges for Re(s) >. If the estimate for M(x) holds, then series converges for every s such that Re(s) > 2. Zeros for ζ(s) in Re(s) > 2 imply poles for ζ(s). Odlyzko and te Riele in 985: Mertens conjecture is FALSE is equivalent to RH. M(x) = O (x 2 +ɛ) Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
30 Littlewood 92: implies the Riemann Hypothesis. ζ(s) = n= µ(n) n s = p ( p s ). Converges for Re(s) >. If the estimate for M(x) holds, then series converges for every s such that Re(s) > 2. Zeros for ζ(s) in Re(s) > 2 imply poles for ζ(s). Odlyzko and te Riele in 985: Mertens conjecture is FALSE is equivalent to RH. M(x) = O (x 2 +ɛ) Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
31 Littlewood 92: implies the Riemann Hypothesis. ζ(s) = n= µ(n) n s = p ( p s ). Converges for Re(s) >. If the estimate for M(x) holds, then series converges for every s such that Re(s) > 2. Zeros for ζ(s) in Re(s) > 2 imply poles for ζ(s). Odlyzko and te Riele in 985: Mertens conjecture is FALSE is equivalent to RH. M(x) = O (x 2 +ɛ) Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
32 Quantity of zeros on the critical line H. Bohr, Landau 94: most of the zeros lie in a strip of width ɛ to the right of the critical line Re(s) = 2. Hardy 94: infinitely many zeros in the critical line. Selberg 942: a positive proportion of zeros in the critical line. Conrey 989: at least 40% of the zeros lie in the critical line. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
33 Computation of zeros year number author Gram Backlund Hutchinson 935,04 Titchmarsh 953,04 Turing ,000 Lehmer ,337 Meller ,000 Lehman 968 3,500,000 Rosser, Yohe, Schoenfeld ,000,000 Brent 979 8,000,00 Brent ,000,00 Brent, van de Lune, te Riele, Winter ,000,00 van de Lune, te Riele 986,500,000,00 van de Lune, te Riele, Winter 200 0,000,000,000 van de Lune (unpublished) ,000,000,000 Wedeniwski ,000,000,000,000 Gourdon and Demichel 2 + i , 2 + i , 2 + i , 2 + i Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
34 Random Matrix Theory Hilbert and Pólya: search for a Hermitian operator whose eigenvalues were the nontrivial zeros of ζ ( 2 + ti). (Eigenvalues are real in Hermitian operators). Montgomery 97: distribution of the gaps between zeros of the Riemann zeta function. likelihood of a gap of length x is proportional to sin2 (πx), (πx) 2 Dyson: the gaps of eigenvalues of certain random Hermitian matrices (the Gaussian Unitary Ensemble) follow the same path. GUE conjecture: all the statistics for the zeros matches the eigenvalues of GUE. numerical evidence found by Odlyzko in the 80s Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
35 A bigger picture RH is not an isolated question. L-functions algebraic-arithmetic objects (variety, number field). Expected to satisfy: Look like Dirichlet series Euler product Functional equation RH Zeros and Poles codify information about the algebraic-arithmetic object Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
36 L(s, χ 3 ) = ( π 3 ) s 2 Γ L(s, χ 3 ) = 2 s + 4 s 5 s + 7 s 8 s +... ( s + 2 p (mod 3) ( p s ) ) ( π ) s 2 L(s, χ 3 ) = Γ 3 p 2 (mod 3) RH (Generalized Riemann Hypothesis) not proved. ( 2 s 2 ( + p s ), ) L( s, χ 3 ). Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
37 Artin, others: Zeta functions of algebraic varieties over finite fields. curve C ζ(s, C) = NA s, A>0 where the A are the effective divisors that are fixed by the action of the Frobenius automorphism, and the norm is given by NA = q deg(a). Euler product, functional equation, RH (Weil conjectures, by Deligne) Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
38 Consequences RH: growth rate of the Möbius function (via Mertens function), growth rate of other multiplicative functions like the sum of divisors, statements about Farey sequences, Landau s function (order of elements in symmetric group), Golbach s weak conjecture, etc GRH: distribution of prime numbers in arithmetic progressions, existence of small primitive roots modulo p, the Miller-Rabin primality test runs in polynomial time, the Shanks-Tonelli algorithm (for finding roots to quadratic equations in modular arithmetic) runs in polynomial time, etc In 2000 the Riemann hypothesis was included in the list of the seven Millennium Prize Problems by the Clay Mathematics Institute. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
39 Enrico Bombieri, The Riemann hypothesis. The millennium prize problems, 07 24, Clay Math. Inst., Cambridge, MA, Borwein, P.; Choi, S.; Rooney, B.; Weirathmueller, A., The Riemann Hypothesis A Resource for the Afficionado and Virtuoso Alike CMS Books in Mathematics 2008, XIV, 538 pp. J. Brian Conrey, The Riemann hypothesis. Notices Amer. Math. Soc. 50 (2003), no. 3, John Derbyshire, Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics. Joseph Henry Press; 448 pages H. M. Edwards, Riemann s zeta function. Pure and Applied Mathematics, Vol. 58. Academic Press, New York-London, 974. xiii+35 pp. Matilde N. Laĺın (U of A) The Riemann Hypothesis March 5, / 28
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