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Introduction The numerical data The Riemann Formula Skewes Numbers Roger Plymen 28 January 2011 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Euler, Riemann, Littlewood John Edensor Littlewood (1885–1977) Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Euler, Riemann, Littlewood John Edensor Littlewood (1885–1977) Cambridge; except for 3 years at Manchester 1907–1910 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Euler, Riemann, Littlewood John Edensor Littlewood (1885–1977) Cambridge; except for 3 years at Manchester 1907–1910 Riemann’s memoir (1859) Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Euler, Riemann, Littlewood John Edensor Littlewood (1885–1977) Cambridge; except for 3 years at Manchester 1907–1910 Riemann’s memoir (1859) Riemann Zeta Function ζ(s) := Roger Plymen P 1/ns with s ∈ C, Skewes Numbers <(s) > 1 Introduction The numerical data The Riemann Formula Euler, Riemann, Littlewood John Edensor Littlewood (1885–1977) Cambridge; except for 3 years at Manchester 1907–1910 Riemann’s memoir (1859) Riemann Zeta Function ζ(s) := P 1/ns with s ∈ C, <(s) > 1 The domain can be extended to the whole of C, a simple pole at s=1 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Euler, Riemann, Littlewood John Edensor Littlewood (1885–1977) Cambridge; except for 3 years at Manchester 1907–1910 Riemann’s memoir (1859) Riemann Zeta Function ζ(s) := P 1/ns with s ∈ C, <(s) > 1 The domain can be extended to the whole of C, a simple pole at s=1 The Riemann Hypothesis: all the zeta zeros in <(s) > 0 are on the line <(s) = 1/2. Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Euler, Riemann, Littlewood John Edensor Littlewood (1885–1977) Cambridge; except for 3 years at Manchester 1907–1910 Riemann’s memoir (1859) Riemann Zeta Function ζ(s) := P 1/ns with s ∈ C, <(s) > 1 The domain can be extended to the whole of C, a simple pole at s=1 The Riemann Hypothesis: all the zeta zeros in <(s) > 0 are on the line <(s) = 1/2. ”Several vain fleeting attempts .....” Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Pre-history of the Zeta Function Theorem (Euler) ζ(2k) ∈ π 2k Q For example: ζ(2) = π 2 /6 ζ(4) = π 4 /90 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Pre-history of the Zeta Function Theorem (Euler) ζ(2k) ∈ π 2k Q For example: ζ(2) = π 2 /6 ζ(4) = π 4 /90 Euler proved the functional equation for the zeta function at all integer points. This involves the zeta function at negative integers! Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Prime Number Theorem π(x) := number of primes up to x Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Prime Number Theorem π(x) := number of primes up to x Rx li(x) := 0 log1 t dt Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Prime Number Theorem π(x) := number of primes up to x Rx li(x) := 0 log1 t dt Theorem The prime number theorem: π(x) ∼ li(x) Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Prime Number Theorem π(x) := number of primes up to x Rx li(x) := 0 log1 t dt Theorem The prime number theorem: π(x) ∼ li(x) The average spacing between primes around x is log x. This is an insight of Gauss Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Prime Number Theorem π(x) := number of primes up to x Rx li(x) := 0 log1 t dt Theorem The prime number theorem: π(x) ∼ li(x) The average spacing between primes around x is log x. This is an insight of Gauss Around 10k the average spacing between primes is k log 10 ≈ 2.3k. If you go from 10k to 10k+1 then the average spacing increases by log 10 ≈ 2.3. Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Prime Number Theorem π(x) := number of primes up to x Rx li(x) := 0 log1 t dt Theorem The prime number theorem: π(x) ∼ li(x) The average spacing between primes around x is log x. This is an insight of Gauss Around 10k the average spacing between primes is k log 10 ≈ 2.3k. If you go from 10k to 10k+1 then the average spacing increases by log 10 ≈ 2.3. The density of primes around t is 1/ log t, so the expected number of primes up to x is li(x) Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The largest known prime Mersenne prime 2p − 1 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The largest known prime Mersenne prime 2p − 1 The largest known prime is the Mersenne prime 243,112,609 − 1. Approx 13 million digits! Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The numerical data Theorem (Kotnik, 2008) π(x) < li(x) for all x ≤ 1014 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Littlewood’s Theorem Theorem (Littlewood, 1914) The difference π(x) − li(x) changes sign infinitely often! Proof. See Christine Lee’s MSc dissertation 2008 for a detailed proof: http://eprints.ma.man.ac.uk/1524 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Consequences This implies that there is a least X for which π(X ) > li(X ). The first crossover. Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Consequences This implies that there is a least X for which π(X ) > li(X ). The first crossover. What is X ? Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Consequences This implies that there is a least X for which π(X ) > li(X ). The first crossover. What is X ? No-one knows. Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Consequences This implies that there is a least X for which π(X ) > li(X ). The first crossover. What is X ? No-one knows. How about an upper bound, i.e. a number Ξ such that (∃x < Ξ) Roger Plymen π(x) > li(x) Skewes Numbers Introduction The numerical data The Riemann Formula Consequences This implies that there is a least X for which π(X ) > li(X ). The first crossover. What is X ? No-one knows. How about an upper bound, i.e. a number Ξ such that (∃x < Ξ) The Skewes number π(x) > li(x) 1043 1010 is the first such upper bound, stated by Stanley Skewes without proof, and conditional on RH Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Riemann Formula Theorem (Riemann) Let J(x) := X1 n n≥1 Then J(x) = li(x) − X li(x ρ ) + π(x) = ∞ Z x ρ and so π(x 1/n ) X µ(n) n≥1 Roger Plymen n u(u 2 du − log 2 − 1) log u J(x 1/n ) Skewes Numbers Introduction The numerical data The Riemann Formula The Riemann Formula One of the great formulas of all time: sum over primes (LHS), sum of zeta zeros (RHS). Reciprocity between primes and zeta zeros. Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Riemann Formula One of the great formulas of all time: sum over primes (LHS), sum of zeta zeros (RHS). Reciprocity between primes and zeta zeros. µ(n) = (−1)k , n = p1 · · · pk , Roger Plymen Skewes Numbers otherwise 0 Introduction The numerical data The Riemann Formula The Riemann Formula One of the great formulas of all time: sum over primes (LHS), sum of zeta zeros (RHS). Reciprocity between primes and zeta zeros. µ(n) = (−1)k , n = p1 · · · pk , The series for J(x) is finite Roger Plymen Skewes Numbers otherwise 0 Introduction The numerical data The Riemann Formula The Riemann Formula One of the great formulas of all time: sum over primes (LHS), sum of zeta zeros (RHS). Reciprocity between primes and zeta zeros. µ(n) = (−1)k , n = p1 · · · pk , otherwise 0 The series for J(x) is finite π(x) comprises increasing/decreasing terms; oscillating terms Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Riemann Formula One of the great formulas of all time: sum over primes (LHS), sum of zeta zeros (RHS). Reciprocity between primes and zeta zeros. µ(n) = (−1)k , n = p1 · · · pk , otherwise 0 The series for J(x) is finite π(x) comprises increasing/decreasing terms; oscillating terms If x = 10316 then there are are 639 terms in the Riemann formula Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Riemann-Ramanujan formula Theorem (Riemann-Ramanujan) Let R(x) := li(x) − 1 li(x 1/2 ) − li(x 1/3 ) − li(x 1/5 ) + li(x 1/6 ) − · · · 2 Then we have π(x) ≈ R(x) Some evidence that this is a good numerical formula. For example, with x = 3, 000, 000, Gauss’ error is 155, Riemann’s error is 0 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Riemann-Ramanujan formula Theorem (Riemann-Ramanujan) Let R(x) := li(x) − 1 li(x 1/2 ) − li(x 1/3 ) − li(x 1/5 ) + li(x 1/6 ) − · · · 2 Then we have π(x) ≈ R(x) Some evidence that this is a good numerical formula. For example, with x = 3, 000, 000, Gauss’ error is 155, Riemann’s error is 0 Suggests that (∀x) π(x) < li(x) Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Riemann-Ramanujan formula Theorem (Riemann-Ramanujan) Let R(x) := li(x) − 1 li(x 1/2 ) − li(x 1/3 ) − li(x 1/5 ) + li(x 1/6 ) − · · · 2 Then we have π(x) ≈ R(x) Some evidence that this is a good numerical formula. For example, with x = 3, 000, 000, Gauss’ error is 155, Riemann’s error is 0 Suggests that (∀x) π(x) < li(x) In Ramanujan’s approach there are no complex zeta zeros (remark of G.H. Hardy) Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The oscillating terms We will need li(z) of a complex number. Z ρ log x u e ρ li(x ) := du u −∞ Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The oscillating terms We will need li(z) of a complex number. Z ρ log x u e ρ li(x ) := du u −∞ The first 100 trillion zeta zeros will be on the line 1/2 + it so let’s consider the curve t 7→ li(x 1/2+it ) Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The oscillating terms We will need li(z) of a complex number. Z ρ log x u e ρ li(x ) := du u −∞ The first 100 trillion zeta zeros will be on the line 1/2 + it so let’s consider the curve t 7→ li(x 1/2+it ) This curve is a (counterclockwise) spiral tending to 0. Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The oscillating terms We will need li(z) of a complex number. Z ρ log x u e ρ li(x ) := du u −∞ The first 100 trillion zeta zeros will be on the line 1/2 + it so let’s consider the curve t 7→ li(x 1/2+it ) This curve is a (counterclockwise) spiral tending to 0. Note that li(x ρ ) + li(x ρ ) = 2< li(x ρ ) The conjugate zero ρ will cancel the imaginary part, leaving (twice) the real part. Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Cancellation or Reinforcement? Littlewood’s Theorem implies that there exists X for which the real parts of the points li(X ρ1 ), . . . , li(X ρn ) on the spiral will reinforce each other to the left of the vertical axis so as to dominate the massive term − li(X 1/2 ) − li(X 1/3 ) − li(X 1/5 ) in the Riemann formula Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Cancellation or Reinforcement? Littlewood’s Theorem implies that there exists X for which the real parts of the points li(X ρ1 ), . . . , li(X ρn ) on the spiral will reinforce each other to the left of the vertical axis so as to dominate the massive term − li(X 1/2 ) − li(X 1/3 ) − li(X 1/5 ) in the Riemann formula How to find X ? Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Skewes Numbers Skewes[1933] ”I propose to postpone the details to a later paper” 1043 Sk1 = 1010 This sketch is conditional on RH Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Skewes Numbers Skewes[1933] ”I propose to postpone the details to a later paper” 1043 Sk1 = 1010 This sketch is conditional on RH Skewes [1955] ”...thanks to Prof Littlewood, but for whose patient profanity this paper could never have become fit for publication” Theorem (Skewes) There exists 3 1010 X < Sk2 = 1010 for which π(X ) > li(X ) Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Lehman’s idea Lehman’s idea: Integrate π(x) − li(x) against a Gaussian kernel f (x) and prove that I (ω, η) > 0 where Z ω+η I (ω, η) := f (e u )(π(e u ) − li(e u ))du ω−η Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Lehmman’s Theorem Theorem (Lehman) Let ρ = 1/2 + iγ and H(T , ω) = −2< X e iγω 2 · e −γ /2α ρ 0<γ≤T then it is enough to prove that H(T , ω) > 1 + with = (A, ω, η, α) Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The single Lehman spiral and cancellation Lehman replaces the many Riemann spirals by a single spiral Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The single Lehman spiral and cancellation Lehman replaces the many Riemann spirals by a single spiral The equidistribution of the numbers {exp(iγω) : ζ(ρ) = 0, γ = =ρ} leads us to expect a fair amount of cancellation. Numerical computation shows that H(T , ω) is very reluctant to exceed 1 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The single Lehman spiral and cancellation Lehman replaces the many Riemann spirals by a single spiral The equidistribution of the numbers {exp(iγω) : ζ(ρ) = 0, γ = =ρ} leads us to expect a fair amount of cancellation. Numerical computation shows that H(T , ω) is very reluctant to exceed 1 The following inequalities must hold: 2/A ≤ 2A/α ≤ η ≤ ω/2 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Numerical Results Lehman [1966] Interval [1.539052 × 101165 , 1.6473479 × 101165 ] Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Numerical Results Lehman [1966] Interval [1.539052 × 101165 , 1.6473479 × 101165 ] te Riele [1987] Interval [6.627990 × 10370 , 6.687911 × 10370 ] Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Numerical Results Lehman [1966] Interval [1.539052 × 101165 , 1.6473479 × 101165 ] te Riele [1987] Interval [6.627990 × 10370 , 6.687911 × 10370 ] Bays-Hudson [2000] Interval [exp(727.95009), exp(727.95409)] Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Numerical Results Lehman [1966] Interval [1.539052 × 101165 , 1.6473479 × 101165 ] te Riele [1987] Interval [6.627990 × 10370 , 6.687911 × 10370 ] Bays-Hudson [2000] Interval [exp(727.95009), exp(727.95409)] Chao-Plymen [2010] Interval [exp(727.951858), exp(727.952178)] Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Main Result Theorem (Chao-Plymen, Int. J. Number Theory 6 (2010) 681–690) Entering 2, 000, 000 zeta zeros, we prove that there exists x in the interval [exp(727.951858), exp(727.952178)] for which π(x) − li(x) > 3.2 × 10151 There are at least 10154 successive integers x in this interval for which π(x) > li(x). This interval is strictly a sub-interval of the interval in Bays & Hudson, and is narrower by a factor of about 12 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Theorem (Saouter-Demichel, Math. Comp. 79 (2010) 2395 – 2405) Entering 22, 000, 000 zeta zeros, we prove that there are at least 6.6587 × 10152 consecutive integers x in the interval [exp(727.95132478), exp(727.95134682)] such that π(x) > li(x). This has been slightly improved by Stefanie Zegowitz, http://eprints.ma.man.ac.uk/1547, by inserting a missing term. Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Numerical Results What about all the other crossovers? There are infinitely many, by Littlewood’s theorem. Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula Numerical Results What about all the other crossovers? There are infinitely many, by Littlewood’s theorem. They remain mysterious Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Riemann Formula Theorem (Riemann) π(x) = X µ(n) n≥1 Roger Plymen n J(x 1/n ) Skewes Numbers Introduction The numerical data The Riemann Formula The Riemann Formula Theorem (Riemann) π(x) = X µ(n) n≥1 n J(x 1/n ) The average spacing between primes around x is log x. Before a crossover interval, the primes are further apart than this. As we enter a crossover interval, the primes become more crowded: the spacing between primes is a little less than 727 in the first crossover interval. As we leave the interval, they start to move further apart Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula The Riemann Formula Theorem (Riemann) π(x) = X µ(n) n≥1 n J(x 1/n ) The average spacing between primes around x is log x. Before a crossover interval, the primes are further apart than this. As we enter a crossover interval, the primes become more crowded: the spacing between primes is a little less than 727 in the first crossover interval. As we leave the interval, they start to move further apart Rubinstein and Sarnak (1994) showed that the proportion of integers for which π(x) > li(x) is about 0.00000026 Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula What do we know about Ξ? Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula What do we know about Ξ? After nearly a century, this is all we know (best to use the log scale): Roger Plymen Skewes Numbers Introduction The numerical data The Riemann Formula What do we know about Ξ? After nearly a century, this is all we know (best to use the log scale): 32.2 < log Ξ < 727.95134682 Roger Plymen Skewes Numbers