Download Skewes Numbers

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