The First 150 Years of the Riemann Zeta-Function
S. M. Gonek
Department of Mathematics
University of Rochester
June 1, 2009/Graduate Workshop on Zeta functions, L-functions
and their Applications
(University of Rochester)
1 / 51
I. Synopsis of Riemann’s paper
Ueber die Anzahl der Primzahlen unter einer
gegebenen Grösse
( On the number of primes less than a given
magnitude )
(University of Rochester)
2 / 51
Figure: Riemann
(University of Rochester)
3 / 51
Figure: First page of Riemann’s paper
(University of Rochester)
4 / 51
What Riemann proves
(University of Rochester)
5 / 51
What Riemann proves
Riemann begins with Euler’s observation that
∞
X
n=1
(University of Rochester)
n−s =
Y
(1 − p−s )−1
(s > 1).
p
5 / 51
What Riemann proves
Riemann begins with Euler’s observation that
∞
X
n=1
n−s =
Y
(1 − p−s )−1
(s > 1).
p
But he lets s = σ + it be complex.
(University of Rochester)
5 / 51
What Riemann proves
Riemann begins with Euler’s observation that
∞
X
n=1
n−s =
Y
(1 − p−s )−1
(s > 1).
p
But he lets s = σ + it be complex.
He denotes the common value by ζ(s) and proves:
(University of Rochester)
5 / 51
What Riemann proves
Riemann begins with Euler’s observation that
∞
X
n=1
n−s =
Y
(1 − p−s )−1
(s > 1).
p
But he lets s = σ + it be complex.
He denotes the common value by ζ(s) and proves:
ζ(s) has an analytic continuation to C, except for a simple pole
at s = 1. The only zeros in σ < 0 are simple zeros at
s = −2, −4, −6, ...
(University of Rochester)
5 / 51
What Riemann proves
Riemann begins with Euler’s observation that
∞
X
n=1
n−s =
Y
(1 − p−s )−1
(s > 1).
p
But he lets s = σ + it be complex.
He denotes the common value by ζ(s) and proves:
ζ(s) has an analytic continuation to C, except for a simple pole
at s = 1. The only zeros in σ < 0 are simple zeros at
s = −2, −4, −6, ...
ζ(s) has a functional equation
π −s/2 Γ(s/2)ζ(s) = π −(1−s)/2 Γ((1 − s)/2)ζ(1 − s)
(University of Rochester)
5 / 51
What Riemann claims
(University of Rochester)
6 / 51
What Riemann claims
ζ(s) has infinitely many nontrivial zeros ρ = β + iγ in the “critical
strip” 0 ≤ σ ≤ 1.
(University of Rochester)
6 / 51
What Riemann claims
ζ(s) has infinitely many nontrivial zeros ρ = β + iγ in the “critical
strip” 0 ≤ σ ≤ 1.
If N(T ) denotes the number of nontrivial zeros ρ = β + iγ with
ordinates 0 < γ ≤ T ,
(University of Rochester)
6 / 51
What Riemann claims
ζ(s) has infinitely many nontrivial zeros ρ = β + iγ in the “critical
strip” 0 ≤ σ ≤ 1.
If N(T ) denotes the number of nontrivial zeros ρ = β + iγ with
ordinates 0 < γ ≤ T , then as T → ∞,
N(T ) =
(University of Rochester)
T
T
T
log
−
+ O(log T ).
2π
2π 2π
6 / 51
What Riemann claims
ζ(s) has infinitely many nontrivial zeros ρ = β + iγ in the “critical
strip” 0 ≤ σ ≤ 1.
If N(T ) denotes the number of nontrivial zeros ρ = β + iγ with
ordinates 0 < γ ≤ T , then as T → ∞,
N(T ) =
T
T
T
log
−
+ O(log T ).
2π
2π 2π
The function ξ(s) = 12 s(s − 1)π −s/2 Γ(s/2)ζ(s) is entire
(University of Rochester)
6 / 51
What Riemann claims
ζ(s) has infinitely many nontrivial zeros ρ = β + iγ in the “critical
strip” 0 ≤ σ ≤ 1.
If N(T ) denotes the number of nontrivial zeros ρ = β + iγ with
ordinates 0 < γ ≤ T , then as T → ∞,
N(T ) =
T
T
T
log
−
+ O(log T ).
2π
2π 2π
The function ξ(s) = 12 s(s − 1)π −s/2 Γ(s/2)ζ(s) is entire and has
the product formula
Y
s
ξ(s) = ξ(0)
1−
.
ρ
ρ
(University of Rochester)
6 / 51
What Riemann claims
ζ(s) has infinitely many nontrivial zeros ρ = β + iγ in the “critical
strip” 0 ≤ σ ≤ 1.
If N(T ) denotes the number of nontrivial zeros ρ = β + iγ with
ordinates 0 < γ ≤ T , then as T → ∞,
N(T ) =
T
T
T
log
−
+ O(log T ).
2π
2π 2π
The function ξ(s) = 12 s(s − 1)π −s/2 Γ(s/2)ζ(s) is entire and has
the product formula
Y
s
ξ(s) = ξ(0)
1−
.
ρ
ρ
Here ρ runs over the nontrivial zeros of ζ(s).
(University of Rochester)
6 / 51
What Riemann claims
(University of Rochester)
7 / 51
What Riemann claims
explicit formula
Let Λ(n) = log p if n = pk and 0 otherwise. Then
ψ(x) =
X
n≤x
(University of Rochester)
Λ(n) = x −
X xρ
ρ
ρ
+
∞
X
x −2n
n=1
2n
−
ζ 0 (0)
ζ(0)
7 / 51
What Riemann claims
explicit formula
Let Λ(n) = log p if n = pk and 0 otherwise. Then
ψ(x) =
X
Λ(n) = x −
X xρ
n≤x
(Riemann states this for π(x) =
(University of Rochester)
ρ
ρ
P
+
p≤x
∞
X
x −2n
n=1
2n
−
ζ 0 (0)
ζ(0)
1 instead.)
7 / 51
What Riemann claims
explicit formula
Let Λ(n) = log p if n = pk and 0 otherwise. Then
ψ(x) =
X
Λ(n) = x −
X xρ
ρ
ρ
n≤x
(Riemann states this for π(x) =
P
+
p≤x
∞
X
x −2n
n=1
2n
−
ζ 0 (0)
ζ(0)
1 instead.)
Note that from this one can see why the Prime Number Theorem,
ψ(x) ∼ x
might be true.
(University of Rochester)
7 / 51
The Riemann Hypothesis
(University of Rochester)
8 / 51
The Riemann Hypothesis
Riemann also makes his famous conjecture.
(University of Rochester)
8 / 51
The Riemann Hypothesis
Riemann also makes his famous conjecture.
Conjecture (The Riemann Hypothesis )
All the zeros ρ = β + iγ in the critical strip lie on the line σ = 1/2.
(University of Rochester)
8 / 51
II. Early developments after the paper
(University of Rochester)
9 / 51
Hadamard 1893
(University of Rochester)
10 / 51
Hadamard 1893
Hadamard developed the theory of entire functions (Hadamard product
formula) and proved the product formula for
ξ(s) =
(University of Rochester)
1
s(s − 1)π −s/2 Γ(s/2)ζ(s).
2
10 / 51
Hadamard 1893
Hadamard developed the theory of entire functions (Hadamard product
formula) and proved the product formula for
ξ(s) =
ξ(s) = ξ(0)
Y
ρ
(University of Rochester)
1
s(s − 1)π −s/2 Γ(s/2)ζ(s).
2
1−
s
ρ
= ξ(0)
Y Imρ>0
1−
s + s2
ρ(1 − ρ)
10 / 51
Hadamard 1893
Hadamard developed the theory of entire functions (Hadamard product
formula) and proved the product formula for
ξ(s) =
ξ(s) = ξ(0)
Y
ρ
1
s(s − 1)π −s/2 Γ(s/2)ζ(s).
2
1−
s
ρ
= ξ(0)
Y Imρ>0
1−
s + s2
ρ(1 − ρ)
To do this he proved the estimate
N(T ) T log T ,
(University of Rochester)
10 / 51
Hadamard 1893
Hadamard developed the theory of entire functions (Hadamard product
formula) and proved the product formula for
ξ(s) =
ξ(s) = ξ(0)
Y
ρ
1
s(s − 1)π −s/2 Γ(s/2)ζ(s).
2
1−
s
ρ
= ξ(0)
Y Imρ>0
1−
s + s2
ρ(1 − ρ)
To do this he proved the estimate
N(T ) T log T ,
which is weaker than Riemann’s assertion about N(T ).
(University of Rochester)
10 / 51
von Mangoldt 1895
(University of Rochester)
11 / 51
von Mangoldt 1895
von Mangoldt proved Riemann’s explicit formula for π(x) and
(University of Rochester)
11 / 51
von Mangoldt 1895
von Mangoldt proved Riemann’s explicit formula for π(x) and
ψ(x) =
X
n≤x
(University of Rochester)
Λ(n) = x −
X xρ
ρ
ρ
+
∞
X
x −2n
n=1
2n
−
ζ 0 (0)
.
ζ(0)
11 / 51
Hadamard and de la Vallée Poussin 1896
(University of Rochester)
12 / 51
Hadamard and de la Vallée Poussin 1896
Hadamard and de la Valleé Poussin independently proved the
asymptotic form of the Prime Number Theorem, namely
(University of Rochester)
12 / 51
Hadamard and de la Vallée Poussin 1896
Hadamard and de la Valleé Poussin independently proved the
asymptotic form of the Prime Number Theorem, namely
ψ(x) ∼ x
(University of Rochester)
12 / 51
Hadamard and de la Vallée Poussin 1896
Hadamard and de la Valleé Poussin independently proved the
asymptotic form of the Prime Number Theorem, namely
ψ(x) ∼ x
To do this, they both needed to prove that
ζ(1 + it) 6= 0
(University of Rochester)
12 / 51
de la Vallée Poussin 1899
(University of Rochester)
13 / 51
de la Vallée Poussin 1899
de la Vallée Poussin proved the Prime Number Theorem with a
remainder term:
√
ψ(x) = x + O xe− c1 log x .
(University of Rochester)
13 / 51
de la Vallée Poussin 1899
de la Vallée Poussin proved the Prime Number Theorem with a
remainder term:
√
ψ(x) = x + O xe− c1 log x .
This required him to prove that there is a zero-free region
σ <1−
(University of Rochester)
c0
log t
13 / 51
von Mangoldt 1905
(University of Rochester)
14 / 51
von Mangoldt 1905
von Mangoldt proved Riemann’s formula for the counting function of
the zeros
(University of Rochester)
14 / 51
von Mangoldt 1905
von Mangoldt proved Riemann’s formula for the counting function of
the zeros
N(T ) =
(University of Rochester)
T
T
T
log
−
+ O(log T )
2π
2π 2π
14 / 51
von Koch 1905
(University of Rochester)
15 / 51
von Koch 1905
von Koch showed that the Riemann Hypothesis implies the Prime
Number Theorem with a “small” remainder term
(University of Rochester)
15 / 51
von Koch 1905
von Koch showed that the Riemann Hypothesis implies the Prime
Number Theorem with a “small” remainder term
RH =⇒ ψ(x) = x + O(x 1/2 log2 x)
(University of Rochester)
15 / 51
III. The order of ζ(s) in the critical strip
(University of Rochester)
16 / 51
ζ(s) in the critical strip
(University of Rochester)
17 / 51
ζ(s) in the critical strip
The critical strip is the most important (and mysterious) region for ζ(s).
(University of Rochester)
17 / 51
ζ(s) in the critical strip
The critical strip is the most important (and mysterious) region for ζ(s).
By the functional equation, it suffices to focus on 1/2 ≤ σ ≤ 1.
(University of Rochester)
17 / 51
ζ(s) in the critical strip
The critical strip is the most important (and mysterious) region for ζ(s).
By the functional equation, it suffices to focus on 1/2 ≤ σ ≤ 1.
A natural question is:
(University of Rochester)
17 / 51
ζ(s) in the critical strip
The critical strip is the most important (and mysterious) region for ζ(s).
By the functional equation, it suffices to focus on 1/2 ≤ σ ≤ 1.
A natural question is: how large can ζ(s) be as t grows?
(University of Rochester)
17 / 51
ζ(s) in the critical strip
The critical strip is the most important (and mysterious) region for ζ(s).
By the functional equation, it suffices to focus on 1/2 ≤ σ ≤ 1.
A natural question is: how large can ζ(s) be as t grows?
This is important because
(University of Rochester)
17 / 51
ζ(s) in the critical strip
The critical strip is the most important (and mysterious) region for ζ(s).
By the functional equation, it suffices to focus on 1/2 ≤ σ ≤ 1.
A natural question is: how large can ζ(s) be as t grows?
This is important because
the growth of an analytic function and the distribution of its zeros
are intimately connected.
(University of Rochester)
17 / 51
ζ(s) in the critical strip
The critical strip is the most important (and mysterious) region for ζ(s).
By the functional equation, it suffices to focus on 1/2 ≤ σ ≤ 1.
A natural question is: how large can ζ(s) be as t grows?
This is important because
the growth of an analytic function and the distribution of its zeros
are intimately connected.
the distribution of primes depends on it.
(University of Rochester)
17 / 51
ζ(s) in the critical strip
The critical strip is the most important (and mysterious) region for ζ(s).
By the functional equation, it suffices to focus on 1/2 ≤ σ ≤ 1.
A natural question is: how large can ζ(s) be as t grows?
This is important because
the growth of an analytic function and the distribution of its zeros
are intimately connected.
the distribution of primes depends on it.
answers to other arithmetical questions depend on it.
(University of Rochester)
17 / 51
Implications of the size of ζ(s)
(University of Rochester)
18 / 51
Implications of the size of ζ(s)
Relation between growth and zeros:
(University of Rochester)
18 / 51
Implications of the size of ζ(s)
Relation between growth and zeros:
Jensen’s Formula. Let f (z) be analytic for |z| ≤ R and f (0) 6= 0. If
z1 , z2 , . . . , zn are all the zeros of f (z) inside |z| ≤ R, then
log
Rn
|z1 z2 · · · zn |
(University of Rochester)
=
1
2π
Z
2π
log |f (Reiθ )|dθ − log |f (0)|.
0
18 / 51
Implications of the size of ζ(s)
Relation between growth and zeros:
Jensen’s Formula. Let f (z) be analytic for |z| ≤ R and f (0) 6= 0. If
z1 , z2 , . . . , zn are all the zeros of f (z) inside |z| ≤ R, then
log
Rn
|z1 z2 · · · zn |
=
1
2π
Z
2π
log |f (Reiθ )|dθ − log |f (0)|.
0
Example of an application to other problems: for 0 < c < 1
(University of Rochester)
18 / 51
Implications of the size of ζ(s)
Relation between growth and zeros:
Jensen’s Formula. Let f (z) be analytic for |z| ≤ R and f (0) 6= 0. If
z1 , z2 , . . . , zn are all the zeros of f (z) inside |z| ≤ R, then
log
Rn
|z1 z2 · · · zn |
=
1
2π
Z
2π
log |f (Reiθ )|dθ − log |f (0)|.
0
Example of an application to other problems: for 0 < c < 1
X
dk (n) = xPk −1 (log x) +
n≤x
(University of Rochester)
1
2πi
Z
c+i∞
c−i∞
ζ k (s)
xs
ds.
s
18 / 51
Estimates at the edge of the strip
(University of Rochester)
19 / 51
Estimates at the edge of the strip
Upper bounds for ζ(s) near σ = 1 allow one to widen the zero-free
region.
(University of Rochester)
19 / 51
Estimates at the edge of the strip
Upper bounds for ζ(s) near σ = 1 allow one to widen the zero-free
region.
This leads to improvements in the remainder term for the PNT.
(University of Rochester)
19 / 51
Estimates at the edge of the strip
Upper bounds for ζ(s) near σ = 1 allow one to widen the zero-free
region.
This leads to improvements in the remainder term for the PNT.
For instance, we saw that de la Vallée Poussin showed that
ζ(σ + it) log t in σ ≥ 1 −
(University of Rochester)
c0
,
log t
19 / 51
Estimates at the edge of the strip
Upper bounds for ζ(s) near σ = 1 allow one to widen the zero-free
region.
This leads to improvements in the remainder term for the PNT.
For instance, we saw that de la Vallée Poussin showed that
ζ(σ + it) log t in σ ≥ 1 −
c0
,
log t
and this implied that the O-term in the PNT is xe−
(University of Rochester)
√
c1 log x
.
19 / 51
Estimates at the edge of the strip
(University of Rochester)
20 / 51
Estimates at the edge of the strip
Littlewood 1922
ζ(σ + it) (University of Rochester)
log t
c log log t
and no zeros in σ ≥ 1 −
log log t
log t
20 / 51
Estimates at the edge of the strip
Littlewood 1922
log t
c log log t
and no zeros in σ ≥ 1 −
log log t
log t
√
=⇒ O-term in PNT xe−c log x log log x
ζ(σ + it) (University of Rochester)
20 / 51
Estimates at the edge of the strip
Littlewood 1922
log t
c log log t
and no zeros in σ ≥ 1 −
log log t
log t
√
=⇒ O-term in PNT xe−c log x log log x
ζ(σ + it) The idea is to approximate
ζ(σ + it) ≈
N
X
1
(University of Rochester)
1
nσ+it
20 / 51
Estimates at the edge of the strip
Littlewood 1922
log t
c log log t
and no zeros in σ ≥ 1 −
log log t
log t
√
=⇒ O-term in PNT xe−c log x log log x
ζ(σ + it) The idea is to approximate
ζ(σ + it) ≈
N
X
1
1
nσ+it
then use Weyl’s method to estimate the exponential sums
(University of Rochester)
20 / 51
Estimates at the edge of the strip
Littlewood 1922
log t
c log log t
and no zeros in σ ≥ 1 −
log log t
log t
√
=⇒ O-term in PNT xe−c log x log log x
ζ(σ + it) The idea is to approximate
ζ(σ + it) ≈
N
X
1
1
nσ+it
then use Weyl’s method to estimate the exponential sums
b
X
a
(University of Rochester)
n−it =
b
X
eif (n) .
a
20 / 51
Estimates at the edge of the strip
(University of Rochester)
21 / 51
Estimates at the edge of the strip
Vinogradov and Korobov 1958 (independently)
ζ(σ + it) log2/3 t and no zeros in σ ≥ 1 −
(University of Rochester)
c
log2/3 t
21 / 51
Estimates at the edge of the strip
Vinogradov and Korobov 1958 (independently)
ζ(σ + it) log2/3 t and no zeros in σ ≥ 1 −
=⇒ O-term in PNT xe−c log
(University of Rochester)
3/5−
c
log2/3 t
x
21 / 51
Estimates at the edge of the strip
Vinogradov and Korobov 1958 (independently)
ζ(σ + it) log2/3 t and no zeros in σ ≥ 1 −
=⇒ O-term in PNT xe−c log
3/5−
c
log2/3 t
x
Where Littlewood used Weyl’s method to estimate the exponential
sums
b
X
n−it ,
a
(University of Rochester)
21 / 51
Estimates at the edge of the strip
Vinogradov and Korobov 1958 (independently)
ζ(σ + it) log2/3 t and no zeros in σ ≥ 1 −
=⇒ O-term in PNT xe−c log
3/5−
c
log2/3 t
x
Where Littlewood used Weyl’s method to estimate the exponential
sums
b
X
n−it ,
a
Vinogradov and Korobov used Vinogradov’s method.
(University of Rochester)
21 / 51
Estimates at the edge of the strip
(University of Rochester)
22 / 51
Estimates at the edge of the strip
Here is a summary:
(University of Rochester)
22 / 51
Estimates at the edge of the strip
Here is a summary:
ζ(1 + it) log t
(University of Rochester)
(de la Vallée Poussin)
22 / 51
Estimates at the edge of the strip
Here is a summary:
ζ(1 + it) log t
log t
ζ(1 + it) log log t
(University of Rochester)
(de la Vallée Poussin)
(Littlewood-Weyl)
22 / 51
Estimates at the edge of the strip
Here is a summary:
ζ(1 + it) log t
log t
ζ(1 + it) log log t
ζ(1 + it) log2/3 t
(University of Rochester)
(de la Vallée Poussin)
(Littlewood-Weyl)
(Vinogradov-Korobov)
22 / 51
Estimates at the edge of the strip
Here is a summary:
ζ(1 + it) log t
log t
ζ(1 + it) log log t
ζ(1 + it) log2/3 t
(de la Vallée Poussin)
(Littlewood-Weyl)
(Vinogradov-Korobov)
What should the truth be?
(University of Rochester)
22 / 51
Estimates at the edge of the strip
Here is a summary:
ζ(1 + it) log t
log t
ζ(1 + it) log log t
ζ(1 + it) log2/3 t
What should the truth be?
(University of Rochester)
(de la Vallée Poussin)
(Littlewood-Weyl)
(Vinogradov-Korobov)
One can show that
22 / 51
Estimates at the edge of the strip
Here is a summary:
ζ(1 + it) log t
log t
ζ(1 + it) log log t
ζ(1 + it) log2/3 t
What should the truth be?
(de la Vallée Poussin)
(Littlewood-Weyl)
(Vinogradov-Korobov)
One can show that
(1 + o(1))eγ log log t ≤i.o. |ζ(1 + it)| ≤RH 2(1 + o(1))eγ log log t.
(University of Rochester)
22 / 51
Estimates inside the strip
(University of Rochester)
23 / 51
Estimates inside the strip
Definition (Lindelöf 1908)
(University of Rochester)
23 / 51
Estimates inside the strip
Definition (Lindelöf 1908)
For a fixed σ let µ(σ) denote the lower bound of the numbers µ such
that
ζ(σ + it) (1 + |t|)µ .
(University of Rochester)
23 / 51
Estimates inside the strip
Definition (Lindelöf 1908)
For a fixed σ let µ(σ) denote the lower bound of the numbers µ such
that
ζ(σ + it) (1 + |t|)µ .
ζ(s) bounded for σ > 1 =⇒ µ(σ) = 0 for σ > 1.
(University of Rochester)
23 / 51
Estimates inside the strip
Definition (Lindelöf 1908)
For a fixed σ let µ(σ) denote the lower bound of the numbers µ such
that
ζ(σ + it) (1 + |t|)µ .
ζ(s) bounded for σ > 1 =⇒ µ(σ) = 0 for σ > 1.
|ζ(s)| ∼ (|t|/2π)1/2−σ |ζ(1 − s)| =⇒ µ(σ) = 1/2 − σ + µ(1 − σ).
(University of Rochester)
23 / 51
Estimates inside the strip
Definition (Lindelöf 1908)
For a fixed σ let µ(σ) denote the lower bound of the numbers µ such
that
ζ(σ + it) (1 + |t|)µ .
ζ(s) bounded for σ > 1 =⇒ µ(σ) = 0 for σ > 1.
|ζ(s)| ∼ (|t|/2π)1/2−σ |ζ(1 − s)| =⇒ µ(σ) = 1/2 − σ + µ(1 − σ).
In particular, µ(σ) = 1/2 − σ for σ < 0.
(University of Rochester)
23 / 51
Lindelöf’s µ-function
(University of Rochester)
24 / 51
Lindelöf’s µ-function
Lindelöf proved that µ(σ) is
(University of Rochester)
24 / 51
Lindelöf’s µ-function
Lindelöf proved that µ(σ) is
continuous
(University of Rochester)
24 / 51
Lindelöf’s µ-function
Lindelöf proved that µ(σ) is
continuous
nonincreasing
(University of Rochester)
24 / 51
Lindelöf’s µ-function
Lindelöf proved that µ(σ) is
continuous
nonincreasing
convex
(University of Rochester)
24 / 51
Lindelöf’s µ-function
Lindelöf proved that µ(σ) is
continuous
nonincreasing
convex
These are in the same circle of ideas as the Phragmen-Lindelöf
theorems.
(University of Rochester)
24 / 51
Lindelöf’s µ-function
Lindelöf proved that µ(σ) is
continuous
nonincreasing
convex
These are in the same circle of ideas as the Phragmen-Lindelöf
theorems.
It follows that µ(1/2) ≤ 1/4, that is,
ζ(1/2 + it) |t|1/4+ .
(University of Rochester)
24 / 51
Lindelöf’s µ-function
Lindelöf proved that µ(σ) is
continuous
nonincreasing
convex
These are in the same circle of ideas as the Phragmen-Lindelöf
theorems.
It follows that µ(1/2) ≤ 1/4, that is,
ζ(1/2 + it) |t|1/4+ .
This is a so called convexity bound .
(University of Rochester)
24 / 51
Breaking convexity
(University of Rochester)
25 / 51
Breaking convexity
Using Weyl’s method of estimating exponential sums, Hardy and
Littlewood showed that
(University of Rochester)
25 / 51
Breaking convexity
Using Weyl’s method of estimating exponential sums, Hardy and
Littlewood showed that
ζ(1/2 + it) |t|1/6+ .
(University of Rochester)
25 / 51
Breaking convexity
Using Weyl’s method of estimating exponential sums, Hardy and
Littlewood showed that
ζ(1/2 + it) |t|1/6+ .
The best results for µ(σ) since have come from exponential sum
methods:
(University of Rochester)
25 / 51
Breaking convexity
Using Weyl’s method of estimating exponential sums, Hardy and
Littlewood showed that
ζ(1/2 + it) |t|1/6+ .
The best results for µ(σ) since have come from exponential sum
methods: van der Corput, Vinogradov, Kolesnik, Bombieri-Iwaniec,
Huxley-Watt.
(University of Rochester)
25 / 51
Breaking convexity
Using Weyl’s method of estimating exponential sums, Hardy and
Littlewood showed that
ζ(1/2 + it) |t|1/6+ .
The best results for µ(σ) since have come from exponential sum
methods: van der Corput, Vinogradov, Kolesnik, Bombieri-Iwaniec,
Huxley-Watt.
Huxley and Watt show that µ(σ) < 9/56.
(University of Rochester)
25 / 51
Breaking convexity
Using Weyl’s method of estimating exponential sums, Hardy and
Littlewood showed that
ζ(1/2 + it) |t|1/6+ .
The best results for µ(σ) since have come from exponential sum
methods: van der Corput, Vinogradov, Kolesnik, Bombieri-Iwaniec,
Huxley-Watt.
Huxley and Watt show that µ(σ) < 9/56.
Conjecture (Lindelöf)
µ(σ) = 0 for σ ≥ 1/2. That is, ζ(1/2 + it) |t| for t large
(University of Rochester)
25 / 51
What we expect the order to be
(University of Rochester)
26 / 51
What we expect the order to be
The LH says that for large |t|
log |ζ(1/2 + it)| ≤ log |t|.
(University of Rochester)
26 / 51
What we expect the order to be
The LH says that for large |t|
log |ζ(1/2 + it)| ≤ log |t|.
It is also known that
(University of Rochester)
26 / 51
What we expect the order to be
The LH says that for large |t|
log |ζ(1/2 + it)| ≤ log |t|.
It is also known that
s
log t
log t
c
≤i.o. log |ζ(1/2 + it)| RH
.
log log t
log log t
(University of Rochester)
26 / 51
What we expect the order to be
The LH says that for large |t|
log |ζ(1/2 + it)| ≤ log |t|.
It is also known that
s
log t
log t
c
≤i.o. log |ζ(1/2 + it)| RH
.
log log t
log log t
Which bound, the upper or the lower, is closest to the truth is one of
the important open questions.
(University of Rochester)
26 / 51
IV. Mean value theorems
(University of Rochester)
27 / 51
Mean value theorems
(University of Rochester)
28 / 51
Mean value theorems
Averages such as
research
(University of Rochester)
RT
0
|ζ(σ + it)|2k dt have been another main focus of
28 / 51
Mean value theorems
Averages such as
research because
(University of Rochester)
RT
0
|ζ(σ + it)|2k dt have been another main focus of
28 / 51
Mean value theorems
Averages such as
research because
RT
0
|ζ(σ + it)|2k dt have been another main focus of
averages as well as pointwise upper bounds tell us about zeros
and have other applications.
(University of Rochester)
28 / 51
Mean value theorems
Averages such as
research because
RT
0
|ζ(σ + it)|2k dt have been another main focus of
averages as well as pointwise upper bounds tell us about zeros
and have other applications.
mean values are easier to prove than point wise bounds.
(University of Rochester)
28 / 51
Mean value theorems
Averages such as
research because
RT
0
|ζ(σ + it)|2k dt have been another main focus of
averages as well as pointwise upper bounds tell us about zeros
and have other applications.
mean values are easier to prove than point wise bounds.
the techniques developed to treat them have proved important in
other contexts.
(University of Rochester)
28 / 51
Mean value theorems
(University of Rochester)
29 / 51
Mean value theorems
Landau 1908
Z
T
|ζ(σ + it)|2 dt ∼ ζ(2σ)T
(σ > 1/2 fixed).
0
(University of Rochester)
29 / 51
Mean value theorems
Landau 1908
Z
T
|ζ(σ + it)|2 dt ∼ ζ(2σ)T
(σ > 1/2 fixed).
0
Hardy-Littlewood 1918
Z
T
|ζ(1/2 + it)|2 dt ∼ T log T .
0
(University of Rochester)
29 / 51
Mean value theorems
Landau 1908
Z
T
|ζ(σ + it)|2 dt ∼ ζ(2σ)T
(σ > 1/2 fixed).
0
Hardy-Littlewood 1918
Z
T
|ζ(1/2 + it)|2 dt ∼ T log T .
0
For this H-L developed the approximate functional equation
X
X
ζ(s) =
n−s + χ(s)
ns−1 + O(...),
√
√
n≤
(University of Rochester)
t/2π
n≤
t/2π
29 / 51
Mean value theorems
Landau 1908
Z
T
|ζ(σ + it)|2 dt ∼ ζ(2σ)T
(σ > 1/2 fixed).
0
Hardy-Littlewood 1918
Z
T
|ζ(1/2 + it)|2 dt ∼ T log T .
0
For this H-L developed the approximate functional equation
X
X
ζ(s) =
n−s + χ(s)
ns−1 + O(...),
√
√
n≤
t/2π
n≤
t/2π
which has proved an extremely important tool ever since.
(University of Rochester)
29 / 51
Mean value theorems
(University of Rochester)
30 / 51
Mean value theorems
Hardy-Littlewood 1918
Z
T
|ζ(σ + it)|4 dt ∼
0
(University of Rochester)
ζ 4 (2σ)
T
ζ(4σ)
(σ > 1/2 fixed).
30 / 51
Mean value theorems
Hardy-Littlewood 1918
Z
T
|ζ(σ + it)|4 dt ∼
0
ζ 4 (2σ)
T
ζ(4σ)
(σ > 1/2 fixed).
Ingham 1926
Z
0
(University of Rochester)
T
|ζ(1/2 + it)|4 dt ∼
T
log4 T .
2π 2
30 / 51
Mean value theorems
Hardy-Littlewood 1918
Z
T
|ζ(σ + it)|4 dt ∼
0
ζ 4 (2σ)
T
ζ(4σ)
(σ > 1/2 fixed).
Ingham 1926
Z
0
T
|ζ(1/2 + it)|4 dt ∼
T
log4 T .
2π 2
This was done by using an approximate functional equation for ζ 2 (s).
(University of Rochester)
30 / 51
Mean value theorems
Hardy-Littlewood 1918
Z
T
|ζ(σ + it)|4 dt ∼
0
ζ 4 (2σ)
T
ζ(4σ)
(σ > 1/2 fixed).
Ingham 1926
T
Z
0
|ζ(1/2 + it)|4 dt ∼
T
log4 T .
2π 2
This was done by using an approximate functional equation for ζ 2 (s).
When k is a positive integer Ramachandra showed that
Z
T
2
|ζ(1/2 + it)|2k dt T logk T .
0
(University of Rochester)
30 / 51
Mean value theorems
Hardy-Littlewood 1918
Z
T
|ζ(σ + it)|4 dt ∼
0
ζ 4 (2σ)
T
ζ(4σ)
(σ > 1/2 fixed).
Ingham 1926
T
Z
0
|ζ(1/2 + it)|4 dt ∼
T
log4 T .
2π 2
This was done by using an approximate functional equation for ζ 2 (s).
When k is a positive integer Ramachandra showed that
Z
T
2
|ζ(1/2 + it)|2k dt T logk T .
0
This is believed to be the correct upper bound as well.
(University of Rochester)
30 / 51
Mean value theorems
(University of Rochester)
31 / 51
Mean value theorems
This suggests the problem of determining constants Ck such that
(University of Rochester)
31 / 51
Mean value theorems
This suggests the problem of determining constants Ck such that
Z
T
2
|ζ(1/2 + it)|2k dt ∼ Ck T logk T .
0
(University of Rochester)
31 / 51
Mean value theorems
This suggests the problem of determining constants Ck such that
Z
T
2
|ζ(1/2 + it)|2k dt ∼ Ck T logk T .
0
Conrey-Ghosh suggested that
Ck =
(University of Rochester)
ak gk
,
Γ(k 2 + 1)
31 / 51
Mean value theorems
This suggests the problem of determining constants Ck such that
Z
T
2
|ζ(1/2 + it)|2k dt ∼ Ck T logk T .
0
Conrey-Ghosh suggested that
Ck =
ak gk
,
Γ(k 2 + 1)
where
ak =
Y p
(University of Rochester)
1
1−
p
k 2 X
∞
r =0
dk2 (pr )
pr
!
31 / 51
Mean value theorems
This suggests the problem of determining constants Ck such that
Z
T
2
|ζ(1/2 + it)|2k dt ∼ Ck T logk T .
0
Conrey-Ghosh suggested that
Ck =
ak gk
,
Γ(k 2 + 1)
where
ak =
Y p
1
1−
p
k 2 X
∞
r =0
dk2 (pr )
pr
!
and gk is an integer.
(University of Rochester)
31 / 51
Mean value theorems
(University of Rochester)
32 / 51
Mean value theorems
In
Z
T
0
(University of Rochester)
|ζ(1/2 + it)|2k dt ∼
2
ak gk
T logk T ,
2
Γ(k + 1)
32 / 51
Mean value theorems
In
Z
T
|ζ(1/2 + it)|2k dt ∼
0
2
ak gk
T logk T ,
2
Γ(k + 1)
g1 = 1 and g2 = 2 are known.
(University of Rochester)
32 / 51
Mean value theorems
In
Z
T
|ζ(1/2 + it)|2k dt ∼
0
2
ak gk
T logk T ,
2
Γ(k + 1)
g1 = 1 and g2 = 2 are known.
Conrey and Ghosh conjectured that g3 = 42.
(University of Rochester)
32 / 51
Mean value theorems
In
Z
T
|ζ(1/2 + it)|2k dt ∼
0
2
ak gk
T logk T ,
2
Γ(k + 1)
g1 = 1 and g2 = 2 are known.
Conrey and Ghosh conjectured that g3 = 42.
Conrey and G conjecured that g4 = 24024.
(University of Rochester)
32 / 51
Mean value theorems
In
Z
T
|ζ(1/2 + it)|2k dt ∼
0
2
ak gk
T logk T ,
2
Γ(k + 1)
g1 = 1 and g2 = 2 are known.
Conrey and Ghosh conjectured that g3 = 42.
Conrey and G conjecured that g4 = 24024.
Keating and Snaith used random matrix theory to conjecture the
value of gk for every value of k > −1/2.
(University of Rochester)
32 / 51
Mean value theorems
In
Z
T
|ζ(1/2 + it)|2k dt ∼
0
2
ak gk
T logk T ,
2
Γ(k + 1)
g1 = 1 and g2 = 2 are known.
Conrey and Ghosh conjectured that g3 = 42.
Conrey and G conjecured that g4 = 24024.
Keating and Snaith used random matrix theory to conjecture the
value of gk for every value of k > −1/2.
Soundararajan has recently shown that on RH
Z
T
|ζ(1/2 + it)|2k dt T logk
2 +
T.
0
(University of Rochester)
32 / 51
V. Zero-density estimates
(University of Rochester)
33 / 51
Zero-density estimates
(University of Rochester)
34 / 51
Zero-density estimates
Let N(σ, T ) denote the number of zeros of ζ(s) with abscissae to the
right of σ and ordinates between 0 and T .
(University of Rochester)
34 / 51
Zero-density estimates
Let N(σ, T ) denote the number of zeros of ζ(s) with abscissae to the
right of σ and ordinates between 0 and T .
Zero-density estimates are bounds for N(σ, T ) when σ > 1/2.
(University of Rochester)
34 / 51
Zero-density estimates
Let N(σ, T ) denote the number of zeros of ζ(s) with abscissae to the
right of σ and ordinates between 0 and T .
Zero-density estimates are bounds for N(σ, T ) when σ > 1/2.
Bohr and Landau 1912 showed that for each fixed σ > 1/2,
N(σ, T ) T .
(University of Rochester)
34 / 51
Zero-density estimates
Let N(σ, T ) denote the number of zeros of ζ(s) with abscissae to the
right of σ and ordinates between 0 and T .
Zero-density estimates are bounds for N(σ, T ) when σ > 1/2.
Bohr and Landau 1912 showed that for each fixed σ > 1/2,
N(σ, T ) T .
Since
N(T ) ∼ (T /2π) log T ,
(University of Rochester)
34 / 51
Zero-density estimates
Let N(σ, T ) denote the number of zeros of ζ(s) with abscissae to the
right of σ and ordinates between 0 and T .
Zero-density estimates are bounds for N(σ, T ) when σ > 1/2.
Bohr and Landau 1912 showed that for each fixed σ > 1/2,
N(σ, T ) T .
Since
N(T ) ∼ (T /2π) log T ,
this says the proportion of zeros to the right of σ > 1/2 tends to 0 as
T → ∞.
(University of Rochester)
34 / 51
Zero-density estimates
(University of Rochester)
35 / 51
Zero-density estimates
Bohr and Landau used Jensen’s formula and
Z T
|ζ(σ + it)|2 dt T (σ > 1/2 fixed)
0
to prove this.
(University of Rochester)
35 / 51
Zero-density estimates
Bohr and Landau used Jensen’s formula and
Z T
|ζ(σ + it)|2 dt T (σ > 1/2 fixed)
0
to prove this.
Today we have much better zero-density estimates of the form
N(σ, T ) T θ(σ) with θ(σ) strictly less than 1.
(University of Rochester)
35 / 51
Zero-density estimates
Bohr and Landau used Jensen’s formula and
Z T
|ζ(σ + it)|2 dt T (σ > 1/2 fixed)
0
to prove this.
Today we have much better zero-density estimates of the form
N(σ, T ) T θ(σ) with θ(σ) strictly less than 1.
The conjecture that N(σ, T ) T 2(1−σ) log T is called the Density
Hypothesis.
(University of Rochester)
35 / 51
Zero-density estimates
Bohr and Landau used Jensen’s formula and
Z T
|ζ(σ + it)|2 dt T (σ > 1/2 fixed)
0
to prove this.
Today we have much better zero-density estimates of the form
N(σ, T ) T θ(σ) with θ(σ) strictly less than 1.
The conjecture that N(σ, T ) T 2(1−σ) log T is called the Density
Hypothesis.
Obviously RH implies the Density Hypothesis.
(University of Rochester)
35 / 51
Zero-density estimates
Bohr and Landau used Jensen’s formula and
Z T
|ζ(σ + it)|2 dt T (σ > 1/2 fixed)
0
to prove this.
Today we have much better zero-density estimates of the form
N(σ, T ) T θ(σ) with θ(σ) strictly less than 1.
The conjecture that N(σ, T ) T 2(1−σ) log T is called the Density
Hypothesis.
Obviously RH implies the Density Hypothesis.
LH implies N(σ, T ) T 2(1−σ)+ .
(University of Rochester)
35 / 51
VI. The distribution of a-values of ζ(s)
(University of Rochester)
36 / 51
The distribution of a-values of ζ(s)
(University of Rochester)
37 / 51
The distribution of a-values of ζ(s)
What can we say about the distribution of non-zero values, a, of the
zeta-function?
(University of Rochester)
37 / 51
The distribution of a-values of ζ(s)
What can we say about the distribution of non-zero values, a, of the
zeta-function?
A lovely theory due mostly to H. Bohr developed around this question.
(University of Rochester)
37 / 51
The distribution of a-values of ζ(s)
What can we say about the distribution of non-zero values, a, of the
zeta-function?
A lovely theory due mostly to H. Bohr developed around this question.
Here are two results.
(University of Rochester)
37 / 51
The distribution of a-values of ζ(s)
What can we say about the distribution of non-zero values, a, of the
zeta-function?
A lovely theory due mostly to H. Bohr developed around this question.
Here are two results.
First, the curve f (t) = ζ(σ + it)
is dense in C.
(University of Rochester)
(1/2 < σ ≤ 1 fixed, t ∈ R)
37 / 51
The distribution of a-values of ζ(s)
What can we say about the distribution of non-zero values, a, of the
zeta-function?
A lovely theory due mostly to H. Bohr developed around this question.
Here are two results.
First, the curve f (t) = ζ(σ + it)
is dense in C. The idea is to
(University of Rochester)
(1/2 < σ ≤ 1 fixed, t ∈ R)
37 / 51
The distribution of a-values of ζ(s)
What can we say about the distribution of non-zero values, a, of the
zeta-function?
A lovely theory due mostly to H. Bohr developed around this question.
Here are two results.
First, the curve f (t) = ζ(σ + it)
(1/2 < σ ≤ 1 fixed, t ∈ R)
is dense in C. The idea is to
Q
show that ζ(σ + it) ≈ p≤N (1 − p−σ−it )−1 for most t.
(University of Rochester)
37 / 51
The distribution of a-values of ζ(s)
What can we say about the distribution of non-zero values, a, of the
zeta-function?
A lovely theory due mostly to H. Bohr developed around this question.
Here are two results.
First, the curve f (t) = ζ(σ + it)
(1/2 < σ ≤ 1 fixed, t ∈ R)
is dense in C. The idea is to
Q
show that ζ(σ + it) ≈ p≤N (1 − p−σ−it )−1 for most t.
use Kronecker’s theorem
to find a t so that the numbers p−it point
Q
in such a way that p≤N (1 − p−σ−it )−1 ≈ a.
(University of Rochester)
37 / 51
The distribution of a-values of ζ(s)
(University of Rochester)
38 / 51
The distribution of a-values of ζ(s)
As a second result, let Na (σ1 , σ2 , T ) be the number of solutions of
ζ(s) = a in the rectangular area σ1 ≤ σ ≤ σ2 , 0 ≤ t ≤ T .
(University of Rochester)
38 / 51
The distribution of a-values of ζ(s)
As a second result, let Na (σ1 , σ2 , T ) be the number of solutions of
ζ(s) = a in the rectangular area σ1 ≤ σ ≤ σ2 , 0 ≤ t ≤ T .
Suppose that 1/2 < σ1 < σ2 ≤ 1.
(University of Rochester)
38 / 51
The distribution of a-values of ζ(s)
As a second result, let Na (σ1 , σ2 , T ) be the number of solutions of
ζ(s) = a in the rectangular area σ1 ≤ σ ≤ σ2 , 0 ≤ t ≤ T .
Suppose that 1/2 < σ1 < σ2 ≤ 1.
Then there exists a positive constant c(σ1 , σ2 ) such that
(University of Rochester)
38 / 51
The distribution of a-values of ζ(s)
As a second result, let Na (σ1 , σ2 , T ) be the number of solutions of
ζ(s) = a in the rectangular area σ1 ≤ σ ≤ σ2 , 0 ≤ t ≤ T .
Suppose that 1/2 < σ1 < σ2 ≤ 1.
Then there exists a positive constant c(σ1 , σ2 ) such that
Na (σ1 , σ2 , T ) ∼ c(σ1 , σ2 )T .
(University of Rochester)
38 / 51
The distribution of a-values of ζ(s)
As a second result, let Na (σ1 , σ2 , T ) be the number of solutions of
ζ(s) = a in the rectangular area σ1 ≤ σ ≤ σ2 , 0 ≤ t ≤ T .
Suppose that 1/2 < σ1 < σ2 ≤ 1.
Then there exists a positive constant c(σ1 , σ2 ) such that
Na (σ1 , σ2 , T ) ∼ c(σ1 , σ2 )T .
Notice that this is quite different from the case a = 0, because modern
zero-density estimates imply
(University of Rochester)
38 / 51
The distribution of a-values of ζ(s)
As a second result, let Na (σ1 , σ2 , T ) be the number of solutions of
ζ(s) = a in the rectangular area σ1 ≤ σ ≤ σ2 , 0 ≤ t ≤ T .
Suppose that 1/2 < σ1 < σ2 ≤ 1.
Then there exists a positive constant c(σ1 , σ2 ) such that
Na (σ1 , σ2 , T ) ∼ c(σ1 , σ2 )T .
Notice that this is quite different from the case a = 0, because modern
zero-density estimates imply
N0 (σ1 , σ2 , T ) T θ
(University of Rochester)
(θ < 1).
38 / 51
VII. Number of zeros on the line as T → ∞
(University of Rochester)
39 / 51
Number of zeros on the line as T → ∞
(University of Rochester)
40 / 51
Number of zeros on the line as T → ∞
n
o
Let N0 (T ) = # 1/2 + iγ ζ(1/2 + iγ) = 0, 0 < γ < T denote the
number of zeros on the critical line up to height T .
(University of Rochester)
40 / 51
Number of zeros on the line as T → ∞
n
o
Let N0 (T ) = # 1/2 + iγ ζ(1/2 + iγ) = 0, 0 < γ < T denote the
number of zeros on the critical line up to height T .
The important estimates were
(University of Rochester)
40 / 51
Number of zeros on the line as T → ∞
n
o
Let N0 (T ) = # 1/2 + iγ ζ(1/2 + iγ) = 0, 0 < γ < T denote the
number of zeros on the critical line up to height T .
The important estimates were
Hardy 1914
N0 (T ) → ∞
(University of Rochester)
(as T → ∞)
40 / 51
Number of zeros on the line as T → ∞
n
o
Let N0 (T ) = # 1/2 + iγ ζ(1/2 + iγ) = 0, 0 < γ < T denote the
number of zeros on the critical line up to height T .
The important estimates were
Hardy 1914
N0 (T ) → ∞
Hardy-Littlewood 1921
(University of Rochester)
(as T → ∞)
N0 (T ) > c T
40 / 51
Number of zeros on the line as T → ∞
n
o
Let N0 (T ) = # 1/2 + iγ ζ(1/2 + iγ) = 0, 0 < γ < T denote the
number of zeros on the critical line up to height T .
The important estimates were
Hardy 1914
N0 (T ) → ∞
Hardy-Littlewood 1921
Selberg 1942
(as T → ∞)
N0 (T ) > c T
N0 (T ) > c N(T )
(University of Rochester)
40 / 51
Number of zeros on the line as T → ∞
n
o
Let N0 (T ) = # 1/2 + iγ ζ(1/2 + iγ) = 0, 0 < γ < T denote the
number of zeros on the critical line up to height T .
The important estimates were
Hardy 1914
N0 (T ) → ∞
Hardy-Littlewood 1921
Selberg 1942
(as T → ∞)
N0 (T ) > c T
N0 (T ) > c N(T )
Levinson 1974
(University of Rochester)
N0 (T ) >
1
3
N(T )
40 / 51
Number of zeros on the line as T → ∞
n
o
Let N0 (T ) = # 1/2 + iγ ζ(1/2 + iγ) = 0, 0 < γ < T denote the
number of zeros on the critical line up to height T .
The important estimates were
Hardy 1914
N0 (T ) → ∞
Hardy-Littlewood 1921
Selberg 1942
N0 (T ) > c T
N0 (T ) > c N(T )
Levinson 1974
Conrey 1989
(as T → ∞)
N0 (T ) >
N0 (T ) >
(University of Rochester)
2
5
1
3
N(T )
N(T )
40 / 51
Number of zeros on the line as T → ∞
n
o
Let N0 (T ) = # 1/2 + iγ ζ(1/2 + iγ) = 0, 0 < γ < T denote the
number of zeros on the critical line up to height T .
The important estimates were
Hardy 1914
N0 (T ) → ∞
Hardy-Littlewood 1921
Selberg 1942
N0 (T ) > c T
N0 (T ) > c N(T )
Levinson 1974
Conrey 1989
(as T → ∞)
N0 (T ) >
N0 (T ) >
2
5
1
3
N(T )
N(T )
These all rely heavily on mean value estimates.
(University of Rochester)
40 / 51
Hardy’s idea
(University of Rochester)
41 / 51
Hardy’s idea
One can write the functional equation as ζ(s) = χ(s)ζ(1 − s),
(University of Rochester)
41 / 51
Hardy’s idea
One can write the functional equation as ζ(s) = χ(s)ζ(1 − s), or as
χ−1/2 (s)ζ(s) = χ1/2 (s)ζ(1 − s).
(University of Rochester)
41 / 51
Hardy’s idea
One can write the functional equation as ζ(s) = χ(s)ζ(1 − s), or as
χ−1/2 (s)ζ(s) = χ1/2 (s)ζ(1 − s).
Then
Z (t) = χ−1/2 (1/2 + it)ζ(1/2 + it)
(University of Rochester)
41 / 51
Hardy’s idea
One can write the functional equation as ζ(s) = χ(s)ζ(1 − s), or as
χ−1/2 (s)ζ(s) = χ1/2 (s)ζ(1 − s).
Then
Z (t) = χ−1/2 (1/2 + it)ζ(1/2 + it)
has the same zeros as ζ(s) on σ = 1/2
(University of Rochester)
41 / 51
Hardy’s idea
One can write the functional equation as ζ(s) = χ(s)ζ(1 − s), or as
χ−1/2 (s)ζ(s) = χ1/2 (s)ζ(1 − s).
Then
Z (t) = χ−1/2 (1/2 + it)ζ(1/2 + it)
has the same zeros as ζ(s) on σ = 1/2 and is real.
(University of Rochester)
41 / 51
Hardy’s idea
One can write the functional equation as ζ(s) = χ(s)ζ(1 − s), or as
χ−1/2 (s)ζ(s) = χ1/2 (s)ζ(1 − s).
Then
Z (t) = χ−1/2 (1/2 + it)ζ(1/2 + it)
has the same zeros as ζ(s) on σ = 1/2 and is real.
If Z (t) had no zeros for t ≥ T0 ,
(University of Rochester)
41 / 51
Hardy’s idea
One can write the functional equation as ζ(s) = χ(s)ζ(1 − s), or as
χ−1/2 (s)ζ(s) = χ1/2 (s)ζ(1 − s).
Then
Z (t) = χ−1/2 (1/2 + it)ζ(1/2 + it)
has the same zeros as ζ(s) on σ = 1/2 and is real.
If Z (t) had no zeros for t ≥ T0 , the integrals
Z
T
T0
Z (t)dt Z
T
|Z (t)|dt
and
T0
would be the same size as T → ∞.
(University of Rochester)
41 / 51
Hardy’s idea
One can write the functional equation as ζ(s) = χ(s)ζ(1 − s), or as
χ−1/2 (s)ζ(s) = χ1/2 (s)ζ(1 − s).
Then
Z (t) = χ−1/2 (1/2 + it)ζ(1/2 + it)
has the same zeros as ζ(s) on σ = 1/2 and is real.
If Z (t) had no zeros for t ≥ T0 , the integrals
Z
T
T0
Z (t)dt Z
T
|Z (t)|dt
and
T0
would be the same size as T → ∞. But they are not.
(University of Rochester)
41 / 51
VIII. Calculations of zeros on the line
(University of Rochester)
42 / 51
Numerical calculations of zeros
(University of Rochester)
43 / 51
Numerical calculations of zeros
Gram 1903
simple.
The zeros up to 50 (the first 15 ) are on the line and
(University of Rochester)
43 / 51
Numerical calculations of zeros
Gram 1903
simple.
The zeros up to 50 (the first 15 ) are on the line and
Backlund 1912 The zeros up to 200 are on the line
(University of Rochester)
43 / 51
Numerical calculations of zeros
Gram 1903
simple.
The zeros up to 50 (the first 15 ) are on the line and
Backlund 1912 The zeros up to 200 are on the line
Hutchison 1925 The zeros up to 300 are on the line
(University of Rochester)
43 / 51
Numerical calculations of zeros
Gram 1903
simple.
The zeros up to 50 (the first 15 ) are on the line and
Backlund 1912 The zeros up to 200 are on the line
Hutchison 1925 The zeros up to 300 are on the line
Titchmarsh, Turing, Lehman, Brent, van de Lune, te Riele,
Odlyzko, Wedeniwski, ...
(University of Rochester)
43 / 51
Numerical calculations of zeros
Gram 1903
simple.
The zeros up to 50 (the first 15 ) are on the line and
Backlund 1912 The zeros up to 200 are on the line
Hutchison 1925 The zeros up to 300 are on the line
Titchmarsh, Turing, Lehman, Brent, van de Lune, te Riele,
Odlyzko, Wedeniwski, ...
Gourdon-Demichel 2004 The first 1013 (ten trillion) zeros are on the
line.
(University of Rochester)
43 / 51
Numerical calculations of zeros
Gram 1903
simple.
The zeros up to 50 (the first 15 ) are on the line and
Backlund 1912 The zeros up to 200 are on the line
Hutchison 1925 The zeros up to 300 are on the line
Titchmarsh, Turing, Lehman, Brent, van de Lune, te Riele,
Odlyzko, Wedeniwski, ...
Gourdon-Demichel 2004 The first 1013 (ten trillion) zeros are on the
line. Moreover, billions of zeros near the 1024 zero are on the line.
(University of Rochester)
43 / 51
IX. More recent developments
(University of Rochester)
44 / 51
Pair correlation
(University of Rochester)
45 / 51
Pair correlation
A major theme of research over the last 35 years has been to
understand the distribution of the zeros on the critical line assuming
that the Riemann Hypothesis is true.
(University of Rochester)
45 / 51
Pair correlation
A major theme of research over the last 35 years has been to
understand the distribution of the zeros on the critical line assuming
that the Riemann Hypothesis is true.
In 1974 Montgomery conjectured that the zeros are distributed like
the eigenvalues of random Hermitian matrices.
(University of Rochester)
45 / 51
Pair correlation
A major theme of research over the last 35 years has been to
understand the distribution of the zeros on the critical line assuming
that the Riemann Hypothesis is true.
In 1974 Montgomery conjectured that the zeros are distributed like
the eigenvalues of random Hermitian matrices.
From 1980 on Odlyzko did a vast amount of numerical calculation
that strongly supported Montgomery’s conjecture.
(University of Rochester)
45 / 51
New mean value theorems
(University of Rochester)
46 / 51
New mean value theorems
G and Conrey, Ghosh, and G proved a number of discrete mean value
theorems of the type
X
X
|ζ(ρ + iα)|2 and
|ζ 0 (ρ)MN (ρ)|2 ,
0<γ≤T
(University of Rochester)
0<γ≤T
46 / 51
New mean value theorems
G and Conrey, Ghosh, and G proved a number of discrete mean value
theorems of the type
X
X
|ζ(ρ + iα)|2 and
|ζ 0 (ρ)MN (ρ)|2 ,
0<γ≤T
0<γ≤T
where ρ = 1/2 + iγ runs over the zeros.
(University of Rochester)
46 / 51
New mean value theorems
G and Conrey, Ghosh, and G proved a number of discrete mean value
theorems of the type
X
X
|ζ(ρ + iα)|2 and
|ζ 0 (ρ)MN (ρ)|2 ,
0<γ≤T
0<γ≤T
where ρ = 1/2 + iγ runs over the zeros.
Assuming RH and sometimes GLH and GRH, Conrey, Ghosh, and G
used these to prove that
(University of Rochester)
46 / 51
New mean value theorems
G and Conrey, Ghosh, and G proved a number of discrete mean value
theorems of the type
X
X
|ζ(ρ + iα)|2 and
|ζ 0 (ρ)MN (ρ)|2 ,
0<γ≤T
0<γ≤T
where ρ = 1/2 + iγ runs over the zeros.
Assuming RH and sometimes GLH and GRH, Conrey, Ghosh, and G
used these to prove that
there are large and small gaps between consecutive zeros.
(University of Rochester)
46 / 51
New mean value theorems
G and Conrey, Ghosh, and G proved a number of discrete mean value
theorems of the type
X
X
|ζ(ρ + iα)|2 and
|ζ 0 (ρ)MN (ρ)|2 ,
0<γ≤T
0<γ≤T
where ρ = 1/2 + iγ runs over the zeros.
Assuming RH and sometimes GLH and GRH, Conrey, Ghosh, and G
used these to prove that
there are large and small gaps between consecutive zeros.
over 70% of the zeros are simple.
(University of Rochester)
46 / 51
Random matrix models
(University of Rochester)
47 / 51
Random matrix models
A major development was Keating and Snaith’s modeling of ζ(s) by the
characteristic polynomials of random Hermitian matrices.
(University of Rochester)
47 / 51
Random matrix models
A major development was Keating and Snaith’s modeling of ζ(s) by the
characteristic polynomials of random Hermitian matrices.
It allowed them to determine the constants gk in
RT
ak gk
2k
k2
0 |ζ(1/2 + it)| dt ∼ Γ(k 2 +1) T log T .
(University of Rochester)
47 / 51
Random matrix models
A major development was Keating and Snaith’s modeling of ζ(s) by the
characteristic polynomials of random Hermitian matrices.
It allowed them to determine the constants gk in
RT
ak gk
2k
k2
0 |ζ(1/2 + it)| dt ∼ Γ(k 2 +1) T log T .
It has had applications to elliptic curves, for example.
(University of Rochester)
47 / 51
Random matrix models
A major development was Keating and Snaith’s modeling of ζ(s) by the
characteristic polynomials of random Hermitian matrices.
It allowed them to determine the constants gk in
RT
ak gk
2k
k2
0 |ζ(1/2 + it)| dt ∼ Γ(k 2 +1) T log T .
It has had applications to elliptic curves, for example.
Hughes,PKeating, O’Connell used it to conjecture the discrete
means 0<γ≤T |ζ 0 (ρ)|2k
(University of Rochester)
47 / 51
Random matrix models
A major development was Keating and Snaith’s modeling of ζ(s) by the
characteristic polynomials of random Hermitian matrices.
It allowed them to determine the constants gk in
RT
ak gk
2k
k2
0 |ζ(1/2 + it)| dt ∼ Γ(k 2 +1) T log T .
It has had applications to elliptic curves, for example.
Hughes,PKeating, O’Connell used it to conjecture the discrete
means 0<γ≤T |ζ 0 (ρ)|2k
Mezzadri used it to study the distribution of the zeros of ζ 0 (s).
(University of Rochester)
47 / 51
Lower order terms and ratios
(University of Rochester)
48 / 51
Lower order terms and ratios
The Keating-Snaith results led to the quest for the lower order terms in
the asymptotic expansion of the moments.
(University of Rochester)
48 / 51
Lower order terms and ratios
The Keating-Snaith results led to the quest for the lower order terms in
the asymptotic expansion of the moments.
This resulted in the discovery of new heuristics for the moments not
involving RMT.
(University of Rochester)
48 / 51
Lower order terms and ratios
The Keating-Snaith results led to the quest for the lower order terms in
the asymptotic expansion of the moments.
This resulted in the discovery of new heuristics for the moments not
involving RMT.
It also led to heuristics for very general moment questions (the so
called “ratios conjecture”).
(University of Rochester)
48 / 51
Lower order terms and ratios
The Keating-Snaith results led to the quest for the lower order terms in
the asymptotic expansion of the moments.
This resulted in the discovery of new heuristics for the moments not
involving RMT.
It also led to heuristics for very general moment questions (the so
called “ratios conjecture”).
(Conrey, Farmer, Keating, Rubenstein, Snaith, Zirnbauer, ...)
(University of Rochester)
48 / 51
A hybrid formula
(University of Rochester)
49 / 51
A hybrid formula
The Keating-Snaith model finds the moment constants gk , but the
arithmetical factors ak have to be inserted after the fact.
(University of Rochester)
49 / 51
A hybrid formula
The Keating-Snaith model finds the moment constants gk , but the
arithmetical factors ak have to be inserted after the fact.
This led to the problem of finding a model for zeta incorporating
characteristic polynomials and arithmetical information.
(University of Rochester)
49 / 51
A hybrid formula
The Keating-Snaith model finds the moment constants gk , but the
arithmetical factors ak have to be inserted after the fact.
This led to the problem of finding a model for zeta incorporating
characteristic polynomials and arithmetical information.
G, Hughes, Keating found an unconditonal hybrid formula for ζ(s).
(University of Rochester)
49 / 51
A hybrid formula
The Keating-Snaith model finds the moment constants gk , but the
arithmetical factors ak have to be inserted after the fact.
This led to the problem of finding a model for zeta incorporating
characteristic polynomials and arithmetical information.
G, Hughes, Keating found an unconditonal hybrid formula for ζ(s).
It says (roughly) that
(University of Rochester)
49 / 51
A hybrid formula
The Keating-Snaith model finds the moment constants gk , but the
arithmetical factors ak have to be inserted after the fact.
This led to the problem of finding a model for zeta incorporating
characteristic polynomials and arithmetical information.
G, Hughes, Keating found an unconditonal hybrid formula for ζ(s).
It says (roughly) that
ζ(s) =
Q
p≤X
1 − p−s
(University of Rochester)
−1 Q
|s−ρ|≤1/ log X
1 − X (ρ−s)e
γ
49 / 51
A hybrid formula
The Keating-Snaith model finds the moment constants gk , but the
arithmetical factors ak have to be inserted after the fact.
This led to the problem of finding a model for zeta incorporating
characteristic polynomials and arithmetical information.
G, Hughes, Keating found an unconditonal hybrid formula for ζ(s).
It says (roughly) that
ζ(s) =
Q
p≤X
1 − p−s
−1 Q
|s−ρ|≤1/ log X
1 − X (ρ−s)e
γ
A heuristic calculation of moments using this leads to ak and gk
appearing naturally.
(University of Rochester)
49 / 51
A hybrid formula
The Keating-Snaith model finds the moment constants gk , but the
arithmetical factors ak have to be inserted after the fact.
This led to the problem of finding a model for zeta incorporating
characteristic polynomials and arithmetical information.
G, Hughes, Keating found an unconditonal hybrid formula for ζ(s).
It says (roughly) that
ζ(s) =
Q
p≤X
1 − p−s
−1 Q
|s−ρ|≤1/ log X
1 − X (ρ−s)e
γ
A heuristic calculation of moments using this leads to ak and gk
appearing naturally.
It also explains why the constant in the moment splits as
(University of Rochester)
ak gk
.
Γ(k 2 +1)
49 / 51
The order of ζ(s) again
(University of Rochester)
50 / 51
The order of ζ(s) again
Finally, the hybrid formula has led to conjectural answers to the deep
question of the exact order of ζ(s) in the critical strip.
(University of Rochester)
50 / 51
The order of ζ(s) again
Finally, the hybrid formula has led to conjectural answers to the deep
question of the exact order of ζ(s) in the critical strip.
Recall that
(1 + o(1))eγ log log t ≤i.o. |ζ(1 + it)| ≤RH 2(1 + o(1))eγ log log t,
(University of Rochester)
50 / 51
The order of ζ(s) again
Finally, the hybrid formula has led to conjectural answers to the deep
question of the exact order of ζ(s) in the critical strip.
Recall that
(1 + o(1))eγ log log t ≤i.o. |ζ(1 + it)| ≤RH 2(1 + o(1))eγ log log t,
so that a factor of 2 is in question.
(University of Rochester)
50 / 51
The order of ζ(s) again
Finally, the hybrid formula has led to conjectural answers to the deep
question of the exact order of ζ(s) in the critical strip.
Recall that
(1 + o(1))eγ log log t ≤i.o. |ζ(1 + it)| ≤RH 2(1 + o(1))eγ log log t,
so that a factor of 2 is in question.
Arguments from the hybrid model suggest that the 2 should be
dropped.
(University of Rochester)
50 / 51
The order of ζ(s) again
(University of Rochester)
51 / 51
The order of ζ(s) again
On the 1/2-line itself recall that
(University of Rochester)
51 / 51
The order of ζ(s) again
On the 1/2-line itself recall that
s
log t
log t
c
≤i.o. log |ζ(1/2 + it)| RH
.
log log t
log log t
(University of Rochester)
51 / 51
The order of ζ(s) again
On the 1/2-line itself recall that
s
log t
log t
c
≤i.o. log |ζ(1/2 + it)| RH
.
log log t
log log t
Here Farmer, G, and Hughes have used the hybrid formula to suggest
that
(University of Rochester)
51 / 51
The order of ζ(s) again
On the 1/2-line itself recall that
s
log t
log t
c
≤i.o. log |ζ(1/2 + it)| RH
.
log log t
log log t
Here Farmer, G, and Hughes have used the hybrid formula to suggest
that
p
log |ζ(1/2 + it)| p
1/2(1 + o(1)) ≤i.o. p
≤ 1/2(1 + o(1)).
log t log log t
(University of Rochester)
51 / 51