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Estimating the Growth Rate of the Zeta Function Using Exponent Pairs

Shreejit Bandyopadhyay July 06, 2015

The Riemann Zeta Function

• The Riemann-zeta function is of prime interest in Number

Theory and also in many other branches of mathematics.

The Riemann Zeta Function

• The Riemann-zeta function is of prime interest in Number

Theory and also in many other branches of mathematics.

• It can be defined either by the infinite summation ζ(s) =
• n∈N

1 ns

• r as the product

p(1 − 1 ps )−1.

The Riemann Zeta Function

• The Riemann-zeta function is of prime interest in Number

Theory and also in many other branches of mathematics.

• It can be defined either by the infinite summation ζ(s) =
• n∈N

1 ns

• r as the product

p(1 − 1 ps )−1.

• The equivalence of these two definitions follows by factoring n as

a product of primes and the fundamental theorem of arithmetic.

The Riemann Zeta Function

• The Riemann-zeta function is of prime interest in Number

Theory and also in many other branches of mathematics.

• It can be defined either by the infinite summation ζ(s) =
• n∈N

1 ns

• r as the product

p(1 − 1 ps )−1.

• The equivalence of these two definitions follows by factoring n as

a product of primes and the fundamental theorem of arithmetic.

• Both the sum and the product converge for σ = Re(s) > 1. We

note that |n−s| = |n−σ|.

The Zeta Function cont.

The Zeta Function cont.

Other representations of the zeta function include

• (Integral form): ζ(s) =

1 Γ(s)

∞

xs−1 ex−1dx,

The Zeta Function cont.

Other representations of the zeta function include

• (Integral form): ζ(s) =

1 Γ(s)

∞

xs−1 ex−1dx,

• (Functional equation):ζ(s) = 2sπs−1sin( πs

2 )Γ(1 − s)ζ(1 − s).

We often write χ(s) = 2sπs−1sin( πs

2 )Γ)(1 − s) to write the

functional equation in a more compact form as ζ(s) = χ(s)ζ(1 − s).

The Zeta Function cont.

Other representations of the zeta function include

• (Integral form): ζ(s) =

1 Γ(s)

∞

xs−1 ex−1dx,

• (Functional equation):ζ(s) = 2sπs−1sin( πs

2 )Γ(1 − s)ζ(1 − s).

We often write χ(s) = 2sπs−1sin( πs

2 )Γ)(1 − s) to write the

functional equation in a more compact form as ζ(s) = χ(s)ζ(1 − s).

• (Approximate functional equation):With s = σ + it and

t = 2πxy, this is ζ(s) =

• n≤x

1 ns + χ(s)

• n≤y

1 n1−s + O(x−σlog|t|) + O(|t|

1 2 −σyσ−1).

Zeta Function in the Critical Strip

Zeta Function in the Critical Strip

• The critical strip is the infinite strip 0 < σ < 1. We wish to

study the rate of growth of the zeta function in this strip.

Zeta Function in the Critical Strip

• The critical strip is the infinite strip 0 < σ < 1. We wish to

study the rate of growth of the zeta function in this strip.

• For a given real part σ, we define

µ(σ) = inf {α > 0|ζ(σ + it) << O(tα)}.

Zeta Function in the Critical Strip

• The critical strip is the infinite strip 0 < σ < 1. We wish to

study the rate of growth of the zeta function in this strip.

• For a given real part σ, we define

µ(σ) = inf {α > 0|ζ(σ + it) << O(tα)}.

• An important hypothesis in this direction is the Lindelof
• hypothesis. It claims that ζ( 1

2 + it) << O(tǫ) for any ǫ > 0.

Zeta Function in the Critical Strip cont.

Zeta Function in the Critical Strip cont.

• It has been shown that the Lindelof hypothesis is equivalent to

showing that the number of zeroes of the zeta function in the critical strip with real part greater than 1

2 and imaginary part

between t and t + 1 is o(log t).

Zeta Function in the Critical Strip cont.

• It has been shown that the Lindelof hypothesis is equivalent to

showing that the number of zeroes of the zeta function in the critical strip with real part greater than 1

2 and imaginary part

between t and t + 1 is o(log t).

• The Riemann hypothesis, of course, claims that there are no

such zeroes at all. So the Lindelof hypothesis is much weaker than the Riemann hypothesis.

Zeta Function in the Critical Strip cont.

• It has been shown that the Lindelof hypothesis is equivalent to

showing that the number of zeroes of the zeta function in the critical strip with real part greater than 1

2 and imaginary part

between t and t + 1 is o(log t).

• The Riemann hypothesis, of course, claims that there are no

such zeroes at all. So the Lindelof hypothesis is much weaker than the Riemann hypothesis.

• But, despite the efforts of many people, even the Lindelof

hypothesis is far from being proved.

Zeta Function as Exponential Sum

Zeta Function as Exponential Sum

• Our goal will be to reduce the sum involved in the zeta function

to an exponential sum.

Zeta Function as Exponential Sum

• Our goal will be to reduce the sum involved in the zeta function

to an exponential sum.

• We see that
• N<n≤M

n−s =

• N<n≤M

n−σ−it << N−σ max

N<u≤M |

• N<n≤u

n−it|.

Zeta Function as Exponential Sum

• Our goal will be to reduce the sum involved in the zeta function

to an exponential sum.

• We see that
• N<n≤M

n−s =

• N<n≤M

n−σ−it << N−σ max

N<u≤M |

• N<n≤u

n−it|.

• This holds because if S(u) =
• N<n≤u

n−it then

• N<n≤M

n−σ−it = S(M)M−σ + σ M

N

S(u)u−σ−1du by partial

• summation. Putting the value of S(u), we get our estimate.

Zeta Function as Exponential Sum

• Our goal will be to reduce the sum involved in the zeta function

to an exponential sum.

• We see that
• N<n≤M

n−s =

• N<n≤M

n−σ−it << N−σ max

N<u≤M |

• N<n≤u

n−it|.

• This holds because if S(u) =
• N<n≤u

n−it then

• N<n≤M

n−σ−it = S(M)M−σ + σ M

N

S(u)u−σ−1du by partial

• summation. Putting the value of S(u), we get our estimate.
• Writing
• n−it =
• e2πi(−logn t

2π ) =

• e(−logn t

2π), we get an exponential sum!

Zeta Function as Exponential Sum cont.

Zeta Function as Exponential Sum cont.

• Using partial summation, we can also show that

|ζ(σ + it)| << |

• n≤t

n−σ−it| + t1−2σlog t.

Zeta Function as Exponential Sum cont.

• Using partial summation, we can also show that

|ζ(σ + it)| << |

• n≤t

n−σ−it| + t1−2σlog t.

• This allows us to look at the zeta function in terms of

exponential sums.

Exponential Sums

Exponential Sums

• An exponential sum is any sum of the form
• n∈I

e(f (n)) where f is a function, I a real interval and e(x)) stands for e2πix. Often, we shall have to assume some nice additional properties of f like continuity, differentiability, etc.

Exponential Sums

• An exponential sum is any sum of the form
• n∈I

e(f (n)) where f is a function, I a real interval and e(x)) stands for e2πix. Often, we shall have to assume some nice additional properties of f like continuity, differentiability, etc.

• By the triangle inequality,

|S| = |

• n∈I

e(f (n))| ≤

• n∈I

|e(f (n))| ≤ |I| if the interval I has integer endpoints. We see that equality holds if f (n) = xn + y and x is an integer and the interval I is closed and has integer

• endpoints. So we need additional conditions on f to improve our

bound.

Exponential Sums cont.

Exponential Sums cont.

• Weyl introduced a trick of dealing with such sums. He wrote

|S|2 =

• a<m,n≤b

e(f (m) − f (n)) =

• |h|<b−a
• n∈Ih

e(f (n + h) − f (n)) where Ir = {n|a < n, n + r ≤ b}. This holds because |S|2 =

• a<m≤b

e(f (m))

• a<n≤b

e(f (n)) =

• a<m≤b

e(f (m))

• a<n≤b

e(−f (n)) =

• a<m,n≤b

e(f (m) − f (n)).

Exponential Sums cont.

• Weyl introduced a trick of dealing with such sums. He wrote

|S|2 =

• a<m,n≤b

e(f (m) − f (n)) =

• |h|<b−a
• n∈Ih

e(f (n + h) − f (n)) where Ir = {n|a < n, n + r ≤ b}. This holds because |S|2 =

• a<m≤b

e(f (m))

• a<n≤b

e(f (n)) =

• a<m≤b

e(f (m))

• a<n≤b

e(−f (n)) =

• a<m,n≤b

e(f (m) − f (n)).

• This helps us because the difference function f (n + h) − f (n) is

usually easier to handle than f (n) itself. For example, if f is a polynomial, it gives a polynomial of one degree less in the

• exponent. Using this repeatedly, we may want to get a linear

polynomial in the exponent.

Exponential Sums cont.

• Weyl introduced a trick of dealing with such sums. He wrote

|S|2 =

• a<m,n≤b

e(f (m) − f (n)) =

• |h|<b−a
• n∈Ih

e(f (n + h) − f (n)) where Ir = {n|a < n, n + r ≤ b}. This holds because |S|2 =

• a<m≤b

e(f (m))

• a<n≤b

e(f (n)) =

• a<m≤b

e(f (m))

• a<n≤b

e(−f (n)) =

• a<m,n≤b

e(f (m) − f (n)).

• This helps us because the difference function f (n + h) − f (n) is

usually easier to handle than f (n) itself. For example, if f is a polynomial, it gives a polynomial of one degree less in the

• exponent. Using this repeatedly, we may want to get a linear

polynomial in the exponent.

• The problem is then trivial because exponential sums of linear

functions are geometric progressions.

Exponential Sums cont.

Exponential Sums cont.

• Another estimate of exponential sums is the van der Corput’s

bound: If f has two continuous derivatives on I and if for some λ > 0, α ≥ 1, λ ≤ |f ′′(x)| ≤ αλ in I, then

• n∈I

e(f (n)) << α|I|λ

1 2 + λ− 1 2 .

Exponent Pairs

Exponent Pairs

• Given a function f (n), we suppose that f ′(n) << n−s. If we take
• ur interval of summation I inside some dyadic interval (N, 2N],

then that means f ′(n) << N−s in the interval. We write f ′(n) ≈ L.

Exponent Pairs

• Given a function f (n), we suppose that f ′(n) << n−s. If we take
• ur interval of summation I inside some dyadic interval (N, 2N],

then that means f ′(n) << N−s in the interval. We write f ′(n) ≈ L.

• For example, if we consider the function f (n) = − t

2πlog n in an

interval I ⊆ (N, 2N], then f ′(n) = − 1

2π t n << N−1 and we have

that L = t

N .

Exponent Pairs

• Given a function f (n), we suppose that f ′(n) << n−s. If we take
• ur interval of summation I inside some dyadic interval (N, 2N],

then that means f ′(n) << N−s in the interval. We write f ′(n) ≈ L.

• For example, if we consider the function f (n) = − t

2πlog n in an

interval I ⊆ (N, 2N], then f ′(n) = − 1

2π t n << N−1 and we have

that L = t

N .

• If for some such function with f ′(n) ≈ L, we are able to prove an

estimate of the form S =

• n∈I

e(f (n)) << LkNl, we call (k, l) an exponent pair.

Exponent Pairs

• Given a function f (n), we suppose that f ′(n) << n−s. If we take
• ur interval of summation I inside some dyadic interval (N, 2N],

then that means f ′(n) << N−s in the interval. We write f ′(n) ≈ L.

• For example, if we consider the function f (n) = − t

2πlog n in an

interval I ⊆ (N, 2N], then f ′(n) = − 1

2π t n << N−1 and we have

that L = t

N .

• If for some such function with f ′(n) ≈ L, we are able to prove an

estimate of the form S =

• n∈I

e(f (n)) << LkNl, we call (k, l) an exponent pair.

• The trivial estimate S << N gives (0, 1) as an exponent pair.

Our goal is to obtain new exponent pairs from known ones and hopefully get better estimates for exponential sums.

A-Process

A-Process

• Van der Corput adapted and modified Weyl’s difference method

and obtained the estimate |S|2 << N2

H + N H

• 1≤h≤H

|S1(h)| for any H ≤ N. Here S1(h) is, of course, the difference function

• a<n,n+h≤b

e(f (n + h) − f (n)).

A-Process

• Van der Corput adapted and modified Weyl’s difference method

and obtained the estimate |S|2 << N2

H + N H

• 1≤h≤H

|S1(h)| for any H ≤ N. Here S1(h) is, of course, the difference function

• a<n,n+h≤b

e(f (n + h) − f (n)).

• If f ′ ≈ L, then heuristically, if f1(n, h) = f (n + h) − f (n),

f ′

1 ≈ L|h|N−1.

A-Process

• Van der Corput adapted and modified Weyl’s difference method

and obtained the estimate |S|2 << N2

H + N H

• 1≤h≤H

|S1(h)| for any H ≤ N. Here S1(h) is, of course, the difference function

• a<n,n+h≤b

e(f (n + h) − f (n)).

• If f ′ ≈ L, then heuristically, if f1(n, h) = f (n + h) − f (n),

f ′

1 ≈ L|h|N−1.

• So, if (k, l) is an exponent pair, |S|2 <<

N2H−1 + NH−1

1≤h≤H

(L|h|N−1)kNl << N2H−1 + HkLkN1−k+l.

A-Process

• Van der Corput adapted and modified Weyl’s difference method

and obtained the estimate |S|2 << N2

H + N H

• 1≤h≤H

|S1(h)| for any H ≤ N. Here S1(h) is, of course, the difference function

• a<n,n+h≤b

e(f (n + h) − f (n)).

• If f ′ ≈ L, then heuristically, if f1(n, h) = f (n + h) − f (n),

f ′

1 ≈ L|h|N−1.

• So, if (k, l) is an exponent pair, |S|2 <<

N2H−1 + NH−1

1≤h≤H

(L|h|N−1)kNl << N2H−1 + HkLkN1−k+l.

• The RHS will be minimal when the two terms there are equal.

We choose the H which makes them equal and see that S << L

k 2k+2 N k+l+1 2k+2 .

A-Process

• Van der Corput adapted and modified Weyl’s difference method

and obtained the estimate |S|2 << N2

H + N H

• 1≤h≤H

|S1(h)| for any H ≤ N. Here S1(h) is, of course, the difference function

• a<n,n+h≤b

e(f (n + h) − f (n)).

• If f ′ ≈ L, then heuristically, if f1(n, h) = f (n + h) − f (n),

f ′

1 ≈ L|h|N−1.

• So, if (k, l) is an exponent pair, |S|2 <<

N2H−1 + NH−1

1≤h≤H

(L|h|N−1)kNl << N2H−1 + HkLkN1−k+l.

• The RHS will be minimal when the two terms there are equal.

We choose the H which makes them equal and see that S << L

k 2k+2 N k+l+1 2k+2 .

• So, from the pair (k, l), we’ve produced a new exponent pair

(

k 2k+2, k+l+1 2k+2 ). This is the so-called A-process.

B-Process

B-Process

• There is another method of obtaining new exponent pairs from
• nes already known. This is called the B−process and depends on

a method of integral approximation.

B-Process

• There is another method of obtaining new exponent pairs from
• nes already known. This is called the B−process and depends on

a method of integral approximation.

• It involves the following results: If f has two continuous

derivatives on [a, b] with f ′ decreasing and if there are two integers H1 and H2 with H1 < f ′(x) < H2 and H = H1 − H2 ≥ 2 then

• n∈I

e(f (n)) =

• H1≤h≤H2

b

a

e(f (x) − hx)dx + O(log h).

B-Process

• There is another method of obtaining new exponent pairs from
• nes already known. This is called the B−process and depends on

a method of integral approximation.

• It involves the following results: If f has two continuous

derivatives on [a, b] with f ′ decreasing and if there are two integers H1 and H2 with H1 < f ′(x) < H2 and H = H1 − H2 ≥ 2 then

• n∈I

e(f (n)) =

• H1≤h≤H2

b

a

e(f (x) − hx)dx + O(log h).

• Suppose that the function f has four continuous derivatives in I

and f ′′ < 0 in I. With I ⊆ (N, 2N], let a, b be the endpoints of I and let α = f ′(b), β = f ′(a). Assume that for some constant F > 0, f 2(x) ≈ FN−2, f 3(x) ≈ FN−3, f 4(x) ≈ FN−4∀x ∈ I. Let xν be such that f ′(xν) = ν and let φ(ν) = −f (xν) + νxν. Then

• n∈I

e(f (n)) =

• α≤ν≤β

e(−φ(ν) − 1

8)

|f ′′(xν)|

1 2

+ error terms.

B-Process cont.

B-Process cont.

• Using the method of stationary phase, we have that

S = e(− 1

8)

• α≤ν≤β

e(−φ(ν)) |f ′′(xν)|

1 2

+ error terms.

B-Process cont.

• Using the method of stationary phase, we have that

S = e(− 1

8)

• α≤ν≤β

e(−φ(ν)) |f ′′(xν)|

1 2

+ error terms.

• Ignoring the error terms as part of our heuristic approach for the

time being, we see that φ′(ν) = xν ≈ N and f ′′(xν) ≈ LN−1. So, by partial summation, we get that

• α≤ν≤β

e(−φ(ν)) |f ′′(xν)|

1 2

<< L− 1

2 N 1 2 min

β′≤β |

• α<ν≤β

e(φ(ν))|.

B-Process cont.

• Using the method of stationary phase, we have that

S = e(− 1

8)

• α≤ν≤β

e(−φ(ν)) |f ′′(xν)|

1 2

+ error terms.

• Ignoring the error terms as part of our heuristic approach for the

time being, we see that φ′(ν) = xν ≈ N and f ′′(xν) ≈ LN−1. So, by partial summation, we get that

• α≤ν≤β

e(−φ(ν)) |f ′′(xν)|

1 2

<< L− 1

2 N 1 2 min

β′≤β |

• α<ν≤β

e(φ(ν))|.

• Assuming that the exponent pair (k, l) can be applied to the last

sum, we get S << L− 1

2 N 1 2 NkLl << Ll− 1 2 Nk+ 1 2 . So, we conclude

that if (k, l) is an exponent pair, so is (l − 1

2, k + 1 2).

B-Process cont.

• Using the method of stationary phase, we have that

S = e(− 1

8)

• α≤ν≤β

e(−φ(ν)) |f ′′(xν)|

1 2

+ error terms.

• Ignoring the error terms as part of our heuristic approach for the

time being, we see that φ′(ν) = xν ≈ N and f ′′(xν) ≈ LN−1. So, by partial summation, we get that

• α≤ν≤β

e(−φ(ν)) |f ′′(xν)|

1 2

<< L− 1

2 N 1 2 min

β′≤β |

• α<ν≤β

e(φ(ν))|.

• Assuming that the exponent pair (k, l) can be applied to the last

sum, we get S << L− 1

2 N 1 2 NkLl << Ll− 1 2 Nk+ 1 2 . So, we conclude

that if (k, l) is an exponent pair, so is (l − 1

2, k + 1 2).

• Starting with the trivial pair (0, 1), we can obtain new pairs by

repeatedly applying the A and B processes. Since applying the B−process twice in succession gets us nowhere, any such sequence

• f transformations look like Ai1BAi2B....

Zeta Function and Exponent Pairs

Zeta Function and Exponent Pairs

• For the zeta function, we use the estimates

|ζ(σ + it)| << |

• n≤t

n−σ−it| + t1−2σlog t and

• N<n≤M

n−σ−it << N−σ max

N<u≤M |

• N<n≤u

n−it| in succession to reduce the problem of showing that ζ( 1

2 + it) << tθ(k,l)log t to

the verification of N− 1

2

• N≤n≤2N

n−it << tθ(k,l) with θ(k, l) a function of the exponent pair (k, l).

Zeta Function and Exponent Pairs

• For the zeta function, we use the estimates

|ζ(σ + it)| << |

• n≤t

n−σ−it| + t1−2σlog t and

• N<n≤M

n−σ−it << N−σ max

N<u≤M |

• N<n≤u

n−it| in succession to reduce the problem of showing that ζ( 1

2 + it) << tθ(k,l)log t to

the verification of N− 1

2

• N≤n≤2N

n−it << tθ(k,l) with θ(k, l) a function of the exponent pair (k, l).

• We split the interval [1, t] into << log t dyadic intervals (N, 2N]

so that in any such interval we can control the constant L to be tN−1. That is, LkNl = tkNl−k.

Zeta Function and Exponent Pairs

• For the zeta function, we use the estimates

|ζ(σ + it)| << |

• n≤t

n−σ−it| + t1−2σlog t and

• N<n≤M

n−σ−it << N−σ max

N<u≤M |

• N<n≤u

n−it| in succession to reduce the problem of showing that ζ( 1

2 + it) << tθ(k,l)log t to

the verification of N− 1

2

• N≤n≤2N

n−it << tθ(k,l) with θ(k, l) a function of the exponent pair (k, l).

• We split the interval [1, t] into << log t dyadic intervals (N, 2N]

so that in any such interval we can control the constant L to be tN−1. That is, LkNl = tkNl−k.

• For each such interval, we get
• N<n≤2N

n−it << tkNl−k for pair (k, l).

Zeta Function and Exponent Pairs cont.

Zeta Function and Exponent Pairs cont.

• Using B(k, l) = (l − 1

2, k + 1 2) in place of (k, l), we also get

• N<n≤2N

n−it << tl− 1

2 Nk−l+ 1 2 .

Zeta Function and Exponent Pairs cont.

• Using B(k, l) = (l − 1

2, k + 1 2) in place of (k, l), we also get

• N<n≤2N

n−it << tl− 1

2 Nk−l+ 1 2 .

• So, ultimately we get that ζ( 1

2 + it) <<

logt[N− 1

2

• N<n≤2N

n−it] << min(tkNl−k− 1

2 , tl− 1 2 Nk−l+ 1 2 )log t.

Zeta Function and Exponent Pairs cont.

• Using B(k, l) = (l − 1

2, k + 1 2) in place of (k, l), we also get

• N<n≤2N

n−it << tl− 1

2 Nk−l+ 1 2 .

• So, ultimately we get that ζ( 1

2 + it) <<

logt[N− 1

2

• N<n≤2N

n−it] << min(tkNl−k− 1

2 , tl− 1 2 Nk−l+ 1 2 )log t.

• Using min(a, b) ≤

√ ab, we get that ζ( 1

2 + it) << t

k+l− 1 2 2

log t << t

k+l− 1 2 2

+ǫ.

Zeta Function and Exponent Pairs cont.

• Using B(k, l) = (l − 1

2, k + 1 2) in place of (k, l), we also get

• N<n≤2N

n−it << tl− 1

2 Nk−l+ 1 2 .

• So, ultimately we get that ζ( 1

2 + it) <<

logt[N− 1

2

• N<n≤2N

n−it] << min(tkNl−k− 1

2 , tl− 1 2 Nk−l+ 1 2 )log t.

• Using min(a, b) ≤

√ ab, we get that ζ( 1

2 + it) << t

k+l− 1 2 2

log t << t

k+l− 1 2 2

+ǫ.

• Similarly, we may show for any σ with 0 < σ < 1 that

ζ(σ + it) << t

k+l−σ 2

log t << t

k+l−σ 2

+ǫ.

Zeta Function and Exponent Pairs cont.

• The exponent pair B(0, 1)=( 1

2, 1 2) gives ζ( 1 2 + it) << t

1 4 , the

bound we can in any case obtain from the functional equation.

Zeta Function and Exponent Pairs cont.

• The exponent pair B(0, 1)=( 1

2, 1 2) gives ζ( 1 2 + it) << t

1 4 , the

bound we can in any case obtain from the functional equation.

• Similarly, the pair AB(0, 1) = ( 1

6, 2 3) gives the bound

ζ( 1

2 + it) << t

1 6 , which can in any case be obtained from the Van

der Corput bound.

Zeta Function and Exponent Pairs cont.

• The exponent pair B(0, 1)=( 1

2, 1 2) gives ζ( 1 2 + it) << t

1 4 , the

bound we can in any case obtain from the functional equation.

• Similarly, the pair AB(0, 1) = ( 1

6, 2 3) gives the bound

ζ( 1

2 + it) << t

1 6 , which can in any case be obtained from the Van

der Corput bound.

• However, other exponent pairs allow us to lower the exponent of

t!

Zeta Function and Exponent Pairs cont.

• The exponent pair B(0, 1)=( 1

2, 1 2) gives ζ( 1 2 + it) << t

1 4 , the

bound we can in any case obtain from the functional equation.

• Similarly, the pair AB(0, 1) = ( 1

6, 2 3) gives the bound

ζ( 1

2 + it) << t

1 6 , which can in any case be obtained from the Van

der Corput bound.

• However, other exponent pairs allow us to lower the exponent of

t!

• For σ = 1

2 and (k, l) = ABABAB(0, 1) = ( 11 82, 57 82), we get

ζ( 1

2 + it) << t

27 164 +ǫ ≈ t0.165.

Recent History

Recent History

• Iwaniec and Bombieri modified Van der Corput’s method to
• btain the bound ζ( 1

2 + it) << t0.161 and there have been further

recent improvements as well.

Recent History

• Iwaniec and Bombieri modified Van der Corput’s method to
• btain the bound ζ( 1

2 + it) << t0.161 and there have been further

recent improvements as well.

• Lindelof hypothesis will be implied if for all ǫ > 0, (ǫ, 1

2 + ǫ) is an

exponent pair.

Recent History

• Iwaniec and Bombieri modified Van der Corput’s method to
• btain the bound ζ( 1

2 + it) << t0.161 and there have been further

recent improvements as well.

• Lindelof hypothesis will be implied if for all ǫ > 0, (ǫ, 1

2 + ǫ) is an

exponent pair.

• The conjecture that for all ǫ > 0, (ǫ, 1

2 + ǫ) is an exponent pair

will also give the conjectured best bounds in other problems like the divisor problem.

Recent History

• Iwaniec and Bombieri modified Van der Corput’s method to
• btain the bound ζ( 1

2 + it) << t0.161 and there have been further

recent improvements as well.

• Lindelof hypothesis will be implied if for all ǫ > 0, (ǫ, 1

2 + ǫ) is an

exponent pair.

• The conjecture that for all ǫ > 0, (ǫ, 1

2 + ǫ) is an exponent pair

will also give the conjectured best bounds in other problems like the divisor problem.

• The A and B processes only give us exponent pairs that lie

above the curve in the figure.

Recent History

• Iwaniec and Bombieri modified Van der Corput’s method to
• btain the bound ζ( 1

2 + it) << t0.161 and there have been further

recent improvements as well.

• Lindelof hypothesis will be implied if for all ǫ > 0, (ǫ, 1

2 + ǫ) is an

exponent pair.

• The conjecture that for all ǫ > 0, (ǫ, 1

2 + ǫ) is an exponent pair

will also give the conjectured best bounds in other problems like the divisor problem.

• The A and B processes only give us exponent pairs that lie

above the curve in the figure.

• Lindelof hypothesis needs that any point in the square is an

exponent pair.

Recent History

• Iwaniec and Bombieri modified Van der Corput’s method to
• btain the bound ζ( 1

2 + it) << t0.161 and there have been further

recent improvements as well.

• Lindelof hypothesis will be implied if for all ǫ > 0, (ǫ, 1

2 + ǫ) is an

exponent pair.

• The conjecture that for all ǫ > 0, (ǫ, 1

2 + ǫ) is an exponent pair

will also give the conjectured best bounds in other problems like the divisor problem.

• The A and B processes only give us exponent pairs that lie

above the curve in the figure.

• Lindelof hypothesis needs that any point in the square is an

exponent pair.

• Recent advances has changed the curve in the square a little, but

Lindelof hypothesis is still far away.

Plot of Exponent Pairs