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Math 229x - Introduction to Analytic Number
Theory
Taught by Hector PastenNotes by Dongryul Kim
Spring 2017
This course was taught by Hector Pasten, and we met on MWF 11-12 inScience Center 411. We followed Problems in Analytic Number Theory by RamMurty during the first half of the semester. There were 4 undergraduates and 4graduate students, and the grading was based on weekly problem sets, a in-classmidterm, and a take-home final. There was no course assistant.
Contents
1 January 23, 2017 51.1 Logistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Playing with sums . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 January 25, 2017 72.1 Chebyshev’s method . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 The Riemann zeta function . . . . . . . . . . . . . . . . . . . . . 9
3 January 27, 2017 103.1 Dirichlet series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4 January 30, 2017 124.1 Dirichlet L-functions . . . . . . . . . . . . . . . . . . . . . . . . . 124.2 Analytic continuation . . . . . . . . . . . . . . . . . . . . . . . . 134.3 Nonvanishing of the L-function at 1 . . . . . . . . . . . . . . . . 13
5 February 1, 2017 155.1 Primes in arithmetic progressions . . . . . . . . . . . . . . . . . . 155.2 Tauberian theorems . . . . . . . . . . . . . . . . . . . . . . . . . 16
6 February 3, 2017 186.1 The Riemann zeta on <(s) = 1 . . . . . . . . . . . . . . . . . . . 186.2 Prime number theorem . . . . . . . . . . . . . . . . . . . . . . . . 19
1 Last Update: August 27, 2018
7 February 6, 2017 207.1 Poisson summation . . . . . . . . . . . . . . . . . . . . . . . . . . 20
8 February 8, 2017 228.1 The Γ-function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228.2 Functional equation of ζ(s) . . . . . . . . . . . . . . . . . . . . . 22
9 February 10, 2017 259.1 Primitive characters . . . . . . . . . . . . . . . . . . . . . . . . . 259.2 Gauss sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
10 February 13, 2017 2710.1 Twisted θ-functions . . . . . . . . . . . . . . . . . . . . . . . . . 27
11 February 15, 2017 2911.1 Growth and zeros of ξ(s) . . . . . . . . . . . . . . . . . . . . . . . 2911.2 Hadamard factorization . . . . . . . . . . . . . . . . . . . . . . . 29
12 February 17, 2017 3112.1 Zero-free region for ζ(s) . . . . . . . . . . . . . . . . . . . . . . . 31
13 February 22, 2017 3313.1 Zero-free region for L(s, χ) . . . . . . . . . . . . . . . . . . . . . . 33
14 February 24, 2017 3514.1 Dealing with quadratic characters . . . . . . . . . . . . . . . . . . 3514.2 Siegel zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
15 February 27, 2017 3815.1 Counting zeros of ζ(s) . . . . . . . . . . . . . . . . . . . . . . . . 38
16 March 1, 2017 4116.1 Integral approximation of ψ . . . . . . . . . . . . . . . . . . . . . 41
17 March 6, 2017 4317.1 Explicit formula for ψ0 . . . . . . . . . . . . . . . . . . . . . . . . 44
18 March 8, 2017 4618.1 Prime number theorem revisited . . . . . . . . . . . . . . . . . . 47
19 March 10, 2017 4819.1 Riemann hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . 4819.2 Lindelof hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . 49
20 March 20, 2017 5020.1 Introduction to sieves . . . . . . . . . . . . . . . . . . . . . . . . 5020.2 Selberg sieve for counting primes . . . . . . . . . . . . . . . . . . 51
2
21 March 22, 2017 5321.1 Selberg sieve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
22 March 24, 2017 5622.1 Application: Counting twin primes . . . . . . . . . . . . . . . . . 56
23 March 27, 2017 5823.1 Analytic large sieve inequality . . . . . . . . . . . . . . . . . . . . 5823.2 Ramanujan sums . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
24 March 31, 2017 6124.1 Arithmetic large sieve . . . . . . . . . . . . . . . . . . . . . . . . 61
25 April 3, 2017 6425.1 Additive problems . . . . . . . . . . . . . . . . . . . . . . . . . . 6425.2 Schnirelmann’s density . . . . . . . . . . . . . . . . . . . . . . . . 64
26 April 5, 2017 6626.1 Using Schnirelmann’s density . . . . . . . . . . . . . . . . . . . . 6626.2 Sums of primes (weak Goldbach) . . . . . . . . . . . . . . . . . . 67
27 April 7, 2017 6927.1 Introduction to the circle method . . . . . . . . . . . . . . . . . . 69
28 April 10, 2017 7228.1 Bounds for exponential sums . . . . . . . . . . . . . . . . . . . . 72
29 April 12, 2017 7529.1 Weyl’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
30 April 14, 2017 7730.1 Hua’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7730.2 Setup for Waring’s problem . . . . . . . . . . . . . . . . . . . . . 78
31 April 17, 2017 8031.1 Minor arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8031.2 Major arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
32 April 19, 2017 8332.1 Global factorization . . . . . . . . . . . . . . . . . . . . . . . . . 8332.2 Main formula for r(N) . . . . . . . . . . . . . . . . . . . . . . . . 84
33 April 21, 2017 8633.1 Evaluation of J(N) . . . . . . . . . . . . . . . . . . . . . . . . . . 8633.2 S(a, q) and S(q) revisited . . . . . . . . . . . . . . . . . . . . . . 87
34 April 24, 2017 8934.1 Formula for T (p) . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
3
35 April 26, 2017 9235.1 Local Waring’s problem . . . . . . . . . . . . . . . . . . . . . . . 92
4
Math 229x Notes 5
1 January 23, 2017
1.1 Logistics
The purpose of this course is to expose you to fundamental methods, results,and problems of analytic number theory. This is not a research course, so wewill not go for the sharpest results. Office hours are Thursdays from 10:00 amto 11:59 am. The text book is Ram Murty, Problems in analytic number theory.For some more advanced material, Iwaniec, Kowalski, Analytic number theoryis a good reference. We will be covering:
• some elementary methods
• prime numbers: prime number theorem, primes in arithmetic progressions
• Dirichlet L-functions
• sieves: Brun’s sieve and Selberg’s Λ2 sieve
• additive problems: Schnirelman’s method and the circle method
There will be weekly assignments, due each monday, an in-class midterm,and a take-home final.
1.2 Playing with sums
You will need to get some practice in changing order of sums and partial sum-mation.
Lemma 1.1. Let a1, a2, . . . be complex numbers and x > 1 be a real number.Let f : [1, x]→ C be a C1
C function. Then∑n≤x
anf(n) =(∑n≤x
an
)f(x)−
∫ x
1
(∑n≤t
an
)f ′(t)dt.
Proof. Exercise.
Example 1.2. Define T (x) =∑n≤x log n. Then
T (x) =(∑n≤x
1)
log x−∫ x
1
(∑n≤t
1)dtt
= x log x−∫ x
1
t− tt
dt = (x+O(1)) log x− (x− 1) +O(log x)
= x log x− x+O(log x).
Example 1.3. Write d(n) = # of divisors of n. Then∑n≤x
d(n) =∑n≤x
∑a·b=n
1.
Math 229x Notes 6
The picture is that we are counting points in a region bounded by two lines anda hyperbola. The big error is coming from the tail. So what Dirichlet did is toadd up to
√n and then multiply by 2 using symmetry. So∑
n≤x
d(n) = 2∑
a≤x1/2
bxac − (x+O(
√x)) = 2
∑a≤x1/2
(xa
+O(1))− x+O(
√x)
= 2x∑
a≤x1/2
1
a− x+O(
√x)
= 2x(log x1/2 + γ +O(x−1/2))− x+O(√x)
= x log x+ (2γ − 1)x+O(√x).
Example 1.4. Write φ(n) = #a mod n is invertible. Also define
µ(n) =
0 if n is not square-free,
(−1)#prime factors on n if n is square-free.
Then
φ(n) = n∑d|n
µ(d)
d.
The sum of φ(n) can be computed as∑n≤x
φ(n) =∑n≤x
n∑d|n
µ(d)
d=∑d≤x
µ(d)
d
∑e≤x/d
de
=∑d≤x
µ(d)∑e≤x/d
e =∑d≤x
µ(d)(1
2
(xd
)2
+O(xd
))=x2
2
∑d≤x
µ(d)
d2+O
(x∑d≤x
1
d
)
=x2
2
( ∞∑n=1
µ(n)
n2+O(x−1)
)+O(x log x) = c · x2 +O(x log x),
where c = 1/2ζ(2) = 3/π2 by a problem in the homework.
Math 229x Notes 7
2 January 25, 2017
Today we will count primes using Chebyshev’s method. I am going to introducea few functions. The von Mangoldt function is defined as
Λ(n) =
log p if n = pα for α ≥ 1 and p prime,
0 otherwise.
Higher prime powers don’t contribute too much, so essentially we are countingprimes. There are the prime counting functions
π(x) = #p ≤ x : p is prime,
θ(x) =∑p≤x
log p,
ψ(x) =∑n≤x
Λ(n).
Clearly ψ(x) ≥ θ(x) and π(x) log x ≥ θ(x). We can also write
ψ(x) = θ(x) + θ(√x) + θ( 3
√x) + · · ·
= θ(x) +O(θ(√x) log x) = θ(x) +O(x1/2(log x)2)
because trivially θ(x) ≤ x log x.
2.1 Chebyshev’s method
Recall that
T (x) =∑n≤x
log n = log(bxc!) = x log x− x+ (log x).
It follows that
T (x)− 2T(x
2
)= (log x)x+O(log x).
The reason we are doing this is because T and Λ can be related in a directway. Recall that
νp(n!) =⌊np
⌋+⌊ np2
⌋+⌊ np3
⌋+ · · · .
Then ∑k≥1
ψ(xk
)=∑k≥1
∑n≤x/k
Λ(n) =∑n≤x
Λ(n)⌊xn
⌋=∑n≤bxc
Λ(n)
⌊bxcn
⌋=∑p
(⌊bxcp
⌋+
⌊bxcp2
⌋+ · · ·
)log p
=∑p
νp(bxc!) log p = log(bxc!) = T (x).
Math 229x Notes 8
It follows that
ψ(x)− ψ(x
2
)+ ψ
(x3
)− · · · = T (x)− 2T (x/2) = (log 2)x+O(log x).
So because ψ is increasing,
ψ(x) ≥ (log 2)x+O(log x).
Corollary 2.1. ψ(x) ∼ θ(x).
Similarly, we have ψ(x)+ψ(x/2) ≤ (log 2)x+O(log x) and so adding up thisinequality for x/2s, we get
ψ(x) ≤ 2(log 2)x+O((log x)2).
Corollary 2.2. There are constants a,A > 0 such that ax < ψ(x) < Ax forx 1. That is, ψ(x) x. Hence θ(x) x.
Because θ(x) ≤ π(x) log x, we have π(x) x/ log x.For the upper bound, we divide the interval into two and compute it sepa-
rately.
π(x) = π( x
(log x)2
)+ #p : x/(log x)2 < p ≤ x
≤ x
(log x)2+
θ(x)
log(x/(log x)2) x
(log x)2+
x
log x.
Corollary 2.3. π(x) x/ log x.
Corollary 2.4. For x 1, there exists a prime between x and 3x.
Proof. We have the estimate (log 2− ε)x < ψ(x) < 2(log 2 + ε)x.
What’s missing in this picture is the asymptotic relation between θ (or ψ)and π. Recall that π(x) log x ≥ θ(x). In the other direction,
θ(x) =∑p≤x
log p = π(x) log x−∫ x
2
π(t)dt
t
and ∫ x
2
π(t)dt
t=
∫ x
2
t
log t
dt
t=
∫ x/ log x
2
dt
log t+
∫ x
x/ log x
dt
log t
1 · x
log x+
1
log x· x x
log x.
It follows that
θ(x) ∼ π(x) log x.
Theorem 2.5. We have:
(1) π(x) log x ∼ ψ(x) ∼ θ(x)
(2) π(x) x/ log x, ψ(x) x, θ(x) x.
This is pretty much it for elementary methods.
Math 229x Notes 9
2.2 The Riemann zeta function
This is not actually Riemann’s function; it is Euler’s. Euler defined
ζ(s) =∑n≥1
1
ns=∏p
(1 +
1
ps+
1
p2s+ · · ·
)=∏p
(1− 1
ps
)−1
which converges absolutely for <(s) > 1. It follows that there are infinitelymany primes, because
∑n−1 = 1.
Now this function is defined only on the half-plane, but I want to do analyticcontinuation. For <(s) > 1, we have
ζ(s) = limx→∞
∑n≤x
1
ns= limx→∞
(bxc · 1
xs+ s
∫ x
1
btcts+1
dt
)= s
∫ ∞1
btcts+1
dt = s
∫ ∞1
dt
ts− s
∫ ∞1
tdtts+1
=s
s− 1− s
∫ ∞1
tdtts+1
.
Note that the fractional part is bounded by 1 and so the integral on the righthand side makes sense for <(s) > 0 and is analytic on that region. And bycomplex analysis, if there is a meromorphic continuation, then it is unique.
Proposition 2.6. ζ(s) has a meromorphic continuation to <(s) > 0 with aunique pole at s = 1. Moreover, the residue is Ress=1 ζ(s) = 1.
Next time we will talk about L-functions.
Math 229x Notes 10
3 January 27, 2017
Last time we discussed Chebyshev’s method for counting primes and definedthe Riemann ζ function. We also showed that ζ(s) has an analytic continuationfor <(s) > 0.
3.1 Dirichlet series
Definition 3.1. A Dirichlet series is a series of form
D(s) =∑n≥1
anns, an ∈ C.
Proposition 3.2. If D(s) converges for s = s0 ∈ C, then for every 0 < α <π/2, D(s) converges uniformly on the region z ∈ C : <(z − s0) ≥ 0, |arg(z −s0)| ≤ α.
Proof. Without loss of generality we may set s0 = 0. This is because we maymake a substitution. Now
∑n≥1 an converges. We want for every M > 0 that
D(s) converges uniformly on s ∈ C : <(s) > 0, |s| ≤M<(s).We have for l > k,
l∑n=k
anns
=
( l∑n=k
an
)1
ls+
l−1∑n=k
( n∑j=k
aj
)( 1
ns− 1
(n+ 1)s
).
Let ε > 0. Take Nε such that for every l > k > Nε, |∑kn=l an| < ε. Then for s
in the region, ∣∣∣∣ l∑n=k
anns
∣∣∣∣ < ε
(1 +
l−1∑n=k
∣∣∣ 1
ns− 1
(n+ 1)s
∣∣∣).We now have the estimate∣∣∣ 1
ns− 1
(n+ 1)s
∣∣∣ =
∣∣∣∣s∫ log(n+1)
logn
e−stdt
∣∣∣∣ ≤ |s|∫ log(n+1)
logn
e−<(s)tdt
=|s|<(s)
( 1
n<(s)− 1
(n+ 1)<(s)
).
Because s is in the region,∣∣∣∣ l∑n=k
anns
∣∣∣∣ < ε(
1 +M( 1
k<(s)− 1
l<(s)
))< ε(1 +M).
Corollary 3.3. There exist σa, σc ∈ [−∞,∞] such that D(s) converges (ab-solutely) for <(s) > σc (<(s) > σa) and does not converge (absolutely) for<(s) < σc (<(s) < σa). (Obviously σc ≤ σa.)
Math 229x Notes 11
Corollary 3.4. The Dirichlet series is D(s) ≡ 0 if and only if an = 0 for alln.
Proof. By uniform convergence,
0 = limσ→∞
D(σ) = limσ→∞
∑n
annσ
=∑n
limσ→∞
annσ
= a1.
Multiply by 2s and repeat this process.
Theorem 3.5 (Landau). Suppose that an ≥ 0. Suppose also that:
(1) D(s) converges for <(s) > σ, where σ ∈ R,
(2) there is a neighborhood of s = σ where D(s) has analytic continuation.
Then there exists an ε > 0 such that D(s) converges for <(s) > σ − ε.
Note that this theorem cannot be applied to the Riemann zeta functionbecause it has a pole at s = 1.
Proof. Without loss of generality let σ = 0. Let ε > 0 such that D(s) hasanalytic continuation to |z − 1| ≤ 1 + ε. By uniform absolute converge near 1,we have
D(k)(1) = (−1)k∑n≥1
an(log n)k
n.
So we have
D(s) =∑k≥0
(−1)k
k!
(∑n≥1
an(log n)k
n
)(s− 1)k.
for |s− 1| ≤ 1 + ε. Then
D(−ε) =∑k≥0
(−1)k
k!
(∑n≥1
an(log n)k
n
)(−1− ε)k
=∑k≥0
1
k!
(∑n≥1
an(log n)k
n
)(1 + ε)
=∑n≥1
ann
exp((log n)(1 + ε)) =∑n≥1
ann−ε
.
That is, the series converge. This shows that it converges for <(s) > −ε.
Math 229x Notes 12
4 January 30, 2017
If you have two Dirichlet series
D(ann, s) =∑n≥1
anns, D(bnn, s) =
∑n≥1
bnns,
that converge for <(s) > σ, then
D(ann, s)D(bnn, s) = D((a ∗ b)nn, s)
where (a ∗ b)n =∑d|n adbn/d for <(s) > σ.
4.1 Dirichlet L-functions
Let N ≥ 1 be an integer and let GN = (Z/NZ)× be the multiplicative group.Define
GN = χ : GN → C× group morph..
By representation theory of abelian groups, we have a non-canonical isomorphicGN ∼= GN . There are orthogonality relations∑
a∈GN
χ(a)ψ(a) =
φ(N), if χ = ψ ∈ GN0 otherwise,∑
χ∈GN
χ(a) · χ(b) =
φ(N) if a = b ∈ GN ,0 otherwise.
Definition 4.1. Extend χ ∈ GN to Z by 0. Then define the Dirichlet L-function of the Dirichlet character χ as
L(s, χ) =∑n≥1
χ(n)
ns.
The series converges absolutely in the region <(s) > 1. We also have apresentation
L(s, χ) =∏p
(1− χ(p)
ps
)−1
which also converges absolutely in <(s) > 1.
Example 4.2. In the case N = 4, there is the principal character χ0 = trivialand χ 6= χ0. We have
L(s, χ0) = 1 +0
2s+
1
3s+ · · · =
(1− 1
2s
)ζ(s).
There are some junk here, and this leads to the notion of primitive characters.The other L-function is
L(s, χ) = 1− 1
3s+
1
5s− 1
7s+ · · · .
Math 229x Notes 13
4.2 Analytic continuation
The first case is χ = χ0. We have
L(s, χ0) =∏p|N
(1− 1
ps
)ζ(s)
and this implies that L(s, χ0) has meromorphic continuation to <(s) > 0. Theunique pole is at s = 1 of order 1 and Ress=1 L(s, χ0) = φ(N)/N .
Now let us look at the case χ 6= χ0. In this case∣∣∣∣∑n≤t
χ(n)
∣∣∣∣ ≤ φ(N),
because there is cancellation on every interval of length N . Then by partialsummation, for <(s) > 1∑
n≤x
χ(n)
ns=
(∑n≤x
χ(n)
)1
xs+ s
∫ x
1
(∑n≤t
χ(n)
)dt
ts+1.
So we get for <(s) > 1,
L(s, χ) = s
∫ ∞1
(∑n≤t
χ(n)
)dt
ts+1.
But the right hand side converges absolutely for <(s) > 0. This gives analyticcontinuation and there is no pole. We moreover have:
Proposition 4.3. The series∑n≥1 χ(n)/ns converges for <(s) > 0, and L(s, χ)
is analytic for <(s) > 0.
The question here is what is L(1, χ)? If it doesn’t go to infinity, what will itbe? For N = 4 and χ 6= χ0, we have
L(1, χ) = 1− 1
3+
1
5− 1
7+ · · · = arctan(1) =
π
4.
4.3 Nonvanishing of the L-function at 1
Fix N > 1. We want to show that if χ 6= χ0 then L(1, χ) 6= 0.
Lemma 4.4. Let p - N and let f(p) = #〈p〉 in GN and let g(p) = #(GN/〈p〉).Then ∏
χ∈GN
(1− χ(p)T ) = (1− T f(p))g(p).
Proof. We have∏χ∈GN
(1− χ(p)T ) =∏
ε∈µf(p)
∏χ(p)=ε
(1− εT ) = (1− T f(p))g(p).
Math 229x Notes 14
Define
FN (s) =∏χ∈GN
L(s, χ).
Lemma 4.5. FN is meromorphic on <(s) > 0. The only possible pole is ats = 1. If some χ 6= χ0 satisfies L(1, χ) = 0, then FN (s) is analytic on <(s) > 0.
Proof. Obvious.
Lemma 4.6. FN (s) can be expressed as a Dirichlet series for <(s) > 1, in theform FN (s) =
∑n≥1 an/n
s with an ≥ 0. Furthermore, the Dirichlet series doesnot converge at s = 1/φ(N).
Proof. Look at the Euler factors. Then
FN (s) =∏p-N
(1− p−sf(p))−g(p) =∏p-N
(1 + f−f(p)s + f−2f(p)s + · · · )g(p).
This immediately proves the first part. Now for s ∈ R, the Euler factor isgreater than 1 + p−φ(N)s + p−2φ(N)s + · · · . A comparison with the zeta functionshows that this cannot converge.
If FN (s) has no pole on <(s) > 0, then Landau imply convergence on <(s) >0. But we have seen that the Dirichlet series does not converge at s = 1/φ(N).Thus there exists a pole.
Theorem 4.7. For χ 6= χ0, L(1, χ) 6= 0.
Math 229x Notes 15
5 February 1, 2017
Last time we showed that if we take the character χ 6= χ0 then L(1, χ) 6= 0.This looks like a random fact, but it has a deeper meaning.
5.1 Primes in arithmetic progressions
Lemma 5.1. Let χ ∈ GN be any character. Then for all s ∈ C with <(s) > 1,we have L(s, χ) 6= 0.
Proof. Look at the Euler product.
Using this fact we can define logL(s, χ) for <(s) > 1 with the condition thatlog 1 = 0. In other words, we want limσ→+∞ logL(σ, χ) = 0.
Lemma 5.2. Let χ ∈ GN . There is gχ(s) holomorphic on <(s) > 1/2 suchthat for all s with <(s) > 1,
logL(s, χ) =∑p
χ(p)
ps+ gχ(s).
Proof. Define
gχ(s) =∑p
∑k≥2
1
k
χ(pk)
pks.
This is holomorphic on <(s) > 1/2. Now
logL(s, χ) =∑p
− log(
1− χ(p)
ps
)=∑p
∑k≥1
1
k
χ(pk)
pks
for <(s) > 1 because Taylor expansion works in the region.
Corollary 5.3. Let χ ∈ GN .(1) If χ = χ0 then
∑p χ0(p)/ps ∼ − log(s− 1) as s→ 1+.
(2) If χ 6= χ0 then there exists a Bχ ∈ C such that
lims→1+
∑p
χ(p)
ps= Bχ.
Proof. (1) Look at the pole.(2) Take Bχ = logL(1, χ)− gχ(1). (This works because L(1, χ) 6= 0.)
Theorem 5.4 (Dirichlet). Let N ≥ 1 and (a,N) = 1. There are infinitely manyprimes p ≡ a (mod N). Furthermore, they have analytic density 1/φ(N):
lims→1+
∑p≡a (N) p
−s∑p p−s =
1
φ(N).
Math 229x Notes 16
Proof. We have
1
φ(N)
∑χ∈GN
χ(a)∑p
χ(p)
ps=∑p
1
ps1
φ(N)
∑χ
χ(p)χ(a)
=∑
p≡a (N)
1
ps
by orthogonality. Now we know the behavior of the sums∑p χ(p)/ps.
5.2 Tauberian theorems
In general, the statements look like this. You have some weak convergence ofsum sum or integral. And you have some analytic property. From this you showthat you have strong convergence. Or you have some estimate for sums withweights and have some analytic property and get estimates without weights.
Theorem 5.5 (Newman’s simple Tauberian theorem). Let F : [0,∞) → C bebounded an locally integrable. For <(s) > 0 define
G(s) =
∫ ∞0
F (t)e−stdt.
(Note that G(s) is analytic on <(s) > 0.) Suppose G(s) has analytic continu-ation to a domain containing s ∈ C : <(s) ≥ 0. Then the improper integral∫∞
0F (t)dt converges and its value is G(0).
Proof. For T > 0 let
GT (s) =
∫ T
0
F (t)e−stdt.
Note GT (s) converges for every s ∈ C and it is entire. We want to show thatlimT→∞GT (0) = G(0).
For an arbitrary R > 0 let δ = δR > 0 such that s ∈ C : <(s) ≥ −δ, |=(s)| ≤R is contained in the domain of analytic extension of G(s). Let
C = ∂s ∈ C : <(s) ≥ −δ, |s| ≤ R.
Also let C + = C ∩ <(s) > 0, C− = C ∩ <(s) < 0. Write
G(0)−GT (0) =1
2πi
∫C
(G(s)−GT (s))esT(
1 +s2
R2
)dss.
We will show that for R fixed, this is “small” as T →∞.On C + let B = ‖F‖∞. For <(s) > 0
|G(s)−GT (s)| =∣∣∣∣∫ ∞T
F (t)e−stdt
∣∣∣∣ ≤ B ∫ ∞T
|e−st|dt =Be−<(s)T
<(s).
Math 229x Notes 17
Note that for w ∈ S1, |1 + w2| = 2|<(w)|. So, for s ∈ C +,∣∣∣esT(1 +( sR
)2)1
s
∣∣∣ =2<(s)e<(s)T
R2.
So ∣∣∣∣ 1
2πi
∫C+
· · ·∣∣∣∣ ≤ 1
2π`(C +)
2B
R2=B
R.
The estimate on C− is more complicated. Look at G and GT separately.Because GT is entire, we have
∫C− GT =
∫γGT where γ = <(s) < 0, |s| = R.
For s ∈ γ,
|GT (s)| =∣∣∣∣∫ T
0
F (t)e−stdt
∣∣∣∣ ≤ B ∫ T
0
|e−st|dt ≤ Be−<(s)T
|<(s)|.
The bound for the other factor is given by 2|<(s)|e<(s)T/R2 and so we get∣∣∣∣ 1
2πi
∫C−
GT (s)(· · · )∣∣∣∣ ≤ B
R.
I will do the other estimate next time.
Math 229x Notes 18
6 February 3, 2017
Proof of Theorem 5.5. We showed
|G(0)−GT (0)| ≤ 1
2π
∣∣∣∣∫C+
(· · · )∣∣∣∣+
∣∣∣∣∫C−
GT (s) · (· · · )∣∣∣∣+
∣∣∣∣∫C−
G(s) · (· · · )∣∣∣∣
≤ B
R+B
R+
1
2π
∣∣∣∣∫C−
G(s)esT(
1 +s2
R2
)dss
∣∣∣∣.We now have to estimate this integral.
Note that on C−, ∣∣∣G(s)(
1 +s2
R2
)1
s
∣∣∣ = OR(1)
as T →∞, i.e., does not depend on T and is actually bounded. Thus
1
2π
∣∣∣∣∫C−
G(s)esT(
1 +s2
R2
)dss
∣∣∣∣R
∫C−
e<(s)T |ds|.
As T →∞, the right hand side converges to 0, when R is fixed.Adding the inequalities up, we get
lim supT→∞
|G(0)−GT (0)| ≤ 2B
R
for any fixed R > 0. Thus limT→∞GT (0) = G(0).
With this theorem our next goal is the prime number theorem.
6.1 The Riemann zeta on <(s) = 1
Lemma 6.1. Let 0 < δ < 1 and w ∈ C with |w| = 1. Then
|(1− δ)−3(1− δw)−4(1− δw2)−1| ≥ 1.
Proof. Since δ is small we can use the Taylor series of − log(1− x). Then
log|· · ·| = <∑k≥1
δk
k(3 + 4wk + w2k) = <
∑k≥1
δk
k((2 + wk)2 − 1) ≥ 0.
This has a deep meaning as you will see later.
Corollary 6.2. Let σ > 1 and t ∈ R. Then |ζ(σ)3ζ(σ + it)4ζ(σ + 2it)| ≥ 1.
Proof. Look at the Euler factors for each p.
Theorem 6.3. ζ(s) does not vanish on <(s) = 1.
Proof. Suppose ζ(1 + it0) = 0. Certainly t0 6= 0 and so 2t0 6= t0. Let α =ord1+it0 ζ(s) ≥ 1 and β = ord1+2it0 ζ(s) ≥ 0. Let σ → 1+ in the expressionζ(σ)3ζ(σ + it0)4ζ(σ + 2it0). The limit is 0 because −3 + 4α+ β ≥ 1. This is acontradiction.
Math 229x Notes 19
6.2 Prime number theorem
We want ψ(x)/x→ 1. This will show π(x) ∼ x/ log x. Recall that
ψ(x) =∑n≤x
Λ(n) x,∑n≥1
Λ(n)
ns= −ζ
′(s)
ζ(s)for <(s) > 1.
Note that for <(s) > 1,∑ Λ(n)
ns= s
∫ ∞1
ψ(s)x−(s+1)dx
by partial summation.
Theorem 6.4. The integral
∫ ∞1
(ψ(x)
x− 1)dxx
converges.
Proof. Define
G(s) =
∫ ∞1
(ψ(x)
x− 1)x−s
dx
x=
∫ ∞0
(ψ(et)e−t − 1)e−stdt.
Note that F (t) = ψ(et)e−t − 1 is a bounded function and clearly locally inte-grable. Also for <(s) > 0,
G(s) =
∫ ∞1
ψ(x)x−(s+2)dx−∫ ∞
1
x−(s+1)dx = − 1
s+ 1
ζ ′(s+ 1)
ζ(s+ 1)− 1
s.
This has analytic continuation to <(s+ 1) = 1.
Math 229x Notes 20
7 February 6, 2017
The midterm will be on Friday, March 3.Our goal is to show π(x) ∼ π/ log x, or equivalently, ψ(x) ∼ x. Together
with the Tauberian theorem and ζ(1 + it) 6= 0, we showed that∫ ∞1
(ψ(x)
x− 1)dxx
converges.
Theorem 7.1 (Prime number theorem). ψ(x) ∼ x.
Proof. Suppose not. Then one of the following holds:
I. There exists λ > 1 and infinitely many T →∞ such that ψ(T ) > λT .In such a case, since ψ is non-decreasing∫ λT
T
(ψ(x)
x− 1)dxx>
∫ λT
T
(λTx− 1)dxx
= c+(λ) > 0.
This contradicts the convergence of the integral.
II. There exists 0 ≤ λ < 1 and infinitely many T →∞ with ψ(T ) < λT .You do the same thing. We have∫ T
λT
(ψ(x)
x− 1)dxx<
∫ T
λT
(λTx− 1)dxx
= c−(λ) < 0.
This again is a contradiction.
The more you understand the behavior of the L-functions to the left, youknow more about primes.
7.1 Poisson summation
Let’s recall some Fourier analysis. Let F : R → C be L1([0, 1]) and 1-periodic.Define
ck(F ) =
∫ 1
0
F (t)e−2πiktdt.
Theorem 7.2. If F is continuous and∑k∈Z|ck(F )| < ∞, then F equals its
Fourier series pointwise:
F (x) =∑k
ck(F )e2πikx.
Now start with a function f : R → C in L1(R). We define its Fouriertransform
f(t) =
∫Rf(x)e−2πixtdx.
Math 229x Notes 21
Example 7.3. The Fourier transform of f(x) = e−πx2
is f(y) = e−πy2
.
Theorem 7.4 (Poisson summation formula). Suppose f : R→ C is continuous,
and there exists a δ > 0 such that |f(x)|, |f(x)| (1 + |x|)−1−δ. Then for everyy ∈ R, ∑
n∈Zf(n+ y) =
∑k∈Z
f(k)e2πiky.
In particular,∑n∈Z f(n) =
∑n∈Z f(n).
Proof. Define F (y) =∑n f(n+y). The bound on f(x) implies uniform absolute
convergence. Thus F is continuous. This is a 1-periodic function. So we cancompute its Fourier coefficients. By uniform and absolute convergence,
ck(F ) =
∫ 1
0
F (x)e−2πikxdx =
∫ 1
0
∑n
f(n+ x)e−2πikxdx
=∑n
∫ 1
0
f(n+ x)e−2πikxdx =∑n
∫ n+1
n
f(x)e−2πikxdx = f(k).
Now analyze F .
Here are some examples of f and f with f, f (1 + |x|)1+δ.
f(t) e−πt2
e−π(t+c)2 e−π(t/r+c)2
f(t) e−πt2
e−πt2+2πit re−π(rt)2+2πicrt
Table 1: Some examples of Fourier transformation
Theorem 7.5. Let α ∈ R. For x > 0 we have∑n
e−π(n+α)2/x = x1/2∑n
e−πn2xe2πiαn,
and in particular, ∑n
e−πn2/x = x1/2
∑n
e−πn2x.
Proof. Take f(t) = e−π(t+α)2/x.
Define
θ(z) =∑n∈Z
eπin2z
for =(z) > 0. Then we have just proved
θ(i/x) = x1/2θ(ix)
for x > 0.
Math 229x Notes 22
8 February 8, 2017
We defined θ(z) =∑n∈Z e
πin2z for z in the upper half plane. We showed that
θ(ix) satisfies a functional equation θ(i/x) = x1/2θ(ix) for x > 0. So we seethat
θ(−1
z
)= (−iz)1/2θ(z)
for all z in the upper half plane, because both sides are holomorphic.
8.1 The Γ-function
Recall that
Γ(s) =
∫ ∞0
e−xxsdx
x
for <(s) > 0. The reason we have a shift is because we are working in the Haarmeasure. The space R>0 has a structure of a multiplicative Lie group, and ithas a locally additive structure. We are in some sense putting these together.
We have a functional equation Γ(s + 1) = sΓ(s) for <(s) > 0. Using thisequation, you can inductively extend Γ to the whole complex plane and get ameromorphic function on C. Here are some facts:
• Γ(s) is holomorphic for <(s) > 0.
• Γ(n) = (n− 1)! for n ∈ Z>0.
• Γ(s) has simple poles at s = 0,−1,−2, . . ..
I will post some additional material on Γ(s) on the webpage.
8.2 Functional equation of ζ(s)
We have
Γ(s
2
)=
∫ ∞0
e−tts/2dt
t=
∫ ∞0
e−πn2x(πn2x)s/2
dx
x
= πs/2ns∫ ∞
0
xs/2e−πn2x dx
x.
So we see the zeta function arising here. We can write for <(s) > 1,
π−s/2Γ(s
2
)ζ(s) =
∑n≥1
∫ ∞0
xs/2e−πn2x dx
x
=
∫ ∞0
xs/2(∑n≥1
e−πn2x
)dx
x=
∫ ∞0
xs/2W (x)dx
x
Math 229x Notes 23
because of absolute convergence, where
W (x) =∑n≥1
e−πn2x =
1
2(θ(ix)− 1).
Now we are going to use the functional equation for θ, but there is a fixed pointof the involution, 1. So∫ ∞
0
xs/2W (x)dx
x=
∫ ∞1
xs/2xs/2W (x)dx
xW (x) +
∫ 1
0
xs/2W (x)dx
x
=
∫ ∞1
xs/2W (s)dx
x+
∫ ∞1
x−s/2W (x−1)dx
x.
We have
W (x−1) =1
2(θ(i/x)− 1) =
1
2(x1/2θ(ix)− 1) = x1/2W (x) +
1
2(x1/2 − 1).
Hence∫ ∞0
xs/2W (x)dx
x=
∫ ∞1
(xs/2 + x(1−s)/2)W (x)dx
x+
1
2
∫ ∞1
x−s/2(x1/2 − 1)dx
x
=1
s(s− 1)+
∫ ∞1
(xs/2 + x(1−s)/2)W (x)dx
x.
Let us define
ξ(s) = s(s− 1)π−s/2Γ(s
2
)ζ(s).
Theorem 8.1 (Riemann). For <(s) > 1,
ξ(s) = 1 + s(s− 1)
∫ ∞1
(xs/2 + x(1−s)/2)W (x)dx
x.
Corollary 8.2. ξ(s) extends to an entire function. In particular, ζ(s) extendsto a meromorphic function on C.
Corollary 8.3. ξ(1− s) = ξ(s). This also gives a functional equation for ζ(s).
Example 8.4. Let us compute ζ(0) = 1 + 1 + · · · . We have
s(s− 1)π−s/2Γ(s
2
)ζ(s) = 1 + s(s− 1)(entire) = 1
at s = 0. The left hand side is 2(s− 1)π−s/2Γ(s/2 + 1)ζ(s). So we can plug ins = 0 and get ζ(0) = −1/2.
Example 8.5 (Trivial zeros). Let n ≥ 1 be a positive integer. Put s = −2n.Then
ξ(−2n) = 1 + 2n(2n+ 1)
∫ ∞1
(positive)dx
x> 0.
Math 229x Notes 24
On the left hand side,
s(s− 1)π−s/2Γ(s/2)ζ(s)
is a product of a nonzero number and a simple pole and ζ(s). So we see thatζ(s) as a simple zero at s = −2n.
Math 229x Notes 25
9 February 10, 2017
9.1 Primitive characters
For a d | N there is a projection πd : GN → Gd because we have Z/NZ→ Z/dZ.Then taking the dual gives
π∗d : Gd → GN .
For any coprime a, d, there exists an r such that (a + rd,N) = 1, because foreach prime factor p | N you can locally find r such that p - a+ rd. So GN → Gdis surjective and π∗d : Gd → GN is injective.
Definition 9.1. χ ∈ GN is imprimitive if there exists a proper divisor d | Nsuch that χ ∈ im(π∗d). Otherwise χ is called primitive.
Lemma 9.2. Let χ ∈ GN and d | N . Then the following are equivalent:
(i) χ ∈ im(π∗d).
(ii) for any a ≡ 1 (mod d), if (a,N) = 1 then χ(a) = 1.
Proof. Both (i) and (ii) are equivalent to: for every a, b, if a ≡ b (mod d) and(a,N) = (b,N) = 1, then χ(a) = χ(b).
Definition 9.3. The conductor cond(χ) of χ is the least Q such that χ ∈im(π∗Q).
Lemma 9.4. Let χ ∈ GN . If for some d | N we have χ ∈ im(π∗d), thencond(χ) | d.
Proof. Write N = pα11 · · · pαrr , where αi ≥ 1 and pi are distinct primes. The
Chinese remainder theorem says that there exist χj ∈ Gpαjj
such that χ =∏rj=1 π
∗pαjj
(χj). Now the result is clear for χj because the divisibility ordering
for prime powers is the same as the usual ordering.Note that for all d | N such that χ ∈ im(π∗d), we have Q =
∏rj=1 cond(χj) | d.
Moreover, χ ∈ im(π∗Q). So by minimality, we have Q = cond(χ).
9.2 Gauss sums
Let χ ∈ GN . The Gauss sum of χ is defined as
τ(χ) =
N∑h=1
χ(h)e( hN
), e(x) = e2πix.
This is kind of a discrete version of Γ because it is the sum of something multi-plicative and additive.
Math 229x Notes 26
Proposition 9.5. Let χ ∈ GN . If (n,N) = 1, then
χ(n)τ(χ) =
N∑h=1
χ(h)e(nhN
).
Moreover, if χ is primitive, then the condition (n,N) = 1 can be dropped.
Proof. For (n,N) = 1,
χ(n)τ(χ) =
N∑k=1
χ(n)χ(k)e(kn
)=
N∑h=1
χ(h)e(nhN
).
For χ primitive, we only need to consider (n,N) = q > 1. Write N = qM andn = qm. Note that χ(n)τ(χ) = 0. So we have to show that the right hand sideis zero. We have
N∑h=1
χ(h)e(hmM
)=
M∑b=1
e(bmM
) q−1∑a=0
χ(aM + b) =
M∑b=1
e(bmM
),
where S(b) =∑q−1a=0 χ(aM + b). It suffices to show that S(b) = 0 for all b.
Note that S is M -periodic. Since χ is primitive and M | N , there exist t ≡ 1(mod M) with (t,N) = 1 and χ(t) 6= 1. Write t = cM + 1. Then
χ(t)S(b) =
q−1∑a=0
χ(taM + cbM + b) =
q−1∑a=0
χ(aM + cbM + b) = S(b).
Because χ(t) 6= 1, we get S(b) = 0.
Theorem 9.6. Let χ ∈ GN be primitive. Then |τ(χ)| =√N .
Proof. For every n
|χ(n)|2|τ(χ)|2 = χ(n)τ(χ)χ(n)τ(χ) =
N∑k=1
N∑h=1
χ(h)χ(k)e(n(h− k)
N
).
Averaging over n gives
φ(N)|τ(χ)|2 =
N∑n=1
|χ(h)|2N = φ(N)N.
The reason we are doing all this is because we want to give a functionalequation for L(s, χ). If we define θ0(z, χ) =
∑n χ(n)e−πn
2z, then this becomes0 if χ(−1) = −1. Anyways, the Poisson summation doesn’t really apply directly.The Gauss sums give way to make this work.
Math 229x Notes 27
10 February 13, 2017
For χ a primitive character, we proved a formula for χ(n)τ(χ) and also provedthat |τ(χ)| = N1/2.
10.1 Twisted θ-functions
Let χ ∈ GN be a primitive character. We have to distinguish two cases: whenχ is even, i.e., χ(−1) = 1, and when χ is odd, i.e., χ(−1) = −1. Suppose nowthat χ is even. For =(z) > 0, we define
θ0(z, χ) =∑n∈Z
χ(n)eiπn2z/N .
The condition χ even ensures that θ0 does not vanish. Since χ is primitive,
τ(χ)θ0(z, χ) =∑n
χ(n)τ(χ)eiπn2z/N =
∑n
N∑h=1
χ(h)eiπn2z/Ne2πihn/N
=
N∑h=1
χ(h)∑n
eiπn2z/Ne2πihn/N .
For x > 0, we get
τ(χ)θ0(ix, χ) =
N∑h=1
χ(h)e−πn2x/Ne2πinh/N
=(Nx
)1/2 N∑h=1
χ(h)∑n
e−π(n+h/N)2N/x
=(Nx
)1/2 N∑h=1
χ(h)∑n
e−π(Nn+h)2/(xN)
=(Nx
)1/2∑m
χ(m)e−πm2/xN =
(Nx
)1/2
θ0(i/x, χ).
We get a χ but that is fine.
Theorem 10.1. Let χ ∈ Gn be primitive and even. Then
θ0(i/x, χ) =τ(χ)
N1/2x1/2θ0(ix, χ).
Again, the property of analytic continuation shows that this actually holds onthe whole upper half plane.
Theorem 10.2. Let N > 1, and let χ ∈ GN be primitive and even. Define
ξ0(s, χ) =(Nπ
)s/2Γ(s
2
)L(s, χ).
Math 229x Notes 28
Then ξ0(s, χ) extends to an entire function. It satisfies the functional equation
ξ0(s, χ) = w0(χ)ξ0(1− s, χ) with w0(χ) =τ(χ)
N1/2.
Let us now look at the case when χ is odd. For =(z) > 0, we define
θ1(z, χ) =∑n
χ(n)neiπn2z/N .
To get a functional equation, we need the Poisson summation formula forneiπn
2z/N . Recall that∑n
e−πn2ye2πinζ = y−1/2
∑n
e−π(n+α)2/y.
By uniform convergence (on α), we can differentiate:
2πi∑n
ne−πn2ye2πinα = −2πy−3/2
∑n
(n+ α)e−π(n+α)2/y.
For h = 1, . . . , N , choose α = h/N and y = x/N . Then∑n
ne−πn2x/Ne2πinh/N = i
(Nx
)3/2∑n
(n+
h
N
)e−π(Nn+h)2/xN
= iN1/2 1
x3/2
∑n
(Nn+ h)e−π(Nn+h)2/xN .
After doing the exactly same thing, we get the following theorem.
Theorem 10.3. Let χ ∈ GN be a odd primitive character. Then for all x > 0,
θ1(i/x, χ) =τ(χ)
iN1/2x3/2θ1(ix, χ).
Because the exponent of x is shifted by 1, we get a slightly different functionalequation.
Theorem 10.4. Let χ ∈ GN be odd and primitive. Define
ξ1(s, χ) =(Nπ
)(s+1)/2
Γ(s+ 1
2
)L(s, χ).
Then ξ1(s, χ) extends to an entire function and satisfies:
ξ1(s, χ) = w1(χ)ξ1(1− s, χ) with w1(χ) =τ(χ)
iN1/2.
In Murty’s book there is a misprint in the formula.
Theorem 10.5. Let N > 1 and χ ∈ GN primitive. Define aχ = a = (1 −χ(−1))/2. Then
ξa(s, χ) =(Nπ
)(s+a)/2
Γ(s+ a
2
)L(s, χ)
is entire and satisfies
ξa(1− s, χ) = εa(χ)ξa(s, χ) with εa(χ) = iaN1/2/τ(χ).
Math 229x Notes 29
11 February 15, 2017
11.1 Growth and zeros of ξ(s)
Recall that
ξ(s) = s(s− 1)π−s/2Γ(s
2
)ζ(s)
= 1 + s(s− 1)
∫ ∞1
(∑n≥1
e−πn2x
)(xs/2 + x(1−s)/2)
dx
x.
This satisfies the functional equation ξ(1− s) = ξ(s). So we are going to focuson <(s) ≥ 1/2. On this region, we have
(1) |s(s− 1)π−s/2| exp(c1|s|).
(2)∣∣∣Γ(s
2
)∣∣∣ =
∣∣∣∣∫ ∞0
e−tts/2−1dt
∣∣∣∣ ≤ ∫ ∞0
e−ttσ/2−1dt < exp(c2|s|(log|σ|+ 1)).
(3) ζ(s) =s
s− 1− s∫ ∞
1
tts+1
dt, and so on the region <(s) ≥ 1/2 and |s| > 2,
we have |ζ(s)| c3 + c4|s|.
This all shows that
|ξ(s)| < exp(c5(|s|+ 1) log(|s|+ 1))
for all s ∈ C. This looks like a crude bound, but this is actually is optimal upto c5, because it is sharp on the real line.
Corollary 11.1. ξ(s) has order 1. So ξ(s) has infinitely many zeros.
Proof. If ξ(s) does not have infinitely many zeros, then ξ(s) has to be a poly-nomial times eA+Bs. But the lower bound does not match this growth.
Theorem 11.2. ξ(s) and ζ(s) have the same zeroes in the critical strip 0 ≤σ ≤ 1, and there are infinitely many zeros, and the other zeros of ζ(s) are thetrivial ones s = −2n.
11.2 Hadamard factorization
In the case of order 1, we have the following factorization:
ξ(s) = eA+Bs∏ρ
(1− s
ρ
)es/ρ,
where ρ runs over the zeros of ξ, repeated according to multiplicities. (Here weare using ξ(0) = 1.) We know that 1 = eA, and so we can take A = 0. Theconstant B has a deeper meaning.
Lemma 11.3. We have:
Math 229x Notes 30
(i)ξ′
ξ(1− s) = −ξ
′
ξ(s).
(ii)ξ′
ξ(s) =
ζ ′
ζ(s) +
1
s− 1− 1
2log π +
1
2
Γ′
Γ
(s2
+ 1)
.
(iii)ξ′
ξ(s) = B +
∑ρ
( 1
s− ρ+
1
ρ
)uniformly and absolutely in every large disc
on C.
Proof. The only thing you need to check is that the Hadamard product is uni-formly and absolutely convergent. This is the whole point of the Hadamardproduct.
Corollary 11.4. B = −p.v.∑ρ
1
ρ, where the principal value means that you
add them for |=ρ| < N and let N →∞.
Proof. This follows from (i) and (iii).
Also, from (ii) and (iii), we get
B = −1
2γ − 1 +
1
2log(4π) ≈ −0.02.
Theorem 11.5. For s ∈ C,
−ζ′
ζ(s) =
1
s− 1−B − 1
2log π +
1
2
Γ′
Γ
(s2
+ 1)−∑ρ
( 1
s− ρ+
1
ρ
),
where ρ runs over the nontrivial zeros of ζ(s).
For σ > 1 and s = σ + it,
<(−ζ′
ζ(s))
=∑n≥1
Λ(n)
nσcos(t log n).
Using the 3-4-1 bound, we get
3<(−ζ′
ζ(σ))
+ 4<(−ζ′
ζ(σ + it)
)+ <
(−ζ′
ζ(σ + 2it)
)≥ 0.
Lemma 11.6. For σ > 1, −ζ′
ζ(σ) <
1
σ − 1+A1.
Lemma 11.7. Let I ⊆ R>0 be a compact interval. Then for σ ∈ I,∣∣∣<Γ′
Γ(σ + it)
∣∣∣I log(|t|+ 2).
Math 229x Notes 31
12 February 17, 2017
12.1 Zero-free region for ζ(s)
Last time we discussed some consequences of the Hadamard product expansion.There is the “partial fraction formula”
−ζ′
ζ(s) =
1
s− 1−B − 1
2log π +
1
2
Γ′
Γ
(s2
+ 1)−∑ρ
( 1
s− ρ+
1
ρ
).
We also had the 3-4-1 bound
3(−ζ′
ζ(σ))
+ 4<(−ζ′
ζ(σ + it)
)+ <
(−ζ′
ζ(σ + 2it)
)≥ 0.
Lemma 12.1. Let I ⊆ R>0 be a compact interval. Then for σ ∈ I,∣∣∣<Γ′
Γ(σ + it)
∣∣∣I log(|t|+ 2).
Proof. Taking the logarithmic derivative of the Weierstrass formula for Γ, weget
<Γ′
Γ(s+ 1) = −γ +
∑n≥1
<( 1
n− 1
s+ n
)= −γ +
∑n≥1
σn+ σ2 + t2
n|s+ n|2.
We have
σn+ σ2 + t2
n|s+ n|2I
1/n2 |t|2 < n
t2/nt2 = 1/n |t|2 ≥ n.
Then the tail is bounded by a constant, and the case when n ≤ |t|2 is small isbounded by log|t|2.
For 1 < σ < 2, and |t| > 1, the partial fraction formula and Lemma 12.1give
<(−ζ′
ζ(s))≤ A2 +A3 +A4 log(|t|+ 2)−
∑ρ
<( 1
s− ρ+
1
ρ
).
Here,
<( 1
s− ρ+
1
ρ
)=
σ − β|s− ρ|2
+β
|ρ|2> 0,
where ρ = β + iτ .We are now going to use the 3-4-1 inequality. We look do the same thing we
did for the non-vanishing of ζ on <(s) = 1. Let us write L (t) = log(|t|+ 2).
Math 229x Notes 32
Lemma 12.2. For 1 < σ < 2 and |t| ≥ 1, we have
<(−ζ′
ζ(s))≤ A5 +A4L (t).
Let ρ0 = β0 + iτ0 be a non-trivial zero. Choosing s = σ + iτ0, we get:
Lemma 12.3. We have for 1 < σ < 2,
<(−ζ′
ζ(σ + iτ0)
)≤ A5 +A4L (τ0)− 1
σ − β0.
From the 3-4-1 inequality and Lemma 11.6, Lemma 12.2, Lemma 12.3, for1 < σ < 2,
0 ≤ 3
σ − 1+A6 +A4L (τ0)− 4
σ − β0+A4L (2τ0),
for any nontrivial β0 + iτ0 with |τ0| ≥ 1. So we have
4
σ − β0<
3
σ − 1+A6 +A7L(τ0).
Choosing σ = 1 + δ/(L (τ0)) gives
β0 < 1− c
L (τ0)
for some absolute constant c.
Theorem 12.4. Let ρ0 = β0 + iτ0 be a non-trivial zero of ζ. Then β0 <1− c/L (τ0) where L (t) = log(|t|+ 2).
Math 229x Notes 33
13 February 22, 2017
We are going to do the same thing for Dirichlet L-functions. In doing this, wemay assume that the characters are primitive, because the two L-functions onlydiffer by a finite number of Euler factors.
13.1 Zero-free region for L(s, χ)
Assume that χ ∈ GN is primitive and N > 1. Let a = aχ. We have
ξa(s, χ) = eAχ+Bχs∏ρ
(1− s
ρ
)es/ρ,
because ξa(s, χ) has order 1 as you proved in the assignment and the only zerosare in the critical strip with ρ 6= 0 (since L(1, χ) 6= 0).
Recall
ξa(s, χ) =(Nπ
)(s+a)/2
Γ(s+ a
2
)L(s, χ).
Theorem 13.1 (Partial fraction).
−L′
L(s, χ) = −Bχ +
1
2log(Nπ
)+
1
2
Γ′
Γ
(s+ a
2
)−∑ρ
( 1
s− ρ+
1
ρ
),
where ρ runs over the nontrivial zeros of L(s, χ).
From the functional equation and L(s, χ) = L(s, χ), we get
<(Bχ) = −1
2
∑ρ
(1
ρ+
1
ρ
)= −
∑ρ
<(1
ρ
).
Therefore, for 1 < σ < 2,
<(−L′
L(s, χ)
)≤ c1 logN + c2 log(|t|+ 2)− [p.v]
∑ρ
σ − β|s− ρ|2
.
(We are using the notation ρ = β + iτ .) These constants c1 and c2 do notdepend on N . It will be convenient to write L (N, t) = log(N(|t|+ 2)). Then
<(−L′
L(s, χ)
)≤ c3L (N, t)− [p.v]
∑ρ
σ − β|s− ρ|2
.
Here the standard 3-4-1 trick doesn’t work, because there is a χ factor. Forany (possibly non-primitive) ψ ∈ GN , we have
<(−L′
L(s, ψ))
)=
∑(n,N)=1
Λ(n)
nσ<(ψ(n)e−it logn).
Math 229x Notes 34
So what we have is
3<(−L′
L(σ, χ0)
)+ 4<
(−L′
L(σ + it, χ)
)+ <
(−L′
L(σ + 2it, χ2)
)≥ 0.
Now here is a problem. If χ2 is not principal, then we can just use this inequality,but if χ2 is principal, then the third term introduces another pole. This wasnot a problem in the case of the Riemann zeta function because we can justwork away from s = 1 and check manually afterwards. But we cannot checkthis conceptually for all L(s, χ). This is an open problem today.
For first term, we have∣∣∣ζ ′ζ
(σ)− L′
L(σ, χ0)
∣∣∣ ≤∑p|N
log p( 1
pσ+
1
p2σ+ · · ·
)≤ logN.
So we are happy as it can be absorbed into L . We also have <(−(ζ ′/ζ)(σ)) <1/(σ − 1) + c4. So
<(−L′
L(σ, χ0)
)<
1
σ − 1+ c5 logN.
For the second term, we again pick a single zero ρ0. Taking t = τ0, we have
<(−L′
L(σ + iτ0, χ)
)< c3L (N, τ0)− 1
σ − β0.
Therefore:
Lemma 13.2. Let 1 < σ < 2, and ρ0 = β0 + iτ0 be some nontrivial zero ofL(s, χ) where χ ∈ GN is primitive with N > 1. Then
4
σ − β0<
3
σ − 1+ c6L (N, τ0) + <
(−L′
L(σ + 2iτ0, χ
2)).
There are two cases: when χ is non-real and when χ is quadratic.Let us look at the case when χ is non-real. Note that χ2 might be imprim-
itive. Let N1 = cond(χ2) where clearly N − 1 | N and let χ2 come from theprimitive ψ. For 1 < σ < 2,∣∣∣L′
L(s, χ2)− L′
L(s, ψ)
∣∣∣ ≤∑p|N
log p( 1
pσ+ · · ·
)≤ logN.
Thus we can use the bound we know:
<(−L′
L(σ + 2it, χ2)
)≤ − logN + c3L (N1, 2t) < c7L (N, τ0).
Hence
4
σ − β0<
3
σ − 1+ c8L (N, τ0).
Doing the same thing we did for the zeta function, we get
β0 > 1− c
L (N, τ0).
This is what we get for a non-real primitive character χ.
Math 229x Notes 35
14 February 24, 2017
Last time we were discussing zero-free regions of L(s, χ), and we have finishedthe case of the non-quadratic characters.
14.1 Dealing with quadratic characters
Lemma 14.1. Let 1 < σ < 2, and ρ0 = β0 + iτ0 be a non-trivial zero of L(s, χ),with χ ∈ GN primitive and N > 1. Then
4
σ − β0<
3
σ − 1+ c1L (N, τ0) + <
(−L′
L(σ + 2iτ0, χ
2)).
We now look at the case when χ is quadratic. Then χ2 = χ0, so for 1 < σ <2,
<(−L′
L(σ + iτ, ξ0)
)≤ <
(−ζ′
ζ(σ + it)
)+ logN
≤ <( 1
s− 1
)+ c2L(N, t),
by the partial fraction decomposition. Now people don’t know what to do whenthe imaginary part of s is too small.
Consider a small absolute constant δ > 0, which we will figure our later.Assume that |τ0| ≥ δ/ logN . Choose
σ = 1 +δ
L (n, τ0).
Then
<( 1
σ + 2iτ0 − 1
)≤∣∣∣ 1
(σ − 1) + 2iτ0
∣∣∣ ≤ L (N, τ0)1
δ√
5.
The constant 1/√
5 is smaller than 1 and this is why everything works out. Nowby the previous bound and the lemma, we get
4
σ − β0<
3
σ − 1+ c3L (N, τ0) +
1
δ√
5L (N, τ0)
= L (N, τ0)(3
δ+
1
δ√
5+ c3
).
So we get
1− β0 < −δ
L (N, τ0)+ δ(3
4+
1
4√
5+δc34
)−1 1
L (N, τ0)
=δ
L (N, τ0)
((3
4+
1
4√
5+δc34
)−1
− 1).
Choosing a very small delta gives the bound we want.
Math 229x Notes 36
14.2 Siegel zeros
What if τ0 is too small? For some 0 < δ 1, suppose that |τ0| < δ/ logN .Then there actually is a pole contribution and there is nothing we can do aboutit.
Proposition 14.2. Let χ ∈ GN with N > 1 be primitive quadratic. For anabsolute constant δII > 0 (independent of N and χ) L(s, χ) can have at mostone nontrivial zero in the region
β0 ≥ 1− δII
L (N, τ).
If such a zero exists, then it is real and simple (and is not 1).
Such a zero, if exists, is called a Siegel zero, or a Siegel–Landau zero.A character χ such that L(s, χ) has a Siegel zero is called exceptional. Thisis not a very well-defined notion since δII is not given. But we can take δII tobe small and assume that there are no Siegel zeros of infinitely many.
Proof. We only need to consider |τ0| < δ/ logN . For 1 < σ < 2, we have
−L′
L(σ, χ) < c6 logN − [p.v]
∑ρ
1
σ − ρ.
Let ρ1 = β1 + iτ1 and ρ2 = β2 + iτ2 be two non-trivial zeros (possibly thesame if ρ1 is multiple). It is enough to consider:
(a) ρ1 is non-real and ρ2 = ρ1,
(b) ρ1 and ρ2 real.
If we can prove that they cannot be both in the region, we are done.We will do (a) first. We end up with
−L′
L(σ, χ) < c6 logN −
( 1
σ − ρ1+
1
σ − ρ1
)= c6 logN − 2(σ − β1)
(σ − β1)2 + τ21
.
Taking σ = 1 + 2δ/ logN , we get
|τ1| <δ
logN=
1
2(σ − 1) ≤ 1
2(σ − β1).
So we have
−L′
L(σ, χ) ≤ c6 logN − 8
5
1
σ − β1.
On the other hand, we have
−L′
L(σ, χ) =
∑n
Λ(n)
nσχ(n) > −
∑n
Λ(n)
nσ=ζ ′
ζ(σ) > − 1
σ − 1− c7.
Math 229x Notes 37
So we get
− 1
2δlogN < c8 logN − 8
5
1
σ − β1
and hence
1− β1 +2δ
logN<
8
5
(c8 +
1
2δ
)−1 1
logN.
So taking δ small, we get the desired bound, because 8/5 > 1.The case (b) can be done very similarly.
Math 229x Notes 38
15 February 27, 2017
Here is the theorem we got so far.
Theorem 15.1. Let χ ∈ GN be primitive with N > 1. There is an absoluteconstant δ > 0 (effective, independent of χ and N) such that if χ is non-real,every nontrivial zero ρ0 = β0 + iτ0 of L(s, χ) satisfies
β0 < 1− δ
L (n, τ0), L (N, t) = log(N(|t|+ 2)). (∗)
Moreover, when χ is quadratic, (∗) also holds, except perhaps for at most oneρ0. If such a ρ0 exists for L(s, χ) then it is real and simple.
For now on up to spring break, we are going to only focus on the Riemannzeta function. There is nothing really interesting happening for Dirichlet L-functions, and I don’t want to go into too much details in an introductorycourse.
We are now going to talk about counting zeros of ζ(s), explicit formula forψ(x), and the prime number theorem with error terms.
15.1 Counting zeros of ζ(s)
We are going to denote
N(T ) =
(# of zeros (with multiplicity) of ζ(s)
in the region 0 < σ < 1 and 0 < t < T.
)Because we are going to do a contour integral, we assume that there is no zeroat exactly height T . In other words, we don’t take T = τ0 for ρ0 a non-trivialzero of ζ(s). The zeros for t < 0 are conjugates.
Consider two contours
R : 2→ 2 + iT → −1 + iT → −1→ 2, R′ : 2→ 2 + iT → 1
2+ iT.
Then clearly
2πN(T ) = ∆R arg ζ(s) = ∆Rξ(s).
Now note that ξ(1− σ + it) = ξ(1− σ + it) = ξ(σ + it). So the bottom line ofξ cancels out and the top two halves are equal. Hence
πN(T ) = ∆R′ arg ξ(s).
Recall
ξ(s) = s(s− 1)π−s/2Γ(s
2
)ζ(s) = 2(s− 1)π−s/2Γ
(s2
+ 1)ζ(s)
So the variation of the argument is the sum of three variations of each compo-nent. We can analyze each component as:
Math 229x Notes 39
(i) ∆R′ arg(s− 1) = π/2 +O(1/T ),
(ii) ∆R′ arg(π−s/2) = −T log π/2,
(iii) by Stirling’s approximation,
∆R′ arg Γ(s
2+ 1)
= = log Γ(5
4+iT
2
)= =
(3
4+iT
2
)log(5
4+iT
2
)− iT
2− 5
4+
1
2log(2π) +O(T−1)
=T
2log(T
2
)− T
2+
3π
8+O(T−1).
We then immediately obtain
N(T ) =T
2πlog
T
2π− T
2π+O(1) +
1
π∆R′ arg ζ(s).
We are now going to estimate the variation of ζ(s), which is erratic. This isthe bound obtained a century ago and hasn’t been improved.
Lemma 15.2. For T →∞ and ρ over non-trivial zeros,∑ρ
1
1 + (T − τ)2 log T.
Proof. For t ≥ 2 and 1 < σ < 3, we have
<(−ζ′
ζ(s))< c1 log t−
∑ρ
<( 1
s− ρ+
1
ρ
).
Take σ = 2 and t = T . Then |(ζ ′/ζ)(s)| 1. This immediately implies∑ρ
<( 1
s− ρ
)≤ c2 log t.
Then
< 1
s− ρ=
2− β(2− β)2 + (T − τ)2
≥ 1
4 + (T − τ)2.
Corollary 15.3. For T large:
(1) There are log T zeros ρ of ζ(s) with |T − τ | < 1.
(2)∑ρ,|T−τ |≥1(T − τ)−2 log T .
Lemma 15.4. Let ε > 0. In the region |s| > 1 and |arg(s)| < π − ε, we have
Γ′
Γ(s) = log s+Oε(|s|−1).
Math 229x Notes 40
Lemma 15.5. For large T and −1 ≤ σ ≤ 2,
ζ ′
ζ(σ + iT ) =
∑ρ,|T−τ |<1
1
s− ρ+O(log T ).
Proof. We can write
ζ ′
ζ(σ + iT ) =
ζ ′
ζ(σ + iT )− ζ ′
ζ(2 + iT ) +O(1)
= O(log T ) +∑ρ
( 1
σ + iT − ρ− 1
2 + iT − ρ
),
by partial fraction and Lemma 15.4. For ρ with |T − τ | ≥ 1, we have∣∣∣ 1
s− ρ− 1
2 + iT − ρ
∣∣∣ =2− σ
|(σ + iT − ρ)(2 + iT − ρ)|≤ 3
|T − τ |2.
Then the corollary implies that these terms contribute at most log T . The otherpart is ∑
ρ,|T−τ |<1
( 1
s− ρ− 1
2 + iT − ρ
)=
∑ρ,|T−τ |<1
1
s− ρ+O(log T ),
by the corollary.
Math 229x Notes 41
16 March 1, 2017
Last time we had an equation
N(t) =T
2πlog( T
2π
)− T
2π+O(1) +
1
π∆R′ arg ζ(s).
Theorem 16.1. For T →∞, we have
N(T ) =T
2πlog( T
2π
)− T
2π+O(log T ).
Proof. It suffices to show that ∆R′ arg ζ(s) = O(log T ). Let us consider anothercontour Q : 1/2 + iT → 1 + iT . We first note that
|ζ(2 + it)− 1| < ζ(2)− 1 =π2
6− 1 < 1.
So ∆R′ arg ζ(s) = O(1)−∆Q arg ζ(s). We also have
∆Q arg ζ(s) = =∫ 2+iT
1/2+iT
ζ ′
ζ(s)ds = =
∫ 2+iT
1/2+iT
( ∑|T−τ |<1
1
s− ρ
)ds+O(log T )
= O(log T ) +
∫|T−τ |<1
∆Q arg(s− ρ)
= O(log T ) + πO(log T ) = O(log T ).
Corollary 16.2. Let us label the nontrivial zeros of ζ(s) with τ > 0 as ρ1, ρ2, . . .such that the imaginary parts are nondecreasing (with multiplicities). Then
τn ∼2πn
log n.
16.1 Integral approximation of ψ
The goal is to approximate
ψ0(x) =
ψ(x) if x > 2 is not a prime power
ψ(x)− 12Λ(x) if x > 2 is a prime power
in terms of the nontrivial zeros of ζ(s).
Theorem 16.3 (Perron’s formula). Let
δ(y) =
0 0 < y < 1
1/2 y = 1
1 y > 1.
Math 229x Notes 42
Define the function
I(y, T ) =1
2πi
∫ c+iT
c−iT
ys
sds, y > 0, c > 0, T > 0.
Then
|δ(y)− I(y, T )| <
yc min1, 1/(T |log y|) y 6= 1,
c/T y = 1.
Proof. Let us first consider the case 0 < y < 1 so that δ(y) = 0. Because y < 1,Cauchy give
I(y, T ) =
(− 1
2πi
∫ ∞+iT
c+iT
+1
2πi
∫ ∞−iTc−iT
)ys
sds.
Then we can estimate∣∣∣∣∫ ∞+iT
c+iT
ysds
s
∣∣∣∣ ≤ 1
T
∫ ∞c
yσdσ =yc
T |log y|.
We also have the contour γ lying on the circle R2 = c2 + T 2. Then
|I(y, T )| ≤ 1
2π
∣∣∣∣∫γ
ysds
s
∣∣∣∣ ≤ 1
2πRycπR < yc.
For y > 1, you can do the same thing but use the rectangle going to the left.Let us now consider the case y = 1. We have
I(1, T ) =1
2πi
∫ T
−T
1
c+ itidt =
1
2π
∫ T
−T
c− itc2 + t2
dt =1
π
∫ T
0
c
c2 + t2dt.
You can just compute this.
Corollary 16.4. For T > 0, c > 1, and x > 2,∣∣∣∣ψ0(x)− 1
2πi
∫ c+iT
c−iT
(−ζ′
ζ(s))xsds
s
∣∣∣∣=
∣∣∣∣∑n≥1
Λ(n)δ(xn
)−∑n≥1
Λ(n)1
2πi
∫ c+iT
c−iT
(x/n)sds
s
∣∣∣∣≤
∑n≥1,n6=x
Λ(n)(xn
)cmin
1,
1
T |log(x/n)|
+c
TΛ(x).
Let c = cx = 1 + 1/ log x. With this choice, define
J(x, T ) =1
2πi
∫ cx+iT
cx−iT
(−ζ′
ζ(s))xssds.
We are going to prove next time that for x > 2 and T > 1,
|ψ0(x)− J(x, T )| x(log x)2
T+ (log x) min
1,
x
〈x〉T
,
where 〈x〉 is its distance to the nearest prime power different to x.
Math 229x Notes 43
17 March 6, 2017
For x ≥ e, we let cx = 1 + (log x)−1 (so that xcx = ex). We defined
J(x, T ) =1
2πi
∫ cx+iT
cx−iT
(−ζ′
ζ(s))xssdx
for T > 0.
Theorem 17.1. For x ≥ e and T > 1 we have
|ψ0(x)− J(x, T )| x(log x)2
T+ (log x) min
1,
x
〈x〉T
.
Proof. We had from Perron’s formula,
|ψ0(x)− J(x, T )| ≤∑
n=1,n6=x
Λ(n)(xn
)cxmin
1,
1
T log|x/n|
+ cxΛ(x)/T.
Note that cxΛ(x)/T log x/T . For the series, we consider the range n ≤ 3x/4or n ≥ 5x/4. In this range, the series is bounded by∑
∑n≥1
Λ(n)x
ncx1
T=x
T
(−ζ′
ζ(cx)
) x log x
T.
Now we look at the range 3x/4 < n < x. Let k be the larges prime powerless than x. (Without loss of generality we may assume 3x/4 < k < x.) Wehave
logx
k= − log
k
x= − log
(1− x− k
x
)≥ x− k
x≥ 〈x〉
x.
So the term at n = k is bounded by
Λ(k)x
T 〈x〉 x log x
T 〈x〉, or
Λ(k) log x.
The other numbers n in this this range is n = k − j for 0 < j < n/4. The sumthen can be bounded as∑
log x
T
∑j<x/4
1
|log(x/(k − j))| log x
T
∑j<4/x
x
j x(log x)2
T.
The other range x < k < 5x/4 can be treated similarly.
So
ψ0(x) =1
2πi
∫ cx+iT
cx−iT
(−ζ′
ζ(s))xssdx+ (Error term),
where the error term goes to 0 as T → ∞. It is very tempting to shift theintegral.
Math 229x Notes 44
17.1 Explicit formula for ψ0
The goal is to prove that for x ≥ e (and 1 < cx ≤ 2),
ψ0(x) = x−p.v.∑ρ
xρ
ρ− ζ ′
ζ(0)− 1
2log(
1− 1
x2
).
This looks like a random formula, but it is not. Consider a rectangle −U−iT →cx − iT → cx + iT → −U + iT → −U − iT . If we assume that the integralson the top, bottom, left vanish as U and T go to infinity, there is only thecontributions of the residues. The poles are at the zeros of ζ, the pole of ζ, ands = 0. The sum of these residues formally is exactly the right hand side.
When we do the contour integral, we would want to stay away from thepoles. So we need some technical assumptions.
• We will take U ≥ 3 an odd integer.
• We will take T ≥ 2 such that if ρ = β+ iτ is any zero of ζ, then |T − τ | 1/ log T . (∗∗)
The second condition can be satisfied because there are at most log T possibleτ in [T − 1, T + 1].
Let us start working. We are going to split the horizontal contours to −U +iT → −1 + iT and −1 + iT → cx + iT .
Lemma 17.2.
∫ cx+iT
−1+iT
(−ζ′
ζ(s))xssds x(log T )2
T log x, and similarly for the con-
jugate path.
Proof. We have ∣∣∣∣∫ cx+iT
−1+iT
(−ζ′
ζ(s))xssds
∣∣∣∣ (log T )2
T
∫ cx
−1
xσdσ.
This is because
−ζ′
ζ(s) =
∑|T−τ |<1
1
s− ρ+O(log T )
and we assume (∗∗).
Lemma 17.3. For <(s) ≤ −1 we have∣∣∣ζ ′ζ
(s)∣∣∣ log(2|s|).
away from radius 1/2 from the trivial zeros s = −2,−4, . . ..
Math 229x Notes 45
Proof. The functional equation gives you
ζ(1− s) = 21−sπ−s cos(πs
2
)Γ(s)ζ(s).
Then taking the logarithmic derivative gives
−ζ′
ζ(1− s) = O(1)− π
2tan(πs
2
)+
Γ′
Γ(s) +
ζ ′
ζ(s)
for <(s) ≥ 2. Now ζ ′/ζ is bounded for <(s) ≥ 2.
Math 229x Notes 46
18 March 8, 2017
We are trying to estimate the integral. Let us write
S : −1 + iT → cx + iT, H : −U + iT → −1 + iT, V : −U − iT → −U + iT.
Last time we proved ∣∣∣∣∫S
(−ζ′
ζ(s))xssds
∣∣∣∣ x(log T )2
T log x
for x ≥ e, T ≥ 2, and under hypothesis (∗∗). There was also the estimate∣∣∣ζ ′ζ
(s)∣∣∣ log(2|s|)
for <(s) ≤ −1 away from the points −2,−4, . . . by distance 1/2. Because ofthis, we let U be an odd integer.
Using this estimate, we have∣∣∣∣∫H
(−ζ′
ζ(s))xsds
s
∣∣∣∣ log(2T )
T
∫ U
1
x−σdσ log T
T
1
x log x
because log(2|s|)/|s| is decreasing in |s|.The same argument gives∣∣∣∣∫
V
(−ζ′
ζ(s))xssds
∣∣∣∣ logU
U
∫ T
−Tx−Udt logU
UTx−U .
If I collect all my bounds,
1
2πi
∫ cx+iT
cx−iT
(−ζ′
ζ(s))xsds
s= (residues) +O
(x(log T )2
T log x+
logU
UTx−U
).
I can make this sum of residues explicit:
(residues) = x−∑|τ |<T
xρ
ρ− ζ ′
ζ(0) +
∑1≤n<U/2
x−2n
2n.
Letting U →∞, we get
1
2πi
∫ cx+iT
cx−iT
(−ζ′
ζ(s))xsds
s= x−
∑|τ |<T
xρ
ρ− ζ ′
ζ(0)− log
(1− 1
x2
)+O
(x(log T )2
T log x
).
We can now use the integral approximation for ψ0(x).
Theorem 18.1 (Explicit formula). For x ≥ e and T ≥ 3 under (∗∗),
ψ0(x) = x−∑|τ |<T
xρ
ρ− ζ ′
ζ(0)− log
(1− 1
x2
)+ E(x, T ),
Math 229x Notes 47
where
|E(x, T )| x(log T )2
T log x+x(log x)2
T+ (log x) min
1,
x
〈x〉T
.
In particular, as T →∞,
ψ0(x) = x−p.v.∑ρ
xρ
ρ− ζ ′
ζ(0)− log
(1− 1
x2
).
18.1 Prime number theorem revisited
Let us estimate the ψ(x) for x ≥ 3 an integer. The first thing to notice is that
ψ(x) = ψ0(x) +O(log x) = x−∑|τ |<T
xρ
ρ+O(log x) + E(x, T ).
This O(log x) is a pretty small error. Also,
|E(x, T )| x
T(log(xT ))2.
Now the goal is to estimate the sum of xρ/ρ for T as large as possible. Wehave ∣∣∣∣ ∑
0<τ<T
xρ
ρ
∣∣∣∣ ≤ xβT ∑0<τ<T
1
|ρ| x1−δ/ log T (log T )2,
because we know how to count zeros, and
βT = max<(ρ) : 0 < τ < T ≤ x1−δ/ log T .
from the zero-free region. I now claim that T = exp(√
log x) is a good choice.You plug things in and you get the following theorem.
Theorem 18.2. ψ(x) = x+O(x exp−c√
log x). So for any A > 1,
ψ(x) = x+OA
( x
(log x)A
).
Math 229x Notes 48
19 March 10, 2017
19.1 Riemann hypothesis
Conjecture 19.1 (Riemann hypothesis). All nontrivial zeros of ζ(s) have realpart 1/2.
Conjecture 19.2 (δ-Riemann hypothesis). For a real number 0 < δ ≤ 1/2,every non-trivial zero ρ of ζ(s) have <(ρ) ≤ 1− δ.
Conjecture 19.3 (Quasi-Riemann hypothesis). The δ-Riemann hypothesis holdsfor some 0 < δ ≤ 1/2.
The explicit formula says that, for x ≥ 3 and T ≥ 3 under hypothesis (∗∗),
ψ(x) = x−∑|τ |<T
xρ
ρ+O
(log x+
x
T(log x)2 + (log T )2
).
Let us assume the δ-Riemann hypothesis for some fixed δ ∈ (0, 1/2]. Then∣∣∣∣ ∑0≤τ<T
xρ
ρ
∣∣∣∣ ≤ x1−δ∑
0<τ<T
x1−δ(log T )2.
Take T = xδ + O(1), where O(1) is there to make T satisfy (∗∗). We get thefollowing theorem.
Theorem 19.4. The δ-Riemann hypothesis implies that
ψ(x) = x+Oδ(x1−δ(log x)2).
Theorem 19.5. Suppose that for some δ > 0 we have
ψ(x) = x+Oε(x1−δ+ε)
for each ε > 0. Then the δ-Riemann hypothesis holds (for the same δ).
Proof. Write E(t) = ψ(x) − x. Our hypothesis is that |E(t)| ε t1−δ+ε. For
<(s) > 1,
−ζ′
ζ(s) =
∑n≥1
Λ(n)
ns= s
∫ ∞1
ψ(x)dx
xs+1=
s
s− 1+ s
∫ ∞1
E(t)dt
ts+1.
The integral is holomorphic for <(s) > 1 − δ. This shows that ζ ′/ζ only has apole at s = 1 in <(s) > 1− δ.
Corollary 19.6. The error term ψ(x) ∼ x cannot be O(xα) for any α < 1/2.
Corollary 19.7. Suppose that for some δ > 0 we have
ψ(x) = x+Oε(x1−δ(log x)2).
Then ψ(x) +O(x1−δ(log x)2).
Math 229x Notes 49
What is the best zero-free region? Vinogradov and Korobov proved in 1958that
σ > 1− δα(log(|t|+ 2))α
,
for any fixed α > 2/3 and δα > 0. This 2/3 hasn’t been improved since.
19.2 Lindelof hypothesis
How large is ζ(s) on the critical strip? An easy exercise is that for 0 < σ < 1and t > 1,
|ζ(s)| t.
Conjecture 19.8 (Lindelof hypothesis). For any ε > 0,∣∣∣ζ(1
2+ it
)∣∣∣ε |t|ε.
Theorem 19.9. The Riemann hypothesis implies the Lindelof hypothesis.
Theorem 19.10 (Lindelof). For σ ∈ R define
µ(σ) = infα : |ζ(σ + it)| σ,α |t|α for |t| > 1.
Then µ : R→ R is convex (hence continuous).
The proof follows from the Phragmen–Lindelof principle. Note that µ(σ) = 0for σ ≥ 1. By the functional equation
ζ(1− s) = 21−sπ−s cos(πs
2
)Γ(s)ζ(s),
we get µ(σ) = 1/2− σ for σ ≤ 0. The convexitiy will give
µ(σ) ≤ 1− σ2
for 0 ≤ σ ≤ 1. Lindelof hypothesis is telling you that µ(σ) = 0 for σ ≥ 1/2 andµ(σ) = 1/2− σ for σ ≤ 1/2.
The best subconvexity bound was proved by Bourgain in 2017:
µ(1/2) ≤ 13
84≈ 0.154.
Math 229x Notes 50
20 March 20, 2017
20.1 Introduction to sieves
Let me start with the concrete problem before going to the general theory. Wewant to estimate the number of primes π(x). There is a combinatorial way ofgetting primes. Starting from 2, you circle 2 and scar out all multiples of 2.Then circle 3 and scar out all multiples of 3. You can do this up to some z, andthis gives an upper bound of π(x). Formally, we define
π(x, z) = #n ≤ x : p - n for all p ≤ z.
Here are some obvious facts:
• π(x) ≤ π(z) + π(x, z)
• π(x) = π(√x) + π(x,
√x) = π(x,
√x) +O(
√x)
Let us define
Pz =∏p≤z
p, δ(m) =
1 m = 1,
0 m 6= 1.
Then
π(x, z) =∑n≤x
δ(gcd(n, Pz)).
There is the combinatorial sieve, that uses∑d|n
µ(d) = δ(n).
Then we have
π(x, z) =∑n≤x
∑d|(n,Pz)
µ(d) =∑d|Pz
µ(d)∑
n≤x,d|n
1 =∑d|Pz
µ(d)(xd
+O(1))
= x∑d|Pz
µ(d)
d+O
(∑d|Pz
1)
= x∏p≤z
(1− 1
p
)+O(2π(z)).
Proposition 20.1. π(x, z) = x∏p≤x
(1− 1
p
)+O(2π(z)).
How good is this? From Chebyshev’s bound we have∏p≤z
(1− 1
p
)=
c
log z+O
( 1
(log z)2
),
Math 229x Notes 51
and we also have 2π(z) ≤ 2z. So
π(x, z) =cx
log z+O
( x
(log z)2+ 2z
).
The best we can do for π is
π(x) ≤ O(log x) +cx
log log x+O
( x
(log log x)2
).
This is worse than Chebyshev, but it can be good for estimating π(x, log x).Maybe we can find another sieve that works well for other problems.
20.2 Selberg sieve for counting primes
Let λ1, λ2, . . . be any sequence of real numbers. Then trivially
δ(m) ≤(∑d|m
λd
)2
.
As we go along, we can impose conditions on (λj). For technical reasons, wewant the following conditions.
(S1) λ1 = 1.
(S2) λd = 0 for d > z or d not square-free.
(S3) |λd| ≤ 1 for all δ.
Under these conditions,
π(x, z) ≤∑n≤x
( ∑d|(n,Pz)
λd
)2
=∑n≤x
∑d1,d2|(n,Pz)
λd1λd2 =∑
d1,d2|Pz
λd1λd2∑n≤x,d1,d2|n
1.
The point is that (S2) tells us that d1, d2 just ranges up to z and square-free.Then
π(x, z) ≤∑
d1,d2≤z
bλd1λd2
( x
[d1, d2]+O(1)
)= x
( ∑d1,d2≤z
b λd1λd2[d1, d2]
)+O(z2),
where Σb means that the we are summing over square-free numbers. Now themain problem is finding λd so that the quadratic form is minimized. We willcall that quadratic form Q(λd).
Let us first diagonalize Q. We have
Q =∑
d1,d2≤z
2λd1λd2d1d2
(d1, d2) =∑
d1,d2≤z
bλd1λd2d1d2
∑k|(d1,d2)
φ(k)
=∑k≤z
b
φ(k)∑
d1,d2≤zk|d1,d2
bλd1d2
λd2d2
=∑k≤z
b
φ(k)α2k,
Math 229x Notes 52
where αk =∑d≤z,k|d
bλd/d. This is the diagonalization and the inverse trans-
form is given by λd = d∑d|k µ(k/d)αk for d, k ≤ z square-free. Note that (S1)
is equivalent to
1 =∑k≤z
b
µ(k)αk.
We want to minimize∑k≤z
bφ(k)α2
k under this condition.Lagrange’s multipliers tells us that for the optimal value ξ, we need
2φ(k)αk = µ(k)ξ.
Then αk = ξµ(k)/(2φ(k)). To find ξ, we use (S1) to get 1 =∑k≤z
b 12φ(k)ξ. This
finally leads us to the choice
αk =1∑
l≤zb 1φ(l)
µ(k)
φ(k).
The conditions (S1) and (S2) are satisfied automatically. We now check (S3).
We have, where we write V =∑k≤z
b1/φ(k),
λd = d∑d|k
αk =d
V
∑d|k,k≤z
bµ(k/d)µ(k)
φ(k)=dµ(d)
V
∑d|k,k≤z
b 1
φ(k)
=dµ(d)
φ(d)V
∑r≤z/d(d,r)=1
b 1
φ(r)=µ(d)
V
∑l|d
1
φ(l)
∑r≤z/d(d,r)=1
b 1
φ(r).
This shows that |λd| ≤ 1.Using this choice, we can compute Q(our choice) = 1/V . Then we get:
Proposition 20.2. π(x, z) ≤ x
V (z)+O(z2), where V (z) =
∑k≤z
b1/φ(k).
Furthermore, choosing z = x1/3 gives π(x) x/ log x.
Math 229x Notes 53
21 March 22, 2017
Last time we considered the problem of estimating
π(x, z) = #n ≤ z : gcd(n, Pz) = 1, Pz =∏p≤z
p.
This is a good upper bound on the prime counting function.
21.1 Selberg sieve
The basic principle for the Selberg sieve is
δ(m) ≤(∑d|m
λd
)2
,
where λd ∈ R and λ1 = 1. You can use more variables than just two. Maynardproved bounded gaps using a multidimensional Selberg sieve.
We consider a fixed sequence sequence of integers a1, a2, . . . and count
N(x, z) = #n ≤ x : (an, Pz) = 1.
We will need to use the data of
Nd(x) = #n ≤ x : d | an.
Last time we used an = n and Nd(x) = x/d+O(1). We are also going to assumethat there are arithmetic functions f(n), R(n), and a set of primes S, such that
(1) f and R are R>0-valued,
(2) f is multiplicative,
(3) f(p) > 1 for any p /∈ S,
(4) for gcd(d, S) = 1, |Nd(x)− x/f(d)| ≤ R(d),
(5) for gcd(d, S) 6= 1, Nd(x) ≤ R(d).
We are going to define g = f ∗ µ,
f(n) =∑d|n
g(d).
For p /∈ S, we have g(p) = f(p) − f(1) = f(p) − 1 > 0. Because g is alsomultiplicative, g(n) > 0 for gcd(n, S) = 1 with n square-free. Using the notation∑Sn =
∑(n,S)=1, we also define
U(z) =
S∑n≤z
b
1
g(n)≥ 1.
This is extremely hard to compute. It is useful to have the following estimate.
Math 229x Notes 54
Lemma 21.1. Let h be the completely multiplicative function defined by h(p) =f(p). Then
U(z) ≥S∑n≤z
1
h(n).
Proof. For n square-free with (n, S) = 1,
h(n)
g(n)=f(n)
g(n)=∏p|n
f(p)
f(p)− 1=∏p|n
(1 +
1
f(p)+
1
f(p)2+ · · ·
)=
∑p|k⇒p|n
1
h(k).
Then
U(z) =
s∑n≤z
b1
g(n)≥
S∑m≤z
1
h(m).
Theorem 21.2. With the previous notation and conditions relative to the se-quence a1, a2, . . ., we have
N(x, z) ≤ x
U(z)+
∑d1,d2≤z
b
R([d1, d2]) ≤ x∑Sn≤z
1h(n)
+∑
d1,d2≤z
R([d1, d2]).
Proof. We are going to pick λ1, λ2, . . . ∈ R satisfying
(1) λ1 = 1,
(2) λd = 0 if d > z or d is not square-free or (d, S) 6= 1,
(3) |λd| ≤ 1.
Then we have
N(x, z) ≤∑n≤x
( ∑d|(an,Pz)
λd
)2
=∑
d1,d2≤z
b
λd1λd2
( ∑d1,d2|ann≤x
1)
=∑
d1,d2≤z
b
λd1λd2N[d1,d2](x)
≤S∑
d1,d2≤z
b
λd1λd2x
f([d1, d2])+
∑d1,d2≤z
b
R([d1, d2])
= x
S∑d1,d2≤z
b
λ1λ2
f([d1, d2])+
∑d1,d2≤z
b
R([d1, d2]).
We now have to analyze the quadratic function. We have
Q =
S∑d1,d2≤z
b
λd1λd2f(d1)f(d2)
f((d1, d2))
=
S∑d1,d2≤z
b
λd1λd2f(d1)f(d2)
∑k|(d1,d2)
g(k) =
S∑k≤z
b
g(k)α2k,
Math 229x Notes 55
where
αk =
S∑k|d≤z
b
λdf(d)
, λd = f(d)
S∑d|k≤z
b
µ(k/d)αk.
Both α and λ are 0 unless the indices are at most z, square-free, and coprimeto S.
We are going to choose
αk =1
U(z)
µ(k)
g(k)
in the support. You can check that |λd| ≤ 1 is satisfied for this choice. Thisgives
Q(z) =
S∑k≤z
b
g(k)1
U(z)2
1
g(k)2=
1
U(z)2
S∑k≤z
b
1
g(k)=
1
U(z).
This finishes the proof.
Here is an interesting case you can apply this theorem. Consider an =n(n+ 2). Let
π2(x) = #p ≤ x : p+ 2 is prime.
This is an upper bound sieve, so we can’t prove that there are infinitely manyprime. You have π2(z) ≤ z +N(x, z) for any z.
Math 229x Notes 56
22 March 24, 2017
Today I want to discuss the application of counting twin primes.
22.1 Application: Counting twin primes
We count the know the size of
π2(x) = #p ⊆ x : p+ 2 is prime .
Heuristically, for n ≤ x the probability that n is prime is around 1/ log x, andsimilarly for n + 2. If the two events are “nearly independent”, then we canexpect
π2(x) x
(log x)2.
The lower bound seems out of reach, but we can prove the upper bound.
Theorem 22.1 (Selberg, after Brun). π2(x) x
(log x)2.
Brun got with an additional (log log x)2 factor, but this is enough to showthat the reciprocal converges.
Proof. We are going to apply Selberg’s sieve to the sequence an = n(n+2). Forthis choice,
π2(x) ≤ N(x, z) + π2(z) ≤ z +N(x, z).
We need estimates on Nd(x). Let me introduce
ρ(d) = #r mod d : r(r + 2) ≡ 0 (mod d).
Then we have ∣∣∣Nd(x)− ρ(d)
dx∣∣∣ ≤ ρ(d).
Also ρ(d) is multiplicative. So we can choose S = ∅, f(d) = d/ρ(d), andR(n) = ρ(n).
We can now apply Selberg’s sieve. We get
N(x, z) ≤ x
U(z)+
∑d1,d2≤z
b
R([d1, d2]), where U(z) =∑k≤z
b 1
g(z)≥∑n≤z
1
h(n).
For n square-free, we claim that ρ(n)ε nε. This is because
ρ(n) =∏p|n
ρ(p) ≤ d(n).
Therefore ∑d1,d2≤z
b
R([d1, d2])ε z2+ε.
Math 229x Notes 57
For U(z), we get1
h(n)=
2Ω(n)−ν2(n)
n,
where Ω(n) is the number of prime factors of n counting repetitions. Then
U(z) ≥∑n≤z
1
h(n)=∑
2-n≤z
2Ω(n)
n≥∑
2-n≤z
d(n)
n
=( ∑
2-n≤z
d(n))1
z+
∫ z
1
( ∑2-n≤t
d(n))dtt2∫ z
1
log t
tdt (log z)2.
ThereforeN(x, z)ε
x
(log z)2+ z2+ε.
Take ε = 1 and z = x1/4.
Corollary 22.2 (Brun’s theorem, 1919). The (possibly finite) series(1
3+
1
5
)+(1
5+
1
7
)+( 1
11+
1
13
)+ · · ·
is convergent.
Math 229x Notes 58
23 March 27, 2017
There will be no lecture this Wednesday.The purpose of the large sieve is to give an upper bound on the size of a
set with restrictions on the reduction modulo p. For a set A ⊆ [1, x], we get aestimate on #A given #A mod p. It is useful when we remove very few residueclasses.
Problem. How large is the least n such that n mod p is not a quadratic residue?
A trivial bound is p/2, because there are p/2 quadratic residues. A non-trivial bound is
√p log p, because the quadratic character is a character and
consecutive sums of a character cannot exceed this. (Polya–Vinogradov in-equality).
Conjecture 23.1. The least non-quadratic number is ε pε.
23.1 Analytic large sieve inequality
Proposition 23.2. Let f : R→ C be C1 and 1-periodic. Then for any z ≥ 1,∑d≤z
∑1≤a≤d(a,d)=1
∣∣∣f(ad
)∣∣∣ ≤ z2
∫ 1
0
|f(t)|dt+
∫ 1
0
|f ′(t)|dt.
Proof. Given any such a/d and t > a/d,
f(t)− f(ad
)=
∫ x
a/d
f ′(t)dx.
Then ∣∣∣f(ad
)∣∣∣ ≤ |f(t)|+∫ t
a/d
|f ′(x)|dx.
Let δ = z−2 and I(a/d) = [a/d, a/d + δ). These intervals are disjoint fora/d, and so integrating over t gives
δ∣∣∣f(a
d
)∣∣∣ ≤ ∫I(a/d)
|f(t)|dt+
∫I(a/d)
∫ t
a/d
|f ′(x)|dx
≤∫I(a/d)
|f(t)|dt+ δ
∫I(a/d)
|f ′(x)|dx.
Adding over d/a gives the desired inequality.
Theorem 23.3 (Analytic large sieve inequality). Let a1, a2, . . . be a sequencein C. For x ≥ 1 define
Sx(t) =∑n≤x
ane(nt), e(α) = e2πiα.
Math 229x Notes 59
Then for z ≥ 1, ∑d≤z
∑1≤a≤d(a,d)=1
∣∣∣Sx(ad
)∣∣∣2 ≤ (z2 + 4πx)∑n≤x
|an|2.
Proof. We use the previous proposition with Sx(t)2. The left hand side is per-fect. Then
(LHS) ≤ z2
∫ 1
0
|Sx(t)|2dt+ 2
∫ 1
0
|S′x(t)Sx(t)|dt.
The first term is just z2∑n≤x|an|2. For the second term, applying Cauchy gives
us∫ 1
0
|S′x(t)Sx(t)|dt ≤(∫ 1
0
|Sα(t)|2dt)1/2(∫ 1
0
|S′α(t)|2dt)1/2
≤ 4π2x2∑n≤x
|an|2.
This has a huge number of applications. For instance, you can show thatthe average error term for the number of primes in arithmetic progressions isvery small.
23.2 Ramanujan sums
Definition 23.4. For d a positive integer and n ∈ Z, the Ramanujan sum isgiven by
cd(n) =∑
1≤a≤d(a,d)=1
e(nad
).
Lemma 23.5. cd(n) is d-periodic on n.
Lemma 23.6. For n fixed, cd(n) is multiplicative on d.
Lemma 23.7. cd(n) =∑r|(n,d)
µ(d/r)r.
Proof. We have
cd(n) =∑
1≤a≤d
e(nad
) ∑k|(a,d)
µ(k) =∑k|d
µ(k)∑
1≤b≤d/k
e(nkbd
)=∑k|d
µ(k)∑
1≤b≤d/k
e( nbd/k
)=∑r|d
µ(d/r)∑r|n
r.
Corollary 23.8. For d ≥ 1 an integer,
(1) cd(n) ∈ Z.
(2) if (d, n) = 1 then cd(n) = µ(d).
Math 229x Notes 60
These Ramanujan sums can be thought of as some kind of a base for Fourierseries. It is periodic on n and it is multiplicative on d. Here are some examplesof Ramanujan–Fourier series.
σ1(n) =π2
6n
∞∑d=1
1
d2cd(n).
Define φ2(n) = n2∏p|n(1− p−2). Then
φ(n) =π2
6n
∞∑d=1
µ(d)
φ2(d)cd(n).
The formula
0 =
∞∑d=1
1
dcd(n)
is equivalent to the prime number theorem. Also Hardy proved
Λ(n) =n
φ(n)
∞∑d=1
µ(d)
φ(d)cd(n).
Math 229x Notes 61
24 March 31, 2017
Last time we discussed the analytic version of the large sieve.
24.1 Arithmetic large sieve
For each prime p, let Ωp ⊆ Z/pZ be a set of “forbidden” residue classes. Let uswrite ω(p) = #Ωp. For x, z ≥ 1, define
S(x, z) = #n ≤ x : for every p ≤ z, n (mod p) /∈ Ωp.
You can even set Ωp = ∅ if you want. But you wouldn’t want to set Ωp = Z/pZ.The question is how large S(x, z) is.
Theorem 24.1. Write
L(z) =∑d≤z
b∏p|d
ω(p)
p− ω(p).
Then
S(x, z) ≤ z2 + 4πx
L(z).
Proof. Extend ω multiplicatively. By the Chinese remainder theorem, we geta forbidden set Ωd ⊆ Z/dZ for each square-free d, such that Ωd/p = Ωp for allp | d and #Ωd = ω(d).
Let n ≤ x such that for every p ≤ z, n (mod p) /∈ Ωp. (We call this condition“∗”.) For d ≤ z square-free,
µ(d) = cd(n− u) =∑
1≤a≤d(a,d)=1
e(a(n− u)
d
),
provided that (u, d) = 1. This is satisfied if u ∈ Ωd ⊆ 1, . . . , d.Add this over n under (∗) and u ∈ Ωd:
µ(d)S(x, z)ω(d) =∑
1≤a≤d(a,d)=1
∑u∈Ωd
e(−au
d
)∑n≤x
(∗)e(aud
).
By Cauchy–Schwarz,
S(x, z)2ω(d)2 ≤( ∑
1≤a≤d(a,d)=1
∣∣∣∣ ∑u∈Ωd
e(−au
d
)∣∣∣∣2)( ∑1≤a≤d(a,d)=1
∣∣∣∣∑n≤x
(∗)e(and
)∣∣∣∣2) = T · S.
Math 229x Notes 62
We can simplify the sums as
T =∑
1≤a≤d(a,d)=1
∑u,v∈Ωd
e(a(u− v)
d
)=
∑u,v∈Ωd
∑1≤a≤d(a,d)=1
e(a(u− v)
d
)
=∑
u,v∈Ωd
cd(u− v) =∑
u,v∈Ωd
∑k|(d,u−v)
µ(d/k)k =∑k|d
kµ(d/k)∑u∈Ωd
∑v∈Ωdk|u−v
1
=∑k|d
kµ(d/k)ω(d)ω(d/k) = dω(d)∑r|d
µ(r)ω(r)
r= dω(d)
∏p|d
(1− ω(p)
p
).
Therefore adding over d ≤ z square-free, we end up with
S(x, z)2L(z) = S(x, z)2∑d≤z
b∏p|d
p− ω(p)
ω(p)≤∑d≤z
b ∑1≤a≤d(a,d)=1
∣∣∣∣∑n≤x
(∗)e(and
)∣∣∣∣2.The right hand side looks like we can apply the analytic large sieve inequality
with
an =
1 if (∗) holds,
0 otherwise.
The inequality gives us
S(x, z)2L(z) ≤ (z2 + 4πx)∑n≤x
|an|2 = (z2 + 4πx)S(x, z).
This finishes the proof.
Example 24.2. Here is a useless example. Let us count n ≤ x, with Ωp = ∅and L(z) = 1. Then
S(x, z) ≤ z2 + 4πx
1.
Taking z = 0 gives S(x, z) ≤ 4πx. Actually you can replace 4πx in all the largesieves, but this takes more work.
Theorem 24.3 (Linnik). Let ε > 0. As x→∞,
#p ≤ x : np > nε ε 1,
where np ≥ 1 is the least quadratic non-residue mod p.
Proof. Define
Ωp =
∅ if there exists n ≤ xε such that (np ) = −1,
r mod p : ( rp ) = −1 otherwise.
Math 229x Notes 63
Then ω(p) = 0 or (p − 1)/2 depending on the condition. Take z =√x. Then
we get
S(x, z) ≤ 14x
L(√x),
and
L(√x) =
∑d≤z
b∏p|d
ω(p)
p− ω(p)≥ 1
2
∑p≤z exceptional
ω(p)6=0
(1− 1
p
)>
1
4
∑p≤z exc.ω(p)6=0
1.
Now let us give a lower bound on S(x, z). We claim that if n ≤ x is xε-smooth, then n is counted in S(x, z). Also, you can prove in an elementary waythat
#n ≤ x : n is xε-smooth ε x.
This shows that L(√x)ε 1.
Conjecture 24.4 (Strong conjecture). np (log p)2.
Math 229x Notes 64
25 April 3, 2017
We are going to start a new topic. We can discuss sieve for a whole semester,but this is an introductory course.
25.1 Additive problems
Let N = 0, 1, . . . and A ⊆ N. Define A [n] = A ∩ [1, n]. We are settingA [0] = ∅ even if 0 ∈ A .
Suppose we are given sets A1, . . . ,Ak ⊆ N. The question is, can every n ∈ Nbe written as n = a1 + · · · + ak with ai ∈ Ai? This is sometimes too strong,so maybe we ask this for n 1. Or maybe we can give some congruenceconditions.
Example 25.1. Here are some examples.
(1) Lagrange’s theorem: Every n ≥ 0 is the some of 4 squares.
(2) Waring’s problem: Let k ≥ 2. Is every n ≥ 1 the sum of at most G(k)k-th powers for a uniform G(k)?
(2′) Get the best G(k), up to finitely many n; we call the bound g(k).
(3) Goldbach conjecture: Is every even integer 1 the sum of two prime?
(4) Ternary Goldbach: Is every odd integer 1 the sum of three primes?
(2) was solved by Hilbert in 1909. Hardy and Littlewood worked on g(k).(3) is open so far. Yitang Zhang’s method gives something on linear relationsbetween primes, but it is not strong enough. (4) was solved by Vinogradov.What Helfgott proved in 2012 is that (4) is true for every odd integer ≥ 7.
In the remaining part of this course, we will study these problems. We willconsider two methods:
(1) Schnirelmann’s density method (with a sieve)
(2) The circle method (invented by Ramanujan; advertised by Hardy andLittlewood; refined by Vinogradov)
We are not going to see this in a very abstract setting, because it is not reallyhelpful. The motivating problems will be:
(1) Every n 1 is the sum of a bounded number of primes.
(2) Waring’s problem.
25.2 Schnirelmann’s density
Here is the heuristic. If A ,B ⊆ N has “positive density”, then we expect thatA + B has larger density. But this fails in general. If A = B = 2N this doesnot seem to work. But maybe the even numbers is not 1/2.
Math 229x Notes 65
Definition 25.2. Let A ⊆ N = 0, 1, . . .. We define the Schnirelmanndensity of A as
δ(A ) = infn≥1
#A [n]
n.
Under this definition δ(N) = 1 and δ(2N) = 0. We also have δ(2N+1) = 1/2.We have δ(primes) = 0 and δ(k-th powers) = 0 if k ≥ 2.
Lemma 25.3. Let A ,B ⊆ N and assume that 0 ∈ A ,B. Let n ≥ 1. If
#A [n] + #B[n] ≥ n
then n ∈ A + B.
Proof. If n ∈ A + B, write n = n + 0. If not, then A [n] = A [n − 1] andB[n] = B[n − 1]. So A [n] and n −B[n] are both contained in [1, n − 1], andthus have an intersection.
Corollary 25.4. Let A ⊆ N with 0 ∈ A and δ(A ) ≥ 1/2. Then A + A = N.
Proof. For n ≥ 1,
#A [n] + #A [n] ≥ δ(A)n+ δ(A)n ≥ n.
Here are some useful remarks we will use without thinking.
(1) 1 /∈ A implies δ(A) = 0.
(2) For all n ≥ 1, #A [n] ≥ δ(A)n.
(3) If #A [n] ≥ cn for all n ≥ 1 then δ(A) ≥ c.(4) δ(A ) always exists and is in [0, 1].
Lemma 25.5. Let A ,B ⊆ N with 0 ∈ A ,B. Then
δ(A + B) ≥ δ(A ) + δ(B)− δ(A )δ(B).
Proof. Let n ≥ 1 and r = #A [n]. We can assume δ(A ), δ(B) > 0, and so1 ∈ A ,B. Let us write
A [n] = 1 = a1 < a2 < . . . < ar.
Let gj = aj+1 − aj − 1 for j = 1, 2, . . . , r − 1. We know A ⊆ A + B. We nowwant to focus on what else is there.
We get at least #B[gj ] elements between aj and aj+1, and B[n− ar] at theend of [1, n]. Now we count
#(A + B)[n] ≥ r +
r−1∑j=1
#B[gj ] + #B[n− ar]
≥ r +
r−1∑j=1
δ(B)gj + δ(B)(n− ar)
= n(δ(A ) + δ(B)− δ(A )δ(B)).
This finishes the proof.
Math 229x Notes 66
26 April 5, 2017
Last time we defined Schnirelmann’s density as
δ(A ) = infn≥1
#A [n]
n, A [n] = A ∩ [1, n].
Let us denote h ·A = A + · · ·+ A where there are h sums.
Corollary 26.1. If δ(A ) ≥ 1/2 and 0 ∈ A then 2 ·A = N.
Lemma 26.2. Let A ,B ⊆ N with 0 ∈ A ∩B. Then
δ(A + B) ≥ δ(A ) + δ(B)− δ(A )δ(B).
Corollary 26.3. Let A1, . . . ,An ⊆ N with 0 ∈ Ai for all i. Then
δ(A1 + · · ·+ An) ≥ 1−n∏i=1
(1− δ(Aj)).
Proof. This follows from the lemma by induction.
Definition 26.4. A set A ⊆ N is a basis of finite order if there exists an hsuch that h ·A = N.
Theorem 26.5. Let A ⊆ N. Suppose 0 ∈ A . If δ(A ) > 0. Then A is a basisof finite order.
Proof. Use Corollary 26.1 and Corollary 26.3.
26.1 Using Schnirelmann’s density
In the case we are interested in, like Waring’s problem or the Goldbach problem,the set has density zero.
Given a set A ⊆ N, we would first need to define A ∗ = A ∪0, 1 and workwith A ∗ instead. Adding 0 in is not much of a problem, because we are lookingat at most h sums, rather than exactly h sums. Adding 1 in is something weneed to take are later on.
Lemma 26.6. Let A ⊆ N. We have δ(A ) > 0 if and only if
(i) 1 ∈ A ,
(ii) lim inf #A [n]/n > 0, i.e, there exists a c > 0 such that #A [x] ≥ cx forx 1.
In most cases, A is too thin to have positive natural density. But maybethe set 2 ·A is not too thin. Here is a way to check this.
Lemma 26.7 (2nd moment trick). Let A ⊆ N. Define
rA (n) = (a, b) ∈ A 2 : a+ b = n.
For x ≥ 1, ( ∑1≤x≤n
rA (n)
)2
≤ (#(2 ·A )[x])∑
1≤n≤x
rA (n)2.
Math 229x Notes 67
26.2 Sums of primes (weak Goldbach)
Let P be the set of primes. Here are some issues: 0 /∈ P, 1 /∈ P, and#P[x] = o(x). The plan is to work with P∗ instead. Then inflate P to 2 ·P.
Let us write γ(n) = rP(n).
Lemma 26.8. For x 1, ∑n≤x
γ(n) x2
(log x)2.
Proof. Just look at the primes p, q ≤ x/2 so that p + q ≤ x. Then applyChebyshev.
In the assignment, using sieves, you showed
Lemma 26.9. γ(n) n
(log n)2
∏p|n
(1 +
1
p
).
Lemma 26.10.∑n≤x
γ(x)2 x3
(log x)4.
Proof. By the previous lemma,
∑n≤x
γ(n)2 x2
(log x)4
∑n≤x
(∏p|n
(1 +
1
p
))2
.
We have∑n≤x
(∏p|n
(1 +
1
p
))2
=∑n≤x
(∑d|n
1
d
)2
=∑n≤x
∑d1,d2|n
1
d1d2
=∑
d1,d2≤x
1
d1d2
⌊ x
[d1, d2]
⌋≤ x
∑d1,d2≤x
1
d1d2
√d1d2
≤ x · ζ(3/2)2.
This proves the proof.
Theorem 26.11. For x 1, #(2 ·P)[x] x.
Proof. This follows from the second moment lemma.
Theorem 26.12. There exists an h such that every integer n ≥ 2 is the sumof at most h primes.
Proof. The previous theorem and Schnirelmann’s theory show that there existsa k such that k ·P∗ = N. We need to get rid of 1.
So given n we can write
n = p1 + · · ·+ pr + 1 + · · ·+ 1 = p1 + · · ·+ pr + `,
Math 229x Notes 68
where r + ` ≤ k.If ` = 0, then we are good. If ` > 1, then we are also good, because then `
can be written as a sum of 2 and 3.What if ` = 1? Look at n − 2. Then n − 2 is good and so n = (n − 2) + 2
or n− 2 uses one 1 and so n = (n− 3) + 3.
Math 229x Notes 69
27 April 7, 2017
We are going to start on the last section of this course, which is the circlemethod.
27.1 Introduction to the circle method
Let A ⊆ N, and consider s ≥ 1 an integer. We can consider the function
f(z) = fA (z) =∑n∈A
zn.
We can work with the formal power series, but we see that this is analytic onD0 = z ∈ C : |z| < 1.
Define
rs(n) = rs,A (n) = #a ∈ A s :
s∑i=1
ai = n.
The key observation is that
f(z)s =∑n≥0
rs(n)zn.
The whole point of the circle method is to recover rs(n) from f(z)s by analyticmeans. For γ ∈ D0 a contour around 0, we have
rs(N) =1
2πi
∫γ
f(z)s
zN+1dz.
Now we have an analytic thing on the right had side and we hope this can beapproximated. The idea of Ramanujan was
(1) approximate near “heave” singularities on S1,
(2) bound the integral away from those singularities,
(3) choose γ in such a that (1) and (2) work.
The main complication is choosing γ. We want to avoid singularities but stayclose to the circle on other parts. In fact, Rademacher used a very complicatedfractal contour to give an exact formula for the partition function. Choosingthe contour really depends on how skillful you are.
This was the Hardy–Littlewood version of Ramanujan’s circle method. Vino-gradov make technical modifications and proved the following theorem.
Theorem 27.1. Every N 1, odd, can be written as sum of 3 primes (andan asymptotic formula for r3,P(N) can be found).
Math 229x Notes 70
What is this technical modification? Let N be fixed. The key observationis that to compute (or approximate) rs(N), we don’t need the whole seriesf(z) =
∑n∈A zn. Instead, we only need fN (z) =
∑n∈A ,n≤N z
n. This is apolynomial and hence entire. So again,
rs(N) =1
2πi
∫γ
fN (z)s
zN+1dz. (1)
Now γ = S1 is allowed, and we can let z = e(α). Then
rs(N) =
∫ 1
0
fN (α)se(−Nα)dα, where FN (α) =∑
n∈A ,n≤N
e(nα). (2)
Now the question is how to estimate the right hand side of (2). For α near arational number of small denominator we cannot expect cancellation in FN (α).In this case, we are going to evaluate (estimate with main term) and otherwisewe are going to bound.
Here is a standard process. Fix some N . (Still, all implicit constants in thearguments should be uniform on N .) We are going to choose two parametersQ = Q(N) a positive integer and δ = δ(N) > 0 a small real number.
Definition 27.2. For 1 ≤ q ≤ Q and 0 ≤ a ≤ q with (a, q) = 1, define
M(a, q) =α ∈ [0, 1] :
∣∣∣α− a
q
∣∣∣ ≤ δ.Then define the major arcs as the union
M =⋃
1≤q≤Q
⋃0≤a≤q(a,q)=1
M(a, q).
The minor arcs is m = [0, 1] \M.
Often, (in practice, always) we choose δ so small that the major arcs aredisjoint, and it is also very common that µ(M)→ 0 as N →∞.
The first step is to bound∫m
using bounds for exponential sums. Recall that
fN (α) =∑
n∈A ,n≤N
e(nα).
We need to estimate the s-th moment of this function. You might have bettercancellation when looking at the s-th moment, and this is the motivation forusing stuff like Weyl’s inequality, Hua’s inequality, or Vinogradov’s inequality.
Then there is the question of how to evaluate the sum on the major arcs.Near α = a/q, write the integrand as
(arithmetic sum) · (analytic sum) + (error),
Math 229x Notes 71
integrate, add over a, q and get
S(N) · (“analytic” function)(N) + (error).
Here S(N) is called the singular series, and this is going to detect the localobstructions. For instance, there aren’t many ways to write an even number asa sum of three odd primes.
After all this, you need to get a clean formula for the analytic part of themain term, and bound S(N) away from 0 (if possible under local conditions).
This is the outline of the method. Next week, we are going to try to getbounds on exponential sums. Then we are going to jump into Waring’s problem.
Math 229x Notes 72
28 April 10, 2017
28.1 Bounds for exponential sums
We write x = bxc+ x, and we write ‖x‖ = minn∈Z|x− n|.
Lemma 28.1 (Dirichlet). Let α ∈ R and x ≥ 1 real. Then there exists arational number a/q ∈ Q with (a, q) = 1, 1 ≤ q ≤ x, such that∣∣∣α− a
q
∣∣∣ ≤ 1
qx≤ 1
q2.
Proof. This is a classical application of the box principle.
Lemma 28.2 (“Technical bound”). Let x, y, α ∈ R. Assume x, y ≥ 1. Considera rational approximation∣∣∣α− a
q
∣∣∣ ≤ 1
q2, (a, q) = 1, 1 ≤ q.
Then ∑n≤x
minxyn,
1
‖αn‖
xy
(1
q+
1
y+
q
xy
)log(2xq).
Proof. We write n = jq + r with j ≥ 0 and 1 ≤ r ≤ q. Then the left hand sideis bound by
(LHS) ≤∑
0≤j≤x/q
q∑r=1
min xy
jq + r,
1
‖α(jq + r)‖
. (0)
Write α− a/q = θ/q2 with |θ| ≤ 1. Let us also focus on yj = bαjq2c. Then
yj + ar
q+αjq2q
+θr
q2=αjq2 + ar
q+(α− a
q
)r = α(jq + r). (1)
Now we can start to estimate ‖α(qj + r)‖.Fix a j ≥ 0. Then for all r with at most 10 exceptions,
‖α(jq + r)‖ ≥∥∥∥yj + ar
q
∥∥∥− 2
q≥exc
1
2
∥∥∥yj + ar
q
∥∥∥ > 0. (2)
For those exceptions, we will want that qj + r q(j + 1) so that we canestimate the other term easily. This is true if j > 0 and also if j = 0 andr > q/2. In the other case (j = 0 and 1 ≤ r ≤ q) we need something betterthan (2) without exceptions. Indeed, we have
‖α(jq + r)‖ ≥∥∥∥arq
∥∥∥− 1
2q≥ 1
2
∥∥∥arq
∥∥∥ > 0 (3)
without exceptions in that case.
Math 229x Notes 73
So by (0),
(LHS) ≤∑
0≤j≤x/q
q∑r=1
min xy
jq + r,
1
‖α(jq + r)‖
≤
∑1≤r≤q/2
2
‖ar/q‖+
∑0≤j≤x/q
∑1≤r≤q
unless j=0q/2<r≤q
min xy
jq + r,
1
‖α(jq + r)‖
q log(2q) +∑
0≤j≤x/q
(xy
q(j + 1)+
∑1≤r≤qq-yj+ar
1
‖(yj + ar)/q‖
)
q log(2q) +xy
qlog(2x) +
x
qq log(2q) log(2xq)xy
( q
xy+
1
q+
1
y
).
This finishes the proof.
The reason we are interested in this is that this shows up in the estimation.When we deal we exponential sums like∑
n∈Ie(αp(n))
for α ∈ R and p(x) ∈ R[x]. Weyl’s differencing method relies on the fact thatthe forward differences of p have smaller degree.
For a real number βj ∈ R and φ : R→ R, we define the difference operators
∆1(φ(x);β) = φ(x+ β)− φ(x),
∆j+1(φ(x);β1, . . . , βj+1) = ∆1(∆j(φ(x);β1, . . . , βj);βj+1).
Example 28.3. For φ(x) = x2,
∆1(φ(x); 1) = (x+ 1)2 − x2 = 2x+ 1, ∆2(φ(x); 1, 1) = 2.
Lemma 28.4 (Weyl’s differencing bound). Let Q ∈ Z≥1, φ : R→ R. Let
T (φ) =
Q∑x=1
e(φ(x)).
Then for all j ≥ 1,
|T (φ)|2j
≤ (2Q)2j−j−1
∣∣∣∣ ∑|h1|<Q
· · ·∑|hj |<Q
Tj
∣∣∣∣,where
Tj =∑x∈Ij
e(∆j(φ(x);h1, . . . , hj))
and Ij = Ij(h1, . . . , hj) are intervals (possibly empty) satisfying
I1(h1) ⊆ [1, Q], Ij(h1, . . . , hj) ⊆ Ij−1(h1, . . . , hj−1).
Math 229x Notes 74
Proof. We use induction on j. In the case j ≥ 1, we have
|T (φ)|2 = T (φ)T (φ) =
Q∑x=1
Q−x∑h1=1−x
e(φ(x+ h1)− φ(x))
=
Q∑x=1
Q−x∑h1=1−x
e(∆1(x)) =
Q−1∑h1=1−Q
∑x∈I1(h1)
e(∆1(x)),
where I1(h1) = [1, Q] ∩ [1− h1, Q− h1].
Next time we will do the inductive step.
Math 229x Notes 75
29 April 12, 2017
We were proving Weyl’s differencing bound.
Lemma 29.1 (Weyl’s differencing bound). Let Q ∈ Z≥1 and φ : R→ R. Let
T (φ) =∑n≤Q
e(φ(n)).
Then for all j ≥ 1,
|T (φ)|2j
≤ (2Q)2j−j−1∑|h1|<Q
· · ·∑|hj |<Q
Tj ,
where
Tj =∑
x∈Ij(h)
e(∆j(φ(x);h))
and Ij(h) are intervals satisfying I1(h1) ⊆ [1, Q] and Ij(h) ⊆ Ij−1(h1, . . . , hj−1).
Proof. Induct on j ≥ 1. We did the case j = 1 last time. Assume case j forsome j ≥ 1. Then
|T (φ)|2 ≤∣∣∣∣(2Q)2j−j−1
∑|h1|<Q
· · ·∑|hj |<Q
Tj
∣∣∣∣2 ≤ (2Q)2j+1−2j−2(2Q)j∑|hi|<Q
|Tj |2.
We note that
|Tj |2 =∑|h|<Q
∑x∈Ij+1(h1,...,hj ,h)
e(∆j(φ(x+ h);h)−∆j(φ(x);h)),
withIj+1(h, h) = Ij(h) ∩ (Ij(h)− h).
So we get the inequality.
29.1 Weyl’s inequality
Theorem 29.2 (Weyl’s inequality). Let φ(x) = αxk+α1xk−1 + · · ·+αk ∈ R[x]
with k ≥ 2. Consider a rational approximation |α − a/q| ≤ q−2 with (a, q) = 1and q ≥ 1. For Q ≥ 1 an integer, let
T (φ) =
Q∑n=1
e(φ(n)).
Then for every ε > 0,
|T (φ)| k,ε Q1+ε(1
q+
1
Q+
q
Qk
)1/2k−1
.
Math 229x Notes 76
Proof. Lemma 29.1 with j = k − 1 gives
|T (φ)|2k−1
≤ (2Q)2k−1−k∑|h1|<Q
· · ·∑
|hk−1|<Q
∑n∈Ik−1(h)
e(h1 · · ·hk−1P (n;h),
where
P (x;h) = k!α(x+
1
2(h1 + · · ·+ hk−1)
)+ (k − 1)!α1.
Since Ik−1(h) ⊆ [1, Q], there are k sums with range of length ≤ 2Q each.So the trivial bound will give k Q
k terms. We need to divide cases, when thetrivial bound is correct, namely when the thing in e is equal to zero, and whenit is not zero, in which we can expect some saving.
The contribution for the sum of terms with h1 · · ·hk−1p(n;h) = 0 is k
Qk−1, because either one of hi is zero or p(n;h) = 0 has at most one solution.Now we estimate the remaining part of the sum. If we denote dt(r) =
#(h1, . . . , ht) : h1 · · ·ht = r,∣∣∣∣ ∑|h1|<Qh1 6=0
· · ·∑
|hk−1|<Qhk−1 6=0
∑n∈Ik−1(h)
e(h1 · · ·hk−1P (n;h))
∣∣∣∣k
k!Qk−1∑r=1
dk−1(r) max1≤R≤Q
∣∣∣ R∑n=1
e(αrn)∣∣∣ε,k Q
ε
k!Qk−1∑r=1
minQ,
1
‖αr‖
.
For r ≤ k!Qk−1, we have Q ≤ k!Qk−1Q/r. This allows us to use the technicalLemma 28.2 with X = k!Qk−1 and Y = Q. This gives the bound
Qεk!Qk−1∑r=1
mink!Qk−1Q
r,
1
‖αr‖
k,ε Q
2εQk(1
q+
1
Q+
q
Qk
).
Therefore
‖T (φ)‖2k−1
ε,k Q2k−1−k
(Qk−1+Qk+ε
(1
q+
1
Q+q
Qk
)) Q2k−1+ε
(1
q+
1
Q+q
Qk
).
This finishes the proof.
Math 229x Notes 77
30 April 14, 2017
30.1 Hua’s lemma
For Q ≥ 1 an integer, k ≥ 2 a fixed integer, we define
f(α) = fQ,k(α) =
Q∑n=1
e(αnk).
Theorem 30.1 (Hua’s lemma). Let 1 ≤ j ≤ k, with the previous notation.Then for any ε > 0, ∫ 1
0
|f(α)|2j
dαε Q2j−j+ε.
Proof. We use induction. For j = 1, we have∫ 1
0
|f(α)|2dα = Q.
Now the inductive step. Assume it for some 1 ≤ j ≤ k − 1 and we prove it forj + 1. By Weyl’s differencing lemma (Lemma 29.1),
|f(α)|2j
≤ (2Q)2j−j−1∑|h1|<Q
· · ·∑|hj |<Q
∑n∈Ij(h)
e(αh1 · · ·hjp(n;h)),
where p(x;h) ∈ Z[x] for any h ∈ Zj and deg p(x;h) = k− j. It’s now importantthat the right hand side is a positive real number, because it came from thesquare modulus of some complex number.
We have to play the same game, but more carefully. For r ∈ Z, let
cr = #
(n, h) ∈ Z1+j :
|hi| < Q for all i and n ∈ Ij(h)h1 · · ·hrp(n;h) = r
.
So we have
|f(α)|2j
≤ (2Q)2j−j−1∑r
cre(αr).
Last time we had the estimates c0 k Qj and for r 6= 0, cr k,ε Q
ε.
But let us analyze |f(α)|2j in a different (trivial) way:
|f(α)|2j−1
= f(α)2j−1
f(−α)2j−1
=∑r
bre(−αr),
where
br = #
(x, y) ∈ Z2j−1+2j−1
:0 < xi, yi ≤ Q for all i
r =∑2j−1
i=1 xki −∑2j−1
i=1 yki
.
Math 229x Notes 78
Finally,∫ 1
0
|f(α)|2j+1
=
∫ 1
0
|f(α)|2j∑
r
bre(−αr)dα
= (2Q)2j−j−1
∫ 1
0
(∑s
cse(αs)
)(∑r
bre(−αr))dα
= (2Q)2j−j−1∑r
brcr = (2Q)2j−j−1
(b0c0 +
∑r 6=0
crbr
).
About the br, we know that∑r
br|f(0)|2j
= Q2j ,
and
b0 =
∫ 1
0
∑r
bre(αr)dα =
∫ 1
0
|f(α)|2j
dαk,ε Q2j−j+ε
by induction. So∫ 1
0
|f(α)|2j+1
dαk,ε Q2j−j−1(Q2j−j+εQj +QεQ2j ).
This finishes the proof.
30.2 Setup for Waring’s problem
Let k ≥ 2 be an integer. This is the exponent for Waring’s problem. Let s bea positive integer, which is the number of k-th powers we add. We will needs > 2k. Then N k,s 1 is the integer that we want to write as xk1 + · · · + xks .
Let ν = 0.01 and δ = δ(k) > 0 such that δ k 1. Define P = b k√Nc and we are
going to use the exponential sum
f(α) =
P∑n=1
e(αnk).
The number that we care about is
r(N) = rk,s(N) = #n ∈ Zs≥1 : N = nk1 + · · ·+ nks.
The circle method is based on the equality
r(N) =
∫ 1
0
f(α)se(−αN)dα.
But then there is half an interval near 0 and the other half near 1. This isannoying and so we will shift it a little bit.
Math 229x Notes 79
The major arcs are, for 1 ≤ a ≤ q ≤ P ν with (a, q) = 1,
M(a, q) =α ∈ R :
∣∣∣α− a
q
∣∣∣ < 1
P k−ν
, M =
⋃M(a, q).
Then
M ⊆ U =( 1
P k−ν, 1 +
1
P k−ν
]and it can be checked that M(a, q) are disjoint. The minor arcs are m = U \M.
Math 229x Notes 80
31 April 17, 2017
Today we are going to deal with the minor arcs. We keep our previous notations
31.1 Minor arcs
Proposition 31.1 (Bound on m). If s ≥ 2k + 1, then∫m
|f(α)|sdαk,s Ns/k−1−δ.
Note that because f(α) is the sum of P exponentials, the trivial bound is P s Ns/k.
Proof. We start with Hua’s lemma. We have∫m
|f(α)|sdα ≤(
supα∈m|f(α)|
)s−2k∫ 1
0
|f(α)|2k
dα
k,ε
(supα∈m|f(α)|
)s−2k
P 2k−k+ε ≤(
supα∈m|f(α)|
)s−2k
N2k/k−1+ε.
Now we estimate the supremum. By Dirichlet, there exists (a, q) = 1 suchthat 1 ≤ q ≤ pk−ν such that ∣∣∣α− a
q
∣∣∣ ≤ 1
qP k−ν.
Because α ∈ m ⊆ U , we have 1 ≤ a ≤ q. But since α /∈ M, we get q > P ν .Weyl’s inequality gives
supα∈m|f(α)| k,ε P
1+ε(1
q+
1
P+
q
P k
)21−k
P 1+ε−ν/2k−1
≤ N1/k+ε−ν/k2n−1
.
Then ∫m
|f(α)|sdαk,s,ε Nsk+2ε−1− ν
k2k−1 .
31.2 Major arcs
The goal is to analyze f(α) for α ∈ M(a, q) ⊆ M. Here we really need anasymptotic expression. To understand it better, we introduce two functions:
• analytic local factor
v(β) =
N∑n=1
1
kn
1k−1e(βn), β ∈ R,
• arithmetic local factor
S(a, q) =
q∑n=1
e(aqnk).
Math 229x Notes 81
Lemma 31.2 (local factorization). Let 1 ≤ a ≤ q ≤ P ν with (a, q) = 1. Letα ∈M(a, q). Then
f(α) =1
qS(a, q)ν(β) +Ok(P 2ν),
where β = α− a/q.
Proof. For t ≥ 1 let T = tk. The partial sums are∑n≤T
1
nn1/k−1 =
∫ T
1
1
kx1/k−1dx+O(1) = t+O(1).
Also ∑n≤t
e(aqnk)
=
q∑r=1
e(aqrk) ∑
n≤tn≡r (q)
1 =t
qS(a, q) +O(q).
The difference that we want to bound is
f(α)− 1
qS(a, q)ν(β) =
∑n≤P
e(αnk)− 1
qS(a, q)
N∑m=1
1
kn1/k−1e(βm)
=
N∑m=1
cme(βm),
where
cm =
e(amq )− 1
qS(a, q) 1km
1/k−1 if m is kth power,
− 1qS(a, q) 1
km1/k−1 otherwise.
We are going to do partial summation, and for this we need control over∑m≤T cm. We have
∑m≤T
cm =∑n≤t
e(ankq
)− 1
qS(a, q)
∑m≤T
1
km1/k−1
=∑n≤t
e(aqnk)−(
1
t
∑n≤t
e(aqnk)
+O(qt
))(t+O(1))
= O(q +
q
t+ 1 · 1
)= O(q).
So by partial summation,
|(difference)| ≤ q · 1 +
∫ N
1
q|β|1dx q(1 + |β|N).
But remember that |β| < P−k+ν because α ∈M(a, q) and q < P ν . So We vet
|(difference)| P ν(
1 +N
P k−ν
) P 2ν .
Math 229x Notes 82
Let us define
V (a, q;α) =1
qS(a, q)v
(α− a
q
).
This is the good approximation for f(α). Explicitly, for α ∈M(a, q),
|f(α)s − V s| s,k |f(α)− V |P s−1 s,k Ps−1+2ν .
So: ∑q≤P ν
∑1≤a≤q(a,q)=1
∫M(a,q)
|f(α)s − V s|dαs,k Ps−k−1+5ν .
Math 229x Notes 83
32 April 19, 2017
Last time we had the local factorization: if we define
V (a, q;α) =1
qS(a, q)v
(α− a
q
)for α ∈M(a, q) then
|f(α)s − V s| s,k Ps−1+2ν .
So we had ∑q≤P ν
∑1≤a≤q(a,q)=1
∫M(a,q)
|f(α)2 − V s|dαs,k Ps−k−1+5ν .
32.1 Global factorization
We have∫M
f(α)se(−αN)dα =∑q≤P ν
∑1≤a≤q(a,q)=1
∫M(a,q)
V (a, q;α)se(−αN)dα
+Os,k(Ns/k−1−δ).
Let us just define this as
r∗(N) =∑q≤P ν
∑1≤a≤q(a,q)=1
∫M(a,q)
V (a, q;α)se(−αN)dα.
Here we have the factorization, and the integrals are the same integrals, withsome shifts. This motivates defining
S(N,Q) =∑q≤Q
∑1≤a≤q(a,q)=1
∫M(a,q)
(1
qS(a, q)
)se(−aqN)
J∗(N) =
∫ pν−k
−pν−kv(β)se(−βN)dβ.
Then it is immediate that
r∗(N) = S (N,P ν) · J∗(N).
We also note that S = Ss,k and J∗ = J∗s,k.
Proposition 32.1 (Global factorization).
r(N) = S(N,P ν)J∗(N) +Ok,s(Ns/k−1−δ)
Proof. Use the previous analysis on M and the analysis on m.
Math 229x Notes 84
32.2 Main formula for r(N)
This S(N,Q) is not a really well-behaved arithmetic function so what we aregoing to do is to let Q→∞ and then analyze the function.
Define
S(q) =∑
1≤a≤q(a,q)=1
(1
qS(a, q)
)se(−aqN)
so that S(N,Q) =∑q≤Q S(q).
Lemma 32.2. |S(q)| k,s q−1−2−k .
Proof. For (a, q) = 1 we use Weyl’s bound for the rational approximation a/q =a/q:
|S(a, q)| k,ε q1+ε(1
q+
1
q+
q
qk
)21−k
k,ε q1+ε−21−k
.
With ε > 0 small,
|S(q)| s,k,ε qq(ε−21−k)s q1−(21−k−ε)(2k+1) s,k,ε q
−1+ε(2k+1)−21−k.
Define the singular series
S(N) =
∞∑q=1
S(q)
which converges absolutely by the lemma. Moreover,
(1) |S(N)| s,k 1,
(2) |S(N,P ν)−S(N)| s,k N−δ.
So we get
r(N) = (S(N) +Ok,s(N−δ))J∗(N) +Ok,s(N
s/k−1−δ).
Now we go to the second part. Recall
J∗(N) =
∫ P ν−k
−P ν−kv(β)se(−βN)dβ.
Lemma 32.3. For |β| ≤ 1/2,
|v(β)| k minN1/k,
1
|β|1/k.
Proof. Recall that v(β) =∑Nn=1 k
−1n1/k−1e(βn). In the case |β| < N−1,
|v(β)| ≤N∑n=1
1
kn1/k−1 = N1/k +Ok(1).
Math 229x Notes 85
In the case 1/N ≤ |β| ≤ 1/2, observe that ‖β‖ = |β|. Write X = b1/|β|c < N .We will have ∣∣∣∣∑
n≤X
1
kn1/k−1e(βn)
∣∣∣∣ ≤ X1/k +Ok(1) 1
|β|1/k.
On the other interval, we do partial summation and get∑X<n≤N
1
kn1/k−1e(βn) =
( ∑X<n≤N
e(βn)
)1
kN1/k−1
+1
k
(1− 1
k
)∫ N
X
( ∑X<n≤t
e(βn)
)dt
t2−1/k
k1
|β|N1/k−1 +
1
|β|X1/k−1 k
1
|β|1/k+
1
|β|1/k.
Therefore we define the singular integral
J(N) =
∫ 1/2
−1/2
v(β)se(−βN)dβ,
for s ≥ 2k + 1 > k. Then
(1) |J(N)| s,k
∫ ∞0
minNs/k,
1
βs/k
dβ s,k N
s/k−1,
(2) |J∗(N)− J(N)| s,k
∫ ∞P ν−k
1
βs/kdβ s,k N
s/k−1−δ.
Theorem 32.4. Assume s ≥ 2k + 1. Then
r(N) = S(N)J(N) +Ok,s(Ns/k−1−δ).
In the remaining lectures, we are going to show that the main term is nottoo small.
Math 229x Notes 86
33 April 21, 2017
Last time we have the main formula: for n ≥ 2k + 1,
r(N) = S(N)J(N) +Ok,s(Ns/k−1−δ).
A better notation would be rk,s, Sk,s, Jk,s, but these are fixed.
33.1 Evaluation of J(N)
Recall that
v(β) =∑
1≤n≤N
1
kn1/k−1e(βn)
and
J(N) = Jk,s(N) =
∫ 1/2
−1/2
v(β)se(−βN)dβ.
The reward for having integration over a unit interval is
Jk,s(N) =∑
n1+···+ns=N
1
ks(n1 · · ·ns)1/k−1.
Lemma 33.1. Let 0 < a ≤ b with a ≤ 1. Then for Q ∈ Z≥1 we have∑1≤n<Q
na−1(Q− n)b−1 =Γ(a)Γ(b)
Γ(a+ b)Qa+b−1 +Oa,b(Q
b−1).
Proof. Let h(x) = xa−1(Q− x)b−1 on (0, Q). You can estimate the error termsand get ∑
1≤n<Q
h(n) =
∫ Q
0
h(x)dx+Oa,b(Qa+b−2 +Qb−1)
= Qa+b−1
∫ 1
0
ta−1(1− t)b−1dt+Oa,b(Qb−1).
Now the integral is the beta function.
Theorem 33.2. Suppose s ≥ 2. Then
Jk,s(N) = Γ(
1 +1
k
)sΓ( sk
)−1
Ns/k−1 +Ok,s(Ns/k−1−1/k).
Proof. We use induction on s ≥ 2. In the case s = 2,
Jk,s(N) =
N−1∑n=1
1
k2(n(N − n))1/k−1
=Γ(1/k)Γ(1/k)
Γ(2/k)
1
k2N2/k−1 +Ok(N2/k−1−1/k).
Math 229x Notes 87
In the inductive step, we have
Jk,s+1(N) =
N−1∑n=1
1
kn1/k−1Jk,s(N − n)
= Γ(
1 +1
k
)sΓ( sk
)−1 1
k
N−1∑n=1
n1/k−1(N − n)s/k−1
+Ok,s
(N−1∑n=1
n1/k−1(N − n)(s−1)/k−1
)= Γ
(1 +
1
k
)sΓ( sk
)−1 1
k
[Γ(1/k)Γ(s/k)
Γ((s+ 1)/k)N (s+1)/k−1
+Ok,s(Ns/k−1)
]+Ok,s(N
s/k − 1)
= Γ(
1 +1
k
)s+1
Γ(s+ 1
k
)−1
N (s+1)/k−1 +Ok,s(N(s+1)/k−1−1/k).
This finishes the proof.
Theorem 33.3. Assume s ≥ 2k + 1. Then
rk,s(N) = S(N)Γ(
1 +1
k
)sΓ( sk
)−1
Ns/k−1 +Ok,s(Ns/k−1−δ).
Moreover, |Sk,s(N)| k,s 1.
We need some finer analysis on the singular series.
33.2 S(a, q) and S(q) revisited
Recall that
S(a, q) =
q∑n=1
e(aqnk), S(q) =
∑1≤a≤q(a,q)=1
(1
qS(a, q)
)se(−aqN).
By definition,
S(N) =
∞∑q=1
S(q)
which converges for s ≥ 2k + 1, because in that case we have the estimate
|S(q)| k,s q−1−2−k uniformly on N .
Lemma 33.4. Let (a, q) = (b, r) = (q, r) = 1. Then
S(ar + bq, qr) = S(a, q)S(b, r).
Proof. Expand the right hand side and use the Chinese remainder theorem.
Math 229x Notes 88
Lemma 33.5. S(q) is a multiplicative function.
Proof. Taking (q, r) = 1, the previous lemma gives
S(q)S(r) =
( ∑1≤a≤q(a,q)=1
(1
qS(a, q)
)se(−aqN))( ∑
1≤b≤r(b,r)=1
(1
rS(b, r)
)se(− brN))
=∑
1≤a≤q(a,q)=1
∑1≤b≤r(b,r)=1
1
(qr)sS(ar + bq, qr)se
(−ar + bq
rqN)
= S(qr)
Then
Sk,s(N) =∑q≥1
S(q).
can be factored into Euler factors, by absolute convergence. For s ≥ 2k + 1 andp a prime number, define
T (p) =∑m≥0
S(pm),
which converges absolutely.
Proposition 33.6. For s ≥ 2k + 1, we have
S(N) =∏p
T (p),
and the convergence is absolute.
In fact, T (p) is a real number, because each S(q) is a real number:
S(q) =∑
1≤a≤q(a,q)=1
(1
qS(a, q)
)se(−aqN)
=∑
1≤a≤q(a,q)=1
(1
q
q∑n=1
e(−aqnk))s
e(−−a
qN)
= S(q).
Corollary 33.7. For pk,s 1, T (p) > 0.
Furthermore, there exists a constant Ck,s, independent of N , such that forp > Ck,s, T (p) > 0 and
∏p>Ck,s
T (p) ∈ [1/2, 3/2]. The remaining lectures will
be about bounding T (p).
Math 229x Notes 89
34 April 24, 2017
Last time we have
S(N) =∏p
T (p)
for s ≥ 2k + 1, where T (p) =∑k≥0 S(pk). This converges absolutely and
uniformly on N , because S is multiplicative on q and |S(q)| s,k q−1−2−k .
Today we analyze T (p).
34.1 Formula for T (p)
We are only going to consider s ≥ 2k + 1, because otherwise the series mightnot even converge.
Definition 34.1. We define
M(q) = #x ∈ (Z/qZ)s : xk1 + · · ·+ xks ≡ N (mod q).
Lemma 34.2.∑d|q
S(d) =1
qs−1M(q).
Proof. We have
M(q) =
q∑x1=1
· · ·q∑
xs=1
1
q
q∑r=1
e(rq
(xk1 + · · ·+ xks −N)).
For each r, let d = (r, q) and a = r/d. Then
qMq =∑d|q
d∑a=1
(a,d)=1
( qd
)s d∑n1=1
· · ·d∑
ns=1
e(ad
(nk1 + · · ·+ nks −N))
= qs∑d|q
d∑a=1
(a,d)=1
(1
d
d∑n=1
e(adnk))s
e(−adN).
The right hand side can be identified as the sum of S(d).
Now for q = ph with h→∞, the sum converges to T (p).
Lemma 34.3. For s ≥ 2k + 1,
T (p) = limh→∞
1
ph(s−1)M(ph).
Corollary 34.4. T (p) ≥ 0.
Math 229x Notes 90
The goal now is to produce enough solutions of xk1 + · · ·+xks ≡ N (mod ph)so that we can give a lower bound for T (p), of course, uniformly on N .
The naıve thing you would do is to find solutions modulo p and then liftthe solutions. But you need a little bit of care: 3 is a square modulo 2 but notmodulo 4.
Definition 34.5. Define τ(p) = νp(k).
Lemma 34.6. The number of invertible kth powers modulo pt (t ≥ 0) is:
(1) ϕ(pt)/(k, ϕ(pt)) if p > 2 or t = 1 or k is odd,
(2) 2t−2/(k, 2t−2) if p = 2 and k is even and t > 1.
Define
γ(p) =
τ(p) + 1 if p > 2 or k is odd or τ = 0,
τ(p) + 2 if p = 2 and k is even and τ > 0.
Lemma 34.7. The number of invertible kth powers mod pγ(p) is
ϕ(pτ(p)+1)/(k, ϕ(pτ(p)+1)).
Lemma 34.8. Let a ∈ Z be coprime to p. The number of solutions to thecongruence xk ≡ a (mod pγ(p)) is either 0 or pγ−τ−1(k, ϕ(pτ(p) + 1)).
Lemma 34.9. Let a ∈ Z be coprime to p. If a is a kth power mod pγ(p) then itis so mod pt for all t.
Proof. For t ≤ γ(p), it is clear. For t > γ(p), the lift of solutions follows fromthe previous lemmas by counting.
Define
M∗(q) = #x ∈ (Z/qZ)s : xk1 + · · ·+ xks ≡ N (mod q), (q, x1) = 1.
Obviously, M(q) ≥M∗(q).
Lemma 34.10. Suppose t ≥ γ(p) and that M∗(pγ) > 0. Then M(pt) ≥p(t−γ)(s−1).
Proof. Let a1, . . . , as ∈ Z be a solution of xk1 + · · · + xks ≡ N (mod pγ) with(a1, p) = 1. Then consider
xk1 ≡ N − (xk2 + · · ·+ xks) (mod pt)
for t ≥ γ. Then any lifting for x2, . . . , xs from pγ to pt is admissible by theprevious lemma, since modulo pγ the right hand side is a kth power.
So any solution counted in M∗(pγ) gives the lower bound for M∗(pt).
The remaining problem is, is M∗(pγ) > 0? We go back to the idea ofdensities.
Math 229x Notes 91
Theorem 34.11 (Cauchy–Davenport). Let q > 1, and let A ,B ⊆ Z/qZ.Suppose:
(1) a = #A > 0 and b = #B,
(2) 0 ∈ B and B \ 0 ∈ (Z/qZ)×.
Then #(A + B) ≥ mina+ b− 1, q.
Math 229x Notes 92
35 April 26, 2017
Last time we had the question of bounding
M∗(q) = #x ∈ (Z/qZ)s : xk1 + · · ·+ xks ≡ N (q), (x1, q) = 1.
We still need to check that for s ≥ 2k + 1, M∗(pγ(p)) > 0 for all p.
35.1 Local Waring’s problem
Theorem 35.1 (Cauchy–Davenport). Let q ≥ 1 and let A ,B ⊆ Z/qZ, suchthat
(1) a = #A > 0, b = #B,
(2) 0 ∈ B and B \ 0 ⊆ (Z/qZ)×.
Then #(A + B) ≥ minq, a+ b− 1.
Proof. We can assume that a + b − 1 ≤ q. Otherwise, remove b − (q − a + q)non-zero elements from B. Now induct on b. For b = 1, we have B = 0 andthis is trivial.
Now let b ≥ 2, in particular, 0 is not the only element. Assume all cases with#B′ < b. We claim that there exist α0 ∈ A and β0 ∈ B such that α0 +β0 ∈ A .Otherwise, given any β0 ∈ B we would have∑
α∈A
(α+ β0) ≡∑α∈A
α (mod q)
and then aβ0 ≡ 0 (mod q). This is not possible because B has invertibleelements.
Let α0 be as in the claim. Let X = β ∈ B : α0 +β /∈ A . Then ∅ 6= X ⊂ B.Let A ′ = A ∪ (α0 +X) and B′ = B \X. If x = #X, then #A ′ = a+ x and#B′ = b− x and x 6= 0. So by induction,
#(A ′ + B′) ≥ minq, (a+ x) + (b− x)− 1 = minq, a+ b− 1.
But note that A ′ + B′ ⊆ A + B.
Proposition 35.2. Let s ≥ 2k+1. Then for every prime p we have M∗(pγ(p)) >0.
Proof. We always consider p - x1. Recall that
τ(p) = νp(k), γ(p) =
τ + 1 p ≥ 3 or k odd,
τ + 2 p = 2 and k even.
For p = 2 and k = 2, we have s ≥ 5 and you can check that x21 + · · ·+x2
5 ≡ ∗(mod 8) has a solution. For p = 2 and k = 3, we have s ≥ 9, and it is easierthat x3
1 + · · ·+ x39 ≡ ∗ (mod 2) has a solution.
Math 229x Notes 93
For p ≥ 3 and k = 2, we have γ = 1. Let A = ((Z/pZ)×)2 and B = A ∪0.Then either A + (s− 1)B = Z/pZ or
#(A + (s− 1)B) ≥ a+ (s− 1)(a+ 1)− (s− 1) = sa ≥ 5(p− 1)/2 > p,
which is not possible.For p ≥ 3 and k ≥ 4, γ = τ + 1 = νp(k) + 1. So pγ ≤ pk. Let A =
((Z/pγZ)×)2 and B = 0 ∪A . Then
a = #A =ϕ(pγ)
(k, ϕ(pγ))=
pτ+1 − pτ
(k, pτ+1 − pτ )=
p− 1
(k, p− 1)≥ p− 1
k.
Either A + (s− 1)B = Z/pγZ or it has size at least
a+ (s− 1)(a+ 1)− (s− 1) = sa ≥ (2k + 1)p− 1
k.
We need (2k + 1)/k2 ≥ p/(p− 1), and this can be checked.For k = 3, you can check.
So what do we get?
Corollary 35.3. T (p) = limh→∞
1
ph(s−1)M(ph) ≥ 1
pγ.
Therefore
Ss,k(N) ≥ 1
2
∏p≤Ck,s
T (p) ≥ 1
2
∏p≤Ck,s
1
pγ(p).
So the singular series is Ss,k(N)s,k 1 uniformly on N . Finally,
Theorem 35.4. For s ≥ 2k + 1, we have
rs,k(N) = κk,sSk,s(N)Ns/k−1 +Ok,s(Ns/k−1−δ),
where
(1) 0 < δ k 1,
(2) κk,s = Γ(
1 +1
k
)sΓ( sk
)−1
k,s 1,
(3) Sk,s(N) is the singular series defined before,
(4) Sk,s k,s 1.
In particular, for N k,s 1, rs,k(N) > 0.
Index
analytic large sieve inequality, 58
basis, 66Brun’s theorem, 57
Cauchy–Davenport inequality, 92combinatorial sieve, 50conductor, 25
δ-Riemann hypothesis, 48Dirichlet L-function, 12Dirichlet series, 10
exceptional character, 36
Fourier transform, 20
Γ-function, 22Gauss sum, 25
Lindelof hypothesis, 49
major arc, 70minor arc, 70
prime number theorem, 20primitive character, 25
quasi-Riemann hypothesis, 48
Ramanujan sum, 59Ramanujan–Fourier series, 60Riemann hypothesis, 48Riemann zeta function, 9
Schnirelmann density, 65Selberg sieve, 51Siegel zero, 36singular integral, 85singular series, 84
θ-function, 21, 27, 28
von Mangoldt function, 7
Weyl’s differencing bound, 73Weyl’s inequality, 75
94