Topics in Analytic Number Theory Spring 2017,
Assignment 2
Deadline: Friday March 17
The total number of points is 100. Grade=(number of points)/10.
We shall try to understand maximal values and averages of certain arithmetic functions
over the integers and special subsets of the integers.
1.
We begin with the divisor function; it is defined through
τ (n) := #{d ∈ N : d divides n}.
10
i) Prove that there exist positive constants c1 < c2 such that for all x > 10 we have
c1
c2
x log log x 6 max{τ (n) : 1 6 n 6 x} 6 x log log x .
Hint. Think about the possible factorisation into primes of an integer n 6 x for
which the divisor function τ (n) is maximised and use the prime number theorem.
15
ii) Prove that the average of τ (n) over all n 6 x is log x in the following precise form:
there exists a constant c0 ∈ R such that
X
√
τ (n) = x log x + c0 x + O( x).
16n6x
Hint. First prove that
τ (n) = a(n) + 2
X
1,
d|n√
16d6 n
where a(n) is −1 if n is an integer square and vanishes otherwise.
2.
We next consider the prime divisor function ω(n); it counts the distinct prime
divisor of n, namely
ω(n) := #{p prime : p divides n}.
1
2
Equivalently, if pi are different primes and αi are positive integers we have
ω(pα1 1 · · · pαk k ) = k.
8
i) Prove that there exist positive constants c3 < c4 such that for all x > 10 we have
c3
log x
log x
6 max{ω(n) : 1 6 n 6 x} 6 c4
.
log log x
log log x
Hint. Think about the possible factorisation into primes of an integer n 6 x for
which the prime divisor function ω(n) is maximised and use the prime number
theorem.
4
ii) Prove that the average of ω(n) over all n 6 x is log log x in the following precise
form,
1 X
ω(n) = log log x + O(1).
x 16n6x
Hint. Use the identity
ω(n) =
X
1,
p|n
as well as the estimate
X1
p6x
p
= log log x + O(1),
which you have to deduce from the prime number theorem in the form
y
y
#{p 6 y} =
+O
.
log y
(log y)2
8
iii) Prove that the average of ω(n)2 over all n 6 x is (log log x)2 in the following precise
form,
1 X
ω(n)2 = (log log x)2 + O (log log x) .
x 16n6x
5
iv) Prove that ω(n) stays very close to log log n for almost all integers n. Specifically
show that for all x > 10 we have that the number of integers 1 6 n 6 x that do
not satisfy
ω(n) − log log n 6 (log log n)3/4
3
is at most O (log logxx)1/10 .
Hint. First show that
X
(ω(n) − log log n)2 = O (x log log x)
16n6x
by combining the results of the previous two parts.
3.
We next consider the Euler function φ(n); it counts the positive integers below n
that are coprime to n. We have
φ(n) Y
1
=
1−
.
n
p
p|n
10
i) Prove that there exist positive constants c5 < c6 such that for all x > 10 we have
n φ(n)
o
c5
c6
6 max
:16n6x 6
.
log log x
n
log log x
Hint. Think about the possible factorisation into primes of an integer n 6 x for
which the prime divisor function ω(n) is maximised and use the Mertens theorem.
15
ii) Prove that
n
φ(n)
is constant on average for 1 6 n 6 x in the following precise form,
Y
1 X n
1
(log x)2
=
1+
+O
,
x 16n6x φ(n)
p(p − 1)
x
p
where the infinite product is taken over all primes. Deduce from this estimate that
!
Y
X 1
1
=
1+
log x + O(1).
φ(n)
p(p
−
1)
p
16n6x
Hint. First prove the identity
X µ(d)2
n
=
.
φ(n) d∈N φ(d)
d|n
4.
We will now consider the average of the divisor function τ (n) over the thin sequence
of integers of the form n = p−1, where p are the prime numbers. If the factorisation
4
of those special integers was similar to that of a general integer then we would expect
that τ (p − 1) is log(p − 1) on average, thus
X
τ (p − 1) (log x)#{p 6 x} x.
p6x
We will show that this is correct up to a correction constant.
5
i) For coprime integers a, d we let
π(x; d, a) := #{p 6 x, p prime : p ≡ a (mod d)}.
Use the identity in the hint of the second part of the first exercise to show that
X
X
τ (p − 1) = 2
π(x; d, 1) + O(x1/2 ).
p6x
5
ii) Use the Brun–Titchmarsh theorem from the first lecture to show that for each
A > 0 one has
X
x(log log x)
π(x; d, 1) .
log x
1/2
1/2
−A
x
15
d6x1/2
(log x)
iii) The Bombieri–Vinogradov
such that
X
max
<d6x
16a6d
d6x1/2 (log x)−A gcd(a,d)=1
theorem states that for each B > 0 there exists A > 0
π(x; d, a) −
1
φ(d)
Z
2
x
du x
.
B
log u
(log x)B
Use it to deduce that
X
ζ(2)ζ(3)
x(log log x)
τ (p − 1) =
x+O
,
ζ(6)
log
x
p6x
where ζ denotes the Riemann zeta function.