Partial Summation
Partial Summation is often no more than an exercise in interchanging
Sums and Integrals.
8. Let x 1 be real, cfw_an n1 a sequence of complex numbers and A (x) =
1nx an .
a) Show that
x
1nx
an (x n) =
A (t) dt.
1
b) Show that
an log
1nx
x
Chapter 5 Sums of Arithmetic Functions.
2014
[2 lectures]
The Prime Number Theorem we proved was a result about the sum of the
von Mangoldt function. In this Chapter we look at sums of other arithmetic
functions.
Let f be an arithmetic function and write
Additional Questions
14. i) To show convergence is uniform in Question 2.
Let an C be a sequence of coecients and set A (x) =
Prove that
A (N )
an
=
+s
s
n
Ns
n=N +1
N
1nx
an .
A (t)
dt.
ts+1
ii) Assume there exists a constant C > 0 such that |A (x)| < C
Solutions to Problem Sheet 2,
1. i) Draw the graphs for [x] and cfw_x.
ii) Show that for R,
+1
+1
[t] dt =
and
1
cfw_t dt = .
2
Hint Split these integrals at the integer which must lie in any interval
of length 1, such as [, + 1] .
iii) Prove that for x
Chapter 1 Additional Questions
9) Recall from notes the denition of the set
N = cfw_n : p|n p N .
Then unique factorisation of integers justies, for real s = > 1, the last
equality in
1
1
1
1
() =
=
1
.
(19)
n nN n pN
p
n=1
This is a rather convoluted
Solutions to Problem Sheet 5
1. a) Start from
= 1,
j
j
(15)
seen in Problem Sheet 4, and use the Composition Method to prove
that
(n)
1
=
x + O (log x) .
n
(2)
nx
Hint The result (15) in the Composition Method gives
(n)
=
n
nx
(a)
a
ax
1.
bx/a
b) Appl
Problem Sheet 3
Solutions
1. From Theorem 3.5 we have
valid for Re s > 0.
1
s
(s) = 1 +
s1
1
cfw_u
du,
u1+s
i) Deduce that
(s) = s
1
[u]
du
u1+s
for Re s > 1.
Note the integral contains [u] in place of cfw_u.
ii) Deduce that
(s) = s
0
cfw_u
du,
u1+s
fo
Solutions to Problem Sheet 4, Part 1
Connections between sq, Q2, 2 and .
1. Show that for all arithmetic functions f, g and h we have
f (g + h) = f g + f h.
So is distributive over +.
Solution For n 1
(f (g + h) (n) =
f (a) (g + h) (b)
ab=n
=
f (a) (g (b)
Solutions to Problem Sheet 1
1) Use Theorem 1.1 to prove that
1
log log x 1
p
px
for all real x 3. This is a version of Theorem 1.1 with the integer N
replaced by the real x.
Hint Given x 3 let N = [x], the largest integer x. Then, importantly,
the sets
Analytic Number Theory
Feedback on the 2011-12 exam
Throughout the paper you may assume that the convolution of two multiplicative functions is multiplicative.
(n) is the number of distinct prime factors of n.
(n) is the number of prime factors of n cou
Chapter 4 Arithmetic Functions and Dirichlet Series.2014
[4 lectures]
Denition 4.1 An arithmetic function is any function f : N C.
Examples
1) The divisor function d (n) (often denoted (n) is the number of divisors of n, i.e.
d (n) =
1.
d|n
Note that if d
MATH61022
Three hours
UNIVERSITY OF MANCHESTER
ANALYTIC NUMBER THEORY
? ? 2012
? ?
Answer THREE out of four questions in Section A. (90 marks in total.) If you answer more
questions then the marks from the three best solutions will be used.
Answer all of
A1 i) Prove that for integer N 1,
1nN
1
log (N + 1) .
n
ii) Justify
1nN
1
1
1
n pN
p
1
,
for any integer N 1.
iii) Prove, by looking at the logarithm as an integral, or otherwise, that
x < log (1 x) <
x
1x
for 0 < x < 1.
iv) Deduce that the sum over all
Chapter 1 Two Proofs of the innitude of Primes
2014
[3 lectures]
Theorem 1.1 The sum over the reciprocals of primes p satises
pN
1
> log log (N + 1) 1
p
for N 2.
Letting N proves the innitude of primes.
Proof Let
N = cfw_n N : p|n p N .
Consider
nN
1
1
1
MATH61022
Three hours
UNIVERSITY OF MANCHESTER
ANALYTIC NUMBER THEORY
? ? 2011
? ?
Answer THREE out of four questions in Section A. (90 marks in total.) If you answer more
questions then the marks from the three best solutions will be used.
Answer all of