Acknowlegements

I would like to thank my supervisor Dr. Terry Tao, for all his help with the writting of this Thesis, both in Los Angeles at the University of California, Los Angeles and in Sydney at the University of New South Wales. I would also like to thank the various members of the academic staff who have been very supportive over my honours year, and have helped me to undertake part of this year on exchange without any problems.

Contents

1 Introduction 1 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Some Basic Theorems and Useful Tools 5

2.1 Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2.1.1 Weyl’s Criterion . . . . . . . . . . . . . . . . . . . . . . 6 2.1.2 Féjer’s Theorem . . . . . . . . . . . . . . . . . . . . . . 8 2.1.3 Van der Corput’s Difference Theorem . . . . . . . . . . 9 2.1.4 Generalised Van Der Corput’s Difference Theorem … 11

2.2 Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.2.1 The Euler Summation Formula . . . . . . . . . . . . . 14 2.2.2 Integration By Parts . . . . . . . . . . . . . . . . . . . 15 2.2.3 Integrating Functions Whose Derivative can be Sim-

plified . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.2.4 Approximating Sequences by Simpler Sequences …. 16 2.2.5 Changing the summation limits . . . . . . . . . . . . . 18

3 Non-Trigonometric Sequences 20 3.1 The distribution of nθ . . . . . . . . . . . . . . . . . . . . . . 21 3.2 The distribution of log n . . . . . . . . . . . . . . . . . . . . . 22 3.3 The distribution of ∑_{j} c_{j}n^{α_{j}} . . . . . . . . . . . . . . . . . . . 23 3.4 The distribution of (∑ c_{j}n^{α_{j}} log^{τ_{j}} n) . . . . . . . . . . . . . . . 26 3.5 Distribution of “slow sequences” . . . . . . . . . . . . . . . . . 27

3.5.1 Outline of Proof . . . . . . . . . . . . . . . . . . . . . . 27 3.5.2 Dealing with terms lying between N_{0} and N_{1} − 1 … 29 3.5.3 Dealing with terms lying between N_{1} + 2 and N_{2} … 30 3.5.4 Dealing with the terms N_{1} and N_{1} + 1 . . . . . . . . . 30 3.5.5 Combining the results . . . . . . . . . . . . . . . . . . 31 3.5.6 Using the remaining terms . . . . . . . . . . . . . . . . 32 3.5.7 Deriving (3.10) . . . . . . . . . . . . . . . . . . . . . . 33

3.6 The Distribution of u_{n} + f(n) . . . . . . . . . . . . . . . . . . 34

4 Trigonometric Sequences 37 4.1 The distribution of (K sin(x_{n})) . . . . . . . . . . . . . . . . . 38 4.2 The distribution of n^{−α} sin n^{β} . . . . . . . . . . . . . . . . . . 40 4.3 The distribution of n^{α} sin n^{−β} . . . . . . . . . . . . . . . . . . 40 4.4 The distribution of n^{α} sin n^{β} . . . . . . . . . . . . . . . . . . . 42

4.4.1 α ≠ k(1 − β) . . . . . . . . . . . . . . . . . . . . . . . 44 4.4.2 α = k(1 − β) . . . . . . . . . . . . . . . . . . . . . . . 49

4.5 The distribution of n^{α} sin n . . . . . . . . . . . . . . . . . . . 52 4.6 The distribution of (f(n) sin n^{β}) . . . . . . . . . . . . . . . . . 57

5 Applying functions to sequences 60 5.1 Adding and multiplying uniformly distributed sequences … 61 5.2 The distribution of log u_{n} . . . . . . . . . . . . . . . . . . . . . 61 5.3 The distribution of u_{n} + sin v_{n} . . . . . . . . . . . . . . . . . . 62 5.4 The distribution of {u_{n}} sin v_{n} . . . . . . . . . . . . . . . . . . 64

A The distribution of n sin n 66

Chapter 1

Introduction

We will consider certain families of sequences and analyse their distribution modulo 1. To do this we will give certain criteria for the uniform distribution of sequences and use these criteria to discuss the distributions of sequences.

1.1 Introduction

In this report we look at sequences of numbers and determine whether their fractional parts are uniformly distributed in the interval [0, 1). The fractional part of a number α (denoted by {α}) is given by α − ⌊α⌋. (Note: we will sometimes work modulo an interval [0, H). In this case, we will let {x}_{H} denote the remainder when x is divided by H.)

To begin we define what it means for a sequence of numbers to be uniformly distributed modulo H, where H ∈ ℝ. An infinite sequence of numbers (α_{i}) is said to be uniformly distributed modulo H if and only if the corresponding sequence of ({α_{i}}_{H}) is uniformly distributed in the interval [0, H). Thus a sequence (α_{i}) is uniformly distributed modulo 1 if its fractional parts are uniformly distributed in the unit interval.

Let us now define what it means to be uniformly distributed in an interval. To do this we first define a function ϕ as follows: Given an infinite sequence of real numbers, (α_{i}) define ϕ_{[m,n]}^{α_{i}}(I) to be the number of i’s with α_{i} ∈ I and m ≤ i ≤ n.

The sequence (α_{i}), A ≤ α_{i} < B is uniformly distributed in the interval [A, B) if and only if for every 0 ≤ a < b ≤ 1,

lim_{N→∞} \frac{ϕ_{[1,N]}([a, b))}{N} = \frac{b − a}{B − A}  (1.1)

What (1.1) is saying is that as N gets very large, the proportion of α_{i}’s lying in a given interval must approach the proportional length of the interval. This is consistent with our intuitive understanding of uniform distribution.

Throughout the course of this report we will often write u.d. mod 1 to mean uniformly distributed modulo 1.

We will consider in very limited detail the distribution of pairs of sequences in R^{2}. If x = (x_{1}, x_{2}) then we write the fractional part of x as {x}= ({x_{1}}, {x_{2}}). Let (x_{n}) be a sequence of vectors in R^{2}. We say that (x_{n}) is uniformly distributed modulo (H_{1}, H_{2}) in R^{2} if

lim_{N→∞} \frac{ϕ_{[1,N]}^{\{x_{n}\}}([a,b))}{N} = ∏_{j=1}^{2} \frac{(b_{j} − a_{j})}{H_{j}}  (1.2)

for all intervals [a,b)= [a_{1}, b_{1}) × [a_{2}, b_{2}) ∈ I^{2}.

Specific families of sequences we will consider include ∑ c_{α}n^{α}, ∑ c_{j}n^{j} log^{τ_{j}} n and n^{α} sin n^{β}.

A large number of the results for non trigonometric sequences are already known and have been included in this report for completeness. As a general rule, it is fairly easy to analyse the behaviour of sequences with a small derivative. Sequences with very small derivatives (e.g. log n) tend not to be u.d. as they are too close to the constant function. The fractional parts of these sequences are not changing quickly enough to be spread out uniformly over the unit interval.

Sequences with slightly larger derivatives (e.g. n^{α}, α < 1) are also fairly easy to analyse. This is because it is quite easy to change the sum which results from Weyl’s Criterion (section 2.1) into an integral and then evaluate this integral.

Sequences with a larger (e.g. n^{α}, α > 1) which can’t be dealt with by evaluating the resulting integral can often be dealt with using Van Der Corput’s Difference Theorem (section 2.1). This theorem essentially lets us replace a sequence with its derivative and analyse its distribution this way. In many cases it will be possible to evaluate the integral associated with one of the higher order derivatives of a sequence and thus form a conclusion about the distribution of that sequence.

Generally speaking sequences with very large derivatives are difficult to analyse, as the fractional part of the terms in the sequences are varying rapidly. In this report we will discuss the distribution of sequences of the form (n^{α} sin n^{β}). These trigonometric functions at times have an extremely fast derivative and so can be difficult to analyse. There has previously been little analysis of the distributions of trigonometric sequences. In [7] we find some analysis of the distributions of families of sequences of the form f(n) sin nx however this analysis is restricted to probabilistic arguments and only works for almost every value of x. The methods used in [?] cannot be used to show the uniform distribution of such a sequence for any particular value of x. An analysis of sequences of the form P (n)f(nα) where P is a polynomial with integer powers, f is a periodic function and α is a constant can be found in [2]. The analysis encompasses functions of the form P (n) sin n however the machinery used is extremely complex. As far as we know these are the only discussions of such families of sequences. In this report an attempt is made to give a general understanding of the behaviour of some trigonometric sequences using less complicated machinery.

We will make a complete analysis of the distribution of sequences of the form (n^{α} sin n^{β}) for β < 1 and α ∈ ℝ. The results are contained in Theorems 4.1.2, 4.2.1, 4.3.1 and 4.4.9.

The restriction β < 1 slows down the derivative of the sequence as n increases and thus makes it easier to analyse the distribution.

A partial analysis is made of the distribution of n^{α} sin n which includes the case when α < \frac{1}{2} and is found in Theorem 4.5.1. In this case we actually discuss the distribution of n^{α} sin 2πan where a is a real number less than 1.

We write a = \frac{p}{q} + δ using a theorem due to Dirichlet, replace n by mq + r and treat n^{α} sin 2πan as a function of m instead. We then show that the extra factor of qδ obtained in the derivative of the sine term will be small in size and so makes the sequence easier to analyse as it now has a (relatively) small derivative.

As an appendix we make a partial analysis of the sequence n sin n using the methods found in [2]. Lack of time has meant that a full analysis of this sequence could not be made.

As far as we know, no analysis has been made for sequences with β > 1. This case is very difficult to deal with, as the sine term is oscillating more rapidly as n increases and so the derivative of the sequence gets very large.

Limited analysis has been made of sequences of the form (α^{n}). Understanding the distribution of this family of sequences would be equivalent to answering the question about whether the digits of π are uniformly distributed. As the derivative of α^{n} is extremely large this is a particularly hard problem and will not be addressed in this report.

1.2 Notation

Throughout the report K will be used to represent a constant. The value of K may vary from line to line.

For a function f and a set of intervals A = ∪_{i=1}^{m}[a_{i}, b_{i}] the notation [f(x)]_{A} will be used to represent f evaluated over the intervals in A. i.e.

∫_{A} f′ = [f(x)]_{A} = ∑_{i=1}^{m}[f(x)]_{a_{i}}^{b_{i}} = ∑_{i=1}^{m}(f(b_{i}) − f(a_{i}))

When writing ϕ_{[m,n]}^{α_{i}}(I) we will often make the following abbreviations.

1. If it is obvious which sequence we are referring to, we will drop the superscript from ϕ.

2. If m=1 we will often not include it in the subscript of ϕ. For example ϕ_{[1,N]}^{α_{i}}(I) = ϕ_{N}^{α_{i}}(I)

3. If I is composed of a single interval we will often not include the brackets around the interval. For example, ϕ_{N}([a, b)) will be written as ϕ_{N}(a, b).

Some Basic Theorems and Useful Tools

This chapter will be devoted to outlining some of the theorems and tools which will be used throughout the report. As a general rule it is both difficult and tedious to evaluate the function ϕ_{N}^{x_{n}}(a, b) for a sequence (x_{n}). For this reason we look for other methods which can be used to determine the distribution of sequences. Many of the proofs in this chapter are standard proofs which have been taken from Chapters 2 and 3 of [6].

2.1 Theorems

2.1.1 Weyl’s Criterion

Weyl’s theorems provide necessary and sufficient conditions for sequences of numbers to be uniformly distributed either in the unit interval or modulo 1 based on their exponential sums.

Theorem 2.1.1 If (α_{n}) is a sequence with 0 ≤ α_{n} ≤ 1 for n = 1, 2, … then a necessary and sufficient condition for (α_{n}) to be uniformly distributed in the unit interval is that

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} f(α_{n}) = ∫_{0}^{1} f(x)dx  (2.1)

for every function f which is Riemann integrable in 0 ≤ x ≤ 1.

Theorem 2.1.2 (Weyl’s Criterion) If (β_{n}) is a sequence not contained in the unit interval Weyl’s Theorem becomes: (β_{n}) is uniformly distributed modulo 1 iff

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πihβ_{n}} = 0 for h = 1, 2, …  (2.2)

Proof of Theorem 2.1.1 Throughout this proof we will let (α_{n}) be a sequence satisfying the hypothesis of Theorem 2.1.1. To prove the sufficiency of (2.1) we assume that (2.1) holds and let f(x) = χ_{[a,b)} be the characteristic function of [a, b). That is

χ_{[a,b)}(x) = \{ \begin{matrix}1 & if a ≤ x < b \\ 0 & otherwise \end{matrix}

Then:

lim_{N→∞} \frac{ϕ_{N}(a, b)}{N} = lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} χ_{[a,b)}(α_{n})

= ∫_{0}^{1} χ_{[a,b)}(α_{n})
= b − a  (using 2.1)

(2.3)

Thus (α_{n}) is uniformly distributed in the unit interval.

To prove the necessity of (2.1) we assume that (α_{n}) is uniformly distributed in the unit interval. (1.1) implies that (2.1) holds for any characteristic function f. Since (2.1) is a linear equation, it will also hold for any step function f in [0, 1).

For a Riemann integrable function f we can always find two step functions, f_{1} and f_{2} with f_{1} ≤ f ≤ f_{2} and ∫_{0}^{1} (f_{2}(x) − f_{1}(x)) dx < ϵ. Since f_{1} satisfies (2.1), we have:

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} f_{1}(α_{n}) = ∫_{0}^{1} f_{1}(x)dx > ∫_{0}^{1} f(x)dx − ϵ

Hence for large enough N,

\frac{1}{N} ∑_{n=1}^{N} f_{1}(α_{n}) > ∫_{0}^{1} f(x)dx − 2ϵ

and since f ≥ f_{1} then:

\frac{1}{N} ∑_{n=1}^{N} f(α_{n}) > ∫_{0}^{1} f(x)dx − 2ϵ  (2.4)

Similarly, using f_{2} and taking N large enough we get:

\frac{1}{N} ∑_{n=1}^{N} f(α_{n}) < ∫_{0}^{1} f(x)dx + 2ϵ  (2.5)

Combining (2.4) and (2.5) shows that for N large enough every Riemann integrable function f satisfies

|\frac{1}{N} ∑_{n=1}^{N} f(α_{n}) − ∫_{0}^{1} f(x)dx| < 2ϵ  (2.6)

which shows the necessity of 2.1. Proof of Theorem 2.1.2 For a sequence (β_{n}) not necessarily in the unit interval let α_{n} denote the fractional part of β_{n}. The necessity of (2.2) will follow directly from (2.1). Assume that (β_{n}) is uniformly distributed modulo 1 and thus (α_{n}) is uniformly distributed in the unit interval and so obeys (2.1). Hence, letting f(x) = e^{2πihx} in (2.1)where h is an integer, h ≠ 0 we have:

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πihβ_{n}} = lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πihα_{n}}

= ∫_{0}^{1} e^{2πihx}dx

= 0 Thus (2.2) is a necessary condition for a sequence (β_{n}) to be u.d. mod 1.

To prove the sufficiency of (2.2) will show that if (2.2) holds for every integer h ≠ 0 then (2.1) is satisfied and so the fractional parts of the β_{n}’s are uniformly distributed in the unit interval. This means that the β_{n}’s are uniformly distributed modulo 1.

Firstly we observe that f(x) = 1 satisfies (2.1) since

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} 1 = 1 = ∫_{0}^{1} dx

Assuming (2.2) we get

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πihβ_{n}} = lim_{N→∞} \frac{1}{N} (∑_{n=1}^{N} cos 2πhα_{n} + i ∑_{n=1}^{N} sin 2πhα_{n})

= 0

(2.7)

Equating real and imaginary parts in (2.7) we see that f_{1h}(x) = cos 2πhx and f_{2h}(x) = sin 2πhx, both satisfy (2.1). Once again using the fact that (2.1) is a linear equation we can deduce that (2.1) holds for any trigonometric polynomial of the form

a_{0} + (a_{1} cos 2πx + b_{1} sin 2πx) + ⋯ + ((a_{m} cos 2πmx + b_{m} sin 2πmx) + ⋯

The Weierstrass approximation theorem tells us that any continuous function f of period 1 can be approximated by a trigonometric polynomial of the form above. So, given ϵ > 0 there exists a trigonometric polynomial f_{ϵ} with |f − f_{ϵ}| < ϵ. We set f_{1} = f_{ϵ} − ϵ and f_{2} = f_{ϵ} + ϵ. Thus f_{1} ≤ f ≤ f_{2} and ∫_{0}^{1}(f_{2}(x) − f_{1}(x))dx < 2ϵ. Through a similar line of argument to that used to show the necessity of (2.1) we can show that f satisfies (2.1) and hence (2.2) is a sufficient condition for a sequence (β_{n}) to be uniformly distributed modulo 1.

2.1.2 Féjer’s Theorem

Theorem 2.1.3 Let f(x) be a function defined for x > 1 that is differentiable for x > x_{0}. If f′(x) → 0 monotonically as x → ∞ and if lim_{x→∞} x|f′(x)| = ∞ then (f(n)) is uniformly distributed modulo 1.

Theorem 2.1.3 actually follows as a corollary from the following theorem:

Theorem 2.1.4 If f(n) is a sequence of real numbers such that Δ f(n) = f(n + 1) − f(n) is monotone as n increases and in addition

lim_{n→∞} Δf(n) = 0 and lim_{n→∞} n|Δf(n)| = ∞

then (f(n)) is uniformly distributed modulo 1.

For a proof of this theorem see [6] page 13.

Proof of Theorem 2.1.3 The mean value theorem implies that if f satisfies the criteria of Theorem 2.1.3 then Δf(n) satisfies the conditions of Theorem 2.1.4 for sufficiently large n. The finitely many exceptional terms do not influence the uniform distribution modulo 1 of the sequence.

2.1.3 Van der Corput’s Difference Theorem

Van der Coput’s Difference Theorem discusses the distribution of a sequence based on the distribution of “differences” between terms of the sequence. It essentially allows us to replace a sequence by its derivative and thus will be a very useful tool in discussing the distribution of a sequences which have one of their higher order derivatives being uniformly distributed.

Theorem 2.1.5 Let (x_{n}) be a sequence of real numbers. If for every positive integer h the sequence (x_{n+h} − x_{n}) is u.d. mod 1 then (x_{n}) is uniformly distributed modulo 1.

To prove the difference theorem we first need the following inequality.

Lemma 2.1.6 Let u_{1} ⋯ u_{N} be complex numbers and let H be an integer with 1 ≤ H ≤ N. Then

H^{2} |∑_{n=1}^{N} u_{n}|^{2} ≤ H(N +H−1) ∑_{n=1}^{N} |u_{n}|^{2}+2(N +H−1) ∑_{h=1}^{H−1}(H−h)ℜ ∑_{n=1}^{N−h} u_{n}ū_{n+h}

Proof of Lemma 2.1.6 Define u_{n} = 0 for all n < 0 and all n > N. Then we have

H ∑_{n=1}^{N} u_{n} = ∑_{p=1}^{N+H−1} ∑_{h=1}^{H−1} u_{p−h}

The Cauchy-Schwartz inequality states that

∑_{n=1}^{n} a_{n}b_{n} ≤ (∑_{p=1}^{P} |a_{p}|^{2})^{\frac{1}{2}} (∑_{p=1}^{P} |b_{p}|^{2})^{\frac{1}{2}}  (2.8)

Letting a_{n} = 1 and b_{n} = ∑_{h=0}^{H−1} u_{p−h} in (2.8) gives:

H^{2} |∑_{n=1}^{N} u_{n}|^{2} = | ∑_{p=1}^{N+H−1} \underbrace{1}_{a_{n}} \underbrace{∑_{h=1}^{H−1} u_{p−h}}_{b_{n}}|^{2}

≤ (N + H − 1) ∑_{p=1}^{N+H−1} |∑_{h=0}^{H−1} u_{p−h}|^{2} by (2.8)

= (N + H − 1) ∑_{p=1}^{N+H−1} (∑_{r=0}^{H−1} u_{p−r}) (∑_{s=0}^{H−1} ū_{p−s})

= (N + H − 1) ∑_{p=1}^{N+H−1} ∑_{h=0}^{H−1} |u_{p−h}|^{2}

+ 2(N + H − 1)ℜ ∑_{p=1}^{N+H−1} ∑_{r,s=0s<r}^{H−1} u_{p−r}ū_{p−s}

= (N + H − 1)(Σ_{1} + 2ℜΣ_{2})

Now Σ_{1} = H ∑_{n=1}^{N} |u_{n}|^{2}. Σ_{2} contains terms of the form u_{n}ū_{n+h} where n = 1 ⋯ N and h = r−s = 1 ⋯ H−1. For fixed n and h the possible choices for the pair (r, s) are given explicitly by ( h , 0 ), (h − 1 , 1 ), ⋯, (H − 1 , H − h − 1 ) and for each such choice the value of p is unique. Hence there are exactly H − h occurrences of u_{n}ū_{n+h} in Σ_{2}. Thus we have

Σ_{2} = ∑_{h=1}^{H−1}(H − h) ∑_{n=1}^{N} u_{n}ū_{n+h}

And since u_{n} = 0 for n > N the summation over n can be restricted to summation over 1 ≤ n ≤ N − h, and Lemma 2.1.6 follows.

Proof of Theorem 2.1.5 Let m be a fixed non zero integer. We apply Lemma 2.1.6 with u_{n} = e^{2πimx_{n}} and dividing by H^{2}N^{2} we get:

\begin{aligned}|\frac{1}{N} ∑_{n=1}^{N} e^{2πimx_{n}} |^{2} & ≤ \frac{N + H − 1}{HN} + \\ & 2 ∑_{h=1}^{H−1} \frac{(N + H − 1)(H − h)(N − h)}{H^{2}N^{2}} |\frac{1}{N − h} ∑_{n=1}^{N−h} e^{2πim(x_{n}−x_{n+h})}| \end{aligned} (2.9)

Now since the sequence (x_{n+h} − x_{n}) is u.d. mod 1 for every h ≥ 1, for such h we have

lim_{N→∞} \frac{1}{N − h} ∑_{n=1}^{N−h} e^{2πim(x_{n}−x_{n+h})} = 0  (2.10)

By (2.9) and (2.10) we get

lim sup_{N→∞} |\frac{1}{N} ∑_{n=1}^{N} e^{2πimx_{n}} |^{2} ≤ \frac{1}{H^{2}}  (2.11)

and since (2.11) holds for every positive integer H we have:

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πimx_{n}} = 0

and thus (x_{n}) is u.d. mod 1.

2.1.4 Generalised Van Der Corput’s Difference Theorem

Although Van der Corput’s Difference theorem is a very powerful tool, it is not particularly useful for sequences whose differences are not u.d. mod 1. There are sequences whose derivative is “almost” u.d. mod 1, about which we would like to say something. We say that a sequence (x_{n}) is “almost” u.d. if (x_{n}) does not satisfy Weyl’s criterion, but has the property that

lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihx_{n}} | → 0

as h → ∞. We develop the generalised Van Der Corput’s Difference theorem stated below to deal with sequences whose differences are “almost” uniformly distributed modulo 1.

Theorem 2.1.7 Let (x_{n}) be a sequence of real numbers. If for every positive integer h we have

lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πim(x_{n+h}−x_{n})}| ≤ \frac{K}{h^{α}}  (2.12)

for some α > 0 then (x_{n}) is uniformly distributed modulo 1.

If α ≥ 1 replace α by α′ with α′ < 1. Certainly (2.12) still holds if α is replaced by α′ and so we can restrict our proof to deal with cases when α < 1. In order to prove the theorem we will first need the following proposition:

Proposition 2.1.8 If N > H and α < 1 then

\frac{1}{N} ∑_{h=1}^{H−1} \frac{(N + H − 1)(H − h)(N − h)}{H^{2}N^{2}h^{α}} ≤ KH^{−α}

Proof Firstly we observe that for α < 1

∑_{h=1}^{H−1} h^{−α} = \underbrace{1 + \frac{1}{2^{α}} + \frac{1}{3^{α}} + ⋯ + \frac{1}{(2^{\frac{1}{α}})^{α}}}_{≤2^{\frac{1}{α}} .1} + \underbrace{\frac{1}{(2^{\frac{1}{α}} + 1)^{α}} + ⋯ + \frac{1}{(2^{\frac{2}{α}})^{α}}}_{≤2^{\frac{2}{α}} . \frac{1}{2}} + ⋯

≤ 2^{\frac{1}{α}} .1 + (2^{\frac{1}{α}})^{2}.\frac{1}{2} + (2^{\frac{1}{α}})^{3}.\frac{1}{4} + ⋯ + (2^{\frac{1}{α}})^{⌈log_{2} H^{α}⌉}.\frac{1}{2^{⌈log_{2} H^{α}⌉−1}}

= 2^{\frac{1}{α}} (1 + 2^{( \frac{1}{α} −1)} + 2^{2( \frac{1}{α} −1)} + ⋯ + 2^{(log_{2} H^{α}−1)( \frac{1}{α} −1)})

= 2^{\frac{1}{α}} ∑_{j=0}^{log_{2} H^{\frac{1}{3}} −1} 2^{j( \frac{1}{α} −1)}

= 2^{\frac{1}{α}} \frac{2^{( \frac{1}{α} −1) log_{2} H^{α}} − 1}{2^{( \frac{1}{α} −1)} − 1}

= \frac{2^{\frac{1}{α}}}{2^{( \frac{1}{α} −1)} − 1} (2^{log^{2} H^{1−α}} − 1)

= K (H^{1−α} − 1)

≤ KH^{1−α} (2.13)

Then

\begin{aligned}∑_{h=1}^{H−1} \frac{(N + H − 1)(H − h)(N − h)}{H^{2}N^{2}h^{α}} & < ∑_{h=1}^{H−1} \frac{2NHN}{H^{2}N^{2}h^{α}} \\ & = \frac{2}{H} ∑_{h=1}^{H−1} \frac{1}{h^{α}} \\ & ≤ \frac{KH^{1−α}}{H} & (Using (2.13)) \\ & = KH^{−α} \end{aligned} (2.14)

And so the proposition is proved.

Proof of Theorem 2.1.7 We now re-work the proof of Van Der Corput’s Difference Theorem. Taking lim sups as N → ∞ equation (2.9) becomes:

lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πimx_{n}} |^{2} ≤ lim sup_{N→∞} (\frac{N + H − 1}{HN}+

2 ∑_{h=1}^{H−1} \frac{(N + H − 1)(H − h)(N − h)}{H^{2}N^{2}} |\frac{1}{N − h} ∑_{n=1}^{N−h} e^{2πim(x_{n}−x_{n+h})}| )

≤ lim sup_{N→∞} (\frac{N + H − 1}{HN} + 2 ∑_{h=1}^{H−1} \frac{(N + H − 1)(H − h)(N − h)}{H^{2}N^{2}h^{α}})

(Using the hypothesis)

≤ lim sup_{N→∞} (\frac{N + H − 1}{HN} + \frac{K}{H^{−α}} )

(Using Proposition 2.1.8)

≤ \frac{K}{H^{−α}}

(2.15)

Thus for every positive integer H we have:

lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πimx_{n}} | ≤ \frac{K}{H^{α/2}}  (2.16)

which means that lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πimx_{n}} = 0 and so (x_{n}) is u.d. mod 1.

2.2 Tools

In this section we outline some of the tools which will be used often in the course of this report. Generally the tools simplify the approximation of exponential sums. They include estimating sums by integrals and then simplifying these integrals (Sections 2.2.1, 2.2.2 and 2.2.3), approximating exponential sums of sequences by exponential sums of nearby simpler sequences (Section 2.2.4) and ignoring the first few terms of a sequence (Section 2.2.5).

2.2.1 The Euler Summation Formula

Euler’s summation formula provides a method for us to change sums into integrals. This is particularly useful as it is far easier for us to evaluate the integrals than it is to evaluate the sums. The formula states that:

∑_{n=M}^{N} F (n) = ∫_{M}^{N} F (x)dx + \frac{1}{2}(F (M) + F (N)) + ∫_{M}^{N} (\{x\} − \frac{1}{2}) F′(x)dx (2.17)

For a proof of this see e.g. [6] page 25.

Given a function f(n) we let F (x) = e^{2πihf(x)}. After dividing both sides of (2.17) by N and taking the limit as N → ∞ we arrive at the following formula:

lim sup_{N→∞} | ∑_{n=M}^{N} e^{2πihf(n)}| = |lim sup_{N→∞} \frac{1}{N} \underbrace{∫_{M}^{N} e^{2πihf(x)}dx}_{First Term} + lim sup_{N→∞} \frac{1}{2N} (e^{2πihf(M)} + e^{2πihf(N)})

+ lim sup_{N→∞} \frac{1}{N} \underbrace{∫_{M}^{N} (\{x\} − \frac{1}{2}) (2πih)f′(x)e^{2πihf(x)}dx}_{Third Term}|

≤ lim sup_{N→∞} \frac{|First Term|}{N} + lim sup_{N→∞} \frac{1}{2N} \underbrace{|e^{2πihf(M)} + e^{2πihf(N)}|}_{<2}

+ lim sup_{N→∞} \frac{|Third Term|}{N}

= lim sup_{N→∞} \frac{|First Term|}{N} + lim sup_{N→∞} \frac{|Third Term|}{N}

(2.18)

Note also that:

lim sup_{N→∞} \frac{|Third Term|}{N} = lim sup_{N→∞} \frac{1}{N} |∫_{M}^{N} (\{x\} − \frac{1}{2}) (2πih)f′(x)e^{2πihf(x)}dx|

≤ lim sup_{N→∞} \frac{K}{N} ∫_{M}^{N} |\{x\} − \frac{1}{2}| |f′(x)| |e^{2πihf(x)}| dx

≤ lim sup_{N→∞} \frac{K}{N} ∫_{M}^{N} |f′(x)|dx

(2.19)

If we can show that lim sup_{N→∞} \frac{|First Term|}{N} = 0 and lim sup_{N→∞} \frac{|Third Term|}{N} = 0 this implies that

lim sup_{N→∞} \frac{1}{N} ∑_{n=M}^{N} e^{2πihf(x)} = 0

2.2.2 Integration By Parts

We will frequently want to estimate the size of an integral of a function of the form e^{2πihf(x)}. To do this we will often use the trick of integrating by parts. Let A be the interval (or set of intervals) over which we are integrating. Then we get

\begin{aligned}|∫_{A} e^{2πihf(x)}dx| & = |∫_{A} \frac{d}{dx} (e^{2πihf(x)}) \frac{dx}{2πihf′(x)}| \\ & ≤ \frac{1}{h} |[\frac{e^{2πihf(x)}}{f′(x)} ]_{A}| + \frac{1}{h} |∫_{A} \frac{e^{2πihf(x)}f′′(x)}{(f′(x))^{2}}dx| \\ & ≤ \frac{1}{h} [\frac{1}{|f′(x)|}]_{A} + \frac{1}{h} ∫_{A} |\frac{f′′(x)}{(f′(x))^{2}} | dx \end{aligned}  (2.20)

It will often be easier for us to approximate this new expression.

2.2.3 Integrating Functions Whose Derivative can be Simplified

Proposition 2.2.1 If f(x) is a function whose derivative can be decomposed into f′(x) = A(x) + B(x) with ϵ|A(x)| ≥ |B(x)| and S is an interval or (set of intervals) over which we are integrating then

\frac{1}{h} |∫_{S} \frac{d}{dx} (e^{2πihf(x)}) \frac{dx}{f′(x)}| ≤ \frac{1}{h} |∫_{S} \frac{d}{dx} (e^{2πihf(x)}) \frac{dx}{A(x)}| + ϵK|S| (2.21)

Proof

\begin{aligned}\frac{1}{h} |∫_{S} \frac{d}{dx} (e^{2πihf(x)}) \frac{dx}{f′(x)}| & = \frac{1}{h} |∫_{S} \frac{d}{dx} (e^{2πihf(x)}) \frac{dx}{A(x) + B(x)}| \\ & = \frac{1}{h} |∫_{S} \frac{d}{dx} (e^{2πihf(x)}) (\frac{1}{A(x)} − \frac{B(x)}{(A(x) + B(x))A(x)}) dx| \\ & \frac{1}{h} ≤ |∫_{S} \frac{d}{dx} (e^{2πihf(x)}) \frac{dx}{A(x)}| \\ & + \frac{1}{h} |∫_{S} \frac{d}{dx} (e^{2πihf(x)}) \underbrace{\frac{1}{A(x) + B(x)}}_{f′(x)} \underbrace{\frac{B(x)}{A(x)}}_{≤ϵ} dx| \\ & ≤ \frac{1}{h} |∫_{S} \frac{d}{dx} (e^{2πihf(x)}) \frac{dx}{A(x)}| \\ & + \frac{1}{h} ∫_{S} |2πihf′(x)e^{2πihf(x)} \frac{ϵ}{f′(x)}| dx \\ & ≤ \frac{1}{h} |∫_{S} \frac{d}{dx} (e^{2πihf(x)}) \frac{dx}{A(x)}| + ϵK|S| \end{aligned} (2.22)

And so the proposition follows.

2.2.4 Approximating Sequences by Simpler Sequences

Theorem 2.2.2 Let (v_{n}) and (u_{n}) be sequences. If for every 0 < ϵ < 1 there is an N_{ϵ} such that v_{n} can be written as v_{n} = u_{n} + ϵ_{n} with ϵ_{n} < ϵ ∀n > N_{ϵ} then

lim_{N→∞} \frac{1}{N} (∑_{n=1}^{N} e^{2πihv_{n}} − ∑_{n=1}^{N} e^{2πihu_{n}}) = 0

Proof Firstly we observe that using Taylor Expansion gives:

\begin{aligned}|e^{2πih(x+ϵ)} − e^{2πihx}| & = |ϵ2πihe^{2πihx} − ϵ^{2}2π^{2}h^{2}e^{2πihx} + ⋯ | \\ & ≤ K_{1}ϵ + K_{2}ϵ^{2} + ⋯ \\ & ≤ Kϵ \\ & = O(ϵ) \end{aligned}  (2.23)

And thus

lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihv_{n}} − ∑_{n=1}^{N} e^{2πihu_{n}} |

= lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N} (e^{2πihv_{n}} − e^{2πihu_{n}}) |

≤ \underbrace{lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N_{ϵ}} (e^{2πihv_{n}} − e^{2πihu_{n}}) |}_{=0} + lim_{N→∞} \frac{1}{N} ∑_{n=N_{ϵ}}^{N} \underbrace{|e^{2πihv_{n}} − e^{2πihu_{n}} |}_{O(ϵ_{n})}

≤ lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} O(ϵ_{n})

≤ O(ϵ)

(2.24)

Since (2.24) holds ∀ ϵ the theorem follows.

The following two corollaries follow as a consequence of the above theorem.

Corollary 2.2.3 Let (u_{n}) and (v_{n}) be sequences which satisfy the hypothesis of Theorem 2.2.2. Then

lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihv_{n}} | = lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihu_{n}} | + O(ϵ)

Proof

lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihv_{n}} | = lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihv_{n}} − ∑_{n=1}^{N} e^{2πihu_{n}} + ∑_{n=1}^{N} e^{2πihu_{n}} |

= \underbrace{lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihv_{n}} − ∑_{n=1}^{N} e^{2πihu_{n}} |}_{=0} + lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihu_{n}} |

= lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihu_{n}} |

(2.25)

Corollary 2.2.4 Let (u_{n}) and (v_{n}) be sequences which satisfy the hypothesis of Theorem 2.2.2. Then

(i) If (u_{n}) is uniformly distributed modulo 1 then (v_{n}) is also uniformly distributed modulo 1. (ii) If (u_{n}) is not uniformly distributed modulo 1 then neither is (v_{n}).

Proof

(i) If (u_{n}) is u.d. mod 1 then the lim sup on the right hand side of (2.25) can be replaced by a limit and we get:

0 = lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihu_{n}} | = lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihv_{n}} |

and thus (v_{n}) is also uniformly distributed modulo 1. (ii) Let w_{n} = u_{n} − ϵ_{n}. By the contrapositive of (i) if (w_{n} + ϵ_{n}) (= (u_{n})) is not u.d. mod 1 then neither is (w_{n}). Hence if (u_{n}) is not u.d. mod 1, then neither is (u_{n} − ϵ_{n}) and similarly, neither is (u_{n} + ϵ_{n}) = (v_{n}).

2.2.5 Changing the summation limits

Proposition 2.2.5

(i) If for every ϵ > 0 and for every h ≠ 0, lim sup_{N→∞} \frac{1}{N} |∑_{ϵN}^{N} e^{2πihf(n)}| = O(ϵ) then (f(n)) is uniformly distributed modulo 1. (ii) Conversely, if for every ϵ > 0 there is an h ≠ 0 such that lim inf_{N→∞} \frac{1}{N} ∑_{ϵN}^{N} e^{2πihf(n)} ≠ 0 then (f(n))is not uniformly distributed modulo 1.

(iii) (f(n)) is uniformly distributed modulo 1 iff lim_{N→∞} ^{\frac{ϕ_{(ϵN,N)}^{f(n)} (a,b)}{N}} = (b − a) (1 − ϵ) .

Proof

(i) Pick ϵ > 0 then using the hypothesis of part (i) we have

lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihf(n)}| ≤ lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{ϵN} e^{2πihf(n)}| + lim sup_{N→∞} \frac{1}{N} | ∑_{n=ϵN}^{N} e^{2πihf(n)}|
= O(ϵ)

Since we can pick ϵ to be arbitrarily small we conclude that

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πihf(n)} = 0

and so (f(n)) is u.d. mod 1. (ii) Firstly we will show that if (f(n)) is u.d. mod 1 then ∀ϵ > 0

lim_{N→∞} \frac{1}{N} ∑_{ϵN}^{N} e^{2πihf(n)} = 0.

lim_{N→∞} | ∑_{n=ϵN}^{N} e^{2πihf(n)}| = | lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πihf(n)} − lim_{N→∞} \frac{1}{N} ∑_{n=1}^{ϵN} e^{2πihf(n)}|

≤ \underbrace{lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihf(n)}|}_{=0} − \underbrace{lim_{N→∞} \frac{1}{N} |∑_{n=1}^{ϵN} e^{2πihf(n)}|}_{=0}

= 0 (2.26)

It should be noted that since all the limits on the right hand side of (2.26) exist then the limit on the left hand side must also exist.

The converse of this statement proves (ii).

The statements of (i) and (ii) combine to give an interesting result. They show that if lim sup_{N→∞} \frac{1}{N} |∑_{n=ϵN}^{N} e^{2πihf(n)}| = O(ϵ) then it must necessarily be zero. (iii) Firstly if (f(n)) is u.d. mod 1 then ∀ϵ > 0

lim_{N→∞} \frac{ϕ_{(ϵN,N)}^{f(n)}(a, b)}{N} = lim_{N→∞} \frac{ϕ_{N}^{f(n)}(a, b)}{N} − lim_{N→∞} \frac{ϵϕ_{ϵN}^{f(n)}(a, b)}{ϵN}
= (b − a)(1 − ϵ)

(2.27)

The converse of (2.27) proves the other direction.

Non-Trigonometric Sequences

The first half of this chapter deals with the distribution of specific families of non-trigonometric sequences such as nθ, log n, n^{α} and combinations of these. The second half of the chapter deals with general families of sequences for which we know something about their derivatives.

3.1 The distribution of nθ

Proposition 3.1.1 If θ is a rational number then (nθ) is not uniformly distributed modulo 1.

Proof Let θ = \frac{p}{q} and let h = q. Then

\begin{aligned}e^{2πihnθ} & = e^{2πinp} \\ & = e^{2πip} \end{aligned}

for all n. Hence,

lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihnθ}| = lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πip}|

= lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N} 1|

= 1 ≠ 0

And thus Weyl’s criterion fails.

Proposition 3.1.2 If θ is irrational then nθ is uniformly distributed modulo 1.

Proof

lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihnθ}| = lim_{N→∞} \frac{1}{N} |∑_{n=1}^{N}(e^{2πihθ})^{n}|

= lim_{N→∞} \frac{1}{N} |\frac{e^{2πihθ}(1 − e^{2πihNθ})}{1 − e^{2πihθ}} |

≤ lim_{N→∞} \frac{1}{N} (\frac{1}{|1 − e^{2πihθ}|} + \frac{|e^{2πihNθ}|}{|1 − e^{2πihθ}|})

= lim_{N→∞} \frac{K}{N}

= 0

Note: The above proof only works as |1 − e^{2πihθ}| ≠ 0 for any value of h, since θ is irrational.

It is not too hard to prove the following Proposition as a consequence of Propositions 3.1.1 and 3.1.2 and Van Der Corput’s difference theorem.

Proposition 3.1.3 If P (n) = a_{0} +a_{1}n+⋯+a_{k}n^{k}, a_{i} ∈ ℝ i = 0, …, k then P (n) is uniformly distributed modulo 1 iff at least one of the a_{i}’s i = 1, …, k is irrational.

Proof For a proof of this see [6] Theorem 3.2 Page 27.

3.2 The distribution of log n

Proposition 3.2.1 (log n) is not uniformly distributed modulo 1.

Proof We will show that (2.2) does not hold for the case when h = 1. To do this we will use Euler’s summation formula.

We look at the terms on the right hand side of equation 2.17.

\frac{First term}{N} = \frac{1}{N} ∫_{1}^{N} e^{2πi log x}dx

= \frac{1}{N} [xe^{2πi log x}]_{1}^{N}

= \frac{K}{N} (Ne^{2πi log N} − 1)

= Ke^{2πi log N}

(3.1)

lim sup_{N→∞} \frac{|Third term|}{N} ≤ lim sup_{N→∞} \frac{1}{N} ∫_{1}^{N} |\frac{2πi}{x}| dx

= lim sup_{N→∞} \frac{K}{N} [log x]_{1}^{N}

= lim sup_{N→∞} \frac{K log N}{N}

= 0

(3.2)

Thus the second and third terms tend to zero as N → ∞, and (3.1) shows that the first term doesn’t converge as N → ∞. This means that lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πi log n} does not exist and so cannot be equal to zero and thus β_{n} = log(n) doesn’t satisfy (2.2). By the necessity of (2.2) we conclude that (log n) is not uniformly distributed modulo 1.

3.3 The distribution of ∑_{j} c_{j}n^{α_{j}}

In this section we will discuss the distribution of (∑_{j} c_{j}n^{α_{j}}). As a result of this discussion we will be able to conclude that (n^{α}) is uniformly distributed modulo 1 for any positive non-integer valued α. (Corollary 3.3.2). It should be noted that the method used to integrate various functions in this section is not necessarily the easiest or most efficient method, however it is the same method as will be necessary in chapter 4. We have used this method here to introduce the reader to arguments which will be used repeatedly later in the report.

For m ∈ ℤ define f_{m}(x) = ∑_{j} c_{j}x^{α_{j}} where the sum consists of finitely many terms, c_{j} ∈ ℝ and α_{j} < m. Let k be the subscript of the largest α_{j} in f_{m}(x). Ensure that α_{k} > 0, α_{k} ∉ ℤ and c_{k} ≠ 0.

Proposition 3.3.1 For f_{m}(x) defined as above, (f_{m}(n)) is uniformly distributed modulo 1.

Proof We will prove this proposition by induction on m. We first consider the case where m = 1. Let k be chosen as above and let k′ be the subscript of the second largest α_{j}. DefineM = max_{j} {|c_{j}α_{j}|} and let T be the number of

terms in the sum of f_{1}(x). Pick N > \frac{1}{ϵ} (\frac{MT}{ϵ|c_{k}α_{k}|})^{\frac{1}{α_{k}−α_{k′}}}. This will mean that for x > ϵN we have ϵc_{k}α_{k}x^{α_{k}−1} < MT x^{α_{k′}−1} which will be necessary during the proof. Then

\begin{aligned}f′_{1}(x) & = ∑_{j} c_{j}α_{j}x^{α_{j}−1} \\ & = \underbrace{c_{k}α_{k}x^{α_{k}−1}}_{A(x)} + \underbrace{∑_{j,j≠k} c_{j}α_{j}n^{α_{j}−1}}_{B(x)} \end{aligned}

And for x > ϵN we have

|B(x)| ≤ ∑_{j,j≠k} |c_{j}α_{j}x^{α_{j}−1}|

≤ ∑_{j,j≠k} |c_{j}α_{j}|x^{α_{k′}−1}

≤ MT x^{α_{k′}−1} ≤ ϵ|c_{k}α_{k}|x^{α_{k}−1}

(Due to our choice of N) = ϵ|A(x)| (3.3)

We will thus be able to apply Proposition 2.2.1.

In order to show that f_{1}(n) is u.d. mod 1 we will consider lim sup_{N→∞} \frac{1}{N} ∑_{n=ϵN}^{N} e^{2πihf_{1}(n)} and show that this is of order ϵ. Invoking Proposition 2.2.5 will then give the desired result. We now look at the terms of the Euler summation formula given in section 2.2.1.

lim sup_{N→∞} \frac{|First Term|}{N} = lim sup_{N→∞} \frac{1}{N} |∫_{ϵN}^{N} e^{2πih ∑_{j} c_{j}x^{α_{j}}} dt|

= lim sup_{N→∞} \frac{1}{N} |∫_{ϵN}^{N} \frac{d}{dx} (e^{2πih ∑_{j} c_{j}x^{α_{j}}} ) (∑_{j} α_{j}c_{j}x^{α_{j}−1})^{−1}|

≤ lim sup_{N→∞} \frac{K}{N} |∫_{ϵN}^{N} \frac{d}{dx} (e^{2πih ∑_{j} c_{j}x^{α_{j}}} ) (α_{k}c_{k}x^{α_{k}−1})^{−1} dx| + lim sup_{N→∞} \frac{ϵK′N}{N}

(Using (3.3) and Proposition 2.2.1)

≤ lim sup_{N→∞} \frac{K}{N} ( |[\frac{1}{α_{k}c_{k}x^{α_{k}−1}} ]_{ϵN}^{N} | + ∫_{ϵN}^{N} |x^{−α_{k}}|dx ) + ϵK′

≤ lim sup_{N→∞} (\frac{K_{1}}{N} [x^{1−α_{k}}]_{M}^{N} + \frac{K_{2}}{N} ∫_{M}^{N} x^{−α_{k}}dx) + ϵK′

≤ lim sup_{N→∞} \frac{K}{N} (N^{1−α_{k}} + N^{1−α_{k}}) + ϵK′

= lim sup_{N→∞} KN^{−α_{k}} + ϵK′

= ϵK′ = O(ϵ)

(3.4)

lim sup_{N→∞} \frac{|Third term|}{N} = lim sup_{N→∞} \frac{K}{N} ∫_{ϵN}^{N} |∑_{j} α_{j}c_{j}x^{α_{j}−1}| dx

≤ lim sup_{N→∞} \frac{KT}{N} ∫_{ϵN}^{N} x^{(α_{k}−1)}dx

= lim sup_{N→∞} \frac{K}{N} [x^{α_{k}}]_{ϵN}^{N}

≤ lim sup_{N→∞} \frac{KN^{α_{k}}}{N}

= 0

(3.5)

Thus combining equations (3.4) and (3.5) and using the results of section 2.2.1 and Proposition 2.2.5 we can conclude that (f_{1}(n)) is u.d. mod 1.

Now assume that (f_{b}(n)) is u.d. mod 1 for all b ≤ m, and consider the distribution of (f_{m}(n)). Once again we let k be the subscript of the largest α_{j} in f_{m}(n). We can assume that the c_{k} ≠ 0 as otherwise we could just use the induction hypothesis to prove the u.d. mod 1 of the sequence (f_{m}(n)). Using Taylor expansion we get:

f_{m}(n + h) − f_{m}(n)

= hf_{m}^{(1)}(n) + \frac{h^{2}}{2} f_{m}^{(2)}(n) + ⋯ + \frac{h^{m−1}}{(m − 1)!}f_{m}^{(m−1)}(n) + O(n^{α_{k}−1})

(Where the constant in the O term depends on the value of h.)

= h ∑_{j} α_{j}c_{j}n^{α_{j}−1} + \frac{h^{2}}{2} ∑_{j}(α_{j} − 1)α_{j}c_{j}n^{α_{j}−2} + ⋯

+ \frac{h^{m−1}}{(m − 1)!}(Kn^{α_{k}} + l.o.t.) + O(n^{α_{k}−1})

= f_{m−1}^{∗}(n) + f_{m−2}^{∗}(n) + ⋯ f_{1}^{∗}(n) + \underbrace{O(n^{α_{k}−1})}_{<ϵ}

(Where we can make the last term as small as we want.) = f_{m−1}^{∗∗}(n) + O(ϵ)

(3.6)

Now the leading co-efficient of f_{m−1}^{∗∗}(n) is α_{k}c_{k} which is non-zero and the leading power of f_{m−1}^{∗∗}(n) is (α_{k} − 1) ∉ ℤ so f_{m−1}^{∗∗}(n) satisfies the induction hypothesis and so is u.d. mod 1. Hence by (2.1.5) (f_{m}(n)) is u.d. mod 1 for all m.

Corollary 3.3.2 If 0 < α, α ∉ ℤ then (n^{α}) is uniformly distributed modulo 1.

Proof Simply let m = ⌈α⌉ and the corollary follows.

3.4 The distribution of (∑ c_{j}n^{α_{j}} log^{τ_{j}} n)

For k > 0 let f_{k}(n) = ∑ c_{j}n^{α_{j}} log^{τ_{j}} n where 0 < α_{j} < k, each α_{j} is unique and at least one α_{j} is positive. In addition we require that c_{j} ≠ 0.

Proposition 3.4.1 For f_{k}(n) defined as above, (f_{k}(n) is uniformly distributed modulo 1 for every k.

Proof We will prove Proposition 3.4.1 by induction on k using a similar argument to that invoked in the previous section. To deal with the distribution of f_{1}(n) we will invoke (Féjer’s Theorem)

f′_{1}(x) = ∑ α_{j}c_{j}x^{α_{j}−1} log^{τ_{j}} x + τ_{j}c_{j}x^{α_{j}−1} log^{τ_{j}−1} x

→ 0 monotonically as x → 0

and

x|f′_{1}(x)| = |∑ α_{j}c_{j}x^{α_{j}} log^{τ_{j}} x + τ_{j}c_{j}x^{α_{j}} log^{τ_{j}−1} x|
→ ∞ as x → 0

So f_{1}(n) satisfies the hypothesis of Theorem 2.1.3 (Féjer’s Theorem) and Proposition 3.4.1 holds for k = 1.

Assume that Proposition 3.4.1 holds for all k ≤ m, and consider the case when k = m + 1. Once again we let f_{k}(n) = ∑ c_{j}n^{α_{j}} log^{τ_{j}} n where α_{j} < k. Let

α = max_{j} α_{j} and τ = max_{j,α_{j}=α} τ_{j}

Finally, let r be the index such that α_{r} = α and τ_{r} = τ. Using Taylor expansion gives:

f_{m+1}(n + h) − f_{m+1}(n)

= hf_{m+1}^{(1)}(n) + \frac{h^{2}}{2} f_{m+1}^{(2)}(n) + ⋯ + \frac{h^{k}}{k!} f_{m+1}^{(k)}(n) + O(n^{s})

where s < 0 and the constant in the O term depends on the value of h

= f_{m}^{∗}(n) + \underbrace{∑_{i=0}^{m−1} ∑_{j} C_{ij}n^{α_{j}−i} log^{τ_{b_{ij}}} n + O(n^{s})}_{<ϵ}

where we can make the last term as small as we want (3.7)

Now the coefficient of n^{α_{r}−1} log^{τ_{r}} n is non-zero and so f_{m}^{∗}(n) satisfies the induction hypothesis and thus is u.d. mod 1. Hence by (2.1.5) (∑ c_{j}n^{α_{j}} log^{τ_{j}} n) is u.d. mod 1

3.5 Distribution of “slow sequences”

In this section we look at sequences which are increasing slowly. The theorem which follows is actually proved more generally and succinctly as a corollary to Theorem 3.6.1. The reason for including this proof is that a very different approach to proving the theorem is used here. Generally when proving the uniform distribution or otherwise of a sequence we try to place either a lower or upper bound on its exponential sum, and the most common way of doing that is by changing the sums into integrals and then integrating. In this section we deal directly with the terms of the exponential sum, and show that the imaginary part of the sum does not go to zero. Thus we obtain a much better understanding of how the exponential sums are behaving.

Theorem 3.5.1 If 0 < f′(x) < \frac{1}{2πx} and f′′(x) < 0 then f(n) is not uniformly distributed modulo 1.

3.5.1 Outline of Proof

To begin with we observe that if lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πihf(n)} = 0 then both the real and imaginary parts of this sum go to zero. In this proof we will show that lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} sin 2πf(n) ≠ 0 and thus Weyl’s criterion fails for h = 1 and hence the sequence is not u.d. mod 1.

To do this we will show that with each rotation about the unit circle the sum of the imaginary terms accrued is strictly negative. This will be done by breaking one rotation around the circle into four components, with the kth component being when θ ∈ [\frac{kπ}{2}, \frac{(k+1)π}{2}], k = 1, 2, 3, 4.

Terms in the 1st and 2nd components have positive imaginary parts, while terms in the 3rd and 4th components have negative imaginary parts. We will show that the negative terms in the 3rd component cancel out with the positive terms in the 1st component and the negative terms in the 4th component cancel out the positive terms in the 2nd one. In addition we will show that there are many extra negative terms left over after this cancellation process, and so the total sum around the circle is negative. We will need to take special care when dealing with terms whose imaginary part is very close to 1 to ensure that they are cancelled out correctly. Proof Pick N_{0} to be large and w.l.o.g. assume f(N_{0}) = k for k ∈ ℤ Let N_{1} be the last n > N_{0} with f(n) − f(N_{0}) < \frac{1}{4} Let N_{2} be the last n > N_{0} with f(n) − f(N_{0}) < \frac{1}{2} Let N_{3} be the last n > N_{0} with f(n) − f(N_{0}) < \frac{3}{4} Let N_{4} be the last n > N_{0} with f(n) − f(N_{0}) < 1

Let d = \frac{3}{2π}. Note: The choice of the number 3 on the numerator of d was an arbitrary choice, we could just have easily have chosen any number less than π. We will now define M_{i}’s, i = 0, …, 3 so that |f(M_{1}) − f(M_{0})| ≃ d and |f(M_{3}) − f(M_{2})| ≃ d. The intervals [f(M_{0}), f(M_{1})] and [f(M_{2}), f(M_{3})] will be placed approximately in the center of the intervals [0, \frac{1}{2}] and [\frac{1}{2}, 1] respectively. Let M_{0} be the last n > N_{0} with f(n) − f(N_{0}) < \frac{1}{2}(\frac{1}{2} − d) Let M_{1} be the first n > N_{0} with f(n) − f(N_{0}) > \frac{1}{2}(\frac{1}{2} + d) Let M_{2} be the last n > N_{0} with f(n) − f(N_{0}) < \frac{3}{2}(\frac{3}{2} − d) Let M_{3} be the first n > N_{0} with f(n) − f(N_{0}) > \frac{3}{2}(\frac{3}{2} + d)

We will show that

ℑ ∑_{n=N_{0}}^{N_{4}} e^{2πif(n)} < 0  (3.8)

and

\frac{1}{M_{3}} ℑ ∑_{n=N_{0}}^{M_{3}} e^{2πif(n)} < −0.01  (3.9)

and that (3.8) and (3.9) combine to give

lim_{N→∞} \frac{1}{N} ℑ∑_{n=1}^{N} e^{2πif(n)} ≠ 0  (3.10)

so (2.2) fails and hence (f(n)) is not uniformly distributed modulo 1. We will need the following lemma

Lemma 3.5.2 let f be a function whcih satisfies the hypothesis of Theorem 3.5.1. Then:

f′ (f^{−1} (f(a) + \frac{1}{2})) < \frac{1}{3}f′(a)

Note: Since f′(x) > 0 ∀x then f has an inverse.

Proof By the MVT and our assumptions about f′(x),

f^{−1} (f(a) + \frac{1}{2}) ≥ |f^{−1} (f(a) + \frac{1}{2}) − a|

= \frac{|f(a) + \frac{1}{2} − f(a)|}{f′(ξ)} (where a ≤ ξ ≤ f^{−1} (f(a) + \frac{1}{2}) )

≥ \frac{1}{2f′(a)}

f′ (f^{−1} (f(a) + \frac{1}{2})) ≤ \frac{1}{f^{−1} (f(a) + \frac{1}{2})}

≤ \frac{2f′(a)}{2π}

< \frac{1}{3}f′(a)

Corollary 3.5.3 If we let a_{j} = N_{0} + j, j = 0, 1, …, M_{1} − N_{0} − 1, then for each j there are at least three points, a_{0j}, a_{1j}, a_{2j} which satisfy a_{ij} ∈ [N_{2} + 1, M_{3}] and f(a_{j}) + \frac{1}{2} ≤ f(a_{ij}) ≤ f(a_{j} +1) + \frac{1}{2}, i = 0, 1, 2.

3.5.2 Dealing with terms lying between N_{0} and N_{1} − 1

Let m = f′(N_{2}), Then for n > N_{2}, |f(n) − f(n − 1)| < m so, ∀j > 0 ∃S_{j} > N_{2} such that (f(S_{j}) − f(N_{0})) ∈ [ frac12 + jm, \frac{1}{2} + (j + 1)m).

Also, for N_{0} ≤ n ≤ N_{1} |f(n) − f(n−1)| ≥ m, so ∀j ≥ 1, k = 1, …, (N_{1} − N_{0} − 1) if (f(N_{0} + k) − f(N_{0})) ∈ [(j − 1)m, jm) then it is the only f(N_{0} + k) which has this property.

Now let j_{k} be the j ≥ 1 such that (f(N_{0} + k)−f(N_{0})) ∈ [(j_{k}−1)m, j_{k}m) and let S_{k} satisfy N_{2} < S_{k} ≤ N_{3} and (f(S_{k}) − f(N_{0})) ∈ [\frac{1}{2} + j_{k}m, \frac{1}{2} + (j_{k} + 1)m).

Remember that sin 2π(f(n) − f(N_{0})) = sin 2πf(n) and in addition sin 2πf(N_{0} + k) < | sin 2πf(S_{k})| so

∑_{n=N_{0}+1}^{N_{1}−1} sin 2πf(n) = ∑_{k=1}^{N_{1}−N_{0}−1} sin 2πf(N_{0} + k)

< ∑_{k=1}^{N_{1}−N_{0}−1} | sin 2πf(S_{k})|

(3.11)

3.5.3 Dealing with terms lying between N_{1} + 2 and N_{2}

Let M = f′(N_{3}), Then for n > N_{3}, |f(n) − f(n − 1)| < M so, ∀j > 0 ∃R_{j}, R_{j} > N_{3} such that (f(R_{j}) − f(N_{0})) ∈ [1 − (j + 1)M, 1 − jM)

Also, for N_{1} ≤ n ≤ N_{2} |f(n) − f(n−1)| ≥ M, so ∀j ≥ 1, k = 0, …, (N_{2} − N_{1} − 2) if (f(N_{2} − k)−f(N_{0})) ∈ [π−jM, π−(j −1)M) then it is the only f(N_{2}−k) which has this property.

Now let j_{k} be the j ≥ 1 such that (f(N_{2} − k)−f(N_{0})) ∈ [\frac{1}{2} − j_{k}M, \frac{1}{2} − (j_{k} − 1) M). Starting with k = N_{2} − N_{1} − 2, and decrementing the value of k by 1 each time let R_{k} be chosen as follows: If (1 − j_{k}M) < \frac{1}{2}(\frac{3}{2}+d) then pick R_{k} so that (f(R_{k})−f(N_{0})) ∈ [1 − (j_{k} + 1)M, 1 − j_{k}M). Otherwise pick R_{k} with M_{2} < R_{k} < M_{3} and R_{k} ≠ R_{j} for j > k, R_{k} ≠ S_{i} for any i.

Then sin 2πf(N_{2} − k) < | sin 2πf(R_{k})| and

∑_{n=N_{1}+2}^{N_{2}} sin 2πf(n) = ∑_{k=0}^{N_{2}−N_{1}−2} sin 2πf(N_{2} − k)

< ∑_{k=0}^{N_{2}−N_{1}−2} | sin 2πf(R_{k})|

(3.12)

3.5.4 Dealing with the terms N_{1} and N_{1} + 1

By Corollary 3.5.3 there are at least three terms a_{i}, i = 0, 1, 2 with f(a_{i}) ∈ [\frac{1}{2} + f(N_{1}), \frac{1}{2} + f(N_{1} + 1)], and these a_{i}’s do not correspond to any S_{k} or R_{k}. Without loss of generality assume a_{0} < a_{1} < a_{2}.

We need to show that

∑_{i=0}^{2} | sin 2πf(a_{i})| > sin 2πf(N_{1}) + sin 2πf(N_{1} + 1).  (3.13)

If f(a_{0}) ≤ \frac{3π}{2} and f(a_{2}) ≥ \frac{3π}{2} then (3.13) clearly holds.

We now consider the cases where (f(a_{0}) − f(N_{0})) > \frac{3}{4} or (f(a_{2}) − f(N_{0})) < \frac{3}{4}. These can essentially be treated as the same case, so w.l.o.g. assume f(a_{0}) > \frac{3}{4}. Then | sin 2πf(a_{2})| ≥ sin 2πf(N_{1} + 1) and it remains to show that | sin 2πf(a_{0})| + | sin 2πf(a_{1})| ≥ sin 2πf(N_{1}).

Let D = f(N_{1}+1)−f(N_{1}). Then f(a_{1}) − f(a_{0}) ≤ \frac{D}{3} and f(a_{2}) − f(a_{1}) ≤ \frac{D}{3}. Now we know that f(N_{1} + 1) − f(N_{1}) < \frac{1}{6N_{1}} , however we will assume that (f(N_{1} + 1) − f(N_{1})) ∼ \frac{1}{2} and thus f(a_{0}) ≃ \frac{5}{6} and f(a_{1}) ≃ \frac{11}{12} as this is the worst possible scenario. Even in this worst case scenario

| sin 2πf(a_{0})| + | sin 2πf(a_{1})| ≥ \frac{\sqrt{3}}{2} + \frac{1}{2}
> 1
> sin 2πf(N_{1})

and so (3.13) holds.

3.5.5 Combining the results

Combining (3.11), (3.12) and (3.13) we get:

∑_{n=N_{0}}^{N_{2}} sin 2πf(n) = ∑_{n=N_{0}+1}^{N_{1}−1} sin 2πf(n) + ∑_{n=N_{1}+1}^{N_{2}} sin 2πf(n) + sin 2πf(N_{1}) + sin 2πf(N_{1} + 1)

< ∑_{k=1}^{N_{1}−N_{0}−1} | sin 2πf(S_{k})| + ∑_{k=1}^{N_{1}−N_{0}−1} | sin 2πf(R_{k})| + ∑_{i=0}^{2} | sin 2πf(a_{i})|

and so

∑_{n=N_{0}}^{N_{2}} sin 2πf(n) + ∑_{k=1}^{N_{1}−N_{0}−1} sin 2πf(S_{k}) + ∑_{k=1}^{N_{1}−N_{0}−1} sin 2πf(R_{k}) + ∑_{i=0}^{2} sin 2πf(a_{i}) < 0

(3.14)

Now since ℑe^{2πif(n)} < 0 for M_{3} ≤ n ≤ N_{4} then

ℑ ∑_{n=N_{0}}^{N_{4}} e^{2πif(n)} ≤ ℑ ∑_{n=N_{0}}^{M_{3}} e^{2πif(n)}

= ∑_{n=N_{0}}^{N_{2}} sin 2πf(n) + ∑_{n=N_{2}+1}^{N_{3}} sin 2πf(n)

≤ ∑_{n=N_{0}}^{N_{2}} sin 2πf(n) + ∑_{k=1}^{N_{1}−N_{0}−1} sin 2πf(S_{k})

+ ∑_{k=1}^{N_{1}−N_{0}−1} sin 2πf(R_{k}) + ∑_{i=0}^{2} sin 2πf(a_{i})

< 0 (by (3.14))

Thus (3.8) holds.

3.5.6 Using the remaining terms

We now let (m_{i}) be the terms with M_{2} < m_{i} ≤ M_{3}, i = 1, 2, …, M_{3} − M_{2} and ask what fraction of the m_{i}’s do not correspond to an S_{k} or R_{k}. No m_{i}’s were used to compensate for terms with N_{0} ≤ n ≤ M_{0}. By the corollary, at most \frac{1}{3} of the m_{i}’s were used to compensate for terms with M_{0} < n ≤ M_{1}. To compensate for terms with M_{1} < n ≤ N_{2} another N_{2} − M_{1} of the a_{i}’s were used

Using the M.V.T. and our assumptions about f′(x) and f′′(x) we see that

\begin{aligned}|M_{3} − M_{2}| & ≥ \frac{1}{f′(M_{2})}|f(M_{3}) − f(M_{2})| \\ & ≥ 2πM_{2}d \\ & ≥ 3M_{2} \end{aligned}  (3.15)

Using (3.15) we get

\frac{1}{3} ≥ \frac{M_{2}}{M_{3} − M_{2}}

≥ \frac{M_{2} − M_{1}}{M_{3} − M_{2}}

≥ \frac{N_{2} − M_{1}}{M_{3} − M_{2}}

and thus at most another \frac{1}{3} of the a_{i}’s were used up here. Hence there are at least \frac{1}{3} of the a_{i}’s left which do not correspond to an S_{k} or R_{k}.

Hence

\frac{1}{M_{3}} ∑_{n=N_{0}}^{M_{3}} sin 2πf(n) = \frac{1}{M_{3}} (∑_{N_{0}}^{N_{2}} sin 2πf(n) + ∑_{k=1}^{N_{1}−N_{0}−1} sin 2πf(S_{k})

+ ∑_{k=1}^{N_{2}−N_{1}−2} sin 2πf(R_{k}) + ∑_{i=0}^{2} sin 2πf(a_{i}))

+ \frac{1}{M_{3}} ∑_{n=N_{2}+1,n≠S_{k},R_{k},a_{i}}^{M_{3}} sin 2πf(n)

< \frac{1}{M_{3}} ∑_{n=M_{2},n≠S_{k},R_{k},a_{i}}^{M_{3}} sin 2πf(n)

(3.16)

Now for n’s appearing in the last term of (3.16) sin 2πf(n) < 0. Also, for M_{2} < n < M_{3} | sin 2πf(n)| ≥ 0.07. Finally using (3.15) again we can deduce that \frac{M_{2}}{M_{3}} ≤ \frac{1}{4}. Using these facts and (3.16) we get:

−\frac{1}{M_{3}} ∑_{n=M_{2},n≠S_{k},R_{k},a_{i}}^{M_{3}} sin 2πf(n) = \frac{1}{M_{3}} ∑_{n=M_{2},n≠S_{k},R_{k},a_{i}}^{M_{3}} | sin 2πf(n)|

> \frac{1}{M_{3}} \frac{0.07}{3}(M_{3} − M_{2})

> 0.02 (1 − \frac{M_{2}}{M_{3}} ) (where C is a constant)

> 0.02 (1 − \frac{1}{4})

> 0.01 (3.17)

Thus combining (3.16) and (3.17) gives (3.9).

3.5.7 Deriving (3.10)

We now slightly change the notation to let N_{0} = N_{00} = f(1), and let N_{i0} = N_{i} and M_{i0} = M_{i}. Now define N_{01} = N_{40} + 1 and take N_{i1} and M_{i1} to be the N_{i}’s and M_{i}’s which we would have gotten if we took N_{0} = N_{01}. Continue on in this manner, so that we get N_{ij} to be the N_{i} obtained on the ”jth” revolution around the circle. Then

\frac{1}{M_{3j}} ℑ ∑_{n=1}^{M_{3j}} e^{2πif(n)} = \frac{1}{M_{3j}} ∑_{n=1}^{M_{3j}} sin 2πf(n)

= \frac{1}{M_{3j}} ∑_{k=0}^{j−1} \underbrace{∑_{N_{0k}}^{N_{4k}} sin 2πf(n)}_{<0} + \underbrace{\frac{1}{M_{3j}} ∑_{n=N_{0j}}^{M_{3j}} sin 2πf(n)}_{<−0.01}

< −0.01

(3.18)

(3.18) shows that as N increases

\frac{1}{N} ℑ ∑_{n=1}^{N} e^{2πif(n)} < −0.01

whenever N = M_{3j}. This happens infinitely often and thus (3.10) holds. Hence the sequence (f(n)) is not uniformly distributed modulo 1.

3.6 The Distribution of u_{n} + f(n)

Theorem 3.6.1 If (u_{n}) is an arbitrary sequence which is uniformly distributed modulo 1 and |f′(x)| < \frac{K}{x} then (u_{n} + f(n)) is uniformly distributed modulo 1.

Proof In this proof we will use Weyl’s criterion as well as the formula below:

∑_{n=m}^{N} b_{n}(a_{n} − a_{n−1}) = b_{N+1}a_{N} − b_{m}a_{m−1} − ∑_{n=m}^{N}(b_{n+1} − b_{n})a_{n}  (3.19)

This technique is known as summation by parts, and can be verified by expanding both sides of (3.19).

Now let

a_{m} = ∑_{j=1}^{m} e^{2πiu_{j}} and b_{m} = e^{2πif(m)}

Then

\begin{aligned}∑_{n=2}^{N} e^{2πih(u_{n}+f(n))} & = ∑_{n=2}^{N} e^{2πihu_{n}}e^{2πihf(n)} \\ & = e^{2πihf(N+1)} ∑_{j=1}^{N} e^{2πihu_{j}} − e^{2πih(u_{1}+f(2))} \\ & − \underbrace{∑_{n=2}^{N} ((e^{2πihf(n+1)} − e^{2πihf(n)}) ∑_{j=1}^{n} e^{2πihu_{j}})}_{S} & (using (3.19)) \end{aligned} (3.20)

Dividing both sides of (3.20) by N and taking the lim sup as N → ∞ we see that the first and second terms tend to 0.

We now consider |e^{2πihf(n)} − e^{2πihf(n−1)}|. Let g(x) = e^{2πihf(x)}. Then |g′(x)| ≤ \frac{K}{x} and for n ≤ x ≤ n + 1, |g′(x)| ≤ \frac{K}{n} ⇒ |g(n + 1) − g(n)| ≤ \frac{K}{n}. That is

|e^{2πihf(n)} − e^{2πihf(n−1)}| ≤ \frac{K}{n}  (3.21)

We now consider S from (3.20)

lim sup_{N→∞} \frac{|S|}{N} = lim sup_{N→∞} \frac{1}{N} |∑_{n=2}^{N}(e^{2πihf(n+1)} − e^{2πihf(n)}) ∑_{j=1}^{n} e^{2πihu_{j}} |

≤ lim sup_{N→∞} \frac{1}{N} ∑_{n=2}^{N} |e^{2πihf(n+1)} − e^{2πihf(n)}| |∑_{j=1}^{n} e^{2πihu_{j}} |

≤ lim sup_{N→∞} \frac{1}{N} ∑_{n=2}^{N} \frac{K}{n} |∑_{j=1}^{n} e^{2πihu_{j}} | (using (3.21))

= lim sup_{N→∞} \frac{1}{N} \frac{K}{N} ∑_{n=2}^{N} |\frac{1}{n} ∑_{j=1}^{n} e^{2πihu_{j}} |

Now (u_{n}) is u.d., so

|\frac{1}{n} ∑_{j=1}^{n} e^{2πiu_{j}} | → 0 as n → ∞

and hence for each ϵ there is a K_{ϵ} such that for n > K_{ϵ},

|\frac{1}{n − 1} ∑_{j=1}^{n} e^{2πiu_{j}} | < ϵ.

Hence

lim sup_{N→∞} \frac{|S|}{N} = lim sup_{N→∞} \frac{K}{N} (∑_{n=2}^{K_{ϵ}} |\frac{1}{n − 1} ∑_{j=1}^{n} e^{2πihu_{j}} | + ∑_{n=K_{ϵ}+1}^{N} |\frac{1}{n − 1} ∑_{j=1}^{n} e^{2πihu_{j}} |)

≤ lim sup_{N→∞} \frac{K}{N} (K′_{ϵ} + ∑_{n=K_{ϵ}+1}^{N} ϵ)

= lim sup_{N→∞} \frac{K}{N} (K′_{ϵ} + (N − K_{ϵ} − 1)ϵ)

≤ ϵ

Since we can pick ϵ to be arbitrarily small we conclude that the term S in (3.20) tends to zero as well. This tells us that Weyl’s criterion holds for the sequence (u_{n} + f(n)) this sequence is uniformly distributed modulo 1.

As a corollary to theorem 3.6.1 we get

Theorem 3.6.2 If |f′(x)| ≤ \frac{K}{x} then (f(n)) is not uniformly distributed modulo 1.

This is a much stronger statement than Theorem 3.5.1. Proof Assume that (f(n)) was u.d. mod 1. Then (−f(n)) would also be u.d. mod 1 and hence letting u_{n} = − f(n) we would arrive at the conclusion that 0 is u.d. mod 1, which is clearly false. Hence (f(n)) cannot be uniformly distributed modulo 1.

Corollary 3.6.3 (log n) is not uniformly distributed modulo 1.

Note: This corollary was already proven by other methods as Proposition 3.2.1. Proof \frac{d}{dx} log x = \frac{1}{x} and so theorem 3.6.2 can be applied to f(x) = log x to yield the desired result.

Trigonometric Sequences

In this chapter we examine the distribution of various families of trigonometric sequences. Many of the arguments in chapter 3 relied on the fact that the sequences we were considering had a monotonic derivative, however this is no longer the case. As a result we have to employ slightly different techniques to analyse the distribution of these sequences. The chapter begins by analysing the most basic family of trigonometric sequences, K sin x_{n} (where x_{n} is u.d. mod 2π) and builds up to analyse more complicated families of sequences. The results combine to give a comprehensive analysis of sequences of the form (n^{α} sin n^{β}) for β < 1, and also include some results for more general families, (f(n) sin n^{β}), β < 1 and (n^{α} sin n), α < \frac{1}{2}.

As a general rule trigonometric sequences are difficult to analyse as their derivative varies wildly, making them hard to integrate. By looking at sequences whose trigonometric part is a function only of n^{β}, β < 1 we at least get to tame this derivative as n gets very large and so the analysis is easier.

As far as we know, there has been no analysis of sequences of the form (n^{α} sin n^{β}) for β > 1.

4.1 The distribution of (K sin(x_{n}))

Theorem 4.1.1 For K > 1 and (x_{n}) u.d. mod 2π, (K sin(x_{n})) is not uniformly distributed modulo 1.

Proof Assume (K sin x_{n}) is u.d. mod 1. Pick 0 < ϵ < \frac{1}{2K^{3}π^{2}} . Now \{K sin x_{n}\} ∈ ({K}, {K} + ϵ) iff

\{x_{n}\}_{2π} ∈ ⋃_{i=0}^{⌊K−1⌋} ((sin^{−1} \frac{i + \{K\}}{K}, sin^{−1} \frac{i + \{K\} + ϵ}{K}) ∪
(π − sin^{−1} \frac{i + \{K\} + ϵ}{K}, π − sin^{−1} \frac{i + \{K\}}{K}))
⋃_{i=1}^{⌊K⌋} ((π + sin^{−1} \frac{i + \{K\} − ϵ}{K}, π + sin^{−1} \frac{i + \{K\}}{K}) ∪
(2π − sin^{−1} \frac{i + \{K\}}{K}, 2π − sin^{−1} \frac{i + \{K\} − ϵ}{K}))

and thus

ϕ_{N}^{K sin x_{n}}(\{K\}, \{K\} + ϵ) = 2 ∑_{i=0}^{⌊K−1⌋} ϕ_{N}^{x_{n}} (sin^{−1} \frac{i + \{K\}}{K}, sin^{−1} \frac{i + \{K\} + ϵ}{K})

+ 2 ∑_{i=i}^{⌊K⌋} ϕ_{N}^{x_{n}} (sin^{−1} \frac{i + \{K\} − ϵ}{K}, sin^{−1} \frac{i + \{K\}}{K})

> 2ϕ_{N}^{x_{n}} (sin^{−1} \frac{K − ϵ}{K}, \frac{π}{2} )

+ 2 ∑_{i=1}^{⌊K−1⌋} ϕ_{N}^{x_{n}} (sin^{−1} \frac{i + \{K\} − ϵ}{K}, sin^{−1} \frac{i + \{K\} + ϵ}{K})

(4.1)

Since (x_{n}) is uniformly distributed modulo 2π we know that lim_{N→∞} \frac{ϕ_{N}^{xn}([a,b))}{N} = \frac{b−a}{2π}. If we divide both sides of (4.1) by N and try to take limits as N → ∞ we see that all the limits on the right hand side exist and thus so must the limit on the left hand side. Hence we get:

lim_{N→∞} \frac{1}{N} ϕ_{N}^{K sin x_{n}}(\{K\}, \{K\} + ϵ) > \frac{1}{π} ((\frac{π}{2} − sin^{−1} \frac{K − ϵ}{K})

+ ∑_{i=1}^{K−1} (sin^{−1} \frac{i + \{K\} + ϵ}{K} − sin^{−1} \frac{i + \{K\} − ϵ}{K}))

> \frac{1}{π} (\frac{π}{2} − sin^{−1} \frac{K − ϵ}{K})

> \frac{ϵ}{πK} \frac{d}{dx} (sin^{−1} \frac{K − ϵ}{K})

(By the MVT)

= \frac{ϵ}{πK} \frac{1}{\sqrt{K^{2} − (K − ϵ)^{2}}}

= \frac{ϵ}{πK} \frac{1}{\sqrt{2Kϵ − ϵ^{2}}}

> \frac{ϵ}{πK^{\frac{3}{2}} \sqrt{2ϵ}}

(4.2)

However if (K sin x_{n}) is u.d. mod 1 then

lim_{N→∞} \frac{1}{N} ϕ_{N}^{K sin x_{n}}(\{K\}, \{K\} + ϵ) = ϵ.  (4.3)

Thus (4.2) and (4.3) combine to give:

π > \frac{1}{K^{\frac{3}{2}} \sqrt{2ϵ}}  (4.4)

but

\begin{aligned}\frac{1}{K^{\frac{3}{2}} \sqrt{2ϵ}} & > \frac{1}{K\sqrt{\frac{2K}{2K^{3}π^{2}}}} \\ & = π \end{aligned}  (4.5)

Which is a contradiction. Thus for K > 1 (K sin x_{n}) cannot be uniformly distributed modulo 1.

The following theorem follows as a corollary from Theorem 4.1.1. Theorem 4.1.2 (sin n^{β}) is not uniformly distributed modulo 1.

Proof The proof of this Theorem breaks into three cases. Case 1: β > 0 Proposition 3.3.1 proved that (2πn^{β}) is uniformly distributed modulo 1. Thus (n^{β}) is uniformly distributed modulo 2π and the result follows from Theorem 4.1.1. Case 2: β = 0 This case is trivial as we are talking about a constant sequence which cannot be u.d. mod 1. Case 3: β < 0 In this case lim_{n→∞} sin n^{β} = 0 and so all the terms of the sequence are positive and tending towards zero and hence their fractional parts are also tending to zero. Thus the sequence cannot be u.d. mod 1.

4.2 The distribution of n^{−α} sin n^{β}

Theorem 4.2.1 If α > 0 then (n^{−α} sin n^{β}) is not uniformly distributed modulo 1.

Proof In this case lim_{n→∞} n^{−α} sin n^{β} = 0 and so all the terms of the sequence are tending towards zero. This means that their fractional parts are either very close to zero or very close to one. Thus the fractional parts cannot be uniformly distributed in the unit interval and so (n^{−α} sin n^{β}) will not be u.d. mod 1.

4.3 The distribution of n^{α} sin n^{−β}

Theorem 4.3.1 For β > 0

(i) If α > β and (α − β) ∉ ℤ (n^{α} sin n^{−β}) is uniformly distributed modulo 1. (ii) If α > β and (α − β) ∈ ℤ Then (n^{α} sin n^{−β}) is uniformly distributed modulo 1 iff β is not a multiple of \frac{1}{2}. (iii) If α ≤ β (n^{α} sin n^{−β}) is not uniformly distributed modulo 1.

Proof We consider the “Taylor” series expansion of n^{α} sin n^{−β}. We know that

sin x = ∑_{k=0}^{∞}(−1)^{k} \frac{x^{2k+1}}{(2k + 1)!}  (4.6)

Substituting x = n^{−β} into (4.6) and multiplying by n^{α} we get:

\begin{aligned}n^{α} sin n^{−β} & = ∑_{k=0}^{∞}(−1)^{k} \frac{n^{α−β(2k+1)}}{(2k + 1)!} \\ & = n^{α−β} ∑_{k=0}^{∞}(−1)^{k} \frac{n^{−2kβ}}{(2k + 1)!} \\ & = n^{α−β} + n^{α−β} ∑_{k=1}^{∞}(−1)^{k} \frac{n^{−2kβ}}{(2k + 1)!} \end{aligned}  (4.7)

(i) Define γ = (α−β). Then γ > 0 and γ is not an integer. Let T = ⌊\frac{γ}{2β} ⌋. We can re-write (4.7) as

n^{α} sin n^{−β} = \underbrace{n^{γ} + ∑_{k=1}^{T}(−1)^{k} \frac{n^{γ−2kβ}}{(2k + 1)!}}_{f(n)} + \underbrace{∑_{k=T +1}^{∞}(−1)^{k} \frac{n^{γ−2kβ}}{(2k + 1)!}}_{ϵ_{n}}  (4.8)

Then f(x) satisfies the hypothesis for Proposition 3.3.1 and is thus uniformly distributed modulo 1. In addition ϵ_{n} → 0 as N → ∞ and so using Corollary 2.2.4 we can conclude that (n^{α} sin n^{−β}) is uniformly distributed modulo 1 for the conditions given in part (i).

(ii) Define m = (α − β). Then m > 0 and m ∈ ℤ. Let T = ⌊\frac{m}{2β} ⌋. Thus (4.7) can be written as:

n^{α} sin n^{−β} = n^{m} + \underbrace{n^{m−2β} + ∑_{k=2}^{T}(−1)^{k} \frac{n^{m−2kβ}}{(2k + 1)!}}_{f(n)} + \underbrace{∑_{k=T +1}^{∞}(−1)^{k} \frac{n^{γ−2kβ}}{(2k + 1)!}}_{ϵ_{n}}(4.9)

where ϵ_{n} → 0 as N → ∞. Since the fractional part of n^{m} is always zero the n^{m} term in (4.9) does not affect the distribution of the sequence. Thus (n^{α} sin n^{−β}) is distributed in the same manner as (f(n) + ϵ_{n}). We now have two cases. Case 1: β is not a multiple of \frac{1}{2}. The leading term of f(n) is n^{m−2β}. Since β is not a multiple of \frac{1}{2} then (m−2β) ∉ ℤ and so f(n) obeys the hypothesis of Proposition 3.3.1 and so is u.d. mod 1. Using this fact and Corollary 2.2.4 we conclude that the reverse direction of (ii) holds. Case 2: β is a multiple of \frac{1}{2}. In this case the powers of n in f(n) are of the form m − 2kβ ∈ ℤ and so f(n) is a polynomial with no irrational coefficients and by Proposition 3.1.3 is not uniformly distributed modulo 1. Combining this with Corollary 2.2.4 we conclude that the forward direction of (ii) holds.

(iii) In this case we have α ≤ β, and so n^{α−β} ≤ 1. In addition the terms in the summation of 4.7 are all approaching zero as n increases. Pick ϵ > 0. Then for large enough n the second term on the right hand side of (4.7) less than ϵ. We are now left with two cases. Either α = β and we get

n^{α} sin n^{−β} = 1 + O(ϵ)  (4.10)

or α < β and choosing n large enough we can make n^{α−β} < ϵ and thus

n^{α} sin n^{−β} = O(ϵ)  (4.11)

Either way (n^{α} sin n^{−β}) cannot be uniformly distributed modulo 1 as it’s fractional parts are all O(ϵ). Thus part (iii) of the theorem is proved.

4.4 The distribution of n^{α} sin n^{β}

The main result of this section is contained in Theorem 4.4.9 and states that for β < 1 and α > 0 then n^{α} sin n^{β} is uniformly distributed modulo 1. Different methods are needed to analyse the sequence depending on whether α is a multiple of 1−β or not. For this reason we break the discussion up into two parts. The first part (section 4.4.1) deals with the case when α is not a multiple of 1 − β and the second part (section 4.4.2) deals with sequences where α is a multiple of 1 − β.

Throughout this section we will make repeated use of the following two lemmas.

Lemma 4.4.1 If f(x) and g(x) are increasing functions with g′(x) < K and A_{Mϵ} is the set defined by A_{Mϵ} = {x | ϵM ≤ x ≤ M, | cos g(x)| ≥ ϵ} then

|[f(x)]_{A_{Mϵ}} | ≤ Kf(M)g(M)

Proof We begin by observing that A_{Mϵ} is broken up into T intervals, where T is the number of times that cos g(x) crosses the x-axis. Let us denote these intervals by I_{t} = (a_{t}, b_{t}) where t = 0 … T. Then certainly

|[f(x)]_{A_{ϵ}} | = |∑_{t=1}^{T} [f(x)]_{I_{t}} |

≤ ∑_{t=1}^{T} |f(b_{t}) − f(a_{t})|

≤ 2T f(M)

(Since f is an increasing function.)

(4.12)

It remains for us to show how many times cos g(x) crosses the x-axis. If cos g(x) = 0 then g(x) = (2t + 1)\frac{π}{2}, t ∈ ℤ. In addition, x ≤ M so we must have

g^{−1} ((2t + 1)\frac{π}{2} ) < M  (4.13)

Note that since g is increasing, g^{−1} exists. It is easy to solve (4.13) to get t ≤ Kg(M). Thus we can conclude that T ≤ Kg(M) and thus the lemma is proved.

Lemma 4.4.2 If f, A_{Mϵ} are defined as in Lemma 4.4.1 and g(x) = x^{β} + ϕ for β < 1 and B_{Mϵ} = [ϵM, M] \ A_{Mϵ} then

lim sup_{M→∞} \frac{1}{M} |∫_{B_{Mϵ}} e^{2πihf(x)}dx| = O(ϵ)

Proof We can define B_{Mϵ} alternatively as B_{Mϵ} = {x | ϵM ≤ x ≤ M, | cos x^{β} + ϕ| < ϵ }. As was the case with A_{Mϵ} this set is divided up into T intervals where once again T is the number of times which cos (x^{β} + ϕ) crosses the x-axis. From above we have that T ≤ KM^{β}. We now ask what the length of each one of these intervals is. This time we denote the intervals of B_{Mϵ} by J_{t} = (a_{t}, b_{t}).

It is not too hard to see that |b_{t}^{β}−a_{t}^{β}| ≤ 2 sin^{−1} ϵ and thus a_{t} ≥ ((b_{t})^{β} − 2 sin^{−1} ϵ)^{\frac{1}{β}}. Thus the length of an interval J_{t} of B_{Mϵ} is given by:

|J_{t}| ≤ b_{t} − ((b_{t})^{β} − 2 sin^{−1} ϵ)^{\frac{1}{β}}
∼ (b_{t}^{β})^{\frac{1}{β} −1} sin^{−1} ϵ + l.o.t.
(Using the M.V.T. since b_{t}^{β} ≫ sin^{−1} ϵ)
≤ K(M^{1−β} sin^{−1} ϵ + l.o.t.)

(Since x^{β} is an increasing function)

(4.14)

Thus we get

lim sup_{M→∞} \frac{1}{M} |∫_{B_{Mϵ}} e^{2πihf(x)}dx| ≤ lim sup_{M→∞} \frac{1}{M} |∫_{ϵM}^{M} χ_{B_{Mϵ}}dx|

≤ lim sup_{M→∞} \frac{KM^{β}M^{1−β} sin^{−1} ϵ + l.o.t.}{M}

= K sin^{−1} ϵ = O(ϵ)

(4.15)

4.4.1 α ≠ k(1 − β)

Theorem 4.4.3 For β < 1, α ∉ ℤ(1−β) (n^{α} sin n^{β}) is uniformly distributed modulo 1.

In order to prove Theorem 4.4.3 we first define f_{mβ}(x) = ∑_{j} c_{j}x^{α_{j}} sin (x^{β} + ϕ_{j}) where the sum consists of finitely many terms, c_{j} ∈ ℝ and α_{j} < m(1 − β). Let k be the subscript of the largest α_{j} in f_{mβ}(x). Ensure that α_{k} > 0 and α_{k} ∉ ℤ(1 − β) and that the α_{j}’s are all different. In addition ensure that c_{k} ≠ 0 and −π < ϕ_{j} ≤ π.

Proposition 4.4.4 (f_{mβ}(n)) is uniformly distributed modulo 1.

Theorem 4.4.3 will then follow directly from Proposition 4.4.4. Proof We will prove Proposition 4.4.4 by induction on m. The proof will follow a similar line of argument to that used to prove Proposition 3.3.1 however extra care will need to be taken due to the additional sine term.

Choose β with 0 < β < 1. Pick ϵ > 0. We now consider the distribution of f_{1β}. Let k be chosen as above and let k′ be the subscript of the second largest α_{j}. Let M be max(max_{j}{c_{j}βϵ}, max_{j}{c_{j}α_{j}}) and let T be the number of terms in the sum of f_{1β}.

1. Pick N_{1} > ϵ^{\frac{1−β}{β}} (So that for x > ϵN_{1}, x^{β} > ϵ).

2. Pick N_{2} > \frac{1}{ϵ} (\frac{2α_{k}}{βϵ^{2}} )^{\frac{1}{β}} (So that for x > ϵN_{2}, and cos (x^{β} + ϕ_{k}) ≥ ϵ |ϵβx^{α_{k}+β−1} cos (x^{β} + ϕ_{k})| > 2|α_{k}x^{α_{k}−1} sin (x^{β} + ϕ_{k})|)

3. Pick N_{3} > \frac{1}{ϵ} (\frac{2T M}{c_{k}α_{k}} )^{\frac{1}{α_{k}−α_{k′} −β}} (So that for x > ϵN_{3},

|c_{k}α_{k}x^{α_{k}−1} sin (x^{β} + ϕ_{k})| > 2T Kx^{α_{k′}−β−1}

> | ∑_{j,j≠k} c_{j}(βx^{α_{j}+β−1} cos (x^{β} + ϕ_{j}) + α_{j}x^{α_{j}−1} sin (x^{β} + ϕ_{j}))|)

Pick N > max(N_{1}, N_{2}, N_{3}). We will consider lim sup_{N→∞} \frac{1}{N} ∑_{n=ϵN}^{N} e^{2πihf_{1β}(n)} and show that this goes to zero by bounding the terms of the Euler summation formula. Let the set A_{Nϵ} and B_{Nϵ} be defined by

\begin{matrix}A_{Nϵ} = \{x | ϵN ≤ x ≤ N, | cos (x^{β} + ϕ_{k})| ≥ ϵ\} \\ B_{Nϵ} = [1, N] \ A_{Nϵ} \end{matrix}

Let us consider the size of the first term in the Euler summation formula. We denote this term by

|First Term| ≤ \underbrace{|∫_{A_{Nϵ}} e^{2πih ∑_{j} c_{j}x^{α_{j}} sin (x^{β}+ϕ_{j})}dx|}_{I_{A}} + \underbrace{|∫_{B_{Nϵ}} e^{2πih ∑_{j} c_{j}x^{α_{j}} sin (x^{β}+ϕ_{j})}dx|}_{I_{B}}

(4.16)

We begin by simplifying this integral using some of the tricks from section 2.2.

Let

A(x) = c_{k}βx^{α_{k}+β−1} cos (x^{β} + ϕ_{k})

and

B(x) = c_{k}α_{k}x^{α_{k}−1} sin (x^{β} + ϕ_{k})+ ∑_{j,j≠k} c_{j}(βx^{α_{j}+β−1} cos (x^{β} + ϕ_{j})+α_{j}x^{α_{j}−1} sin (x^{β} + ϕ_{j}))

Then

\frac{d}{dx} (∑_{j} c_{j}x_{j}^{α} sin (x^{β} + ϕ_{j})) = A(x) + B(x)

We have:

ϵ|A(x)| = ϵc_{k}β|x^{α_{k}+β−1} cos (x^{β} + ϕ_{k})|

> 2c_{k}α|x^{α_{k}−1} sin (x^{β} + ϕ_{k})|

(Since N > N_{2})

> c_{k}α|x^{α_{k}−1} sin (x^{β} + ϕ_{k})| + | ∑_{j,j≠k} c_{j}(βx^{α_{j}+β−1} cos (x^{β} + ϕ_{j}) + α_{j}x^{α_{j}−1} sin (x^{β} + ϕ_{j}))|

(Since N > N_{3})

≥ |B(x)| (4.17) so we will be able to apply Proposition 2.2.1. We now use (2.20) to bound |I_{A}| from (4.16).

|I_{A}| = |∫_{A_{Nϵ}} \frac{d}{dx} (\frac{e^{2πih ∑_{j} c_{j}x_{j}^{α} sin (x^{β}+ϕ_{j})}}{2πih} ) ×

(∑_{j} c_{j}(βx^{α_{j}+β−1} cos (x^{β} + ϕ_{j}) + α_{j}x^{α_{j}−1} sin (x^{β} + ϕ_{j})))^{−1} dx|

≤ K |∫_{A_{Nϵ}} \frac{d}{dx} (e^{2πih ∑_{j} c_{j}x_{j}^{α} sin (x^{β}+ϕ_{j})}) \frac{dx}{c_{k}βx^{α_{k}+β−1} cos (x^{β} + ϕ_{k})}| + ϵK′N

(Using 2.21)

≤ [\frac{K}{|c_{k}βx^{α_{k}+β−1} cos (x^{β} + ϕ_{k})|}]_{A_{Nϵ}}

+ K ∫_{A_{Nϵ}} |\frac{(α_{k} + β − 1)x^{α_{k}+β−2} cos (x^{β} + ϕ_{k}) + βx^{α_{k}+2β−2} sin (x^{β} + ϕ_{k})}{c_{k}βx^{2α_{k}+2β−2} cos^{2} (x^{β} + ϕ_{k})}| dx + ϵK′N

(Using 2.20)

≤ [\frac{K}{|x^{α_{k}+β−1}ϵ|}]_{A_{Nϵ}} + \underbrace{∫_{A_{Nϵ}} \frac{K_{1}ϵx^{α_{k}+β−2} + K_{2}x^{α_{k}+2β−2}}{x^{2α_{k}+2β−2}ϵ^{2}} dx}_{J} + ϵK′N

(4.18)

Now

[\frac{K}{|x^{α_{k}+β−1}ϵ|}]_{A_{Nϵ}} ≤ \frac{KN^{1−α_{k}−β}N^{β}}{ϵ}
(Using Lemma 4.4.1 with M = N and g(x) = x^{β} + ϕ_{k})

= \frac{KN^{1−α_{k}}}{ϵ}

(4.19)

We now consider |J| which was deferred from above.

\begin{aligned}|J| & = ∫_{A_{Nϵ}} \frac{K_{1}ϵx^{α_{k}+β−2} + K_{2}x^{α_{k}+2β−2}}{x^{2α_{k}+2β−2}ϵ^{2}} dx \\ & ≤ ∫_{A_{Nϵ}} \frac{Kx^{α_{k}+2β−2}}{x^{2α_{k}+2β−2}ϵ^{2}} dx \\ & (Since N > N_{1}) \\ & ≤ \frac{K}{ϵ^{2}} ∫_{A_{Nϵ}} x^{−α_{k}}dx \\ & = \frac{K}{ϵ^{2}} [x^{1−α_{k}}]_{A_{Nϵ}} \\ & ≤ \frac{K}{ϵ^{2}} [x^{1−α_{k}}]_{1}^{N} \\ & ≤ \frac{K}{ϵ^{2}} N^{1−α_{k}} \end{aligned}  (4.20)

Thus combining (4.18), (4.19) and (4.20) gives

lim sup_{N→∞} \frac{1}{N} |∫_{A_{Nϵ}} e^{2πih ∑_{j} c_{j}x^{α_{j}} sin (x^{β}+ϕ_{j})}dx| ≤ lim sup_{N→∞} (\frac{KN^{−α_{k}}}{ϵ} + \frac{K′N^{−α_{k}}}{ϵ^{2}} + ϵK′′)
= O(ϵ)

(4.21)

Now let B_{Nϵ} = [ϵN, N] \ A_{Nϵ}. Then using Lemma 4.4.2 with M = N we get

lim sup_{N→∞} \frac{1}{N} |∫_{B_{Nϵ}} e^{2πih ∑_{j} c_{j}x^{α_{j}} sin (x^{β}+ϕ_{j})}dx| = O(ϵ)  (4.22)

Substituting (4.21) and (4.22) into (4.16) shows us that lim sup_{N→∞} \frac{|First Term|}{N} = O(ϵ). We now look at the third term in the Euler summation formula.

|Third term| ≤ ∫_{ϵN}^{N} ∑_{j} c_{j} (K_{1j}|x^{α_{j}+β−1} cos (x^{β} + ϕ_{k})| + K_{2j}|x^{α_{j}−1} sin (x^{β} + ϕ_{k})|) dx

≤ ∑_{j} c_{j} (K_{1j} ∫_{ϵN}^{N} x^{α_{j}+β−1}dx + K_{2j} ∫_{ϵN}^{N} x^{α_{j}−1}dx)

≤ [∑_{j} c_{j}K_{1j}x^{α_{j}+β} + K_{2j}x^{α_{j}})]_{ϵN}^{N}

≤ ∑_{j} c_{j}(K_{1j}N^{α_{j}+β} + K_{2j}N^{α_{j}})

(4.23)

And since α_{j} < (1 − β) for all j Then (α_{j} + β) < 1 and dividing by N and taking the lim sup as N → ∞ we see that the third term of the Euler summation formula goes to 0 as N → ∞. Making use of (2.17) we have

lim sup_{N→∞} \frac{1}{N} |∑_{ϵN}^{N} e^{2πihf_{1β}(n)}| = O(ϵ)

and using Proposition 2.2.5 (f_{1β}(n)) is u.d. mod 1.

Now let us assume that (f_{bβ}(n)) is uniformly distributed modulo 1 for all b ≤ m and consider the distribution of (f_{(m+1)β}(n)). Using Taylor expansion gives:

f_{mβ}(n + h) − f_{mβ}(n) = hf_{(m+1)β}^{(1)}(n) + \frac{h^{2}}{2} f_{(m+1)β}^{(2)}(n) + ⋯ + O(n^{−δ})

(Where δ > 0 and the constant in the O term depends on h)

= hf_{mβ}^{∗}(n) + \frac{h^{2}}{2} f_{(m−1)β}^{∗}(n) + ⋯ + \underbrace{O(n^{−δ})}_{<ϵ}

= f_{mβ}^{∗∗}(n) + ϵ_{n} (4.24)

Now the leading coefficient f_{mβ}^{∗∗}(n) is c_{k}β ≠ 0 and the leading power of f_{mβ}^{∗∗}(n) is α_{k} − (1 − β) ∉ ℤ(1 − β). Thus f_{(α+β−1)β}^{∗∗}(n) satisfies the induction hypothesis and thus is u.d. mod 1. Hence by the difference theorem Proposition 4.4.4 holds.

Thus if α and β satisfy the hypothesis of Theorem 4.4.3 then (n^{α} sin n^{β}) is uniformly distributed modulo 1.

4.4.2 α = k(1 − β)

Theorem 4.4.5 If 0 < β < 1 and k ∈ ℤ then (n^{k(1−β)} sin n^{β}) is uniformly distributed modulo 1.

We let g_{kβ}(n) = c_{k}n^{k(1−β)} sin (n^{β} + ϕ)+f_{kβ}(n) where f_{k(1−β)β}(n) is defined as above and ϕ is as above. Theorem 4.4.5 will follow as a direct result of the following Proposition.

Proposition 4.4.6 (g_{kβ}(n)) is uniformly distributed modulo 1.

Proof We will prove Proposition 4.4.6 by induction on k. We will also need a number of other propositions.

Proposition 4.4.7

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πim(g_{1β}(n+h)−g_{1β}(n))} = lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πimKh cos (n^{β}+ϕ)}

Proof Using Taylor Expansion gives:

g_{1β}(n + h) − g_{1β}(n) =

c_{1}hβ cos (n^{β} + ϕ) + \underbrace{c_{1}hn^{−β} sin (n^{β} + ϕ)}_{<ϵ}

+ \underbrace{c_{1}h^{2}(n^{β−1} sin (n^{β} + ϕ) − βn^{−β−1} sin (n^{β} + ϕ) + βn^{−1} sin (n^{β} + ϕ)) + ⋯}_{<ϵ}

+ \underbrace{h ∑_{j} (c_{j}α_{j}n^{α_{j}−1} sin (n^{β} + ϕ) + c_{j}βn^{α_{j}−(1−β)} cos (n^{β} + ϕ)) + ⋯}_{<ϵ}

(4.25)

and so by Corollary 2.2.4

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πim(g_{1β}(n+h)−g_{1β}(n))} = lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} e^{2πimKh cos (n^{β}+ϕ)} + O(ϵ)

and since we can choose ϵ to be as small as we want Proposition 4.4.7 follows.

Proposition 4.4.8

lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πimKh cos (n^{β}+ϕ)}| ≤ \frac{K}{h^{1/3}}

Proof Once again we use Euler’s summation formula. We let

\begin{matrix}A_{Nh} = \{x|1 ≤ x ≤ N, | cos x^{β}| ≥ h^{−1/3}\} \\ B_{Nh} = [1, N]/A_{Nh} \end{matrix}

lim sup_{N→∞} \frac{|First Term|}{N} ≤ lim sup_{N→∞} \frac{1}{N} \underbrace{|∫_{A_{Nh}} e^{2πimKh cos (x^{β}+ϕ)}dx|}_{I_{A}}

+ lim sup_{N→∞} \frac{1}{N} \underbrace{|∫_{B_{Nh}} e^{2πimKh cos (x^{β}+ϕ)}dx|}_{I_{B}}

(4.26)

Now

lim sup_{N→∞} \frac{|I_{A}|}{N} = lim sup_{N→∞} \frac{1}{N} |∫_{A_{Nh}} \frac{d}{dx} (e^{2πimKh cos x^{β}}) (2πimKhβx^{β−1} cos x^{β})^{−1} dx|

≤ lim sup_{N→∞} \frac{K}{Nh} |[\frac{1}{|x^{β−1} cos (x^{β} + ϕ)|}]_{A_{Nh}} | +

lim sup_{N→∞} \frac{1}{Nh} ∫_{A_{Nh}} \frac{|K_{1}x^{2β−2} sin (x^{β} + ϕ) − K_{2}x^{β−2} cos (x^{β} + ϕ)|}{|x^{2β−2} cos^{2} (x^{β} + ϕ)|}
(Using 2.20)

≤ lim sup_{N→∞} \frac{K}{Nh} [\frac{x^{1−β}}{cos (x^{β} + ϕ)}]_{A_{Nh}} + lim sup_{N→∞} \frac{K_{1}}{Nh} ∫_{A_{Nh}} |\begin{matrix}sin & (x^{β} + ϕ) \\ cos & 2 & (x^{β} + ϕ) \end{matrix}| + l.o.t. dx

≤ lim sup_{N→∞} \frac{KN^{1−β}N^{β}}{Nh.h^{−1/3}} + lim sup_{N→∞} \frac{K_{1}}{Nh.h^{−2/3}} ∫_{A_{Nh}} dx

(Using Lemma 4.4.1 with )M = N and g(x) = x^{β} + ϕ)

= lim sup_{N→∞} (\frac{KN}{Nh^{2/3}} + \frac{K_{1}N}{Nh^{1/3}} )

≤ Kh^{−1/3} (4.27)

and using Lemma 4.4.2 with M = N gives

lim sup_{N→∞} \frac{|I_{B}|}{N} = O(h^{−1/3})  (4.28)

Finally

lim sup_{N→∞} \frac{|Third Term|}{N} ≤ lim sup_{N→∞} \frac{K}{N} ∫_{1}^{N} |x^{β−1} cos (x^{β} + ϕ)| dx

≤ lim sup_{N→∞} \frac{K}{N} ∫_{1}^{N} x^{β−1}dx

≤ lim sup_{N→∞} KN^{β−1}

= 0

(4.29)

Combining equations (4.26) - (4.29) proves the proposition.

Using Propositions 4.4.7 and 4.4.8 we see that

lim sup_{N→∞} \frac{1}{N} |∑_{ϵN}^{N} e^{2πih(g_{1β}(n+h)−g_{1β}(n))}| ≤ \frac{K}{h^{1/3}}

and so we can use the generalised Van Der Corput’s Difference Theorem to deduce that (g_{1β}(n)) is u.d. mod 1.

Now assume that (g_{mβ}(n)) is u.d. mod 1 for all m < k. Using Taylor expansion gives:

g_{kβ}(n + h) − g_{kβ}(n) = hg_{kβ}^{(1)}(n) + \frac{h^{2}}{2} g_{kβ}^{(2)}(n) + ⋯ + O(n^{−δ})

(Where δ > 0 and the constant in the O term depends on h)

= hg_{(k−(1−β))β}^{∗}(n) + \frac{h^{2}}{2} g_{(k−2(1−β))β}^{∗}(n) + ⋯ + \underbrace{O(n^{−δ})}_{<ϵ}

= g_{(k−(1−β))β}^{∗∗}(n) + ϵ_{1} (4.30)

Now the leading coefficient g_{(k−(1−β))β}^{∗∗}(n) is c_{k}β ≠ 0 and the leading power of g_{(k−(1−β))β}^{∗∗}(n) is (k − 1)(1 − β). Thus g_{kβ}^{∗∗}(n) satisfies the induction hypothesis and so is u.d. mod 1. Hence by the difference theorem Proposition 4.4.6 holds.

Theorem 4.4.9 For 0 < β < 1 and 0 < α, (n^{α} sin n^{β}) is uniformly distributed modulo 1.

Proof Combining Theorems 4.4.3 and 4.4.5 proves the theorem.

4.5 The distribution of n^{α} sin n

Theorem 4.5.1 If α < \frac{1}{2} then (n^{α} sin n) is uniformly distributed modulo 1.

In order to prove the Theorem 4.5.1 we will actually prove the following more general theorem.

Theorem 4.5.2 If α < \frac{1}{2} and a ∈ ℝ then n^{α} sin 2πan is uniformly distributed modulo 1.

Proof We will make use of the following theorem due to Dirichlet.

Theorem 4.5.3 For any number a and integer M ∃ a rational \frac{p}{q} with |q| <

M and |a − \frac{p}{q} | < \frac{1}{qM}.

Proof It suffices to find a non-zero integer q such that |q| < M and ||aq|| < \frac{1}{M}, where ||aq|| is the distance from aq to the nearest integer.

Consider the numbers {na}, n = 1, ⋯, M These are M numbers in the interval [0, 1]. By the pigeonhole principle, there must be a pair of them, say \{ia\} and \{ja\} with |\{ia\} − \{ja\}| < \frac{1}{M}. Setting q = |i − j| proves the theorem.

Pick β with α < β < \frac{1}{2}, pick ϵ > 0 and let N_{i}, i = 1, 2 be chosen as follows

1. Pick N_{1} > \frac{1}{4ϵ^{4}}

2. Pick N_{2} > (\frac{α}{2πϵ^{3}} )^{\frac{1}{1−β}}, so that for x > ϵ(N^{1−β}q + r) we have 2πϵ^{2}x^{α} > αx^{α−1}

Pick N > max(N_{1}, N_{2}) and pick M ∼ N^{β}.

Using Theorem 4.5.2 we write a = \frac{p}{q} + δ noting that 1 < q < N^{β}, qδ < N^{−β} and make the change of variables n = mq + r, m ∈ ℤ, 0 ≤ r < q. We then have

∑_{n=ϵN}^{N} e^{2πihn^{α} sin 2πan} ≤ ∑_{r=0}^{q−1} ∑_{m=⌈ \frac{ϵN}{q} ⌉}^{⌊ \frac{N}{q} ⌋} e^{2πih(mq+r)^{α} sin (2πa(mq+r))}  (4.31)

But

sin (2πa(mq + r)) = sin (2π (\frac{p}{q} + δ) (mq + r))

= sin (2π (mp + mqδ + \frac{rp}{q} + rδ))

= sin (2π (mqδ + \frac{rp}{q} + rδ))

(Since mp ∈ ℤ)

(4.32)

and substituting (4.32) into (4.31) gives

| ∑_{n=ϵN}^{N} e^{2πihn^{α} sin 2πan}| ≤ |∑_{r=0}^{q−1} ∑_{m=⌈ϵN/q⌉}^{⌊N/q⌋} e^{2πih(mq+r)^{α} sin(2π(mqδ+ \frac{rp}{q} +rδ))}| + O(q) (4.33)

The advantage of this change of variables is that it is now easier for us to apply the Euler summation formula. This is because when differentiating the exponent we now either reduce the power of n out the front or we pull down a qδ which is of size ∼ N^{−β}.

We now look at the inner sum of (4.33). Given our choice of M this sum becomes:

∑_{m=ϵq/q}^{N/q} e^{2πih(mq+r)^{α}sin(2π(mqδ+ \frac{rp}{q} +rδ))}

As we have done previously we will estimate the size of the sum by examining the Euler summation formula. Now let g(x) = 2π (δx + \frac{rp}{q}), C(x) = 2πδx^{α} and D(x) = αx^{α−1}. Define the sets A_{Nϵr} and B_{Nϵr} as follows:

A_{Nϵr} = \{x | ϵN ≤ x ≤ N, |C(x) cos g(x)+D(x) sin g(x)| ≥ ϵ\sqrt{C(x)^{2} + D(x)^{2}}\}

Using the fact that A sin x + B cos x = \sqrt{A^{2} + B^{2}} sin x + tan^{−1} B/A we see that A_{Nϵr} corresponds to the set of x where

\sqrt{C(x)^{2} + D(x)^{2}}| sin g(x) + θ(x) + D(x)| ≥ ϵ\sqrt{C(x)^{2} + D(x)^{2}}

Where θ(x) = tan^{−1} \frac{D(x)}{C(x)} = tan^{−1} \frac{2πδx}{α} This is the same as the set of x where | sin g(x) + θ(x)| ≥ ϵ. Thus an equvalent definition of the set A_{Nϵr} is:

A_{Nϵr} = {x | ϵN ≤ x ≤ N, | sin g(x) + θ(x)| ≥ ϵ}

B_{Nϵr} = [ϵN, N] \ A_{Nϵr}

We will soon show the sum over r of the size of all the B_{Nϵr} sets is small. The first term of the Euler summation formula is given by:

|First Term| = |∫_{ϵN/q}^{N/q} e^{2πih(tq+r)^{α} sin(2π(tqδ+ \frac{rp}{q} +rδ))}dt|

≤ |\frac{1}{q} ∫_{ϵN}^{N} e^{2πihx^{α} sin (2π(δx+ \frac{rp}{q}))}dx| + O(1)

(letting x = tq + r )

≤ \underbrace{|\frac{1}{q} ∫_{A_{Nϵr}} e^{2πihx^{α} sin (2π(δx+ \frac{rp}{q}))}dx|}_{I_{A}} + \underbrace{|\frac{1}{q} ∫_{B_{Nϵr}} e^{2πihx^{α} sin (2π(δx+ \frac{rp}{q}))}dx|}_{I_{B}} +O(1)

(4.34)

|I_{A}| = \frac{K}{q} |∫_{A_{Nϵr}} \frac{d}{dx} (e^{2πihx^{α} sin g(x)}) (αx^{α−1} sin g(x) + 2πδx^{α} cos g(x))^{−1} dx|

= \frac{K}{q} |[\frac{1}{ϵ\sqrt{(αx^{α−1})^{2} + (2πδx^{α})^{2}}} ]_{A_{Nϵr}} |

+ \frac{K}{q} |\underbrace{∫_{A_{Nϵr}} e^{2πihx^{α} sin g(x)} \frac{d}{dx} ((αx^{α−1} sin g(x) + 2πϵx^{α} cos g(x))^{−1}) dx}_{J}|

(4.35)

We now attempt to invoke as similar argument to Lemma 4.4.1 to evaluate the first term of (4.35). ?? So for each x ∈ A_{Nϵr} we have

\frac{1}{\sqrt{(αx^{α−1})^{2} + (2πδx^{α})^{2}}} < k(N) < N^{−α}

We now ask how many intervals there are in the set A_{Nϵr}. We are concerned with the set where | sin g(x) + θ(x)| ≥ ϵ. Let the number of intervals in this set be T As in Lemma 4.4.1 we find that if | sin g(x) + θ(x)| = 0 then g(x) + θ(x) = tπ, t ∈ ℤ. Now θ(x) is positive, and is no bigger than \frac{π}{2}, and the \frac{rp}{q} term of g(x) can be treated as a constant. Thus we are essentially concerned with how many times 2πδx = tπ. In addition x < N. Thus we solve the inequality

\frac{πt}{2πδ} < N

to find that t < 2δN and can thus conclude that T < KδN. Thus we find that

\frac{K}{q} |[\frac{1}{ϵ\sqrt{(αx^{α−1})^{2} + (2πδx^{α})^{2}}} ]_{A_{Nϵr}} | ≤ \frac{KδNN^{−α}}{qϵ}  (4.36)

We now look at the integral J.

|J| ≤ \frac{K}{q} ∫_{A_{Nϵr}} |\frac{d}{dx} ((αx^{α−1} sin g(x) + 2πϵx^{α} cos g(x))^{−1})| dx

≤ \frac{K}{q} ∫_{A_{Nϵr}} |\frac{α(α − 1)x^{α−2} sin g(x) + 4πδαx^{α−1} cos g(x) + 4π^{2}δ^{2}x^{α} sin g(x)}{(αx^{α−1} sin g(x) + 2πϵx^{α} cos g(x))^{2}} | dx

≤ \frac{K}{q} ∫_{A_{Nϵr}} \frac{K_{1}x^{α−2} + K_{2}δx^{α−1} + K_{3}δ^{2}x^{α−2}}{ϵ^{2}(δ^{2}x^{2α} + x^{2α−2})} dx

(4.37)

Now by the arithmetic mean, geometric mean inequality we have

δx^{α−1} <= \frac{δ^{2}x^{α}}{2} + \frac{x^{α−2}}{2} .

And so we can further bound |J| by:

\begin{aligned}|J| & ≤ \frac{K}{q} ∫_{A_{Nϵr}} \frac{K_{1}x^{α−2} + K_{2}δ^{2}x^{α−2}}{ϵ(δ^{2}x^{2α} + x^{2α−2})} \\ & ≤ \frac{K}{q} ∫_{ϵN}^{N} x^{−α}dx \\ & ≤ \frac{KN^{1−α}}{q} \end{aligned}  (4.38)

Our next aim is to show that ∑_{r=1}^{q} |B_{Nϵr}| = O(ϵNq). We fix x and let C_{Nϵx} = \{r | 1 ≤ r ≤ q, | sin (2π (δx + \frac{rp}{q}) + θ(x)) | < ϵ\} ask how big C_{Nϵr} is.

Let ϵ_{1} = \frac{sin^{−1} ϵ}{2π}. Now if sin (2πy + ϕ) < ϵ then we must have \{y\} ∈ (K, K+ϵ_{1}) for some K ∈ [0, 1).

Letting y = \frac{rp}{q} above we find that the size of C_{nϵx} is given by ϕ_{q}^{\{ \frac{kp}{q} \}}(K, K+ ϵ_{1}). Now \{\frac{kp}{q}\} takes on q evenly spaced values in the interval [0, 1). Thus there will be approximately ϵ_{1}q of them in the interval [K, K + ϵ_{1}). So the size of C_{Nϵx} is O(qϵ_{1}) = O(qϵ).

Now using Fubini’s theorem we can conclude that

\begin{aligned}∑_{r=1}^{q} |B_{Nϵr}| & = ∑_{x=ϵN}^{N} |C_{Nϵx}| \\ & = O(ϵNq) \end{aligned}  (4.39)

Thus combining (4.35), (4.37) and (4.39) gives:

\frac{1}{N} ∑_{r=1}^{q} |First Term| ≤ \frac{KqN^{1−α}}{Nq} + \frac{1}{qN} O(ϵNq)
≤ KN^{−α} + O(ϵ)

(4.40)

We now consider the third term.

|Third term| = K ∫_{ϵN/q}^{N/q} |q(tq + r)^{α−1} sin (2π (tqδ + \frac{rp}{q} + rδ))

+qδ(tq + r)^{α} cos (2π (tqδ + \frac{rp}{q} + rδ))| dt

≤ K ∫_{ϵN/q}^{N/q} (q(tq + r)^{α−1} + qδ(tq + r)^{α}) dt

= K_{1} [(tq + r)^{α}]_{ϵN/q}^{N/q} + \frac{K_{2}qδ}{q} [(tq + r)^{α+1}]_{ϵN/q}^{N/q}

≤ K_{1}N^{α} + \frac{K_{2}qδ}{q} N^{α+1}

(Since r < q < N^{β} the r gets absorbed in the constant.)

≤ \frac{K_{1}N}{q} (qN^{α−1}) + \frac{K_{2}N}{q} (qδN^{α})

≤ \frac{KN}{q} (N^{(α+β)−1} + N^{α−β})

(4.41)

Note that since α < β < \frac{1}{2} we have (α − β) < 0 and (α + β) − 1 < 0. Now combining (4.33), (4.40) and (4.41) gives

lim sup_{N→∞} \frac{1}{N} | ∑_{n=ϵN}^{N} e^{2πihn^{α} sin 2πan}| ≤ lim sup_{N→∞} \frac{1}{q} ∑_{r=0}^{q} \frac{q}{N} | ∑_{m=⌈ϵN/q⌉}^{⌊N/q⌋} e^{2πih(mq+r)^{α} sin(2π(mqδ+ \frac{rp}{q} +rδ))}|

≤ lim sup_{N→∞} \frac{1}{q} ∑_{r=0}^{q}(KN^{−α} + K′ (N^{(α+β)−1} + N^{α−β}) + O(ϵ))

≤ lim sup_{N→∞} K(N^{−α} + N^{(α+β)−1} + N^{α−β}) + O(ϵ)

= O(ϵ)

(4.42)

Thus n^{α} sin 2πan is u.d. mod 1. for 0 < α < 1 and a irrational.

Proof of Theorem 4.5.1 Letting a = \frac{1}{2π} in Theorem 4.5.2 proves this theorem.

4.6 The distribution of (f(n) sin n^{β})

For a sufficiently slow function f we find that we can extend the results of Theorem 4.4.3 to include more general families of sequences.

Theorem 4.6.1 For 0 < β < 1, if f(x) is a function such that f′(x) ≤ \frac{K}{x^{γ}} with γ > 1 − β and f(x) → ∞, then (f(n) sin n^{β}) is uniformly distributed modulo 1.

Proof Pick 0 < β < 1 and assume f(x) meets the conditions above. Pick 0 < ϵ < 1 and choose N_{i}, i = 1, 2 as follows:

1. N_{1} (So that f(ϵN_{2}) > 1.)

2. N_{3} > N_{2} and N_{3} > \frac{1}{ϵ} (\frac{β}{ϵ^{2}} )^{\frac{1}{β+γ−1}} (So that ϵβx^{β−1}f(x)ϵ > x^{−γ} > f′(x) sin x^{β}.)

Then choose N > N_{2}.

Once again we use the Euler summation formula and define

\begin{matrix}A_{Nϵ} = \{x|ϵN ≤ x ≤ N, | cos x^{β}| ≥ ϵ\} \\ B_{Nϵ} = [ϵN, N]/A_{Nϵ} \end{matrix}

We look at the size fo the first term.

|First Term| ≤ \underbrace{|∫_{A_{Nϵ}} e^{2πihf(x) sin x^{β}}dx|}_{I_{A}} + \underbrace{|∫_{B_{Nϵ}} e^{2πihf(x) sin x^{β}}dx|}_{I_{B}}  (4.43)

|I_{A}| = K |∫_{A_{Nϵ}} \frac{d}{dx} (e^{2πihf(x) sin x^{β}}) (f′(x) sin x^{β} + βx^{β−1}f(x) cos x^{β})^{−1}dx|

≤ K |∫_{A_{Nϵ}} \frac{d}{dx}e^{2πihf(x) sin x^{β}}(2βx^{β−1}f(x) cos x^{β})^{−1}dx| + ϵK′N

(Since N > N_{3} we can apply (2.21))

≤ K [\frac{1}{|x^{β−1}f(x) cos x^{β}|}]_{A_{Nϵ}} +

∫_{A_{Nϵ}} |\frac{K_{1}x^{2β−2}f(x) sin x^{β} + K_{2}x^{β−2}f(x) cos x^{β} + K_{3}x^{β−1}f′(x) cos x^{β}}{x^{2β−2}f^{2}(x) cos^{2} x^{β}} | dx + ϵK′N

(Using (2.20))

(4.44)

Now observe that since f′(x) < Kx^{−γ} < Kx^{β−1} then

|K_{1}x^{2β−2}f(x) sin x^{β} + K_{2}x^{β−2}f(x) cos x^{β} + K_{3}x^{β−1}f′(x) cos x^{β}| ≤ |K_{1}x^{2β−2}f(x)| + |K_{2}x^{β−2}f(x)| + |K_{3}x^{β−1}f′(x)| ≤ Kx^{2β−2}|f(x)| + K′x^{2β−2} ≤ Kx^{2β−2}|f(x)| (Since N > N_{2})

Thus we have:

|I_{A}| ≤ \frac{KN}{ϵf(N)} + K ∫_{A_{Nϵ}} \frac{1}{|f(x) cos^{2} x^{β}|}dx + l.o.t. + ϵK′N

(Using Lemma 4.4.1 with M = N and g(x) = x^{β})

≤ \frac{KN}{ϵf(N)} + \frac{K′′}{ϵ^{2}} ∫_{A_{Nϵ}} \frac{dx}{f(x)} + l.o.t. + ϵK′N

≤ \frac{KN}{ϵf(N)} + \frac{K′′N}{ϵ^{2}f(ϵN)} + l.o.t. + ϵK′N

(4.45)

Once again using Lemma 4.4.2 with M = N we have

lim sup_{N→∞} \frac{|I_{B}|}{N} = lim sup_{N→∞} \frac{1}{N} |∫_{B_{Nϵ}} e^{2πihf(x) sin x^{β}}dx|

= O(ϵ)

(4.46)

Combining equations 4.43, 4.45 and 4.46 gives:

lim sup_{N→∞} \frac{|First Term|}{N} ≤ lim sup_{N→∞} (\frac{K}{ϵf(N)} + \frac{K′′}{ϵ^{2}f(ϵN)} + O(ϵ) + \underbrace{l.o.t.}_{→0})

= O(ϵ)

(4.47)

lim sup_{N→∞} \frac{|Third Term|}{N} ≤ lim sup_{N→∞} \frac{K}{N} ∫_{ϵN}^{N} |x^{β−1}f′(x) cos x^{β}| dx

≤ lim_{N→∞} \frac{K}{N} ∫_{ϵN}^{N} x^{−γ+β−1}dx

= lim_{N→∞} \frac{K}{N} [x^{β−γ}]_{ϵN}^{N}

≤ lim_{N→∞} \frac{KN^{β−γ}}{N}

= lim_{N→∞} KN^{−(γ+(1−β))}

= 0

(4.48)

Combining equations (4.47) and (4.48) and using Proposition 2.2.5 we see that

lim sup_{N→∞} \frac{1}{N} |∑_{n=1}^{N} e^{2πihf(x) sin x^{β}} | = O(ϵ)

and so (f(x) sin x^{β}) is u.d. mod 1.

Applying functions to sequences

In the previous chapters we have analysed the distribution of particular families of sequences. In this chapter we take a different approach and attempt to apply functions to families of sequences about which we already know something. We have already seen an attempt at this in section 4.1 when we discussed the distribution of sin x_{n} where (x_{n}) was uniformly distributed modulo 2π. Most of the time however we will not be able to arrive at particularly conclusive results about the distribution of such sequences. This is because we can generally find fairly contrived yet valid sequences which serve as a counter example to anything we would like to conclude.

5.1 Adding and multiplying uniformly distributed sequences

Let (u_{n}) and (v_{n}) be sequences which are uniformly distributed modulo 1 and consider the distribution of (u_{n} + v_{n}). Very little can be can be said about the above sequence. In some situations it will be u.d. and in others it will not. To illustrate the first of these two cases, let u_{n} = \sqrt{2}n and v_{n} = \sqrt{3}n then u_{n} + v_{n} = (\sqrt{2} + \sqrt{3}) n which is u.d. since \sqrt{2} + \sqrt{3} is irrational.

To illustrate the second case, let u_{n} be as above, and let v_{n} = − \sqrt{2}n. Then u_{n} + v_{n} = 0 ∀n and hence is clearly not u.d.

Similarly we consider the distribution of (u_{n}v_{n}) and again arrive at inconclusive results.

When u_{n} = v_{n} = \sqrt{2}n then u_{n}v_{n} = 2n^{2} which is clearly not u.d. mod 1 However if u_{n} is as above and v_{n} = \sqrt{3}n Then u_{n}v_{n} = \sqrt{6}n^{2} which is u.d. mod 1 using Proposition 3.1.3.

5.2 The distribution of log u_{n}

We now ask whether the distribution of (log u_{n}) is related to the distribution of (u_{n}).

Certainly if (u_{n}) is not u.d. mod 1, then we can have (log u_{n}) not u.d. mod 1. To see this simply let u_{n} = n.

Now let u_{n} = nθ where θ is irrational. Then u_{n} is u.d. mod 1 and

∑_{n=1}^{N} e^{2πi log nθ} = ∑_{n=1}^{N} e^{2πi(log n+log θ)}

= e^{2πi log θ} ∑_{n=1}^{N} e^{2πi log n}

However dividing by N and attempting to take the limit as N approaches infinity we find that the limit on the right hand side does not exist (see proof of Proposition 3.2.1) and thus the limit on the left hand side does not exist and therefore is certainly not zero, hence log nθ is not u.d. mod 1. Thus if (u_{n}) is u.d. mod 1, then we can have (log u_{n}) not u.d. mod 1.

Now suppose that 0 ≤ log u_{n} < 1 and that (log u_{n}) is uniformly distributed in the unit interval and so (log u_{n}) is u.d. mod 1. Then 1 ≤ u_{n} < e.

Let v_{n} = {u_{n}} and consider the distribution of v_{n}

lim_{N→∞} \frac{ϕ_{N}^{v_{n}}(0, e − 2)}{N} = lim_{N→∞} \frac{ϕ_{N}^{u_{i}}(1, e − 1) + ϕ_{N}^{u_{n}}(2, e)}{N}

= lim_{N→∞} \frac{ϕ_{N}^{log u_{n}}(0, log (e − 1)) + ϕ_{N}^{log u_{n}}(log 2, log e)}{N}

= log (e − 1) + 1 − log 2

≃ 0.848 > e − 2

Hence (v_{n}) = ({u_{n}}) does not satisfy (1.1) and so is not uniformly distributed in the unit interval. Hence (u_{n}) is not uniformly distributed modulo 1.

It is thus possible to have (log u_{n}) u.d. mod 1, and (u_{n}) not u.d. mod 1.

Finally we let u_{n} = e^{n\sqrt{2}} − \{e^{n\sqrt{2}}\} + \{n\sqrt{2}\}. Then \{u_{n}\} = \{n\sqrt{2}\} and so u_{n} is uniformly distributed modulo 1. We now consider the distribution of (log u_{n}). To do this we will use the following fact: Pick ϵ > 0 and let η_{0} = \frac{1}{ϵ}. If η_{0} < a, b and |b − a| ≤ 1 using the mean value theorem we have:

| log b − log a| < ϵ  (5.1)

Pick N_{0} so that e^{N_{0}\sqrt{2}} > η_{0} + 1. Then since |u_{n} − e^{n\sqrt{2}}| < 1 we can use (5.1) to show that for n > N_{0}

| log u_{n} − log e^{n\sqrt{2}}| < ϵ

This means that for n > N_{0}, log u_{n} = n\sqrt{2} + ϵ_{n} where |ϵ_{n}| < ϵ Since ϵ can be chosen to be arbitrarily small we can conclude that for n > N_{0} (log u_{n}) is uniformly distributed modulo 1. This means that (log u_{n}) is uniformly distributed modulo 1 and so it is possible to have both (u_{n}) and (log u_{n}) u.d. mod 1.

¿From the above analysis we can conclude that the distribution of (log u_{n}) is not at all dependent on the distribution of (u_{n}).

5.3 The distribution of u_{n} + sin v_{n}

Proposition 5.3.1 If (u_{n}, v_{n}) is u.d. mod (1, 2π) then (u_{n} + sin v_{n}) is uniformly distributed modulo 1.

Proof Let y_{k} = u_{k}+sin v_{k} Now {y_{k}} ∈ [0, h) iff z_{k} = {u_{k}}+{sin v_{k}} ∈ [0, h) ∪ [ 1 , 1 + h).

Our aim is to show that lim_{N→∞} \frac{ϕ_{N}^{\{y_{k}\}}(0,h)}{N} = h In order to do this we will first consider the joint distribution of ({u_{n}}, {sin v_{n}}).

For ϵ > 0, a, b ∈ [0, 1) let θ_{1} = sin^{−1} b, γ_{1} = sin^{−1} (b + ϵ), θ_{2} = sin^{−1} (1 − b) and γ_{2} = sin^{−1} (1 − (b + ϵ)). If r_{k} = ({u_{n}}, {sin v_{n}}) ∈ (a, a + ϵ) x (b, b + ϵ) then

s_{k} = ({u_{n}}, {v_{n}}_{2π})
∈ (a, a + ϵ) × [(θ_{1}, γ_{1}) ∪ (π − γ_{1}, π − θ_{1}) ∪
(π + γ_{2}, π + θ_{2}) ∪ (2π − θ_{2}, 2π − γ_{2})]

And so we have

lim_{N→∞} \frac{1}{N} ϕ_{N}^{r_{k}}((a, a + ϵ) × (b, b + ϵ))

= lim_{N→∞} \frac{1}{N} (ϕ_{N}^{s_{k}}((a, a + ϵ) × (θ_{1}, γ_{1})) + ϕ_{N}^{s_{k}}((a, a + ϵ) × (π − γ_{1}, π − θ_{1}))

+ϕ_{N}^{s_{k}}((a, a + ϵ) × (π + γ_{2}, π + θ_{2})) + ϕ_{N}^{s_{k}}((a, a + ϵ) × (2π − θ_{2}, 2π − γ_{2})))

= (a + ϵ − a)\frac{(γ_{1} − θ_{1}) + (π − θ_{1} − π + γ_{1}) + (π + θ_{2} − π − γ_{2}) + (2π − γ_{2} − 2π + θ_{2})}{2π}

(Since (u_{n}, v_{n}) is u.d. mod (1, 2π))

= \frac{2ϵ}{2π} ((sin^{−1} (b + ϵ) − sin^{−1} b) + (sin^{−1} (1 − (b + ϵ)) − sin^{−1} (1 − b)))

= \frac{ϵ^{2}}{π} (\frac{1}{\sqrt{1 − b^{2}}} + \frac{1}{\sqrt{1 − (1 − b)^{2}}} ) + O(ϵ^{3})

(by Taylor expansion) (5.2)

We now consider ϕ_{N}^{{y_{k}}}(0, h) for N large.

lim_{N→∞} \frac{1}{N} ϕ_{N}^{\{y_{k}\}}(0, h) = lim_{N→∞} \frac{1}{N} (ϕ_{N}^{z_{k}}(0, h) + ϕ_{N}^{z_{k}}(1, 1 + h))

= lim_{N→∞} \frac{1}{N} (lim_{ϵ→0} ∑_{β=0}^{\frac{1}{ϵ} −1} ∑_{α= \frac{1}{ϵ} −β−1}^{\frac{1+h}{ϵ} −β−1} ϕ_{N}^{r_{k}}(E_{αβϵ}))

where E_{αβϵ} = (ϵα, ϵ(α + 1)) × (ϵβ, ϵ(β + 1))

(5.3)

Now when ever r_{k} is in ∑β=−1^{\frac{1}{ϵ}} ∑_{α= \frac{1}{ϵ} −β−2}^{\frac{1+h}{ϵ} −β} E_{αβϵ} Then z_{k} is in (0, h)∪(1, 1+

h) and whenever z_{k} is in (0, h) ∪ (1, 1 + h) then r_{k} is in ∑_{β=2}^{\frac{1}{ϵ} −2} ∑_{α= \frac{1}{ϵ} −β}^{\frac{1+h}{ϵ} −β−2} E_{αβϵ} and so we can evaluate lim_{N→∞} \frac{1}{N} ϕ_{N}^{\{y_{k}\}}(0, h) by evaluating the integral associated with the right hand side of (5.3), making use of (5.2). We get:

lim_{N→∞} \frac{1}{N} ϕ_{N}^{\{y_{k}\}}(0, h) = \frac{1}{π} ∫_{0}^{1} ∫_{1−y}^{1+h−y} (\frac{1}{\sqrt{1 − y^{2}}} + \frac{1}{\sqrt{1 − (1 − y)^{2}}} ) dxdy

= \frac{h}{π} ∫_{0}^{1} (\frac{1}{\sqrt{1 − y^{2}}} + \frac{1}{\sqrt{1 − (1 − y)^{2}}} ) dy

= \frac{h}{π} [sin^{−1} y − sin^{−1} (1 − y)]_{0}^{1}

= \frac{h}{π} [\frac{π}{2} − 0 − 0 + \frac{π}{2} ]

= h (5.4)

and so (y_{k}) is u.d. mod 1.

5.4 The distribution of {u_{n}} sin v_{n}

Proposition 5.4.1 If (u_{n}) is uniformly distributed modulo 1 and (v_{n}) is uniformly distributed modulo 2π then ({u_{n}} sin v_{n}) is not uniformly distributed modulo 1.

Proof Let S_{1N} = {n : n < N, {u_{n}} < ϵ} and S_{2N} = {n : n < N, 0 < sin v_{n} < ϵ} Now assume that ({u_{n}} sin v_{n}) is u.d. mod 1, then

ϵ = lim_{N→∞} \frac{ϕ_{N}^{\{u_{n}\} sin v_{n}}(0, ϵ)}{N}

> lim_{N→∞} \frac{|S_{1N} ∪ S_{2N}|}{N}

= lim_{N→∞} \frac{|S_{1N}| + |S_{2N}| − |S_{1N} ∩ S_{2N}|}{N}

> ϵ + f(ϵ) − ϵ^{2}

(5.5)

Where f(ϵ) = lim_{N→∞} \frac{ϕ_{N}^{sin vn}(0,ϵ)}{N}. And thus from (5.5) we get

f(ϵ) < ϵ^{2}  (5.6)

Now let γ = sin^{−1} ϵ then

f(ϵ) = lim_{N→∞} \frac{ϕ_{N}^{sin v_{n}}(0, ϵ)}{N}

= lim_{N→∞} \frac{ϕ_{N}^{v_{n}}(0, γ) + ϕ_{N}^{v_{n}}(π − γ, π)}{N}

= \frac{(γ − 0) + (π − π + γ)}{2π}

(Since (v_{n}) is u.d. mod 2π.)

= \frac{γ}{π}

(5.7)

Taking ϵ = 0.1 we get f(ϵ) = 0.032 > ϵ^{2} = 0.01 which contradicts (5.6). Thus ({u_{n}} sin v_{n}) cannot be u.d. mod 1.

Appendix A

The distribution of n sin n

In this appendix we attempt simplify a proof given in [2] to show the uniform distribution of the sequence (n sin n). The proof of [2] actually falls down in one particular case. While an attempt was made to understand this case, lack of time meant that we could not write it up in time for this report.

Theorem A.0.2 For α ∈ ℝ \ ℚ the sequence (n sin 2παn) is uniformly distributed modulo 1.

In the course of the proof we will make use of the following Lemma which is known as Van Der Corput’s estimate.

Lemma A.0.3 Let g ∈ C^{2}[X_{1}, X_{2}] and assume that 0 < λ ≤ g′′(x) ≤ \frac{λ}{ϵ} then

| ∑_{x=X_{1}}^{X_{2}} e^{2πig(x)}| ≤ \frac{1}{ϵ} (Xλ^{\frac{1}{2}} + λ^{− \frac{1}{2}} )  (A.1)

where X = X_{2} − X_{1}.

Proof (See [9]) If λ > 1 then we just estimate the left-hand side trivially by X. So we may assume that λ ≤ 1.

We shall shortly show the estimate

∑_{n=A}^{B} e^{2πig(n)} = 2λ^{−1/2}  (A.2)

whenever there exists an integer M such that

M − \frac{1}{2} ≤ g′(n) ≤ M + \frac{1}{2}

for all A ≤ n ≤ B.

Assume (A.2) for the moment. From the hypotheses on g′′ we see that g′ is increasing and that g′(X_{2}) − g′(X_{1}) ≤ \frac{Xλ}{ϵ}. Thus we can partition the left-hand side of (A.1) into \frac{Xλ+1}{ϵ} summations of the form (A.2), so that we can estimate the left-hand side of (A.1) by

\frac{(Xλ + 1)λ^{−1/2}}{ϵ^{2}} ≤ \frac{1}{ϵ^{2}} (Xλ^{1/2} + λ^{−1/2})

as desired.

Now we prove (A.2). We may set M = 0 by the trick of by replacing g(n) by g(n) − nM (note that this does not affect (A.2)). Thus we have

−\frac{1}{2} ≤ g′(n) ≤ \frac{1}{2}

for all A ≤ n ≤ B.

By dividing the sum into three smaller sums if necessary, we can assume that one of the following three statements is true:

Case 1: −1/2 ≤ g′(n) ≤ −λ^{1/2} for all A ≤ n ≤ B.

Case 2: −λ^{1/2} < g′(n) < λ^{1/2} for all A ≤ n ≤ B.

Case 3: λ^{1/2} ≤ g′(n) ≤ 1/2 for all A ≤ n ≤ B.

We will not consider Case 1 as it is very similar to Case 3.

Consider Case 2. By the hypothesis on g′′ there are at most 2λ^{−1/2} terms in the sum, so we may trivially bound this sum by 2λ^{−1/2}, which is OK.

Now consider Case 3. We shall use a summation by parts trick.

¿From the MVT we have

g(n + 1) = g(n) + g′(n + θ_{n})

for some 0 ≤ θ_{n} ≤ 1. Thus

e^{2πig(n)} = e^{2πig(n+1)} − e^{2πig(n)} \frac{1}{e^{2πig′(n+θ_{n})} − 1}.

Since

\frac{1}{e^{iθ} − 1} = \frac{1}{2} \frac{e^{iθ} + 1}{e^{iθ} − 1} − \frac{1}{2} = \frac{i}{2} cot(θ/2) − \frac{1}{2}

we have

\begin{aligned}∑_{n=A}^{B} e^{2πig(n)} & = ∑_{n=A}^{B−1} e^{2πig(n)} + O(1) \\ & = ∑_{n=A}^{B−1} (e^{2πig(n+1)} − e^{2πig(n)} (\frac{i}{2} cot(πg′(n + θ_{n})) − \frac{1}{2})) + O(1). \end{aligned}

By the summation by parts formula previously seen as (3.19) we can rewrite this as

e^{2πig(B)} (\frac{i}{2} cot (πg′(B − 1 + θ_{B−1})) − \frac{1}{2}) − e^{2πig(A)} (\frac{i}{2} cot (πg′(A + θ_{A})) − \frac{1}{2})

− ∑_{n=A+1}^{B−1} e^{2πig(n)} (\frac{i}{2} cot (πg′(n + θ_{n})) − \frac{i}{2} cot (πg′(n − 1 + θ_{n−1})) + O(1)

The first two terms are no bigger than λ^{−1/2} since we are in Case 3 and cot(θ) = \frac{1}{θ} for 0 < θ < \frac{π}{2}. Now look at the third term. The absolute value of this term is less than

∑_{n=A+1}^{B−1} | cot(πg′(n + θ_{n})) − cot(πg′(n − 1 + θ_{n−1})|.

But in Case 3, the sequence cot(πg′(n + θ_{n})) is decreasing in n, so we can telescope this series as

cot(πg′(A + 1 + θ_{A+1})) − cot(πg′(B − 1 + θ_{B−1})

which is no bigger than λ^{−1/2} since we are in Case 3.

Lemma A.0.3 can actually be generalised to work for all deriavtives of g. It’s general form is stated below:

Lemma A.0.4 Let g ∈ C^{2}[X_{1}, X_{2}] and assume that 0 < λ_{j} ≤ g^{(j)}(x) ≤ Kλ_{j} then lettign J = 2^{j} we have

| ∑_{x=X_{1}}^{X_{2}} e^{2πig(x)}| ≤ K (Xλ^{1/(2−J)} + 1 + X^{1−2/J} + X(λX^{4−8/J})^{−2/J}) (A.3)

where X = X_{2} − X_{1}.

We will use this Lemma for the case when j = 3 without proof. Proof of Theorem A.0.2 As usual we pick ϵ > 0. Let ϵ_{2} = sin^{−1} ϵ and pick b ∈ ℤ with b ≠ 0. We now choose N to be “large enough” depending on b and ϵ. Let M ∼ N^{\frac{5}{8}} and invoke Theorem 4.5.3 to write α = \frac{p}{q} + \frac{1}{Q} with q < N^{\frac{5}{8}} and \frac{q}{Q} < N^{\frac{−5}{8}} . Without loss of generality let us assume that Q, q > 0.

Throughout the proof we let

S = | ∑_{n= \frac{N}{2}}^{N} e^{2πibn sin 2παn}|  (A.4)

Our aim will be to show that S < KNϵ. This will mean that dividing by N and taking the lim sup as N → ∞ will give lim sup_{N→∞} \frac{S}{N} = 0 and so making use of Proposition 2.2.5 we can conclude that (n sin 2παn) is uniformly distributed modulo 1.

The proof now breaks into two cases, the first where Q ≥ \frac{N}{ϵ_{2}} and so α is close to a rational, and the second where Q ≤ \frac{N}{ϵ_{2}} and so α is far from being a rational. Case 1: Q > \frac{N}{ϵ_{2}} We now write n = mq + k with 0 ≤ k ≤ q . Then

sin 2παn = sin (2π (\frac{p}{q} + \frac{1}{Q}) (mq + k)) = sin (2π (\frac{p}{q} + \frac{1}{Q}) (mq + k))

For fixed k let f(m) = 2π (\frac{kp}{q} + \frac{mq+k}{Q}) and g(m) = b(mq + k) sin f(m) and note that

g′′(m) = K \frac{q^{2}}{Q} cos f(m) + K′ (\frac{q}{Q})^{2} (mq + k) sin f(m)

Let us now denote by B_{k} the set of k’s for which either sin f(m) < ϵ or cos f(m) < ϵ and so one of the terms of g′′ is very small. These are k’s for which either

0 < \{\frac{f(m)}{2π}\} < sin^{−1} ϵ or cos^{−1} ϵ < \{\frac{f(m)}{2π}\} < 1

Now \frac{mq+k}{Q} < \frac{N}{Q} < ϵ_{2} = sin^{−1} ϵ and so if 0 < \{\frac{kp}{q}\} < 2 sin^{−1} ϵ then sine condition is satisfied, and a similar argument works for the cosine condition. Hence for q large enough the size of B_{k} can be approximated by

ϕ_{q}^{\{ \frac{2πkp}{q} \}}(0, 2ϵ_{2})+ϕ_{q}^{\{ \frac{2πkp}{q} \}}(1−2ϵ_{2}, 1). Now \{\frac{kp}{q}\} takes on q evenly spaced values in the interval [0, 1). Thus for large enough q there will be approximately 2ϵ_{2}q of them in the interval [0, 2ϵ_{2}) and the interval [1 − 2ϵ_{2}, 1). Thus the size of B_{k} is O(qϵ). Let A_{k} denote the set of all other k’s. Then we have

\begin{aligned}S & ≤ ∑_{A_{k}} |∑_{m=1}^{\frac{N}{q}} e^{2πig(m)}| + ∑_{B_{k}} |∑_{m=1}^{\frac{N}{q}} e^{2πig(m)}| \\ & ≤ ∑_{A_{k}} |∑_{m=1}^{\frac{N}{q}} e^{2πig(m)}| + Kqϵ . \frac{N}{q} \\ & ≤ ∑_{A_{k}} |∑_{m=1}^{\frac{N}{q}} e^{2πig(m)}| + KNϵ \end{aligned}  (A.5)

Now for k ∈ A_{k} we have:

\begin{aligned}|g′′(m)| & = |K \frac{q^{2}}{Q} cos f(m) + K′ (\frac{q}{Q})^{2} (mq + k) sin f(m)| \\ & ≤ K \frac{q^{2}}{Q} + Kϵ_{2} \frac{q^{2}}{Q} \\ & ≤ K \frac{q^{2}}{Q} \end{aligned}  (A.6)

In addition,

\begin{aligned}|g′′(m)| & = |K \frac{q^{2}}{Q} cos f(m) + K′ (\frac{q}{Q})^{2} (mq + k) sin f(m)| \\ & ≥ K \frac{q^{2}}{Q} ϵ_{2} − K (\frac{q}{Q})^{2} (mq + k)ϵ_{2} \\ & ≥ K \frac{q^{2}}{Q} ϵ_{2} − K \frac{q^{2}}{Q} ϵ_{2}^{2} \\ & ≥ K \frac{q^{2}}{Q} ϵ_{2} \end{aligned}  (A.7)

This case now splits into three subcases. Case 1a: \frac{N^{5/4}}{ϵ^{2}ϵ_{2}^{2}} < Q < ϵ^{2}ϵ_{2}^{2}N^{2}

In this case we can apply Lemma A.0.3 with λ = \frac{q^{2}}{Q} ϵ_{2} . Note that:(\frac{q^{2}}{ϵ^{2}} ) <

\frac{N^{5/4}}{ϵ^{2}} < Q so \frac{q^{2}}{Q} < ϵ^{2}. We get:

\begin{aligned}|∑_{m} e^{2πig(m)}| & ≤ \frac{1}{ϵ_{2}} (\frac{N}{q} λ^{\frac{1}{2}} + λ^{− \frac{1}{2}} ) \\ & = \frac{N}{qϵ_{2}} ((\frac{q^{2}}{Q} )^{1/2} + \frac{q\sqrt{Q}}{Nq}) \\ & ≤ \frac{N}{qϵ_{2}} (ϵϵ_{2} + ϵϵ_{2}) \\ & ≤ \frac{2Nϵ}{q} \end{aligned}  (A.8)

Case 1b: Q < \frac{N^{5/4}}{ϵ^{2}ϵ_{2}^{2}} In this case we are required to look at the third deriavtive of g and use the general form of Van Der Corput’s estimate. We have:

g^{(3)}(m) = K \frac{q^{3}}{Q^{2}} sin f(m) + K′ (\frac{q}{Q})^{3} cos f(m)

As previously we can bound g^{(3)} form above and below via:

\begin{aligned}|g^{(3)}(m)| & = |K \frac{q^{3}}{Q^{2}} sin f(m) + K′ (\frac{q}{Q})^{2} (mq + k) cos f(m)| \\ & ≤ K \frac{q^{3}}{Q^{2}} + Kϵ_{2} \frac{q^{2}}{Q} \\ & ≤ K \frac{q^{3}}{Q^{2}} \end{aligned}  (A.9)

and

\begin{aligned}|g^{(3)}(m)| & = |K \frac{q^{3}}{Q^{2}} sin f(m) + K′ (\frac{q}{Q})^{3} (mq + k) cos f(m)| \\ & ≥ K \frac{q^{3}}{Q^{2}} ϵ_{2} − K (\frac{q}{Q})^{3} (mq + k)ϵ_{2} \\ & ≥ K \frac{q^{3}}{Q^{2}} ϵ_{2} − K \frac{q^{3}}{Q^{2}} ϵ_{2}^{2} \\ & ≥ K \frac{q^{3}}{Q^{2}} ϵ_{2} \end{aligned}  (A.10)

We can apply Lemma A.0.4 with λ_{3} = \frac{q^{3}}{Q^{2}} . Observe that \frac{q^{3}}{Q^{2}} < \frac{N^{15/8}}{N^{2}} = N^{−1/8}. We get:

|∑_{m} e^{2πig(m)}| ≤ K (\frac{N}{q} (\frac{q^{3}}{Q^{2}} )^{1/6} + 1 + (\frac{N}{q} )^{3/4} + \frac{N}{q} (\frac{Q^{2}}{q^{3}} (\frac{N}{q} )^{3})^{−1/4})

= \frac{KN}{q} (N^{−1/48} + \frac{q}{N} + (\frac{q}{N} )^{3/4} + (\frac{Q^{2}}{N^{3}} )^{−1/4})

≤ \frac{KN}{q} (N^{−1/48} + N^{−3/8} + N^{−9/32} + (\frac{N^{−3/4}}{ϵ^{2}ϵ_{2}^{2}} )^{1/4})

≤ \frac{KNN^{−1/48}}{q} (For N large enough)

≤ \frac{KNϵ}{q} (For N large enough)

(A.11)

We noe joinc Cases 1a and 1b together and show that we can put an appropriate bound on S.

Using (A.4) and (A.8) we get

\begin{aligned}S & ≤ ∑_{A_{k}} |∑_{m} e^{2πig(m)}| + ∑_{B_{k}} |∑_{m} e^{2πig(m)}| \\ & ≤ ∑_{A_{nk}} \frac{Hϵ}{q} + KNϵ \\ & ≤ q . \frac{Nϵ}{q} + KNϵ \\ & = KNϵ \end{aligned}  (A.12)

This is the appropriate bound for S and so if \frac{N}{ϵ} < Q, ϵN^{2} then the sequence is u.d. mod 1. Case 1c: Q > ϵN The proof of this case in [2] actually falls down for this case when considering the sequence (n sin 2παn). It is possible to find a proof that S = O(Nϵ) in this case, however limited time meant that we could not write it up.

Case 2: Q ≤ \frac{N}{ϵ_{2}} We begin by setting H = ϵ^{3}ϵ_{2}^{5}Q. We now break the sum from (A.4) up into smaller but overlapping sums of length H and sum these up. The result is that each term from (A.4) gets summed H times, and in addition a few extra terms at the end are added. We find that

S ≤ \frac{1}{H} | ∑_{n= \frac{N}{2} −H}^{N} ∑_{h=H}^{2H} e^{2πib(n+h) sin 2πα(n+h)} − \underbrace{extra terms which are added at the end}_{<H}|

≤ \frac{1}{H} ∑_{n= \frac{N}{2} −3H}^{N} |∑_{h=H}^{2H} e^{2πib(n+h) sin 2πα(n+h)}| + Nϵ

(Since H < ϵ^{3}ϵ_{2}^{5}Q < cϵ^{3}ϵ_{2}^{4}N < Nϵ)

(A.13)

We now write h = mq + k and observe that: sin 2πα(n + h) = sin (2παn + 2π (\frac{p}{q} + \frac{1}{Q}) (mq + k)) = sin (2παn + 2π (\frac{kp}{q} + \frac{h}{Q})).

For fixed n, k let f(m) = 2παn + 2π (\frac{kp}{q} + \frac{mq+k}{Q}) and let g(m) = b(n + mq + k) sin f(m). We see that

g′′(m) = K (\frac{q}{Q})^{2} (n + h) sin f(m) + K′ \frac{q^{2}}{Q} cos f(m)  (A.14)

In addition we can now write S as:

S ≤ \frac{1}{H} ∑_{n,k} |∑_{m} e^{2πig(m)}| + Nϵ  (A.15)

Where n ranges from 1 to N, k ranges from 1 to q and m ranges from \frac{H}{q} to \frac{2H}{q}. We would like to examine the size of the inner sum of (A.15). In order to to this we will attempt to find a λ such that λ < |g′′| < Kλ and then invoke Lemma A.0.3. Unfortunately there are problem values of (n, k) for which g′′(m) is very small or even vanishes. In order to avoid this we will break the set of (n, k) pairs up into the “good” set and the “problem” set and show that the “problem” set is not particularly large.

There are now two sub cases. Either Q < KNϵ and so the dominant term in the second derivative is the first one, or Q ≥ KNϵ in which case the two terms of the second derivative are of roughly the same size. We will consider these two cases separately. Case 2a: Q < KNϵ Let us denote by B_{nk} the set of pairs (n, k) which have sin f(m) < ϵ_{2} and by A_{nk} the remaining pairs. Now \frac{mq+k}{Q} < \frac{H}{Q} = ϵ^{3}ϵ_{2}^{5} and so this term of f(m) makes very little difference to sin f(m). Thus the size of B_{nk} is essentially

the same as ∑_{k=1}^{q} ϕ_{N}^{\{nα+ \frac{kp}{q} \}}(0, ϵ). And since (nα) is uniformly distributed modulo 1, for N large enough this is about qNϵ. Thus we get

\begin{aligned}\frac{1}{H} ∑_{B_{nk}} |∑_{m} e^{2πig(m)}| & ≤ \frac{1}{H} . qNϵ . \frac{H}{q} \\ & = Nϵ \end{aligned}  (A.16)

Now over the set A_{nk} we have ϵ_{2} ≤ sin f(m) ≤ 1 and so since the first term of g′′ is the dominant one and since (n+h) > \frac{N}{2} −3H = O(N) we have:

K (\frac{q}{Q})^{2} Nϵ_{2} < |g′′(m)| < K (\frac{q}{Q})^{2} N and we can apply Lemma A.0.3 with

λ = (\frac{q}{Q})^{2} N.

Case 2b: Q ≥ KNϵ In this case we rewrite the g′′ as follows.

g′′(m) = K (\frac{q}{Q})^{2} \sqrt{(n + h)^{2} + K′Q^{2}} sin (f(m) + 2πθ(n))  (A.17)

Where θ(n) = \frac{1}{2π} tan^{−1} (\frac{KQ}{n+h}).

Let us denote by B_{nk} the set of pairs (n, k) which have sin (f(m) + 2πθ(n)) < ϵ_{2} and by A_{nk} the remaining pairs. As before the \frac{mq+k}{Q} term of f(m) makes very little difference to sin f(m) + θ(n). Thus the size of B_{nk} is essentially

the same as ϕ_{N}^{\{n^{α}+ \frac{kp}{q} +θ(n)\}}(0, ϵ). Now θ(n) is a decreasing function ranging over the domain [1, n]. We can break this domain up into subintervals I_{i} = [n_{i−1}, n_{i}], i = 1 … \frac{1}{4ϵ} with 1 = n_{0} < n_{1} < ⋯ < n_{\frac{1}{ϵ}} such that for n ∈ I_{i} we have θ(n) = θ(n_{i}) + O(ϵ). Now for |I_{i}| > M_{0} where M_{0} depends only on epsilon, we certainly have

\begin{aligned}ϕ_{[n_{i−1},n_{i}]}^{\{n^{α}+ \frac{kp}{q} +θ(n)\}}(0, ϵ) & = ϕ_{[n_{i−1},n_{i}]}^{\{n^{α}+ \frac{kp}{q} +θ(n_{i})+O(ϵ)\}}(0, ϵ) \\ & < Kϵ(n_{i} − n_{i−1}) \end{aligned}  (A.18)

and for |I_{i}| < M_{0} then we at least have

ϕ_{[n_{i−1},n_{i}]}^{\{n^{α}+ \frac{kp}{q} +θ(n)\}}(0, ϵ) < M_{0}  (A.19)

Combining (A.18) and (A.19) we get

ϕ_{N}^{\{n^{α}+ \frac{kp}{q} +θ(n)\}}(0, ϵ) = ∑_{i=1}^{\frac{1}{4ϵ}} ϕ_{[n_{i−1},n_{i}]}^{\{n^{α}+ \frac{kp}{q} +θ(n)\}}(0, ϵ)

≤ ∑_{i=1}^{\frac{1}{4ϵ}} (Kϵ(n_{i} − n_{i−1}) + M_{0})

= KNϵ + \frac{K′M_{0}}{ϵ}

≤ KNϵ

(For N large enough since M_{0} does not depend on N) (A.20)

Thus the size of B_{nk} is essentially the same as it was above and we find that (A.16) holds in this case as well.

Now over the set A_{nk} we have ϵ_{2} ≤ sin f(m) ≤ 1. In addition we observe that KNϵ < Q < \sqrt{(n + h)^{2} + K′Q^{2}} < N and so

K (\frac{q}{Q})^{2} Nϵϵ_{2} < |g′′(m)| < K (\frac{q}{Q})^{2} N

and as in the first sub case we can apply Lemma A.0.3 with λ = (\frac{q}{Q})^{2} N.

We now join both sub cases back together and apply Lemma A.0.3 over the good k’s. Observe that

λ < N^{− \frac{10}{8}} N = N^{− \frac{1}{4}}

and

\frac{q}{H} = \frac{q}{ϵ^{3}ϵ_{2}^{5}Q} < \frac{N^{− \frac{5}{8}}}{ϵ^{3}ϵ_{2}^{5}} = O(N^{− \frac{1}{2}} )

We get

|∑_{m} e^{2πig(m)}| ≤ K (\frac{H}{q} λ^{\frac{1}{2}} + 1 + λ^{− \frac{1}{2}} )

= \frac{KH}{q} (λ^{\frac{1}{2}} + (\frac{q}{H} ) + \frac{q}{Hλ^{\frac{1}{2}}} )

≤ \frac{KH}{q} (N^{− \frac{1}{4}} + N^{− \frac{1}{2}} + \frac{q}{ϵ^{3}ϵ_{2}^{5}Q} \frac{Q}{q\sqrt{N}} )

≤ \frac{KHN^{− \frac{1}{4}}}{q}

(For N large enough)

≤ \frac{KHϵ}{q}

(For N large enough)

(A.21)

And using (A.15), (A.16) and (A.21) we get

\begin{aligned}S & ≤ \frac{1}{H} ∑_{A_{nk}} |∑_{m} e^{2πig(m)}| + \frac{1}{H} ∑_{B_{nk}} |∑_{m} e^{2πig(m)}| + KNϵ \\ & ≤ \frac{1}{H} ∑_{A_{nk}} \frac{Hϵ}{q} + KNϵ \\ & ≤ \frac{1}{H} . Nq . \frac{Hϵ}{q} + KNϵ \\ & = KNϵ \end{aligned}  (A.22)

This is the bound we were after, and so we can conclude that for Q < \frac{N}{ϵ_{2}} then the sequence is uniformly distributed modulo 1.

Bibliography

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[2] D. Berend, G. Kolesnik, “Distribution modulo 1 of some oscillating sequences”, Israel J. Math. 71 (1990), vol. 2, 161-179

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[4] K. Chandrasekharan, “Introduction to Analytic Number Theory”, Springer-Verlay 1968

[5] Hardy and Wright, “An Introduction to the Theory of Numbers”, Oxford University Press, 4th ed. 1960

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[7] Leveque, “The distribution modulo 1 of trigonometric sequences”, Duke Math Journal 20 (1953) 367-374

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