### 数学代写|数论代写Number theory代考|MATH 2068

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论代写Number theory代考|COUNTABLE AND UNCOUNTABLE SETS

Definition 3.6. A set $S$ is said to be countable if there exists a one-to-one function from $S$ to $\mathbb{N}$, the set of natural numbers, and uncountable if it is not countable.

Lemma 3.28. Any finite set is countable; the set of integers is countable.
Lemma 3.29. Let $f$ be a function from $A$ to $B$ and $g$ a function from $B$ to $C$. If $f$ and $g$ are both one-to-one, so is the composite function $g \circ f$. If $g \circ f$ is one-to-one, then so is $f$.

Corollary 3.30. If $T$ is countable and there is a one-to-one function from $S$ to $T$, then $S$ is countable.

Exercise. Give an example to show that if $g \circ f$ is one-to-one, $g$ need not be one-to-one.

Theorem 3.31. If $S$ and $T$ are countable, then so are $S \times T$ and $S \cup T$. If $S$ is countable and $R \subseteq S$, then $R$ is countable.
Proof. Let $f: S \rightarrow \mathbb{N}$ and $g: T \rightarrow \mathbb{N}$ be one-to-one functions. Then
$$h: S \times T \rightarrow \mathbb{N}, \quad h(s, t)=2^{f(s)} 3^{g(t)}$$
and
$$k: S \cup T \rightarrow \mathbb{N} \times \mathbb{N}, \quad k(x)= \begin{cases}(0, f(x)) & \text { if } x \in S \ (1, g(x)) & \text { if } x \notin S\end{cases}$$
are one-to-one. If $R \subseteq S$, then the restriction $\left.f\right|{R}$ of $f$ to $R$ is one-one. The above constructions can easily be generalised to prove the following. Theorem 3.32. Suppose that the sets $S{0}, S_{1}, S_{2}, \ldots$ are countable. Then

• $S_{0} \times S_{1} \times \cdots \times S_{n}$ is countable for each natural number $n$;
• $S_{0} \cup S_{1} \cup S_{2} \cup \cdots$ is countable.
Exercise. Explain why we need to restrict the Cartesian product in this result to finitely many terms, but do not need to restrict the union in the same way. The second of the above properties can also be extended a little further.
Theorem 3.33. Let $T$ be countable and suppose that for every $t \in T$, the set $S_{t}$ is countable. Then
$$S=\bigcup_{t \in T} S_{t}$$
is a countable set.

## 数学代写|数论代写Number theory代考|THE PRIME NUMBER THEOREM

For any real number $x$, we denote by $\pi(x)$ the number of primes less than or equal to $x$. For example, $\pi(1)=0$ and $\pi(10)=4$ and $\pi(100)=\pi(100.5)=25$. The symbol $\pi$ is customary and is used because it is the Greek equivalent of “p”, the first letter of the word “prime” – it has, of course, nothing to do with the trigonometric constant $\pi$. The following result, which gives an estimate for $\pi(x)$ in terms of elementary functions, was proved independently and more or less simultaneously in 1896 by Hadamard [27] and de la Vallée Poussin [65]. A proof is given by Hardy and Wright [29].

Theorem 3.37. The Prime Number Theorem. The number of primes not exceeding $x$ satisfies
$$\frac{\pi(x)}{x / \log x} \rightarrow 1 \text { as } x \rightarrow \infty$$
where log denotes the natural (base e) logarithm.

We may use the Prime Number Theorem to estimate the least common multiple of the first $n$ positive integers. Given a prime $p$ and a positive integer $n$, there is a unique non-negative integer $\alpha$ such that $p^{\alpha} \leq n<p^{\alpha+1}$; the least common multiple will be the product of all these powers $p^{\alpha}$. Therefore
$$L_{n}=\operatorname{lcm}(1,2, \ldots, n)=\prod_{\substack{p \leq n \ p \text { prime }}} p^{\alpha} \leq \prod_{\substack{p \leq n \ p \text { prime }}} n=n^{\pi(n)}$$
the last equality being true because the product consists of $\pi(n)$ equal factors. Let $c$ be a constant greater than $e$. It follows from the Prime Number Theorem that if $n$ is sufficiently large then
$$\frac{\pi(n)}{n / \log n}<\log c$$
and hence
$$L_{n} \leq n^{\pi(n)}=e^{\pi(n) \log n}<c^{n} .$$
In section $3.4$ we chose $c=3$ to keep things simple. We could have taken a slightly smaller value, which would have resulted in a slightly larger value for $s$; however, this would have made no significant difference to the result.

## 数学代写|数论代写Number theory代考|DEFINITION AND BASIC PROPERTIES

Definition 4.1. A finite or infinite expression of the form
$$a_{0}+\frac{b_{1}}{a_{1}+\frac{b_{2}}{a_{2}+\frac{b_{3}}{a_{3}+\cdots}}}$$
is called a continued fraction. A simple continued fraction is one in which every $b_{k}$ is 1 , every $a_{k}$ is an integer, and every $a_{k}$ except possibly $a_{0}$ is positive. For a (finite or infinite) simple continued fraction we shall also use the notations
$$a_{0}+\frac{1}{a_{1}+} \frac{1}{a_{2}+} \frac{1}{a_{3}+\ldots} \quad \text { and } \quad\left[a_{0}, a_{1}, a_{2}, a_{3}, \ldots\right] .$$
A finite simple continued fraction is said to represent the number obtained by performing the arithmetic in the obvious way; an infinite simple continued fraction $\left[a_{0}, a_{1}, a_{2}, a_{3}, \ldots\right]$ represents the real number $\alpha$ if
$$\alpha=\lim {n \rightarrow \infty}\left[a{0}, a_{1}, a_{2}, a_{3}, \ldots, a_{n}\right]$$ Let $k \in \mathbb{N}$. The integer $a_{k}$ is called the $k$ th partial quotient of the continued fraction $\left[a_{0}, a_{1}, a_{2}, \ldots\right]$, or of the number $\alpha$ it represents; the continued fraction $\alpha_{k}=\left[a_{k}, a_{k+1}, a_{k+2}, \ldots\right]$ is the kth complete quotient of $\alpha$; and the continued fraction $\left[a_{0}, a_{1}, \ldots, a_{k}\right]$ is the kth convergent to $\alpha$.

## 数学代写|数论代写Number theory代考|COUNTABLE AND UNCOUNTABLE SETS

H:小号×吨→ñ,H(s,吨)=2F(s)3G(吨)

ķ:小号∪吨→ñ×ñ,ķ(X)={(0,F(X)) 如果 X∈小号 (1,G(X)) 如果 X∉小号

• 小号0×小号1×⋯×小号n对每个自然数都是可数的n;
• 小号0∪小号1∪小号2∪⋯是可数的。
锻炼。解释为什么我们需要将这个结果中的笛卡尔积限制为有限多个项，但不需要以相同的方式限制并集。上述第二个属性也可以进一步扩展。
定理 3.33。让吨是可数的并且假设对于每个吨∈吨, 集合小号吨是可数的。然后
小号=⋃吨∈吨小号吨
是可数集。

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