## 数学代写|解析数论作业代写Analytic Number Theory代考|MATH4307

statistics-lab™ 为您的留学生涯保驾护航 在代写解析数论Analytic Number Theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写解析数论Analytic Number Theory代写方面经验极为丰富，各种代写解析数论Analytic Number Theory相关的作业也就用不着说。

• 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|解析数论作业代写Analytic Number Theory代考|The Möbius function

Definition 2.1. A real or complex-valued function defined on the set of positive integers is called an arithmetical function.

Arithmetical functions play an important role in the study of numbers. Let us now introduce one of the most important arithmetical functions, namely, the Möbius function $\mu(n)$.
Definition 2.2. Let $\mu(1)=1$. If $n=p_1^{\alpha_1} \cdots p_k^{\alpha_k}$, then define
$$\mu(n)= \begin{cases}(-1)^k & \text { if } \alpha_1=\alpha_2=\cdots=\alpha_k=1, \ 0 & \text { otherwise. }\end{cases}$$
The function $\mu(n)$ is known as the Möbius function.
We observe that by the above definition, the Möbius function $\mu(n)$ is identically zero if and only if $n$ has a square factor greater than 1.
Definition 2.3. We write
$$\sum_{d \mid n} f(d)$$
to denote the sum of the values of $f$ over divisors $d$ of $n$.
Remark 2.1. Suppose $d_1, d_2, \cdots, d_k$ are the divisors of $n$, then
$$\sum_{d \mid n} f(d)=f\left(d_1\right)+f\left(d_2\right)+\cdots+f\left(d_k\right) .$$

Now, each $d_j$ can be written in the form $n / d_j^{\prime}$, where $d_j^{\prime}$ is the conjugate divisor of $d_j$. Hence,
$$\sum_{d \mid n} f(d)=f\left(\frac{n}{d_1^{\prime}}\right)+f\left(\frac{n}{d_2^{\prime}}\right)+\cdots+f\left(\frac{n}{d_k^{\prime}}\right)=\sum_{d \mid n} f\left(\frac{n}{d}\right) .$$
Therefore
$$\sum_{d \mid n} f(d)=\sum_{d \mid n} f\left(\frac{n}{d}\right)$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|The Euler totient function

Definition 2.5. The Euler totient $\varphi(n)$ is defined to be the number of positive integers not exceeding $n$ which are relatively prime (see Definition 1.5) to $n$.
It is sometimes convenient to write $\varphi(n)$ as
$$\varphi(n)=\sum_{\substack{k=1 \(k, n)=1}}^n 1 .$$
The first important result for $\varphi(n)$ is the following theorem:
Theorem 2.2. Let $n$ be any positive integer. Then
$$\sum_{d \mid n} \varphi(d)=n .$$
Proof. Let
$$S={k \in \mathbf{Z} \mid 1 \leq k \leq n}$$
and
$$A(d)={k \in \mathbf{Z} \mid(k, n)=d, 1 \leq k \leq n} .$$
Since every integer $k \leq n$ has a unique $(k, n)$, we conclude that $S$ is a disjoint union of $A(d)$ and we deduce that
$$\sum_{d \mid n} f(d)=n,$$
where $f(d)$ be the number of elements in $A(d)$.
Let
$$B(d)={1 \leq q \leq n / d \mid(q, n / d)=1} .$$
We claim that there is a one to one correspondence between the elements in $A(d)$ and $B(d)$. If $1 \leq k \leq n$, then the element $k / d \in B(d)$ since
$$(k / d, n / d)=1,$$
by Theorem $1.10$ (c).

# 解析数论代写

## 数学代写|解析数论作业代写Analytic Number Theory代考|The Möbius function

$$\mu(n)=\left{(-1)^k \quad \text { if } \alpha_1=\alpha_2=\cdots=\alpha_k=1,0 \quad\right. \text { otherwise. }$$

$$\sum_{d \mid n} f(d)$$

$$\sum_{d \mid n} f(d)=f\left(d_1\right)+f\left(d_2\right)+\cdots+f\left(d_k\right)$$

$$\sum_{d \mid n} f(d)=f\left(\frac{n}{d_1^{\prime}}\right)+f\left(\frac{n}{d_2^{\prime}}\right)+\cdots+f\left(\frac{n}{d_k^{\prime}}\right)=\sum_{d \mid n} f\left(\frac{n}{d}\right)$$

$$\sum_{d \mid n} f(d)=\sum_{d \mid n} f\left(\frac{n}{d}\right)$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|The Euler totient function

$$\varphi(n)=\sum_{k=1 \backslash(k, n)=1}^n 1$$

$$\sum_{d \mid n} \varphi(d)=n .$$

$$S=k \in \mathbf{Z} \mid 1 \leq k \leq n$$

$$A(d)=k \in \mathbf{Z} \mid(k, n)=d, 1 \leq k \leq n .$$

$$\sum_{d \mid n} f(d)=n,$$

$$B(d)=1 \leq q \leq n / d \mid(q, n / d)=1$$

$$(k / d, n / d)=1$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|解析数论作业代写Analytic Number Theory代考|MATH3170

statistics-lab™ 为您的留学生涯保驾护航 在代写解析数论Analytic Number Theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写解析数论Analytic Number Theory代写方面经验极为丰富，各种代写解析数论Analytic Number Theory相关的作业也就用不着说。

• 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|解析数论作业代写Analytic Number Theory代考|The Euclidean Algorithm

In this section, we prove a result that allows us to compute the greatest common divisor of two integers. First, we need a lemma.
Lemma 1.11. Let a, $b, q$, and $r$ be integers such that
$$a=b q+r,$$
then
$$(a, b)=(b, r) .$$
Proof. Let $d=(a, b)$ and $d^{\prime}=(b, r)$. Note that since $d \mid a$ and $d \mid b$, we find that $d \mid(a-b q)$ by Theorem $1.6$ (c). Hence, $d \mid r$ and $d$ is a common divisor of $b$ and $r$. By Definition $1.4$ (c), $d \mid d^{\prime}$ since $d^{\prime}=(b, r)$. Similarly, $d^{\prime} \mid b$ and $d^{\prime} \mid r$ implies that $d^{\prime} \mid(b q+r)$ by Theorem $1.6$ (c) and consequently, $d^{\prime} \mid a$. By Definition $1.4$ (c), $d^{\prime} \mid d$ since $d=(a, b)$. Therefore, by Theorem $1.6$ (i), $d=d^{\prime}$.

Theorem 1.12 (The Euclidean Algorithm). Given positive integers a and $b$, where $b \nmid a$. Let $r_0=a, r_1=b$, and apply the division algorithm repeatedly to obtain a set of remainders $r_2, r_3, \ldots, r_n, r_{n+1}$ defined successively by the relations
$$\begin{array}{ccc} r_0 & =r_1 q_1+r_2 & 0<r_2<r_1 \ r_1 & =r_2 q_2+r_3 & 0<r_3<r_2 \ \vdots & \ r_{n-2}=r_{n-1} q_{n-1}+r_n & 0<r_n<r_{n-1} \ r_{n-1}=r_n q_n+r_{n+1} & r_{n+1}=0 . \end{array}$$
Then $r_n$, the last nonzero remainder in this process is $(a, b)$, the greatest common divisor of $a$ and $b$.

Proof. There is a stage at which $r_{n+1}=0$ because the $r_i$ are decreasing and non-negative. Next, applying Lemma 1.11, we find that
$$(a, b)=\left(r_0, r_1\right)=\left(r_1, r_2\right)=\cdots=\left(r_n, r_{n+1}\right)=\left(r_n, 0\right)=r_n .$$
This completes the proof of Theorem 1.12.

## 数学代写|解析数论作业代写Analytic Number Theory代考|Fundamental Theorem of Arithmetic

Theorem $1.17$ (Fundamental Theorem of Arithmetic). Every positive integer $n>1$ can be expressed as a product of primes; this representation is unique apart from the order in which the factors occur.

Proof. We first show that $n$ can be expressed as a prime or a product of primes. We use induction on $n$. The statement is clearly true for $n=2$ since 2 is a prime. Suppose $m$ is a prime or a product of primes for $2 \leq m \leq n-1$. If $n$ is a prime then we are done. Suppose $n$ is composite then $n=a b$, where $11$ is a prime or a product of primes.

To prove uniqueness, we use induction on $n$ again. If $n=2$ then the representation of $n$ as a product of primes is clearly unique. Assume, then that it is true for all integers greater than 1 and less than $n$. We shall prove that it is also true for $n$. If $n$ is prime, then there is nothing to prove. Assume, then, that $n$ is composite and that $n$ has two factorizations, say,
$$n=p_1 p_2 \cdots p_s=q_1 q_2 \cdots q_t .$$
Since $p_1$ divides the product $q_1 q_2 \cdots q_t$, it must divide at least one factor by Corollary 1.16. Relabel $q_1, q_2, \ldots, q_t$ so that $p_1 \mid q_1$. Then $p_1=q_1$ since both $p_1$ and $q_1$ are primes. In (1.4), we may cancel $p_1$ on both sides to obtain
$$n / p_1=p_2 \cdots p_s=q_2 \cdots q_t .$$
Now the induction hypothesis implies that the two factorizations of $n / p_1$ must be the same, apart from the order of the factors. Therefore, $s=t$ and the factorizations in (1.4) are also identical, apart from order. This completes the proof.

# 解析数论代写

## 数学代写|解析数论作业代写Analytic Number Theory代考|The Euclidean Algorithm

$$a=b q+r$$

$$(a, b)=(b, r) .$$

(C)。因此， $d \mid r$ 和 $d$ 是公约数 $b$ 和 $r$. 根据定义 $1.4$ (C) ， $d \mid d^{\prime}$ 自从 $d^{\prime}=(b, r)$. 相似地， $d^{\prime} \mid b$ 和 $d^{\prime} \mid r$ 暗示 $d^{\prime} \mid(b q+r)$ 通过定理1.6(c) 因此， $d^{\prime} \mid a$. 根据定义 $1.4$ (C)， $d^{\prime} \mid d$ 自从 $d=(a, b)$. 因 此，由定理 $1.6$ (一世)， $d=d^{\prime}$.

$$r_0=r_1 q_1+r_2 \quad 0<r_2<r_1 r_1=r_2 q_2+r_3 \quad 0<r_3<r_2 \vdots \quad r_{n-2}=r_{n-1} q_{n-1}+r_n$$

$$(a, b)=\left(r_0, r_1\right)=\left(r_1, r_2\right)=\cdots=\left(r_n, r_{n+1}\right)=\left(r_n, 0\right)=r_n .$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|Fundamental Theorem of Arithmetic

$$n=p_1 p_2 \cdots p_s=q_1 q_2 \cdots q_t$$

$$n / p_1=p_2 \cdots p_s=q_2 \cdots q_t$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|解析数论作业代写Analytic Number Theory代考|MAST90136

statistics-lab™ 为您的留学生涯保驾护航 在代写解析数论Analytic Number Theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写解析数论Analytic Number Theory代写方面经验极为丰富，各种代写解析数论Analytic Number Theory相关的作业也就用不着说。

• 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|解析数论作业代写Analytic Number Theory代考|Least Integer Axiom and Mathematical Induction

Let
$$\mathbf{Z}={0, \pm 1, \pm 2, \cdots}$$
be the set of integers. The Least Integer Axiom (see [10]), also known as the Well Ordering Principle, states that there is a smallest integer in every nonempty subset of non-negative integers. It is useful in establishing the following result.

Theorem 1.1. Let $S(1), S(2), \cdots, S(n), \cdots$ be statements, one for each integer $n \geq 1$. If some of these statements are false, then there is a first false statement.
Proof. Set
$$T=\left{k \in \mathbf{Z}^{+} \mid S(k) \text { is false }\right} .$$
Since at least one statement is false, $T$ is nonempty. By the Least Integer Axiom, there exists a smallest integer $n$ in $T$. This implies that $S(n)$ is the first false statement.

From Theorem 1.1, we deduce the Principle of Mathematical Induction.
Theorem 1.2. Let $S(n)$ be statements, one for each $n \geq 1$. Suppose that the following conditions are satisfied by $S(n)$ :
(a) The statement $S(1)$ is true.
(b) If $S(n)$ is true, then $S(n+1)$ is true.
Then $S(n)$ is true for all integers $n \geq 1$.

## 数学代写|解析数论作业代写Analytic Number Theory代考|Greatest common divisors

Definition 1.3. A common divisor of integers $a$ and $b$ is an integer $c$ with $c \mid a$ and $c \mid b$.

Definition 1.4. A greatest common divisor of integers $a$ and $b$ is a number $d$ with the following properties:
(a) The integer $d$ is non-negative.
(b) The integer $d$ is a common divisor of $a$ and $b$.
(c) If $e$ is any common divisor of $a$ and $b$, then $e \mid d$.

Note that if $d$ and $d^{\prime}$ are both greatest common divisors of $a$ and $b$, then $d$ is a common divisor of $a$ and $b$ and $d^{\prime}$ is a greatest common divisor, we note that $d^{\prime} \mid d$ using Definition $1.4$ (c). Similarly, since $d^{\prime}$ is a common divisor and $d$ is a greatest common divisor, $d \mid d^{\prime}$. By Theorem $1.6$ (i), $|d|=\left|d^{\prime}\right|$ and by Definition $1.4$ (a), we deduce that $d=d^{\prime}$. This shows that the greatest common divisor of $a$ and $b$ is unique.
The notation for the greatest common divisor of $a$ and $b$ is
$$(a, b) \text {. }$$
Remark 1.2. When $a$ and $b$ are zeros, then $(0,0)=0$. If $a=0$ and $b$ is nonzero, then $(0, b)=b$.

We will next show that the greatest common divisor of two integers exists. By Remark 1.2, it suffices to consider the case when both $a$ and $b$ are nonzero.

Theorem 1.7. Let $a$ and $b$ be nonzero integers. Then the smallest positive integer in the set
$$P:={s a+t b \mid s, t \in \mathbf{Z} \text { and } s a+t b>0}$$
is $(a, b)$.
Proof. If $a$ is positive then $a \in P$ since
$$a=1 \cdot a+0 \cdot b .$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|Least Integer Axiom and Mathematical Induction

$$\mathbf{Z}=0, \pm 1, \pm 2, \cdots$$

$\mathrm{T}=\backslash \mathrm{left}{\mathrm{k} \backslash$ in $\backslash m a t h b f{Z} \wedge{+} \backslash \mathrm{mid} \mathrm{S}(\mathrm{k}) \backslash \mathrm{text}{$ 为假 $} \backslash$ ight $}$ 。

(a) 声明 $S(1)$ 是真的。
(b) 如果 $S(n)$ 是真的，那么 $S(n+1)$ 是真的。

## 数学代写|解析数论作业代写Analytic Number Theory代考|Greatest common divisors

(a) 整数 $d$ 是非负的。
(b) 整数 $d$ 是公约数 $a$ 和 $b$.
(c) 如果 $e$ 是任何公约数 $a$ 和 $b$ ，然后 $e \mid d$.

$$(a, b) \text {. }$$

$$P:=s a+t b \mid s, t \in \mathbf{Z} \text { and } s a+t b>0$$

$$a=1 \cdot a+0 \cdot b$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|解析数论作业代写Analytic Number Theory代考|MATH4307

statistics-lab™ 为您的留学生涯保驾护航 在代写解析数论Analytic Number Theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写解析数论Analytic Number Theory代写方面经验极为丰富，各种代写解析数论Analytic Number Theory相关的作业也就用不着说。

• 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|解析数论作业代写Analytic Number Theory代考|The Euclidean algorithm

Iheorem $1.12$ provides a practical method for computing the $\operatorname{gcd}(a, b)$ when the prime-power factorizations of $a$ and $b$ are known. However, considerable calculation may be required to obtain these prime-power factorizations and it is desirable to have an alternative procedure that requires less computation. There is a useful process, known as Euclid’s algorithm, which does not require the factorizations of $a$ and $b$. This process is based on successive divisions and makes use of the following theorem.

Theorem 1.14 The division algorithm. Given integers $a$ and $b$ with $b>0$, there exists a unique pair of integers $q$ and $r$ such that
$$a=b q+r, \text { with } 0 \leq r<b .$$
Moreover, $r=0$ if, and only if, $b \mid a$.
Note. We say that $q$ is the quotient and $r$ the remainder obtained when $b$ is divided into $a$.

## 数学代写|解析数论作业代写Analytic Number Theory代考|A product formula for n

The sum for $\varphi(n)$ in Theorem $2.3$ can also be expressed as a product extended over the distinct prime divisors of $n$.
Theorem $2.4$ For $n \geq 1$ we have
$$\varphi(n)=n \prod_{p \mid n}\left(1-\frac{1}{p}\right) .$$
Proof. For $n=1$ the product is empty since there are no primes which divide 1 . In this case it is understood that the product is to be assigned the value $1 .$

Suppose, then, that $n>1$ and let $p_{1}, \ldots, p_{r}$ be the distinct prime divisors of $n$. The product can be written as
(4)
\begin{aligned} \prod_{p \mid n}\left(1-\frac{1}{p}\right) &=\prod_{i=1}^{r}\left(1-\frac{1}{p_{i}}\right) \ &=1-\sum \frac{1}{p_{i}}+\sum \frac{1}{p_{i} p_{j}}-\sum \frac{1}{p_{i} p_{j} p_{k}}+\cdots+\frac{(-1)^{r}}{p_{1} p_{2} \cdots p_{r}} . \end{aligned}
On the right, in a term such as $\sum 1 / p_{i} p_{j} p_{k}$ it is understood that we consider all possible products $p_{i} p_{j} p_{k}$ of distinct prime factors of $n$ taken three at a time. Note that each term on the right of (4) is of the form $\pm 1 / d$ where $d$ is a divisor of $n$ which is either 1 or a product of distinct primes. The numerator $\pm 1$ is exactly $\mu(d)$. Since $\mu(d)=0$ if $d$ is divisible by the square of any $p_{i}$ we see that the sum in (4) is exactly the same as
$$\sum_{d \mid n} \frac{\mu(d)}{d} .$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|The Euclidean algorithm

$$a=b q+r, \text { with } 0 \leq r<b .$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|A product formula for n

$$\varphi(n)=n \prod_{p \mid n}\left(1-\frac{1}{p}\right)$$

(4)
$$\prod_{p \mid n}\left(1-\frac{1}{p}\right)=\prod_{i=1}^{r}\left(1-\frac{1}{p_{i}}\right) \quad=1-\sum \frac{1}{p_{i}}+\sum \frac{1}{p_{i} p_{j}}-\sum \frac{1}{p_{i} p_{j} p_{k}}+\cdots+\frac{(-1)^{r}}{p_{1} p_{2} \cdots p_{r}}$$

$$\sum_{d \mid n} \frac{\mu(d)}{d} .$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|解析数论作业代写Analytic Number Theory代考|MATH4304

statistics-lab™ 为您的留学生涯保驾护航 在代写解析数论Analytic Number Theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写解析数论Analytic Number Theory代写方面经验极为丰富，各种代写解析数论Analytic Number Theory相关的作业也就用不着说。

• 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|解析数论作业代写Analytic Number Theory代考|The fundamental theorem of arithmetic

Theorem 1.10 Fundamental theorem of arithmetic. Every integer $n>1$ can be represented as a product of prime factors in only one way, apart from the order of the factors.

Proof. We use induction on $n$. The theorem is true for $n=2$. Assume, then, that it is true for all integers greater than 1 and less than $n$. We shall prove it is also true for $n$. If $n$ is prime there is nothing more to prove. Assume, then, that $n$ is composite and that $n$ has two factorizations, say
$$n=p_{1} p_{2} \cdots p_{s}=q_{1} q_{2} \cdots q_{t} .$$
We wish to show that $s=t$ and that each $p$ equals some $q$. Since $p_{1}$ divides the product $q_{1} q_{2} \cdots q_{t}$ it must divide at least one factor. Relabel $q_{1}, q_{2}, \ldots, q_{t}$ so that $p_{1} \mid q_{1}$. Then $p_{1}=q_{1}$ since both $p_{1}$ and $q_{1}$ are primes. In (2) we may cancel $p_{1}$ on both sides to obtain
$$n / p_{1}=p_{2} \cdots p_{s}=q_{2} \cdots q_{t} .$$
If $s>1$ or $t>1$ then $1<n / p_{1}<n$. The induction hypothesis tells us that the two factorizations of $n / p_{1}$ must be identical, apart from the order of the factors. Therefore $s=t$ and the factorizations in (2) are also identical, apart from order. This completes the proof.

Note. In the factorization of an integer $n$, a particular prime $p$ may occur more than once. If the distinct prime factors of $n$ are $p_{1}, \ldots, p_{r}$ and if $p_{i}$ occurs as a factor $a_{i}$ times, we can write
$$n=p_{1}{ }^{a_{1}} \cdots p_{r}^{a_{r}}$$ or, more briefly,
$$n=\prod_{i=1}^{r} p_{i}^{a_{i}} .$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|The series of reciprocals of the primes

Theorem 1.13 The infinite series $\sum_{n=1}^{\infty} 1 / p_{n}$ diverges.
Proof. The following short proof of this theorem is due to Clarkson [11]. We assume the series converges and obtain a contradiction. If the series converges there is an integer $k$ such that
$$\sum_{m=k+1}^{\infty} \frac{1}{p_{m}}<\frac{1}{2} .$$
Let $Q=p_{1} \cdots p_{k}$, and consider the numbers $1+n Q$ for $n=1,2, \ldots$ None of these is divisible by any of the primes $p_{1}, \ldots, p_{k}$. Therefore, all the prime factors of $1+n Q$ occur among the primes $p_{k+1}, p_{k+2} \ldots$ Therefore for each $r \geq 1$ we have
$$\sum_{n=1}^{r} \frac{1}{1+n Q} \leq \sum_{t=1}^{x}\left(\sum_{m=k+1}^{x} \frac{1}{p_{m}}\right)^{t},$$
since the sum on the right includes among its terms all the terms on the left. But the right-hand side of this inequality is dominated by the convergent geometric series
$$\sum_{t=1}^{x}\left(\frac{1}{2}\right)^{t} .$$
Therefore the series $\sum_{n=1}^{x} 1 /(1+n Q)$ has bounded partial sums and hence converges. But this is a contradiction because the integral test or the limit comparison test shows that this series diverges.

## 数学代写|解析数论作业代写Analytic Number Theory代考|The fundamental theorem of arithmetic

$$n=p_{1} p_{2} \cdots p_{s}=q_{1} q_{2} \cdots q_{t} .$$

$$n / p_{1}=p_{2} \cdots p_{s}=q_{2} \cdots q_{t} .$$

$$n=p_{1}^{a_{1}} \cdots p_{r}^{a_{r}}$$

$$n=\prod_{i=1}^{r} p_{i}^{a_{i}} .$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|The series of reciprocals of the primes

$$\sum_{m=k+1}^{\infty} \frac{1}{p_{m}}<\frac{1}{2}$$

$$\sum_{n=1}^{r} \frac{1}{1+n Q} \leq \sum_{t=1}^{x}\left(\sum_{m=k+1}^{x} \frac{1}{p_{m}}\right)^{t}$$

$$\sum_{t=1}^{x}\left(\frac{1}{2}\right)^{t}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|解析数论作业代写Analytic Number Theory代考|MAST90136

statistics-lab™ 为您的留学生涯保驾护航 在代写解析数论Analytic Number Theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写解析数论Analytic Number Theory代写方面经验极为丰富，各种代写解析数论Analytic Number Theory相关的作业也就用不着说。

• 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|解析数论作业代写Analytic Number Theory代考|Greatest common divisor

If $d$ divides two integers $a$ and $b$, then $d$ is called a common divisor of $a$ and $b$. Thus, 1 is a common divisor of every pair of integers $a$ and $b$. We prove now that every pair of integers $a$ and $b$ has a common divisor which can be expressed as a linear combination of $a$ and $b$.

Theorem $1.2$ Given any two integers $a$ and $b$, there is a common divisor $d$ of $a$ and $b$ of the form
$$d=a x+b y,$$ where $x$ and $y$ are integers. Moreover, every common divisor of $a$ and $b$ divides this $d$.

Proof. First we assume that $a \geq 0$ and $b \geq 0$. We use induction on $n$, where $n=a+b$. If $n=0$ then $a=b=0$ and we can take $d=0$ with $x=y=0$. Assume, then, that the theorem has been proved for $0,1,2, \ldots$, $n-1$. By symmetry, we can assume $a \geq b$. If $b=0$ take $d=a, x=1$, $y=0$. If $b \geq 1$ apply the theorem to $a-b$ and $b$. Since $(a-b)+b=$ $a=n-b \leq n-1$, the induction assumption is applicable and there is a common divisor $d$ of $a-b$ and $b$ of the form $d=(a-b) x+b y$. This $d$ also divides $(a-b)+b=a$ so $d$ is a common divisor of $a$ and $b$ and we have $d=a x+(y-x) h$, a linear combination of $a$ and $b$. To complete the proof we need to show that every common divisor divides $d$. But a common divisor divides $a$ and $b$ and hence, by linearity, divides $d$.

## 数学代写|解析数论作业代写Analytic Number Theory代考|Prime numbers

Definition An integer $n$ is called prime if $n>1$ and if the only positive divisors of $n$ are 1 and $n$. If $n>1$ and if $n$ is not prime, then $n$ is called composite.

EXAMPLES The prime numbers less than 100 are 2, 3, 5, 7, 11, 13, 17, 19,23, $29,31,37,41,43,47,53,59,61,67,71,73,79,83,89$, and 97 .
Notation Prime numbers are usually denoted by $p, p^{\prime}, p_{i}, q, q^{\prime}, q_{i}$.
Theorem 1.6 Every integer $n>1$ is either a prime number or a product of prime numbers.

Proof. We use induction on $n$. The theorem is clearly true for $n=2$. Assume it is true for every integer $1$ so each of $c, d$ is a product of prime numbers, hence so is $n$.
Theorem 1.7 Euclid. There are infinitely many prime numbers.

EUCLID’s Proof. Suppose there are only a finite number, say $p_{1}, p_{2}, \ldots, p_{n}$. Let $N=1+p_{1} p_{2} \cdots p_{n}$. Now $N>1$ so either $N$ is prime or $N$ is a product of primes. Of course $N$ is not prime since it exceeds each $p_{i}$. Moreover, no $p_{i}$ divides $N$ (if $p_{i} \mid N$ then $p_{i}$ divides the difference $N-p_{1} p_{2} \cdots p_{n}=1$ ). This contradicts Theorem 1.6.

## 数学代写|解析数论作业代写Analytic Number Theory代考|Greatest common divisor

$$d=a x+b y,$$

$a=n-b \leq n-1$, 归纳假设适用且存在公约数 $d$ 的 $a-b$ 和 $b$ 形式的 $d=(a-b) x+b y$. 这个 $d$ 也分
$(a-b)+b=a$ 所以 $d$ 是的公约数 $a$ 和 $b$ 我们有 $d=a x+(y-x) h$, 的线性组合 $a$ 和 $b$. 为了完成证明，我们需 要证明每个公约数都可以除 $d$. 但是一个公约数除以 $a$ 和 $b$ 因此，通过线性，除以 $d$.

## 数学代写|解析数论作业代写Analytic Number Theory代考|Prime numbers

EUCLID 的证明。假设只有一个有限的数字，比如说 $p_{1}, p_{2}, \ldots, p_{n}$. 让 $N=1+p_{1} p_{2} \cdots p_{n}$. 现在 $N>1$ 所 以要么 $N$ 是素数或 $N$ 是素数的乘积。当然 $N$ 不是素数，因为它超过了每个 $p_{i}$. 此外，没有 $p_{i}$ 划分 $N$ (如果 $p_{i} \mid N$ 然后 $p_{i}$ 划分差异 $N-p_{1} p_{2} \cdots p_{n}=1$ ）。这与定理 $1.6$ 相矛盾。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。