## 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。