### 数学代写|数论作业代写number theory代考|PMATH650

statistics-lab™ 为您的留学生涯保驾护航 在代写数论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。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论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|数论作业代写number theory代考|SOME ELEMENTARY NUMBER THEORY

This section contains some basic number-theoretic definitions and results which you ought to know. Proofs in this section are abbreviated or omitted, and you should be able to supply proofs for yourself. If necessary, this material can be found in any work on elementary number theory. The most popular of the classic texts are regularly revised, thereby offering a proven exposition together with additions which bring the content and presentation up to date. From a very crowded field we mention Hardy and Wright $[28],[29]$, Niven and Zuckerman $[45],[46]$ and Baker [10].

Lemma 1.10. The division algorithm. If $a$ and $b$ are integers with $b>0$, then there exist integers $q$ and $r$ such that $a=b q+r$ and $0 \leq r<b$.

Using the division algorithm recursively gives the Euclidean algorithm for computing the greatest common divisor of two integers, not both zero.

Lemma 1.11. The Bézout property. If $a$ and $b$ are integers, not both zero, and $g$ is the greatest common divisor of $a$ and $b$, then there exist integers $x$ and $y$ such that $a x+b y=g$.

Given specific $a$ and $b$, you should know how to use the Euclidean algorithm to find $g, x$ and $y$.

Lemma 1.12. If $a$ and $m$ have no common factor and $a \mid m n$, then $a \mid n$.
Definition 1.4. Let $m$ be a positive integer. We say that integers a and b are congruent modulo $m$, written $a \equiv b(\bmod m)$, if $m \mid a-b$.

To “reduce an integer $a$ modulo $m$ ” means to find an integer $b$ such that $a \equiv b(\bmod m)$ and $b$ lies in a “suitable” range, usually $0 \leq b<m$. That this can always be done is a consequence of the division algorithm. Although congruence notation is just another way of expressing a divisibility relation, and in that sense “nothing new”, it is very useful because congruence shares many of the basic properties of equality.

## 数学代写|数论作业代写number theory代考|IRRATIONALITY OF er

In the actual details of the final proof, Hermite’s method is (at least for the earlier results) not too difficult. However, the motivation behind the proof can be obscure. Therefore, instead of giving the proofs straight away, we shall start by trying to explain the aims and ideas behind a relatively simple case. We wish to generalise results of Chapter 1 by showing that if $r$ is rational then $e^{r}$ is irrational, with the obvious exception that $e^{0}=1$.

As usual we seek a proof by contradiction: take $r=a / b$ with $a \neq 0$, and suppose that $e^{r}=p / q$. Following the method of Theorem $1.9$, we try to obtain a contradiction by constructing an integer that lies between 0 and 1 . Hermite’s idea, which originated in his study of approximations to $e^{x}$, was to consider the definite integral
$$\int_{0}^{r} f(x) e^{x} d x$$
and to identify a function $f$ which will give us what we want. Integrating by parts yields
$$\int_{0}^{r} f(x) e^{x} d x=\left(f(r) e^{r}-f(0)\right)-\int_{0}^{r} f^{\prime}(x) e^{x} d x$$
and since the integral on the right-hand side has very much the same form as that on the left, we may apply the same procedure repeatedly to obtain
$$\int_{0}^{r} f(x) e^{x} d x=\left(f(r)-f^{\prime}(r)+f^{\prime \prime}(r)-\cdots\right) e^{r}-\left(f(0)-f^{\prime}(0)+f^{\prime \prime}(0)-\cdots\right)$$
Here the right-hand side purports to contain two infinite series and therefore must be treated with caution, but if we choose $f$ to be a polynomial, then the sums will actually involve a finite number of terms only, and we shall have no convergence problems. We write
$$F(x)=f(x)-f^{\prime}(x)+f^{\prime \prime}(x)-f^{\prime \prime \prime}(x)+\cdots,$$
so that
$$\int_{0}^{r} f(x) e^{x} d x=F(r) e^{r}-F(0)$$
and the next step is to make some sort of evaluation of the right-hand side. An idea that will help is to notice that $F(0)$ will be simple if $f$ has a large number of derivatives that vanish at $x=0$; that is, $f(x)$ should have many factors of $x$. Similarly, $f(x)$ should have many factors of $r-x$ in order to keep $F(r)$ simple. So we set
$$f(x)=c x^{n}(r-x)^{n}$$
where $c$, a constant, and $n$ are yet to be chosen. Now
\begin{aligned} f^{(k)}(0) &=k ! \times\left{\text { coefficient of } x^{k}\right} \ &=k ! c\left(\begin{array}{c} n \ k-n \end{array}\right) r^{2 n-k}(-1)^{k-n} \end{aligned}
if $n \leq k \leq 2 n$, and $f^{(k)}(0)=0$ otherwise. Recall that our aim is to make the integral (2.1), or something similar, an integer. The expression for $f^{(k)}(0)$ contains a factor $r^{2 n-k}$, and this could have a denominator as big as $b^{n}$. Therefore, we choose $c=b^{n} ;$ then $f^{(k)}(0)$ is always an integer, and so is $F(0)$. Either by invoking the symmetry of $f$ or by direct calculation, we note that
\begin{aligned} f(r-x)=f(x) & \Rightarrow(-1)^{k} f^{(k)}(r-x)=f^{(k)}(x) \ & \Rightarrow f^{(k)}(r)=(-1)^{k} f^{(k)}(0) \end{aligned}
Therefore, $F(r)$ is an integer too, and so is
$$q \int_{0}^{r} f(x) e^{x} d x=p F(r)-q F(0)$$

## 数学代写|数论作业代写number theory代考|SOME ELEMENTARY NUMBER THEORY

$0 \leq b<m$. 总能做到这一点是除法算法的结果。尽管全等表示法只是表示可分关系的另一种方式，并且在这个意

## 数学代写|数论作业代写number theory代考|IRRATIONALITY OF er

$$\int_{0}^{r} f(x) e^{x} d x$$

$$\int_{0}^{r} f(x) e^{x} d x=\left(f(r) e^{r}-f(0)\right)-\int_{0}^{r} f^{\prime}(x) e^{x} d x$$

$$\int_{0}^{r} f(x) e^{x} d x=\left(f(r)-f^{\prime}(r)+f^{\prime \prime}(r)-\cdots\right) e^{r}-\left(f(0)-f^{\prime}(0)+f^{\prime \prime}(0)-\cdots\right)$$

$$F(x)=f(x)-f^{\prime}(x)+f^{\prime \prime}(x)-f^{\prime \prime \prime}(x)+\cdots,$$

$$\int_{0}^{r} f(x) e^{x} d x=F(r) e^{r}-F(0)$$

$$f(x)=c x^{n}(r-x)^{n}$$

\begin{aligned } f \wedge { ( k ) } ( 0 ) \& = k \text { ! \timesไleft } { \backslash \text { text } { } \mathrm { x } ^ { \wedge } { k } \backslash \text { right } } \text { 的系数 } \backslash \& = k \mathrm { ~ ! ~ c l

$$f(r-x)=f(x) \Rightarrow(-1)^{k} f^{(k)}(r-x)=f^{(k)}(x) \quad \Rightarrow f^{(k)}(r)=(-1)^{k} f^{(k)}(0)$$

$$q \int_{0}^{r} f(x) e^{x} d x=p F(r)-q F(0)$$

## 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。