数学代写|数论代写Number theory代考|MAST90136

如果你也在 怎样代写数论Number theory这个学科遇到相关的难题,请随时右上角联系我们的24/7代写客服。


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。

我们提供的数论Number theory及其相关学科的代写,服务范围广, 其中包括但不限于:

  • Statistical Inference 统计推断
  • Statistical Computing 统计计算
  • Advanced Probability Theory 高等概率论
  • Advanced Mathematical Statistics 高等数理统计学
  • (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|MAST90136


It is known that $\pi$ is irrational: we shall prove this in the next chapter. It is not hard to see that at least one of the numbers $\pi+e$ and $\pi e$ must be irrational (in fact, at least one must be transcendental – see Chapter 3 ); although, most likely, both are irrational, this has not been proved for either one individually. As a consequence of a difficult result due to Gelfond and Schneider (Theorem $5.18$ ) we know that $e^{\pi}$ is irrational; however it is still unknown whether or not $\pi^{e}$ is irrational. It can also be shown that various numbers such as, for

example, $e^{\sqrt{2}}$ and $2^{\sqrt{2}}$ are irrational. However, the irrationality of $\pi^{\sqrt{2}}$ and $2^{e}$, and that of the Euler-Mascheroni constant
\gamma=\lim {n \rightarrow \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\log n\right)=0.57721 \cdots $$ remain undecided. Another problem which has attracted much attention is to investigate the irrationality of the numbers $\zeta(n)$. Here $n \geq 2$ is an integer and $\zeta$ is the Riemann zeta function defined by $$ \zeta(s)=\sum{k=1}^{\infty} \frac{1}{k^{s}}=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\cdots
for $s>1$. By methods of complex integration we can show that if $n$ is even then $\zeta(n)$ is a rational number times $\pi^{n}$, and this is known to be irrational. On the other hand, it is much harder to find out anything of interest about $\zeta(n)$ for odd $n$. In 1978 the French mathematician $\mathrm{R}$. Apéry sensationally proved that $\zeta(3)$ is irrational. His complicated argument had the appearance of being completely unmotivated, and all of the techniques he had used would have been available two centuries earlier: for these reasons, few people believed that the proof could possibly be correct. Nevertheless it was found possible eventually to confirm all of Apéry’s assertions and thereby establish what has been called “a proof that Euler missed”. A brief (but not easy!) account of Apéry’s work is given in [66].

数学代写|数论代写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|数论代写Number theory代考|IRRATIONALITY OF e r

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.

数学代写|数论代写Number theory代考|MAST90136



众所周知圆周率是非理性的:我们将在下一章证明这一点。不难看出,至少有一个数字圆周率+和和圆周率和必须是非理性的(事实上,至少有一个必须是超验的——见第 3 章);尽管很可能两者都是不合理的,但这还没有单独证明。由于 Gelfond 和 Schneider 的困难结果(定理5.18) 我们知道和圆周率是不合理的;但是否与否仍是未知数圆周率和是不合理的。还可以证明,各种数字,例如,对于

例子,和2和22是不合理的。然而,非理性圆周率2和2和, 和欧拉-马斯切罗尼常数

C=林n→∞(1+12+13+⋯+1n−日志⁡n)=0.57721⋯仍未决定。另一个备受关注的问题是调查数字的不合理性G(n). 这里n≥2是一个整数并且G是由定义的黎曼 zeta 函数

为了s>1. 通过复积分的方法,我们可以证明,如果n就在那时G(n)是有理数次圆周率n,这被认为是不合理的。另一方面,要找到任何感兴趣的东西要困难得多G(n)对于奇数n. 1978年法国数学家R. 阿佩里轰动地证明了这一点G(3)是不合理的。他复杂的论证看起来完全没有动机,而他使用的所有技术在两个世纪前就可以使用:由于这些原因,很少有人相信这个证明可能是正确的。然而,最终发现有可能证实阿佩里的所有断言,从而建立所谓的“欧拉遗漏的证据”。[66] 中给出了 Apéry 工作的简要(但不容易!)说明。

数学代写|数论代写Number theory代考|SOME ELEMENTARY NUMBER THEORY

本节包含一些您应该知道的基本数论定义和结果。本节中的证明被缩写或省略,您应该能够自己提供证明。如有必要,可以在任何有关初等数论的著作中找到该材料。最受欢迎的经典文本会定期进行修订,从而提供经过验证的阐述以及使内容和演示保持最新的补充。在一个非常拥挤的领域,我们提到了 Hardy 和 Wright [28]、[29]、Niven 和 Zuckerman[45],[46]和贝克[10]。

引理 1.10。除法算法。如果一个和b是整数b>0, 那么存在整数q和r这样一个=bq+r和0≤r<b.


引理 1.11。Bézout 财产。如果一个和b是整数,不都是零,并且G是的最大公约数一个和b, 那么存在整数X和是这样一个X+b是=G.


引理 1.12。如果一个和米没有公因数并且一个∣米n, 然后一个∣n.
定义 1.4。让米为正整数。我们说整数 a 和 b 是模数全等的米, 写一个≡b(反对米), 如果米∣一个−b.

为了“减少一个整数一个模块米”的意思是找一个整数b这样一个≡b(反对米)和b位于“合适”的范围内,通常0≤b<米. 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,并且在这个意义上“没什么新意”,但它非常有用,因为全等具有许多相等的基本属性。

数学代写|数论代写Number theory代考|IRRATIONALITY OF e r

在最终证明的实际细节中,Hermite 的方法(至少对于早期的结果)并不太难。然而,证明背后的动机可能是模糊的。因此,与其直接给出证明,不如从尝试解释一个相对简单的案例背后的目的和想法开始。我们希望通过证明如果r那么是理性的和r是非理性的,但明显的例外是和0=1.

像往常一样,我们寻求反证法:取r=一个/b和一个≠0,并假设和r=p/q. 遵循定理的方法1.9,我们试图通过构造一个介于 0 和 1 之间的整数来获得矛盾。Hermite 的想法,起源于他对近似值的研究和X, 是考虑定积分







数学代写|数论代写Number theory代考 请认准statistics-lab™

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。







术语 广义线性模型(GLM)通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归,以及方差分析和方差分析(仅含固定效应)。



有限元是一种通用的数值方法,用于解决两个或三个空间变量的偏微分方程(即一些边界值问题)。为了解决一个问题,有限元将一个大系统细分为更小、更简单的部分,称为有限元。这是通过在空间维度上的特定空间离散化来实现的,它是通过构建对象的网格来实现的:用于求解的数值域,它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统,以模拟整个问题。然后,有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。





随机过程,是依赖于参数的一组随机变量的全体,参数通常是时间。 随机变量是随机现象的数量表现,其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值(如1秒,5分钟,12小时,7天,1年),因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中,往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录,以得到其自身发展的规律。


多元回归分析渐进(Multiple Regression Analysis Asymptotics)属于计量经济学领域,主要是一种数学上的统计分析方法,可以分析复杂情况下各影响因素的数学关系,在自然科学、社会和经济学等多个领域内应用广泛。


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。