### 数学代写|数论代写Number theory代考|PMTH 338

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|数论代写Number theory代考|IRRATIONALITY OF π

Similar ideas can be used to prove the irrationality of $\pi$, and more generally, of $\cos r$ if $r$ is rational and non-zero. Our aim will be to integrate $f(x) \sin x$ for suitable functions $f$; we shall find that the polynomial used in previous proofs is not always suitable.
Theorem 2.4. $\pi$ is irrational.
Proof. Suppose that $\pi=a / b$; define
$$f(x)=\frac{\left(a x-b x^{2}\right)^{n}}{n !}=\frac{x^{n}(a-b x)^{n}}{n !}$$
and once again integrate by parts:
\begin{aligned} \int_{0}^{\pi} f(x) \sin x d x &=(f(\pi)+f(0))+\int_{0}^{\pi} f^{\prime}(x) \cos x d x \ &=(f(\pi)+f(0))-\int_{0}^{\pi} f^{\prime \prime}(x) \sin x d x \end{aligned}
In the second integration, we have used the fact that $\sin \pi=\sin 0=0$. Continuing to integrate in the same way we obtain
\begin{aligned} \int_{0}^{\pi} f(x) \sin x d x &=(f(\pi)+f(0))-\left(f^{\prime \prime}(\pi)+f^{\prime \prime}(0)\right)+\cdots \ &=F(\pi)+F(0) \end{aligned}
where
$$F(x)=f(x)-f^{\prime \prime}(x)+f^{(\mathrm{iv})}(x)-f^{(\mathrm{vi})}(x)+\cdots$$

Using Lemma $2.1$ (derivatives of polynomials), and recalling that by assumption $\pi=a / b$, we see that both $F(\pi)$ and $F(0)$ are integers. But $f(x) \sin x$ is always positive for $0<x<\pi$ and we have
$$0<\int_{0}^{\pi} f(x) \sin x d x \leq \pi \frac{(a \pi)^{n}}{n !} .$$
If $n$ is sufficiently large, the right-hand side is less than 1 , and we have a contradiction in the usual manner. Therefore, $\pi$ is irrational.

Comment. We might reasonably expect to obtain a similar proof by using the integral
$$\int_{0}^{\pi} f(x) \cos x d x$$
with the same polynomial $f(x)$. In fact, the attempt fails utterly! (Exercise. Explain why.) We can, however, prove the irrationality of $\pi$ by considering
$$\int_{-\pi}^{\pi} \frac{(a-b x)^{n}(a+b x)^{n}}{n !} \cos x d x$$
though in this case, the integrand is not always positive for $-\pi<x<\pi$, and the fact that the integral is non-zero (while possibly “obvious” from figure 2.1) is slightly tricky to prove carefully.
Figure 2.1 The graph of $y=\left(a^{2}-b^{2} x^{2}\right)^{n} \cos x$.

## 数学代写|数论代写Number theory代考|SOME RESULTS OF ELEMENTARY CALCULUS

Theorem 2.5. Irrationality of cosines. If $r$ is rational and not zero, then $\cos r$ is irrational.

Proof. Let $r=a / b$ be a non-zero rational; assume that $\cos r=p / q$. Without loss of generality assume that $a$ and $b$, and hence $r$, are positive. Choose
$$f(x)=x^{n}(a-b x)^{2 n}(2 a-b x)^{n} ;$$
in this case we find it more convenient not to include $n$ ! in the denominator. Integrating twice by parts yields
$$\int_{0}^{r} f(x) \sin x d x=f(0)-f(r) \cos r+f^{\prime}(r) \sin r-\int_{0}^{r} f^{\prime \prime}(x) \sin x d x$$
repeating the procedure and writing
$$F(x)=f(x)-f^{\prime \prime}(x)+f^{(\mathrm{iv})}(x)-f^{(\mathrm{vi})}(x)+\cdots$$
gives eventually
$$\int_{0}^{r} f(x) \sin x d x=F(0)-F(r) \cos r+F^{\prime}(r) \sin r .$$
Now observe that $f(x)$ is a polynomial in $(a-b x)^{2}$, since
$$f(x)=\frac{(a-b x)^{2 n}\left(a^{2}-(a-b x)^{2}\right)^{n}}{b^{n}} ;$$
if we set $g(x)=x^{2 n}\left(a^{2}-x^{2}\right)^{n}$, then $g$ is an even function and so $g^{(k)}(0)=0$ whenever $k$ is odd. But we have
\begin{aligned} f(x)=b^{-n} g(a-b x) & \Rightarrow f^{(k)}(x)=(-b)^{k} b^{-n} g^{(k)}(a-b x) \ & \Rightarrow \quad f^{(k)}(r)=-b^{k-n} g^{(k)}(0)=0 \quad \text { for odd } k, \end{aligned}
and so $F^{\prime}(r)=0$. Therefore, we can rewrite (2.6) as
$$q \int_{0}^{r} f(x) \sin x d x=q F(0)-p F(r) .$$
Now applying the lemma on derivatives of polynomials shows that $f^{(k)}(r)$ is a multiple of $(2 n) !$, and hence also a multiple of $(n+1)$ !, for all $k$. In the case of $f^{(k)}(0)$, we need a little more information than is given by the lemma. Since
$$f^{(k)}(0)=k ! \times\left{\text { coefficient of } x^{k}\right}$$
we see that $f^{(k)}(0)$ is zero for $kn$. Moreover, for $k=n$ we have
$$f^{(n)}(0)=n ! 2^{n} a^{3 n}$$

The proof of Theorem $2.4$ can be viewed in a slightly different light. Taking $r=\pi$, we assumed that $r$ is rational and used the fact that $\cos r$ is rational to reach a contradiction. However, the proof relied vitally on the fact that $\sin \pi=0$, so one could not expect exactly the same proof to work for arbitrary rational $r$. Nevertheless, it turns out that by modifying the proof somewhat, we can prove the following result.

## 数学代写|数论代写Number theory代考|SOME RESULTS OF ELEMENTARY CALCULUS

Proofs of the following results are again left up to the reader. If assistance is required, any basic calculus text should suffice.
Lemma 2.7. For any real constants $\beta$ and $\gamma$, we have
$$\frac{\beta \gamma^{n}}{n !} \rightarrow 0 \quad \text { as } n \rightarrow \infty$$
Corollary 2.8. Comparison of exponentials and factorials. If $\beta$ and $\gamma$ are real constants, then $\left|\beta \gamma^{n}\right|<n$ ! for all sufficiently large integers $n$.

Lemma 2.9. Derivatives of even and odd functions. Let $f$ be a differentiable function from $\mathbb{R}$ to $\mathbb{R}$. Then

• $f$ is even if and only if $f^{\prime}$ is odd;
• if $f$ is odd then $f^{\prime}$ is even.
Exercise. Why is the converse of the second result not true? Fill in the gap and then prove the statement:
“if $f^{\prime}$ is even and … then $f$ is odd”.
Corollary 2.10. If $g$ is an odd function, then $g^{(k)}(0)=0$ for all even $k$. If $g$ is an even function, then $g^{(k)}(0)=0$ for all odd $k$.
Lemma 2.11. Derivatives of a product (Leibniz’ rule). If $k \geq 0$, then
$$\frac{d^{k}(u v)}{d x^{k}}=\sum_{j=0}^{k}\left(\begin{array}{l} k \ j \end{array}\right) \frac{d^{j} u}{d x^{j}} \frac{d^{k-j} v}{d x^{k-j}}$$

## 数学代写|数论代写Number theory代考|IRRATIONALITY OF π

F(X)=(一个X−bX2)nn!=Xn(一个−bX)nn!

∫0圆周率F(X)罪⁡XdX=(F(圆周率)+F(0))+∫0圆周率F′(X)因⁡XdX =(F(圆周率)+F(0))−∫0圆周率F′′(X)罪⁡XdX

∫0圆周率F(X)罪⁡XdX=(F(圆周率)+F(0))−(F′′(圆周率)+F′′(0))+⋯ =F(圆周率)+F(0)

F(X)=F(X)−F′′(X)+F(一世在)(X)−F(在一世)(X)+⋯

0<∫0圆周率F(X)罪⁡XdX≤圆周率(一个圆周率)nn!.

∫0圆周率F(X)因⁡XdX

∫−圆周率圆周率(一个−bX)n(一个+bX)nn!因⁡XdX

## 数学代写|数论代写Number theory代考|SOME RESULTS OF ELEMENTARY CALCULUS

F(X)=Xn(一个−bX)2n(2一个−bX)n;

∫0rF(X)罪⁡XdX=F(0)−F(r)因⁡r+F′(r)罪⁡r−∫0rF′′(X)罪⁡XdX

F(X)=F(X)−F′′(X)+F(一世在)(X)−F(在一世)(X)+⋯

∫0rF(X)罪⁡XdX=F(0)−F(r)因⁡r+F′(r)罪⁡r.

F(X)=(一个−bX)2n(一个2−(一个−bX)2)nbn;

F(X)=b−nG(一个−bX)⇒F(ķ)(X)=(−b)ķb−nG(ķ)(一个−bX) ⇒F(ķ)(r)=−bķ−nG(ķ)(0)=0 对于奇数 ķ,

q∫0rF(X)罪⁡XdX=qF(0)−pF(r).

f^{(k)}(0)=k ！\times\left{\text { } x^{k}\right} 的系数f^{(k)}(0)=k ！\times\left{\text { } x^{k}\right} 的系数

F(n)(0)=n!2n一个3n

## 数学代写|数论代写Number theory代考|SOME RESULTS OF ELEMENTARY CALCULUS

bCnn!→0 作为 n→∞

• F是即使且仅当F′很奇怪；
• 如果F那么奇怪F′甚至。
锻炼。为什么第二个结果的反面不成立？填补空白，然后证明陈述：
“如果F′是偶数，然后……然后F很奇怪”。
推论 2.10。如果G是奇函数，那么G(ķ)(0)=0甚至所有人ķ. 如果G是偶函数，那么G(ķ)(0)=0对于所有奇怪的ķ.
引理 2.11。产品的衍生物（莱布尼茨规则）。如果ķ≥0， 然后
dķ(在在)dXķ=∑j=0ķ(ķ j)dj在dXjdķ−j在dXķ−j

## 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。