### 数学代写|matlab代写|ENES206

MATLAB是一个编程和数值计算平台，被数百万工程师和科学家用来分析数据、开发算法和创建模型。

statistics-lab™ 为您的留学生涯保驾护航 在代写matlab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matlab代写方面经验极为丰富，各种代写matlab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|matlab代写|TAYLOR AND LAURENT EXPANSIONS AND SINGULARITIES

In the previous section we showed what a crucial role singularities play in complex integration. Before we can find the most general way of computing a closed complex integral, our understanding of singularities must deepen. For this, we employ power series.

One reason why power series are so important is their ability to provide locally a general representation of a function even when its arguments are complex. For example, when we were introduced to trigonometric functions in high school, it was in the context of a right triangle and a real angle. However, when the argument becomes complex, this geometrical description disappears and power series provide a formalism for defining the trigonometric functions, regardless of the nature of the argument.

Let us begin our analysis by considering the complex function $f(z)$, which is analytic everywhere on the boundary, and the interior of a circle whose center is at $z=z_{0}$. Then, if $z$ denotes any point within the circle, we have from Cauchy’s integral formula that
$$f(z)=\frac{1}{2 \pi i} \oint_{C} \frac{f(\zeta)}{\zeta-z} d \zeta=\frac{1}{2 \pi i} \oint_{C} \frac{f(\zeta)}{\zeta-z_{0}}\left[\frac{1}{1-\left(z-z_{0}\right) /\left(\zeta-z_{0}\right)}\right] d \zeta,$$
where $C$ denotes the closed contour. Expanding the bracketed term as a geometric series, we find that
$$f(z)=\frac{1}{2 \pi i}\left[\oint_{C} \frac{f(\zeta)}{\zeta-z_{0}} d \zeta+\left(z-z_{0}\right) \oint_{C} \frac{f(\zeta)}{\left(\zeta-z_{0}\right)^{2}} d \zeta+\cdots+\left(z-z_{0}\right)^{n} \oint_{C} \frac{f(\zeta)}{\left(\zeta-z_{0}\right)^{n+1}} d \zeta+\cdots\right]$$
Applying Cauchy’s integral formula to each integral in Equation 1.7.2, we finally obtain
$$f(z)=f\left(z_{0}\right)+\frac{\left(z-z_{0}\right)}{1 !} f^{\prime}\left(z_{0}\right)+\cdots+\frac{\left(z-z_{0}\right)^{n}}{n !} f^{(n)}\left(z_{0}\right)+\cdots$$
or the familiar formula for a Taylor expansion. Consequently, we can expand any analytic function into a Taylor series. Interestingly, the radius of convergence ${ }^{6}$ of this series may be shown to be the distance between $z_{0}$ and the nearest nonanalytic point of $f(z)$.

## 数学代写|matlab代写|EVALUATION OF REAL DEFINITE INTEGRALS

One of the important applications of the theory of residues consists of the evaluation of certain types of real definite integrals. Similar techniques apply when the integrand contains a sine or cosine.

• Example 1.9.1
Let us evaluate the integral
$$\int_{0}^{\infty} \frac{d x}{x^{2}+1}=\frac{1}{2} \int_{-\infty}^{\infty} \frac{d x}{x^{2}+1} .$$
This integration occurs along the real axis. In terms of complex variables, we can rewrite Equation $1.9 .1$ as
$$\int_{0}^{\infty} \frac{d x}{x^{2}+1}=\frac{1}{2} \int_{C_{1}} \frac{d z}{z^{2}+1},$$
where the contour $C_{1}$ is the line $\Im(z)=0$. However, the use of the residue theorem requires an integration along a closed contour. Let us choose the one pictured in Figure 1.9.1. Then
$$\oint_{C} \frac{d z}{z^{2}+1}=\int_{C_{1}} \frac{d z}{z^{2}+1}+\int_{C_{2}} \frac{d z}{z^{2}+1},$$
where $C$ denotes the complete closed contour and $C_{2}$ denotes the integration path along a semicircle at infinity. Clearly we want the second integral on the right side of Equation $1.9 .3$ to vanish; otherwise, our choice of the contour $C_{2}$ is poor. Because $z=R c^{\theta i}$ and $d z=i R e^{\theta i} d \theta$
$$\left|\int_{C_{2}} \frac{d z}{z^{2}+1}\right|=\left|\int_{0}^{\pi} \frac{i R \exp (\theta i)}{1+R^{2} \exp (2 \theta i)} d \theta\right| \leq \int_{0}^{\pi} \frac{R}{R^{2}-1} d \theta$$
which tends to zero as $R \rightarrow \infty$. On the other hand, the residue theorem gives
$$\oint_{C} \frac{d z}{z^{2}+1}=2 \pi i \operatorname{Res}\left(\frac{1}{z^{2}+1} ; i\right)=2 \pi i \lim _{z \rightarrow i} \frac{z-i}{z^{2}+1}=2 \pi i \times \frac{1}{2 i}=\pi .$$

## 数学代写|matlab代写|TAYLOR AND LAURENT EXPANSIONS AND SINGULARITIES

$$f(z)=\frac{1}{2 \pi i} \oint_{C} \frac{f(\zeta)}{\zeta-z} d \zeta=\frac{1}{2 \pi i} \oint_{C} \frac{f(\zeta)}{\zeta-z_{0}}\left[\frac{1}{1-\left(z-z_{0}\right) /\left(\zeta-z_{0}\right)}\right] d \zeta$$

$$f(z)=\frac{1}{2 \pi i}\left[\oint_{C} \frac{f(\zeta)}{\zeta-z_{0}} d \zeta+\left(z-z_{0}\right) \oint_{C} \frac{f(\zeta)}{\left(\zeta-z_{0}\right)^{2}} d \zeta+\cdots+\left(z-z_{0}\right)^{n} \oint_{C} \frac{f(\zeta)}{\left(\zeta-z_{0}\right)^{n+1}} d \zeta+\cdots\right]$$

$$f(z)=f\left(z_{0}\right)+\frac{\left(z-z_{0}\right)}{1 !} f^{\prime}\left(z_{0}\right)+\cdots+\frac{\left(z-z_{0}\right)^{n}}{n !} f^{(n)}\left(z_{0}\right)+\cdots$$

## 数学代写|matlab代写|EVALUATION OF REAL DEFINITE INTEGRALS

• 示例 $1.9 .1$
让我们评估积分
$$\int_{0}^{\infty} \frac{d x}{x^{2}+1}=\frac{1}{2} \int_{-\infty}^{\infty} \frac{d x}{x^{2}+1}$$
这种整合沿实轴发生。就复变量而言，我们可以重写方程1.9.1作为
$$\int_{0}^{\infty} \frac{d x}{x^{2}+1}=\frac{1}{2} \int_{C_{1}} \frac{d z}{z^{2}+1}$$
轮廓在哪里 $C_{1}$ 是线 $\Im(z)=0$. 然而，使用余数定理需要沿闭合轮廓进行积分。让我们选择图 $1.9 .1$ 中所示的那 个。然后
$$\oint_{C} \frac{d z}{z^{2}+1}=\int_{C_{1}} \frac{d z}{z^{2}+1}+\int_{C_{2}} \frac{d z}{z^{2}+1},$$
在哪里 $C$ 表示完整的闭合轮廓，并且 $C_{2}$ 表示沿无限远半圆的积分路径。显然我们想要方程右边的第二个积分 1.9.3消失; 否则，我们选择的轮廓 $C_{2}$ 很穷。因为 $z=R c^{\theta i}$ 和 $d z=i R e^{\theta i} d \theta$
$$\left|\int_{C_{2}} \frac{d z}{z^{2}+1}\right|=\left|\int_{0}^{\pi} \frac{i R \exp (\theta i)}{1+R^{2} \exp (2 \theta i)} d \theta\right| \leq \int_{0}^{\pi} \frac{R}{R^{2}-1} d \theta$$
趋向于零 $R \rightarrow \infty$. 另一方面，剩余定理给出
$$\oint_{C} \frac{d z}{z^{2}+1}=2 \pi i \operatorname{Res}\left(\frac{1}{z^{2}+1} ; i\right)=2 \pi i \lim _{z \rightarrow i} \frac{z-i}{z^{2}+1}=2 \pi i \times \frac{1}{2 i}=\pi .$$

## 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。