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• 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|复分析作业代写Complex function代考|Complex Power Series

The theory of Taylor series in real-variable calculus associates to each infinitely differentiable function $f$ from $\mathbb{R}$ to $\mathbb{R}$ a formal power series expansion at each point of $\mathbb{R}$, namely
$$\sum_{n=0}^{\infty} \frac{f^{(n)}(p)}{n !}(x-p)^{n}, \quad p \in \mathbb{R} .$$
There is no general guarantee that this series converges for any $x$ other than $x=p$. Moreover, there is also no general guarantee that, even if it does converge at some $x \neq p$, its sum is actually equal to $f(x)$. An instance of this latter phenomenon is the function
$$f(x)=\left{\begin{array}{lll} e^{-1 / x^{2}} & \text { if } & x \neq 0 \ 0 & \text { if } & x=0 . \end{array}\right.$$
This function can be easily checked to be $C^{\infty}$ on $\mathbb{R}$ with $0=f(0)=$ $f^{\prime}(0)=f^{\prime \prime}(0)=\ldots$ (use l’Hôpital’s rule to verify this assertion). So the Taylor expansion of $f$ at 0 is
$$0+0 x+0 x^{2}+0 x^{3}+\cdots,$$
which obviously converges for all $x$ with sum $\equiv 0$. But $f(x)$ is 0 only if $x=0$. (An example of the phenomenon that the Taylor series need not even converge except at $x=p$ is given in Exercise 64.) The familiar functions of calculus – $\sin , \cos , e^{x}$, and so forth-all have convergent power series. But most $C^{\infty}$ functions on $\mathbb{R}$ do not. Real functions $f$ that have, at each point $p \in \mathbb{R}$, a Taylor expansion that converges to $f$ for all $x$ near enough to $p$ are called real analytic on $\mathbb{R}$.

## 数学代写|复分析作业代写Complex function代考|The Power Series Expansion

As previously discussed, we first demonstrate that a holomorphic function has a convergent complex power series expansion (locally) about any point in its domain. Note that since a holomorphic function is defined on an arbitrary open set $U$ while a power series converges on a disc, we cannot expect a single power series expanded about a fixed point $P$ to converge to $f$ on all of $U$.

Theorem 3.3.1. Let $U \subseteq \mathbb{C}$ be an open set and let $f$ be holomorphic on $U$. Let $P \in U$ and suppose that $D(P, r) \subseteq U$. Then the complex power series
$$\sum_{k=0}^{\infty} \frac{\left(\partial^{k} f / \partial z^{k}\right)(P)}{k !}(z-P)^{k}$$
has radius of convergence at least $r$. It converges to $f(z)$ on $D(P, r)$.
Proof. Recall that from Theorem 3.1.1 we know that $f$ is $C^{\infty}$. So the coefficients of the power series expansion make sense. Given an arbitrary $z \in D(P, r)$, we shall now prove convergence of the series at this $z$. Let $r^{\prime}$ be a positive number greater than $|z-P|$ but less than $r$ so that
$$z \in D\left(P, r^{\prime}\right) \subseteq \bar{D}\left(P, r^{\prime}\right) \subseteq D(P, r)$$
Assume without loss of generality that $P=0$ (this simplifies the notation considerably, but does not change the mathematics) and apply the Cauchy integral formula to $f$ on $D\left(P, r^{\prime}\right)$. Thus for $z \in D\left(P, r^{\prime}\right)=D\left(0, r^{\prime}\right)$ we have
$$f(z)=\frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta-z} d \zeta$$

$=\frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta} \frac{1}{1-z \cdot \zeta^{-1}} d \zeta$
$=\frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta} \sum_{k=0}^{\infty}\left(z \cdot \zeta^{-1}\right)^{k} d \zeta .$

# 复分析代写

## 数学代写|复分析作业代写Complex function代考|Complex Power Series

$$\sum_{n=0}^{\infty} \frac{f^{(n)}(p)}{n !}(x-p)^{n}, \quad p \in \mathbb{R} .$$

$\sin , \cos , e^{x}$ ，等等一都有收敛的幂级数。但最 $C^{\infty}$ 上的功能 $\mathbb{R}$ 不要。实函数 $f$ 在每一点都有 $p \in \mathbb{R}$ ，泰勒展开式收敛到 $f$ 对所有人 $x$ 足够接近 $p$ 被称为实分析 $\mathbb{R}$.

## 数学代写|复分析作业代写Complex function代考|The Power Series Expansion

$$\sum_{k=0}^{\infty} \frac{\left(\partial^{k} f / \partial z^{k}\right)(P)}{k !}(z-P)^{k}$$

$$z \in D\left(P, r^{\prime}\right) \subseteq \bar{D}\left(P, r^{\prime}\right) \subseteq D(P, r)$$

$$f(z)=\frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta-z} d \zeta$$
\begin{aligned} &=\frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta} \frac{1}{1-2 \cdot \zeta^{-1}} d \zeta \ &=\frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta} \sum_{k=0}^{\infty}\left(z \cdot \zeta^{-1}\right)^{k} d \zeta . \end{aligned}

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## 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。

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