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# 数学代写|复分析代写Complex analysis代考|MATH3979 Taylor’s theorem

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## 数学代写|复分析代写Complex analysis代考|Taylor’s theorem

Recall once more that a function differentiable at every point of an open set $G$ in the complex plane is said to be regular on $G$. The central insight of this chapter is that, for an open disc $D$, the concepts ‘function regular on $D$ ‘ and ‘power series convergent absolutely at each point of $D^{\prime}$ are virtually interchangeable: if you have one of these, you are guaranteed to have the other also. Much of the usefulness of this is due to the fact that we can carry out term-by-term differentiation on any power series. To revise briefly:

• A complex power series $\sum c_n(z-a)^n$ will generally have a radius of convergence (let us denote it by $r$ ) such that the series converges absolutely at every point interior to its circle of convergence $C(a, r)$ but diverges at every point exterior to that circle. The exceptional types are (i) certain such series converge only at $z=a$ and are said to have radius of convergence 0 , while (ii) certain important such series converge absolutely at every point of the complex plane, and for these the radius of convergence is said to be infinite.
• Term-by-term differentiation works for complex power series just as it does for real series: that is, if $\sum c_n(z-a)^n$ has non-zero radius of convergence $r$ and $f(z)$ denotes its sum at the typical point $z$ interior to $C(a, r)$, then $f$ is differentiable there, and $f^{\prime}(z)=\sum n c_n(z-a)^{n-1}$. So:
if $f(z)=c_0+c_1(z-a)+c_2(z-a)^2+c_3(z-a)^3+c_4(z-a)^4+\cdots$
then $f^{\prime}(z)=\quad c_1+2 c_2(z-a)+3 c_3(z-a)^2+4 c_4(z-a)^3+\cdots$.

## 数学代写|复分析代写Complex analysis代考|Taylor series

We’ll start with a tiny lemma that is going to save us time in the main proof.
8.2.1 Lemma Suppose that $a$ is some particular point in the complex plane, and that $w$ and $z$ are distinct from $a$ and from one another, and that we know which of them is further away from $a$ : say, $|z-a|<|w-a|$. Then
$$\frac{1}{w-z}=\sum_0^{\infty} \frac{(z-a)^n}{(w-a)^{n+1}}$$
the series being an (absolutely convergent) geometric series.
Proof Write $w-z$ as $(w-a)-(z-a)$ and, in turn, as $(w-a)\left(1-\frac{z-a}{w-a}\right)$. Then we have
$$\frac{1}{w-z}=(w-a)^{-1}\left(1-\frac{z-a}{w-a}\right)^{-1}=\frac{1}{w-a} \sum_0^{\infty}\left(\frac{z-a}{w-a}\right)^n$$
$$=\sum_0^{\infty} \frac{(z-a)^n}{(w-a)^{n+1}} \quad \text { as predicted. }$$
The series is absolutely convergent because the modulus of its common ratio
$$\left|\frac{z-a}{w-a}\right|=\frac{|z-a|}{|w-a|}$$
is less than 1. (The reader who, perfectly reasonably, wonders why on earth anyone would want to replace such a simple piece of algebra as $\frac{1}{w-z}$ by a whole infinite series will see at least part of the answer in the upcoming proof.)

## 数学代写|复分析代写复分析代考|泰勒定理

if $f(z)=c_0+c_1(z-a)+c_2(z-a)^2+c_3(z-a)^3+c_4(z-a)^4+\cdots$
then $f^{\prime}(z)=\quad c_1+2 c_2(z-a)+3 c_3(z-a)^2+4 c_4(z-a)^3+\cdots$ .

## 数学代写|复分析代写复杂分析代考|泰勒级数

$$\frac{1}{w-z}=\sum_0^{\infty} \frac{(z-a)^n}{(w-a)^{n+1}}$$

$$\frac{1}{w-z}=(w-a)^{-1}\left(1-\frac{z-a}{w-a}\right)^{-1}=\frac{1}{w-a} \sum_0^{\infty}\left(\frac{z-a}{w-a}\right)^n$$
$$=\sum_0^{\infty} \frac{(z-a)^n}{(w-a)^{n+1}} \quad \text { as predicted. }$$

$$\left|\frac{z-a}{w-a}\right|=\frac{|z-a|}{|w-a|}$$

## 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。