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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|微积分代写Calculus代写|Non-Entire Functions

A non-entire function is not analytic across the whole complex plane $\mathbb{C}$. Many important functions may be analytic in some subset of the complex plane but have singularities-points at which the function is not analytic-at other spots. These points form barriers through which a given power series’ disk of convergence cannot cross and analytic continuation fails. Non-entire functions also include analytic functions that could be extended as entire but whose domains are not all of $\mathbb{C}$. For example, each of the functions
$$f(z)=\frac{\sin z}{z}, \quad g(z)=\frac{\mathrm{e}^{z}}{z^{2}}, \quad \text { and } \quad h(z)=\mathrm{e}^{1 / z}$$
has a lone singularity at $z=0$ (each written intentionally so as not to include 0 in its domain) but is analytic for $|z|>0$. A function’s set of singularities cannot only be places where the function is not analytic but also not even defined. The three functions just mentioned are simply not even defined at $z=0$.

What can we say about such functions? A lot. They have a Taylor series representation at each point in their domain of analyticity, but the expansion cannot be about any singularity. Plus, any Taylor series is representative only in the corresponding disk of convergence and not throughout the domain of analyticity, since the disk of convergence is restricted to extend only out to the nearest singularity. For functions with only one singularity, is there a series representation, different from the Taylor series, that could be valid everywhere but the one bad singularity spot? The answer turns out to be yes. For example, using the Taylor series for $\sin z$ and $\mathrm{e}^{z}$,

• $f(z)=\frac{\sin z}{z}=\frac{1}{z} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} z^{2 n+1}=1-\frac{z^{2}}{3 !}+\cdots+\frac{(-1)^{n} z^{2 n}}{(2 n+1) !}+\cdots$
$\cdot g(z)=\frac{\mathrm{e}^{z}}{z^{2}}=\frac{1}{z^{2}} \sum_{n=0}^{\infty} \frac{z^{n}}{n !}=\frac{1}{z^{2}}+\frac{1}{z}+\frac{1}{2}+\frac{z}{3 !}+\cdots \frac{z^{n-2}}{n !}+\cdots$, and
• $h(z)=\mathrm{e}^{1 / z}=\sum_{n=0}^{\infty} \frac{(1 / z)^{n}}{n !}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}=1+\frac{1}{z}+\frac{1}{2 z^{2}}+\frac{1}{3 ! z^{3}}+\cdots \frac{1}{n ! z^{n}}+\cdots$

## 数学代写|微积分代写Calculus代写|Singularities

This section explores personalities of large categories of non-entire functions. The examples
$$f(z)=\frac{\sin z}{z}, \quad g(z)=\frac{\mathrm{e}^{z}}{z^{2}}, \quad \text { and } \quad h(z)=\mathrm{e}^{1 / z}$$
each have an isolated singularity at 0 , where the function is not holomorphic but can be surrounded by a disk across which (elsewhere from the singularity) the function is holomorphic. But cach funetion has a peculiar nuance that makes them distinct:

• You could choose to additionally define $f(0)=1$, and the resulting function is not only continuous but differentiable at 0 , and $f^{\prime}(0)=0$. Mathematicians say you’ve removed the singularity.
• You can’t do that for $g(0)$, but you could if you multiply $g(z)$ by $z^{2}$.
• But $h(z)$ is different, because no power you might multiply $h(z)$ by allows you to remove it.

## 数学代写|微积分代写Calculus代写|Non-Entire Functions

$$f(z)=\frac{\sin z}{z}, \quad g(z)=\frac{\mathrm{e}^{z}}{z^{2}}, \quad \text { and } \quad h(z)=\mathrm{e}^{1 / z}$$

• $f(z)=\frac{\sin z}{z}=\frac{1}{z} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} z^{2 n+1}=1-\frac{z^{2}}{3 !}+\cdots+\frac{(-1)^{n_{2} 2 n}}{(2 n+1) !}+\cdots$ $g(z)=\frac{\mathrm{e}^{z}}{z^{2}}=\frac{1}{z^{2}} \sum_{n=0}^{\infty} \frac{z^{n}}{\frac{n^{n}}{n}}=\frac{1}{z^{2}}+\frac{1}{z}+\frac{1}{2}+\frac{z}{3 !}+\cdots \frac{z^{n}-2}{n !}+\cdots, \quad$ 和
• $h(z)=\mathrm{e}^{1 / z}=\sum_{n=0}^{\infty} \frac{(1 / z)^{n}}{n !}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}=1+\frac{1}{z}+\frac{1}{2 z^{2}}+\frac{1}{3 ! z^{3}}+\cdots \frac{1}{n ! z^{n}}+\cdots$

## 数学代写|微积分代写Calculus代写|Singularities

$$f(z)=\frac{\sin z}{z}, \quad g(z)=\frac{\mathrm{e}^{z}}{z^{2}}, \quad \text { and } \quad h(z)=\mathrm{e}^{1 / z}$$

• 您可以选择另外定义 $f(0)=1$ ，得到的函数不仅是连续的，而且在 0 处可微分，并且 $f^{\prime}(0)=0$. 数学家说你已经消除了奇点。
• 你不能妏样做 $g(0)$, 但如果你乘以你可以 $g(z)$ 经过 $z^{2}$.
• 但 $h(z)$ 是不同的，因为没有力量可以乘 $h(z)$ by 允许您删除它。

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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。

assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师