# 数学代写|组合学代写Combinatorics代考|CS586

#### Doug I. Jones

Lorem ipsum dolor sit amet, cons the all tetur adiscing elit

couryes-lab™ 为您的留学生涯保驾护航 在代写组合学Combinatorics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。我们的专家在代写组合学Combinatorics代写方面经验极为丰富，各种代写组合学Combinatorics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
couryes™为您提供可以保分的包课服务

## 数学代写|组合学代写Combinatorics代考|The zeros of the derivatives of e

These points are determined by writing
$$e^{i z}=t_k^{(n)}$$
for appropriate $k$ and $n$. Taking logarithm carefully $\left(t_k^{(n)}\right.$ is a negative real number, so we must use the complex logarithm with complex branches) we get that
$$z=-i \log \left|t_k^{(n)}\right|+(2 l+1) \pi \quad(l=0, \pm 1, \pm 2, \ldots) .$$
Let us introduce now a notion which originates from the work of $\mathrm{Pólya}^5$ [468]. The final set of a $\left(\right.$ complex, entire $\left.{ }^6\right)$ function $f$ is constituted by those points $z$ which, in any neighbor of themselves contain zeros of infinitely many derivatives of $f$.

It is, in general, very hard to determine the final set of an entire function. A result from $1980[210]$ is the following: the final set of the entire function $e^{-e^z}$ consists of the horizontal lines $y=2 l \pi$ ( $l$ runs through the integers). One can see that $e^{e^{i z}}=e^{-i(z+\pi)}$, thus the final set of $e^{e^{i z}}$ is the collection of the vertical lines with abscissa $(2 l+1) \pi$. Now recall (3.9). We see that the abscissas agree, so the logarithms of the absolute values of the elements of $\mathcal{T}$ must be dense on the real line. Since the logarithm is a continuous function, it must be true that $\mathcal{T}$ is dense in $]-\infty, 0[$.

## 数学代写|组合学代写Combinatorics代考|The sum of the reciprocals of the zeros

We deduce the formulas in this section for general polynomials. Let
$$p(x)=a_n x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0$$
be a polynomial, and let $\alpha_1, \ldots, \alpha_n$ be its zeros (multiple zeros appear in the list as many times as their multiplicity). Then it is possible to write $p_n(x)$ as
$$\begin{gathered} p_n(x)=\left(\alpha_1-x\right) \cdots\left(\alpha_2-x\right) \cdots\left(\alpha_n-x\right)= \ \alpha_1 \alpha_2 \cdots \alpha_n\left(1-\frac{x}{\alpha_1}\right)\left(1-\frac{x}{\alpha_2}\right) \cdots\left(1-\frac{x}{\alpha_n}\right) . \end{gathered}$$
Let us denote the product of the zeros by $\alpha$ :
$$\alpha=\alpha_1 \alpha_2 \cdots \alpha_n$$
(Remember that this product is easy to determine by (3.7).) Thus,
$$\frac{1}{\alpha} p_n(x)=\left(1-\frac{x}{\alpha_1}\right)\left(1-\frac{x}{\alpha_2}\right) \cdots\left(1-\frac{x}{\alpha_n}\right) .$$
The coefficient of $x$ on the right-hand side is
$$-\frac{1}{\alpha_1}-\frac{1}{\alpha_2}-\cdots-\frac{1}{\alpha_n}$$
while on the left-hand side it is $a_1 / \alpha$. Thus,
$$\sum_{k=1}^n \frac{1}{\alpha_k}=-\frac{a_1}{\alpha}$$
If we want to apply this to the Bell polynomials, we should be careful a bit, because one zero of the Bell polynomials is $\alpha_1=0$. Therefore, in place of $B_n(x)$, we apply the above to $B_n(x) / x$, and we finally get that
$$\sum_{k=2}^n \frac{1}{t_k^{(n)}}=-\left{\begin{array}{l} n \ 2 \end{array}\right}=1-2^{n-1}$$

# 组合学代考

## 数学代写|组合学代写Combinatorics代考|The sum of the reciprocals of the zeros

$\$ \$$p(x)=a_{-} n x^{\wedge} n+a_{-}{n-1} x^{\wedge}{n-1}+I c d o t s+a_{-} 1 x+a_{-} 0 \ \$$ 为

Ibegin{gathered $}$
p_n(x)=lleft(lalpha_1-x|right) \cdotslleft(Ialpha_2-
$x \backslash r i g h t) \backslash c d o t s \backslash$ 左(lalpha_n-x\right) $=1$
IcdotsVleft(1-Ifrac{x}{|alpha_n}|right) 。
lend{gathered $}$
$\$ \$$让我们用 \।alpha\ 表示零的乘积: \ \$$
Ialpha=lalpha_1 Ialpha_2 Icdots Ialpha_n
$\$ \$$（请记住，这个乘积很容易由 (3.7) 确定。）因此， \ \$$
Ifrac ${1} \backslash$ |alpha} p_n(x)=|left(1-|frac ${x}{$ lalpha_1}|right) {lalpha_n}|right)。
$\$ \$$右边 \ \times 的系数为 \ \$$

$\$ \$$\sum_{k \left.=2}^{\wedge} n \backslash f r a c{1} t _k \wedge{(n)}\right}=-\backslash left \backslash begin { array } {\mid} n \backslash 2 \backslash lend { array } \backslash r i g h t }$$
1-2^{\wedge}{n-1}
$$\ \$$

## 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。

Days
Hours
Minutes
Seconds

# 15% OFF

## On All Tickets

Don’t hesitate and buy tickets today – All tickets are at a special price until 15.08.2021. Hope to see you there :)