## 数学代写|组合学代写Combinatorics代考|NWI-IBC016

statistics-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。

## 数学代写|组合学代写Combinatorics代考|Generating Function of Permutations by Inversions

In Section 1.1, we looked at descents of permutations. That is, we studied instances in which an entry in a permutation was larger than the entry directly following it. A more “global” permutation statistic is that of inversions. This statistic will look for instances in which an entry of a permutation is smaller than some entry following it (not necessarily directly).

DEFINITION 2.1 Let $p=p_1 p_2 \cdots p_n$ be a permutation. We say that $\left(p_i, p_j\right)$ is an inversion of $p$ if $ip_j$.
Example 2.2
Permutation 31524 has four inversions, namely $(3,1),(3,2),(5,2)$, and $(5,4)$.
This line of research started as early as 1901 [254]. In this section, we survey some of the most interesting results in this area. The number of inversions of $p$ will be denoted by $i(p)$, though some authors prefer $i n v(p)$. It is clear that $0 \leq i(p) \leq\left(\begin{array}{c}n \ 2\end{array}\right)$ for all $n$-permutations, and that the two extreme values are attained by permutations $12 \cdots n$ and $n(n-1) \cdots 1$, respectively. It is relatively easy to find the generating function enumerating all permutations of length $n$ with respect to their number of inversions.

THEOREM 2.3
For all positive integers $n \geq 2$,
$$\sum_{p \in S_n} z^{i(p)}=I_n(z)=(1+z)\left(1+z+z^2\right) \cdots\left(1+z+z^2+\cdots+z^{n-1}\right) .$$

PROOF We prove the statement by induction on $n$. In fact, we prove that each of the $n$ ! expansion terms of the product $I_n(z)$ corresponds to exactly one permutation in $S_n$. Moreover, the expansion term $z^{a_1} z^{a_2} \cdots z^{a_{n-1}}$ will correspond to the unique permutation in which, for each $i \in[n]$, the entry $i+1$ precedes exactly $a_i$ entries that are smaller than itself.

If $n=2$, then there are two permutations to count, $p=12$ has no inversions, and $p^{\prime}=21$ has one inversion. So $\sum_{p \in S_2} z^{i(p)}=1+z$ as claimed. Furthermore, $p=12$ is represented by the expansion term 1 , and $p^{\prime}=21$ is represented by the expansion term $z$.

## 数学代写|组合学代写Combinatorics代考|Explicit Definition of Determinants

There are several undergraduate mathematics courses and textbooks that only give a recursive definition of the determinant of a square matrix. That is, $\operatorname{det}\left(\begin{array}{ll}a & b \ c & d\end{array}\right)$ is defined to be equal to $a d-b c$, and then the determinant of the $n \times n$ matrix $A=\left(a_{i j}\right)$ is defined to be
$$\operatorname{det} A=\sum_{j=1}^n(-1)^{j-1} a_{1 j} A_{1 j}$$
where $A_{1 j}$ is the $(n-1) \times(n-1)$ matrix obtained from $A$ by removing the first row and the $j$ th column.

If that is the only definition of determinants the reader has seen, he may find the following result interesting.
THEOREM 2.21
Let $A=\left(a_{i j}\right)$ be an $n \times n$ matrix. Then we have
$$\operatorname{det} A=\sum_{p \in S_n}(-1)^{i(p)} a_{1 p_1} a_{2 p_2} \cdots a_{n p_n} .$$
That is, $\operatorname{det} A$ is obtained by taking all $n$ ! possible $n$-tuples of entries so that there is exactly one of the $n$ entries in each row and each column, multiplying the elements of each such $n$-tuple together, finally taking a signed sum of these $n$ ! products, where the sign is determined by the parity of $i(p)$, and $p$ is the permutation determined by each chosen $n$-tuple.

In other words, the $n$-tuples correspond to all possible placements of $n$ rooks on an $n \times n$ chessboard so that no two of them hit each other.

# 组合学代考

## 数学代写|组合学代写Combinatorics代考|Generating Function of Permutations by Inversions

$$\sum_{p \in S_n} z^{i(p)}=I_n(z)=(1+z)\left(1+z+z^2\right) \cdots\left(1+z+z^2+\cdots+z^{n-1}\right) .$$

## 数学代写|组合学代写Combinatorics代考|Explicit Definition of Determinants

$$\operatorname{det} A=\sum_{j=1}^n(-1)^{j-1} a_{1 j} A_{1 j}$$

$$\operatorname{det} A=\sum_{p \in S_n}(-1)^{i(p)} a_{1 p_1} a_{2 p_2} \cdots a_{n p_n} .$$

## 有限元方法代写

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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。

## 数学代写|组合学代写Combinatorics代考|MAT21018

statistics-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。

## 数学代写|组合学代写Combinatorics代考|Alternating Runs and Alternating Subsequences

The length of the longest alternating subsequence of a permutation is closely connected to the number of alternating runs as shown by the following proposition.
PROPOSITION 1.53
Let $n \geq 2$. Then $a_k(n)=\frac{1}{2}(G(n, k-1)+G(n, k))$.
PROOF If an $n$-permutation $p$ has $i$ alternating runs and starts in a descent, then $\operatorname{as}(\mathrm{p})=\mathrm{i}+1$ as can be seen by considering the entries of $p$ that are peaks or valleys, as well as the first and last entry of $p$. It follows from the pigeon-hole principle that $p$ cannot contain a longer alternating subsequence. If $p$ starts in an ascent, then $\operatorname{as}(\mathrm{p})=\mathrm{i}$ by similar considerations.

Therefore, the $n$-permutations $p$ satisfying $\operatorname{as}(\mathrm{p})=\mathrm{k}$ are precisely the $n$ permutations with $k-1$ alternating runs starting in a descent and the $n$ permutations with $k$ alternating runs starting in an ascent.

Proposition 1.53 implies that if $T_n(z)=\sum_{p \in S_n} z^{\text {as(p) }}$, then
$$T_n(z)=\frac{1}{2}(1+z) G_n(z) .$$
So the polynomials $T_n(z)$ have real roots only, and $\lfloor n / 2\rfloor$ of their roots are equal to -1 .
The first few polynomials $T_n(z)$ are shown below.

1. $T_1(z)=z$,
2. $T_2(z)=z+z^2$,
3. $T_3(z)=z+3 z^2+2 z^3$,
4. $T_4(z)=z+7 z^2+11 z^3+5 z^4$,
5. $T_5(z)=z+15 z^2+43 z^3+45 z^4+16 z^5$,
6. $T_6(z)=z+31 z^2+148 z^3+268 z^4+211 z^5+61 z^6$,
7. $T_7(z)=z+63 z^2+480 z^3+1344 z^4+1767 z^5+1113 z^6+272 z^7$.

## 数学代写|组合学代写Combinatorics代考|Alternating Permutations

Sometimes an entire permutation is an alternating sequence, leading to the following definition.

DEFINITION $\mathbf{1 . 5 4}$ We say that the $n$-permutation $p$ is alternating if the longest alternating subsequence of $p$ is of length $n$. Similarly, we say that $p$ is reverse alternating if the longest reverse alternating subsequence of $p$ is of length $n$.

For instance, 312 and 5241736 are alternating permutations. Clearly, $p$ is alternating if and only if its complement, that is, the $n$-permutation whose $i$ th entry is $n+1-p_i$, is reverse alternating.

The number of alternating $n$-permutations is called an Euler number (not to be confused with the Eulerian numbers $A(n, k)$ ) and is denoted by $E_n$. The reader is invited to verify that $E_2=1, E_3=2, E_4=5$, and $E_5=16$. The Euler numbers have a very interesting exponential generating function. This is the content of the next theorem.
THEOREM 1.55
Set $E_0=1=E_1$. Then the equality
$$E(z)=\sum_{n \geq 0} E_n \frac{z^n}{n !}=\sec z+\tan z$$

holds.
PROOF Let $L(n+1) R$ be an alternating or reverse alternating permutation of length $n+1$. So $L$ is the string on the left of the maximal entry, and $R$ is the string on the right of the maximal entry. Then $R$ is reverse alternating, and so is $L^r$, that is, the reverse of $L$.
This observation leads to the recurrence relation
$$2 E_{n+1}=\sum_{k=0}^n\left(\begin{array}{l} n \ k \end{array}\right) E_k E_{n-k},$$
for $n \geq 1$. In terms of generating functions, this is equivalent to
$$2 E^{\prime}(z)=E^2(z)+1,$$
with $E(0)=1$.

# 组合学代考

## 数学代写|组合学代写Combinatorics代考|Alternating Runs and Alternating Subsequences

$$T_n(z)=\frac{1}{2}(1+z) G_n(z) .$$

$T_1(z)=z$，

$T_2(z)=z+z^2$，

$T_3(z)=z+3 z^2+2 z^3$，

$T_4(z)=z+7 z^2+11 z^3+5 z^4$，

$T_5(z)=z+15 z^2+43 z^3+45 z^4+16 z^5$，

$T_6(z)=z+31 z^2+148 z^3+268 z^4+211 z^5+61 z^6$，

$T_7(z)=z+63 z^2+480 z^3+1344 z^4+1767 z^5+1113 z^6+272 z^7$．

## 数学代写|组合学代写Combinatorics代考|Alternating Permutations

$$E(z)=\sum_{n \geq 0} E_n \frac{z^n}{n !}=\sec z+\tan z$$

hold住。

$$2 E_{n+1}=\sum_{k=0}^n\left(\begin{array}{l} n \ k \end{array}\right) E_k E_{n-k},$$

$$2 E^{\prime}(z)=E^2(z)+1,$$

## 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。