数学代写|计算线性代数代写Computational Linear Algebra代考|MATHS 7104

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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|计算线性代数代写Computational Linear Algebra代考|Algorithms for Triangular Systems

A nonsingular triangular linear system $\boldsymbol{A x}=\boldsymbol{b}$ is easy to solve. By Lemma $2.5 \boldsymbol{A}$ has nonzero diagonal elements. Consider first the lower triangular case. For $n=3$ the system is
$$\left[\begin{array}{ccc} a_{11} & 0 & 0 \ a_{21} & a_{22} & 0 \ a_{31} & a_{32} & a_{33} \end{array}\right]\left[\begin{array}{l} x_{1} \ x_{2} \ x_{3} \end{array}\right]=\left[\begin{array}{l} b_{1} \ b_{2} \ b_{3} \end{array}\right]$$
From the first equation we find $x_{1}=b_{1} / a_{11}$. Solving the second equation for $x_{2}$ we obtain $x_{2}=\left(b_{2}-a_{21} x_{1}\right) / a_{22}$. Finally the third equation gives $x_{3}=\left(b_{3}-a_{31} x_{1}-\right.$ $\left.a_{32} x_{2}\right) / a_{33}$. This process is known as forward substitution. In general
$$x_{k}=\left(b_{k}-\sum_{j=1}^{k-1} a_{k, j} x_{j}\right) / a_{k k}, \quad k=1,2, \ldots, n .$$
When $A$ is a lower triangular band matrix the number of arithmetic operations necessary to find $\boldsymbol{x}$ can be reduced. Suppose $\boldsymbol{A}$ is a lower triangular $d$-banded, so that $a_{k, j}=0$ for $j \notin\left{l_{k}, l_{k}+1, \ldots, k\right.$ for $k=1,2, \ldots, n$, and where $l_{k}:=\max (1, k-d)$, see Fig. 3.2. For a lower triangular $d$-band matrix the calculation in (3.7) can be simplified as follows
$$x_{k}=\left(b_{k}-\sum_{j=l_{k}}^{k-1} a_{k, j} x_{j}\right) / a_{k k}, \quad k=1,2, \ldots, n .$$
Note that (3.8) reduces to $(3.7)$ if $d=n$. Letting $A\left(k, l_{k}:(k-1)\right) * x\left(l_{k}:(k-1)\right)$ denote the sum $\sum_{j=l_{k}}^{k-1} a_{k j} x_{j}$ we arrive at the following algorithm, where the initial ” $\mathrm{r} “$ in the name signals that this algorithm is row oriented. The algorithm takes a nonsingular lower triangular $d$-banded matrix $A \in \mathbb{C}^{n \times n}$, and $\boldsymbol{b} \in \mathbb{C}^{n}$, as input, and returns an $\boldsymbol{x} \in \mathbb{C}^{n}$ so that $\boldsymbol{A} \boldsymbol{x}=\boldsymbol{b}$. For each $k$ we take the inner product of a part of a row with the already computed unknowns.

数学代写|计算线性代数代写Computational Linear Algebra代考|Counting Operations

It is useful to have a number which indicates the amount of work an algorithm requires. In this book we measure this by estimating the total number of (complex) arithmetic operations. We count both additions, subtractions, multiplications and divisions, but not work on indices. As an example we show that the LU factorization of a full matrix of order $n$ using Gaussian elimination requires exactly
$$N_{L U}:=\frac{2}{3} n^{3}-\frac{1}{2} n^{2}-\frac{1}{6} n$$
operations. Let $M, D, A, S$ be the number of (complex) multiplications, divisions, additions, and subtractions. In (3.2) the multiplications and subtractions occur in the calculation of $a_{i j}^{k+1}=a_{i j}^{(k)}-l_{i k}^{(k)} a_{k j}^{(k)}$ which is carried out $(n-k)^{2}$ times. Moreover,

each calculation involves one subtraction and one multiplication. Thus we find $M+$ $S=2 \sum_{k=1}^{n-1}(n-k)^{2}=2 \sum_{m=1}^{n-1} m^{2}=\frac{2}{3} n(n-1)\left(n-\frac{1}{2}\right)$. For each $k$ there are $n-k$ divisions giving a sum of $\sum_{k=1}^{n-1}(n-k)=\frac{1}{2} n(n-1)$. Since there are no additions we obtain the total
$$M+D+A+S=\frac{2}{3} n(n-1)\left(n-\frac{1}{2}\right)+\frac{1}{2} n(n-1)=N_{L U}$$
given by $(3.9)$
We are only interested in $N_{L U}$ when $n$ is large and for such $n$ the term $\frac{2}{3} n^{3}$ dominates. We therefore regularly ignore lower order terms and use number of operations both for the exact count and for the highest order term. We also say more loosely that the number of operations is $O\left(n^{3}\right)$. We will use the number of operations counted in one of these ways as a measure of the complexity of an algorithm and say that the complexity of LU factorization of a full matrix is $O\left(n^{3}\right)$ or more precisely $\frac{2}{3} n^{3}$.

We will compare the number of arithmetic operations of many algorithms with the number of arithmetic operations of Gaussian elimination and define for $n \in \mathbb{N}$ the number $G_{n}$ as follows:

数学代写|计算线性代数代写Computational Linear Algebra代考|Pivoting

Interchanging two rows (and/or two columns) during Gaussian elimination is known as pivoting. The element which is moved to the diagonal position $(k, k)$ is called the pivot element or pivot for short, and the row containing the pivot is called the pivot row. Gaussian elimination with row pivoting can be described as follows.

1. Choose $r_{k} \geq k$ so that $a_{r_{k}, k}^{(k)} \neq 0$.
2. Interchange rows $r_{k}$ and $k$ of $A^{(k)}$.
3. Eliminate by computing $l_{i k}^{(k)}$ and $a_{i j}^{(k+1)}$ using (3.2).
To show that Gaussian elimination can always be carried to completion by using suitable row interchanges suppose by induction on $k$ that $A^{(k)}$ is nonsingular. Since $A^{(1)}=A$ this holds for $k=1$. By Lemma $2.4$ the lower right diagonal block in $A^{(k)}$ is nonsingular. But then at least one element in the first column of that block must be nonzero and it follows that $r_{k}$ exists so that $a_{r_{k}, k}^{(k)} \neq 0$. But then $A^{(k+1)}$ is nonsingular since it is computed from $A^{(k)}$ using row operations preserving the nonsingularity. We conclude that $A^{(k)}$ is nonsingular for $k=1, \ldots, n$.

数学代写|计算线性代数代写Computational Linear Algebra代考|Algorithms for Triangular Systems

[一个1100 一个21一个220 一个31一个32一个33][X1 X2 X3]=[b1 b2 b3]

Xķ=(bķ−∑j=1ķ−1一个ķ,jXj)/一个ķķ,ķ=1,2,…,n.

数学代写|计算线性代数代写Computational Linear Algebra代考|Counting Operations

ñ大号在:=23n3−12n2−16n

数学代写|计算线性代数代写Computational Linear Algebra代考|Pivoting

1. 选择rķ≥ķ以便一个rķ,ķ(ķ)≠0.
2. 交换行rķ和ķ的一个(ķ).
3. 通过计算消除l一世ķ(ķ)和一个一世j(ķ+1)使用（3.2）。
为了证明高斯消元总是可以通过使用适当的行交换来完成，假设通过归纳ķ那一个(ķ)是非奇异的。自从一个(1)=一个这适用于ķ=1. 引理2.4右下角块一个(ķ)是非奇异的。但是，该块的第一列中至少有一个元素必须是非零的，并且它遵循rķ存在使得一个rķ,ķ(ķ)≠0. 但是之后一个(ķ+1)是非奇异的，因为它是从一个(ķ)使用保留非奇异性的行操作。我们得出结论一个(ķ)是非奇异的ķ=1,…,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。