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assignmentutor-lab™ 为您的留学生涯保驾护航 在代写机器学习 machine learning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。我们的专家在代写机器学习 machine learning代写方面经验极为丰富，各种代写机器学习 machine learning相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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

计算机代写|机器学习代写machine learning代考|P A=L U Factorization

The following easy proposition shows that, in principle, $A$ can be premultiplied by some permutation matrix $P$, so that $P A$ can be converted to upper-triangular form without using any pivoting. Permutations are discussed in some detail in Section 30.3, but for now we just need this definition. For the precise connection between the notion of permutation (as discussed in Section 30.3) and permutation matrices, see Problem $8.16$.

Definition 8.3. A permutation matrix is a square matrix that has a single 1 in every row and every column and zeros everywhere else.

It is shown in Section $30.3$ that every permutation matrix is a product of transposition matrices (the $P(i, k) \mathrm{s}$ ), and that $P$ is invertible with inverse $P^{\top}$.

Proposition 8.4. Let $A$ be an invertible $n \times n$-matrix. There is some permutation matrix $P$ so that $(P A)(1: k, 1: k)$ is invertible for $k=1, \ldots, n$.

Proof. The case $n=1$ is trivial, and so is the case $n=2$ (we swap the rows if necessary). If $n \geq 3$, we proceed by induction. Since $A$ is invertible, its columns are linearly independent; in particular, its first $n-1$ columns are also linearly independent. Delete the last column of A. Since the remaining $n-1$ columns are linearly independent, there are also $n-1$ linearly independent rows in the corresponding $n \times(n-1)$ matrix. Thus, there is a permutation of these $n$ rows so that the $(n-1) \times(n-1)$ matrix consisting of the first $n-1$ rows is invertible. But then there is a corresponding permutation matrix $P_1$, so that the first $n-1$ rows and columns of $P_1 A$ form an invertible matrix $A^{\prime}$. Applying the induction hypothesis to the $(n-1) \times(n-1)$ matrix $A^{\prime}$, we see that there some permutation matrix $P_2$ (leaving the $n$th row fixed), so that $\left(P_2 P_1 A\right)(1: k, 1: k)$ is invertible, for $k=1, \ldots, n-1$. Since $A$ is invertible in the first place and $P_1$ and $P_2$ are invertible, $P_1 P_2 A$ is also invertible, and we are done.

计算机代写|机器学习代写machine learning代考|Dealing with Roundoff Errors; Pivoting Strategies

Let us now briefly comment on the choice of a pivot. Although theoretically, any pivot can be chosen, the possibility of roundoff errors implies that it is not a good idea to pick very small pivots. The following example illustrates this point. Consider the linear system
\begin{aligned} 10^{-4} x+y &=1 \ x+y &=2 . \end{aligned}
Since $10^{-4}$ is nonzero, it can be taken as pivot, and we get
$$\begin{array}{rlr} 10^{-4} x+\frac{y}{y} & = & 1 \ \left(1-10^4\right) y & = & 2-10^4 . \end{array}$$
Thus, the exact solution is
$$x=\frac{10^4}{10^4-1}, \quad y=\frac{10^4-2}{10^4-1} .$$
However, if roundoff takes place on the fourth digit, then $10^4-1=9999$ and $10^4-2=9998$ will be rounded off both to 9990 , and then the solution is $x=0$ and $y=1$, very far from the exact solution where $x \approx 1$ and $y \approx 1$. The problem is that we picked a very small pivot. If instead we permute the equations, the pivot is 1 , and after elimination we get the system
$$\begin{array}{ccc} x+2 & = & 2 \ \left(1-10^{-4}\right) y & = & 1-2 \times 10^{-4} . \end{array}$$

This time, $1-10^{-4}=0.9999$ and $1-2 \times 10^{-4}=0.9998$ are rounded off to $0.999$ and the solution is $x=1, y=1$, much closer to the exact solution.

To remedy this problem, one may use the strategy of partial pivoting. This consists of choosing during Step $k(1 \leq k \leq n-1)$ one of the entries $a_{i k}^{(k)}$ such that
$$\left|a_{i k}^{(k)}\right|=\max {k \leq p \leq n}\left|a{p k}^{(k)}\right| .$$
By maximizing the value of the pivot, we avoid dividing by undesirably small pivots.

机器学习代考

计算机代写|机器学习代写machine learning代考|Dealing with Roundoff Errors; Pivoting Strategies

$$10^{-4} x+y=1 x+y \quad=2 .$$

$$10^{-4} x+\frac{y}{y}=1\left(1-10^4\right) y=2-10^4 .$$

$$x=\frac{10^4}{10^4-1}, \quad y=\frac{10^4-2}{10^4-1} .$$

$$x+2=2\left(1-10^{-4}\right) y=1-2 \times 10^{-4} .$$

$$\left|a_{i k}^{(k)}\right|=\max k \leq p \leq n\left|a p k^{(k)}\right| .$$

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。

assignmentutor™您的专属作业导师
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