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## 数学代写数值分析代写Numerical analysis代考|Gauss-Seidel Method and SOR

Closely related to the Jacobi Method is an iteration called the Gauss-Seidel Method. The only difference between Gauss-Seidel and Jacobi is that in the former, the most recently updated values of the unknowns are used at each step, even if the updating occurs in the current step. Returning to Example 2.19, we see that Gauss-Seidel looks like this:
\begin{aligned} & {\left[\begin{array}{l} u_0 \ v_0 \end{array}\right]=\left[\begin{array}{l} 0 \ 0 \end{array}\right]} \ & {\left[\begin{array}{l} u_1 \ v_1 \end{array}\right]=\left[\begin{array}{l} \frac{5-v_0}{3} \ \frac{5-u_1}{2} \end{array}\right]=\left[\begin{array}{c} \frac{5-0}{3} \ \frac{5-5 / 3}{2} \end{array}\right]=\left[\begin{array}{c} \frac{5}{3} \ \frac{5}{3} \end{array}\right]} \ & {\left[\begin{array}{l} u_2 \ v_2 \end{array}\right]=\left[\begin{array}{l} \frac{5-v_1}{3} \ \frac{5-u_2}{2} \end{array}\right]=\left[\begin{array}{c} \frac{5-5 / 3}{3} \ \frac{5-10 / 9}{2} \end{array}\right]=\left[\begin{array}{c} \frac{10}{9} \ \frac{35}{18} \end{array}\right]} \ & {\left[\begin{array}{l} u_3 \ v_3 \end{array}\right]=\left[\begin{array}{l} \frac{5-v_2}{3} \ \frac{5-u_3}{2} \end{array}\right]=\left[\frac{5-35 / 18}{3}\right]=\left[\begin{array}{c} \frac{55}{54} \ \frac{5-55 / 54}{2} \end{array}\right] .} \end{aligned}
Note the difference between Gauss-Seidel and Jacobi: The definition of $v_1$ uses $u_1$, not $u_0$. We see the approach to the solution $[1,2]$ as with the Jacobi Method, but somewhat more accurately at the same number of steps. Gauss-Seidel often converges faster than Jacobi if the method is convergent. Theorem 2.11 verifies that the GaussSeidel Method, like Jacobi, converges to the solution as long as the coefficient matrix is strictly diagonally dominant.

Gauss-Seidel can be written in matrix form and identified as a fixed-point iteration where we isolate the equation $(L+D+U) x=b$ as
$$(L+D) x_{k+1}=-U x_k+b$$

## 数学代写|数值分析代写Numerical analysis代考|Convergence of iterative methods

In this section we prove that the Jacobi and Gauss-Seidel Methods converge for strictly diagonally dominant matrices. This is the content of Theorems 2.10 and 2.11 .
The Jacobi Method is written as
$$x_{k+1}=-D^{-1}(L+U) x_k+D^{-1} b$$
Theorem A.7 of Appendix A governs convergence of such an iteration. According to this theorem, we need to know that the spectral radius $\rho\left(D^{-1}(L+U)\right)<1$ in order to guarantee convergence of the Jacobi Method. This is exactly what strict diagonal dominance implies, as shown next.

Proof of Theorem 2.10. Let $R=L+U$ denote the nondiagonal part of the matrix. To check $\rho\left(D^{-1} R\right)<1$, let $\lambda$ be an eigenvalue of $D^{-1} R$ with corresponding eigenvector $v$. Choose this $v$ so that $|v|_{\infty}=1$, so that for some $1 \leq m \leq n$, the component $v_m=1$ and all other components are no larger than 1 . (This can be achieved by starting with any eigenvector and dividing by the largest component. Any constant multiple of an eigenvector is again an eigenvector with the same eigenvalue.) The definition of eigenvalue means that $D^{-1} R v=\lambda v$, or $R v=\lambda D v$.

Since $r_{m m}=0$, taking absolute values of the $m$ th component of this vector equation implies
\begin{aligned} & \left|r_{m 1} v_1+r_{m 2} v_2+\cdots+r_{m, m-1} v_{m-1}+r_{m, m+1} v_{m+1}+\cdots+r_{m n} v_n\right| \ & \quad=\left|\lambda d_{m m} v_m\right|=|\lambda|\left|d_{m m}\right| . \end{aligned}
Since all $\left|v_i\right| \leq 1$, the left-hand side is at most $\sum_{j \neq m}\left|r_{m j}\right|$, which, according to the strict diagonal dominance hypothesis, is less than $\left|d_{m m}\right|$. This implies that $|\lambda|\left|d_{m m}\right|<$ $\left|d_{m m}\right|$, which in turn forces $|\lambda|<1$. Since $\lambda$ was an arbitrary eigenvalue, we have shown $\rho\left(D^{-1} R\right)<1$, as desired. Now Theorem A.7 from Appendix A implies that Jacobi converges to a solution of $A x=b$. Finally, since $A x=b$ has a solution for arbitrary $b$, $A$ is a nonsingular matrix.
Putting the Gauss-Seidel Method into the form of (2.43) yields
$$x_{k+1}=-(L+D)^{-1} U x_k+(L+D)^{-1} b$$

## 数学代写数值分析代写Numerical analysis代考|Gauss-Seidel Method and SOR

$$\left[\begin{array}{ll} u_0 & v_0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \end{array}\right] \quad\left[u_1 v_1\right]=\left[\frac{5-v_0}{3} \frac{5-u_1}{2}\right]=\left[\frac{5-0}{3} \frac{5-5 / 3}{2}\right]=\left[\frac{5}{3} \frac{5}{3}\right]\left[u_2 v_2\right]=\left[\frac{5-v_1}{3} \frac{5-u_2}{2}\right]=\left[\frac{5-5 / 3}{3} \frac{5-10 / 9}{2}\right]=\left[\frac{10}{9} \frac{35}{18}\right]$$

Gauss-Seidel 可以写成矩阵形式并确定为定点迭代，我们在其中分离方程 $(L+D+U) x=b$ 作为
$$(L+D) x_{k+1}=-U x_k+b$$

## 数学代写|数值分析代写Numerical analysis代考|Convergence of iterative methods

$$x_{k+1}=-D^{-1}(L+U) x_k+D^{-1} b$$

$$\left|r_{m 1} v_1+r_{m 2} v_2+\cdots+r_{m, m-1} v_{m-1}+r_{m, m+1} v_{m+1}+\cdots+r_{m n} v_n\right| \quad=\left|\lambda d_{m m} v_m\right|=|\lambda|\left|d_{m m}\right| .$$

$$x_{k+1}=-(L+D)^{-1} U x_k+(L+D)^{-1} b$$

## 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。

Posted on Categories:LU分解代写, Numerical analysis, 多项式插值方法代写, 数值分析, 数值积分代写, 数学代写, 最小二乘法代写

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## 数学代写数值分析代写Numerical analysis代考|Partial pivoting

At the start of classical Gaussian elimination of $n$ equations in $n$ unknowns, the first step is to use the diagonal element $a_{11}$ as a pivot to eliminate the first column. The partial pivoting protocol consists of comparing numbers before carrying out each elimination step. The largest entry of the first column is located, and its row is swapped with the pivot row, in this case the top row.

In other words, at the start of Gaussian elimination, partial pivoting asks that we select the $p$ th row, where
$$\left|a_{p 1}\right| \geq\left|a_{i 1}\right|$$
for all $1 \leq i \leq n$, and exchange rows 1 and $p$. Next, elimination of column 1 proceeds as usual, using the “new” version of $a_{11}$ as the pivot. The multiplier used to eliminate $a_{i 1}$ will be
$$m_{i 1}=\frac{a_{i 1}}{a_{11}}$$
and $\left|m_{i 1}\right| \leq 1$
The same check is applied to every choice of pivot during the algorithm. When deciding on the second pivot, we start with the current $a_{22}$ and check all entries directly below. We select the row $p$ such that
$$\left|a_{p 2}\right| \geq\left|a_{i 2}\right|$$
for all $2 \leq i \leq n$, and if $p \neq 2$, rows 2 and $p$ are exchanged. Row 1 is never involved in this step. If $\left|a_{22}\right|$ is already the largest, no row exchange is made.

## 数学代写|数值分析代写Numerical analysis代考|Permutation matrices

Before showing how row exchanges can be used with the $L U$ factorization approach to Gaussian elimination, we will discuss the fundamental properties of permutation matrices.

A permutation matrix is an $n \times n$ matrix consisting of all zeros, except for a single 1 in every row and column.

Equivalently, a permutation matrix $P$ is created by applying arbitrary row exchanges to the $n \times n$ identity matrix (or arbitrary column exchanges). For example,
$$\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \ 1 & 0 \end{array}\right]$$
are the only $2 \times 2$ permutation matrices, and
\begin{aligned} & {\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right],\left[\begin{array}{lll} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{array}\right],\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{array}\right],} \ & {\left[\begin{array}{lll} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{array}\right],\left[\begin{array}{lll} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{array}\right],\left[\begin{array}{lll} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{array}\right]} \end{aligned}
are the $\operatorname{six} 3 \times 3$ permutation matrices.
The next theorem tells us at a glance what action a permutation matrix causes when multiplied on the left of another matrix.

## 数学代写数值分析代写Numerical analysis代考|Partial pivoting

$$\left|a_{p 1}\right| \geq\left|a_{i 1}\right|$$

$$m_{i 1}=\frac{a_{i 1}}{a_{11}}$$

$$\left|a_{p 2}\right| \geq\left|a_{i 2}\right|$$

## 数学代写|数值分析代写Numerical analysis代考|Permutation matrices

$$\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \ 1 & 0 \end{array}\right]$$
are the only $2 \times 2$ permutation matrices, and
\begin{aligned} & {\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right],\left[\begin{array}{lll} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{array}\right],\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{array}\right],} \ & {\left[\begin{array}{lll} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{array}\right],\left[\begin{array}{lll} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{array}\right],\left[\begin{array}{lll} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{array}\right]} \end{aligned}

## 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。

Posted on Categories:LU分解代写, Numerical analysis, 多项式插值方法代写, 数值分析, 数值积分代写, 数学代写, 最小二乘法代写

## avatest™帮您通过考试

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## 数学代写数值分析代写Numerical analysis代考|Operation count for the elimination step of Gaussian elimination

The elimination step for a system of $n$ equations in $n$ variables can be completed in $\frac{2}{3} n^3$ $+\frac{1}{2} n^2-\frac{7}{6} n$ operations.

Normally, the exact operation count is less important than order-of-magnitude estimates, since the details of implementation on various computer processors differ. The main point is that the number of operations is approximately proportional to the execution time of the algorithm. We will commonly make the approximation of $\frac{2}{3} n^3$ operations for elimination, which is a reasonably accurate approximation when $n$ is large.
After the elimination is completed, the tableau is upper triangular:
$$\left[\begin{array}{cccc:c} a_{11} & a_{12} & \ldots & a_{1 n} & b_1 \ 0 & a_{22} & \ldots & a_{2 n} & b_2 \ \vdots & \vdots & \ddots & \vdots & \vdots \ 0 & 0 & \ldots & a_{n n} & b_n \end{array}\right]$$
In equation form,
\begin{aligned} a_{11} x_1+a_{12} x_2+\cdots+a_{1 n} x_n & =b_1 \ a_{22} x_2+\cdots+a_{2 n} x_n & =b_2 \ & \vdots \ a_{n n} x_n & =b_n, \end{aligned}
where, again, the $a_{i j}$ refer to the revised, not original, entries. To complete the computation of the solution $x$, we must carry out the back-substitution step, which is simply a rewriting of $(2.8)$ :
\begin{aligned} x_1 & =\frac{b_1-a_{12} x_2-\cdots-a_{1 n} x_n}{a_{11}} \ x_2 & =\frac{b_2-a_{23} x_3-\cdots-a_{2 n} x_n}{a_{22}} \ & \vdots \ x_n & =\frac{b_n}{a_{n n}} . \end{aligned}

## 数学代写|数值分析代写Numerical analysis代考|Operation count for the back-substitution step of Gaussian elimination

The back-substitution step for a triangular system of $n$ equations in $n$ variables can be completed in $n^2$ operations.

The two operation counts, taken together, show that Gaussian elimination is made up of two unequal parts: the relatively expensive elimination step and the relatively cheap back-substitution step. If we ignore the lower order terms in the expressions for the number of multiplication/divisions, we find that elimination takes on the order of $2 n^3 / 3$ operations and that back substitution takes on the order of $n^2$.

We will often use the shorthand terminology of “big-O” to mean “on the order of,” saying that elimination is an $O\left(n^3\right)$ algorithm and that back substitution is $O\left(n^2\right)$. This usage implies that the emphasis is on large $n$, where lower powers of $n$ become negligible by comparison. For example, if $n=100$, only about 1 percent or so of the calculation time of Gaussian elimination goes into the back-substitution step. Overall, Gaussian elimination takes $2 n^3 / 3+n^2 \approx 2 n^3 / 3$ operations. In other words, for large $n$, the lower order terms in the complexity count will not have a large effect on the estimate for running time of the algorithm and can be ignored if only an estimated time is required.

Estimate the time required to carry out back substitution on a system of 500 equations in 500 unknowns, on a computer where elimination takes 1 second.

Since we have just established that elimination is far more time consuming than back substitution, the answer will be a fraction of a second. Using the approximate number $2(500)^3 / 3$ for the number of multiply/divide operations for the elimination step, and $(500)^2$ for the back-substitution step, we estimate the time for back substitution to be
$$\frac{(500)^2}{2(500)^3 / 3}=\frac{3}{2(500)}=0.003 \mathrm{sec} .$$

## 数学代写数值分析代写Numerical analysis代考|Operation count for the elimination step of Gaussian elimination

$$a_{11} x_1+a_{12} x_2+\cdots+a_{1 n} x_n=b_1 a_{22} x_2+\cdots+a_{2 n} x_n \quad=b_2 \vdots a_{n n} x_n \quad=b_n,$$

$$x_1=\frac{b_1-a_{12} x_2-\cdots-a_{1 n} x_n}{a_{11}} x_2=\frac{b_2-a_{23} x_3-\cdots-a_{2 n} x_n}{a_{22}} \vdots x_n \quad=\frac{b_n}{a_{n n}} .$$

## 数学代写|数值分析代写Numerical analysis代考|Operation count for the back-substitution step of Gaussian elimination

$$\frac{(500)^2}{2(500)^3 / 3}=\frac{3}{2(500)}=0.003 \mathrm{sec}$$

## 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。