### 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Jacobi and Gauss–Seidel Iterations

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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

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Jacobi and Gauss–Seidel Iterations

Both the Jacobi and Gauss-Seidel methods are splitting methods and can be used whenever $A$ has nonzero diagonal elements. We write $A$ in the form $A=L+D+U$, where $L$ is the subdiagonal (or strictly lower triangular), $D$ is the diagonal, and $U$ is the superdiagonal (or strictly upper triangular) portion of $A$.
Jacobi method
We choose $A-B=D$, the diagonal part of $A$, or in other words we let $B=L+U$. The iteration step is given by
$$D \mathbf{x}^{(k+1)}=-(L+U) \mathbf{x}^{(k)}+\mathbf{b}$$

Gauss-Seidel method
We set $A-B=L+D$, the lower triangular portion of $\mathrm{A}$, or in other words $B=U$. The sequence $\mathbf{x}^{(k)}, k=1, \ldots$, is generated by
$$(L+D) \mathbf{x}^{(k+1)}=-U \mathbf{x}^{(k)}+\mathbf{b} .$$
Note that there is no need to invert $L+D$; we calculate the components of $x^{(k+1)}$ in sequence using the components we have just calculated by forward substitution:
$$A_{i, i} x_{i}^{(k+1)}=-\sum_{ji} A_{i, j} x_{j}^{(k)}+b_{i}, \quad i=1, \ldots, n .$$
As we have seen in the previous section, the sequence $\mathbf{x}^{(k)}$ converges to the solution if the spectral radius of the iteration matrix, $H_{J}=-D^{-1}(L+U)$ for Jacobi or $H_{G S}=-(L+D)^{-1} U$ for Gauss-Seidel, is less than one. We will show this is true for two important classes of matrices. One is the class of positive definite matrices and the other is given in the following definition.
Definition 2.7 (Strictly diagonally dominant matrices). A matric $A$ is called strictly diagonally dominant by rows if
$$\left|A_{i, i}\right|>\sum_{\substack{j=1 \ j \neq i}}^{n}\left|A_{i, j}\right|$$
for $i, 1, \ldots, n$.
For the first class, the following theorem holds:

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Relaxation

The efficiency of the splitting method can be improved by relaxation. Here, instead of iterating $(A-B) \mathbf{x}^{(k+1)}=-B \mathbf{x}^{(k)}+\mathbf{b}$, we first calculate $(A-$ $B) \tilde{\mathbf{x}}^{(k+1)}=-B \mathbf{x}^{(k)}+\mathbf{b}$ as an intermediate value and then let
$$\mathbf{x}^{(k+1)}=\omega \tilde{\mathbf{x}}^{(k+1)}+(1-\omega) \mathbf{x}^{(k)}$$
for $k=0,1, \ldots .$, where $\omega \in \mathbb{R}$ is called the relaxation parameter. Of course $\omega=1$ corresponds to the original method without relaxation. The parameter $\omega$ is chosen such that the spectral radius of the relaxed method is smaller. The smaller the spectral radius, the faster the iteration converges. Letting $\mathbf{c}=(A-B)^{-1} \mathbf{b}$, the relaxation iteration matrix $H_{\omega}$ can then be deduced from
$$\mathbf{x}^{(k+1)}=\omega \tilde{\mathbf{x}}^{(k+1)}+(1-\omega) \mathbf{x}^{(k)}=\omega H \mathbf{x}^{(k)}+(1-\omega) \mathbf{x}^{(k)}+\omega \mathbf{c}$$
as
$$H_{\omega}=\omega H+(1-\omega) I .$$
It follows that the eigenvalues of $H_{\omega}$ and $H$ are related by $\lambda_{\omega}=\omega \lambda+(1-\omega)$. The best choice for $\omega$ would be to minimize $\max \left{\left|\omega \lambda_{i}+(1-\omega)\right|, i=1, \ldots, n\right}$ where $\lambda_{1}, \ldots, \lambda_{n}$ are the eigenvalues of $H$. However, the eigenvalues of $H$ are often unknown, but sometimes there is information (for example, derived from the Gerschgorin theorem), which makes it possible to choose a good if not optimal value for $\omega$. For example, it might be known that all the eigenvalues are real and lie in the interval $[a, b]$, where $-1<a<b<1$. Then the interval containing the eigenvalues of $H_{\omega}$ is given by $[\omega a+(1-\omega), \omega b+(1-\omega)]$. An optimal choice for $\omega$ is the one which centralizes this interval around the origin:
$$-[\omega a+(1-\omega)]=\omega b+(1-\omega) .$$
It follows that
$$\omega=\frac{2}{2-(a+b)} .$$
The eigenvalues of the relaxed iteration matrix lie in the interval $\left[-\frac{b-a}{2-(a+b)}, \frac{b-a}{2-(a+b)}\right]$. Note that if the interval $[a, b]$ is already symmetric about zero, i.e., $a=-b$, then $\omega=1$ and no relaxation is performed. On the other hand consider the case where all eigenvalues lie in a small interval close to 1. More specifically, let $a=1-2 \epsilon$ and $b=1-\epsilon$; then $\omega=\frac{2}{3 \epsilon}$ and the new interval is $\left[-\frac{1}{3}, \frac{1}{3}\right]$.

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Steepest Descent Method

In this section we look at an alternative approach to construct iterative methods to solve $A \mathbf{x}=\mathbf{b}$ in the case where $A$ is symmetric and positive definite. We consider the quadratic function
$$F(\mathbf{x})=\frac{1}{2} \mathbf{x}^{T} A \mathbf{x}-\mathbf{x}^{T} \mathbf{b}, \quad \mathbf{x} \in \mathbb{R}^{n}$$
It is a multivariate function which can be written as
$$F\left(x_{1}, \ldots, x_{n}\right)=\frac{1}{2} \sum_{i, j=1}^{n} A_{i j} x_{i} x_{j}-\sum_{i=1}^{n} b_{i} x_{i}$$
Note that the first sum is a double sum. A multivariate function has an extremum at the point where the derivatives in each of the directions $x_{i}$, $i=1, \ldots, n$ vanish. The vector of derivatives is called the gradient and is denoted $\nabla F(\mathbf{x})$. So the extremum occurs when the gradient vanishes, or in other words when $\mathbf{x}$ satisfies $\nabla F(\mathbf{x})=0$. The derivative of $F(\mathbf{x})$ in the direction of $x_{k}$ is
\begin{aligned} \frac{d}{d x_{k}} F\left(x_{1}, \ldots, x_{n}\right) &=\frac{1}{2}\left(\sum_{i=1}^{n} A_{i k} x_{i}+\sum_{j=1}^{n} A_{k j} x_{j}\right)-b_{k} \ &=\sum_{j=1}^{n} A_{k j} x_{j}-b_{k} \end{aligned}
where we used the symmetry of $A$ in the last step. This is one component of the gradient vector and thus
$$\nabla F(\mathbf{x})=A \mathbf{x}-\mathbf{b}$$

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Jacobi and Gauss–Seidel Iterations

Jacobi 和 Gauss-Seidel 方法都是分裂方法，可以在任何时候使用一种具有非零对角元素。我们写一种在表格中一种=大号+D+在， 在哪里大号是次对角线（或严格的下三角形），D是对角线，并且在是上对角线（或严格上三角形）的部分一种.
Jacobi方法

DX(ķ+1)=−(大号+在)X(ķ)+b

Gauss-Seidel 方法

(大号+D)X(ķ+1)=−在X(ķ)+b.

|一种一世,一世|>∑j=1 j≠一世n|一种一世,j|

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Relaxation

X(ķ+1)=ωX~(ķ+1)+(1−ω)X(ķ)

X(ķ+1)=ωX~(ķ+1)+(1−ω)X(ķ)=ωHX(ķ)+(1−ω)X(ķ)+ωC

Hω=ωH+(1−ω)一世.

−[ω一种+(1−ω)]=ωb+(1−ω).

ω=22−(一种+b).

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Steepest Descent Method

F(X)=12X吨一种X−X吨b,X∈Rn

F(X1,…,Xn)=12∑一世,j=1n一种一世jX一世Xj−∑一世=1nb一世X一世

ddXķF(X1,…,Xn)=12(∑一世=1n一种一世ķX一世+∑j=1n一种ķjXj)−bķ =∑j=1n一种ķjXj−bķ

∇F(X)=一种X−b

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