Matlab solves linear equations (2) steepest descent method

1. Algorithm principle

 From some initial point $X^{\left(0\right)}$ , along$f\left(X\right)$ at point$X^{}$Negative gradient direction at${}^{(}$${}^0$${}^{)}$

$r^{(0)}=-\nabla f(X^{(0)})=b-A{X}^{\left(0\right)}$

Find f ( X ) f(X)The minimum point of$f$$($$X$$)$$X^{\left(1\right)}$ , then from$X^{\left(1\right)}$ start, repeat the above process to get$X^{\left(2\right)}$ . Going on like this, get the sequence$X^{(k)}$。

 It can be proved that from any initial point $X^{Starting\; from\; (0)}$ , the sequence X ( k ) { {X}^{(k)}}obtained by the steepest descent method$X^{\left(k\right)}$ all converge to make$X$ minimizes$The\; solution\; of\; f\left(X\right)$ , that is, the equation$AX=b$ solution. The iteration format of the steepest descent method is: given initial value$X^{(0)}$，$X^{\left(k\right)}$ is determined as follows

$r^{(k)}=-\nabla f(X^{(k)})=b-A{X}^{\left(k\right)}$ $l_k=\frac{<r^{(k)T},r^{(k)}>}{<r^{(k)T},A{r}^{(k)}>}$ $X^{(k+1)}=X^{(k)}+\lambda r^{\left(k\right)}$

Second, the source program

The source program of the steepest descent method:

%最速下降法
%     'A';系数矩阵
%     'b':右端项
%     'e0':求解精度
%     'x';方程的解
%     'k':迭代次数
%     'tol':总误差
function [x,k,tol]=fastest(A,b,e0)
x0=zeros(length(b),1);
x=x0;
k=0;
i=1;
r=b-A*x0;
while norm(r)>=e0 
r=b-A*x0;
q=dot(r,r)/dot(A*r,r);
x=x0+q*r;
k=k+1;
tol(i)=norm(x-x0);%误差
x0=x;
i=i+1;
    end
end  

Three, case analysis

Also for the calculation example P113 in the textbook, solve the linear equation system Ax=b, and

enter the following code: (the source code of the build function is in the first article of this series)

clc;
clear;
n=input('Please input n:');	%输入矩阵A的阶数
e0=input('Please input the accuracy:');
[A,b]=build(n);			%建立课本例3.2方程组系数矩阵及右端项
[x,k,error]=fastest(A,b,e0); 
q=log(error);
plot(q,'linewidth',1.5)
xlabel('迭代次数');
ylabel('log(error)');
title('最速下降法迭代误差变化曲线');

When n=100, when the accuracy is 1*10 to the -6th power, it takes 19465 iterations to meet the accuracy requirement