線性方程組的迭代解法——最速下降法

　　1.代碼

%%最速下降法（用於求解正定對稱方程組）
%%線性方程組M*X = b，M是方陣，X0是初始解向量,epsilon是控制精度
function TSDM = The_steepest_descent_method(M,b,X0,epsilon)
m = size(M);up = 1000;e = floor(abs(log(epsilon)));
X(:,1) = X0;
r(:,1) = b-M*X0;
for k = 1:up
    alpha = Inner_product(r(:,k),r(:,k))/Inner_product(M*r(:,k),r(:,k));
    X(:,k+1) = X(:,k)+alpha*r(:,k);
    r(:,k+1) = b-M*X(:,k+1);
    X_delta(:,k) = X(:,k+1)-X(:,k);
    if sqrt(Inner_product(X_delta(:,k),M*X_delta(:,k))) < epsilon
        break;
    end
end
disp('迭代次數為：');
k-1
TSDM = vpa(X(:,k),e);
    %%內積
    function IP = Inner_product(M1,M2)
        MAX = max(size(M1));
        sum = 0;
        for i = 1:MAX
            sum = sum+M1(i)*M2(i);
        end
        IP = sum;
    end
end

　　2.例子

clear all
clc
for i = 1:4
    for j = 1:4
        if i == j
            M(i,j) = 2.1;
        else 
            M(i,j) = 1.5;
        end
    end
end
b = [1 2 3 4]';
X0 = [1 1 1 1]';
epsilon = 1e-4;

S = The_steepest_descent_method(M,b,X0,epsilon)

M\b

　　結果為

迭代次數為：
ans =
    21
S =
  -2.12110743
 -0.454511872
   1.21208369
   2.87867925
ans =
   -2.1212
   -0.4545
    1.2121
    2.8788
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