数值分析实验报告6

1、实验名称 :线性方程组迭代解法1）实验目的：1. 熟悉 matlab 编程。2. 学习线性方程组迭代解法的程序设计算法。2）实验题目：第一题 ：研究解线性方程组ax=b 迭代法收敛速度。a 为 20 阶五对角距阵要求：(1)选取不同的初始向量x0 及右端向量b，给定迭代误差要求，用雅可比迭代和高斯-赛德尔迭代法求解，观察得到的序列是否收敛？若收敛，记录迭代次数， 分析计算结果并得出你的结论。(2)用 sor 迭代法求解上述方程组，松弛系数取 1 =1 for i=1:n m=0; % 此步也可以用ifj=i条件判定一下。for j=1:(i-1) m=m+a(i,j)*x(k,j); endf

2、or j=(i+1):n m=m+a(i,j)*x(k,j); end x(k+1,i)=(b(i)-m)/a(i,i); end l=0; % 判定满足条件使循环停止迭代。for i=1:n l=l+abs(x(k+1,i)-x(k,i); endif l=1 for i=1:n p=0;q=0; for j=1:(i-1) p=p+a(i,j)*x(k+1,j); endfor j=(i+1):n q=q+a(i,j)*x(k,j); end x(k+1,i)=(b(i)-q-p)/a(i,i); end l=0; % 判定满足条件使循环停止迭代。for i=1:n l=l+abs(x(k

3、+1,i)-x(k,i); endif l=1 if w=2|w=1 请重新输入 w的值， w在1与2之间 ; breakendfor i=1:n p=0;q=0; for j=1:(i-1) p=p+a(i,j)*x(k+1,j); endfor j=i:n q=q+a(i,j)*x(k,j); end x(k+1,i)=x(k,i)+w*(b(i)-q-p)/a(i,i); end l=0; % 判定满足条件使循环停止迭代。for i=1:n l=l+abs(x(k+1,i)-x(k,i); endif le breakend k=k+1; end% 输出所有的 x的值。 x(k+1,:)

4、 k 第二题： 代码：雅克比迭代法实现的函数function x = p211_1_jj(n) hn = get_hn(n); b = get_b(n); temp = 0; x0 = zeros(1, n); x_old = zeros(1, n); x_new = zeros(1, n); disp(now jacobi method!); disp(start with the vector that (0, 0, 0, .)t); for i = 1:n for k = 1:n x_old = x_new; temp = 0; for j = 1:n if(j = i) temp =

5、temp + hn(i, j) * x_old(j); end end x_new(i) = (b(i) - temp) / hn(i, i); end end x = x_new; end sor 迭代法实现的函数function x = p211_1_sor(n, w) hn = get_hn(n); b = get_b(n); temp01 = 0; temp02 = 0; x0 = zeros(1, n); x_old = zeros(1, n); x_new = zeros(1, n); disp(now successive over relaxtion method!); dis

6、p(start with the vector that (0, 0, 0, .)t); for i = 1:n for k = 1:n x_old = x_new; temp01 = 0; temp02 = 0; for j = 1:n if(j i) temp02 = temp02 + hn(i, j) * x_old(j); end end end x_new(i) = w * (b(i) - temp01 - temp02) / hn(i, i) + x_old(i); end x = x_new; end 5）实验结果：第一题雅克比（ jacobi）迭代法此时初值全取 1； b=2.

7、2 1.7 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.7 2.2; yakebi(0.00001) ans = columns 1 through 9 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 columns 10 through 18 1.0000 1.0000 1.0000 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 columns 19 through 20 0.9787 0.

8、9793 k = 12 此时初值全取 1； b=2.5 1.9 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.9 2.5; yakebi(0.00001) ans = columns 1 through 12 1.0969 1.0707 1.0219 1.0103 1.0039 1.0016 1.0006 1.0003 1.0001 1.0001 1.0001 1.0001 columns 13 through 20 1.0003 1.0006 1.0016 1.0039 1.0103 1.0219 1.07

9、07 1.0969 k = 14 高斯赛得尔迭代法此时初值全取 1； b=2.2 1.7 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.7 2.2; gaoshisaideer(0.00001) ans = columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 columns 13 through 20 0.9999 0.9998 0.9995 0.998

10、9 0.9970 0.9941 0.9787 0.9793 k = 14 此时初值全取 1； b=2.5 1.9 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.9 2.5; gaoshisaideer(0.00001) ans = columns 1 through 12 1.0969 1.0707 1.0219 1.0103 1.0039 1.0016 1.0006 1.0003 1.0001 1.0001 1.0001 1.0001 columns 13 through 20 1.0003 1.0006

11、1.0016 1.0039 1.0103 1.0219 1.0707 1.0969 k = 14 超松驰法 caosongci(0.00001,1.5) ans = columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793 k = 25 caosongci(0.00001,1.4) ans

12、= columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793 k = 19 caosongci(0.00001,1.3) ans = columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.

13、0000 1.0000 1.0000 1.0000 columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793 k = 15 caosongci(0.00001,1.6) ans = columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.997

14、0 0.9941 0.9787 0.9793 k = 34 caosongci(0.00001,1.7) ans = columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793 k = 47 caosongci(0.00001,1.9) ans = columns 1 through 12 0

15、.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793 k = 150 第二题：对于雅克比迭代法，通过执行以下代码 p211_1_jj(6) p211_1_jj(8) p211_1_jj(10) 可以分别得到：now jacobi method! start with the vector that (0, 0, 0, .)t an

16、s = 2.4500 1.1036 0.6265 0.4060 0.2831 0.2071 now jacobi method! start with the vector that (0, 0, 0, .)t ans = 2.7179 1.4101 0.8524 0.5809 0.4221 0.3198 0.2497 0.1995 now jacobi method! start with the vector that (0, 0, 0, .)t ans = columns 1 through 9 2.9290 1.6662 1.0517 0.7423 0.5554 0.4315 0.34

17、45 0.2807 0.2325 column 10 0.1951 对于 sor 迭代法，执行相对应代码n=6, =1,1.25,1.5的时候 p211_1_sor(6, 1) now successive over relaxtion method! start with the vector that (0, 0, 0, .)t ans = 2.4500 1.1036 0.6265 0.4060 0.2831 0.2071 p211_1_sor(6, 1.25) now successive over relaxtion method! start with the vector that

18、 (0, 0, 0, .)t ans = 3.0625 0.2310 0.8704 0.3389 0.3141 0.2097 p211_1_sor(6, 1.5) now successive over relaxtion method! start with the vector that (0, 0, 0, .)t ans = 3.6750 -1.1009 2.0106 -0.3994 0.7670 -0.0384 与n=8, =1,1.25,1.5的时候 p211_1_sor(8, 1) now successive over relaxtion method! start with t

19、he vector that (0, 0, 0, .)t ans = 2.7179 1.4101 0.8524 0.5809 0.4221 0.3198 0.2497 0.1995 p211_1_sor(8, 1.25) now successive over relaxtion method! start with the vector that (0, 0, 0, .)t ans = 3.3973 0.4887 1.0898 0.5062 0.4501 0.3203 0.2573 0.2042 p211_1_sor(8, 1.5) now successive over relaxtion method! start with the vector that (0, 0, 0, .)t ans = 4.0768 -0.9424 2.2923 -0.2753 0.9252 0.0578 0.4071 0.1275 与n=10, =1,1.25,1.5的时候 p211_1_sor(10, 1) now successive over relaxtion method! start with the vector that (0, 0, 0, .)t ans = columns 1 through 9 2.9290 1.6662 1