西南交大数值分析第一次大作业答案

1、数值分析大作业1、证明：1-x-sinx=0在0,1内有一个根，使用二分法求误差不大于0.5*10-4的根要迭代多少次，并输出每一步的迭代解和迭代误差证明：令f(x= 1-x-sinx;f(0=1,f(1=-sin1f(0*f(1<0f(x=1-cosx<0在0,1内恒成立所以1-x-sinx=0在0,1内恒有一个根程序：function chap2bisecta = 0;b = 1;fprintf('n | a | b | c | r n'for k=1:15c = (a+b/2;r=(b-a/2;fa =1-a-sin(a;fb =1-b-sin(b;fc =1

2、-c-sin(c;fprintf('%d | %f | %f | %f n',k,a,b,c,r;if abs(fc<0.5*10(-4 r=c; sprintf('the root is: %d' , r;elseif fa*fc<0 b=c;elseif fb*fc<0 a=c;endendroot = (a+b/2结果：n | a | b | c | r 1 | 0.000000 | 1.000000 | 0.500000 |5.000000e-001 | 2 | 0.500000 | 1.000000 | 0.750000 |2.500

3、000e-001 | 3 | 0.500000 | 0.750000 | 0.625000 |1.250000e-001 | 4 | 0.500000 | 0.625000 | 0.562500 |6.250000e-002 | 5 | 0.500000 | 0.562500 | 0.531250 |3.125000e-002 | 6 | 0.500000 | 0.531250 | 0.515625 |1.562500e-002 | 7 | 0.500000 | 0.515625 | 0.507813 |7.812500e-003 | 8 | 0.507813 | 0.515625 | 0.5

4、11719 |3.906250e-003 | 9 | 0.507813 | 0.511719 | 0.509766 |1.953125e-003 | 10 | 0.509766 | 0.511719 | 0.510742 |9.765625e-004 | 11 | 0.510742 | 0.511719 | 0.511230 |4.882813e-004 | 12 | 0.510742 | 0.511230 | 0.510986 |2.441406e-004 | 13 | 0.510742 | 0.511230 | 0.510986 |2.441406e-004 | 14 | 0.510742

5、 | 0.511230 | 0.510986 |2.441406e-004 | 15 | 0.510742 | 0.511230 | 0.510986 |2.441406e-004 | root =0.510986328125000。由运行结果可知，迭代12次时即可求出解。8、给出方程组(1）、研究SOR方法求解的收敛性；(2）、用Gauss-Seidel法及及=1.25的SOR法求解，取，迭代到(1SOR方法收敛的充分必要条件是，,其中。(21）Gauss-Seidel法程序：functionx,k,index=G_S(A,b,ep,it_maxif nargin<4 it_max=1

6、00;endif nargin<3 ep=1e-5;endn=length(A;k=0;x=zeros(n,1;y=zeros(n,1;index=1;while 1y=x;for i=1:nz=b(i;for j=1:nif j=iz=z-A(i,j*x(j;endendif abs(A(i,i<1e-10|k=it_maxindex=0;return;endz=z/A(i,i;x(i=z;endif norm(y-x,inf break;endk=k+1endA=4 3 0;3 4 -1;0 -1 4;b=24 30 -24'ep=1e-5x,k,index=sor(A

7、,b,ep,100运行结果：x =3.0000151461293813.999987378225516-5.000003155443621k = 24即迭代了24次，得到方程的解。2）SOR法程序：functionx,k,index=SOR(A,b,ep,w,it_maxif nargin<5it_max=100;endif nargin<4 w=1;endif nargin<3 ep=1e-5;endn=length(A;k=0;x=zeros(n,1;y=zeros(n,1;index=1;while 1y=x;for i=1:nz=b(i;for j=1:nif j=iz=z-A(i,j*x(j;endendif abs(A(i,i<1e-10|k=it_maxindex=0;return;endz=z/A(i,i;x(i=(1-w*x(i+w*z;endif norm(y-x,inf break;endk=k+1endA=4 3 0;3 4 -1;0 -1 4;b=24 30 -24'ep=1e-5x,k,index=s