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1、数值分析实习报告- 1 -数值分析实习报告上机实习题一一、题目： b与已知A12.38412,2.115237,-1.061074,1.112336,-0.113584,0.718719,1.742382,3.067813,-2.0317432.115237,19.141823,-3.125432,-1.012345,2.189736,1.563849,-0.784165,1.112348,3.123124-1.061074,-3.125432,15.567914,3.123848,2.031454,1.836742,-1.056781,0.336993,-1.0101 031.112336,
2、-1.012345,3.123848,27.108437,4.101011,-3.741856,2.101023,-0.71828,-0.037585A=-0.113584,2.189736,2.031454,4.101011,19.897918,0.431637,-3.111223,2.121314,1.7841370.718719,1.563849,1.836742,-3.741856,0.431637,9.789365,-0.103458,-1.103456,0.2384171.742382,-0.784165,-1.056781,2.101023,-3.111223,-0.103458
3、,14.7138465,3.123789,-2.2134743.067813,1.112348,0.336993,-0.71828,2.121314,-1.103456,3.123789,30.719334,4.446782-2.031743,3.123124,-1.010103,-0.037585,1.784317,0.238417,-2.213474,4.446782,40.00001b=2.1874369,33.992318,-25.173417,0.84671695,1.784317,-86.612343,1.1101230,4.719345,-5.67843921用 Househol
4、d 变换，把A 化为三对角阵B （并打印(0)=0, 迭代9 次） =1.4,X 。 2用超松弛法求解BX=bB ）。（取松弛因子3用列主元素消去法求解BX=b 。二、解题方法的理论依据:1 、用 Householder变换的理论依据 1令 A0=A,a(ij)1=a(ij),已知 Ar_1 即 Ar_1=a(ij)r 2 Sr=sqrt(pow(a,2) 3 a(r)=Sr*Sr+abs(a(r+1,r)*Sr 4 y(r)=A(r_1)*u?/a? 5 Kr=(/2)*Ur的转置 *Yr/a? 6 Qr=Yr-Kr*Ur 7 Ar=A(r-1)-(Qr*Ur的转置 +Ur*Qr 的转置 )
5、 r=1,2, ,n-22 、用超松弛法求解(m)(m-1)其基本思想：在高斯方法已求出x,x的基础上，组合新的序列，从而加快收敛速度。其算式：Bi i 1 X i 1Bi i X iBi i1 X i 1 biBi i 1B i i 1b i X iX i 1X 0i1Bi iBi iB i i X i wX i X 0iX 0i X i 其中是超松弛因子，当1 时，可以加快收敛速度3 、用消去法求解用追赶消去法求Bx=b 的方法：d1i 1 bi a1i 2 Bi 1 i ,- 2 -数值分析实习报告1i i1i1 Bib 1 1 Bi i c ,q10=0 , u10=0 ,8 ,i),
6、i 1,2ib1a1i q11q1ici(u1i(d1ia1iu1i)( b1i a1iq1i1),i 1,2,9x9=u19xi q1i x i1 u1i, i 8,7,1三、 1计算程序：#include math.h#include stdio.h#define ge 8void main()int sign(double x);double a9=12.38412,2.115237, -1.061074,1.112336, -0.113584,0.718719,1.742382,3.067813, -2.031743, 2.115237,19.141823, -3.125432, -1
7、.012345,2.189736,1.563849, -0.784165,1.112348,3.123124, -1.061074, -3.125432,15.567914,3.123848,2.031454,1.836742, -1.056781,0.336993, -1.010103, 1.112336, -1.012345,3.123848,27.108437,4.101011, -3.741856,2.101023, -0.71828,-0.037585, -0.113584,2.189736,2.031454,4.101011,19.897918,0.431637, -3.11122
8、3,2.121314,1.784317, 0.718719,1.563849,1.836742, -3.741856,0.431637,9.789365, -0.103458, -1.103456,0.238417, 1.742382, -0.784165, -1.056781,2.101023, -3.111223,-0.103458,14.713846,3.123789, -2.213474, 3.067813,1.112348,0.336993, -0.71828,2.121314, -1.103456,3.123789,30.719334,4.446782, -2.031743,3.1
9、23124, -1.010103, -0.037585,1.784317,0.238417, -2.213474,4.446782,40.00001;double k,h,s,w;int i,j,n,m,g;double u9,x19,y9,q9,b1910,x9;double b9=2.1874369,33.992318, -25.173417,0.84671695,1.784317, -86.612343,1.1101230,4.719345, -5.6784392;for(j=0;j7;+j)/*Household变换*/s=0.0;for(i=j+1;i0)?(s*s+s*aj+1j)
10、:(s*s-s*aj+1j);for(g=0;g9;+g)- 3 -数值分析实习报告if (g=j)ug=0;else if (g=j+1)ug=aj+1j+s*sign(aj+1j);else ug=agj;for(m=0;m9;+m)ym=0;for(n=0;n9;+n)ym=ym+amn*un;ym=ym/h;k=0;for(i=0;i9;+i)k=k+ui*yi;k=0.5*k/h;for(i=0;i9;+i)qi=yi-k*ui;for(n=0;n9;+n)for(m=0;m9;+m)amn=amn-(qm*un+um*qn);printf(Household:n);for(i=0;
11、i9;+i)for(j=0;j9;+j)if (j%9=0);printf(% -9.5f,aij););w=1.4; /* 超松弛法 */ for(i=0;i9;i+)x1i=0;for(i=0;i9;i+)for(j=0;j9;j+)if(i=j)b1ij=0;else b1ij= -aij/aii;for(i=0;i9;i+)b1i9=bi/aii;- 4 -数值分析实习报告for(n=0;n9;n+)for(i=0;i9;i+)s=0;for(j=0;j9;j+)s=s+b1ij*x1j;s=s+b1i9;x1i=x1i*(1-w)+w*s;for(i=0;i9;i+)if (i=5);牰湩晴尨 ?屦椬砬嬱嵩 ?);u0=a00;/* 以下是消去法*/y0=b0;for(i=1;i=0;i - )xi=(yi-aii+1*xi+1)/ui;for(i=0;i=(1e -6)?1: -1);r
