数值分析编程题c语言概要.doc

上机实习题一 一、题目： 已知A与b 12.38412,2.115237,-1.061074,1.112336,-0.113584,0.718719,1.742382,3.067813,-2.031743 2.115237,19.141823,-3.125432,-1.012345,2.189736,1.563849,-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 A= -0.113584,2.189736,2.031454,4.101011,19.897918,0.431637,-3.111223,2.121314,1.784137 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,-0.103458,14.7138465,3.123789,-2.213474 3.067813,1.112348,0.336993,-0.71828,2.121314,-1.103456,3.123789,30.719334,4.446782 -2.031743,3.123124,-1.010103,-0.037585,1.784317,0.238417,-2.213474,4.446782,40.00001 b={2.1874369,33.992318,-25.173417,01.784317,-86.612343,1.1101230,4.719345,-5.6784392} 1．用Household变换，把A化为三对角阵B（并打印B）。 2．用超松弛法求解BX=b（取松弛因子ω=1.4,X(0)=0,迭代9次）。 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) r=1,2,,……,n-2 2 、用超松弛法求解 其基本思想：在高斯方法已求出x(m),x(m-1)的基础上，组合新的序列，从而加快收敛速度。 其算式： 其中ω是超松弛因子，当ω1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： , , , , q1[0]=0 , u1[0]=0 , x[9]=u1[9] 三、1．计算程序： #include math.h #include stdio.h #define ge 8 void main() { int sign(double x); double a[][9]= { {12.38412,2.115237,-1.061074,1.112336,-0.113584,0.718719,1.742382,3.067813,-2.031743}, { 2.115237,19.141823,-3.125432,-1.012345,2.189736,1.563849,-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.