北师大数值分析作业

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数值分析大作业

数值分析上机作业（一）一、算法的设计方案1、幂法求解λ1、λ501幂法主要用于计算矩阵的按模最大的特征值和相应的特征向量，即对于|λ1|≥|λ2|≥.....≥|λn|可以采用幂法直接求出λ1，但在本题中λ1≤λ2≤……≤λ501，我们无法判断按模最大的特征值。

但是由矩阵A的特征值条件可知|λ1|和|λ501|之间必然有一个是最大的，通过对矩阵A使用幂法迭代一定次数后得到满足精度ε=10−12的特征值λ0，然后在对矩阵A做如下的平移：B=A-λ0I由线性代数(A-PI)x=(λ-p)x可得矩阵B的特征值为：λ1-λ0、λ2-λ0…….λ501-λ0。

对B矩阵采用幂法求出B矩阵按模最大的特征值为λ∗=λ501-λ0，所以λ501=λ∗+λ0，比较λ0与λ501的大小，若λ0>λ501则λ1=λ501，λ501=λ0；若λ0<λ501，则令t=λ501，λ1=λ0，λ501=t。

求矩阵M按模最大的特征值λ的具体算法如下：任取非零向量u0∈R nηk−1=u T(k−1)∗u k−1y k−1=u k−1ηk−1u k=Ay k−1βk=y Tk−1u k(k=1,2,3……)当|βk−βk−1||βk|≤ε=10−12时，迭终终止，并且令λ1=βk2、反幂法计算λs和λik由已知条件可知λs是矩阵A 按模最小的特征值，可以应用反幂法直接求解出λs。

使用带偏移量的反幂法求解λik，其中偏移量为μk=λ1+kλ501−λ140(k=1,2,3…39)，构造矩阵C=A-μk I，矩阵C的特征值为λik−μk，对矩阵C使用反幂法求得按模最小特征值λ0，则有λik=1λ0+μk。

求解矩阵M按模最小特征值的具体算法如下：任取非零向量u 0∈R n ηk−1= u T (k−1)∗u k−1y k−1=u k−1ηk−1 Au k =y k−1βk =y T k−1u k (k=1,2,3……)在反幂法中每一次迭代都要求解线性方程组Au k =y k−1，当K 足够大时，取λn =1βk 。

数值分析作业(完整版)

的逆阵 A ，用左除命令 A \ E 检验你的结果。
clc clear close all A=[1 1 1 1 1;1 2 3 4 5;1 3 6 10 15;1 4 10 20 35;1 5 15 35 70]; fprintf('对上述矩阵进行列主元素分解：\n') for i=1:1:r-1 [mx,ro]=max(abs(A(i:r,i))); % 寻找a阵第i列的最大值 [A(i,:),A(ro+i-1,:)]=exchange(A(i,:),A(ro+i-1,:)); % 进行行与行交换 for j=i+1:1:r A(j,:)=A(j,:)-A(j,i)/A(i,i)*A(i,:); end A End %--矩阵A的逆阵 A1=inv(A) %--左除验证 E=eye(5); A2=A\E % 5x5单位阵 % A阵的逆矩阵 % 输出每次交换后的A
第一章
1、计算积分 I n
Code： clc clear close all n=9; %--梯形积分法 x=0:0.01:1; y=(x.^n).*exp(x-1); In = trapz(x,y); In2=vpa(In,6) % 6位有效数字 %--高精度积分法 F = @(x1)(x1.^n).*exp(x1-1); s = quad(F,0,1); s1=vpa(s,6)
0
0, 0, 0, 0, 0 。
T
if abs(er(:,i-1))<=e fprintf('在迭代 %d 次之后，满足精度要求，x向量的值如下：\n',i); fprintf('x1=%.5f, x2=%.5f, x3=%.5f, x4=%.5f, x5=%.5f\n',x(1,i),x(2,i),x(3,i),x(4,i),x(5,i)); break end end %--绘图 figure(1) plot(1:1:i,x(1,:),'b',1:1:i,x(2,:),'k',1:1:i,x(3,:),'g',1:1:i,x(4,:), 'r',1:1:i,x(5,:),'c') legend('x1','x2','x3','x4','x5') grid on title('Jacobi迭代法——x值随迭代次数变化曲线') figure(2) plot(1:1:i-1,er(1,:),'b',1:1:i-1,er(2,:),'k',1:1:i-1,er(3,:),'g',1:1: i-1,er(4,:),'r',1:1:i-1,er(5,:),'c') legend('△x1','△x2','△x3','△x4','△x5') grid on title('Jacobi迭代法——△x值随迭代次数变化曲线') %% fprintf('\n-------------Gauss-Seidel迭代法---------------------\n'); U=-(A1-D); L=-(A2-D); DL_1=inv(D-L); M1=DL_1*U; b2=DL_1*b; x1(:,1)=M1*x0+b2; for j=2:1:100 x1(:,j)=M1*x1(:,j-1)+b2; er1(:,j-1)=x1(:,j)-x1(:,j-1); if abs(er1(:,j-1))<=e fprintf('在迭代 %d 次之后，满足精度要求，x向量的值如下：\n',j); fprintf('x1=%.5f, x2=%.5f, x3=%.5f, x4=%.5f, x5=%.5f\n',x1(1,j),x1(2,j),x1(3,j),x1(4,j),x1(5,j)); break end end %--绘图 figure(3) plot(1:1:j,x1(1,:),'b',1:1:j,x1(2,:),'k',1:1:j,x1(3,:),'g',1:1:j,x1(4 ,:),'r',1:1:j,x1(5,:),'c') legend('x1','x2','x3','x4','x5')

数学八年级上册第6章数据的分析作业课件北师大版

原价(元)
10 10 15 20 25
现价(元)
5 5 15 25 30
平均日人数(千人) 1 1 2 3 2
(1)该风景区称调整前后这5个景点门票的平均收费不变，平均日总收入持平，风景区是怎样计算的？
(2)另一方面，游客认为调整收费后风景区的平均日总收入相对于调价前，实际上增加了约9.4%，问游客是怎样计算的？
第将一组数据按照由小到大(或由大到小)的顺序排列．如果数据的个数是奇数，则处于 _______ 位置中的间数就是这组数据的中位数；如果数据的个数是偶数，则
____________中__间__两__个__数__据就的是平这均组数据的中位数．
数是6，则这组数据的中位数是( )
C
A．5 B．5.5 C．6 D．7
12．有5个从小到大排列的正整数，中位数是3，唯一的众数是8，则这5个数的和为
____．
22
13．某校九(1)班40名同学中，14岁的有1人，15岁的有211人5 ，16岁的有16人，17 岁的有2人，则这个班同学年龄的中位数是____岁．
第六章数据的分析
6．1 平均数
1．一般地，对于 n 个数 x1，x2，…xn，我们把n1(x1＋x2＋…＋xn) 叫做这 n 个数的_算__术_平__均__数_____，记作__x____．
练习 1：(2016·桂林)数据 7，8，10，12，13 的平均数是(C )
A．7
B．9
C．10
D．12
3．一组数据2，4，x，－1的平均数为3，则x的值是____. 7
4．(2016·深圳)已知一组数据x1，x2，x3，x4的平均数是5，则数据x1＋3，x2＋3，

数值分析练习题加答案(一)

数值分析期末考试一、设80~=x ，若要确保其近似数的相对误差限为0.1%，则它的近似数x 至少取几位有效数字？（4分）解：设x 有n 位有效数字。

因为98180648=<<=，所以可得x 的第一位有效数字为8（1分）又因为21101011000110821--⨯=<⨯⨯≤n ε，令321=⇒-=-n n ，可知x 至少具有3位有效数字（3分）。

二、求矩阵A 的条件数1)(A Cond （4分）。

其中⎥⎦⎤⎢⎣⎡=4231A 解：⎥⎦⎤⎢⎣⎡--=-5.05.1121A （1分） 1A =7（1分） 2711=-A （1分）249)(1=A Cond （1分）三、用列主元Gauss 消元法法求解以下方程组（6分）942822032321321321=++-=++--=+-x x x x x x x x x解：→⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---→⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡----→⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡----5.245.2405.35.230914220321821191429142821120321 ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---→⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---8175835005,245.24091425.33.2305.245.2409142(4分) 等价三角方程组为：⎪⎪⎩⎪⎪⎨⎧-=-=+-=++,8175835,5.245.24,942332321x x x x x x （1分）回代得1,3,5123==-=x x x （1分）四、设.0,2,3,1,103)(3210234=-===-+-=x x x x x x x x f 1）求以3210,,,x x x x 为节3次Lagrange 多项式；（6分） 2）求以3210,,,x x x x 为节3次Newton 多项式；（6分）3）给出以上插值多项式的插值余项的表达式（3分）解：由0,2,3,13210=-===x x x x 可得10)(,34)(,1)(,11)(3210-==-=-=x f x f x f x f即得： +------+------=))()(())()(()())()(())()(()()(312101320130201032103x x x x x x x x x x x x x f x x x x x x x x x x x x x f x L=------+------))()(())()(()())()(())()(()(23130321033212023102x x x x x x x x x x x x x f x x x x x x x x x x x x x f+-+--+-⨯-+-+--+-⨯-)03)(23)(13()0)(2)(1()1()01)(21)(31()0)(2)(3(11x x x x x x326610.)20)(30)(10()2)(3)(1()10()02)(32)(12()0)(3)(1(34x x x x x x x x x -+--=+--+--⨯-+---------⨯2）计算差商表如下：i x )(i x f 一阶差商二阶差商三阶差商1 -11 3 -1 5 -2 34 -7 4 0-10-225-1则=+-----+-+-=)2)(3)(1()3)(1(4)1(511)(3x x x x x x x N326610x x x -+--3）)2)(3)(1())()()((!4)()(3210)4(3+--=----=x x x x x x x x x x x x f x R ξ五、给定方程组b Ax =，其中⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=100131w w w w A 。

数值分析试题

数值分析考试题一、填空题(每小题3分，共15分) 1.已知x =62.1341是由准确数a 经四舍五入得到的a 的近似值，试给出x 的绝对误差界_______________.2. 已知矩阵1221A ⎡⎤=⎢⎥⎣⎦，则A 的奇异值为 _________. 3. 设x 和y 的相对误差均为0.001，则xy 的相对误差约为____________. 4. 424()53,,()_____.i i f x xx x i f x =+-∆=若=则5. 下面Matlab 程序所描述的数学表达式为________________________.a =[10,3,4,6];t=1/(x -1);n=length(a )();1:1:1*();y a n for k n y t y a k end==--=+二、(10分)设32()()f x x a =-。

（1）写出解()0f x =的Newton 迭代格式；（2）证明此迭代格式是线性收敛的。

三、 (15分)已知矛盾方程组Ax=b ，其中21110,1101211A b ⎡⎤-⎡⎤⎢⎥⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎣⎦-⎢⎥⎣⎦，（1）用Householder 方法求矩阵A 的正交分解，即A=QR 。

（2）用此正交分解求矛盾方程组Ax=b 的最小二乘解。

四、(15分) 给出数据点：012343961215i i x y =⎧⎨=⎩（1）用1234,,,x x x x 构造三次Newton 插值多项式3()N x ，并计算 1.5x =的近似值3(1.5)N 。

（2）用事后误差估计方法估计3(1.5)N 的误差。

五、(15分)（1）设012{(),(),()}ϕϕϕx x x 是定义于[-1，1]上关于权函数2()x x ρ=的首项系数为1的正交多项式组，若已知01()1,()x x x ϕϕ==，试求出2()x ϕ。

（2）利用正交多项式组012{(),(),()}ϕϕϕx x x ，求()f x x =在11[,]22-上的二次最佳平方逼近多项式。

数值分析大作业四

《数值分析》大作业四一、算法设计方案：复化梯形积分法，选取步长为1/500=0.002，迭代误差控制在E ≤1.0e-10①复化梯形积分法：11()[()()2()]2n bak hf x dx f a f b f a kh -=⎰≈+++∑，截断误差为：322()''()''(),[,]1212T b a b a R f h f a b n ηηη--=-=-∈其中。

复化Simpson 积分法，选取步长为1/50=0.02，迭代误差控制在E ≤1.0e-10②Simpson 积分法：121211()[()()4()2()]3m m bi i a i i hf x dx f a f b f x f x --==≈+++∑∑⎰，截断误差为：4(4)(),[,]180s b a R h f a b ηη-=-∈。

③Guass积分法选用Gauss-Legendre 求积公式：111()()ni i i f x dx A f x -=≈∑⎰截断误差为：R= ()()n 2n 422n!2×(2[2!]2n 1f n n ⨯（2）η（））＋ η∈(1,1)。

选择9个节点：-0.9681602395,-0.8360311073,-0.6133714327,-0.3242534234,0,0.3242534234,0.6133714327,0.8360311073,0.9681602395，对应的求积系数依次为：0.0812743884,0.1806481607,0.2606106964,0.3123470770,0.3302393550,0.3123470770,0.2606106964,0.1806481607,0.0812743884。

二、程序源代码：#include<stdio.h>#include<math.h>#include<stdlib.h>#define E 1.0e-10/****定义函数g和K*****/double g(double a){double b;b=exp(4*a)+(exp(a+4)-exp(-a-4))/(a+4);return b;}double K(double a,double b){double c;c=exp(a*b);return c;}/******复化梯形法******/void Tixing( ){double u[1001],x[1001],h,c[1001],e;int i,j,k;FILE *fp;fp=fopen("f:/result0. xls ","w");h=1.0/1500;for(i=0;i<3001;i++){x[i]=i*h-1;u[i]=g(x[i]);}for(k=0;k<100;k++){e=0;for(i=0;i<1001;i++){for(j=1,c[i]=0;j<N-1;j++)c[i]+=K(x[i],x[j])*u[j];u[i]=g(x[i])-h*c[i]-h/2*(K(x[i],x[0])*u[0]+K(x[i],x[N-1])*u[N-1]);e+=h*(exp(4*x[i])-u[i])*(exp(4*x[i])-u[i]);}if(e<=E) break;}for(i=0;i<1001;i++)fprintf(fp,"%.12lf,%.12lf\n",x[i],u[i]);fclose(fp);}/******复化Simpson法******/void simpson( ){double u[101],x[101],h,c[101],d[101],e;int i,j,k;FILE *fp;fp=fopen("f:/result1.xls","w");h=1.0/50;for(i=0;i<101;i++){x[i]=i*h-1;u[i]=g(x[i]);}for(k=0;k<50;k++){e=0;for(i=0;i<101;i++){for(j=1,c[i]=0,d[i]=0;j<51;j++){c[i]+=K(x[i],x[2*j-1])*u[2*j-1];if(j<50)d[i]+=K(x[i],x[2*j])*u[2*j];}u[i]=g(x[i])-4*h/3*c[i]-2*h/3*d[i]-h/3*(K(x[i],x[0])*u[0]+K(x[i],x[M-1])*u[M-1]);e+=h*(exp(4*x[i])-u[i])*(exp(4*x[i])-u[i]);}if(e<=E) break;}for(i=0;i<101;i++)fprintf(fp,"%.12lf,%.12lf\n",x[i],u[i]);fclose(fp);}/******Gauss积分法******/void gauss( ){double x[9]={-0.9681602395,-0.8360311073,-0.6133714327,-0.3242534234,0,\0.3242534234,0.6133714327,0.8360311073,0.9681602395},A[9]={0.0812743884,0.1806481607,0.2606106964,0.3123470770,0.3302393550,\0.3123470770,0.2606106964,0.1806481607,0.0812743884},u[9],c[9],e;int i,j,k;FILE *fp;fp=fopen("f:/result2. xls ","w");for(i=0;i<9;i++)u[i]=g(x[i]);for(k=0;k<50;k++){e=0;for(i=0;i<9;i++){for(j=0,c[i]=0;j<9;j++)c[i]+=A[j]*K(x[i],x[j])*u[j];u[i]=g(x[i])-c[i];e+=A[i]*(exp(4*x[i])-u[i])*(exp(4*x[i])-u[i]);}if(e<=E) break;}for(i=0;i<9;i++)fprintf(fp,"%.12lf,%.12lf\n",x[i],u[i]);fclose(fp);}/******主函数******/main(){Tixing ( );Simpson( );Gauss( );return 0;}三、运算结果复化梯形数据-10.018323-0.920.02523-0.9980.018471-0.9180.025433-0.9960.018619-0.9160.025637-0.9940.018768-0.9140.025843-0.9920.018919-0.9120.026051-0.990.019071-0.910.02626-0.9880.019224-0.9080.026471-0.9860.019378-0.9060.026683-0.9840.019534-0.9040.026897-0.9820.019691-0.9020.027113-0.980.019849-0.90.027331-0.9780.020008-0.8980.02755-0.9760.020169-0.8960.027772-0.9740.020331-0.8940.027995-0.9720.020494-0.8920.028219-0.970.020658-0.890.028446-0.9680.020824-0.8880.028674-0.9660.020992-0.8860.028905-0.9640.02116-0.8840.029137-0.9620.02133-0.8820.029371-0.960.021501-0.880.029607-0.9580.021674-0.8780.029844-0.9560.021848-0.8760.030084-0.9540.022023-0.8740.030326-0.9520.0222-0.8720.030569-0.950.022378-0.870.030815-0.9480.022558-0.8680.031062-0.9460.022739-0.8660.031311-0.9440.022922-0.8640.031563-0.9420.023106-0.8620.031816-0.940.023291-0.860.032072-0.9380.023478-0.8580.032329-0.9360.023667-0.8560.032589-0.9340.023857-0.8540.032851-0.9320.024048-0.8520.033114-0.930.024241-0.850.03338-0.9280.024436-0.8480.033648-0.9260.024632-0.8460.033918-0.9240.02483-0.8440.034191-0.9220.025029-0.8420.034465-0.840.034742-0.760.047841-0.8380.035021-0.7580.048225-0.8360.035302-0.7560.048613 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0.74219.452890.82226.78914 0.74419.609140.82427.00431 0.74619.766640.82627.22121 0.74819.925410.82827.43985 0.7520.085450.8327.66025 0.75220.246780.83227.88242 0.75420.409410.83428.10638 0.75620.573340.83628.33213 0.75820.738580.83828.5597 0.7620.905160.8428.78909 0.76221.073070.84229.02033 0.76421.242330.84429.25342 0.76621.412950.84629.48839 0.76821.584940.84829.72524 0.7721.758310.8529.964 0.77221.933080.85230.20467 0.77422.109250.85430.44728 0.77622.286830.85630.69184 0.77822.465840.85830.93836 0.7822.646290.8631.18686 0.78222.828190.86231.43735 0.78423.011550.86431.68986 0.78623.196380.86631.9444 0.78823.382690.86832.20098 0.7923.570510.8732.45962 0.79223.759830.87232.72034 0.79423.950670.87432.98315 0.79624.143040.87633.24807 0.79824.336960.87833.51513 0.824.532440.8833.78432 0.80224.729490.88234.05568 0.80424.928110.88434.32922 0.80625.128340.88634.60496 0.80825.330170.88834.882910.8935.163090.94643.99154 0.89235.445520.94844.344880.89435.730220.9544.701070.89636.017210.95245.060110.89836.306510.95445.422040.936.598120.95645.786870.90236.892080.95846.154630.90437.188410.9646.525350.90637.487110.96246.899050.90837.788210.96447.275750.9138.091730.96647.655470.91238.397680.96848.038240.91438.70610.9748.424090.91639.016990.97248.813040.91839.330380.97449.205110.9239.646280.97649.600330.92239.964720.97849.998720.92440.285720.9850.400320.92640.60930.98250.805140.92840.935480.98451.213210.9341.264280.98651.624560.93241.595720.98852.039210.93441.929820.9952.45720.93642.26660.99252.878540.93842.606090.99453.303270.9442.948310.99653.73140.94243.293270.99854.162980.94443.64101154.59802复化Simpson数据：-1 0.018319929 -0.34 0.256658088 0.32 3.596641805 -0.98 0.0198445 -0.32 0.278035042 0.34 3.896195298-0.96 0.021494322 -0.3 0.301192133 0.36 4.220697765-0.94 0.023283225 -0.28 0.326278124 0.38 4.572227037-0.92 0.025220379 -0.26 0.353453177 0.4 4.95303418-0.9 0.027320224 -0.24 0.382891765 0.42 5.365557596-0.88 0.029594431 -0.22 0.41478194 0.44 5.812438891-0.86 0.032059069 -0.16 0.527292277 0.54 8.671138204-0.84 0.034728638 -0.14 0.571209036 0.56 9.39333156-0.82 0.037621263 -0.12 0.61878367 0.58 10.17567433-0.8 0.040754615 -0.1 0.670320427 0.6 11.02317608-0.78 0.044149394 -0.08 0.726149698 0.62 11.94126383-0.76 0.047826844 -0.06 0.78662861 0.64 12.93581634-0.74 0.051810827 -0.04 0.85214479 0.66 14.01320231-0.72 0.056126648 -0.02 0.92311742 0.68 15.1803205-0.7 0.060802006 0 1.0000013 0.7 16.44464467 -0.68 0.065866854 0.02 1.083288424 0.72 17.81427057 -0.66 0.071353499 0.04 1.173512427 0.74 19.29796874 -0.64 0.077297255 0.06 1.271250748 0.76 20.90523965 -0.62 0.083735917 0.08 1.377129533 0.78 22.64637562 -0.6 0.090711017 0.1 1.491826493 0.8 24.53252554 -0.58 0.098266855 0.12 1.616076341 0.82 26.57576756 -0.56 0.106452202 0.14 1.750674449 0.84 28.78918506 -0.54 0.11531904 0.16 1.896482943 0.86 31.18695183 -0.52 0.12492459 0.18 2.054435268 0.88 33.78442141 -0.5 0.135329888 0.2 2.225543071 0.9 36.59822683 -0.48 0.14660204 0.22 2.410901825 0.92 39.64638571 -0.46 0.158812728 0.24 2.611698647 0.94 42.94841704 -0.44 0.17204064 0.26 2.829219145 0.96 46.52546475 -0.42 0.18636997 0.28 3.064856356 0.98 50.40043451 -0.4 0.201892977 0.3 3.320119013 1 54.59813904 -0.38 0.218708553 0.46 6.296539601-0.36 0.236924875 0.48 6.820959636-0.2 0.449328351 0.5 7.389057081-0.18 0.486751777 0.52 8.0044696750102030405060四、讨论①在满足相同精度要求的情况下复化梯形积分法比复化Simpson 积分法计算所需节点数多，计算量大。

数值分析大作业之2

数值分析大作业二一、算法设计方案(一)算法流程1.将要求解的矩阵进行Householder变换，进行拟上三角化，并输出拟上三角化的结果；2.将拟上三角化后的矩阵进行带双步位移的QR分解(最大迭代次数1000次)，逐个求出特征值，输出QR法结束后得到的Q、R、RQ阵，输出求出的特征值（用实部和虚部表示）3.对于求出来的实特征值，使用带原点平移的反幂法求出对应的特征向量，输出这些特征向量。

（二）程序设计流程1. 定义精度和最大迭代次数；2. 使用容器vector进行编程（方便增减元素），使用传引用调用（提高执行效率）；3. 将各个步骤用到的数学算法封装成函数，方便调用。

具体需要的函数如下：double VectMod(vector< double > &p)：求向量的模vector< double > NumbMultiVect(vector< double > &vect, double a)：数乘向量double VectMultiVect(vector< double > &y, vector< double > &u)：求两个向量的积vector< double > ConveArraMultiVect(vector< vector< double > > &B, vector<double > &u)：矩阵的转置乘向量vector< double > ArraMultiVect(vector< vector< double > > &B, vector< double > &u)：矩阵乘向量vector< vector< double > > VectMultiCovVect(vector< double > &a, vector< double >&b)：向量乘向量转置得到矩阵void ArraSubs(vector< vector< double > > &C, vector< vector < double > > D) ：两个矩阵相减Vector< double > VectSubs(vector< double > &a, vector< double > &b)：两个向量相减vector<vector<double>> GetM(vector<vector<double>> &A)：求解矩阵M （用于双部位移QR迭代用）double GetMode(vector < vector < double > > &B, const int r)：求解矩阵B的r列向量的模double GetModeH(vector < vector < double > > &B, const int r)：求解矩阵B的第r列向量的模，用于拟对角化vector< vector< double > > NumbMultiArra(vector< vector< double > > &D, double a)：一个实数乘矩阵bool IsBirZeroH(vector< vector< double > > &B, const int r)：判断B[i][r]对角线下是否为零void GausElim(vector< vector< double > > a)：列主元高斯消元法求齐次方程解向量void Stop(vector< vector< double > > &Ar)：停止，结束程序void SolutS1S2(complex< double > &s1, complex< double > &s2, vector< vector<double > > &A)：求解二阶子阵的特征值s1，s2；void Save2(complex< double > &s1, complex< double > &s2)：保存两个特征值void Save1(complex< double > &s)：保存一个特征值void JudgemBelow2(vector< vector< double > > &A, vector< vector< double > > Abk)：对于m == 1 及m == 0 的处理void Hessenberg(vector< vector< double > > &A)：矩阵拟上三角化void QRMethod(vector< vector< double > > A)：矩阵QR分解void CalculatAk(vector< vector < double > > &Ak)：带双步位移QR迭代法二、源程序#include "stdafx.h"#include <vector>#include <iostream>#include <math.h>#include <complex>#include <fstream>using namespace std;const double epsion = 1e-12;const int L = 1000;int m,n;int k = 1;vector< vector< double > > I;vector< complex< double > > Lambda;///////////////////////////以下为自定义的算法流程中用到的函数double VectMod(vector< double > &p) //求向量的模{double value = 0.0;vector< double >::size_type i,j;j = p.size();for (i=1; i<j; i++){value += p[i] * p[i];}value = sqrt(value);return value;}vector< double > NumbMultiVect(vector< double > &vect, double a) //数乘向量{int j = vect.size();vector< double > b(j, 0);for (int i=1; i<j; i++){b[i] = a * vect[i];}return b;}double VectMultiVect(vector< double > &y, vector< double > &u)//两个向量相乘{vector< double >::size_type a = y.size();double value = 0;for (vector< double >::size_type i=1; i<a; i++){value += y[i] * u[i];}return value;}vector< double > ConveArraMultiVect(vector< vector< double > > &B, vector< double > &u)//矩阵的转置乘向量{int a = B.size();int b = u.size();vector< double > vec(a, 0);if (a != b){cerr << "Array and Vector not match in size!";}else{for (int i=1; i<a; i++){for (int j=1; j<a; j++){vec[i] += B[j][i] * u[j];}}return vec;}}vector< double > ArraMultiVect(vector< vector< double > > &B,vector< double > &u) //矩阵乘向量{int a = B.size();int b = u.size();vector< double > vec(a, 0);if (a != b){cerr << "Array and Vector not match in size!";}else{for (int i=1; i<a; i++){for (int j=1; j<a; j++){vec[i] += B[i][j] * u[j];}}return vec;}}vector< vector<double> > GetM(vector< vector <double> > &A){int a = A.size();double s = A[a-2][a-2] + A[a-1][a-1];double t = A[a-2][a-2] * A[a-1][a-1] - A[a-1][a-2] * A[a-2][a-1];vector<vector<double>> D(a, vector< double >(a, 0));for (int i=1; i<a; i++){{double sum = 0;for (int k=1; k<a; k++){sum += A[i][k] * A[k][j];}D[i][j] = sum - s * A[i][j] + t *(i==j ? 1.0 : 0);}}return D;}double GetMode(vector < vector < double > > &B, const int r){double value = 0;int a = B.size();for (int k=r; k<a; k++){value += B[k][r] * B[k][r];}value = sqrt(value);return value;}double GetModeH(vector < vector < double > > &B, const int r){double value = 0;int a = B.size();for (int k=r+1; k<a; k++){value += B[k][r] * B[k][r];}value = sqrt(value);return value;}vector< vector< double > > NumbMultiArra(vector< vector< double > > &D, double a)//数乘//向量{int b = D.size();vector< vector< double > > U(b, vector< double >(b, 0));for (int i=1; i<b; i++){{U[i][j] = a *D[i][j];}}return U;}bool IsBirZero(vector< vector< double > > &B, const int r){bool b = true;int a = B.size();for (int i=r+1; i<a; i++){if(abs(B[i][r]) > epsion){b = false;}}return b;}bool IsBirZeroH(vector< vector< double > > &B, const int r){bool b = true;int a = B.size();for (int i=r+2; i<a; i++){if(abs(B[i][r]) > epsion){b = false;}}return b;}vector< vector< double > > VectMultiCovVect(vector< double > &a,vector< double > &b)//向量乘向量的转置得到矩阵{int s1 = a.size();int s2 = b.size();if (s1 != s2){cerr << "Vectors not match in size ! ";}else{vector< vector< double > > U(s1, vector< double >(s1, 0));for (int i=1; i<s1; i++){for (int j=1; j<s1; j++){U[i][j] = a[i] * b[j];}}return U;}}void ArraSubs(vector< vector< double > > &C,vector< vector < double > > D){int a = C.size();int b = D.size();int c = C[0].size();int d = D[0].size();if (a!=b || c!=d){cerr << "Vectors not match in size !";}else{for (int i=1; i<a; i++){for (int j=1; j<a; j++){C[i][j] -= D[i][j];}}}}vector< double > VectSubs(vector< double > &a,vector< double > &b){int s1 = a.size();int s2 = b.size();if(s1 != s2){cerr << "Vectors not match in size !";}else{vector< double > value(s1,0);for (int i=1;i<s1;i++){value[i] = a[i] - b[i];}return value;}}void GausElim(vector< vector< double > > a) //高斯消元{int sz = a.size();vector< int > fx, ufx,ufxp, record;vector< double > x(sz, 0);vector< vector< double > > ret;//*****消元过程********for (int k = 1; k < sz-1; k++){double max = a[k][k];int p=0;for (int i=k+1; i<sz; i++){if (abs(a[i][k]) > abs(max)){p =i;max = a[i][k];}}//选出主元行if (abs(max) >= epsion){if (p != 0){for (int j=k; j<sz; j++){double temp = a[k][j];a[k][j] = a[p][j];a[p][j] = temp;}//交换主元行}for (int i=k+1; i<sz; i++){double tt = a[i][k] / a[k][k];for (int j=k+1; j<sz; j++){a[i][j] = a[i][j] - tt * a[k][j];}}}// end of if (abs(max) >= epsion)}// end of for (int k = 1; k < sz; k++)for (int i=1; i<sz; i++){int p=0;for (int j=1; j<=i;j++){if(abs(a[j][i]) >= 10*epsion)//不为零{p=j;}}record.push_back(p);}for (int i=1; i<sz; i++){int p=0;for (int j=sz-2; j>=0; j--){if (record[j] == i){p=j + 1;}}if (p != 0){ufxp.push_back(p);ufx.push_back(i);}}for (int i=1; i<sz; i++){int p = 0;for (int j=0; j < ufxp.size(); j++){if (ufxp[j] == i){p = 1;}}if (p == 0){fx.push_back(i);}}//end of for (int i=1; i<sz; i++)int c = fx.size();//***************************************************** for (int i=0; i<c; i++){for (int j=0; j<c; j++){if (i == j){x[fx.at(j)] = 1.0 + epsion;}else{x[fx.at(j)] = 0;}}int b = ufxp.size();for (int s=b-1; s>=0; s--){double temp=0;for(int t=ufxp.at(s)+1; t<sz; t++){temp += x[t] * a[ufx.at(s)][t];}x[ufxp.at(s)] = (0 - temp) / a[ufx.at(s)][ufxp.at(s)];}ret.push_back(x);}//end of for (int i=0; i<c; i++)//******************************************************* int sz1 = ret.size();int sz2 = ret[0].size();ofstream result2("CharactVector .txt", ios::app);result2.precision(12);for (int i=0; i<sz1; i++){for (int j=1;j<sz2; j++){result2 << scientific << ret[i][j] << endl;}result2 << endl << endl;}result2 << "下一个特征值的特征向量！" << endl;}void Stop(vector< vector< double > > &Ar){int a = Lambda.size();for (int i=0; i<a; i++){if (abs(Lambda[i].imag())<=epsion){vector< vector< double > > An=Ar;ArraSubs(An,NumbMultiArra(I,Lambda[i].real()));GausElim(An);}}ofstream result("result.txt");result.precision(12);vector<complex< double > >::iterator i,j;j = Lambda.end();for (i = Lambda.begin(); i<j; i++){result << scientific <<(*i) << endl;}exit(0);}void SolutS1S2(complex< double > &s1, complex< double > &s2,vector< vector< double > > &A){double s = A[m-1][m-1] + A[m][m];double t = A[m-1][m-1] * A[m][m] - A[m][m-1] * A[m-1][m];double det = s *s - 4.0 * t;if (det > 0){s1.imag(0);s1.real ((s + sqrt(det)) / 2);s2.imag (0);s2.real((s - sqrt(det)) / 2);}else{s1.imag(sqrt(0 - det) / 2);s1.real(s / 2);s2.imag(0 - sqrt(0 - det) / 2);s2.real(s / 2);}}void Save2(complex< double > &s1, complex< double > &s2)//存储特征值s1，s2 {Lambda.push_back(s1);Lambda.push_back(s2);}void Save1(complex< double > &s){Lambda.push_back(s);}void JudgemBelow2(vector< vector< double > > &A,vector< vector< double > > Abk){if (m == 1){complex< double > lbd;lbd.imag(0);lbd.real(A[1][1]);Save1(lbd);Stop(Abk);}else if (m == 0){Stop(Abk);}}//拟上三角化void Hessenberg(vector< vector< double > > &A){int a = A.size();vector< double > u(a,0);vector< double > p(a,0);vector< double > q(a,0);vector< double > w(a,0);double d, c, h, t;for (int r=1; r<a-2; r++){if (!IsBirZeroH(A, r)){d = GetModeH(A,r);c = (A[r+1][r] > 0) ?(0-d) : d;h = c * c - c * A[r+1][r];for (int j=1; j<a; j++){if (j < r+1){u[j] = 0;}else if (j == r+1){u[j] = A[j][r] - c;}else{u[j] = A[j][r];}}// end of for (int j=1; j<=m; j++)p = NumbMultiVect(ConveArraMultiVect(A, u), 1 / h);q = NumbMultiVect(ArraMultiVect(A, u), 1 / h);t = VectMultiVect(p, u) / h;w = VectSubs(q, NumbMultiVect(u, t));ArraSubs(A, VectMultiCovVect(w, u));ArraSubs(A, VectMultiCovVect(u, p));}// end of if (!IsBirZero(B, r))}// end of for (int r=1; r<m; r++)ofstream result("NISHANGDANJIAO.txt");result.precision(12);for (int i=1; i<a; i++){for (int j=1; j<a; j++){result << scientific << A[i][j] << " ";}result << endl;}}void QRMethod(vector< vector< double > > A){int a = A.size();vector< vector< double > > Q(a,vector < double > (a,0));for (int i=1; i<a; i++){Q[i][i] = 1.0;}vector< vector< double > > RQ(A);//vector< double > u(a,0);vector< double > v(a,0);vector< double > p(a,0);vector< double > q(a,0);vector< double > w(a,0);vector< double > l(a,0);double d, c, h, t;for (int r=1; r<a-1; r++){if (!IsBirZero(A, r)){d = GetMode(A,r);c = (A[r][r] > 0) ?(0-d) : d;h = c * c - c * A[r][r];for (int j=1; j<a; j++){if (j < r){u[j] = 0;}else if (j == r){u[j] = A[j][j] - c;}else{u[j] = A[j][r];}}// end of for (int j=1; j<=m; j++)l = ArraMultiVect(Q, u);ArraSubs(Q, VectMultiCovVect(l,NumbMultiVect(u, 1/h)));v = NumbMultiVect(ConveArraMultiVect(A, u), 1 / h);ArraSubs(A, VectMultiCovVect(u, v));p = NumbMultiVect(ConveArraMultiVect(RQ, u), 1 / h);q = NumbMultiVect(ArraMultiVect(RQ, u), 1 / h);t = VectMultiVect(p, u) / h;w = VectSubs(q, NumbMultiVect(u, t));ArraSubs(RQ, VectMultiCovVect(w, u));ArraSubs(RQ, VectMultiCovVect(u, p));}// end of if (!IsBirZero(B, r))}// end of for (int r=1; r<m; r++)ofstream result("QR.txt");result.precision(12);for (int i=1; i<a; i++){for (int j=1; j<a; j++){result << scientific << Q[i][j] << " ";}result << endl;}result << endl << endl;for (int i=1; i<a; i++){for (int j=1; j<a; j++){result << scientific << A[i][j] << " ";}result << endl;}result << endl << endl;for (int i=1; i<a; i++){for (int j=1; j<a; j++){result << scientific << RQ[i][j] << " ";}result << endl;}}void CalculatAk(vector< vector < double > > &Ak){int a = Ak.size();vector< vector < double > > B(a,vector < double > (a,0));vector< double > u(a,0);vector< double > v(a,0);vector< double > p(a,0);vector< double > q(a,0);vector< double > w(a,0);double d, c, h, t;B = GetM(Ak);for (int r=1; r<a-1; r++){if (!IsBirZero(B, r)){d = GetMode(B,r);c = (B[r][r] > 0) ?(0-d) : d;h = c * c - c * B[r][r];for (int j=1; j<a; j++){if (j < r){u[j] = 0;}else if (j == r){u[j] = B[j][j] - c;}else{u[j] = B[j][r];}}// end of for (int j=1; j<=m; j++)v = NumbMultiVect(ConveArraMultiVect(B, u), 1 / h);ArraSubs(B, VectMultiCovVect(u, v));p = NumbMultiVect(ConveArraMultiVect(Ak, u), 1 / h);q = NumbMultiVect(ArraMultiVect(Ak, u), 1 / h);t = VectMultiVect(p, u) / h;w = VectSubs(q, NumbMultiVect(u, t));ArraSubs(Ak, VectMultiCovVect(w, u));ArraSubs(Ak, VectMultiCovVect(u, p));}// end of if (!IsBirZero(B, r))}// end of for (int r=1; r<m; r++)}int _tmain(int argc, _TCHAR* argv[]){vector< vector< double > > A(11, vector< double >(11,0));vector< vector< double > > Abk;I=A;for (int i=1; i<11; i++){vector< double > temp;for (int j=1; j<11; j++){if(i != j){A[i][j] = sin(0.5 * i + 0.2 * j);I[i][j] = 0;}else{A[i][j] = 1.5 * cos(i + 1.2 *j);I[i][j] = 1.0;}}}Abk = A;n = A.size() - 1;m = n;//初始化问题Hessenberg(A);QRMethod(A);while (1){if (abs(A[m][m-1]) <= epsion){complex< double > lbdm;lbdm.imag(0);lbdm.real(A[m][m]);Save1(lbdm);m -= 1;//对A进行降维处理！！！！A.pop_back();int a = A.size();for (int i=0; i<a; i++){A[i].pop_back();}JudgemBelow2(A, Abk);}else{complex< double > va1, va2;SolutS1S2(va1, va2, A);if (m == 2){Save2(va1,va2);Stop(Abk);}//end of if (m == 2)if ( abs(A[m-1][m-2]) <= epsion){Save2(va1,va2);m = m - 2;//矩阵降维A.pop_back();A.pop_back();int a = A.size();for (int i=0; i<a; i++){A[i].pop_back();A[i].pop_back();}JudgemBelow2(A, Abk);}else{if (k == L){cerr << "Stop without solution";exit(-1);}else{CalculatAk(A);k += 1;}}}// end of if (abs(A[m][m-1]) >= epsion)}}三、实验结果（1）A经过拟上三角化后得到的矩阵-8.827516758830e-001 -9.933136491826e-002 -1.103349285994e+000-7.600443585637e-001 1.549101079914e-001 -1.946591862872e+000-8.782436382928e-002 -9.255889387184e-001 6.032599440534e-0011.518860956469e-001-2.347878362416e+000 2.372370104937e+000 1.819290822208e+0003.237804101546e-001 2.205798440320e-001 2.102692662546e+0001.816138086098e-001 1.278839089990e+000 -6.380578124405e-001-4.154075603804e-0011.0556********e-016 1.728274599968e+000 -1.171467642785e+000-1.243839262699e+000 -6.399758341743e-001 -2.002833079037e+0002.924947206124e-001 -6.412830068395e-001 9.783997621285e-0022.557763574160e-001-5.393383812774e-017 -3.165873865181e-017 -1.291669534130e+000-1.111603513396e+000 1.171346824096e+000 -1.307356030021e+0001.803699177750e-001 -4.246385358369e-001 7.988955239304e-0021.608819928069e-0011.533464996622e-017 5.963321406181e-017 0.000000000000e+0001.560126298527e+000 8.125049397515e-001 4.421756832923e-001-3.588616128137e-002 4.691742313671e-001 -2.736595050092e-001 -7.359334657750e-0021.300562737229e-016 -3.097060010889e-017 0.000000000000e+0000.000000000000e+000 -7.707773755194e-001 -1.583051425742e+000-3.042843176799e-001 2.528712446035e-001 -6.709925401449e-0012.544619929082e-0011.610216724767e-016 -2.211571837369e-016 0.000000000000e+0000.000000000000e+000 6.483933100712e-017 -7.463453456938e-001-2.708365157019e-002 -9.486521893682e-001 1.195871081495e-0011.929265617952e-0021.368550186199e-016 7.151513190805e-017 0.000000000000e+0000.000000000000e+000 -2.522454384246e-017 1.072074718358e-016-7.701801374364e-001 -4.697623990618e-001 4.988259468008e-0011.137691603776e-001-2.780851300718e-017 -6.708630788363e-017 0.000000000000e+0000.000000000000e+000 -3.282796821065e-017 4.879774287475e-0171.854829286220e-016 7.013167092107e-001 1.582180688475e-0013.862594614233e-001-2.124604440055e-017 -1.707979758930e-016 0.000000000000e+0000.000000000000e+000 1.013496750890e-016 -4.153758325241e-0171.222621691887e-016 -5.551115123126e-017 4.843807602783e-0013.992777995177e-001（2）Q-3.519262579534e-001 4.427591982245e-001 -6.955982513607e-0016.486200753651e-002 3.709718861896e-001 1.855847143605e-001-1.628942319628e-002 -1.181053169648e-001 -5.255375383724e-002 -5.486582943568e-002-9.360277287361e-001 -1.664679186545e-001 2.615299548560e-001 -2.438671728934e-002 -1.394774360893e-001 -6.977585391241e-0026.124472142963e-003 4.440505443139e-002 1.975907909728e-0022.062836970533e-002-4.208697095111e-017 -8.810520554692e-001 -3.989762796959e-0013.720308728479e-002 2.127794064090e-001 1.064463557221e-001-9.343171079758e-003 -6.774200464527e-002 -3.014340698675e-002 -3.146955080444e-002-2.150178169911e-017 4.009681353156e-017 -5.371806806439e-001 -1.234945854205e-001 -7.063151608719e-001 -3.533456368496e-0013.101438948264e-002 2.248676491598e-001 1.000601783527e-0011.044622748702e-0016.113458775639e-018 -3.721194648197e-0177.068175586628e-0189.892235468621e-001 -1.239411731211e-001 -6.200358589825e-0025.442272839461e-003 3.945881637235e-002 1.755813350011e-0021.833059462907e-0025.184948268593e-017 -4.198303559531e-017 3.255071298412e-017-1.527665297025e-017 5.323610690264e-001 -6.733900344896e-0015.910581205868e-002 4.285425323867e-001 1.906901343193e-0011.990794495295e-0016.419444583601e-017 4.121668945107e-017 3.096964582901e-017-5.050360323651e-017 -7.078737686239e-017 -6.0597********e-001 -9.165783032818e-002 -6.645586508974e-001 -2.957110877580e-001 -3.0872********e-0015.455993559780e-017 -9.724896332186e-017 3.956603694870e-0171.913857001710e-018 -4.316583456405e-0172.797376665557e-0179.933396625117e-001 -9.690440311939e-002 -4.311990584470e-002-4.501694411183e-002-1.108640877071e-017 4.655237056115e-017 -1.0722********e-017 -9.470236121745e-018 4.277993227986e-017 8.866601870176e-017 -2.516725033065e-016 5.410088006061e-001 -5.817540838226e-001 -6.0734********e-001-8.470152033092e-018 9.650816729410e-017 -1.429362211392e-017 -2.789222052040e-017 -3.690793770141e-017 -8.090765462521e-017 -1.964050295697e-016 -6.772274683437e-017 -7.221591336735e-0016.917269588876e-001（3）R2.508342744917e+000 -2.185646885493e+000 -1.314609070786e+000-3.558787493836e-002 -2.609857850388e-001 -1.283121847090e+000 -1.390878610606e-001 -8.712897972161e-001 3.849367902971e-0013.353802899665e-0012.100627753398e-016 -1.961603277854e+000 2.407523727633e-0017.054714572823e-001 5.957204318279e-001 5.526978774676e-001-3.268209924413e-001 -5.769498668364e-002 2.871129330189e-001 -8.895128754189e-002-3.300197935770e-016 -1.872873410421e-016 2.404534601993e+0001.706758096328e+000 -4.239566704091e-001 3.405332305815e+000-1.050017655852e-001 1.462257102734e+000 -6.684487469283e-001 -4.027*********e-0013.0773********e-017 1.746386351950e-017 0.000000000000e+0001.577122080722e+000 6.399535133956e-001 3.468127872427e-001-5.701786649768e-002 4.014788054433e-001 -2.222476176311e-001 -6.317059236442e-0021.760039865880e-016 9.988285339980e-017 0.000000000000e+0000.000000000000e+000 -1.447846997770e+000 -1.415724007744e+000-2.806139044665e-001 -2.817910521892e-001 -4.611434881851e-0021.996629079956e-0018.804885435596e-017 4.996802050991e-017 0.000000000000e+0000.000000000000e+000 8.831633340975e-017 1.231641451542e+0001.619701003419e-001 1.962638275504e-001 5.350035621760e-001-1.509273424767e-001-7.728357669400e-018 -4.385868928757e-018 0.000000000000e+0000.000000000000e+000 -7.751835246841e-018 -1.279231078565e-017-7.753441914209e-001 -3.464514508821e-001 4.312226803504e-0011.234643696237e-001-5.603391361152e-017 -3.179943413314e-017 0.000000000000e+0000.000000000000e+000 -5.620413613517e-017 -9.274974944455e-0170.000000000000e+000 1.296312940612e+000 -4.288053318338e-0012.737334158165e-001-2.493361499851e-017 -1.414991023727e-017 0.000000000000e+0000.000000000000e+000 -2.500935953597e-017 -4.127119443934e-0170.000000000000e+000 0.000000000000e+000 -6.707396440648e-001-4.842320121884e-001-2.603055667460e-017 -1.477242832192e-017 0.000000000000e+0000.000000000000e+000 -2.610963355436e-017 -4.308689959101e-0170.000000000000e+000 0.000000000000e+000 -1.110223024625e-0167.168323926323e-002（4）RQ1.163074414164e+0002.632670934508e+000 -1.772796003272e+000-8.668899138521e-002 3.300503471047e-001 1.455162371214e+000 -9.730650448593e-001 -4.873031174655e-001 -7.756411630489e-001 -3.249201979113e-0011.836115060851e+000 1.144286420080e-001 -9.880381403133e-0015.589725694767e-001 4.694190067101e-002 -2.978478237007e-0011.617130577649e-002 6.936977702522e-001 1.367670571405e-0011.419099231519e-0025.598200524418e-016 -2.118520153533e+000 -1.876189745783e+000-5.407071940597e-001 1.171538359721e+000 -2.550323020223e+0001.691577936540e+000 1.229951613262e+000 1.387947777212e+0008.667502917242e-001-3.179059303049e-017 -5.261714527977e-017 -8.471995127808e-0014.382910468318e-001 -1.008632199185e+000 -7.959374261495e-0014.769258865577e-001 4.072683083890e-001 4.096390493527e-0013.363378940862e-001-3.514195999269e-016 3.391949386044e-017 -2.160938214545e-016 -1.432244342447e+000 -5.742284908055e-001 1.213151477723e+000 -3.457508625575e-001 -4.749853573124e-001 -3.176158274191e-001 -4.294507015032e-002-3.704735750656e-017 -1.0998********e-016 -1.953241363386e-0178.269089741494e-017 6.556779598004e-001 -9.275250974463e-0012.529079844053e-001 6.905949216976e-001 -2.374430675823e-002-2.429781119781e-001-6.420051822287e-017 3.865817713597e-017 -3.806718328740e-0172.129734126613e-017 7.853*********e-017 4.698400884876e-001-2.730776009527e-001 7.821296259798e-001 -9.580964936399e-0027.846239841323e-0021.478496020372e-016 -8.383952267131e-017 9.577540382744e-017-7.120338911078e-017 -1.220876056602e-016 -1.854471832043e-0161.287679058937e+000 -3.576058900348e-001 -4.116725408807e-0033.914268216423e-0014.477158378610e-017 -6.204017427568e-017 3.360322998507e-017-1.133882337819e-017 -2.759056827456e-017 -9.217582575819e-0172.214639994444e-016 -3.628760503545e-001 7.398980975354e-0017.241608309576e-0023.408891482791e-017 2.353495494255e-017 1.932283926688e-017-3.456910153081e-017 -2.015177337156e-017 -8.084422691634e-017 -5.839476980893e-017 5.551115123126e-017 -5.176670596524e-0024.958522909877e-002（5）特征值(6.360627875745e-001,0.000000000000e+000)(5.650488993501e-002,0.000000000000e+000)(9.355889078188e-001,0.000000000000e+000)(-9.805309562902e-001,1.139489127430e-001)(-9.805309562902e-001,-1.139489127430e-001)(1.577548557113e+000,0.000000000000e+000)(-1.484039822259e+000,0.000000000000e+000)(-2.323496210212e+000,8.930405177200e-001)(-2.323496210212e+000,-8.930405177200e-001)(3.383039617436e+000,0.000000000000e+000)（6）实特征值的特征向量4.745031936539e+0003.157868541750e+0001.729946912420e+001-1.980049331455e+000-3.187521973524e+0017.794009603201e+000-1.004255685845e+0011.670757770480e+0011.310524273054e+0011.000000000001e+000下一个实特征值对应的特征向量：-5.105003830625e+000-4.886319842360e+0009.505161576151e+000-6.788331788214e-001-9.604334110499e+000-3.0457********e+0001.574873885602e+001-7.395037126277e+000-7.109654943661e+0001.000000000001e+000下一个实特征值对应的特征向量：2.792418944529e+0001.598236841512e+000-5.207507040911e-001-1.667886451702e+000-1.225708535859e+0017.241214790777e+000-5.398214501433e+0002.841008912974e+001-1.216518754416e+0011.000000000001e+000下一个实特征值对应的特征向量：6.217350824581e-001-1.115111815226e-001-2.483773580804e+000-1.306860840421e+000-3.815605442533e+0008.117305509395e+000-1.239170883675e+000-6.800309586197e-0012.691900144840e+0001.000000000001e+000下一个实特征值对应的特征向量：-1.730784582112e+0012.408192534967e+0014.133852365119e-001-8.572044074538e+0009.287334657928e-002-7.832726042776e-002-6.374274025716e-001-3.403204760832e-001。

BUAA数值分析大作业三

北京航空航天大学2020届研究生《数值分析》实验作业第九题院系：xx学院学号：姓名：2020年11月Q9:方程组A.4一、算法设计方案（一）总体思路1.题目要求∑∑===k i kj s r rsy x cy x p 00),(对f(x, y) 进行拟合，可选用乘积型最小二乘拟合。

),(i i y x 与),(i i y x f 的数表由方程组与表A-1得到。

2.),(**j i y x f 与1使用相同方法求得，),(**j i y x p 由计算得出的p(x,y)直接带入),(**j i y x 求得。

1. ),(i i y x 与),(i i y x f 的数表的获得对区域D ={ （x,y)|1≤x ≤1.24,1.0≤y ≤1.16}上的f (x , y )值可通过xi=1+0.008i ，yj=1+0.008j ，得到),(i i y x 共31×21组。

将每组带入A4方程组，即可获得五个二元函数组，通过简单牛顿迭代法求解这五个二元数组可获得z1~z5有关x,y 的表达式。

再将),(i i y x 分别带入z1~z5表达式即可获得f(x,y)值。

2.乘积型最小二乘曲面拟合2.1使用乘积型最小二乘拟合，根据k 值不用，有基函数矩阵如下：⎪⎪⎪⎭⎫ ⎝⎛=k i i k x x x x B 0000 ， ⎪⎪⎪⎭⎫ ⎝⎛=k j jk y y y y G 0000数表矩阵如下：⎪⎪⎪⎭⎫⎝⎛=),(),(),(),(0000j i i j y x f y x f y x f y x f U记C=[rs c ]，则系数rs c 的表达式矩阵为：11-)(-=G G UG B B B C T TT ）（通过求解如下线性方程，即可得到系数矩阵C 。

UG B G G C B B T T T =)()(2.2计算),(),,(****j i j i y x p y x f (i =1,2,…,31 ; j =1,2,…,21) 的值),(**j i y x f 的计算与),(j i y x f 相同。

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4、设 x0, x1, , xn 为 n+1 个互异节点， l j (x)( j 0,1, ,n) 为这组节点上的 n 次 Lagrange
插值基函数，试证：
n
(1)
x
k j
l
j
(
x)
xk
j0
(k 0,1, , n)
n
(2) ( xj x)k l j (x) 0
j0
(k 1, , n)
(3)设P(x)是任意一个最高次系数为 1的n+1次多项式，则
（ 2）（ 2）取 a=1，对矩阵 A 进行 Cholesky 分解，并用平方根法求解上述方程组，计算过程保留 2 位小数。
7、用追赶法解下列方程组
2 1 0 0 x1
0
1 2 1 0 x2
1
0 1 2 1 x3
0
0 0 1 2 x4 2.5
8、已知
6 13 17
A 13 29 38
17 38 50
1、分别用 Gauss 消去法、列主元素法和全主元素法解下列方程组， 2 x1 x2 2 x3 6 4 x1 3x2 x3 11 6 x1 x2 5x3 13
计算过程保留 3 位小数。
2、用三角分解法求解题 1 中的方程组。
3、用紧凑格式解下列方程组，并写出 L,U 矩阵。
1 2 3 4 x1
1、设下列各数均为经过四舍五入后得到的近似值，试求各数的绝对误差限和相对误差限。 a 3580,b 0.00476,c 2958 10 2, d 0.1430 10 8
2、已知 a 1.2031,b 0.978 是经过四舍五入后得到的近似值，问 a b, a b 有几位有效
数字？ 3、计算球的体积，为使其相对误差限为 1%，测量半径 R 时，相对误差最大为多少？
l31 l32 1
ln1 l n2 ln3
1 lnn 1 1
5、用三角分解法求下列矩阵的逆矩阵。
11 1 21 0 1 10
6、设有方程组 Ax=b ，其中 A
210 12a 0a2
x ( x1, x2, x3 )T ,b (3,3,1) T
（ 1）（ 1）求出 A 能进行 Cholesky 分解，即 A=LL T（其中 L 为下三角矩阵）的 a 取值范围。
6、已知函数表为
1
2
3
4
xi
3
5
9
15
y f ( xi )
分别用 Newton 向前、向后插值公式计算 f(1.5) ， f(3.7) 的近似值。
7、设 f (x) x7 5x5 1 ，求差商
(1)f [20 ,21, ,2 7 ]
(2) f [2 0 ,21, ,2 k ]
(k 8)
8、设 f (x) 是一个 n 次多项式，
2、已知函数表为
0． 527
xi
0． 727
0． 807
0． 927
0． 01075 0． 01219 0． 01188 0． 01426
yi
用二次插值计算 y(0.7) 的近似值。
3、已知函数表为
1
3
4
6
xi
-7
5
8
14
y f ( xi )
试求其 3 阶 Lagrange 插值多项式，并以此计算 f(2) 的近似值。
求 cond( A)1 及 cond( A) ，并说明方程组 Ax=b 是否病态。
9、已知方程组
x1 0.99x2 1 0.99x1 0.98x2 1 的解为 x1 100, x2 100
(1) 计算系数矩阵的条件数。
(2) 取 x1* (1,0)T , x2* (100.5, 99.5)T ，分别计算残量 ri b
f (x)
n k
ak x (an
k0
0) 试证：
n m 1次多项式， m n 1
f [ x0 , x1, , xm , x] an 0
,m n 1 ,m n 1
9、设 f ( x) 3xex 2e2x 节点 x0 1, x1 1.05 ，求的 3 次 Hermite 插值多项式及 f(1.03)
的近似值，并估计误差。
Ax*i (i
1,2) 。
10、求解超定方程组
的最小二乘解。
2x1 4x2 11 3x1 5x2 3 x1 2x2 6 2x1 x2 7
1、已知函数表为
-1
0
1
xi
yi 2xi
0．5 1
2
（ 1）（ 1）利用线性插值计算 20.3 的近似值并估计误差。（ 2）（ 2）利用二次插值计算 20.3 的近似值并估计误差。
2
1 4 9 16 x2 10
1 8 24 64 x3 44
1 16 81 256 x4 90
1
0
1
4、若 Lk
lk 1k 1
Байду номын сангаасlk 2k
0
0
(k 1,2, , n 1)
l nk
1
1
0
求证： (1)
L
1 k
2I
Lk
1
l k 1k 1
l k 2k
1
0
0
l nk
1
1
l21 1
0
(2)
L1
1L
1 2
Ln1 1
n
P( x) P( xj )l j ( x)
j0
n 1( x)
n
(x xj)
j0
5、已知函数表为
1． 615 1．634 1． 702 1． 828
xi
2． 41450 2．46259 2． 65271 3． 03035
y f ( xi )
试求其 3 阶 Newton 插值多项式，并以此计算 f(1.682) 的近似值。