数值分析大作业模板
数值分析大作业三四五六七
数值分析大作业三四五六七数值分析大作业三四五六七Document number【SA80SAB-SAA9SYT-SAATC-SA6UT-SA18】大作业三1. 给定初值0x 及容许误差，编制牛顿法解方程f (x )=0的通用程序. 解：Matlab 程序如下：函数m 文件：fu.mfunction Fu=fu(x)Fu=x^3/3-x;end函数m 文件：dfu.mfunction Fu=dfu(x)Fu=x^2-1;end用Newton 法求根的通用程序Newton.mclear;x0=input('请输入初值x0:');ep=input('请输入容许误差:');flag=1;while flag==1x1=x0-fu(x0)/dfu(x0);if abs(x1-x0)<ep< p="">flag=0;endx0=x1;endfprintf('方程的一个近似解为：%f\n',x0);寻找最大δ值的程序：Find.mcleareps=input('请输入搜索精度：');ep=input('请输入容许误差：');flag=1;k=0;x0=0;while flag==1sigma=k*eps;x0=sigma;k=k+1;m=0;flag1=1;while flag1==1 && m<=10^3x1=x0-fu(x0)/dfu(x0);if abs(x1-x0)endm=m+1;x0=x1;endif flag1==1||abs(x0)>=epflag=0;endendfprintf('最大的sigma 值为：%f\n',sigma);2．求下列方程的非零根5130.6651()ln 05130.665114000.0918x x f x x +??=-= ?-解：Matlab 程序为：（1）主程序clearclcformat longx0=765;N=100;errorlim=10^(-5);x=x0-f(x0)/subs(df(),x0);n=1;while n<n< p="">x=x0-f(x0)/subs(df(),x0);if abs(x-x0)>errorlimn=n+1;elsebreak;endx0=x;enddisp(['迭代次数: n=',num2str(n)])disp(['所求非零根: 正根x1=',num2str(x),' 负根x2=',num2str(-x)])（2）子函数非线性函数ffunction y=f(x)y=log((513+0.6651*x)/(513-0.6651*x))-x/(1400*0.0918);end（3）子函数非线性函数的一阶导数dffunction y=df()syms x1y=log((513+0.6651*x1)/(513-0.6651*x1))-x1/(1400*0.0918);y=diff(y);end运行结果如下：迭代次数: n=5所求非零根: 正根x1=767.3861 负根x2=-767.3861大作业四试编写MATLAB 函数实现Newton 插值，要求能输出插值多项式. 对函数21()14f x x=+在区间[-5，5]上实现10次多项式插值.分析：（1）输出插值多项式。
数值分析大作业一
数值分析大作业一一、算法设计方案1、求λ1和λ501的值：思路：采用幂法求出按模最大特征值λmax，该值必为λ1或λ501，若λmax小于0，则λmax=λ1；否则λmax=λ501。
再经过原点平移，使用幂法迭代出矩阵A-λmax I的特征值，此时求出的按模最大特征值即为λ1和λ501的另一个值。
2、求λs的值：采用反幂法求出按模最小的特征值λmin即为λs，其中的方程组采用LU分解法进行求解。
3、求与μk最接近的特征值：对矩阵A采用带原点平移的反幂法求解最小特征值，其中平移量为：μk。
4、A的条件数cond(A)=| λmax/λmin|；5、A的行列式的值：先将A进行LU分解，再求U矩阵对角元素的乘积即为A 行列式的值。
二、源程序#include<iostream>#include<iomanip>#include<math.h>#define N 501#define E 1.0e-12 //定义精度常量#define r 2#define s 2using namespace std;double a[N];double cc[5][N];void init();double mifa();double fmifa();int max(int aa,int bb);int min(int aa,int bb);int max_3(int aa,int bb,int cc);void LU();void main(){double a1,a2,d1,d501=0,ds,det=1,miu[39],lamta,cond;int i,k;init();/*************求λ1和λ501********************/a1=mifa();if(a1<0)d1=a1; //若小于0则表示λ1的值elsed501=a1; //若大于0则表示λ501的值for(i=0;i<N;i++)a[i]=a[i]-a1;a2=mifa()+a1;if(a2<0)d1=a2; //若小于0则表示λ1的值elsed501=a2; //若大于0则表示λ501的值cout<<"λ1="<<setiosflags(ios::scientific)<<setprecision(12)<<d1<<"\t";cout<<"λ501="<<setiosflags(ios::scientific)<<setprecision(12)<<d501<<endl;/**************求λs*****************/init();ds=fmifa();cout<<"λs="<<setiosflags(ios::scientific)<<setprecision(12)<<ds<<endl;/**************求与μk最接近的特征值λik**************/cout<<"与μk最接近的特征值λik:"<<endl;for(k=0;k<39;k++){miu[k]=d1+(k+1)*(d501-d1)/40;init();for(i=0;i<N;i++)a[i]=a[i]-miu[k];lamta=fmifa()+miu[k];cout<<"λi"<<k+1<<"\t\t"<<setiosflags(ios::scientific)<<setprecision(12)<<lamta<<en dl;}/**************求A的条件数**************/cout<<"矩阵A的条件式";cond=abs(max(abs(d1),abs(d501))/ds);cout<<"cond="<<setiosflags(ios::scientific)<<setprecision(12)<<cond<<endl;/**************求A的行列式**************/cout<<"矩阵A的行列式";init();LU();for(i=0;i<N;i++){det*=cc[2][i];}cout<<"det="<<setiosflags(ios::scientific)<<setprecision(12)<<det<<endl;system("pause");}/**************初始化函数，给a[N]赋值*************/void init(){int i;for(i=1;i<=501;i++)a[i-1]=(1.64-0.024*i)*sin((double)(0.2*i))-0.64*exp((double)(0.1/i)); }/**************幂法求最大绝对特征值**************/double mifa(){int i,k=0;double u[N],y[N]={0},b=0.16,c=-0.064,Beta_=0,error;for(i=0;i<501;i++)u[i]=1; //令u[N]=1for(k=1;k<2000;k++) //控制最大迭代次数为2000{/***求y(k-1)***/double sum_u=0,gh_sum_u;for(i=0;i<N;i++){sum_u+=u[i]*u[i]; }gh_sum_u=sqrt(sum_u);for(i=0;i<N;i++){y[i]=u[i]/gh_sum_u;}/****求新的uk****/u[0]=a[0]*y[0]+b*y[1]+c*y[2];u[1]=b*y[0]+a[1]*y[1]+b*y[2]+c*y[3]; //前两列和最后两列单独拿出来求中D间的循环求for(i=2;i<N-2;i++){u[i]=c*y[i-2]+b*y[i-1]+a[i]*y[i]+b*y[i+1]+c*y[i+2];}u[N-2]=c*y[N-4]+b*y[N-3]+a[N-2]*y[N-2]+b*y[N-1];u[N-1]=c*y[N-3]+b*y[N-2]+a[N-1]*y[N-1];/***求beta***/double Beta=0;for(i=0;i<N;i++){Beta+=y[i]*u[i];}//cout<<"Beta"<<k<<"="<<Beta<<"\t"; 输出每次迭代的beta /***求误差***/error=abs(Beta-Beta_)/abs(Beta);if(error<=E) //若迭代误差在精度水平内则可以停止迭代{return Beta;} //控制显示位数Beta_=Beta; //第个eta的值都要保存下来，为了与后个值进行误差计算 }if(k==2000){cout<<"error"<<endl;return 0;} //若在最大迭代次数范围内都不能满足精度要求说明不收敛}/**************反幂法求最小绝对特¬征值**************/double fmifa(){int i,k,t;double u[N],y[N]={0},yy[N]={0},b=0.16,c=-0.064,Beta_=0,error;for(i=0;i<501;i++)u[i]=1; //令u[N]=1for(k=1;k<2000;k++){double sum_u=0,gh_sum_u;for(i=0;i<N;i++){sum_u+=u[i]*u[i]; }gh_sum_u=sqrt(sum_u);for(i=0;i<N;i++){y[i]=u[i]/gh_sum_u;yy[i]=y[i]; //用重新赋值，避免求解方程组的时候改变y的值}/****LU分解法解方程组Au=y，求新的***/LU();for(i=2;i<=N;i++){double temp_b=0;for(t=max(1,i-r);t<=i-1;t++)temp_b+=cc[i-t+s][t-1]*yy[t-1];yy[i-1]=yy[i-1]-temp_b;}u[N-1]=yy[N-1]/cc[s][N-1];for(i=N-1;i>=1;i--){double temp_u=0;for(t=i+1;t<=min(i+s,N);t++)temp_u+=cc[i-t+s][t-1]*u[t-1];u[i-1]=(yy[i-1]-temp_u)/cc[s][i-1];}double Beta=0;for(i=0;i<N;i++){Beta+=y[i]*u[i];}error=abs(Beta-Beta_)/abs(Beta);if(error<=E){return (1/Beta);}Beta_=Beta;}if(k==2000){cout<<"error"<<endl;return 0;} }/**************求两数最大值的子程序**************/int max(int aa,int bb){return(aa>bb?aa:bb);}/**************求两数最小值的子程序**************/int min(int aa,int bb){return(aa<bb?aa:bb);}/**************求三数最大值的子程序**************/int max_3(int aa,int bb,int cc){ int tt;if(aa>bb)tt=aa;else tt=bb;if(tt<cc) tt=cc;return(tt);}/**************LU分解**************/void LU(){int i,j,k,t;double b=0.16,c=-0.064;/**赋值压缩后矩阵cc[5][501]**/for(i=2;i<N;i++)cc[0][i]=c;for(i=1;i<N;i++)cc[1][i]=b;for(i=0;i<N;i++)cc[2][i]=a[i];for(i=0;i<N-1;i++)cc[3][i]=b;for(i=0;i<N-2;i++)cc[4][i]=c;for(k=1;k<=N;k++){for(j=k;j<=min(k+s,N);j++){double temp=0;for(t=max_3(1,k-r,j-s);t<=k-1;t++)temp+=cc[k-t+s][t-1]*cc[t-j+s][j-1];cc[k-j+s][j-1]=cc[k-j+s][j-1]-temp;}//if(k<500){for(i=k+1;i<=min(k+r,N);i++){double temp2=0;for(t=max_3(1,i-r,k-s);t<=k-1;t++)temp2+=cc[i-t+s][t-1]*cc[t-k+s][k-1];cc[i-k+s][k-1]=(cc[i-k+s][k-1]-temp2)/cc[s][k-1];}}}}三、程序结果。
北航数值分析大作业三
一、题目：关于x, y, t, u, v, w 的下列方程组0.5cos 2.670.5sin 1.070.5cos 3.740.5sin 0.79t u v w x t u v w y t u v w x t u v w y +++-=⎧⎪+++-=⎪⎨+++-=⎪⎪+++-=⎩1、试用数值方法求出f(x, y)在区域 {(,)|00.8,0.5 1.5}D x y x y =≤≤≤≤上的一个近似表达式,0(,)kr s rsr s p x y cx y ==∑要求(,)p x y 一最小的k 值达到以下的精度10202700((,)(,))10i j i j i j f x y p x y σ-===-≤∑∑其中，0.08,0.50.05i j x i y j ==+。
2、计算****(,),(,)i j i j f x y p x y （i = 1, 2, …，8；j = 1, 2,…，5）的值，以观察(,)p x y 逼近(,)f x y 的效果，其中，*i x =0.1i , *j y =0.5+0.2j 。
说明：1、用迭代方法求解非线性方程组时，要求近似解向量()k x 满足()(1)()12||||/||||10k k k x x x --∞∞-≤2、作二元插值时，要使用分片二次代数插值。
3、要由程序自动确定最小的k 值。
4、打印以下内容：●算法的设计方案。
●全部源程序（要求注明主程序和每个子程序的功能）。
●数表：,,i j x y (,)i j f x y （i = 0,1,2,…,10；j = 0,1,2,…,20）。
●选择过程的,k σ值。
●达到精度要求时的,k σ值以及(,)p x y 中的系数rs c （r = 0,1,…,k;s = 0,1,…,k ）。
●数表：**,,i j x y ****(,),(,)i j i j f x y p x y （i = 1, 2, ...,8；j = 1, 2, (5)。
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 。
北航数值分析大作业一
《数值分析B》大作业一SY1103120 朱舜杰一．算法设计方案：1.矩阵A的存储与检索将带状线性矩阵A[501][501]转存为一个矩阵MatrixC[5][501] .由于C语言中数组角标都是从0开始的，所以在数组MatrixC[5][501]中检索A的带内元素a ij的方法是：A的带内元素a ij=C中的元素c i-j+2,j2.求解λ1，λ501，λs①首先分别使用幂法和反幂法迭代求出矩阵按摸最大和最小的特征值λmax和λmin。
λmin即为λs；如果λmax>0,则λ501=λmax；如果λmax<0,则λ1=λmax。
②使用带原点平移的幂法（mifa（）函数），令平移量p=λmax，求出对应的按摸最大的特征值λ，max，如果λmax>0,则λ1=λ，max+p；如果λmax<0,则λ501=λ，max+p。
3.求解A的与数μk=λ1+k（λ501-λ1）/40的最接近的特征值λik （k=1,2，…，39）。
使用带原点平移的反幂法，令平移量p=μk，即可求出与μk最接近的特征值λik。
4.求解A的（谱范数）条件数cond（A）2和行列式d etA。
①cond（A）2=|λ1/λn|，其中λ1和λn分别是矩阵A的模最大和最小特征值。
②矩阵A的行列式可先对矩阵A进行LU分解后，detA等于U所有对角线上元素的乘积。
二．源程序#include<stdio.h>#include<iostream.h>#include<stdlib.h>#include<math.h>#include<float.h>#include<iomanip.h>#include<time.h>#define E 1.0e-12 /*定义全局变量相对误差限*/int max2(int a,int b) /*求两个整型数最大值的子程序*/{if(a>b)return a;elsereturn b;}int min2(int a,int b) /*求两个整型数最小值的子程序*/{if(a>b)return b;elsereturn a;}int max3(int a,int b,int c) /*求三整型数最大值的子程序*/{ int t;if(a>b)t=a;else t=b;if(t<c) t=c;return(t);}void assignment(double array[5][501]) /*将矩阵A转存为数组C[5][501]*/{int i,j,k;//所有元素归零for(i=0;i<=4;){for(j=0;j<=500;){array[i][j]=0;j++;}i++;}//第0,4行赋值for(j=2;j<=500;){k=500-j;array[0][j]=-0.064;array[4][k]=-0.064;j++;}//第1,3行赋值for(j=1;j<=500;){k=500-j;array[1][j]=0.16;array[3][k]=0.16;j++;}//第2行赋值for(j=0;j<=500;){ k=j;j++;array[2][k]=(1.64-0.024*j)*sin((double)(0.2*j))-0.64*exp((double)(0.1/j));}}double mifa(double u[501],double array[5][501],double p) /*带原点平移的幂法*/ {int i,j; /* u[501]为初始迭代向量*/double a,b,c=0; /* array[5][501]为矩阵A的转存矩阵*/double y[501]; /*p为平移量*/for(;;){a=0;b=0;/*选用第一种迭代格式*///求ηk-1for(i=0;i<=500;i++){a=a+u[i]*u[i];}a=sqrt(a);//求y k-1for(i=0;i<=500;i++){y[i]=u[i]/a;}//求u kfor(i=0;i<=500;i++){u[i]=0;for(j=max2(i-2,0);j<=min2(i+2,500);j++){u[i]+=array[i-j+2][j]*y[j];}u[i]=u[i]-p*y[i]; /*引入平移量*/}//求βkfor(i=0;i<=500;i++){b+=y[i]*u[i];}if(fabs((b-c)/b)<=E) /*达到精度水平，迭代终止*/break;c=b;}return (b+p); /*直接返回A的特征值*/}void chuzhi(double a[]) /*用随机数为初始迭代向量赋值*/ {int i;srand((int)time(0));for(i=0;i<=500;i++){a[i]=(10.0*rand()/RAND_MAX); /*生成0~10的随机数*/}}void chuzhi2(double a[],int j) /*令初始迭代向量为e i*/{int i;for(i=0;i<=500;i++){a[i]=0;}a[j]=1;}void LU(double array[5][501]) /*对矩阵A进行Doolittle分解*/{ /*矩阵A转存在C[5][501]中*/int j,k,t; /*分解结果L，U分别存在C[5][501]的上半部与下半部*/ for(k=0;k<=500;k++){for(j=k;j<=min2((k+2),500);j++){for(t=max3(0,k-2,j-2);t<=(k-1);t++){array[k-j+2][j]-=array[k-t+2][t]*array[t-j+2][j];}}if(k<500)for(j=k+1;j<=min2((k+2),500);j++){for(t=max3(0,k-2,j-2);t<=(k-1);t++){array[j-k+2][k]-=array[j-t+2][t]*array[t-k+2][k];}array[j-k+2][k]=array[j-k+2][k]/array[2][k];}}}double fmifa(double u[501],double array[5][501],double p){ /*带原点平移的反幂法*/ int i,j;double a,b,c=0;double y[501];//引入平移量for(i=0;i<=500;i++){array[2][i]-=p;}//先将矩阵Doolittle分解LU(array);for(;;){a=0;b=0;//求ηk-1for(i=0;i<=500;i++){a=a+u[i]*u[i];}a=sqrt(a);//求y k-1for(i=0;i<=500;i++){y[i]=u[i]/a;}//回带过程，求解u kfor(i=0;i<=500;i++){u[i]=y[i];}for(i=1;i<=500;i++){for(j=max2(0,(i-2));j<=(i-1);j++){u[i]-=array[i-j+2][j]*u[j];}}u[500]=u[500]/array[2][500];for(i=499;i>=0;i--){for(j=i+1;j<=min2((i+2),500);j++){u[i]-=array[i-j+2][j]*u[j];}u[i]=u[i]/array[2][i];}//求βkfor(i=0;i<=500;i++){b+=y[i]*u[i];}if(fabs((b-c)/b)<=E) /*达到精度要求，迭代终止*/break;c=b;}return (p+(1/b)); /*直接返回距离原点P最接近的A的特征值*/ }//主函数main(){ int i;double d1,d501,ds,d,a;double u[501];double MatrixC[5][501];printf(" 《数值分析》计算实习题目第一题\n");printf(" SY1103120 朱舜杰\n");//将矩阵A转存为MatrixCassignment(MatrixC);//用带原点平移的幂法求解λ1，λ501chuzhi(u);d=mifa(u,MatrixC,0);chuzhi(u);a=mifa(u,MatrixC,d);if(d<0){d1=d;d501=a;}else{d501=d;d1=a;}printf("λ1=%.12e\n",d1);printf("λ501=%.12e\n",d501);//用反幂法求λschuzhi(u);ds=fmifa(u,MatrixC,0);printf("λs=%.12e\n",ds);//用带原点平移的反幂法求λikfor(i=1;i<=39;i++){a=d1+(i*(d501-d1))/40;assignment(MatrixC);chuzhi(u);d=fmifa(u,MatrixC,a);printf("与μ%02d=%+.12e最接近的特征值λi%02d=%+.12e\n",i,a,i,d);}//求A的条件数d=fabs((d1/ds));printf("A的（谱范数）条件数cond<A>2=%.12e\n",d);//求detAassignment(MatrixC);LU(MatrixC);a=1;for(i=0;i<=500;i++){a*=MatrixC[2][i];}printf("行列式detA=%.12e\n",a);//测试不同迭代初始向量对λ1计算结果的影响。
上海大学_王培康_数值分析大作业
数值分析大作业（2013年5月）金洋洋（12721512），机自系1.下列各数都是经过四舍五入得到的近似值，试分别指出它 们的绝对误差限, 相对误差限和有效数字的位数。
X1 =5.420, x 2 =0.5420, x 3=0.00542, x 4 =6000, x 5=50.610⨯解：根据定义：如果*x 的绝对误差限 不超过x 的某个数位的半个单位，则从*x 的首位非零数字到该位都是有效数字。
显然根据四舍五入原则得到的近视值，全部都是有效数字。
因而在这里有：n1=4, n2=4, n3=3, n4=4, n5=1 （n 表示x 有效数字的位数） 对x1：有a1=5, m1=1 （其中a1表示x 的首位非零数字，m1表示x1的整数位数） 所以有绝对误差限 14311(1)101022x ε--≤⨯=⨯相对误差限 31()0.510(1)0.00923%5.4201r x x x εε-⨯=== 对x2：有a2=5, m2=0 所以有绝对误差限 04411(2)101022x ε--≤⨯=⨯相对误差限 42()0.510(2)0.00923%0.54202r x x x εε-⨯=== 对x3：有a3=5, m3=-2 所以有绝对误差限 23511(3)101022x ε---≤⨯=⨯相对误差限 53()0.510(3)0.0923%0.005423r x x x εε-⨯=== 对x4：有a4=0, m4=4 所以有绝对误差限 4411(4)1022x ε-≤⨯=相对误差限 4()0.5(4)0.0083%60004r x x x εε=== 对x5：有a5=6, m5=5 所以有绝对误差限 51411(5)101022x ε-≤⨯=⨯相对误差限 45()0.510(5)8.3%600005r x x x εε⨯===2.对矩阵A 进行LU 分解， 并求解方程组Ax b =其中211132122A ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦，465b ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦解：A=LU 代入方程Ax b = 可转化为L y bU x y⎧=⎪⎨=⎪⎩先对矩阵A 进行LU 分解，如下11121311121321222321112112222113233132333111311232223113322333111u u u u u u A LU l u u l u l u u l u u l l u l u l u l u l u l u u ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥===++⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥+++⎣⎦⎣⎦⎣⎦根据系数相应相等有：第一行：112u =，121u =，131u = 第二行：21112121l u l ==，可得210.5l =211222220.513l u u u +=⨯+=，可得22 2.5u = 211323230.512l u u u +=⨯+=，可得23 1.5u =第三行：31113121l u l ==，可得310.5l =31123222320.51 2.52l u l u l +=⨯+=，可得320.6l =3113322333330.510.6 1.52l u l u u u ++=⨯+⨯+=，可得330.6u =所以有：12110.51 2.5 1.50.50.610.6A LU ⎡⎤⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦解方程如下123140.5160.50.615y y y ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦，可得123440.6y y y ⎡⎤⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦ 11223321142.5 1.540.60.6x y x y x y ⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦，可得123111x x x ⎡⎤⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦3. 用 J 迭代法和 G-S 迭代法求解方程组 123123123202324812231530x x x x x x x x x ++=⎧⎪++=⎨⎪-+=⎩时， 若取初始解向量(0)(0,0,0)T x = ， 问各需迭代多少次才能使误差()*610k x x-∞-≤ 。
北航数值分析大作业一
北京航空航天大学数值分析大作业一学院名称自动化专业方向控制工程学号ZY*******学生姓名许阳教师孙玉泉日期2021 年11月26 日设有501501⨯的实对称矩阵A ，⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡=5011A a b c b c c b c b a其中，064.0,16.0),501,,2,1(64.0)2.0sin()024.064.1(1.0-==⋅⋅⋅=--=c b i e i i a ii 。
矩阵A 的特征值为)501,,2,1(⋅⋅⋅=i i λ，并且有||min ||,501150121i i s λλλλλ≤≤=≤⋅⋅⋅≤≤1λ，501λ和s λ的值。
A 的与数4015011λλλμ-+=kk 最接近的特征值)39,,2,1(⋅⋅⋅=k k i λ。
A 的(谱范数)条件数2)A (cond 和行列式detA 。
一 方案设计1 求1λ，501λ和s λ的值。
s λ为按模最小特征值，||min ||5011i i s λλ≤≤=。
可使用反幂法求得。
1λ，501λ分别为最大特征值及最小特征值。
可使用幂法求出按模最大特征值，如结果为正，即为501λ，结果为负，那么为1λ。
使用位移的方式求得另一特征值即可。
2 求A 的与数4015011λλλμ-+=kk 最接近的特征值)39,...,2,1(=k k i λ。
题目可看成求以k μ为偏移量后，按模最小的特征值。
即以k μ为偏移量做位移，使用反幂法求出按模最小特征值后，加上k μ，即为所求。
3 求A 的(谱范数)条件数2)(A cond 和行列式detA 。
矩阵A 为非奇异对称矩阵，可知，||)(min max2λλ=A cond(1-1)其中m ax λ为按模最大特征值，min λ为按模最小特征值。
detA 可由LU 分解得到。
因LU 均为三角阵，那么其主对角线乘积即为A 的行列式。
二 算法实现1 幂法使用如下迭代格式：⎪⎪⎩⎪⎪⎨⎧⋅===⋅⋅⋅=------||max |)|sgn(max ||max /),,(111111)0()0(10k k k k k k k k Tn u u Ay u u u y u u u β任取非零向量 (2-1)终止迭代的控制理论使用εβββ≤--||/||1k k k ， 实际使用εβββ≤--||/||||||1k k k(2-2)由于不保存A 矩阵中的零元素，只保存主对角元素a[501]及b,c 值。
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。