数值分析上机第四次作业

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。

数值分析上机作业（四）常微分方程数值解一、问题描述考虑以下问题2000()999.75()1000.25()()du u t v t dtdv u t v t dt=−++=−初始条件为(0)0,(0)2u v ==−。

其精确解为0.52000.50.52000.5() 1.4998750.4998751() 2.999750.000251t t t t u t e e v t e e −−−−=−++=−−+请分别用古典四级四阶显式Runge-Kutta 方法和隐式二级四阶Runge-Kutta 方法计算，计算区间取成[0,20]，并与精确解比较。

二、方法描述2.1古典四级四阶显式Runge-Kutta 方法考虑一阶常微分方程的初值问题000'(,),[,]()y f x y x x b y x y =∈=（1）经典四级四阶Runge-Kutta 方法的计算公式为112341213243(22)6(,)(,)22(,)22(,)n n n n n n n n n n h y y K K K K K f x y h h K f x y K h h K f x y K K f x h y hK +=++++==++=++=++四阶Runge-Kutta 方法的每一步需要计算四次函数值f ，可以证明其阶段误差为5()O h 。

再考虑一阶常微分方程组的初值问题0000'(,(),())'(,(),())()()x f t x t y t y g t x t y t x t x y t y ====（2）四阶Runge-Kutta 公式为1123411234(22)6(22)6n n n n h x x f f f f h y y g g g g ++=++++=++++其中11211211322322433433(,,)(,,)(,,)222(,,)222(,,)222(,,)222(,,)(,,)n n n n n n n n n n n n n n n n n n n n n n n n f f t x y g g t x y h h h f f t x f y g h h h g g t x f y g h h h f f t x f y g h h h g g t x f y g f f t h x hf y hg g g t h x hf y hg ===+++=+++=+++=+++=+++=+++2.2隐式二级四阶Runge-Kutta 方法考虑一阶常微分方程的初值问题（1），其计算公式为112112212()2111((,()244111((,()244m m m m m m h y y K K K f x h y hK hK K f x h y hK hK +=++=+++=++++再考虑一阶常微分方程组的初值问题（2），其计算公式为112112()2()2n n n n h x x f f h y y g g ++=++=++112121121221212211111((,(,()2444411111((,(,()2444411111((,(,()2444411((,(24m m m m m m m m m m m f f t h x hf hf y hg hg g g t h x hf hf y hg hg f f t h x hf hf y hg hg g g t h x hf =+++++−=+++++−=++++++=+++1212111,()444m hf y hg hg +++三、方案设计3.1给定函数将给定的初值问题写成通用形式，即编写函数将这个函数保存为f.m。

第一章第二题(1) 截断误差为104-时：k=1;n=0;m=0;x=0;e=1e-4;while k==1x1=x+(-1)^n/(2*n+1);if abs(x-x1)<ey=4*x1;m=n+1;break;endx=x1;k=1;n=n+1;endformat longy,my =3.141792613595791m =5001（2）截断误差为108-时：k=1;n=0;m=0;x=0;e=1e-8;while k==1x1=x+(-1)^n/(2*n+1);if abs(x-x1)<ey=4*x1;m=n+1;break;endx=x1;k=1;n=n+1;endformat longy,my =3.141592673590250m =50000001由以上计算可知，截断误差小于104-时，应取5001项求和，π=3.141792613595791；截断误差小于108-时，应取50000001项求和，π=3.141592673590250。

第二章第二题a=[0 -2 -2 -2 -2 -2 -2 -2];b=[2 5 5 5 5 5 5 5];c=[-2 -2 -2 -2 -2 -2 -2 0];v=220;r=27;d=[v/r 0 0 0 0 0 0 0];n=8;for i=2:na(i)=a(i)/b(i-1);b(i)=b(i)-c(n-1)*a(i);d(i)=d(i)-a(i)*d(i-1);end;d(n)=d(n)/b(n);for i=n-1:-1:1d(i)=(d(i)-c(i)*d(i+1));end;I=d'I =1.0e+002 *1.490717294184090.704617906351300.311568212434910.128623612390290.049496991380330.017168822994210.004772412363470.00047741598598第三章第一题(1)Jacobi迭代法：b=[12;-27;14;-17;12]x = [0;0;0;0;0;]k = 0;r = 1;e=0.000001A=[10,1,2,3,4;1,9,-1,2,-3;2,-1,7,3,-5;3,2,3,12,-1;4,-3,-5,-1,15;] D = diag(diag(A));B = inv(D)*(D-A);f = inv(D)*b;p = max(abs(eig(B)));if p >= 1'迭代法不收敛'returnendwhile r >ex0 = x;x = B*x0 + f;k = k + 1;r = norm (x-x0,inf);endxk计算结果：x =1.0000-2.00003.0000-2.00001.0000k =65(2) 高斯赛德尔迭代：A=[10,1,2,3,4;1,9,-1,2,-3;2,-1,7,3,-5;3,2,3,12,-1;4,-3,-5,-1,15;]x=[0;0;0;0;0];b=[12;-27;14;-17;12]c=0.000001L=-tril(A,-1)U=-triu(A,1)D=(diag(diag(A)))X=inv(D-L)*U*x+inv(D-L)*b;k=1;while norm(X-x,inf)>= cx=X;X=inv(D-L)*U*x+inv(D-L)*b;k=k+1;endXk计算结果：X =1.0000-2.00003.0000-2.00001.0000k =37(3) SOR：A=[10,1,2,3,4;1,9,-1,2,-3;2,-1,7,3,-5;3,2,3,12,-1;4,-3,-5,-1,15] x=[0;0;0;0;0];b=[12;-27;14;-17;12]e=0.000001w=1.44;L=-tril(A,-1)U=-triu(A,1)D=(diag(diag(A)))X=inv(D-w*L)*((1-w)*D+w*U)*x+w*inv(D-w*L)*bn=1;while norm(X-x,inf)>=ex=X;X=inv(D-w*L)*((1-w)*D+w*U)*x+w*inv(D-w*L)*b;n=n+1;endXn计算结果：X =1.0000-2.00003.0000-2.00001.0000n =22由以上可知，共轭梯度法收敛速度明显快于Jacobi法和G-S法。

第一章（1）按从大到小的顺序计算S的程序为：N#include<stdio.h>void main(){float n,j,s=0;printf("请输入n：");scanf("%f",&n);for(j=2;j<=n;j++)s=s+1/(j*j-1);printf("s=%f\n",s);}(2)按从小到大的顺序计算S的程序为：N#include<stdio.h>void main(){float n,j,s=0;printf("请输入n：");scanf("%f",&n);for(j=n;j>=2;j--)s=s+1/(j*j-1);printf("s=%f\n",s);}(3)精确值，从大到小，从小到大的计算程序为：#include<stdio.h>void main(){float n,j,s1,s2=0,s3=0;printf("请输入n：");scanf("%f",&n);s1=(3.0/2-1/n-1/(n+1))/2;for(j=2;j<=n;j++)s2=s2+1/(j*j-1);for(j=n;j>=2;j--)s3=s3+1/(j*j-1);printf("精确算法的结果为：%f\n",s1);printf("从大到小顺序的结果为：%f\n",s2); printf("从小到大顺序的结果为：%f\n",s3); }结果如下：精确值从大到小的值从小到大的值有效位数从大到小从小到大210S0.740049 0.740049 0.74005 6 5410S0.7499 0.749852 0.7499 4 4610S0.749999 0.749852 0.749999 3 6(4)通过本上机题，我们可以看出：当n较小时，两种算法的结果都很接近精确值，而当n较大时，两种算法的结果差别较大，按从大到小的顺序计算的值与精确值有较大的误差，而按从小到大的顺序计算的值与精确值吻合。

数值分析上机第四次作业实验项目：共轭梯度法求解对称正定的线性方程组实验内容：用共轭梯度法求解下面方程组(1) 123421003131020141100155x x x x -⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪--- ⎪ ⎪ ⎪= ⎪ ⎪ ⎪-- ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭ 迭代20次或满足()(1)1110k k x x --∞-<时停止计算。

(2) Ax b =,A 是1000阶的Hilbert 矩阵或如下的三对角矩阵，A[i,i]=4，A[i,i-1]=A[i-1,i]=-1，i=2,3,..,nb[1]=3, b[n]=3, b[i]=2，i=2,3,…,n-1迭代10000次或满足()()710k k r b Ax -=-≤时停止计算。

（3）*考虑模型问题，方程为222222(),(,)(0,1)(0,1)(0,)1,(1,),01(,0)1,(,1),01xy y x u u x y e x y D x yu y u y e y u x u x e x ∂∂+=+∈=⨯∂∂==≤≤==≤≤用正方形网格离散化，若取1/,10h N N ==，得到100n =的线性方程组，并用共轭梯度法（CG 法）求解，并对解作图。

实验要求：迭代初值可以取01(,1,...,)ij u i j N ==，计算到32||||10k r -≤停止．本题有精确解(,)xy u x y e =，这里k u 表示以k ij u 为分量的向量，u 表示在相应点(,)i j 上取值作为分量的向量．实验一：（1）编制函数子程序CGmethod 。

function [x,k]=CGmethod(A,b)n=length(A);x=zeros(n,1);r=b-A*x;rho=r'*r;k=0;while rho>10^(-12) & k<20k=k+1;if k==1p=r;elsebeta=rho/rho1;p=r+beta*p;endw=A*p;alpha=rho/(p'*w);x=x+alpha*p;r=r-alpha*w;rho1=rho;rho=r'*r;end编制主程序shiyan1_1：clear,clcA=[2,-1,0,0;-1,3,-1,0;0,-1,4,1;0,0,-1,5]; b=[3,-2,1,5]';[x,k]=CGmethod(A,b)运行结果为：x =1.3882-0.2855-0.02220.9367k =20（2）编制函数子程序CGmethod_1function [x,k]=CGmethod_1(A,b)n=length(A);x(1:n,1)=0;r=b-A*x;r1=r;k=0;while norm(r1,1)>=10^(-7)&k<10^4k=k+1;if k==1p=r;elsebeta=(r1'*r1)/(r'*r);p=r1+beta*p;endr=r1;w=A*p;alpha=(r'*r)/(p'*w);x=x+alpha*p;r1=r-alpha*w;end编制主程序shiyan1_2:clear,clcn=1000;A=hilb(n);b=sum(A')';[x,k]=CGmethod_1(A,b)运行结果为：x 的值,均接近1，迭代次数k=32实验二实验目的：用复化Simpson 方法、自适应复化梯形方法和Romberg 方法求数值积分。

第一章一、题目设∑=-=Nj N j S 2211，其精确值为)11123(21+--N N 。

（1）编制按从大到小的顺序11131121222-+⋯⋯+-+-=N S N ，计算SN 的通用程序。

（2）编制按从小到大的顺序1211)1(111222-+⋯⋯+--+-=N N S N ，计算SN 的通用程序。

（3）按两种顺序分别计算64210,10,10S S S ,并指出有效位数。

（编制程序时用单精度） （4）通过本次上机题，你明白了什么？ 二、MATLAB 程序N=input('请输入N(N>1)：');AccurateValue=single((0-1/(N+1)-1/N+3/2)/2); %single 使其为单精度 Sn1=single(0); %从小到大的顺序 for a=2:N; Sn1=Sn1+1/(a^2-1); endSn2=single(0); %从大到小的顺序 for a=2:N; Sn2=Sn2+1/((N-a+2)^2-1); endfprintf('Sn 的值 (N=%d)\n',N);disp('____________________________________________________') fprintf('精确值 %f\n',AccurateValue); fprintf('从大到小计算的结果 %f\n',Sn1); fprintf('从小到大计算的结果 %f\n',Sn2);disp('____________________________________________________')三、结果请输入N(N>1)：100Sn的值(N=100)____________________________________________________精确值0.740049从大到小计算的结果0.740049从小到大计算的结果0.740050____________________________________________________请输入N(N>1)：10000Sn的值(N=10000)____________________________________________________精确值0.749900从大到小计算的结果0.749852从小到大计算的结果0.749900____________________________________________________请输入N(N>1)：1000000Sn的值(N=1000000)____________________________________________________精确值0.749999从大到小计算的结果0.749852从小到大计算的结果0.749999____________________________________________________四、结果分析可以得出，算法对误差的传播又一定的影响，在计算时选一种好的算法可以使结果更为精确。

数值分析上机作业（1、2、3、4、6章）第一章17. 舍入误差与有效数设2211NN j S j ==-∑，其精确值为1311221N N ⎛⎫-- ⎪+⎝⎭。

（1）编制按从大到小的顺序22211121311N S N =+++---，计算N S 的通用程序； （2）编制按从小到大的顺序2221111(1)121N S N N =+++----，计算N S 的通用程序；（3）按两种顺序分别计算210S ，410S ，610S ，并指出有效位数（编制程序时使用单精度）；（4）通过本上机题你明白了什么？运行结果：按从大到小的顺序计算得：N N S误差e有效位数2⨯8 100.7400495 94.95049501392230710-4⨯ 4 100.7498521 54.79049995000258010-6⨯ 3 100.7498521 41.46900000499994310-按从小到大的顺序计算得：N N S误差e有效位数2⨯84.95049501392230710-100.7400495 944.99950003618465610-⨯8 100.7499000 965.00044450291170510-⨯11 100.7499990 13（4）通过本题可以看出，不同算法造成的误差是不同的，好的算法可以让计算结果精度更高。

对于本题，当采用从大到小的顺序累加计算时，计算误差随着N的增大而增大；当采用从小到大的顺序累加计算时，计算误差随着N的增大而减小。

因此在N比较大时宜采用从小到达的顺序累加计算。

第二章20.Newton 迭代法（1）给定初值0x 及容许误差ε，编制Newton 法解方程()0f x =根的通用程序；（2）给定方程3()03x f x x =-=，易知其有三个根*1x =，*20x =，*3x =①由Newton 方法的局部收敛性可知存在0δ>，当0(,)x δδ∈-时Newton 迭代序列收敛于根*2x ，试确定尽可能大的δ；②试取若干初始值，观察当0(,1)x ∈-∞-，(1,)δ--，(,)δδ-，(,1)δ，(1,)+∞时Newton 序列是否收敛以及收敛于哪一个根；（3）通过本上机题，你明白了什么？本实验取610ε-=，找到的最大的0.774597δ=②当0(,1)x ∈-∞-时，计算结果如下：x0 xend -1000 -1.732051 -500 -1.732051 -100 -1.732051 -10 -1.732051 -1.5-1.732051Newton 序列收敛于-1.732051当0(1,)xδ∈--时，计算结果如下：x0 xend-0.95 1.732051-0.90 1.732051-0.85 1.732051-0.80 -1.732051-0.78 -1.732051 Newton序列收敛于1.732051或-1.732051当0(,)xδδ∈-时，计算结果如下：x0 xend-0.77 0.000000-0.5 0.000000-0.1 0.0000000.3 0.0000000.77 0.000000 Newton序列收敛于0当0(,1)xδ∈时，计算结果如下：x0 xend0.78 1.7320510.80 1.7320510.85 -1.7320510.90 -1.7320510.95 -1.732051 Newton序列收敛于1.732051或-1.732051当0(1,)x∈+∞时，计算结果如下：x0 xend1.5 1.73205110 1.732051100 1.732051500 1.7320511000 1.732051Newton序列收敛于1.732051（3）通过本题发现，Newton迭代法解方程初始值的选取非常重要，不同的初始值会收敛于方程不同的根，且有些区间是全局收敛，有些区间是局部收敛。

《数值分析》上机习题1.用Newton法求方程X7 - 28X4 + 14=0在（0.1，1.9）中的近似根（初始近似值取为区间端点，迭代6次或误差小于0.00001）。

#include<stdio.h>#include<math.h>int main(){ float x1,x,f1,f2;static int count=0;x1=0.1 ;//定义初始值do{x=x1;f1=x*(x*x*x*x*x*x-28*x*x*x)+14;f2=7*x*x*x*x*x*x-4*28*x*x*x;//对函数f1求导x1=x-f1/f2; count++; }while(fabs(x1-x)>=1e-5);printf("%8.7f\n",x1); printf("%d\n",count);return 0;}2.已知函数值如下表：试用三次样条插值求f(4.563)及f’(4.563)的近似值。

include <stdio.h>#include <math.h>#define N 11main(){double B[N+1][N+1],m,x,u[N+1],y[N+1],c[N+1],d[N+1];double e[N+1]={2,0,4.15888308,6.5916738,8.3177664,9.6566268,10.750557,11.675460 6,12.47667,13.1833476,13.8155106,14.0155106};int i;x=4.563;B[0][0]=-2;B[0][1]=-4;B[N][N-1]=4;B[N][N]=2;for(i=1;i<N;i++){B[i][i-1]=1;B[i][i]=4;B[i][i+1]=1;}u[0]=B[0][0];y[0]=e[0];for(i=1;i<N;i++){m=B[i][i-1]/u[i-1];u[i]=B[i][i]-m*B[i-1][i];y[i]=e[i]-m*y[i-1];}c[N]=y[N]/u[N];for(i=N-1;i>=0;i--)c[i]=(y[i]-B[i][i+1]*c[i+1])/u[i];for(i=0;i<12;i++){m=fabs(x-i);if(m>=2)d[i]=0;else if(m<=1)d[i]=0.5*fabs(pow(m,3))-m*m+2.0/3;elsed[i]=(-1.0/6.0)*fabs(pow(m,3))+m*m-2*fabs(m)+4/3.0;} m=0;for(i=0;i<12;i++) m=m+c[i]*d[i];printf("f(%4.3f)=%f\n",x,m);printf("f'(4.563)=%lf\n",(c[4]-c[2])/2); }3.用Romberg 算法求)00001.0(sin )75(32314.1=+⎰ε允许误差dx x x x x . #include "stdafx.h" #include<stdio.h> #include<math.h> float f(float x) {float f=0.0;f=pow(3.0,x)*pow(x,1.4)*(5*x+7)*sin(x*x); return (f); } main() {int i=1,j,k,n=12;float T[12],a=1.0,b=3.0,s=0.0; T[0]=0.5*(b-a)*(f(a)+f(b));for(j=1;j<n-1;j++){ for(k=1;k<=pow(2,j-1);k++) s+=f(a+(2*k-1)*(b-a)/pow(2,j)); T[j]=0.5*(T[j-1]+(b-a)*s/pow(2,j-1)); s=0.0; }T[11]=(4*T[1]-T[0])/(float)3;for(;fabs(T[11]-T[0])>0.00001;i++) {T[0]=T[11];for(j=1;j<n-1-i;j++) T[j]=(pow(4,i)*T[j+1]-T[j])/(pow(4,i)-1); T[11]=(pow(4,i+1)*T[1]-T[0])/(pow(4,i+1)-1); }printf("%f\n",T[11]); }4. 用定步长四阶Runge-Kutta 求解一、程序要求用定步长四阶法求解 y1’=1y2’=y3 y3’=1000-1000y2-100y3(y1(0)=y2(0)=y3(0)=0) h=0.0005,打印yi(0.025),yi(0.045),yi(0.085),yi(0.1),(i=1,2,3)h =0.0005，打印y i (0.025) ，⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧===--===0)0(0)0(0)0(10010001000//1/321323321y y y y y dt dy ydt dy dt dyy i(0.045) ，y i(0.085) ，y i(0.1) ，(i=1，2，3)#include "stdafx.h"#include <stdio.h>#include <math.h>double F(double x,double y[4],double f[4]){f[1]=0*x+0*y[1]+0*y[2]+0*y[3]+1;f[2]=0*x+0*y[1]+0*y[2]+1*y[3]+0;f[3]=0*x+0*y[1]-1000*y[2]-100*y[3]+1000;return(1);}void main(){double F(double x,double y[4],double f[4]);doubleh=0.0005,x=0,Y[4],k[5][4],s[4],f[4],det,m[4]={0.025,0.045,0.085,0.1};int i,j,t; for(t=0;t<=3;t++)/*龙格-库塔算法*/{for(j=0;j<=3;j++)Y[j]=0; //每求一组值后将初值条件还原为0 for(i=1;i<=int(m[t]/h);i++){for(j=1;j<=3;j++)s[j]=Y[j];det=F(x,s,f);for(j=1;j<=3;j++)k[1][j]=h*f[j]; /*四阶古典公式中的k值和求和的计算*/ for(j=1;j<=3;j++)s[j]=Y[j]+0.5*k[1][j];det=F(x+0.5*h,s,f);for(j=1;j<=3;j++)k[2][j]=h*f[j];for(j=1;j<=3;j++)s[j]=Y[j]+0.5*k[2][j];det=F(x+0.5*h,s,f);for(j=1;j<=3;j++)k[3][j]=h*f[j];for(j=1;j<=3;j++)s[j]=Y[j]+k[3][j];det=F(x+h,s,f);for(j=1;j<=3;j++)k[4][j]=h*f[j];for(j=1;j<=3;j++)Y[j]=Y[j]+(k[1][j]+2*k[2][j]+2*k[3][j]+k[4][j])/6;x+=h;} for(j=1;j<=3;j++)printf("y[%d](%f)=%f ",j,m[t],Y[j]); printf("\n"); } } 5.⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡=40.00001 4.446782 2.213474- 0.238417 1.784317 0.037585- 1.010103- 3.123124 2.031743- 4.446782 30.719334 3.123789 1.103456- 2.121314 0.71828- 0.336993 1.112348 3.067813 2.213474- 3.123789 14.7138465 0.103458- 3.111223- 2.101023 1.056781- 0.784165- 1.7423820.238417 1.103456- 0.103458- 9.789365 0.431637 3.741856- 1.836742 1.563849 0.718719 1.7843172.1213143.111223- 0.431637 19.8979184.101011 2.031454 2.189736 0.113584-0.037585- 0.71828- 2.101023 3.741856- 4.101011 27.108437 3.123848 1.012345- 1.112336 1.010103- 0.336993 1.056781- 1.836742 2.031454 3.123848 15.567914 3.125432- 1.061074- 3.123124 1.112348 0.784165- 1.563849 2.189736 1.012345- 3.125432- 19.141823 2.115237 2.031743- 3.067813 1.742382 0.718719 0.113584- 1.112336 1.061074- 2.115237 12.38412A Tb )5.6784392- 4.719345 1.1101230 86.612343- 1.784317 0.84671695 25.173417- 33.992318 2.1874369(= 用列主元消去法求解Ax=b 。

如有帮助欢迎下载支持数值分析上机题姓名：陈作添学号： 040816习题 120．（上机题）舍入误差与有效数N11 3 1 1设S N，其精确值为 。

22 2 NN 1j 2j1（1）编制按从大到小的顺序111 ，计算 S 的通用程序。

S N1 321N 21 N22（2）编制按从小到大的顺序111，计算 S 的通用程序。

S N1(N 1)2 122 1NN 2（3）按两种顺序分别计算S 102 ， S 104 ， S 106 ，并指出有效位数。

（编制程序时用单精度）（4）通过本上机题，你明白了什么？按从大到小的顺序计算 S N 的通用程序为： 按从小到大的顺序计算 S N 的通用程序为：#include<iostream.h> #include<iostream.h> float sum(float N) float sum(float N) {{float j,s,sum=0; float j,s,sum=0; for(j=2;j<=N;j++) for(j=N;j>=2;j--) {{s=1/(j*j-1); s=1/(j*j-1); sum+=s;sum+=s;}}return sum;return sum;}}从大到小的顺序的值从小到大的顺序的值精确值有效位数从大到小从小到大0.7400490.740050.74004965 S 1020.7498520.74990.74994 4S 1040.7498520.7499990.74999936S 106通过本上机题， 看出按两种不同的顺序计算的结果是不相同的，按从大到小的顺序计算的值与精确值有较大的误差， 而按从小到大的顺序计算的值与精确值吻合。

从大到小的顺序计算得到的结果的有效位数少。

计算机在进行数值计算时会出现“大数吃小数”的现象，导致计算结果的精度有所降低，我们在计算机中进行同号数的加法时，采用绝对值较小者先加的算法，其结果的相对误差较小。

《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 b ak h f x dx f a f b f a kh -=⎰≈+++∑，截断误差为：322()''()''(),[,]1212Tb a b a R f h f a b nηηη--=-=-∈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 ai i h f x dx f a f b f x f x --==≈+++∑∑⎰，截断误差为：4(4)(),[,]180sb a R h fa 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 1fn 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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0.89235.445520.94844.344880.89435.730220.9544.701070.89636.017210.95245.060110.89836.306510.95445.422040.936.598120.95645.786870.90236.892080.95846.154630.90437.188410.9646.525350.90637.487110.96246.899050.90837.788210.96447.275750.9138.091730.96647.655470.91238.397680.96848.038240.91438.70610.9748.424090.91639.016990.97248.813040.91839.330380.97449.205110.9239.646280.97649.600330.92239.964720.97849.998720.92440.285720.9850.400320.92640.60930.98250.805140.92840.935480.98451.213210.9341.264280.98651.624560.93241.595720.98852.039210.93441.929820.9952.45720.93642.26660.99252.878540.93842.606090.99453.303270.9442.948310.99653.73140.94243.293270.99854.162980.94443.64101154.59802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 8.004469675Gauss积分数据：0102030405060四、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。