中南大学数值分析作业完整版~
《数值分析》作业参考答案-2

《数值分析》作业参考答案一. 选择题1. A ; 2.B ; 3.B ; 4.D; 5.C; 6.D; 7.C; 8.B; 9.D; 10.C ; 11.B ; 12.A ; 13.A; 14.C; 15.A; 16.B; 17.D; 18.A 19.D,20.C,21.A,22.D,23.C,24.C. 二. 填空题1. 3,3,3 ;2. 1,2/3 ;3. 100!2^100 ;4. (-1 ,1);5.)())((102010n x x x x x x ---Λ ; 6. xxx x x g sin 1cos )(+--=,2;7. (-1, 1); 8. x ; 9. 4 ; 10. 5,9 ; 11. )(211nn n x cx x +=+, 2; 12. 31x x ++ ; 13. 10/9, 4; 14. 10, 55, 550; 15. 3ab b -+ 16. 312-x 17. 3118.431,431,21i i +-- 19.x x +22 20. 3b a a -+ . 三.1. 12)(2++=x x x p 2. )()(x f x p = 3.12292.512,916,910====a B C A , 代数精确度为5 3.证明:||)'(||||'||)'(1-⋅=A A A A A A cond设}'m ax {的特征值的模A A =λ,})'m ax {(11的特征值的模--=A A β,则 上式=2212))((||||||||A cond AA =⋅=⋅-βλ4.1. (12分)⎪⎪⎪⎪⎪⎭⎫⎝⎛---=1111433221L ,⎪⎪⎪⎪⎪⎭⎫ ⎝⎛---=4534231112U ,⎪⎪⎪⎪⎪⎭⎫ ⎝⎛-=0101X 2. (8分)Seidel 收敛,因为A 实正定对称阵. 迭代格式⎪⎪⎩⎪⎪⎨⎧+=++-=+=+=+++++++2/)1(2/)2(2/)(2/)2()1(3)1(4)(4)1(2)1(3)(3)1(1)1(2)(2)1(1k k k k k k k k k k x x x x x x x x x x 5. 14)(22+-=x x p π,余项6*54|)2(|61)(cos 322ππ≤-≤-x x x p x 6.6. 证明:当0=A 时,结论显然成立;当0≠A 时,因222||||||||||||||X A Y AX Y T<,故 2220,0||||||||||||||sup A Y X AX Y T Y X ≤≠≠;又A A T是实对称矩阵,故存在正交阵),,,(21n p p p P Λ=使得⎪⎪⎪⎭⎫ ⎝⎛==n TT D AP A P λλO 1, i P 是特征值i λ对应的特征向量。
数值分析作业及参考答案

数值分析第一次作业及参考答案1. 设212S gt =,假定g 是准确的,而对t 的测量有0.1±秒的误差,证明当t 增加时S 的绝对误差增加,而相对误差却减少。
解:2**22211()0.122()0.10.2()1122,(),().r r e S S S gt gt gt e S gt e S t gt gt t e S e S =-=-====∴↑↑↓2. 设2()[,]f x C a b ∈且()()0f a f b ==,求证2''1max ()()max ().8a x ba xb f x b a f x ≤≤≤≤≤-解:由112,0),(,0)()()0()00.a b L x l x l x =⨯+⨯=(两点线性插值 插值余项为"111()()()()()()[,]2R x f x L x f x a x b a b ξξ=-=--∈ [,].x a b ∴∀∈有12211()()"()()()max "()[()()]221()()1max "()[]()max "().228a x ba xb a x b f x R x f x a x b f x x a b x x a b x f x b a f x ξ≤≤≤≤≤≤==--≤---+-≤=-21max ()()max "()8a xb a x b f x b a f x ≤≤≤≤∴≤-3. 已测得函数()y f x =的三对数据:(0,1),(-1,5),(2,-1),(1)用Lagrange 插值求二次插值多项式。
(2)构造差商表。
(3)用Newton 插值求二次插值多项式。
解:(1)Lagrange 插值基函数为0(1)(2)1()(1)(2)(01)(02)2x x l x x x +-==-+-+-同理 1211()(2),()(1)36l x x x l x x x =-=+ 故2202151()()(1)(2)(2)(1)23631i i i p x y l x x x x x x x x x =-==-+-+-++=-+∑(2)令0120,1,2x x x ==-=,则一阶差商、二阶差商为0112155(1)[,]4,[,]20(1)12f x x f x x ---==-==-----0124(2)[,,]102f x x x ---==-22()1(4)(0)1*(0)(1)31P x x x x x x =+--+-+=-+4. 在44x -≤≤上给出()xf x e =的等距节点函数表,若用二次插值求x e 的近似值,要使截断误差不超过610-,问使用函数表的步长h 应取多少?解:()40000(),(),[4,4],,,, 1.x k x f x e f x e e x x h x x h x x th t ==≤∈--+=+≤考察点及(3)200044343()()[(()]()[()]3!(1)(1)(1)(1)3!3!.(4,4).6f R x x x h x x x x h t t t e t h th t h e h e ξξ=----+-+≤+⋅⋅-=≤∈-则436((1)(1)100.006.t t t h --+±<< 在点 得5. 求2()f x x =在[a,b ]上的分段线性插值函数()h I x ,并估计误差。
中南大学数值分析试题1

习题一1、已知e=2.71828L 问下列近似值A x 有几位有效数字?相对误差界是多少?(1), 2.7A x e x == (2), 2.718A x e x ==(3)/100,0.027A x e x ==(4)/100,0.02718A x e x ==2、设原始数据的下列近似值每位都是有效数字:*1 1.1021x = *20.031x = *356.430x =试计算(1)***123x x x ++,(2)**23/x x ,并估计它们的相对误差界。
3、设x 的相对误差界是δ,求n x 的相对误差界。
4、正方形的边长为10cm,问测量边长的误差界多大时才能保证面积误差不超过20.1cm5、为了使计算球体体积时的相对误差不超过1%,问测量半径R 时允许的相对误差界是多少?6、三角函数值取四位有效数字,怎样计算1cos 2−o 才能保证精度?7、设0Y =28,按递推公式11,2,n n Y Y n −==L 计算,27.982≈时,100Y 将有多大的误差?8、下列公式是否要做变换才能避免有效数字的损失?如何变换?(1)sin sin x y − (2)arctan arctan x y −2− (4)(2(1)/2x e − 9、[]n M R 且非奇异,又设x 为n R 上的一个向量范数,定义P xPx =试证明P x 也是n R 上的一个向量范数。
10、为对称正定矩阵,定义试证明A x 也是n R 上的一个向量范数。
11、证明向量范数的等价性:(1)2x x ∞≤≤ (2) 1x x n x ∞∞≤≤(3) 21x x ≤≤12、设矩阵0.60.50.10.3A ⎛⎞=⎜⎟⎝⎠计算A 的行范数、列范数、2范数。
13、设A 为n 阶方阵,U 为n 阶正交阵,试证222||||||||||||AU UA A ==14、对算子范数 ||||⋅,设 ||||1B <,求证 11||()||.1||||I B B −≤±+ 15、举例说明矩阵的谱半径不是矩阵范数。
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')
中南大学数值分析试题7

2 2 −1
0
3 1 −2 , (2) A = 1 4 2 .
−2
1
2 1
10. 设矩阵A ∈ Rn×n 为Hessenberg形, 对QR变换 A = QR, B = QT AQ = RQ 证明矩阵Q 和B 都是Hessenberg形矩阵. 2
2=
1) 是A 的一个特征值及对应的特征
.
向量. 试证: 若有正交矩阵P 使得P x = e1 , 则有 P AP T = (b) 已知矩阵
λ
0
0 B
2 A= 10
10 5 −8
2
−8 .
2
11
的 一 个 特 征 值λ = 9和 对 应 的 特 征 向 量x = (2/3, 1/3, 2/3)T . 试 求 镜 面 反 射 矩 阵P 使得P x = e1 , 并计算P AP T . 7. 用正交相似变换将下列矩阵化为对称三对角矩阵:
1.0 1.0
0 1 3
0.5 0.25
0.25 .
0.5
2.0ห้องสมุดไป่ตู้
5. 设x = (1, 1, 1, 1)T , 用下列两种方法分别求正交矩阵P , 使得P x = ± (a) P 为平面旋转矩阵的乘积. (b) P 为镜面反射矩阵.
x
2
e1 .
1
6. (a) 设矩阵A ∈ Rn×n 为对称矩阵, λ 和x( x
习题七
1. 用幂法求下列矩阵的主特征值和主特征向量:
3 −2 −4 A= −2 6 −2 .
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

数值分析大作业数值分析大作业姓名:黄晨晨学号:S1*******学院:储运与建筑工程学院学院班级:储建研17-2实验3.1 Gauss消去法的数值稳定性实验实验目的:理解高斯消元过程中出现小主元即很小时引起方程组解数值不定性实验内容:求解方程组Ax=b,其中(1)A1=0.3×10?1559.14315.291?6.130?1211.29521211,b1=59.1746.7812;(2)A2=10?7013 2.099999999999625?15?10102,b2=85.90000000000151;实验要求:(1)计算矩阵的条件数,判断系数矩阵是良态的还是病态的(2)用Gauss列主元消去法求得L和U及解向量x1,x2∈R4(3)用不选主元的高斯消去法求得L和U及解向量x1,x2∈R4(4)观察小主元并分析对计算结果的影响(1)计算矩阵的条件数,判断系数矩阵是良态的还是病态的代码:format longeformat compactA1=[0.3*10^-15,59.14,3,1;5.291,-6.130,-1,2;11.2,9,5,2;1,2,1,1] b1=[59.17;46.78;1;2]n=4C1=cond(A1,1) %C1为A1矩阵1范数下的条件数C2=cond(A1,2) %C2为A1矩阵2范数下的条件数C3=cond(A1,inf) %C3为1矩阵谱范数下的条件数结果:C1 =1.362944708720448e+02C2 =6.842955771253409e+01C3 =8.431146*********e+01显然A1矩阵为病态矩阵将矩阵A2,b2输入上述代码中求得A2矩阵的条件数为:C1 =1.928316831682894e+01C2 =8.993938090170119e+00C3 =1.835643564356072e+01显然A2矩阵也为病态矩阵(2)用Gauss列主元消去法求得L和U及解向量x1,x2∈R4代码:for k=1:n-1a=max(abs(A1(k:n,k)))if a==0returnendfor i=k:nif abs(A1(i,k))==ay=A1(i,:)A1(i,:)=A1(k,:)A1(k,:)=yx=b1(i,:)b1(i,:)=b1(k,:)b1(k,:)=xbreakendendif A1(k,k)~=0A1(k+1:n,k)=A1(k+1:n,k)/A1(k,k)A1(k+1:n,k+1:n)=A1(k+1:n,k+1:n)-A1(k+1:n,k)*A1(k,k+1:n) elsebreakendendL=tril(A1,0);for i=1:nL(i,i)=1;endLU=triu(A1,0)y1=L\b1x1=U\y1得到如下结果:x1 =3.845714853511634e+001.609517394778522e+00-1.547605454206655e+011.041130489899787e+01将A2=[10,-7,0,1;-3,2.0999********,6,2;5,-1,5,-1;0,1,0,2]b2=[8;5.900000000001;5;1]代入上述代码求得结果如下:X2 =4.440892098500626e-16-9.999999999999993e-019.999999999999997e-011.000000000000000e+00(3)用不选主元的高斯消去法求得L和U及解向量x1,x2∈R4代码:format longeformat compactA1=[0.3*10^-15,59.14,3,1;5.291,-6.130,-1,2;11.2,9,5,2;1,2,1,1] b1=[59.17;46.78;1;2][L,U]=lu(A1)y1=L\b1x1=U\y1求得如下结果:x1=3.845714853511634e+001.609517394778522e+00-1.547605454206655e+011.041130489899787e+01将A2=[10,-7,0,1;-3,2.0999********,6,2;5,-1,5,-1;0,1,0,2] b2=[8;5.900000000001;5;1]代入上述代码,求得结果如下:x 2 =4.440892098500626e-16 -9.999999999999993e-01 9.999999999999997e-01 9.999999999999999e-01(2)(3)求得结果相同,可验证结果正确。
数值分析作业

数值分析课后作业:习题一1.在字长为3的十进制计算机上计算f (3.33)和g (3.33),其中f(x)=x 4-x 3+3x 2+x-2,g(x)=(((x-1)x+3)x+1)x-2解: m=3; f=@(x)digit(digit(x^4,m)- digit(x^3,m)+ digit(3*x^2,m)+ digit(x-2,m),m); g=@(x)digit(digit(digit( digit(digit(digit( (x-1)*x,m)+3,m)*x,m)+1,m)*x,m)-2,m); f(3.33) g(3.33) 有ans = 121 ans =121 2.下列各近似值的绝对误差限都是1021⨯-3,试指出它们各有几位有效数字:x=1.00052, y=0.05, z=0.00052.解:当 x=1.00052时, 由丨X*—X 丨 ≤0.5×10-3 得 x=1.00052 有四位有效数字; 同理 y=0052 有两位有效数字 Z=0.00052有零位有效数字 3,计算圆的面积,要使其相对误差限为1%,问测量半径r 允许的相对误差限是多少? 解:设圆的面积为S , 由题意有|e(S)|≤1%。
又S=πr 2 dS=2πr dr 所以 dS/S=(2πrdr)/(πr 2)=2(dr/r)∴|e(r)|≈21|e(S)|≤0.5×1%=0.5% 11.数组与矩阵是Matlab 编程的基础,试学习Matlab 的数组与矩阵的表示方法,并举例介绍数组、矩阵的常见运算. 解:>> syms a b c d; >> a=[1 2 3];>> b=[4 5 6];>> a+bans =5 7 9>> b-aans =3 3 3>> a.*bans =4 10 18 >> a.^2 ans = 1 4 9>> c=[1 2 3;1 2 3;1 2 3];>> d=[4 5 6;4 5 6;4 5 6];>> cc = 1 2 3 1 2 3 1 2 3d = 4 5 6 4 5 6 4 5 6 >> c+dans =5 7 9 5 7 9 5 7 9>> d-cans = 3 3 33 3 33 3 3 12.学习使用Matlab 命令help 和doc 学习自己感兴趣的Matlab 的运算、函数或命令的用法,并对于任意给定的实数a,b,c,编写Matlab 程序求方程ax 2+bx+c=0的根. 解:x 1=a ac b b b 24)sgn(2---, x 2=1ax c1 x>0 其中 sgn = 0 x=0 -1 x<0 disp('Please input the coefficients of');disp('quadratic equation ax^2+bx+c=0, respectively') a=input('a='); b=input('b='); c=input('c=');m=3; if abs(a)<eps & abs(b)<eps error End if abs(a)<eps disp('Since a=0, quadrtic equation degen erates into a linear equation.') disp('The only solution of the linear equtio n is')x=digit(-c/b,m) return Enddelta=b^2-4*a*c; temp=sqrt(delta); x 1=(-b+temp)/(2*a) ; x 2=(-b-temp)/(2*a) ;err1=abs(a*x 1^2+b*x 1+c) ; err2=abs(a*x 2^2+b*x 2+c) ; if b>0x 1=(-b-temp)/(2*a) End if b<0x 1=(-b+temp)/(2*a) End if b=0x 1=temp/(2*a) Endx 2=c/(a*x 1)err1=abs(a*x 1^2+b*x 1+c) err2=abs(a*x 2^2+b*x 2+c) if abs(a)<epsdisp('Since a=0, quadrtic equation degen erates into a linear equation.')disp('The only solution of the linear equtio n is')x=digit(-c/b,m) return Enddelta=digit(digit(b^2,m)-digit(4*digit(a*c,m),m),m);temp=digit(sqrt(delta),m);x 1=digit(digit(-b+temp,m)/digit(2*a,m),m); x 2=digit(digit(-b-temp,m)/digit(2*a,m),m); err1=abs(a*x 1^2+b*x 1+c); err2=abs(a*x 2^2+b*x 2+c); if b>0x 1=digit(digit(-b-temp,m)/digit(2*a,m),m) ; End if b<0x 1=digit(digit(-b+temp,m)/digit(2*a,m),m); End if b=0x 1=digit(temp/digit(2*a,m),m); Endx 2=digit(digit(c/a,m)/x1,m) ; err1=abs(a*x 1^2+b*x 1+c) ; err2=abs(a*x 2^2+b*x 2+c) ; 14分别利用ln (1+x)=11,)1(11≤<--+∞=∑x nx nn n 和ln11...),12...53(2111253<<-++++++=-++x n x x x x x x n ,给出计算ln2的近似方法,编写相应的Matlab 程序,并比较算法运行情况. 解:方法一: x=1; s=0;for k=1:100s=s+(-1)^(k+1)*(x^k)/k; end sq=log(2)err=abs(t-q) ans= t =0.6882 q =0.6931 err = 0.0050方法二x=1/3; s=0;for k=1:2:100 s=s+(x^k)/k; end t=2*s q=log(2)err=abs(t-q) Ans= t =0.6931 q =0.6931 err =2.2204e-16所以方法二较方法一好。
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数值分析
数学实验报告
姓名:XX
学号:xx
指导老师:***
专业班级:xx
目录
1. 高斯消去法 (3)
2. LU分解 (6)
3. 用牛顿法求积分 (10)
4. 用复化梯形法求积分 (12)
5. 用复化辛普森法、复化辛普森变步长法求积分 (13)
6. 节点加密复化梯形公式 (16)
7. 龙贝格积分 (17)
8. 欧拉方法、休恩方法、泰勒方法、龙格-库塔方法 (20)
一.高斯消去法
x(i)=(b(i)-sum)/A(i,i);
end
4:实验结果:
(1)高斯消去法
(2)高斯列主元消去法
5:实验总结
这两个程序让我对高斯消去法有了更深刻的理解,能更灵活的运用各种基础函数与矩阵的运算来进行求解,参考了书上的已有程序流程图,程思想需要进一步完善,需要对函数的进一步熟悉。
姓名:xx
2012年3月25日
二. LU分解
m=j;
end
end
if m~=i
for k=1:n
c(k)=A(i,k);
A(i,k)=A(m,k);
A(m,k)=c(k);
end
t=b(i);
b(i)=b(m);
b(m)=t;
end
for j=i:n
for k=1:i-1
M(k)=L(i,k)*U(k,j);
end
U(i,j)=A(i,j)-sum(M);
end
for j=i+1:n
for k=1:i-1
M(k)=L(j,k)*U(k,i);
end
L(j,i)=(A(j,i)-sum(M))/U(i,i);
end
end
x=U\(L\b);
4:实验结果:
(1)普通LU分解
三. 用牛顿法求积分
(2)列主元LU 分解
5:实验总结
L U 分解在上学期已经学习过,这次的实验让我对LU 分解有了更深的了解,又掌握了一种解线性方程组的好方法。
姓名:XX
2012年3月29日 学号
XX 班级 XX 姓名
XX 指导教师 易昆南 实验题目 用牛顿法求积分
评 分
1、设计(实习)目的:
1. 进一步了解牛顿法及其应用 2.
进一步理解牛顿法求积分的思想
2、实验内容:
用牛顿法求函数x x x x f ++=2
3
)(的积分 3.详细设计:
function y=newton(a,b,n) x=a:(b-a)/n:b; %插值节 y=x.^3+x.^2+x;
四.用复化梯形法求积分
五. 用复化辛普森法、复化辛普森变步长法求积分
六.节点加密复化梯形公式
七.龙贝格积分
八.欧拉方法、休恩方法、泰勒方法、龙格-库塔方法
4.龙格-库塔方法
>> [x1 y1] =lungkuta(1)
[x2 y2] =lungkuta(1/2)
[x3 y3] =lungkuta(1/4)
[x4 y4] =lungkuta(1/8)
plot(x1,y1,'-',x2,y2,'r',x3,y3,'g',x4,y4,'b')
x1 =
1 2 3
y1 =
0.769531250000000 1.043746948242188 1.615647614002228
x2 =
0.500000000000000 1.000000000000000 1.500000000000000 2.000000000000000 2.500000000000000 3.000000000000000
y2 =
0.935424804687500 0.466060072183609 0.450289419204637 0.558880619571974 0.701568099555500 0.853603753857624
x3 =
Columns 1 through 6
5:实验总结
用数值分析中的方法编程求积分它能帮助我们简化繁琐又难以计算的数学问题。
在设计过程中我发现原来以为枯燥无味的数学其实是趣味无穷的,这让我更加坚定了学习这门课程的信心与动力,相信通过深入地学习,今后定能有所收获。
.设计过程中,我深刻体会到与同学交流是十分重要的。
有时候一个小小的错误自己总是检查不出来,但让同学帮忙看一下马上就查出来了。
姓名:XX
2012年4月12日。