数值分析实验报告

时间：2024.3.31

第一次作业

题目

[作业1]. 试使用subplot 命令,生成[-1,1]上的第二个至第五个Chebyshev 多项

式，并按照2*2 方式多窗口显示.

[选做题1-2] 应用秦九韶算法计算多项式

0 1 1 0 ( ) n n ( 0)

n n p x a x a x + a x a a

- = + +L+ + ¹

在x = x*处的函数值和导数值。试编程实现，并举例运行给出计算结果。

程序以及结果

第一题

x=-1:0.05:1;

t0=1.0+0*x;

t1=x;

t2=2*x.*t1-t0;

t3=2*x.*t2-t1;

t3=2*x.*t3-t2;

t3=2*x.*t2-t1;

t4=2*x.*t3-t2;

subplot(2,2,1)

plot(x,t1,'-k')

xlabel('x');ylabel('y')

title('t1')

subplot(2,2,2)

plot(x,t2,'-k')

xlabel('x');ylabel('y')

title('t2')

subplot(2,2,3)

plot(x,t3,'-k')

xlabel('x');ylabel('y')

title('t3')

subplot(2,2,4)

plot(x,t4,'-k')

xlabel('x');ylabel('y')

title('t4')

第二题

秦九韶函数、

做题1-2]

应用秦九韶算法计算多项式

在x = x*处的函数值和导数值。试编程实现，并举例运行给出计算结果。

function jieguo=qinjiushao(A,x)

n=length(A);

B=zeros(n);

B(1)=A(1);

for k=1:n-1

B(k+1)=B(k)*x+A(k+1)

end

例子

A=[2,0,-3,3,-4];

x=-2;

qinjiushao(A,x)

jieguo =

ans =

单个例子A=[2,0,-3,3,-4]

B=zeros(1,5)

B(1)=A(1)

L=length(A)

x=-2

for k=2:L

B(k)=B(k-1)*x+A(k)

end

C=zeros(1,3)

C(1)=B(1)

for k=2:L-1

C(k)=C(k-1)*x+B(k)

end

A =

2 0 -3 3 -4

B =

0 0 0 0 0

B =

2 0 0 0 0

L =

x =

-2

B =

2 -4 0 0 0

B =

2 -4 5 0 0

B =

2 -4 5 -7 0

B =

2 -4 5 -7 10

C =

0 0 0

C =

2 0 0

C =

2 -8 0

C =

2 -8 21

C =

2 -8 21 -49

第二次作业

作业2

题目自己编程用基于Newton形式插值多项式求近似解

程序以及结果

牛顿函数

function W=newton(X,Y)

m=length(X);

M=zeros(m-1,1);

for i=1:m-1

M(i)=(Y(i)-Y(i+1))/(X(i)-X(i+1));

end

W=zeros(m,1);

W(1)=Y(1);

W(2)=M(1);

l=3;

for k=2:m-1

N=zeros(m-k,1);

for j=1:m-k

N(j)=(M(j)-M(j+1))/(X(j)-X(j+k));

end

W(l)=N(1);

l=l+1;

M=N;

end

例子

X=[0.4,0.5,0.6,0.7,0.8,0.9]

Y=[-0.916291, -0.693147, -0.510826, -0.357765, -0.223144, -0.105361]

X =

0.4000 0.5000 0.6000 0.7000 0.8000 0.9000

Y =

-0.9163 -0.6931 -0.5108 -0.3578 -0.2231 -0.1054

newton(X,Y)

ans =

-0.9163

2.2314

-2.0412

1.9272

-0.3096

-7.0625

x=log(0.78)

y=ans(1)+ans(2)*(x-0.4)+ans(3)*(x-0.4)*(x-0.5)+ans(4)*(x-0.4)*(x-0.5)*(x-0.6)+ans(5)*(x-0.4)*(x-0.5)*(x-0.6)*(x-0.7)+ans(6)*(x-0.4)*(x-0.5)*(x-0.6)*(x-0.7)*(x-0.8)

x =

-0.2485

这是ln0.78的近似值

y =

-0.6158

作业2’

给定函数f x= 1+1/（1+ x2）， x Î [ 5,5] ，节点 x = ?5 + k (k = 0,1,?,10) ，请同时显示三次样条插值与Lagrange插值，要求分别给出图例。

拉格朗日函数/

function v = polyinterp(x,y,u)

%POLYINTERP Polynomial interpolation.

% v = POLYINTERP(x,y,u) computes v(j) = P(u(j)) where P is the

% polynomial of degree d = length(x)-1 with P(x(i)) = y(i).

% Use Lagrangian representation.

% Evaluate at all elements of u simultaneously.

n = length(x);

v = zeros(size(u));

for k = 1:n

w = ones(size(u));

for j = [1:k-1 k+1:n]

w = (u-x(j))./(x(k)-x(j)).*w;

end

v = v + w*y(k);

end

例子

x=zeros(1,11);

for k=0:10

x(k+1)=-5+k;

y(k+1)=1/(1+x(k+1)^2);

end

u=-5:0.1:5;

polyinterp(x,y,u)

hold on

plot(u,ans,'-r')

axis([-5,5,-0.5,2])

w=spline(x,y,u)

plot(u,w,'-b')

legend('拉格朗日','三次样条')

hold off

拉格朗日插值结果

ans =

Columns 1 through 10

0.0385 1.2303 1.8044 1.9590 1.8458 1.5787 1.2402 0.8881 0.5604 0.2802

Columns 11 through 20

0.0588 -0.1007 -0.2013 -0.2496 -0.2546 -0.2262 -0.1743 -0.1083 -0.0363 0.0349

Columns 21 through 30

0.1000 0.1554 0.1987 0.2292 0.2472 0.2538 0.2510 0.2414 0.2278 0.2131

Columns 31 through 40

0.2000 0.1912 0.1888 0.1946 0.2099 0.2353 0.2710 0.3165 0.3708 0.4326

Columns 41 through 50

0.5000 0.5710 0.6432 0.7142 0.7817 0.8434 0.8971 0.9409 0.9733 0.9933

Columns 51 through 60

1.0000 0.9933 0.9733 0.9409 0.8971 0.8434 0.7817 0.7142 0.6432 0.5710

Columns 61 through 70

0.5000 0.4326 0.3708 0.3165 0.2710 0.2353 0.2099 0.1946 0.1888 0.1912

Columns 71 through 80

0.2000 0.2131 0.2278 0.2414 0.2510 0.2538 0.2472 0.2292 0.1987 0.1554

Columns 81 through 90

0.1000 0.0349 -0.0363 -0.1083 -0.1743 -0.2262 -0.2546 -0.2496 -0.2013 -0.1007

Columns 91 through 100

0.0588 0.2802 0.5604 0.8881 1.2402 1.5787 1.8458 1.9590 1.8044 1.2303

Column 101

0.0385

三次样条插值结果

w =

Columns 1 through 10

0.0385 0.0406 0.0427 0.0446 0.0465 0.0484 0.0503 0.0522 0.0543 0.0565

Columns 11 through 20

0.0588 0.0614 0.0642 0.0673 0.0707 0.0745 0.0786 0.0832 0.0883 0.0939

Columns 21 through 30

0.1000 0.1067 0.1140 0.1220 0.1307 0.1401 0.1504 0.1614 0.1734 0.1862

Columns 31 through 40

0.2000 0.2149 0.2314 0.2503 0.2720 0.2973 0.3269 0.3613 0.4011 0.4472

Columns 41 through 50

0.5000 0.5597 0.6242 0.6909 0.7573 0.8205 0.8782 0.9275 0.9661 0.9911

Columns 51 through 60

1.0000 0.9911 0.9661 0.9275 0.8782 0.8205 0.7573 0.6909 0.6242 0.5597

Columns 61 through 70

0.5000 0.4472 0.4011 0.3613 0.3269 0.2973 0.2720 0.2503 0.2314 0.2149

Columns 71 through 80

0.2000 0.1862 0.1734 0.1614 0.1504 0.1401 0.1307 0.1220 0.1140 0.1067

Columns 81 through 90

0.1000 0.0939 0.0883 0.0832 0.0786 0.0745 0.0707 0.0673 0.0642 0.0614

Columns 91 through 100

0.0588 0.0565 0.0543 0.0522 0.0503 0.0484 0.0465 0.0446 0.0427 0.0406

Column 101

0.0385

第三次作业

题目

程序以及结果

作业 3第二题

用Romberg积分法求积分10sin ydyy∫ ,要求T T (0) 61(0) 10??? <k k .结果要求显出中间步骤的结果。

clc;

clear;

a=0;b=1;

fa=a;

fb=1/b;

Err=10e-5;

N=6;

T=zeros(N+1);

T(1,1)=(b-a)*(fb+fa)/2;

j=1;

for i=j+1:N

n=2^(i-2);

h=(b-a)*2^(2-i);

for k=1:n

x(k)=a+(2*k-1)*h/2;

end

y=sin(x)/x;

H=h*sum(y);

T(i,j)=(T(i-1,j)+H)/2;

end

for n=j:N

for m=n:N-1

T(m+1,n+1)=(4^n*T(m+1,n)-T(m,n))/(4^n-1)

end

if abs(T(n+1,n+1)-T(n,n))<Err

break;

end

format long

disp(T(n,n))

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0 0 0 0

0.410326958533918 0 0 0 0

0.261684547893063 0 0 0 0

0.159083327234799 0 0 0 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0 0 0

0.410326958533918 0 0 0 0

0.261684547893063 0 0 0 0

0.159083327234799 0 0 0 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0 0 0

0.410326958533918 0.349121013633706 0 0 0

0.261684547893063 0 0 0 0

0.159083327234799 0 0 0 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0 0 0

0.410326958533918 0.349121013633706 0 0 0

0.261684547893063 0.212137077679444 0 0 0

0.159083327234799 0 0 0 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0 0 0

0.410326958533918 0.349121013633706 0 0 0

0.261684547893063 0.212137077679444 0 0 0

0.159083327234799 0.124882920348711 0 0 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0.531643466553942 0 0

0.410326958533918 0.349121013633706 0 0 0

0.261684547893063 0.212137077679444 0 0 0

0.159083327234799 0.124882920348711 0 0 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0.531643466553942 0 0

0.410326958533918 0.349121013633706 0.335810111557420 0 0

0.261684547893063 0.212137077679444 0 0 0

0.159083327234799 0.124882920348711 0 0 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0.531643466553942 0 0

0.410326958533918 0.349121013633706 0.335810111557420 0 0

0.261684547893063 0.212137077679444 0.203004815282493 0 0

0.159083327234799 0.124882920348711 0 0 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0.531643466553942 0 0

0.410326958533918 0.349121013633706 0.335810111557420 0 0

0.261684547893063 0.212137077679444 0.203004815282493 0 0

0.159083327234799 0.124882920348711 0.119065976526662 0 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0.531643466553942 0 0

0.410326958533918 0.349121013633706 0.335810111557420 0.332701645605094 0

0.261684547893063 0.212137077679444 0.203004815282493 0 0

0.159083327234799 0.124882920348711 0.119065976526662 0 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0.531643466553942 0 0

0.410326958533918 0.349121013633706 0.335810111557420 0.332701645605094 0

0.261684547893063 0.212137077679444 0.203004815282493 0.200896794706701 0

0.159083327234799 0.124882920348711 0.119065976526662 0 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0.531643466553942 0 0

0.410326958533918 0.349121013633706 0.335810111557420 0.332701645605094 0

0.261684547893063 0.212137077679444 0.203004815282493 0.200896794706701 0

0.159083327234799 0.124882920348711 0.119065976526662 0.117733614006728 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0.531643466553942 0 0

0.410326958533918 0.349121013633706 0.335810111557420 0.332701645605094 0

0.261684547893063 0.212137077679444 0.203004815282493 0.200896794706701 0.200379912938472

0.159083327234799 0.124882920348711 0.119065976526662 0.117733614006728 0

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0.531643466553942 0 0

0.410326958533918 0.349121013633706 0.335810111557420 0.332701645605094 0

0.261684547893063 0.212137077679444 0.203004815282493 0.200896794706701 0.200379912938472

0.159083327234799 0.124882920348711 0.119065976526662 0.117733614006728 0.117407483886336

0 0 0 0 0

Columns 6 through 7

0 0

T =

Columns 1 through 5

0.500000000000000 0 0 0 0

0.729425538604203 0.805900718138937 0 0 0

0.593944793234554 0.548784544778004 0.531643466553942 0 0

0.410326958533918 0.349121013633706 0.335810111557420 0.332701645605094 0

0.261684547893063 0.212137077679444 0.203004815282493 0.200896794706701 0.200379912938472

0.159083327234799 0.124882920348711 0.119065976526662 0.117733614006728 0.117407483886336

0 0 0 0 0

Columns 6 through 7

0 0

0.117326376917566 0

0 0

n =

0.117326376917566

第四次作业

题目

程序以及结果

作业 4

第一题

作业4 求解下列三对角线性方程组： ????????????????????

[说明]本题可以选择多种求解方式：

先判断A是否可以进行LU分解，如果可以，则进行Doolittle分解，并通过解两个三角形方程组（用左除的方法）得到原问题的解；否则显示不能分解并直接用左除求解；

先判断A是否是对称正定矩阵，如果是，则进行进行Cholesky分解，并进一步求解；否则显示不能分解并直接用左除求解；

应用追赶法求解，注意考虑存贮空间的合理设置。

function [n]=panduan(A)

l=det(A);

p=rank(A);

if l==0

n=-1;

else

for j=p:1

A(j,:)=[];

A(:,j)=[];

if det(A)==0

n=-1;

break

end

n=0;

end

clc

clear

A=[4 -1 0 0 0; -1 4 -1 0 0; 0 -1 4 -1 0; 0 0 -1 4 -1 ;0 0 0 -1 4]

B=[100;200;200;200;100]

panduan(A)

A =

4 -1 0 0 0

-1 4 -1 0 0

0 -1 4 -1 0

0 0 -1 4 -1

0 0 0 -1 4

B =

100

200

100

ans =

% 由函数panduan 判断出A矩阵顺序主子式大于零，所以A可以进行LU分解

[L,U]=lu(A)

%解LY=B

Y=L\B

%解UX=Y

X=U\Y

L =

1.0000 0 0 0 0

-0.2500 1.0000 0 0 0

0 -0.2667 1.0000 0 0

0 0 -0.2679 1.0000 0

0 0 0 -0.2679 1.0000

U =

4.0000 -1.0000 0 0 0

0 3.7500 -1.0000 0 0

0 0 3.7333 -1.0000 0

0 0 0 3.7321 -1.0000

0 0 0 0 3.7321

Y =

100.0000

225.0000

260.0000

269.6429

172.2488

X =

46.1538

84.6154

92.3077

84.6154

46.1538

第二题

% 由第一题得出A为正定矩阵

n=length(A)

m=0

for i=1:n

for j=1:n

if A(i,j)~=A(j,i)

m=-1

else

m=0

end

n =

m =

% 所以A矩阵为对称阵

%进行进行Cholesky分解

% 所以A矩阵为对称阵

A=[4 -1 0 0 0;-1 4 -1 0 0;0 -1 4 -1 0;0 0 -1 4 -1;0 0 0 -1 4];

U=chol(A)

U =

2.0000 -0.5000 0 0 0

0 1.9365 -0.5164 0 0

0 0 1.9322 -0.5175 0

0 0 0 1.9319 -0.5176

0 0 0 0 1.9319

B=[100;200;200;200;100];

Y=U'\B

X=U\Y

Y =

50.0000

116.1895

134.5628

139.5757

89.1625

X =

46.1538

84.6154

92.3077

84.6154

46.1538

第三题

function[x]=zhuigan(A,B)

n=length(A)

A(1,2)=A(1,2)/A(1,1)

for i=2:n-1

A(i,i+1)=A(i,i+1)/(A(i,i)-A(i,i-1)*A(i-1,i))

end

y=zeros(n,1);

y(1)=B(1)/A(1,1)

for j=2:n

y(j)=(B(j)-A(j,j-1)*y(j-1))/(A(j,j)-A(j,j-1)*A(j-1,j))

end

x=ones(n,1);

x(n)=y(n)

for k=n-1:-1:1

x(k)=y(k)-A(k,k+1)*x(k+1)

end

clear

A=[4 -1 0 0 0; -1 4 -1 0 0; 0 -1 4 -1 0; 0 0 -1 4 -1 ;0 0 0 -1 4]

B=[100;200;200;200;100]

zhuigan(A,B)

A =

4 -1 0 0 0

-1 4 -1 0 0

0 -1 4 -1 0

0 0 -1 4 -1

0 0 0 -1 4

B =

100

200

100

n =

A =

4.0000 -0.2500 0 0 0

-1.0000 4.0000 -1.0000 0 0

0 -1.0000 4.0000 -1.0000 0

0 0 -1.0000 4.0000 -1.0000

0 0 0 -1.0000 4.0000

A =

4.0000 -0.2500 0 0 0

-1.0000 4.0000 -0.2667 0 0

0 -1.0000 4.0000 -1.0000 0

0 0 -1.0000 4.0000 -1.0000

0 0 0 -1.0000 4.0000

A =

4.0000 -0.2500 0 0 0

-1.0000 4.0000 -0.2667 0 0

0 -1.0000 4.0000 -0.2679 0

0 0 -1.0000 4.0000 -1.0000

0 0 0 -1.0000 4.0000

A =

4.0000 -0.2500 0 0 0

-1.0000 4.0000 -0.2667 0 0

0 -1.0000 4.0000 -0.2679 0

0 0 -1.0000 4.0000 -0.2679

0 0 0 -1.0000 4.0000

y =

25.0000

60.0000

69.6429

y =

25.0000

60.0000

69.6429

72.2488

y =

25.0000

60.0000

69.6429

72.2488

46.1538

x =

1.0000

46.1538

x =

1.0000

84.6154

46.1538

x =

1.0000

92.3077

84.6154

46.1538

x =

1.0000

84.6154

92.3077

84.6154

46.1538

x =

46.1538

84.6154

92.3077

84.6154

46.1538

ans =

46.1538

84.6154

92.3077

84.6154

46.1538

% 由于作业是分两次做的，所以第三题是从另一个notebook复制来的

第五次作业

题目

作业5 分别用Jacobi迭代法和Gauss-Seidel迭代法解以下方程组：（不必单独上交，包含在实验报告中即可） ??????????????????

[说明] 可选择

两种方法均应用矩阵形式迭代求解；

两种方法均应用分量形式迭代求解。

程序以及结果

% 矩阵A由Variable Editor 生成的

A =

4 -1 0 -1 0 0

-1 4 -1 0 -1 0

0 -1 4 0 0 -1

-1 0 0 4 -1 0

0 -1 0 -1 4 -1

0 0 -1 0 -1 4

b=[0;5;0;6;-2;6]

b =

-2

jacobi(A,b,[0;0;0;0;0;0])

ans =

1.0000

2.0000

1.0000

2.0000

1.0000

2.0000

高斯方法

function y=g(A,b,x0)

%jacobi iteration

D=diag(diag(A));

U=triu(A,1);

L=tril(A,-1);

BG=-(D+L)\U;

fG=(D+L)\b;

y=BG*x0+fG;

n=1;

while norm(y-x0)>=1.0e-6

x0=y;

y=BG*x0+fG;

n=n+1;

end

A=[4 -1 0 -1 0 0;-1 4 -1 0 -1 0;0 -1 4 0 0 -1 ; -1 0 0 4 -1 0; 0 -1 0 -1 4 -1;0 0 -1 0 -1 4];

b=[0;5;0;6;-2;6];

g(A,b,[0;0;0;0;0;0])

ans =

1.0000

2.0000

1.0000

2.0000

1.0000

2.0000

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