數(shù)值分析第一次作業(yè)

1、數(shù)值分析第一次作業(yè)班級 學號 姓名 習題24、用Newton法求方程f(x)=x3-2*x2-4*x-7=0在3，4中的根。代碼： functionx_star,k=Newton1fname,dfname,x0,ep,Nmax if nargin<5 Nmax=500; end if nargin<4 ep=1e-5;end x=x0;x0=x+2*ep;k=0; while abs(x0-x)>ep&k<Nmax k=k+1 x0=x;x=x0-feval(fname,x0)/feval(dfname,x0); end x_star=x; if k=Nmax

2、warning(已迭代上限次數(shù));endfname=inline('x3-2*x2-4*x-7');dfname=inline('3*x2-4*x-4');x_star,k=Newton1(fname,dfname,3.5)x_star = 3.6320k = 4方法二：24用割線法求方程的根function x_star,k=Gline(fun,x0,x1,ep,Nmax)if nargin<5 Nmax=500;endif nargin<4 ep=1e-5;endk=0;while abs(x1-x0)>ep&k<Nmax k

3、=k+1; x2=x1-feval(fun,x1)*(x1-x0)/(feval(fun,x1)-feval(fun,x0) x0=x1; x1=x2;endx_star=x1;if k=Nmax warning('已迭代上限次數(shù)');endfun=inline('x3-2*x2-4*x-7');x_star,k=Gline(fun,3,4)x2 = 3.5263x2 = 3.6168x2 = 3.6327x2 = 3.6320x2 = 3.6320x_star = 3.6320k = 5習題33、用列主元消去法解方程組 -1 2 -2; 3 -1 4; 2 -

4、3 -2x1 x2 x3=-1 7 0代碼：function x=Gauss_x1(A,b) A=A;b,n=length(b); for k=1:n-1 s=A(k,k); p=k; for i=l+1:n if abs(s)<abs(A(i,k) s=A(I,k); p=I; end endAfor i=k+1:n m=A(i,k)/A(k,k); fprintf(m%d%d=%fn,i,k,m); for j=k:n+1 A(i,j)=A(i,j)-m*A(k,j);end end fprintf(A%d=n,k+1); Aend A(n,n+1)=A(n,n+1)/A(n.n);

5、 for i=n-1:-1:1 s=0 for j=i+1:n s=s+A(i,j)*A(j,n+1); end A(i,n+1)=(A(i,n+1)-s)/A(i,i); end A(:,n+1) A=-1,2,-2;3,-1,4;2,-3,-2;b=-1;7;0;x=Gauss_x1(A,b)A = 3.0000 -1.0000 4.0000 7.0000 0 1.6667 -0.6667 1.3333 0 -2.3333 -4.6667 -4.6667A= 3.0000 -1.0000 4.0000 7.0000 0 -2.3333 -4.6667 -4.6667 0 0 -4.0000

6、 -2.0000x = 2.0000 1.0000 0.5000 4、用追趕法解三對角方程 2 -1 0 0 0;-1 2 -1 0 0;0 -1 2 -1 0;0 0 -1 2 -1;0 0 0 -1 2x1 x2 x3 x4 x5=1 0 0 0 0 代碼： function x=zhuigan(A,B,C,D) n=length(B);Xzeros(1,n);U=zeros(1,n); Q=zeros(1,n);U(1)=C(1)/B(1); Q(1)=D(1)/B(1); for i=2:n-1 U(i)=C(i)/(B(i)-U(i-1)*A(i-1); end for i=2:n

7、Q(i)=(D(i)-Q(i-1)*A(i-1)/(B(i)-U(i-1)*A(i-1); end X(n)=Q(n); for i=n-1:-1:1 X(i)=Q(i)-U(i)*X(i+1); end X A=-1，-1，-1，-1;B=2,2,2,2,2;C=-1,-1,-1,-1;D=1;0;0;0;0;X=zhuigan(A,B,C,D)X= 0.8333 0.6667 0.5000 0.3333 0.16676、用三角分解法解方程組 -2 4 8;-4 18 -16;-6 2 -20x1 x2 x3=5 8 7代碼functiony,x=LU_s(A,b)b=b'A=A&#

8、39;b',n=length(b');x=zeros(n,1);y=zeros(n,1);U=zeros(n);L=eye(n);for k=1:n U(1,k)=A(1,k); L(k,1)=A(k,1)/U(1,1);endfor i=2:n for k=i:n lu=0; lu1=0; for j=1:i-1 lu=lu+L(i,j)*U(j,k); lu1=lu1+L(k,j)*U(j,i); end U(i,k)=A(i,k)-lu; L(k,i)=(A(k,i)-lu1)/U(i,i); endendLUfor i=1:n ly=0; for j=1:i ly=ly

9、+L(i,j)*y(j); end y(i)=b(i)-ly;endfor i=n:-1:1 ly1=0; for j=i+1:n ly1=ly1+U(i,j)*x(j); end x(i)=(y(i)-ly1)/U(i,i);endA=-2,4,8;-4,18,-16;-6,2,-20;b=5;8;7;y,x=LU_s(A,b)A = -2 4 8 5 -4 18 -16 8 -6 2 -20 7L = 1 0 0 2 1 0 3 -1 1U = -2 4 8 0 10 -32 0 0 -76y = 5 -2 -10x = -1.5316 0.2211 0.13163-8用LU分解法解線性方

10、程組5,7,9,10;6,8,10,9;7,10,8,7;5,7,6,5x1 x2 x3 x4=1 1 1 1代碼functiony,x=LU_s(A,b)b=b'A=A'b',n=length(b');x=zeros(n,1);y=zeros(n,1);U=zeros(n);L=eye(n);for k=1:n U(1,k)=A(1,k); L(k,1)=A(k,1)/U(1,1);endfor i=2:n for k=i:n lu=0; lu1=0; for j=1:i-1 lu=lu+L(i,j)*U(j,k); lu1=lu1+L(k,j)*U(j,i)

11、; end U(i,k)=A(i,k)-lu; L(k,i)=(A(k,i)-lu1)/U(i,i); EndendLUfor i=1:n ly=0; for j=1:i ly=ly+L(i,j)*y(j); end y(i)=b(i)-ly;endfor i=n:-1:1 ly1=0; for j=i+1:n ly1=ly1+U(i,j)*x(j); end x(i)=(y(i)-ly1)/U(i,i);endA=5,7,9,10;6,8,10,9;7,10,8,7;5,7,6,5;b=1;1;1;1;y,x=LU_s(A,b)A = 5 7 9 10 1 6 8 10 9 1 7 10 8 7 1 5 7 6 5 1L = 1.0000 0 0 0 1.2000 1.0000 0 0 1.4000 -0.5000 1.0000 0 1.00