数值分析作业 顾磊磊第三、四章.doc

姓名：吴军 专业：化学工程（专第三章 1、解： Matlab 程序： function [A,Y]=lagrange(x,y,x0) %检验输入参数 if nargin 2 || nargin 3 error(Incorrect Number of Inputs); end if length(x)~=length(y) error(The length of x must be equal to it of y); end m=length(x);n=m-1;L=zeros(m,m); %计算基本插值多项式的系数 for i=1:n+1 C=1; for j=1:n+1 if i~=j if abs(x(i)-x(j))eps error(there are two same nodes); end C=conv(C,poly(x(j)))/(x(i)-x(j)); end end L(i,:)=C; end %计算Lagrange插值多项式的系数 A=y*L; %计算f(x0)的近似值 if nargin==3 Y=polyval(A,x0); end A=fliplr(A); Return x = [π/8,3π/8]; y = cos(x); x0 = π/3; [A,Y] = lagrange(x,y,x0); P1 = vpa(poly2sym(A),3) Y P1 =1.19x - 0.689 Y =0.4729 x0 = π/3; [A,Y] = lagrange(x,y,x0); P2=vpa(poly2sym(A),3) Y P2 = x2 - 0.109x - 0.336 Y =0.5174 x = [0,π/8,π/4,3π/8,π/2]; y= cos(x); x0 = π/3; [A,Y]=lagrange(x,y,x0); P4=vpa(poly2sym(A),3) Y P4 =x4 + 0.00282x3 - 0.514x2 + 0.0232x + 0.0287 Y =0.5001 解： Matlab 程序 function [T,y0]=aitken(x,y,x0,T0) if nargin==3 T0=[]; end n0=size(T0,1); m=max(size(x)); n=n0+m; T=zeros(n,n+1); T(1:n0,1:n0+1)=T0; T(n0+1:n,1)=x; T(n0+1:n,2)=y; if n0==0 i0=2; else i0=n0+1; end for i=i0:n for j=3:i+1 T(i,j)=fun(T(j-2,1),T(i,1),T(j-2,j-1),T(i,j-1),x0); end end y0=T(n,n+1); return function [y]=fun(x1,x2,y1,y2,x) y=y1+(y2-y1)*(x-x1)/(x2-x1); return %选取0、1、3、4四个节点，求三次插值多项式 x=[0,1,3,4]; y=[0.5,1.25,3.5,2.75]; x0=2.8; [T,y0]=aitken(x,y,x0) T = 0 0.5000 0 0 0 1.0 1.2500 2.6000 0 0 3.0 3.5000 3.2999 3.2300 0 4.0 2.7500 2.0750 2.2850 3.4190 y0 =3.419000000000000 解： Matlab 程序 function [C,D,Y]=newpoly(x0,y0,x) %检验输入参数 if nargin 2 | nargin 3 error(Incorrect Number of Inputs); end if length(x0)~=length(y0) error(The length of x0 must be equal to it of y0); end n=length(x0);D=zeros(n,n);D(:,1)=y0; %计算差商表 for j=2:n for k=j:n if abs(x0(k)-x0(k-j+1))eps error(Divided by Zero,there are two nodes are the