常州大學數值分析第三章

第三章作業(yè)1.設節(jié)點xo=0,x1=n/8,x2=n/4,x3=3n/8,x4=n/2,試適當選取上述節(jié)點，用拉格朗日插值法分別構造cosx在區(qū)間［0,n/2］上的一次、二次、四次差值多項式P1（x）,P2（x）和P4（x），并分別計算P1（n/3）,P2（n/3）和P4（n/3）.解：xOx1x2x3x484x4P(兀/3)=0.500147.解：選取x=0，x=1,x=2，x=3為節(jié)點0123>>T0=[0.00.5];x=[123]';y=[1.252.753.5]';x0=2.8;T=aitken(x,y,x0,T0)0.00000.50000.00000.00000.00001.00001.25002.60000.00000.00002.00002.75000.00000.50000.00000.00000.00001.00001.25002.60000.00000.00002.00002.75003.65004.49000.00003.00003.50003.30003.23003.4820161）拉格朗日差值.選取函數點八、、P(x)^―0.636620x+11P(兀/3)?0.3333331(2)選擇x0=0,x2=n/4,x4=n/2的節(jié)點y0=cosx0=1,y2=cosx2=0.707106,y4=cosx4=0,可得、(x-x)(x-x)(x-x)(x-x)P(x)=y12+y0210(x-x)(x-x)0(x-x)1(x-x)(x-0(x-x)1(x-x)+(x-x)(x-x)2(x-x)(x-x)即P(x)?-1.482067x2+5.511124x+1'%(兀/3)?1.145968(3)選擇x0=0,,x1=n/8,x2=n/4,x3=3n/8，x4=n/2的節(jié)點y0=cosx0=1,y1=cosx1=0.923879,y2=cosx2=0.707106,y3=cosx3=0.382683,y4=cosx4=0可得P4(x)=￡y(H)i=0j=0,聲%Xj，得P3(x)=1+0.0031x-0.5154x0110+0.0242x3+0.02y=cos(x)+sin(x),xg[一兀，兀]x0二-pi:0.5*pi:pi;y0=cos(x0);x=-pi:0.05*pi:pi;iflength(x0)~=length(y0)error('Thelengthofx0mustbeequaltoitofy0');endw=length(x0);n=w-1;L=zeros(w,w);fork=1:n+10110V=1;forj=1:n+1ifk~二jifabs(x0(k)-x0(j))<epserror('DividedbyZero,therearetwonodesarethesame');endV=conv(V,poly(x0(j)))/(x0(k)-x0(j));endendL(k,:)=V;

endC=y0*L;Y=polyval(C,x);y=cos(x)+sin(x);r=y-Y;plot(x,Y,'r—，,x,y,，b-，,x,r,，k-.','LineWidth',2);legend(，Lagrangepolynomial，,，Theoriginalf(x)，,，Error，,0)2)牛頓差值12)牛頓差值.選取函數y=1+擔，]日一5,5]>>x0=-5:1:5;〉〉y0=1./(1+x0.*x0);>>x=-5:0.1:5;>>iflength(x0)~=length(y0)>>error('Thelengthofx0mustbeequaltoitofy0');end>>n=length(x0);>>D=zeros(n,n);>>D(:,1)=y0';>>forj=2:nfork=j:nifabs(x0(k)-x0(k-j+1))<epserror('DividedbyZero,therearetwonodesarethesame');end>>D(k,j)=(D(k,j-1)-D(k-1,j-1))/(x0(k)-x0(k-j+1));endendC=D(n,n);fork=(n-1):-1:1C=conv(C,poly(x0(k)));m=length(C);C(m)=C(m)+D(k,k);endY=polyval(C,x);y=1./(1+x.*x+20*x);r=y-Y;>plot(x,Y,'r--',x,y,'b-',x,r,'k-.','LineWidth',2);>legend('Newtonpolynomial','Th