數(shù)值分析(第五版)計算實習題第三章

1、數(shù)值分析計算實習題第三章第二次作業(yè)：題一：x=-1:0.2:1;y=1./(1+25.*x.2);f1=polyfit(x,y,3)f=poly2sym(f1)y1=polyval(f1,x)x2=linspace(-1,1,10)y2=interp1(x,y,x2)plot(x,y,'r*-',x,y1,'b-')holdonplot(x2,y2,'k')legend('數(shù)據(jù)點','3次擬合曲線','3次多項式插值')xlabel('X'),ylabel('Y')輸

2、出：f1 =0.0000 -0.5752 0.0000 0.4841f =(4591875547102675*x3)/81129638414606681695789005144064 - (1469057404776431*x)/20282409603651670423947251286016 4360609662300613/9007199254740992y1 =-0.0911 0.1160 0.2771 0.3921 0.4611 0.4841 0.2771 0.1160 -0.0911(3305*x2)/5746 + + 0.4611 0.3921x2 =-1.0000 -0.7778

3、 -0.5556 -0.3333 -0.1111 0.1111 0.3333 0.5556 0.7778 1.0000y2 =0.0385 0.0634 0.1222 0.3000 0.7222 0.7222 0.3000 0.1222 0.0634 0.0385題二：X=0.0 0.1 0.2 0.3 0.5 0.8 1.0;Y=1.0 0.41 0.50 0.61 0.91 2.02 2.46;p1=polyfit(X,Y,3)p2=polyfit(X,Y,4)Y1=polyval(p1,X)Y2=polyval(p2,X)plot(X,Y,'r*',X,Y1,'b

4、-.',X,Y2,'g-')p3=polyfit(X,Y,2)Y3=polyval(p3,X)f1=poly2sym(p1)f2=poly2sym(p2)f3=poly2sym(p3)plot(X,Y,'r*',X,Y1,'b-.',X,Y2,'g-',X,Y3,'m-')legend('數(shù)據(jù)點','3次多項式擬合','4次多項式擬合','2次多項式擬合') xlabel('X軸'),ylabel('Y軸')輸出

5、：p1 =-6.6221 12.8147 -4.6591 0.9266p2 =2.8853 -12.3348 16.2747 -5.2987 0.9427Y1 =0.9266 0.5822 0.4544 0.5034 0.9730 2.0103 2.4602Y2 =0.9427 0.5635 0.4399 0.5082 1.0005 1.9860 2.4692p3 =3.1316 -1.2400 0.7356Y3 =0.7356 0.6429 0.6128 0.6454 0.8984 1.7477 2.6271f1 =- (7455778416425075*x3)/11258999068426

6、24 + (1803512222945435*x2)/140737488355328 - (40981580032809*x)/8796093022208 + 8345953784399011/9007199254740992f2 =(1624271450198125*x4)/562949953421312 - (3471944732519173*x3)/281474976710656 + (4580931990070659*x2)/281474976710656 - (1491459232922115*x)/281474976710656 + 1061409433081293/1125899

7、906842624f3 =(18733*x2)/5982 - (74179*x)/59820 + 73337/99700題三：建立三角插值函數(shù)的m文件function A,B,Y1,Rm=sanjiaobijin(X,Y,X1,m)%A B分別是m階三角多項式Tm（x）的系數(shù)aj，bj（j=1,2,.,m）的系數(shù)矩陣，Y1是Tm（x）在X1處的值，X Y數(shù)據(jù)點 ,Rm為均方誤差n=length(X)-1;max1=fix(n-1)/2);if m>max1m=max1;endA=zeros(1,m+1);B=zeros(1,m+1);Ym=(Y(1)+Y(n+1)/2;Y(1)=Ym;Y

8、(n+1)=Ym;A(1)=2*sum(Y)/n;fori=1:mB(i+1)=sin(i*X)*Y'A(i+1)=cos(i*X)*Y'endA=2*A/n;B=2*B/n;A(1)=A(1)/2;Y1=A(1);for k=1:mY1=Y1+A(k+1)*cos(k*X1)+B(k+1)*sin(k*X1);Tm=A(1)+A(k+1).*cos(k*X)+B(k+1).*sin(k*X);k=k+1;endY,Tm,Rm=(sum(Y-Tm).2)/n輸出：>> X=-pi:2*pi/33:pi;>> Y=X.2.*cos(X);A,B,Y1,Rm

9、=sanjiaobijin(X,Y,X1,16)輸出：A =1 至 12 列-0.1397 4.4002 -2.8326 1.2355 -0.9128 0.6982 -0.6773 0.6635 -0.6541 0.647413 至 17 列-0.6426 0.6393 -0.6370 0.6355 -0.6348B =1.0e-15 *1 至 12 列0 -0.0194 -0.0150 -0.0044 -0.0300 -0.0821 -0.0599 -0.0133 -0.0211 0.029713 至 17 列0.0178 0.0962 -0.1049 0.0328 -0.0122 即可得16插值多項式的值X1=-pi:0.001:pi;A,B,Y1,Rm=sanjiaobijin(X,Y,X1,16)plot(X,Y,&