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1、《矩阵与数值分析》实验报告学院：土木市政工程姓名：徐博书学号：20806075教师：张宏伟班级：2班一、为了逼近飞行中的野鸭的顶部轮廓曲线，已经沿着这条曲线选择了一组点。见下表。1．对这些数据构造三次自然样条插值函数，并画出得到的三次自然样条插值曲线；2．对这些数据构造Lagrang插值多项式，并画出得到的Lagrang插值多项式曲线。x0.91.31.92.12.63.03.94.44.75.06.0f(x)1.31.51.852.12.62.72.42.152.052.12.25x7.08.09.210.511.311.612.012.613.013.3f(x)2.32.251.951.
2、40.90.70.60.50.40.25解:1.三次样条插值函数程序代码及运行结果clearall;clc;x=[0.91.31.92.12.63.03.94.44.75.06.07.08.09.210.511.311.612.012.613.013.3];y=[1.31.51.852.12.62.72.42.152.052.12.252.32.251.951.40.90.70.60.50.40.25];n=length(y);h=zeros(1,n-1);fori=1:n-1h(i)=x(i+1)-x(i);endr=zeros(1,n-2);fori=2:n-1r(i)=h(i)/(h(
3、i)+h(i-1));endu=zeros(1,n-2);fori=2:n-1u(i)=h(i-1)/(h(i)+h(i-1));endg=zeros(1,n);g(1)=3*(y(2)-y(1))/h(1);g(n)=3*(y(n)-y(n-1))/h(n-1);fori=2:n-1g(i)=3*(u(i)*(y(i-1)-y(i))/h(i)+r(i)*(y(i)-y(i-1))/h(i-1));endA=zeros(n,n);A(1,1)=2;A(1,2)=1;A(n,n)=2;A(n,n-1)=1;fori=2:n-1forj=1:nifi==jA(i,j)=2;A(i,j-1)=r
4、(i);A(i,j+1)=u(i);endendendm=Ag';fori=1:n-1z=x(i):0.01:x(i+1);s=((h(i)+2.*(z-x(i)))/h(i).^3).*((z-x(i+1)).^2).*y(i)+((h(i)-2.*(z-x(i+1)))/h(i).^3).*((z-x(i)).^2).*y(i+1)+(z-x(i)).*((z-x(i+1)).^2).*m(i)/(h(i).^2)+(z-x(i+1)).*((z-x(i)).^2).*m(i+1)/(h(i).^2);plot(x(i),y(i),'kp',z,s);holdonendgridon;程
5、序运行结果：M=[0.4217,0.6566,-2.6655,2.222,-1.0192,0.5411,0.1305,0.5379,-0.1994,0.2599,-0.0709,0.0238,-0.0243,0.0234,-0.1913,1.8850,-0.8150,0.1235,-0.0199,0.4008,-0.9504]；2.Lagrange插值多项式程序代码及运行结果clearall;clc;x=[0.91.31.92.12.63.03.94.44.75.06.07.08.09.210.511.311.612.012.613.013.3];y=[1.31.51.852.12.62.7
6、2.42.152.052.12.252.32.251.951.40.90.70.60.50.40.25];n=length(y);l=ones(1,n);fori=1:nforj=1:nifi==jl(i)=l(i);elsel(i)=l(i)/(x(i)-x(j));endendendl=l.*y;fori=1:n-1p=zeros();z=x(i):0.01:x(i+1);forj=1:ns=ones();forq=1:nifj~=qs=s.*(z-x(q));endendp=p+l(j).*s;endplot(x(i),y(i),'ko',z,p);holdonendgridon;程序
7、运行结果：二、对于问题将h=0.025的Euler法，h=0.05的改进的Euler法和h=0.1的4阶经典的Runge-Kutta法在这些方法的公共节点0.1，0.2，0.3，0.4和0.5处进行比较。精确解为：。1.h=0.025的Euler法完整MATLAB程序新建M-文件，建立euler1函数，保存至系统默认路径。function[x,y]=euler1(dyfun,xspan,y0,h)x=xspa
