newtonchazhi.rar_MATLAB 求解线性方程_newton_newton插值_求解线性方程组

function [C,D]=newpoly1(X,Y) %INPUT X is a vector that contains a list of abscissas % Y is a vector that contains a list of ordinates %OUTPUT C is a vector that contains the coefficients of the Newton % intepolatory polynomial % D is the divided-difference table n=length(X); D=zeros(n,n); D(:,1)=Y'; for j=2:n for k=j:n D(k,j)=(D(k,j-1)-D(k-1,j-1))/(X(k)-X(k-j+1)); end end C=D(n,n); for k=(n-1):-1:1 C=conv(C,poly(X(k))); m=length(C); C(m)=C(m)+D(k,k); end 以上是M文件，下面是程序实现的例子： >> X=[-2.15,-1.00,0.01,1.02,2.03,3.25]; >> Y=[17.03,7.24,1.05,2.03,17.06,23.05]; >> [C,D]=newpoly(X,Y) C = -0.2169 0.0648 2.1076 3.3960 -4.5745 1.0954 D = 17.0300 0 0 0 0 0 7.2400 -8.5130 0 0 0 0 1.0500 -6.1287 1.1039 0 0 0 2.0300 0.9703 3.5144 0.7604 0 0 17.0600 14.8812 6.8866 1.1129 0.0843 0 23.0500 4.9098 -4.4715 -3.5056 -1.0867 -0.2169