function p=Newton1(x1,y,x2)%p为多项式估计出的插值syms xn = length(x1);%差商的求法for i=2:n   f1(i,1)=(y(i)-y(i-1))/(x1(i)-x1(i-1));endfor i=2:n    for j=i+1:n       f1(j,i)=(f1(j,i-1)-f1(j-1,i-1))/(x1(j)-x1(j-i));   endendf1=[y‘,f1]%  输出带0阶差商的差商表格%Newton插值函数Newton=f1(1,1);for i=2:n    tt=1;    for j=1:i-1        tt=tt*(x-x1(j));    end    Newton=Newton+f1(i,i)*tt;endfprintf(‘Newton插值函数为\n‘)expand(Newton)     % 将连乘多项式合并展开x = x2;p = eval(Newton);   % 代入值计算%fprintf(‘Newton插值函数在所求点x2的函数值为\n‘)%p