用牛顿法matlab程序解题!

Newton-Raphson 求解非线性方程组matlab源程序matlab程序如下：function hom[P,iter,err]=newton('f','JF',[7。8e-001;4。9e-001; 3。7e-001],0。01,0。001,1000);disp(P);disp(iter);disp(err);function Y=f(x,y,z)Y=[x^2+y^2+z^2-1;2*x^2+y^2-4*z;3*x^2-4*y+z^2];function y=JF(x,y,z)f1='x^2+y^2+z^2-1';f2='2*x^2+y^2-4*z';f3='3*x^2-4*y+z^2';df1x=diff(sym(f1),'x');df1y=diff(sym(f1),'y');df1z=diff(sym(f1),'z');df2x=diff(sym(f2),'x');df2y=diff(sym(f2),'y');df2z=diff(sym(f2),'z');df3x=diff(sym(f3),'x');df3y=diff(sym(f3),'y');df3z=diff(sym(f3),'z');j=[df1x,df1y,df1z;df2x,df2y,df2z;df3x,df3y,df3z];y=(j);function [P,iter,err]=newton(F,JF,P,tolp,tolfp,max)%输入P为初始猜测值,输出P则为近似解%JF为相应的Jacobian矩阵%tolp为P的允许误差%tolfp为f(P)的允许误差%max：循环次数Y=f(F,P(1),P(2),P(3));for k=1:maxJ=f(JF,P(1),P(2),P(3));Q=P-inv(J)*Y;Z=f(F,Q(1),Q(2),Q(3));err=norm(Q-P);P=Q;Y=Z;iter=k;if (e