急求！ 用牛顿法matlab程序解题！

用牛顿法求f=2×x1^2+x2^2-4×x1+2   的最小值
	           

答案：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:max
    J=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 (err<tolp)||(abs(Y)<tolfp||abs(Y)<0.0001)
        break
    end
end  
   <pre lang="matlab" line="1" file="test.m">
function   homework4
[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:max
    J=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 (err<tolp)||(abs(Y)<tolfp||abs(Y)<0.0001)
        break
    end

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