牛顿法求解非线性方程组matlab源程序

1、牛顿法求解非线性方程组 matlab源程序Newton-Raphson求解非线性方程组 matlab源程序 matlab程序如下：fun ctio nhomP,iter,err=newt on ('f,'JF',7.8e-001;4.9e-001; 3.7e-001,0.01,0.001,1000);disp (P);disp (iter);disp (err);fun cti on Y=f(x,y,z)Y=xA2+yA2+zA2-1;2*xA2+yA2-4*z;3*xA2-4*y+zA2;fun cti on y=JF(x,y,z)f1='xA2+yA2+zA

2、2-1'f2='2*xA2+yA2-4*z'f3='3*xA2-4*y+zA2'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= dif

3、f (sym(f3),'y');df3z= diff (sym(f3),'z');j=df1x,df1y,df1z;df2x,df2y,df2 z; df3x,df3y,df3z; y=(j);maxfunction P,iter,err=newton(F,JF,P,tolp,tolfp, %俞入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-

4、nv (J)*Y;Z=f (F,Q(1),Q(2),Q(3);err= n orm(Q-P);P=Q;Y=Z; iter=k;if (err<tolp)|( abs(Y)<tolfp|abs(Y)<0.0001) breakendend<pre Ian g="matlab"li ne ="1" file="test.m">fun cti onhomework4P,iter,err=newt on ('f,'JF',7.8e-001;4.9e-001;3.7e-001,0.01,0.

5、001,1000);disp (P);disp (iter);disp (err);fun cti on Y=f(x,y,z) Y=xA2+yA2+zA2-1;2*xA2+yA2-4*z;3*xA2-4*y+zA2;fun cti on y=JF(x,y,z)f1='xA2+yA2+zA2-1'f2='2*xA2+yA2-4*z'f3='3*xA2-4*y+zA2'df1x= diff (sym(f1),'x');df1y= diff (sym(f1),'y');df1z= diff (sym(f1),'z

6、');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,df2 z; df3x,df3y,df3z; y=(j);maxfunction P,iter,err=newton(F,JF,P,tolp,tolfp,%俞入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亘(JF,P(1),P (2),P (3);Q=P-nv (J)*Y;Z=f (F,Q(1