Matlab-5:牛顿迭代法工具箱

 function [f,L]=Newton(f,a)
 %this is newton teration whic is used for solving implicit One-dimensional Euler method
 %users can used it directly for solve equation.
 %the code was writen by HD.dong in january 8 2017.
 %--------------------------------
 % syms x;
 %  %         h='[x^4-4*x^2+4]';
 % %     h='[x^3+2*x^2+10*x-20]';
 %        h='[x^3-x-1]';
 % %  h='[x^3+4*x^2-10]';
 % x0=0.6;%users can set any value except zero,because diff(h,x) is Singular when x is zero.
 % [X L]=Newton(h,x0);
 %--------------------------------------------------------------------
 lambda=1;%newton downhill factor
 L(1)=lambda;
 x0=a;
 x1=x0-Jacoi(f,x0)\F(f,a)*lambda;
 tol=1e-5;
 ttol=1e-8;
 i=1;
  while norm(x1-x0,1)>=tol
          lambda=1;
  while abs(F(f,x1))>=abs(F(f,x0)) & lambda>=ttol
  lambda=lambda/2;
  x1=x0-Jacoi(f,x0)\F(f,x0)*lambda;
  end
    x0=x1;
        x1=x0-Jacoi(f,x0)\F(f,x0)*lambda;
        i=i+1;
        L(i)=lambda;
  end
  f=x1;
 function G=Jacoi(f,x0)
 syms x;
 G=vpa(subs(diff(f,x),'x',x0));
 function H=F(f,x0)
 H=vpa(subs(f,'x',x0));

算法推导：