% Another Fixed point method to solve the nonlinear system of equation
% (x-0.01)^2 + y^2 = 1
% x^2 - (y-0.01)^2 = .25
% in the first quadrant (near (1/4, 1/4) := x_0
% instead of Jacobian(x_{n-1}), use Jacobian(x_0)
% should be equivalent to the method in fixed_point_1.m

clear;
format long e;

x = [.5; .5]

% repeat the following command till convergence (about 20+ iterations)

x = x - [ 2*(0.5-0.01) 8*0.5; 2*0.5 -2*(0.5-0.01)]\ ...
    [ x(2)^2+4*( x(2)-0.01 )^2-1; x(1)^2-(x(2)-0.01)^2-0.25 ]