% Solve A - 1/x = 0 using Newton's method
% Does not require any divisions

A = 3.0;

x0 = 2.0;
%x0 = 0.5;
%x0 = 2/3;


e0 = abs(x0-1/A);

for i=1:15,
    x1 = x0*(2-A*x0);
    e1 = abs(x1-1/A);
    
    fprintf('Iter: %3d, x: %15.10e, e1/e0: %e\n', i, x1, e1/e0);
    
    if abs(x0-x1)/max(1.0E-06,abs(x0)) < 1.0E-09,
        % Success!
        fprintf('\nConvergence!\n');        
        break;
    end
    
    x0 = x1;
    e0 = e1;    
end