% NEWTONDEMO Script demonstrates use of Newton's method
%            to solve  x^3 - 2*x - 5 = 0.
%            
x = 2;         % Initial guess for root
xtol = 1e-14;  % Tolerance for change in x
ftol = 1e-14;  % Tolerance for residual
itmx = 10;     % Max # of iterations allowed

% Print a table header

disp('Iter   Approximate Root x    | fx |      | dx |')
fx = 1 + ftol; % Enforce 1 iteration
dx = 1 + xtol;
it = 0;
while ( abs(dx) >= xtol & abs(fx) >= ftol & it < itmx )
  fx = x^3 - 2*x -5;    % Residual
  df = 3*x^2 - 2;       % Derivative
  dx = fx/df;           % Newton correction
  x  = x - dx;          % New approximate root
  it = it + 1;          % Bump counter
  disp(sprintf(' %2d  %21.17f  %8.2e    %8.2e',it,x,abs(fx),abs(dx)))
end
disp(sprintf('The computed root is %21.17f',x))