function[x_k,iter_hist] = newton(x_0,step_tol,fn_tol,maxiter,cost_fn)

% This matlab function uses Newton's method to compute
% a solution of the equation f(x)=0.

% Initialize and check to see if x_0 is a solution.
x_k = x_0;
[f,J] = feval(cost_fn,x_k);
fprintf('iter=%d |f(x_0)|=%2.3e |e|=%2.3e\n',0,norm(f),0) 
if f == 0
    disp(' f(x_0) = 0. ')
    return
end

% Save iteration history.
% iter_hist = [iteration, |f|, |step|];
iter_hist = [0, norm(f), 0];

% Newton's Method
for k = 1:maxiter
      e = J\f;
      x_k = x_k - e;
      [f,J] = feval(cost_fn,x_k);
      % Save iteration info and output to screen.
      nf = norm(f);
      ne = norm(e);
      iter_hist = [iter_hist; [k, nf, ne]];
      fprintf('iter=%d |f(x_k)|=%2.3e |e|=%2.3e\n',k,nf,ne) 
      
      % Check the stopping criteria
      if norm(f) < fn_tol
          disp(' |f(x_k)| stopping tolerance met.')
          return
      elseif norm(e) < step_tol
           disp(' error stopping tolerance met.')
          return        
      elseif k == maxiter
          disp(' Max number of iterations reached.')
          return
      end
  end