Page - 198 - in Programming for Computations – Python - A Gentle Introduction to Numerical Simulations with Python 3.6, Volume Second Edition

198 7 SolvingNonlinearAlgebraicEquations # Here, either a solution is found, or too many iterations if abs(F_norm) > eps: iteration_counter = -1 return x, iteration_counter Wecantest thefunctionNewton_systemwiththe2×2system(7.10)and(7.11): def test_Newton_system1(): from numpy import cos, sin, pi, exp def F(x): return np.array( [x[0]**2 - x[1] + x[0]*cos(pi*x[0]), x[0]*x[1] + exp(-x[1]) - x[0]**(-1.)]) def J(x): return np.array( [[2*x[0] + cos(pi*x[0]) - pi*x[0]*sin(pi*x[0]), -1], [x[1] + x[0]**(-2.), x[0] - exp(-x[1])]]) expected = np.array([1, 0]) tol = 1e-4 x, n = Newton_system(F, J, x=np.array([2, -1]), eps=0.0001) print(n, x) error_norm = np.linalg.norm(expected - x, ord=2) assert error_norm < tol, ’norm of error ={:g}’.format(error_norm) print(’norm of error ={:g}’.format(error_norm)) Here, the testing is based on the L2 norm6 of the error vector. Alternatively, we could test against the values of x that the algorithm finds, with appropriate tolerances. For example, as chosen for the error norm, ifeps=0.0001, a tolerance of10−4 canbeused forx[0]andx[1]. 7.7 Exercises Exercise7.1:UnderstandWhyNewton’sMethod CanFail The purpose of this exercise is to understand when Newton’s method works and fails. To this end, solve tanhx=0 byNewton’smethodandstudy the intermediate details of the algorithm. Start withx0 = 1.08. Plot the tangent in each iteration of Newton’s method. Then repeat the calculations and the plotting when x0 = 1.09. Explainwhatyouobserve. Filename:Newton_failure.*. Exercise7.2:SeeIf theSecantMethod Fails Does the secant method behave better than Newton’s method in the problem describedin Exercise7.1?Try the initial guesses 1. x0 =1.08andx1 =1.09 2. x0 =1.09andx1 =1.1 6 https://en.wikipedia.org/wiki/Norm_(mathematics)#Euclidean_norm.