Page - 206 - in Programming for Computations – Python - A Gentle Introduction to Numerical Simulations with Python

206 6 SolvingNonlinearAlgebraicEquations F_value = F(x) F_norm = np.linalg.norm(F_value, ord=2) # l2 norm of vector iteration_counter = 0 while abs(F_norm) > eps and iteration_counter < 100: delta = np.linalg.solve(J(x), -F_value) x = x + delta F_value = F(x) F_norm = np.linalg.norm(F_value, ord=2) iteration_counter += 1 # Here, either a solution is found, or too many iterations if abs(F_norm) > eps: iteration_counter = -1 return x, iteration_counter Wecan test the functionNewton_systemwith the2 2 system(6.11)-(6.12): 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’ % error_norm print ’norm of error =%g’ % error_norm Here, the testing is based on theL2normof the error vector. Alternatively,we could test against the values of x that the algorithmfinds, with appropriate toler- ances. For example, as chosen for the error norm, if eps=0.0001, a tolerance of 10 4 canbeused forx[0]andx[1]. 6.7 Exercises Exercise6.1:UnderstandwhyNewton’smethodcanfail The purpose of this exercise is to understandwhenNewton’smethodworks and fails. To this end, solve tanhxD0byNewton’smethodandstudy the intermediate details of the algorithm. Startwithx0 D 1:08. Plot the tangent in each iterationof Newton’smethod. Then repeat the calculations and the plottingwhenx0 D 1:09. Explainwhatyouobserve. Filename:Newton_failure.*.