I have this code to solve the Newton method, and in the last iteration, the output values are wrong. These values should not be negative since I checked them manually on paper. As far as I know, the code is correct, but I can’t understand why it displays negative values, and also the final value of u should ideally be a positive value between 0 and 1. Here is the code:
import copy import math tlist = [0.0, 0.07, 0.13, 0.15, 0.2, 0.22] # list of start time for the phonemes w = 1.0 def time() : t_u = 0.0 for i in range(len(tlist)- 1) : t_u = t_u + 0.04 # regular time interval print t_u print tlist[i], ' ' , tlist[i+1], ' ', tlist[i -1] if t_u >= tlist[i] and t_u <= tlist[i + 1] : poly = poly_coeff(tlist[i], tlist[i + 1], t_u) Newton(poly) else : poly = poly_coeff(tlist[i - 1], tlist[i], t_u) Newton(poly) def poly_coeff(start_time, end_time, t_u) : """The equation is k6 * u^3 + k5 * u^2 + k4 * u + k0 = 0. Computing the coefficients for this polynomial.""" """Substituting the required values we get the coefficients.""" t0 = start_time t3 = end_time t1 = t2 = (t0 + t3) / 2 w0 = w1 = w2 = w3 = w k0 = w0 * (t_u - t0) k1 = w1 * (t_u - t1) k2 = w2 * (t_u - t2) k3 = w3 * (t_u - t3) k4 = 3 * (k1 - k0) k5 = 3 * (k2 - 2 * k1 + k0) k6 = k3 - 3 * k2 + 3 * k1 -k0 print k0, k1, k2, k3, k4, k5, k6 return [[k6,3], [k5,2], [k4,1], [k0,0]] def poly_differentiate(poly): """ Differentiate polynomial. """ newlist = copy.deepcopy(poly) for term in newlist: term[0] *= term[1] term[1] -= 1 return newlist def poly_substitute(poly, x): """ Apply value to polynomial. """ sum = 0.0 for term in poly: sum += term[0] * (x ** term[1]) return sum def Newton(poly): """ Returns a root of the polynomial""" x = 0.5 # initial guess value epsilon = 0.000000000001 poly_diff = poly_differentiate(poly) while True: x_n = x - (float(poly_substitute(poly, x)) / float(poly_substitute(poly_diff, x))) if abs(x_n - x) < epsilon : break x = x_n print x_n print "u: ", x_n return x_n if __name__ == "__main__" : time()
The output for the last iteration is as follows:
where k6 = -0.02, k5 = 0.03, k4 = -0.03 and k0 = 0.0
0.2 0.2 0.22 0.15 0.0 -0.01 -0.01 -0.02 -0.03 0.03 -0.02 -0.166666666667 -0.0244444444444 -0.000587577730193 -3.45112269878e-07 -1.19102451449e-13 u: -1.42121180685e-26
The initial value of the assumption is 0.5, so if it is replaced in the polynomial, then the output is -0.005.
Then, again, the same initial value is replaced in the differentiated polynomial. The result of this is -0.015.
Now these values are replaced in the Newton equation, then the answer should come as 0.166666667. But the actual answer is no.
Thanks.