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# -*- coding: utf-8 -*- """ Created on Mon Oct 09 01:50:41 2017 @author: Vinícius """ import numpy as np def g(x): return np.exp(-x) def ipf(g, p0, TOL, NMAX = 1000, output = False): ER = TOL + 1.0 n = 0 d = np.ones(1000) while (ER > TOL) and (n < NMAX): n = n + 1 p1 = g(p0) ER = abs((p1 - p0)/p0) d[n] = ER p0 = p1 if (output): print n, p1, d[n]/d[n-1] assert ( n != NMAX ), "Numero Maximo de Iteraçoes!!!" return p1 p = ipf(g, 1.0, 1E-5, 1000, True) """ 1 0.367879441171 0.632120558829 2 0.692200627555 1.39466494994 3 0.500473500564 0.314182342811 4 0.606243535086 0.763009566743 5 0.545395785975 0.474914946606 6 0.579612335503 0.625067670643 7 0.560115461361 0.536170523232 8 0.57114311508 0.585299550725 9 0.564879347391 0.557038402984 10 0.568428725029 0.572935609611 11 0.566414733147 0.563878047343 12 0.567556637328 0.569001514915 13 0.566908911921 0.566091460531 14 0.567276232176 0.567740487448 15 0.567067898391 0.566804806106 16 0.567186050099 0.567335327681 17 0.567119040057 0.56703439969 18 0.567157044001 0.567205054106 19 0.567135490206 0.567108263815 20 0.56714771426 0.567163156236 21 0.567140781458 0.567132023867 22 0.567144713347 0.567149680215 """
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