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61 lines (46 loc) · 1.6 KB
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# Approximant de Padé
import matplotlib.pyplot as plt
import numpy as np
# Affiche le graphe de la fonction f sur [a,b] avec N points
def trace(f, a, b, N=100, color='blue', label=''):
list_x, list_y = [], []
for x in np.linspace(a, b, N):
list_x.append(x)
list_y.append(f(x))
plt.plot(list_x, list_y, color=color, label=label)
return
# Exemple Padé pour f = exp(x)
def exemple1():
f = lambda x: np.exp(x)
trace(f, -2, 3, color='red', label='f(x) = exp(x)')
g4 = lambda x: 1 + x + 1/2*x**2 + 1/6*x**3 + 1/24*x**4 # Dl ordre 4
trace(g4, -2, 3.3, color='green', label="DL d'ordre 4")
p22 = lambda x: (12 + 6*x + x**2)/(12 - 6*x + x**2) # padé ordre (2,2)
trace(p22, -2, 4, color='orange', label="Padé d'ordre (2,2)")
plt.legend()
# plt.axis('equal')
plt.xlim(-2, 4)
# plt.ylim(0.5, 1.5)
plt.legend()
plt.grid(True, which='both')
plt.tight_layout()
# plt.savefig('approx-pade-01.png',dpi=600)
plt.show()
exemple1()
# Exemple erreur f = exp(x)
def exemple2():
g4 = lambda x: np.exp(x) - (1 + x + 1/2*x**2 + 1/6*x**3 + 1/24*x**4) # Dl ordre 4
trace(g4, -1, 1, color='green', label="erreur DL d'ordre 4")
p22 = lambda x: np.exp(x) - (12 + 6*x + x**2)/(12 - 6*x + x**2) # padé ordre (2,2)
trace(p22, -1, 1, color='orange', label="erreur Padé d'ordre (2,2)")
plt.legend()
# plt.axis('equal')
plt.xlim(-1, 1)
# plt.ylim(0.5, 1.5)
plt.legend()
plt.grid(True, which='both')
plt.tight_layout()
# plt.savefig('approx-pade-02.png',dpi=600)
plt.show()
# exemple2()
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