Fix: Correct dichotimie

ex3.py

@@ -72,11 +72,8 @@ def relaxation(A, b, omega=1.0, epsilon=1e-6, max_iter=100_000):
     # Pré-calculer (D - ωL)^(-1) une seule fois
     inv_D_omega_L = np.linalg.inv(D - omega * L)
 
-    if omega == 1:
-        return gauss_seidel(A, b)
-
     for _ in range(max_iter):
-        x_new = inv_D_omega_L @ ((1 - omega) * D @ x + omega * (U @ x + b))
+        x_new = inv_D_omega_L @ (omega * (b - U @ x) - omega * L @ x + (1 - omega) * D @ x)
         if np.linalg.norm(x_new - x, ord=2) < epsilon:
             return x_new
         x = x_new

ex5.py

@@ -1,29 +1,31 @@
 import matplotlib.pyplot as plt
 import numpy as np
 
-dich_steps = []
 new_raph_steps = []
-
+dich_steps = []
 
 def f(x) -> float:
     return np.exp(x) - 2 * np.cos(x)
 
 
-def dichotomie(x_0, a, b, epsilon, n=0) -> float:
+def dichotomie(a, b, epsilon=1e-6, n=0, steps=None) -> float:
-    delta = f(a) * f(x_0)
+    if steps is None:
-    x_1 = (a + b) / 2
+        steps = []
 
-    dich_steps.append(x_1)
+    x = (a + b) / 2
+
+    delta = f(a) * f(x)
+    dich_steps.append(x) # si on veut afficher les étapes
 
     if np.abs(b - a) <= epsilon:
-        return x_1
+        return x
 
     if delta < 0:
-        return dichotomie(x_1, a, x_0, epsilon, n+1)
+        return dichotomie(a, x, epsilon, n+1)
     elif delta == 0:
-        return x_1
+        return x
     else:
-        return dichotomie(x_1, x_0, b, epsilon, n+1)
+        return dichotomie(x, b, epsilon, n+1)
 
 
 def f_prime(x) -> float:
@@ -72,7 +74,7 @@ def ex1() -> None:
     plt.grid()
     plt.show()
 
-    dichotomie_res = dichotomie(0, 0, 1, 10e-6)
+    dichotomie_res = dichotomie(0, 1, 10e-6)
     print(dichotomie_res)
 
     plt.plot(dich_steps)