Actualización de calculo Pi()

solución con menos iteraciones, utilizando los promedios entre pares e impares se aproxima mas rápido al numero exacto

Chapter1/calculating_pi.py

@@ -14,18 +14,37 @@
 # See the License for the specific language governing permissions and
 # limitations under the License.
 
+# solucion con menos iteraciones
+# tengo otra version mas simplificada pero la estoy probando, que calcula a la iteracion 30 y otra a la 10
 
 def calculate_pi(n_terms: int) -> float:
     numerator: float = 4.0
     denominator: float = 1.0
     operation: float = 1.0
     pi: float = 0.0
+
+    pi_ant1: float = 4  # mem_pi_prev
+    pi_ant2: float = 0  # men_pi_prev_prev
+    pi_prom1: float = 0 # pi_mean_1
+    pi_prom2: float = 0 # pi_mean_2
+
     for _ in range(n_terms):
+        pi_ant2 = pi_ant1       # backup1
+        pi_ant1 = pi            # backup2
         pi += operation * (numerator / denominator)
         denominator += 2.0
         operation *= -1.0
-    return pi
 
+    pi_prom1 = (pi + pi_ant1) /2    # pi_mean_1
+    pi_prom2 = (pi_ant1 + pi_ant2) /2   # pi_mean_2
+    pi = (pi_prom1 + pi_prom2) /2   # pi_mean(pi_means)
+
+    return pi
 
 if __name__ == "__main__":
-    print(calculate_pi(1000000))
+    #print(calculate_pi(350)) # PAR:POR ABAJO  IMPAR:POR ARIBA
+    pi1 = calculate_pi(99)  # impar
+    pi2 = calculate_pi(100) # par
+    print(pi1)
+    print(pi2)
+    print( (pi1 + pi2) /2 ) # pi_mean(means)