"""
Berechnung von pi über arctan(x)
(c) Jürgen Meier
www.3d-meier.de
26.06.2024
"""

import math
import decimal


# Anzahl Kommastellen
decimal.getcontext().prec = 105


a1 = decimal.Decimal(1)
a2 = decimal.Decimal(2)
a3 = decimal.Decimal(3)
a4 = decimal.Decimal(4)
a6 = decimal.Decimal(6)
a9 = decimal.Decimal(9)
a32 = decimal.Decimal(32)
a73 = decimal.Decimal(73)
a2943 = decimal.Decimal(2943)

m1 = decimal.Decimal(-1)

x1 = a1/a9
x2 = a1/a32
x3 = a1/a73
x4 = a1/a2943

# Reihenentwicklung für arctan(x)
k = 300

t1 = 0
for n in range(0, k + 1):
    a = (((m1)**n)*(x1**(a2*n + a1)))/(a2*n + a1)
    t1 = t1 + a

t2 = 0
for n in range(0, k + 1):
    a = (((m1)**n)*(x2**(a2*n + a1)))/(a2*n + a1)
    t2 = t2 + a

t3 = 0
for n in range(0, k + 1):
    a = (((m1)**n)*(x3**(a2*n + a1)))/(a2*n + a1)
    t3 = t3 + a

t4 = 0
for n in range(0, k + 1):
    a = (((m1)**n)*(x4**(a2*n + a1)))/(a2*n + a1)
    t4 = t4 + a
    
p = a6*t1 + a3*t2 + a2*t3 + t4
print(4*p)

print(((complex(9, 1)/complex(9, -1))**6)*
((complex(32, 1)/complex(32, -1))**3) *
((complex(73, 1)/complex(73, -1))**2) *
((complex(2943, 1)/complex(2943, -1))))