The mathematical constant pi can be approximated to at least 11 decimals by usin

Question

The mathematical constant pi can be approximated to at least 11 decimals by using the mathematical algorithm: pi = squareroot 12 (1/1 middot 3^- 1/3 middot 3^+ 1/5 middot 3^- 1/7 middot 3^+ 1/9 middot 3^..) Notice that the sign alternates from minus (-)to plus (+) for each term. This approximation becomes more accurate as you increase the number of terms used. You need to evaluate this calculation based on a varying number of terms, (i.e.: how accurate is the approximation with 50, 100, 500, 1000 or more terms)

Explanation / Answer

import math
from decimal import Decimal
pie = Decimal(float(22.0/7))
def oddgenerator(n):
odd = []
for i in range(n):
if i%2==0:
pass
else:
odd.append(i)
return odd
  
def Pie(n=100):
start = Decimal(math.sqrt(12))
odd = oddgenerator(n)
sum = 0
for i in range(len(odd)):
if i%2==0:
sum += 1/Decimal(odd[i]*(3**i))
else:
sum -= 1/Decimal(odd[i]*(3**i))
total = start * sum
return Decimal(total)
print "This program approximate value of PI"
print "More Elements increase accuracy"
print "This program support any limit of number"
n=int(raw_input("Enter number of elements: "))
print "Number of Elements"+str(n);
p=Pie(n)
print "Approx PI: "+str(p)
print "Correct PI: "+str(pie)
print "Difference :"+ str(Decimal(pie-p))

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