Prevent Rounding to Zero in Python

Question

I have a program meant to approximate pi using the Chudnovsky Algorithm, but a term in my equation that is very small keeps being rounded to zero.

Here is the algorithm:

import math
from decimal import *
getcontext().prec = 100

pi = Decimal(0.0)
C = Decimal(12/(math.sqrt(640320**3)))
k = 0
x = Decimal(0.0)
result = Decimal(0.0)
sign = 1
while k<10:
    r = Decimal(math.factorial(6*k)/((math.factorial(k)**3)*math.factorial(3*k)))
    s = Decimal((13591409+545140134*k)/((640320**3)**k))
    x += Decimal(sign*r*s)
    sign = sign*(-1)
    k += 1
result = Decimal(C*x)
pi = Decimal(1/result)


print Decimal(pi)

The equations may be clearer without the "decimal" terms.

import math

pi = 0.0
C = 12/(math.sqrt(640320**3))
k = 0
x = 0.0
result = 0.0
sign = 1
while k<10:
    r = math.factorial(6*k)/((math.factorial(k)**3)*math.factorial(3*k))
    s = (13591409+545140134*k)/((640320**3)**k)
    x += sign*r*s
    sign = sign*(-1)
    k += 1
result = C*x
pi = 1/result

print pi

The issue is with the "s" variable. For k>0, it always comes to zero. e.g. at k=1, s should equal about 2.1e-9, but instead it is just zero. Because of this all of my terms after the first =0. How do I get python to calculate the exact value of s instead of rounding it down to 0?

Answer 1

Try:

s = Decimal((13591409+545140134*k)) / Decimal(((640320**3)**k))

The arithmetic you're doing is native python - by allowing the Decimal object to perform your division, you should eliminate your error.

You can do the same, then, when computing r.

Answer 2

A couple of comments.

If you are using Python 2.x, the / returns an integer result. If you want a Decimal result, you convert at least one side to Decimal first.

math.sqrt() only return ~16 digits of precision. Since your value for C will only be accurate to ~16 digits, your final result will only be accurate to 16 digits.