'How to perform binomial-coefficient and factorial calculation with more precision?
I was comparing the result of my following python calculation with Mathematica: https://www.wolframalpha.com/input?i=sum+%28500+choose+r+%29%28-1%29%5Er+%2F%28r%21%29+%2C+r%3D0+to+500
import numpy as np
from decimal import *
import scipy.special
from scipy.special import factorial
getcontext().prec = 30
i = 500
sum(np.array([scipy.special.comb(Decimal(i), (r), exact=True)*pow(-1, r)/Decimal(factorial(r, exact=False)) for r in range(i+1)]))
Both calculations are giving almost same value until i = 400
but failing to converge after that even though I set an arbitrary precision in my python code via the decimal
module. Calculation with Mathematica seems correct. May I know, how can we get same result in python as that of Mathematica for larger i
?
Solution 1:[1]
The problem is that you have exact=False in the factorial.
>>> import numpy as np
>>> from decimal import *
>>> import scipy.special
>>> from scipy.special import factorial
>>> getcontext().prec = 30
>>>
>>> i = 500
>>> sum(np.array([scipy.special.comb(Decimal(i), (r), exact=True)*pow(-1, r)/Decimal(factorial(r, exact=False)) for r in range(i+1)]))
Decimal('-7.13859062099388393889008217957')
>>> sum(np.array([scipy.special.comb(Decimal(i), (r), exact=True)*pow(-1, r)/Decimal(factorial(r, exact=True)) for r in range(i+1)]))
Decimal('0.196589352363439561009074161963')
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | VRehnberg |