How to perform binomial-coefficient and factorial calculation with more precision?-CodePudding

I was comparing the result of my following python calculation with Mathematica: https://www.wolframalpha.com/input?i=sum (500 choose r )(-1)^r /(r!) , r=0 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?

CodePudding user response：

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')