Le samedi 24 février 2024 à 18:11:25 UTC+1, Gareth Ma a écrit :
Note that you can wrap it in `Decimal` or `Fraction`, which are both builtin Python libraries. I stand corrected. Python ints are bignums, *not* 32-bits integers. and fractions.Fractions are rationals. Ordinary integer arithmetic wroks with these rationals. We just need a binomial coefficient function returning an int (math.comb returns a float) : >>> from math import factorial >>> from fractions import Fraction >>> def ibin(n, k): return Fraction(factorial(n), factorial(k)*factorial(n-k)) if n>=k else 0 ... >>> def g(n, k, r): return (-1)**k*ibin(n,k)*Fraction((n-k)**r, n**r) ... >>> def f(n, r): return float(sum([g(n,k,r) for k in range(n+1)])) ... >>> f(365, 2000) 0.21611945163321844 A tad more muscular than using Sage… Thanks a lot ! I learned something useful today… On Saturday 24 February 2024 at 13:54:51 UTC Emmanuel Charpentier wrote: Le vendredi 23 février 2024 à 23:23:20 UTC+1, Dima Pasechnik a écrit : [ Snip…] the normal Python way, without any symbolic sum, would be like this: sage: sage: g(n,k,r)=(-1)^(k)*binomial(n,k)*(n-k)^r/n^r ....: sage: def f(n,r): return math.fsum([1.0*g(n,k,r) for k in range(n+1)]) ....: sage: f(365,2000) 0.21611945163321847 This works *in Sagemath*. It wouldn’t work in Python : the range of the magnitudes of the terms of the *alternating* sum are way too large for the precision of Python’s floats, necessary if you want to use math.comb. Programming this in Python would need some serious analytical work, or using a multiple-precision integer library, which Sage does for you… [ Re-snip... ] HTH, -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-support+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/sage-support/4dfa86e9-d3c8-42cc-8bda-826640da9cf8n%40googlegroups.com.
