如何计算写成乘积的数学家的幂和？

Question

I'm a beginning Python programmer with no experience with sympy - just for fun. I'm trying to generate the formulas for sums of powers of integers as described in the latest Mathologer video (Power Sums). The following program generates the set of equations that follow. While correct, I would like the polynomials in the numerator to be factored where appropriate and individual terms to actually be summed (ie, a common denominator). I've tried a lot of "simplifiers", but nothing seems to work. Any ideas?

Program :

from sympy import S, Matrix, Symbol, simplify, pprint
import math

def bn(n, k):

    sign = n%2+k%2  # used to negate odd (0 based) diagonals
    n = n+1         # shift triangle up and left
    bc = math.factorial(n)//math.factorial(k)//math.factorial(n-k)
    if sign == 1: bc = -bc
    return bc


if __name__ == "__main__": 

    N = 5

    n = Symbol('n')
    v = Matrix([n**i for i in range(1,N+1)])

    M = [[bn(i,k) for k in range(0,i+1)]+(N-i-1)*[S(0)] for i in range(0,N)]
    eqs = simplify(Matrix(M).inv()*v)
    print(eqs.__repr__())
    # pprint(eqs, use_unicode=True)      

Output (note; had to replace double *'s with ^):

Matrix([[                           n],
        [                 n*(n + 1)/2],
        [       n*(2*n^2 + 3*n + 1)/6],
        [       n^2*(n^2 + 2*n + 1)/4],
        [n^5/5 + n^4/2 + n^3/3 - n/30]])

Answer 1

Does this do what you want?

In [8]: eqs.applyfunc(factor)                                                                                                                                 
Out[8]: 
⎡                 n                  ⎤
⎢                                    ⎥
⎢             n⋅(n + 1)              ⎥
⎢             ─────────              ⎥
⎢                 2                  ⎥
⎢                                    ⎥
⎢        n⋅(n + 1)⋅(2⋅n + 1)         ⎥
⎢        ───────────────────         ⎥
⎢                 6                  ⎥
⎢                                    ⎥
⎢             2        2             ⎥
⎢            n ⋅(n + 1)              ⎥
⎢            ───────────             ⎥
⎢                 4                  ⎥
⎢                                    ⎥
⎢                    ⎛   2          ⎞⎥
⎢n⋅(n + 1)⋅(2⋅n + 1)⋅⎝3⋅n  + 3⋅n - 1⎠⎥
⎢────────────────────────────────────⎥
⎣                 30                 ⎦

Answer 2

Note that as an alternative, and to check the Mathologer formulas, you may also calculate these sums as:

from sympy import symbols, Sum, factor, expand, Poly

j, n = symbols("j n", integer=True, positive=True)
for k in range(11):
    print(k, ":", factor(Sum(j**k, (j, 1, n)).doit()))

giving:

 0 : n
 1 : n*(n + 1)/2
 2 : n*(n + 1)*(2*n + 1)/6
 3 : n**2*(n + 1)**2/4
 4 : n*(n + 1)*(2*n + 1)*(3*n**2 + 3*n - 1)/30
 5 : n**2*(n + 1)**2*(2*n**2 + 2*n - 1)/12
 6 : n*(n + 1)*(2*n + 1)*(3*n**4 + 6*n**3 - 3*n + 1)/42
 7 : n**2*(n + 1)**2*(3*n**4 + 6*n**3 - n**2 - 4*n + 2)/24
 8 : n*(n + 1)*(2*n + 1)*(5*n**6 + 15*n**5 + 5*n**4 - 15*n**3 - n**2 + 9*n - 3)/90
 9 : n**2*(n + 1)**2*(n**2 + n - 1)*(2*n**4 + 4*n**3 - n**2 - 3*n + 3)/20
10 : n*(n + 1)*(2*n + 1)*(n**2 + n - 1)*(3*n**6 + 9*n**5 + 2*n**4 - 11*n**3 + 3*n**2 + 10*n - 5)/66

If you substitute factor() by expand() , you get all the powers separated:

 0 : n
 1 : n**2/2 + n/2
 2 : n**3/3 + n**2/2 + n/6
 3 : n**4/4 + n**3/2 + n**2/4
 4 : n**5/5 + n**4/2 + n**3/3 - n/30
 5 : n**6/6 + n**5/2 + 5*n**4/12 - n**2/12
 6 : n**7/7 + n**6/2 + n**5/2 - n**3/6 + n/42
 7 : n**8/8 + n**7/2 + 7*n**6/12 - 7*n**4/24 + n**2/12
 8 : n**9/9 + n**8/2 + 2*n**7/3 - 7*n**5/15 + 2*n**3/9 - n/30
 9 : n**10/10 + n**9/2 + 3*n**8/4 - 7*n**6/10 + n**4/2 - 3*n**2/20
10 : n**11/11 + n**10/2 + 5*n**9/6 - n**7 + n**5 - n**3/2 + 5*n/66

With Poly(Sum(j**k, (j, 1, n)).doit()).all_coeffs() you can get just the coefficients:

 0 : [1, 0]
 1 : [1/2, 1/2, 0]
 2 : [1/3, 1/2, 1/6, 0]
 3 : [1/4, 1/2, 1/4, 0, 0]
 4 : [1/5, 1/2, 1/3, 0, -1/30, 0]
 5 : [1/6, 1/2, 5/12, 0, -1/12, 0, 0]
 6 : [1/7, 1/2, 1/2, 0, -1/6, 0, 1/42, 0]
 7 : [1/8, 1/2, 7/12, 0, -7/24, 0, 1/12, 0, 0]
 8 : [1/9, 1/2, 2/3, 0, -7/15, 0, 2/9, 0, -1/30, 0]
 9 : [1/10, 1/2, 3/4, 0, -7/10, 0, 1/2, 0, -3/20, 0, 0]
10 : [1/11, 1/2, 5/6, 0, -1, 0, 1, 0, -1/2, 0, 5/66, 0]