The investigation below suggests that the found polynomial families are appell sequences <=> are have translation properties <=> have interesting derivative properties

Q: are $p_m(n, k)$ linearly independent? Sol: show $p_m(0, k) = k^m$. Then this uses the fact the $x^m$ are linearly independent polynomials. Simple using the recurrence definition.

Q: given a polynomial $p$ that is BH, is it a linear comb. of $\{p_m\}$? Given any polynomial in (n,k), is it a linear combination of $\{p_m\}$? Scratch: the bernoulli polynomials have an inversion formula, take a look at that.

In [1]:

from sympy import *

In [66]:

x = symbols('x')
k = symbols('k')
a = symbols('a')
n = symbols('n')

In [161]:

import functools


def bh(m):
    if m < 0:
        return lambda n, k: 0
    if m == 0:
        return lambda n, k: 1
    elif m == 1:
        return lambda n, k: k
    else:
        
        def alternating_tangent(n):
            # https://mathworld.wolfram.com/TangentNumber.html 
            return 2 ** (2*n) * (2**(2*n)-1) * Abs(bernoulli(2*n)) / (2*n)
        
        def R(n, k):
            running_sum = 0
            for j in range(1, round(m/2) + 1):
                i = m - 2*j
                running_sum += (-1)**(j) * alternating_tangent(j) * binomial(m-1, (2 * j - 1)) * bh(i)(n, k)
            return running_sum
        
        def fm(n, k):
            return k * bh(m-1)(n,k) + n * R(n, k)
        
        return fm
    
bh_ = lambda m: (lambda x: bh(m)(n, x)) # fixing n

In [167]:

def f(m):
    def fm(n, k):
        return (k + n) ** (m)
    return fm

f_ = lambda m: (lambda x: f(m)(n, x)) # fixing n

In [169]:

for s in [f_, bh_]:
    assert expand(s(2)(x-2)) == expand(s(2)(x) - 2 * s(1)(x) * 2 + s(0)(x) * 4)
    assert expand(s(3)(x+1)) == expand(s(3)(x) + 3 * s(2)(x) + 3 * s(1)(x) + s(0)(x))

In [181]:

# satifies appell seq. 
assert expand(diff(bh_(3)(x), x)) == expand(3 * bh_(2)(x))
assert expand(diff(bh_(4)(x), x)) == expand(4 * bh_(3)(x))
assert expand(diff(bh_(5)(x), x)) == expand(5 * bh_(4)(x))

In [228]:

collect( (expand(bh(4)(1, k+1) + bh(4)(1, k-1)))/2, n)

Out[228]:

$\displaystyle k^{4}$

In [230]:

expand(f(3)(1, k + 1) + f(3)(1, k - 1)) 

Out[230]:

$\displaystyle 2 k^{3} + 6 k^{2} + 12 k + 8$

In [203]:

expand(2 *binomial(3,0)*bh(3)(n, k) + 2 *binomial(3,1)*bh(1)(n, k))

Out[203]:

$\displaystyle k^{3} - 3 k n + 3 k$

In [231]:

expand(bh(4)(0, k))

Out[231]:

$\displaystyle k^{4}$

In [234]:

expand(bh(4)(n+1, k+1) + bh(3)(n+1, k+1) +  bh(4)(n+1, k-1) + bh(3)(n+1, k-1) )

Out[234]:

$\displaystyle 2 k^{4} + 2 k^{3} - 12 k^{2} n - 6 k n + 6 n^{2} + 4 n$

In [237]:

expand(2 * bh(4)(n, k) + 2*bh(3)(n, k))

Out[237]:

$\displaystyle 2 k^{4} + 2 k^{3} - 12 k^{2} n - 6 k n + 6 n^{2} + 4 n$

In [238]:

expand(bh(4)(n+1, k+1) + bh(3)(n+1, k+1))

Out[238]:

$\displaystyle k^{4} + 5 k^{3} - 6 k^{2} n + 3 k^{2} - 15 k n - 8 k + 3 n^{2} - n - 2$

In [240]:

expand(bh(4)(n+1, k-1) + bh(3)(n+1, k-1))

Out[240]:

$\displaystyle k^{4} - 3 k^{3} - 6 k^{2} n - 3 k^{2} + 9 k n + 8 k + 3 n^{2} + 5 n + 2$

In [242]:

expand(bh(4)(n, k) )

Out[242]:

$\displaystyle k^{4} - 6 k^{2} n + 3 n^{2} + 2 n$

In [243]:

expand(bh(6)(n, k))

Out[243]:

$\displaystyle k^{6} - 15 k^{4} n + 45 k^{2} n^{2} + 30 k^{2} n - 15 n^{3} - 30 n^{2} - 16 n$

In [246]:

for m in range(7):
    print(m, expand(bh(m)(n, k)))

0 1
1 k
2 k**2 - n
3 k**3 - 3*k*n
4 k**4 - 6*k**2*n + 3*n**2 + 2*n
5 k**5 - 10*k**3*n + 15*k*n**2 + 10*k*n
6 k**6 - 15*k**4*n + 45*k**2*n**2 + 30*k**2*n - 15*n**3 - 30*n**2 - 16*n

In [252]:

expand(bh(6)(n, k))

Out[252]:

$\displaystyle k^{6} - 15 k^{4} n + 45 k^{2} n^{2} + 30 k^{2} n - 15 n^{3} - 30 n^{2} - 16 n$

In [ ]: