(HP15C)(HP67)(HP41C) Bernoulli Polynomials

[Albert Chan]

Code:

B(m,x) := sum(1/(n+1) * sum((-1)^k * comb(n,k) * (x+k)^m, k = 0 .. n), n = 0 .. m)

Another way to get B_m(x) is by synthetic division, for falling factorial form.

We can think of B as "derivative" of sum of power function, symbol B_m(x) == B^m

Σ(k^m, k=0 .. x-1) = ∫(B^m) = B^m+1 / (m+1)       + C

(08-05-2019 12:21 AM)Albert Chan Wrote:  From problem 12, Σ(x^m, x=0 to n-1) = n^m+1 / (m+1)

Except for symbol names, the 2 sums look very similar!
We can easily get x^m to its falling factorial form, via synthetic division.

The only issue is to get the integration constant, B_m(0) = B(m)
B(1) = -1/2, B(2k+1) = 0, we only need to worry about B(2k)

(07-28-2019 12:02 AM)Albert Chan Wrote:  Sum of k^6 formula = \(1\binom{n}{1}+63\binom{n}{2}+602\binom{n}{3}+2100\binom{n}{4}+3360\binom{n}{5}​+2520\binom{n}{6}+720\binom{n}{7}\)

B(6) = Linear term coefficient = 1/1 - 63/2 + 602/3 - 2100/4 + 3360/5 - 2520/6 + 720/7 = 1/42

Example, lets do formula for B₆(x)

Synthetic Division, x^m to falling factorial form, for B_m formula

Code:


    1   0   0   0   0   0   0  // x^6
1>  1   1   1   1   1   1   1
2>  1   3   7  15  31  63
3>  1   6  25  90 301
4>  1  10  65 350
5>  1  15 140
6>  1  21

Constant of integration, adjusted for using falling factorial numbers.

B(6) = 1*0!/1 - 63*1!/2 + 301*2!/3 - 350*3!/4 + 140*4!/5 - 21*5!/6 + 1*6!/7 = 1/42

x^5 = x^4 * x, we pick numbers 3rd entries from the right.

x^5 = [1, 15, 25, 10, 1] • [x^1, x^2, x^3, x^4, x^5]

(B^6-B(6))/6 = [1/2, 15/3, 25/4, 10/5, 1/6] • [x^2, x^3, x^4, x^5, x^6]      // "integrate"

B^6 = 1/42 + [3, 30, 75/2, 12, 1] • [x^2, x^3, x^4, x^5, x^6]

Or, in efficient "horner" form:

B₆(x) = 1/42 + x*(x-1)*(3 + (x-2)*(30 + (x-3)*(75/2 + (x-4)*(12 + (x-5)))))