Wikipedia:Reference desk/Archives/Mathematics/2016 November 25

= November 25 =

Polynomial coefficients generated by a sum of powers
Is there a general algorithm to generate the coefficients of the expansion of $$f_k(n) = \sum_{i=1}^n i^k $$? For example, $$f_7(n) = \frac{n^8}{8} + \frac{n^7}{2} + \frac{7 n^6}{12} - \frac{7 n^4}{24} + \frac{n^2}{12}$$. I could use polynomial interpolation on the first $$k+2$$ terms of the series, but that gets impractical quickly if $$k$$ is large. 24.255.17.182 (talk) 21:47, 25 November 2016 (UTC)


 * One interesting thing I noticed is that $$\frac{n^{k-i} (i+1)!}{\operatorname{C}(k, i-1)}$$ is invariant with respect to $$k$$, and starts out $$1, \frac{1}{2}, 0, -1, 0, 20, 0, -1512, 0, 302400, 0, -131345280, 0, 108972864000, 0, 157661939635200, 0, 371498581149696000...$$, but I don't see an obvious pattern here and this series isn't in OEIS. 24.255.17.182 (talk) 22:13, 25 November 2016 (UTC)


 * (ec)Use Binomial_coefficient and the Hockey-stick identity.
 * $$\sum_{i=1}^n i^k = \sum_{i=1}^n \sum_{m=0}^k a_m \binom i m = \sum_{m=0}^k a_m \sum_{i=1}^n \binom i m = \sum_{m=0}^k a_m \binom {n+1}{m+1}$$
 * Bo Jacoby (talk) 22:24, 25 November 2016 (UTC).


 * Sorry if it's obvious but could you explain how $$a_m$$ is to be computed? 24.255.17.182 (talk) 23:03, 25 November 2016 (UTC)
 * I think what you're looking for is Faulhaber's formula. A generalization is the Euler–Maclaurin formula. --RDBury (talk) 01:43, 26 November 2016 (UTC)
 * The formula is found in the link I gave you.
 * $$a_m = \sum_{i=0}^m (-1)^{m-i} \binom{m}{i} i^k.$$
 * Bo Jacoby (talk) 07:18, 26 November 2016 (UTC).