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

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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, $f_7(n)=\frac{n^8}{8}+\frac{n^7}{2}+\frac{7n^6}{12}-\frac{7n^4}{24}+\frac{n^2}{12}$. 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 $\frac{n^{k-i}(i+1)!}{\mathrm{C}(k,i-1)}$ 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#Binomial_coefficients_as_a_basis_for_the_space_of_polynomials and the Hockey-stick identity.

$\sum _{i=1}^n{i}^k=\sum _{i=1}^n\sum _{m=0}^k{a}_m(\genfrac{}{}{0ex}{}{i}{m})=\sum _{m=0}^k{a}_m\sum _{i=1}^n(\genfrac{}{}{0ex}{}{i}{m})=\sum _{m=0}^k{a}_m(\genfrac{}{}{0ex}{}{n+1}{m+1})$

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.

$a_m=\sum _{i=0}^m(-1{)}^{m-i}(\genfrac{}{}{0ex}{}{m}{i})i^k.$

Bo Jacoby (talk) 07:18, 26 November 2016 (UTC).