Lesson 7

The first few power-sum polynomials are: \[\begin{align*}S_{1}(n)&=\frac{1}{2}n^{2}+\frac{1}{2}n,\\ S_{2}(n)&=\frac{1}{3}n^{3}+\frac{1}{2}n^{2}+\frac{1}{6}n,\\ S_{3}(n)&=\frac{1}{4}n^{4}+\frac{1}{2}n^{3}+\frac{1}{4}n^{2},\\ S_{4}(n)&=\frac{1}{5}n^{5}+\frac{1}{2}n^{4}+\frac{1}{3}n^{3}-\frac{1}{30}n,\\ S_{5}(n)&=\frac{1}{6}n^{6}+\frac{1}{2}n^{5}+\frac{5}{12}n^{4}-\frac{1}{12}n^{2}.\end{align*}\] For \(k \ge 2\), the leading terms follow the pattern \[S_{k}(n)=\frac{1}{k+1}n^{k+1}+\frac{1}{2}n^{k}+\frac{k}{12}n^{k-1}+0\cdot n^{k-2}+\dotsb .\] In fact, the coefficients of \(n^{k-2}, n^{k-4}, \dotsc\) are zero.

One way to see the symmetry behind these zero coefficients is the identity \[S_{k}(-n)=(-1)^{k+1}S_{k}(n-1),\] understood as an identity of polynomials. Equivalently, \(S_{k}(x-\frac{1}{2})\) is even for odd \(k\) and odd for even \(k\). In particular, for odd \(k\), the polynomial \(S_{k}(n)\) can be written as a polynomial in \[u=\frac{n(n+1)}{2}.\] For example, \(S_{3}(n)=u^{2}\). The remaining coefficients are usually encoded by the Bernoulli numbers, but we will not need this general formula here.