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Given a polynomial $f(x)$ of degree $k$ on field $\mathbb{F}$, one can find a polynomial $S_f(x)$ of degree $\leq k+1$ such that for each positive integer $n$, $$ S_f(n)=\sum_{j=0}^{n-1}f(j). $$

There exists such a polynomial that the condition holds for $n=1,\cdots,k+2$. But how to prove for each $n\geq1$, the condition also holds?

Jyrki Lahtonen
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Knt
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  • By linearity it suffices to prove this for all monomials $f(x)=x^\ell$, $0\le\ell\le k$. In characteristic zero we need the power sum. Search for Faulhaber's formula. In characteristic $p>0$ the same works, if $k<p$. If $k\ge p$ the sums $\sum_{j=0}^{n-1} j^k$ reduce to lower powers by Little Fermat. – Jyrki Lahtonen Oct 04 '19 at 16:00
  • See, for example, 1 and 2. – Jyrki Lahtonen Oct 04 '19 at 16:03
  • You know a bit of linear algebra, right? It is enough to find a basis of the vector space of degree-$\leq k$ polynomials such that the claim is true whenever $f$ belongs to this basis. The simplest basis is $\left(x^0, x^1, \ldots, x^k\right)$, but it is not very easy to prove the claim on this basis. There is a better basis for that: $\left(f_0, f_1, \ldots, f_k\right)$, where $f_i = x\left(x-1\right)\left(x-2\right)\cdots\left(x-i+1\right)$. – darij grinberg Oct 04 '19 at 16:29
  • Actually, your claim does not hold without qualification. I believe you want to require $\mathbb{F}$ to have characteristic $0$. – darij grinberg Oct 04 '19 at 16:30

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