Define the forward difference operator $$\Delta f(x) = f(x+1) - f(x)$$ I believe that if $f(x)$ is a polynomial with integer coefficients, $\Delta^k f(x)$ is divisible by k!. By linearity it suffices to consider a single monomial $f(x) = x^n$. I've checked this for small values of $n$ and $k$, and believe that a simple proof should exist, but am unable to find it.

In particular, brute force gives $$\Delta^k x^n = \sum_j x^{n-j} \left[ \binom{n}{j} \sum_i (-1)^{k-i} \binom{k}{i} i^j \right]$$ but the terms in brackets appear to have no closed form solution (see (20)-(25) of http://mathworld.wolfram.com/BinomialSums.html).

Motivation: I have an unknown integer coefficient polynomial of degree $n$ sampled at $x = 0, 1, \ldots, n$, and want to prove that all intermediate results in the classical divided difference algorithm are integers.

Geoffrey Irving
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  • This may be helpful: [link](http://www.trans4mind.com/personal_development/mathematics/series/polynomialEquationDifferences.htm) – Milind Hegde Jun 07 '13 at 07:05
  • That's an interesting link, but doesn't seem to establish the desired result. In particular, it is easy to prove that the leading coefficient of $\Delta^k x^n$ is $k!\binom{n}{k}$ by induction (indeed, it is the same as the coefficient of the $k$th derivative), but I'm not sure how to extend that result to the other terms. – Geoffrey Irving Jun 07 '13 at 07:09
  • More data: Mathematica confirms that the result holds for $k \le 100$ and $n \le 200$. – Geoffrey Irving Jun 07 '13 at 07:22

2 Answers2


By linearity, it suffices to prove this for the polynomials $x(x - 1)\cdots(x - (n-1))$. This is just $n! {x \choose n}$. A basic property of the forward difference operator is that $\Delta {x \choose n} = {x \choose n-1}$, from which it follows that

$$\Delta^k x(x - 1)\cdots(x - (n-1)) = n! {x \choose n-k} = k! {n \choose k} x(x - 1) \cdots(x - (n-k-1))$$

and the conclusion follows.

Geoffrey Irving
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Qiaochu Yuan
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    This is a nice illustration of the general principle that if you ever choose a basis (of a vector space or more generally a free module), choose the one best suited to your needs rather than the first one that comes to mind. – Qiaochu Yuan Jun 07 '13 at 07:31
  • +1. I saw this question just before going to lunch, and was about to come and post this answer. This is also an illustration of the fact that (see the "By linearity it suffices to consider a single monomial $f(x)=x^n$" in the question) sometimes deciding to prove the simplest case isn't necessarily the best strategy -- the general case might be easier to prove, or a different special case as in this example. – ShreevatsaR Jun 07 '13 at 08:00
  • Thanks! Quite foolish of me to miss the Newton basis, especially since the motivating algorithm is interpolation into that basis. :) – Geoffrey Irving Jun 07 '13 at 18:04

This is really the same answer as that of Qiaochu Yuan, but I find the "binomial coefficients of $x$", much as I approve the notation, a bit distracting when next to ordinary binomial coefficients. One can do without them, using falling factorial powers instead: $x^\underline n=x(x-1)\ldots(x-n+1)$, which is of course the same as $n!\binom xn$. Elementarily $$ (x+1)^\underline n-x^\underline n=x^\underline{n-1}((x+1)-(x-n+1)) =nx^\underline{n-1}, $$ in other words $\Delta(x^\underline n)=nx^\underline{n-1}$ (which should remind of ordinary calculus), and $$ \Delta^k(x^\underline n)=n(n-1)\ldots(n-k+1)x^\underline{n-k} =k!\binom nkx^\underline{n-k}, $$ giving what we want. Note that I resisted the temptation to write the coefficient as $n^\underline k$ to avoid the kind of distraction I mentioned in the first sentence. Also every factor in the expression is in $\Bbb Z[x]$.

Marc van Leeuwen
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