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Question: For every polynomial $P(x)$ of degree at n we have $$\sum_{i=0}^{n+1}(-1)^i\binom{n+1}{i}P(i)=0$$

Well I was inducting on degree of the polynomial, for $n=0$ its true and I got a bit confused in the inductive step, actually I cant figure the inductive step.

Is this property (if true) a famous one?

Thank you for your help.

2 Answers2

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Let us consider the (forward difference) operator $\delta$ acting on $\mathbb{R}[x]$ in the following way: $$ \forall p\in\mathbb{R}[x],\qquad (\delta p)(x) = p(x+1)-p(x).\tag{1} $$ The following properties are straightforward to prove:

  • (P1) If $p(x)$ is a non-costant polynomial with degree $d$,
    $\quad\;\;(\delta p)(x)$ is a polynomial with degree $d-1$;
  • (P2) If the leading term of $p(x)$ is $ax^d$, the leading term of $(\delta p)(x)$ is $ad x^{d-1}$,
    $\quad\;\;$ i.e. the operator $\delta$ acts on the leading term like the derivative $\tfrac{d}{dx};$
  • (P3) By induction, $$ (\delta^{(n)}p)(x) = \sum_{k=0}^{n}(-1)^k \binom{n}{k}p(x+k). $$ Due to $(P3),(P2),(P1)$ we have that for every $p(x)\in\mathbb{R}[x]$ with degree $n$ $$ \sum_{k=0}^{n}(-1)^k\binom{n}{k}p(k) = n!\tag{2} $$ holds, and for every $m\geq n+1$ $$ \sum_{k=0}^{m}(-1)^k\binom{m}{k}p(k) = 0\tag{3}$$ holds as well.
Jack D'Aurizio
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Here is one approach:

Let $Lp = \sum_{k=0}^{n+1} \binom{n+1}{k} (-1)^k p(k)$. Note that $L$ is linear, so if we can establish that $L(x \mapsto x^m) = 0$ (for $m =0,...,n$), then the result follows.

Let $\phi(x) = (1-x)^{n+1}$ and note that $\phi^{(m)}(1) = 0$ for $m=0,...,n$.

Since $\phi^{(m)}(x) = \sum_{k=0}^{n+1} \binom{n+1}{k} (-1)^k k(k-1)\cdots(k-m+1)x^{k-m}$, we see that $L(x \mapsto x(x-1)\cdots(x-m+1)) = \phi^{(m)}(1) = 0$ (for $m =0,...,n$).

Since we can write $x(x-1)\cdots(x-m+1) = x^m + \text{ terms in }1,x,...,x^{m-1} $, we have the desired result.

copper.hat
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