Relation between two polynomials

Question

I am stuck on the following exercise. Let $P \in \mathbb{C}[X]$ be a polynomial of degree $n \in \mathbb{N}$, $n > 0$. We write :

$$ P(X) = \sum_{k=0}^{n} a_k X^k $$

with $a_0, \ldots, a_n \in \mathbb{C}^{n+1}$. I would like to prove that :

$$ \sum_{k=0}^{n} (-1)^{n-k} \begin{pmatrix} n \\ k \end{pmatrix} P(k) = n! ~ a_n. $$

Could someone, please, suggest me ways/ideas to prove this equality ?

Answer 1

Here is an idea. Both sides equal $(\Delta^nP)(X)$. Where $(\Delta P)(X)=P(X+1)-P(X)$. We claim that $$ (\Delta^nP)(X)=n!a_{n}. $$ To see this write $$ P(X) = \sum_{k=0}^{n} a_k X^k= \sum_{k=0}^{n} a_k \sum_{m=0}^{k}S(k,m)X^{\underline{m}}\tag{1} $$ where $S(k,m)$ is the stirling number of the second kind. and $X^{\underline{M}}$ is the falling factorial. Since $\Delta(X^{\underline m})=mX^{\underline{m-1}}$, $\Delta^{n}(X^{\underline {m}})=m^{\underline n}X^{\underline{m-n}}$. If $m<n$, then $\Delta^{n}(X^{\underline {m}})=0$. Hence taking $\Delta ^{n}$ of both sides of (1) yields that $$ (\Delta^nP)(X)=n!S(n,n)a_{n}=n!a_{n}\tag{2}. $$ Now write $$ P(X)=b_{0}\binom{X}{0}+b_{1}\binom{X}{1}+\dotsb+b_{n}\binom{X}{n}\tag{3} $$ Observe that $$ \Delta\binom{X}{k}=\binom{x}{k-1}. $$ Hence it follows that $$ (\Delta^nP)(X)=b_{n}=\sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} P(k). \tag{4} $$ You can prove (4) by induction using the relationship. $$ b_{n}=P(n)-b_{0}-b_{1}\binom{n}{1}-\dotsb-b_{n-1}\binom{n}{n-1}. $$ Equations (2) and (4) yield the result.

Answer 2

The polynomials $\binom{X}{u}:u=0,\ldots,n$ are linearly independent over $\Bbb C$ and generates the set of all polynomials of degree $n$. Thus there exists $b_u\in\Bbb C$ such that $$P(X)=\sum_{u=0}^nb_u\binom Xu$$ note that $n!a_n=b_n$.

Let $\mathrm D$ denote the formal derivative respect to $X$, then \begin{align} f_u(X) &:=\sum_{k=0}^n\binom nk(-1)^{n-k}\binom ku X^{k-u}\\ &=\sum_{k=0}^n\binom nk(-1)^{n-k}\frac 1{u!}\mathrm D^u X^k\\ &=\frac 1{u!}\mathrm D^u\sum_{k=0}^n\binom nk(-1)^{n-k} X^k\\ &=\frac 1{u!}\mathrm D^u(X-1)^n\\ \end{align} hence $f_u(1)=0$ for $u<n$ while $f_n(1)=1$. Consequently, \begin{align} \sum_{k=0}^n(-1)^{n-k}\binom nkP(k) &=\sum_{u=0}^n b_uf_u(1)\\ &=b_n\\ &=a_nn! \end{align}