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Let $\mathbb R_n[X]$ be the vector space of polynomials of degree at most $n$. Let $u$ the endomorphism sending $P$ to $P(X+1)-P(X)$. I want to show that there exists a unique family of polynomials $Q_0, \cdots,Q_n$ such that $Q_0=1$ and $u(Q_{k})=Q_{k-1}$ and $Q_k(0)=0$ for $k=1..n$.

I know so far that the map sending $P$ to $P(X+1)$ is an automorphism and that $\ker(u)=\mathbb R_{0}[X]$ and $Im(u)=R_{n-1}[X]$ but don't know how to go further. Thank you for your help!

palio
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    It might remind you of the identity of the Pascal triangle, $\binom{n+1}{k+1}-\binom{n}{k+1}=\binom{n}{k}$. Also related is the Newton interpolation method. – Lutz Lehmann Mar 02 '14 at 14:10
  • I can't see how to use this identity to show the existence of these polynomials – palio Mar 02 '14 at 15:20

1 Answers1

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Existence: $Q_k(X) := \binom{X}{k} := \frac{X(X-1)\dotsc(X-k+1)}{k!}$ does the job.

Uniqueness: If $Q_0,\dotsc,Q_n$ satisfy the requirements, then since $\ker(u)=\mathbb{R}$ we easily get $Q_k = c \binom{X}{k}$ for some constant $c$. But then $Q_0=1$ gives $c=1$.

Martin Brandenburg
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    Can we show the existence and uniqueness without knowing the exact form of $Q_k(X)$. Because the next question after the existence and unicity we ask to show that $deg(Q_k)=k$ and that ${0,\cdots,k-1}$ are roots of $Q_k$ and then show that the $Q_k$ has the expression you gave? Thank you for your help!!! – palio Mar 02 '14 at 14:36