階乗冪と関数の離散化

$$\begin{array}{rl}\mathrm{\Delta }{x}^{\underset{―}{n}}& =(x+1{)}^{\underset{―}{n}}-x^{\underset{―}{n}}\\ & =(x+1)\cdot x\cdots (x-n+3)\cdot (x-n+2)-x\cdot (x-1)\cdots (x-n+2)(x-n+1)\\ & =(x+1)\cdot x^{\underset{―}{n-1}}-x^{\underset{―}{n-1}}\cdot (x-n+1)\\ & =\{(x+1)-(x-n+1)\}\cdot x^{\underset{―}{n-1}}\\ & =n{x}^{\underset{―}{n-1}}\end{array}$$

$f(x)$の次数に関する帰納法で示す.
$$\begin{array}{r}\mathrm{deg}f=0\Rightarrow f(x)=f(0)=\frac{{\mathrm{\Delta }}^{0}f(0)}{0!}{x}^{\underset{―}{0}}=\sum _{n=0}^{\mathrm{deg}f}\frac{{\mathrm{\Delta }}^{n}f(0)}{n!}{x}^{\underset{―}{n}}\end{array}$$
$\mathrm{deg}f=k$で成立すると仮定.
$f(x)\in \mathbb{C}[x]$が$\mathrm{deg}f=k+1$を満たすとき, $\mathrm{deg}\mathrm{\Delta }f=k$より,
$$\begin{array}{rl}\mathrm{\Delta }f(x)& =\sum _{n=0}^k\frac{{\mathrm{\Delta }}^n\mathrm{\Delta }f(0)}{n!}{x}^{\underset{―}{n}}\\ & =\sum _{n=0}^k\frac{{\mathrm{\Delta }}^{n+1}f(0)}{n!}\cdot \frac{1}{n+1}\mathrm{\Delta }{x}^{\underset{―}{n+1}}\\ & =\mathrm{\Delta }\sum _{n=0}^k\frac{{\mathrm{\Delta }}^{n+1}f(0)}{(n+1)!}{x}^{\underset{―}{n+1}}\\ & =\mathrm{\Delta }\sum _{n=1}^{k+1}\frac{{\mathrm{\Delta }}^{n}f(0)}{n!}{x}^{\underset{―}{n}}\end{array}$$
よって,
$$\begin{array}{rl}& \mathrm{\Delta }(f(x)-\sum _{n=1}^{k+1}\frac{{\mathrm{\Delta }}^{n}f(0)}{n!}{x}^{\underset{―}{n}})=0\\ \Rightarrow & f(x)-\sum _{n=1}^{k+1}\frac{{\mathrm{\Delta }}^{n}f(0)}{n!}{x}^{\underset{―}{n}}=\mathrm{const}.\\ \Rightarrow & f(x)-\sum _{n=1}^{k+1}\frac{{\mathrm{\Delta }}^{n}f(0)}{n!}{x}^{\underset{―}{n}}=f(0)\\ \Rightarrow & f(x)=f(0)+\sum _{n=1}^{k+1}\frac{{\mathrm{\Delta }}^{n}f(0)}{n!}{x}^{\underset{―}{n}}=\sum _{n=0}^{k+1}\frac{{\mathrm{\Delta }}^{n}f(0)}{n!}{x}^{\underset{―}{n}}\end{array}$$
帰納法により示された.

$\subset$ は定理より従う.
任意の$x,n\in \mathbb{N}$に対して,
$$\begin{array}{r}\frac{x^{\underset{―}{n}}}{n!}=\{\begin{array}{ll}0\in \mathbb{Z}& (x<n)\\ (\genfrac{}{}{0ex}{}{x}{n})\in \mathbb{Z}& (x\ge n)\end{array}\end{array}$$
よって,$\supset$も分かる.