A comparison of the derivative of powers $\frac {d}{dx}x^n=nx^{n-1}$ known from calculus with the result of the difference operator of powers $\Delta x^n=\sum_{k=1}^n \binom {n}{k} x^{n-k}$ shows that the latter does not have such a simple compact expression. It turns out that a function leading to a similar compact expression when using the difference operator is the so-called falling factorial power $x^{\underline n},$ and we will see soon that $\Delta x^{\underline n}=nx^{\underline {n-1}}.$ We first define the falling factorial power and its counterpart - the rasing factorial power and study some of their basic properties.
For a real or complex number \(x\) and a positive integer $n\ge 1$ we define the following operators:
falling factorial power of \(x\) (also read "\(x\) to the \(n\) falling"):
\[x^{\underline{n}}:= x(x-1)(x-2) \cdots (x-n+1).\]rising factorial power of \(x\) (also read "\(x\) to the \(n\) rising"):
\[x^{\overline{n}}:= x(x+1)(x+2) \cdots (x+n-1).\]
Moreover, for $x\not\in\{-1,-2,\ldots,-n\}$ we define $$x^{\underline 0}:=1,\quad x^{\underline{-n}}:=\frac{1}{(x+n)^{\underline{n}}}=\frac{1}{(x+n)(x+n-1)\cdots(x+2)(x+1)},$$ and for $x\not\in\{1,2,\ldots,n\}$ $$x^{\overline 0}:=1,\quad x^{\overline{-n}}:=\frac{1}{(x-n)^{\overline{n}}}=\frac{1}{(x-n)(x-n+1)\cdots(x-2)(x-1)}.$$
Corollaries: 1
Definitions: 2 3
Lemmas: 4
Motivations: 5
Parts: 6
Proofs: 7 8 9 10 11 12
Propositions: 13 14 15
Solutions: 16 17 18
Theorems: 19
