Proof
(related to Proposition: Factorial Polynomials have a Unique Representation)
- Assume, a given factorial polynomial $\phi(x)$ of degree $n$ has the two representations $$\begin{align}\phi(x)&=b_{n+k}x^{\underline{n+k}}+b_{n+k-1}x^{\underline{n+k-1}}+\ldots+b_{n+1}x^{\underline{n+1}}+\nonumber \\
&\quad+b_nx^{\underline{n}}+b_{n-1}x^{\underline{n-1}}+\ldots+b_1x^{\underline{1}}+b_0\tag{1}\\
\phi(x)&=a_nx^{\underline{n}}+a_{n-1}x^{\underline{n-1}}+\ldots+a_1x^{\underline{1}}+a_0,\quad a_n\neq 0\tag{2}.\end{align}$$
- This means that both equations $(1)$ and $(2)$ hold for all complex numbers $x\in\mathbb C.$
- Subtracting $(2)$ from $(1)$ gives us $$\begin{align}0&=b_{n+k}x^{\underline{n+k}}+b_{n+k-1}x^{\underline{n+k-1}}+\ldots+b_{n+1}x^{\underline{n+1}}+\nonumber \\
&\quad+(b_n-a_n)x^{\underline{n}}+(b_{n-1}-a_{n-1})x^{\underline{n-1}}+\ldots+(b_1-a_1)x^{\underline{1}}+(b_0-a_0)\tag{3}\end{align}$$
for all $x\in\mathbb C.$
- Now, note that if for all $x\in\mathbb C$ $$\begin{align}0&=\alpha_mx^{\underline{m}}+\alpha_{m-1}x^{\underline{m-1}}+\ldots+\alpha_1x^{\underline{1}}+\alpha_0\tag{4}\end{align}$$ then $$\begin{align}0&=\alpha_m=\alpha_{m-1}=\cdots=a_1=\alpha_0.\tag{5}\end{align}$$
- For if we let $x=0,$ then $\alpha_0=0.$
- The difference operator of falling factorial powers of $(4)$ gives us $$\begin{align}0&=m\alpha_mx^{\underline{m-1}}+(m-1)\alpha_{m-1}x^{\underline{m-1}}+\ldots+2\alpha_2x^{\underline{2}}+\alpha_1\tag{6}\end{align}$$
- For $x=0$ it follows $\alpha_1=0.$
- Repeating this procedure, gives us $(5)$.
- From $(3)$ and $(5)$ it follows $b_{n+k}=b_{n+k-1}=\cdots=b_{n+1}=0$ and $b_r=a_r$ for $r=0,1,\ldots,n.$
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References
Bibliography
- Miller, Kenneth S.: "An Introduction to the Calculus of Finite Differences And Difference Equations", Dover Publications, Inc, 1960
