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## Proof: (related to "Factorial Polynomials vs. Polynomials")

Proof of the statement by induction.

### $”\Rightarrow”$

• Let $\phi(x)=a_nx^{\underline{n}}+a_{n-1}x^{\underline{n-1}}+\ldots+a_1x^{\underline{1}}+a_0$ be a factorial polynomial of degree $n\ge 0.$
• Base case $n=0:$
• If $\phi(x)=a_0$ then set $b_0:=a_0$ and we have $\phi(x)=b_0.$
• Induction Step $n\to n+1$
• Assume, $n\ge 0$ and any factorial polynomial $$\phi(x)=a_mx^{\underline{m}}+a_{m-1}x^{\underline{m-1}}+\ldots+a_1x^{\underline{1}}+a_0$$ of degree $m$ equals some polynomial $$\phi(x)=b_mx^{m}+b_{m-1}x^{m-1}+\ldots+b_1x^{1}+b_0$$ of degree $m$ for all $m\le n.$
• By definition of the falling factorial power, we have $$x^\underline{n+1}=x(x-1)\cdots(x-m)=x^{n+1}+p(x),$$ where $p(x)$ is some polynomial of degree $n.$
• By induction hypothesis, $\phi(x)=a_{n+1}x^\underline{n+1}+q(x)$ with a polynomial $q(x)$ of degree $n.$
• Thus, $\phi(x)=a_{n+1}(x^{n+1}+p(x))+q(x)=a_{n+1}x^{m+1}+s(x),$ where $s(x)=a_{n+1}p(x)+q(x)$ is a polynomial of degree $n.$
• Thus, $\phi(x)$ can be written as a polynomial of degree $n+1$.

### $”\Leftarrow”$

• analogously
q.e.d

| | | | created: 2020-03-30 17:18:34 | modified: 2020-03-30 17:18:35 | by: | references: [8404]

### Bibliography (further reading)

[8404] Miller, Kenneth S.: “An Introduction to the Calculus of Finite Differences And Difference Equations”, Dover Publications, Inc, 1960