Proof
(related to Theorem: Taylor's Formula Using the Difference Operator)
- By hypothesis, $f:D\to\mathbb C$ is a factorial polynomial, i.e. $f(x)=a_nx^{\underline{n}}+a_{n-1}x^{\underline{n-1}}+\ldots+a_1x^{\underline{1}}+a_0$ for some complex numbers $a_n,\ldots,a_1x^{\underline{1}}a_0\in\mathbb C.$
- For $x=0$, we have $f(0)=a_0.$
- Taking the difference operator yields with the formula for the difference operator of falling factorial powers $$\Delta f(x)=na_nx^{\underline{n-1}}+(n-1)a_{n-1}x^{\underline{n-2}}+\ldots+a_1$$ and $\Delta f(0)=a_1.$
- Proceeding in this fashion, we get with the kth difference operator $\Delta^k f(0)=k! \cdot a_k$ (with $k!$ being the factorial)
- Hence,
$$f(x)=\frac{\Delta^n f(0)}{n !} x^{\underline{n}}+\frac{\Delta^n f(0)}{(n-1) !} x^{\underline{n-1}}+ \ldots + \frac{\Delta^2 f(0)}{2 !} x^{\underline{2}} + \Delta f(0) x^{\underline{1}}+ f(0).$$
∎
Thank you to the contributors under CC BY-SA 4.0! 
- Github:
-
References
Bibliography
- Miller, Kenneth S.: "An Introduction to the Calculus of Finite Differences And Difference Equations", Dover Publications, Inc, 1960
