I am reading a book on difference equations by Samuel Goldberg. In the book, he states: $$\Delta^p y(x)= 0, \forall p>2$$ for a quadratic. So I checked the pattern and I think that the following may be true: $$\Delta^p y(x)= 0, \forall p>n$$ for a polynomial of degree $n$ $ (a_0+a_1x+a_2x^2+ ...+a_nx^n).$
But I don't know how to prove this. My best method so far has been to repeat it for a few polynomials and then say "similarly, this is true for the rest...." but that's not really a proof.
I suspect that induction will be the method and I found one on here that vaguely explains it but I would really appreciate a more detailed thought process.
Furthermore, is this a double implication?
Note that $\Delta y(x)=y(x+h)-y(x)$.
Thanks in advance.
Use induction to prove that $\Delta^p x^n$ is a polynomial of degree $n - p$.
You could also try your hand at $\Delta x^{\underline{n}} = n x^{\underline{n - 1}}$, where $x^{\underline{n}} = x (x - 1) \dotsm (x - n + 1)$ (i.e., $n$ factors); if you apply $\Delta$ more than $n$ times the result is zero; and see that you can write your polynomial in terms of such terms.