Proof of a neat pattern in polynomials Let $f_1:\mathbb{R}\to\mathbb{R}$ such that $$f(x) = ax + b\space\space \forall\space x \in \mathbb{R}$$
It can be easily verified that $$f(x)-2f(x-1)+f(x-2)=0 \space \forall \space x \in 
\mathbb{R}---(I)$$
Now,$\space$let $f_2:\mathbb{R}\to\mathbb{R}$ such that $$f(x) = ax^2+bx+c \space \forall\space x\space \in \mathbb{R}$$
After some trial and error, I found that $$f(x)-3f(x-1)+3f(x-2)-f(x-3) = 0 \space \forall\space x\space \in \mathbb{R}---(II)$$
One could see the coefficients of terms in the expansion of $(1-x)^2$ being the coefficients of terms in $(I)$ in the same order, and the coefficients of terms in the expansion of $(1-x)^3$ being the coefficients of terms in $(II)$ in the same order.
I tried to continue the above observation:
Let $f_{4}:\mathbb{R}\to\mathbb{R}$ such that $$f(x) = ax^3 + bx^2 + cx + d\space\space \forall\space x \in \mathbb{R}$$
It can be verified that $$f(x)-4f(x-1)+6f(x-2)-4f(x-3)+f(x-4) = 0\space \forall\space x\space \in \mathbb{R}---(III)$$
And voila, the coefficients of terms in the expansion of (1-x)^4 are the coefficient of terms in $(III)$ in the same order.
I have been trying to extend this result in general to $f_{n}$ where $n \in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers, have tried to approach this problem via several methods but couldn't get anywhere.
Any hint towards proving the general result(if at all the result is true) will be appreciated
Thank You
PS: $a,b,c,d \in \mathbb{R}$ and $a \neq 0$
 A: This article on forward differences will prove helpful, or at least it's my initial take. It will require a little comfort with basic calculus and summation notation, however; I don't know if there's a more elementary angle, this is simply what came to mind.

Recall from calculus: we can define the derivative $f'$ of a function $f$ by
$$f'(x) := \lim_{x \to 0} \frac{f(x+h) - f(x)}{h}$$
We can define a discretized version of this (one someone might use for sequences, for instance), by taking the argument of the limit for $h=1$. We define this "forward difference" operation to be $\newcommand{\D}{\Delta} \D f$:
$$(\D f)(x) \equiv \D f(x) := f(x+1) - f(x)$$
One may ask: what happens if we apply this twice? Well, first, one can show that $\D$ is linear, i.e. for any $\alpha,\beta \in \mathbb{R}$, we have
$$\D (\alpha f+\beta g) = \alpha \D f + \beta \D g$$
(Try proving it for yourself!) Then
$$\begin{align*}
\D^2 f(x) 
&= \D (\D f)(x) \\
&= \D \Big( f(x+1) - f(x) \Big)\\
&= \D f(x+1) - \D f(x) \\
&= f(x+2) - f(x+1) - \Big( f(x+1) - f(x) \Big) \\
&= f(x+2) - 2 f(x+1) - f(x)
\end{align*}$$
You might be seeing the relation to your equation. The pattern holds in general: applied $n$ times,
$$\D^n f(x) = \sum_{k=0}^n \binom n k (-1)^{n-k} f(x+k)$$
(Try proving this one for yourself too! Induction may be helpful.)

Now, with this in mind, define $p$ a polynomial of degree $n$:
$$p(x) = \sum_{k=0}^n a_k x^k$$
Find $\D p$:
$$\D p(x) = \sum_{k=0}^n a^k \Big( (x+1)^k - x^k \Big)$$
Note that, using the binomial theorem
$$(x+1)^k - x^k 
= -x^k + \sum_{j=0}^k \binom k j x^j$$
If we extract out the $j=k$ term of the sum we get
$$(x+1)^k - x^k 
= -x^k + x^k+ \sum_{j=0}^{k-1} \binom k j x^j
= \sum_{j=0}^{k-1} \binom k j x^j$$
Notice what this means: $\D p$ has degree $n-1$.
Thus, $\D^m p$ has degree $n-m$.
If we reach the case $m=n$, then $\D^n p$ is a degree $0$ (constant) polynomial. Of course, if constant, it is easy to see that its forward difference is zero. That is,
$$\D^{\deg(p)+1}p \equiv 0$$
for any polynomial $p$.
(You can frame this in a more formal induction proof if you'd like, but you can see the core of the argument here.)

Your observations are in line with this:

*

*Taking $f_1$ linear, $\D^2 f_1 \equiv 0$

*Taking $f_2$ quadratic, $\D^3 f_2 \equiv 0$

*Taking $f_3$ cubic, $\D^4 f_3 \equiv 0$
and so on.
A: A generating function also works: take $f(x) = x^k$ with $k<n$.  $$\sum_{r=0}^{n} \binom{n}{r} (-1)^r(x-r)^k = \left[\left(\frac{d}{dt}\right)^{\! k} \,  \sum_{r=0}^{n} \binom{n}{r} (-1)^r e^{t(x-r)}\right]_{t=0}=\left[\left(\frac{d}{dt}\right)^{\! k} \, e^{tx}(1-e^{-t})^n\right]_{t=0}.$$ But as $k<n$, this expression will be zero, as at least one of the factors of $(1-e^{-t})$ will survive in every term of the differentiation.
A: By the binomial theorem,
$$f(x)-4f(x-1)+6f(x-2)-4f(x-3)+f(x-4) = ((1 - T^{-1})^4f)(x),$$
where $T$ is the translation operator given by $(Tg)(x) = g(x + 1)$. In the general case, by the binomial theorem, $(1 - T^{-1})^n = \sum_{i = 0}^{n}\binom{n}{i}(-1)^iT^{-i}$, so the $n$-th thing you will write down is $(1 - T^{-1})^nf(x) = \sum_{i = 0}^{n}\binom{n}{i}(-1)^{i}f(x - i)$, and you conjecture that if $f$ is of degree at most $n - 1$, then $(1 - T^{-1})^nf = 0$. This is true because if $f(x)$ is of degree at most $k$, then $(1 - T^{-1})f(x) = f(x) - f(x - 1)$ is of degree at most $k - 1$ (to prove this, it suffices to check this for $f(x) = x^k$, which is easy by the binomial theorem).
