<s>[INST] Suppose $f(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n-1}x+a_n$. Then $\Delta^nf(x)=n!h^n$ and $\Delta^{n+r}f(x)=0,$ for $r=1,2,3,4,5... \infty$. Suppose $f(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n-1}x+a_n$. Then $\Delta^nf(x)=n!h^n$ and $\Delta^{n+r}f(x)=0,$ for $r=1,2,3,4,5... \infty$.(Where we have equally spaced points $x,x+h,...,x+nh$ with corresponding values $f(x),f(x+h),...f(x+nh)$. Forward difference operator is defined as $\Delta f(x)=f(x+h)-f(x)$)
My attempt:-
When $f(x)=x,f(x)=x^2+a_1x+a_2$ , I could prove, How do I prove generally?
Is the argument true?
$\Delta f(x)=f(x+h)-f(x)=(x+h)^n+a_1(x+h)^{n-1}+a_2(x+h)^{n-2}+...+a_{n-1}(x+h)+a_n-(x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n-1}x+a_n).$ By binomial expansion, we get a polynomial of degree $n-1$. Following the step n times, we landed upon $\Delta^nf(x)=$constant. But I am not able to prove it is $\Delta^nf(x)=n!h^n$.
 [/INST] Let $e_n(x) = x^n$. We can write a polynomial $p(x) = \sum_{k=0}^n a_k x^k$ as $p = \sum_{k=0}^n a_k e_k$
Note that $\Delta^r$ is a linear operator, so
$\Delta^r p = \sum_{k=0}^n a_k \Delta^r e_k$, in particular, we
need only focus on computing $\Delta^n e_m$ for $m =n,n+1,...$.
Note that $(\Delta^1 e_1 )(x) = h$ and $\Delta e_0 = 0$ (that is $\Delta $ applied to constants results in zero).
So, suppose that for $k=1,...,n$ that $(\Delta^n e_n)(x) = n! h^n$
and $\Delta^n e_r = 0$ for $r=1,...,n-1$.
Suppose $n \ge 1$, then
\begin{eqnarray}
(\Delta e_{n+1}) (x) &=& \sum_{k=1}^{n+1} \binom{n+1}{k} x^{n+1-k} h^k \\
&=&  \sum_{k=1}^{n+1} \binom{n+1}{k} e_{n+1-k} (x) h^k \\
&=&  (n+1)h e_{n}(x)+\sum_{k=2}^{n+1} \binom{n+1}{k} e_{n+1-k} (x) h^k
\end{eqnarray}
Then $\Delta^{n+1} e_{n+1} = \Delta^{n} (\Delta e_{n+1}) = (n+1)\Delta^{n} e_n = (n+1)! h^{n+1}$.
Note that since $\Delta^n e_n$ is a constant, it follows that $\Delta^{n+r} e_n = \Delta^r \Delta^{n} e_n =0$ for $r=1,2,..$.
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