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 $r=1,2,3,4,5... \infty$.

Solution 1:

Yes, it is true. You can prove it by induction.

Assume $\Delta^n (x^n)=n!h^n$ and $\Delta^m (x^n) = 0$, if $m> n$ hold.

Note that $$ \Delta^{n+1}(x^{n+1}) = \Delta^n((x+h)^{n+1}) - \Delta^n(x^{n+1}) $$

and apply the assumptions to simplify the right-hand side. This completes induction.

Solution 2:

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 $r=1,2,..$.