# Proving for sequences that “$K$th difference is constant” implies "$K$th degree polynomial

Here's something that strikes me as clear but I couldn't prove it (I must be missing something simple):

Take a sequence $\{ a_n \}$. The first differences are the sequence $\{a^{(1)}_n\}$ where $a^{(1)}_n = a_n - a_{n-1} = a^{(0)}_n - a^{(0)}_{n-1}$, and so on, with

$$a^{(k)}_n = a^{(k-1)}_n - a^{(k-1)}_{n-1}$$

It makes sense to me that if $a^{(K)}_n$ is constant (and $a^{(k)}_n$ is nonconstant $\forall k < K$), then $a_n$ is an $K$th degree polynomial, first and foremost because of the corresponding relationship with derivatives.

But I couldn't manage to prove this (don't know where to start, really). Any hints for how to proceed (a full solution is also fine)

• What is $a_n$ a number or a function? – user251257 Mar 17 '16 at 16:56
• number, let's say reals. – MichaelChirico Mar 17 '16 at 17:02
• Then, what do you mean by $a_n$ is an $K$th degree polynomial? – user251257 Mar 17 '16 at 17:05
• it's a real-valued sequence, the closed form formula for which is a polynomial – MichaelChirico Mar 17 '16 at 17:14
• You should image it like the mean value theorem. Just because that the derivative is somewhere zero, it needs not vanish everywhere. – user251257 Mar 17 '16 at 17:18

## 1 Answer

Hint: The formulation of your question is not sound.

When considering a (let's say real-valued) sequence \begin{align*} (a_n)_{n\geq 0} \end{align*} we should keep in mind that this is a function which maps the natural numbers (i.e. the indices) to real numbers.

\begin{align*} a:\mathbb{N}\rightarrow\mathbb{R}\\ a(n)=a_n \end{align*}

On the other hand a real-valued polynomial $p(x)$ is (usually) defined from the real numbers to the real numbers. \begin{align*} p:\mathbb{R}\rightarrow\mathbb{R}\\ p(x)=\ldots \end{align*}

The statement $a_n$ is a $K$-th degree polynomial is not correct. Try to reformulate the problem and maybe this will also help to solve it.