Lemma 10.58.5. Suppose that $f: n \mapsto f(n) \in A$ is defined for all $n$ sufficiently large and suppose that $n \mapsto f(n) - f(n-1)$ is a numerical polynomial. Then $f$ is a numerical polynomial.
Proof. Let $f(n) - f(n-1) = \sum \nolimits _{i = 0}^ r \binom {n}{i} a_ i$ for all $n \gg 0$. Set $g(n) = f(n) - \sum \nolimits _{i = 0}^ r \binom {n + 1}{i + 1} a_ i$. Then $g(n) - g(n-1) = 0$ for all $n \gg 0$. Hence $g$ is eventually constant, say equal to $a_{-1}$. We leave it to the reader to show that $a_{-1} + \sum \nolimits _{i = 0}^ r \binom {n + 1}{i + 1} a_ i$ has the required shape (see remark above the lemma). $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: