The Stacks project

Lemma 43.15.4. Let $P$ be a polynomial of degree $r$ with leading coefficient $a$. Then

\[ r! a = \sum \nolimits _{i = 0, \ldots , r} (-1)^ i{r \choose i} P(t - i) \]

for any $t$.

Proof. Let us write $\Delta $ the operator which to a polynomial $P$ associates the polynomial $\Delta (P) = P(t) - P(t - 1)$. We claim that

\[ \Delta ^ r(P) = \sum \nolimits _{i = 0, \ldots , r} (-1)^ i {r \choose i} P(t - i) \]

This is true for $r = 0, 1$ by inspection. Assume it is true for $r$. Then we compute

\begin{align*} \Delta ^{r + 1}(P) & = \sum \nolimits _{i = 0, \ldots , r} (-1)^ i {r \choose i} \Delta (P)(t - i) \\ & = \sum \nolimits _{n = -r, \ldots , 0} (-1)^ i {r \choose i} (P(t - i) - P(t - i - 1)) \end{align*}

Thus the claim follows from the equality

\[ {r + 1 \choose i} = {r \choose i} + {r \choose i - 1} \]

The lemma follows from the fact that $\Delta (P)$ is of degree $r - 1$ with leading coefficient $ra$ if the degree of $P$ is $r$. $\square$

Comments (0)

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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AZY. Beware of the difference between the letter 'O' and the digit '0'.