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 $r$. $\square$
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Your email address will not be published.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: