User:Virginia-American/Sandbox/Allen

$$\mbox{Lemma 4:}\ $$

$$\mbox{Let }P(x) = a_0 x^{\mu} + a_1 x^{\mu-1} + \dots + a_{\mu-1}x + a_{\mu}\mbox{ be a polynomial of degree }\mu > 0$$$$\mbox{ where } a_0\ne0\mbox{ and } a_\mu\ne0.\;\;$$ $$\;\;\mbox{Let }\;$$ $$(\xi_\mu)\; $$ $$\mbox{ be the sequence }\;$$ $$\mbox {consisting of the negatives } \;$$ $$\mbox {of the roots of }P(x).\;$$

$$\mbox{ Then for a nonnegative integer }\; n,\; $$ $$\;\;P(n)=a_\mu\frac{((1+\xi_\mu))_n}{((\xi_\mu))_n}.$$

$$\mbox{Proof:}\;$$

$$\mbox{From the factorization }\;$$ $$P(x)=a_0(x+\xi_1)(x+\xi_2)\dots(x+\xi_\mu)\;$$ $$\mbox{we see that }\;$$
 * $$P(0)=a_\mu=a_0\xi_1\xi_2\dots\xi_\mu.\;$$

$$\mbox{ Now}\;$$

\begin{align} n+\xi &=\frac{\Gamma(\xi+n+1)}{\Gamma(\xi+n)} \\ &=\frac{\Gamma(1+\xi+n)}{\Gamma(1+\xi)} \frac{\Gamma(\xi+1)}{\Gamma(\xi+n)} \\ &=\frac{\Gamma(1+\xi+n)}{\Gamma(1+\xi)} \xi\frac{\Gamma(\xi)}{\Gamma(\xi+n)} \\ &=\xi \frac{(1+\xi)_n}{(\xi)_n}. \end{align} $$

$$\mbox{ Therefore, }\;$$



\begin{align} P(n) &=a_0(n+\xi_1)(n+\xi_2)\dots(n+\xi_\mu)\\

&=a_0\xi_1 \frac{(1+\xi_1)_n}{(\xi_1)_n}  \xi_2\frac{(1+\xi_2)_n}{(\xi_2)_n}\dots\xi_\mu  \frac{(1+\xi_\mu)_n}{(\xi_\mu)_n}\\ &=a_\mu\frac{((1+\xi_\mu))_n}{((\xi_\mu))_n}. \end{align} $$

$$\mbox{ }\;$$ $$\mbox{ }\;$$ $$\mbox{ }\;$$ $$\mbox{ }\;$$ $$\mbox{ }\;$$