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}\neq 0{\mbox{ and }}a_{\mu }\neq 0.\;\;$   $\;\;{\mbox{Let }}\;$ $(\xi _{\mu })\;$ ${\mbox{ be the sequence }}\;$ ${\mbox{consisting of the negatives }}\;$ ${\mbox{of the roots of }}P(x).\;$

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

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

${\mbox{From the factorization }}\;$ $P(x)=a_{0}(x+\xi _{1})(x+\xi _{2})\dots (x+\xi _{\mu })\;$ ${\mbox{we see that }}\;$

$P(0)=a_{\mu }=a_{0}\xi _{1}\xi _{2}\dots \xi _{\mu }.\;$

${\mbox{ Now}}\;$

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

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

${\begin{aligned}P(n)&=a_{0}(n+\xi _{1})(n+\xi _{2})\dots (n+\xi _{\mu })\\&=a_{0}\xi _{1}{\frac {(1+\xi _{1})_{n}}{(\xi _{1})_{n}}}\xi _{2}{\frac {(1+\xi _{2})_{n}}{(\xi _{2})_{n}}}\dots \xi _{\mu }{\frac {(1+\xi _{\mu })_{n}}{(\xi _{\mu })_{n}}}\\&=a_{\mu }{\frac {((1+\xi _{\mu }))_{n}}{((\xi _{\mu }))_{n}}}.\end{aligned}}$

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