let be the function

$$ \frac{\Gamma (m+1)}{\Gamma(m-2r+1)} $$

where m and r are integers...

then my question is if for $ m<0 $ but $ r$ always positive integer the function above or its derivatives turn singular

i think that $$ \frac{\Gamma (m+1)}{\Gamma(m-2r+1)} $$ could be realted to POCHAMMER polynomials if m and r are integers (no matter positive or negative) but i am not sure

for m and r non-integers i have no problem but for m integer and r positive integers i think this function is related to some POchammer polynomials


  • $\begingroup$ If $r$ is a positive integer, then $$\frac{\Gamma(z+1)}{\Gamma(z-2r+1)} = z(z-1)\dotsc (z-2r+1)$$ is a polynomial. I guess it's a Pochhammer polynomial, since that is also written $z^{\underline{2r}}$ or $(z)_{2r}$ as Pochhammer symbols. $\endgroup$ – Daniel Fischer Nov 6 '14 at 12:31
  • $\begingroup$ OK tahnks i need it for any formula involivng Euler-Maclaurin summation formula thanks Daniel :) $\endgroup$ – Jose Garcia Nov 6 '14 at 12:34

I suspect your answer may come from analysiing the hypergeometric function; namely (for $ \lvert z \rvert = 1$) \begin{equation} \sideset{_2}{_1}F(q, b;c;z) = \sum_{n=0}^{\infty}\frac{q_{(n)}b_{(n)}}{c_{(n)}}\frac{z^{n}}{n!} \end{equation} Which is undefined for $c$ a non-positive integer. Here, $q$ is your Pochammer symbol; which for $n>0$ is defined as \begin{equation} q_{(n)} = q(q+1)(q+2) \cdots (q+n-1) \end{equation} and is equal to $1$ when $n=0$. Your probelm will relate your Pochammer symbol to these rational Gamma functions may be a study in power series solutions to the Hypergeometric Equation.

EDIT: In response to the comment below by Jose; then why not consider this further comment.

If $q$ is the Pochhammer symbol, then it's derivative can be composed as: \begin{equation} \frac{d}{dx} q_{(n)} = q_{(n)}\left(\Psi_{0}(n+x) - \Psi_{0}(x)\right) \end{equation} Where $\Psi_{0}$ is the so-called digamma function; \begin{eqnarray} \Psi (z) & \equiv & \frac{d}{dz} \ln \Gamma(z) \\ & = & \frac{\Gamma'(z)}{\Gamma(z)} \end{eqnarray}

  • $\begingroup$ my probelm was most related to see if the derivatives of $$ \frac{\Gamma (z+1)}{\Gamma (z-2r+1)} $$ for m negative and r postive turned to be $\infty$ $\endgroup$ – Jose Garcia Nov 6 '14 at 13:30

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