I have been thinking about a specific problem for quite some time. Imagine we have the following product, where $x \in \mathbb{N}$ and $a \in \mathbb{Z^+}$ then we now that the following holds: \begin{align} \prod_{j=k}^{n-1} (x +a +j) = \frac{(x+a+n-1)!}{(x+a+k-1)!} = \frac{\Gamma(x+a+n)}{\Gamma(x+a+k)}. \end{align} What if this $a$ were not to be in $\mathbb{Z}^+$, but rather in $\mathbb{R}^+$, could we then still have a similar expression in terms of the Gamma function?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Yes, $\Gamma(z+1)=z\Gamma(z)$ holds even for complex numbers. $\endgroup$– user65203Jul 10, 2019 at 14:36
-
$\begingroup$ You can group $x+a$ in a single variable, for clarity. $\endgroup$– user65203Jul 10, 2019 at 14:38
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Absolutely. This type of product is common enough in mathematics to have its own notation, the Pochammer symbol, defined as $$ (a)_n \equiv \prod_{j=0}^{n-1}(a+j). $$ From the Gamma recursion identity $\Gamma(z+1) = z\Gamma(z)$, we can write the symbol in terms of the Gamma function: \begin{multline} \frac{\Gamma(a+n)}{\Gamma(a)} = (a+n-1)\frac{\Gamma(a+n-1)}{\Gamma(a)} = (a+n-1)(a+n-2)\frac{\Gamma(a+n-2)}{\Gamma(a)}=\,\,... \\ = (a+n-1)(a+n-2)...(a+1)a\frac{\Gamma(a)}{\Gamma(a)} = \prod_{j=0}^{n-1}(a+j) = (a)_n. \end{multline} Since your product can be written as the Pochhammer symbol $(x+a+k)_{n-k}$, we see that indeed, $$ \prod_{j=k}^{n-1}(x+a+j) = (x+a+k)_{n-k} = \frac{\Gamma(x+a+n)}{\Gamma(x+a+k)}. $$
