Expanding $\frac{\Gamma(n)}{\Gamma(n-k)}$ as a polynomial

Question

I want to expand $\frac{\Gamma(n)}{\Gamma(n-k)}$ as a polynomial, where $\Gamma$ is the gamma function.

For $k\in\mathbb{N}$, it can be "simplified" as

$$\frac{\Gamma(n)}{\Gamma(n-k)}=(n-1)(n-2)(n-3)\dots(n-k)$$

I was wondering if it were possible to expand that into $\sum_{i=0}^ka_in^{k-i}$ form.

Then, there is the harder case of $k\in\mathbb R$.

I imagined that it would be of the form $\Pi_{i=0}^\infty(n-r_i)$ where $r_n$ is the $n$th root.

If this were the case, we see that

$$\Gamma(n)\ne0\implies\frac1{\Gamma(n-k)}=0$$

This occurs at the poles

$$\implies r_i-k=-i$$

$$r_i=i+k$$

And put into product form

$$\frac{\Gamma(n)}{\Gamma(n-k)}=\Pi_{i=0}^\infty(n-r_i)=\Pi_{i=0}^\infty(n-k-i)$$

Sadly, I don't think it is possible to multiple the product out because it diverges.

• Q1 Is there a polynomial/expanded form for $k\in\mathbb N$ using summations if needed?

• Q2 What about $k\in\mathbb R$?

EDIT

From below, you can see I have worked the case for $k\in\mathbb N$, but I still need $k\in\mathbb R$.

$$\frac{\Gamma(u)}{\Gamma(u-n)}=\sum_{k=0}^n(-1)^{n-k}s(n,k)(u-n)^k$$

My assumption for $n\in\mathbb R$ is that the formula becomes

$$\frac{\Gamma(u)}{\Gamma(u-n)}=\sum_{k=0}^\infty(-1)^{n-k}s(n,k)(u-n)^k$$

For similar reasons for why Euler extended binomial expansion the way he did. I note that my proposed formula holds true if $n\in\mathbb N$, but doesn't make much sense for $n\in\mathbb R$.

Answer 1

As WolframAlpha states:

$$\frac{\Gamma(x+n)}{\Gamma(x)}=x(x+1)(x+2)(x+3)\dots(x+n-1)=\sum_{k=0}^n(-1)^{n-k}s(n,k)x^k$$

where $s(n,k)$ is a Stirling number of the first kind.

Rewritable as

$$x(x+1)(x+2)(x+3)\dots(x+n-1)=(x+n-1)(x+n-2)(x+n-3)\dots x=\frac{\Gamma(x+n)}{\Gamma(x+n-n)}$$

Use substitution $x+n=u$

$$=\frac{\Gamma(u)}{\Gamma(u-n)}$$

Answer 2

Since $\frac{\Gamma(x+n)}{\Gamma(x)}=(x)_n$, the answer is simply given by the generating function for Stirling numbers of the first kind.

Answer 3

In particular, for $\displaystyle k=\frac{1}{2}$,

\begin{align*} \Gamma \left( n-\frac{1}{2} \right) &= \frac{(2n-3)!!}{2^{n-1}} \sqrt{\pi} \\ \frac{\Gamma \left( n \right)}{\Gamma \left( n-\frac{1}{2} \right)} &= \frac{(n-1)!}{(2n-3)!!} \frac{2^{n-1}}{\sqrt{\pi}} \\ &= \frac{(n-1)!(2n-2)!!}{(2n-2)!!(2n-3)!!} \frac{2^{n-1}}{\sqrt{\pi}} \\ &= \frac{2^{n-1} [(n-1)!]^{2}}{(2n-2)!} \frac{2^{n-1}}{\sqrt{\pi}} \\ &= \frac{2^{2(n-1)}}{\binom{2n-2}{n-1}\sqrt{\pi}} \\ &= \frac{2^{2(n-1)}(n-1)!}{(2n-2)(2n-3) \ldots n \sqrt{\pi}} \end{align*}

which is not a polynomial in $n$.

By $\displaystyle \Gamma \left( \frac{1}{3} \right)= 2^{7/9}3^{-1/12} \sqrt[3]{\pi K(\sin 15^{\circ})}$, we can do the something similar for $k=\frac{1}{3}$.