Is there any way I can calculate this ratio more quickly than evaluating the (potentially very large) numerator and denominator?

$$ \frac{\Gamma(z+\alpha)}{\Gamma(z)} $$

I'm specifically interested in the case where $0 < \alpha < 1$

I derived a formula for the special case where $\alpha = \frac{1}{2}$:

$$ \Gamma(n + \frac{1}{2}) = \binom{n - \frac{1}{2}}{n}n!\sqrt{\pi} $$ $$ \frac{\Gamma(n + \frac{1}{2})}{\Gamma(n + 1)} = \binom{n - \frac{1}{2}}{n}\sqrt{\pi} $$ $$ \frac{\Gamma(n + 1)}{\Gamma(n + \frac{1}{2})} = \frac{1}{\binom{n - \frac{1}{2}}{n}\sqrt{\pi}} $$ $$ \frac{\Gamma(z + \frac{1}{2})}{\Gamma(z)} = \frac{1}{\binom{z - 1}{z - \frac{1}{2}}\sqrt{\pi}} $$

and the obvious case where $\alpha = 1$:

$$ \frac{\Gamma(z+1)}{\Gamma(z)} = z $$

Is there a more general formula that encompasses any $\alpha$?

  • $\begingroup$ a way to go is defining $G(z,\alpha ):=\frac{\Gamma (z)}{\Gamma (z+\alpha )}$, then you want to estimate such function $G$ for $|\alpha |<1$ and some fixed $z$. For this task I will try to search about some of the many representations of the $\Gamma $ function that simplifies/makes efficient enough this task $\endgroup$ – Masacroso Oct 29 '19 at 16:08

There is no general form, since that would imply $\Gamma$ had a "closed form" for any values. It is possible to approximate the ratio as:

$$\frac{\Gamma(z+\alpha)}{\Gamma(z)}\underset{z\to\infty}\sim z^\alpha$$

which holds regardless of $\alpha$ and may be a direct consequence of the definition of $\Gamma$ depending on how you choose to define it.

Of course better asymptotics may be found from Stirling approximations.


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