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I am interested in knowing the value of the fraction $y = \frac{\Gamma(\frac{x+1}{2})}{\Gamma(\frac{x}{2})}$ for different non-negative values of $x$.

Plotting $y$ suggests that the value of the fraction follows a power function of the form $ax^p$.

enter image description here

Using power regression in Excel, I find $a=0.6445$ and $p=0.5179$. Although, the $R^2$ value of $0.9987$ is quite high, I am wondering:

If it exists, what is the closed-form expression to describe the relationship between $y = \frac{\Gamma(\frac{x+1}{2})}{\Gamma(\frac{x}{2})}$ and $x$?

Although I am primarily interested in the particular fraction presented in this question, more general answers are also welcome.

Blue
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Gamal
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  • @metamorphy Thank you for your comment. If you can turn your comment into an answer, I will be happy to accept it. :) – Gamal Sep 14 '20 at 12:08

2 Answers2

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There's no known closed form for this. Any alternative expression still involves two gammas; say, using the duplication formula, you can rewrite it in terms of $\Gamma(x)$ and $\Gamma(x/2)$, but it doesn't really help. The "power law" holds only in the asymptotic sense, because of $\lim\limits_{x\to+\infty}\Gamma(x+a)/[x^a\Gamma(x)]=1$ for any real $a$ (in our case $a=1/2$).

metamorphy
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That fraction has an exact meaning in terms of Rising and Falling Factorials. $$ \eqalign{ & {{\Gamma \left( {x/2 + 1/2} \right)} \over {\Gamma \left( {x/2} \right)}} = \left( {x/2} \right)^{\,\overline {\,1/2\;} } = \cr & = \left( {x/2 - 1/2} \right)^{\,\underline {\,1/2\,} } = \Gamma \left( {3/2} \right)\left( \matrix{ x/2 - 1/2 \cr 1/2 \cr} \right) = \cr & = {{\sqrt \pi } \over 2}\left( \matrix{ x/2 - 1/2 \cr 1/2 \cr} \right) \cr} $$

For non-negative integral exponents, the Rising and Falling factorials would be polynomials in $x$ of corresponding degree.
Here, being the exponent $1/2$, it will be asymtotic to $(x/2)^{1/2}$.

G Cab
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