What periodic function does $\frac{\Gamma(x+1)}{2^x \Gamma(\frac{x}{2}+1)^2}$ tend to?

Question

If one graphs $$ y= \frac{\Gamma(x+1)}{2^x \Gamma(\frac{x}{2}+1)^2} = \frac{x!}{2^x \left( \left( \frac{x}{2}\right)!\right)^2} $$

One immediately notices that as $x \rightarrow -\infty$ the function tends to some periodic function that looks an awful lot like $\tan(x)$. Is there a way to make this idea precise and state what that function is? I was thinking something like, there exists a periodic function $f(x)$ such that $f(x) - y(x) \rightarrow 0$ as $x \rightarrow -\infty$.

See this image:

Which was produced using desmos.com/calculator

Answer 1

Asymptotic Behavior

Euler's Reflection Formula says $$ \Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)\tag1 $$ and the Legendre Duplication Formula says $$ \Gamma(x)\,\Gamma\!\left(x+\tfrac12\right)=\frac{2\sqrt\pi}{4^x}\Gamma(2x)\tag2 $$ Gautschi's Inequality says that for $0\lt s\lt1$, $$ x^{1-s}<\frac{\Gamma(x+1)}{\Gamma(x+s)}<(x+1)^{1-s}\tag3 $$ which can be extended to show that $$ \lim_{x\to\infty}\frac{\Gamma(x+s)}{\Gamma(x)\,x^s}=1\tag4 $$ Thus, $$ \begin{align} \frac{\Gamma(x+1)}{2^x\Gamma\left(\frac x2+1\right)^2} &=\frac1{2^x}\frac{\pi\csc(-\pi x)}{\Gamma(-x)}\frac{\Gamma\left(-\frac{x}2\right)^2}{\pi^2\csc^2\left(-\frac{\pi x}2\right)}\tag{5a}\\ &=\frac1{2\pi}\frac1{2^x}\tan\left(-\frac{\pi x}2\right)\frac{\Gamma\!\left(-\frac{x}2\right)\Gamma\!\left(-\frac{x}2+\frac12\right)}{\Gamma(-x)}\frac{\Gamma\!\left(-\frac{x}2\right)}{\Gamma\!\left(-\frac{x}2+\frac12\right)}\tag{5b}\\ &=\frac1{\sqrt\pi}\tan\left(-\frac{\pi x}2\right)\frac{\Gamma\!\left(-\frac{x}2\right)}{\Gamma\!\left(-\frac{x}2+\frac12\right)}\tag{5c}\\[3pt] &\sim\frac{\tan(-\pi x/2)}{\sqrt{-\pi x/2}}\tag{5d} \end{align} $$ Explanation:
$\text{(5a):}$ $(1)$ says $\Gamma(x+1)=\frac{\pi\csc(-\pi x)}{\Gamma(-x)}$
$\text{(5b):}$ $\frac{\csc(-\pi x)}{\csc^2\left(-\frac{\pi x}2\right)}=-\frac12\tan\left(\frac{\pi x}2\right)$
$\text{(5c):}$ $(2)$ says $\frac{\Gamma\left(-\frac{x}2\right)\Gamma\left(-\frac{x}2+\frac12\right)}{\Gamma(-x)}=\frac{2\sqrt\pi}{2^{-x}}$
$\text{(5d):}$ $(4)$ says $\frac{\Gamma\left(-\frac{x}2\right)}{\Gamma\left(-\frac{x}2+\frac12\right)}\sim\frac1{\sqrt{-x/2}}$

---

Not Really Like Tangent

$\dfrac{\tan(-\pi x/2)}{\sqrt{-\pi x/2}}$ is not periodic, though it appears to be over short spans. As $x\to-\infty$, the tangent-like shapes get flattened out.

Here is how they look near $x=-15$:

and here is how they look near $x=-1005$:

Answer 2

The Legendre duplication formula can be written as $$ \frac{\Gamma(2z)}{2^{2z-1}\Gamma(z+\tfrac12)} = \frac{\Gamma(z)}{\sqrt\pi}. $$ Applying this with $z=\frac x2+\frac12$ gives $\displaystyle \frac{\Gamma(x+1)}{2^x\Gamma(\tfrac x2+1)} = \frac{\Gamma(\tfrac x2+\tfrac12)}{\sqrt\pi} $, and so $$ \frac{\Gamma(x+1)}{2^x \Gamma(\frac{x}{2}+1)^2} = \frac1{\sqrt\pi} \frac{\Gamma(\tfrac x2+\tfrac12)}{\Gamma(\frac{x}{2}+1)}. $$ Applying the reflection formula $\Gamma(z) = \dfrac\pi{\Gamma(1-z)\sin\pi z}$ twice yields $$ \frac1{\sqrt\pi} \frac{\Gamma(\tfrac x2+\tfrac12)}{\Gamma(\frac{x}{2}+1)} = \frac1{\sqrt\pi} \frac{\Gamma(-\frac x2)\sin\pi(\frac x2+1)}{\Gamma(\frac12-\frac x2)\sin\pi(\frac x2+\frac12)} = -\frac1{\sqrt\pi} \frac{\Gamma(-\frac x2)}{\Gamma(\frac12-\frac x2)} \tan\tfrac{\pi x}2. $$ Finally, Stirling's formula implies that $\Gamma(y+\frac12)\sim \Gamma(y)\sqrt y$ as $y\to\infty$ (where $\sim$ means that the quotient of the two sides tends to $1$). Therefore $$ \frac{\Gamma(x+1)}{2^x \Gamma(\frac{x}{2}+1)^2} = -\frac1{\sqrt\pi} \frac{\Gamma(-\frac x2)}{\Gamma(\frac12-\frac x2)} \tan\tfrac{\pi x}2 \sim -\sqrt{\frac2{\pi|x|}}\tan\tfrac{\pi x}2 $$ as $x\to-\infty$. So there is a tangent function involved, but also a slight decay.

Answer 3

For negative values of $x$, $$y= \frac{\Gamma(x+1)}{2^x\, \Gamma(\frac{x}{2}+1)^2} $$ is closer and closer to $$\sin ^2\left(\frac{\pi x}{2}\right) \csc (\pi (x+1))=-\frac{1}{2} \tan \left(\frac{\pi x}{2}\right)$$ This comes from Euler's reflection formula of the gamma function.

Edit

As @GregMartin commented, I forgot the next term of the expansion. So, give all credit to him.

Using more terms, for negative values of $x$ $$y=- \sqrt{-\frac{2}{\pi x}} \tan \left(\frac{\pi x}{2}\right)\Bigg[ 1-\frac{1}{4 x}+\frac{1}{32 x^2}+O\left(\frac{1}{x^3}\right)\Bigg]$$

It could even be better if the series expansion is made a simple Padé approximant such as $$\frac {64 x^2-8 x+11 }{64 x^2+8 x+11 }+O\left(\frac{1}{x^5}\right)$$

In order to check how good (or bad) is the approximation, I computed $$I_n=\int_{-2n-0.99}^{-2n+0.99} \Bigg[\frac{\Gamma(x+1)}{2^x\, \Gamma(\frac{x}{2}+1)^2}+ \sqrt{-\frac{2}{\pi x}}\, \tan \left(\frac{\pi x}{2}\right)\,\,\frac {64 x^2-8 x+11 }{64 x^2+8 x+11 }\Bigg]^2\, dx$$

$$\left( \begin{array}{cc} n & I_n \\ 10 & 5.371 \times 10^{-16} \\ 9 & 1.766\times 10^{-15} \\ 8 & 6.733\times 10^{-15} \\ 7 & 3.105\times 10^{-14} \\ 6 & 1.846\times 10^{-13} \\ 5 & 1.567\times 10^{-12} \\ 4 & 2.271\times 10^{-11} \\ 3 & 7.961\times 10^{-10} \\ 2 & 1.562\times 10^{-07} \end{array} \right)$$

What could be interesting is to see if we can approximate the roots of $$\frac{\Gamma(x+1)}{2^x\, \Gamma(\frac{x}{2}+1)^2}=k $$