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What is the value of the product of Gamma functions \begin{align} \prod_{k=1}^{10} \Gamma\left(\frac{k}{10}\right) \end{align} and can it be shown that \begin{align} \prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) \approx \frac{\pi^{9}}{54} \end{align} and \begin{align} \prod_{k=1}^{40} \Gamma\left(\frac{k}{10}\right) \approx \left( 6 + \frac{625}{4501}\right) \pi^{18}. \end{align}

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Check the multiplication formula: en.wikipedia.org/wiki/… –  Yves Daoust Jun 6 at 21:12

2 Answers 2

up vote 10 down vote accepted

Original Question

As shown at the end of this answer, $$ \Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)\tag{1} $$ As shown at the beginning of this answer, $$ \prod_{k=1}^{n-1}\sin(k\pi/n)=\frac{n}{2^{n-1}}\tag{2} $$ Using $(1)$ and $(2)$, we get $$ \begin{align} \left[\prod_{k=1}^{10}\Gamma\left(\frac{k}{10}\right)\right]^2 &=\prod_{k=1}^9\Gamma\left(\frac{k}{10}\right)\Gamma\left(1-\frac{k}{10}\right)\\ &=\prod_{k=1}^9\pi\csc(k\pi/10)\\ &=\frac{2^9\pi^9}{10}\tag{3} \end{align} $$ Thus, $$ \prod_{k=1}^{10}\Gamma\left(\frac k{10}\right)=16\pi^4\sqrt{\frac\pi5}\tag{4} $$


Gauss's Multiplication Formula

Define $$ f(x)=\prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right)\tag{5} $$ then $f$ is log-convex and $$ \begin{align} x\,f(x) &=x\,\prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right)\\ &=x\,\Gamma(x)\,\prod_{k=1}^{n-1}\Gamma\left(x+\frac kn\right)\\ &=\Gamma(x+1)\,\prod_{k=0}^{n-2}\Gamma\left(x+\frac{k+1}n\right)\\ &=\prod_{k=1}^{n-1}\Gamma\left(x+\frac1n+\frac kn\right)\\[6pt] &=f\left(x+\frac1n\right)\tag{6} \end{align} $$ Plugging $\frac xn$ into $(6)$ gives $$ \frac xnf\left(\frac xn\right)=f\left(\frac{x+1}n\right)\tag{7} $$ $(7)$ and log-convexity implies that $$ f\left(\frac xn\right)=C_n\frac{\Gamma(x)}{n^x}\tag{8} $$ Using $(1)$ and $(2)$ yield $$ \begin{align} f\left(\frac1n\right)^2 &=\prod_{k=1}^{n-1}\Gamma\left(\frac kn\right)\Gamma\left(1-\frac kn\right)\\ &=\prod_{k=1}^{n-1}\pi\csc(k\pi/n)\\ &=\frac1n2^{n-1}\pi^{n-1}\tag{9} \end{align} $$ $(8)$ and $(9)$ yield $$ C_n=\sqrt{n2^{n-1}\pi^{n-1}}\tag{10} $$ Therefore, $(8)$ and $(10)$ give Gauss's Multiplication Formula $$ \prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right) =\sqrt{n2^{n-1}\pi^{n-1}}\frac{\Gamma(nx)}{n^{nx}}\tag{11} $$


Second Question

Using $(11)$, $$ \begin{align} \prod_{k=1}^{20}\Gamma\left(\frac k{10}\right) &=\prod_{k=0}^{9}\Gamma\left(\frac1{10}+\frac k{10}\right)\prod_{k=0}^{9}\Gamma\left(\frac{11}{10}+\frac k{10}\right)\\ &=5120\pi^9\frac{\Gamma(1)}{10^1}\frac{\Gamma(11)}{10^{11}}\\ &=\frac{\pi^9\,9!}{10\cdot5^9}\tag{12} \end{align} $$


Third Question

Using $(11)$, $$ \begin{align} &\prod_{k=1}^{40}\Gamma\left(\frac k{10}\right)\\ &=\small\prod_{k=0}^{9}\Gamma\left(\frac1{10}+\frac k{10}\right)\prod_{k=0}^{9}\Gamma\left(\frac{11}{10}+\frac k{10}\right)\prod_{k=0}^{9}\Gamma\left(\frac{21}{10}+\frac k{10}\right)\prod_{k=0}^{9}\Gamma\left(\frac{31}{10}+\frac k{10}\right)\\ &=\left(5120\pi^9\right)^2\frac{\Gamma(1)}{10^1}\frac{\Gamma(11)}{10^{11}}\frac{\Gamma(21)}{10^{21}}\frac{\Gamma(31)}{10^{31}}\\ &=\frac{2^{18}\pi^{18}}{10^{62}}10!\,20!\,30!\tag{13} \end{align} $$

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Nice answers. +1 I knew of 3 proofs of the gamma multiplication formula, but that wasn't one of them. –  Random Variable Jun 7 at 1:39
    
Nice solution. I added some notes about the approximations in a follow-up solution. –  Leucippus Jun 7 at 2:02

Since @robjohn has done an excellent job at providing an answer in detail I will add my comments as an additional solution. The results presented here follow the numbering in robjohn's work.

In robjohn's equation (12) the factor \begin{align} \frac{9! \ \pi^{9}}{2 \cdot 5^{10}} \end{align} has been provided. Numerically this can be seen as $.018579456\cdots \ \pi^{9}$. By comparing this to that of $1/54 = .0185185\cdots$ one can make a good approximation by \begin{align} \prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) = \frac{9! \ \pi^{9}}{2 \cdot 5^{10}} \approx \frac{\pi^{9}}{54}. \tag{14} \end{align} Another possible value is $20/1077 = .0185701021\cdots$ which leads to the statement \begin{align} \prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) = \frac{9! \ \pi^{9}}{2 \cdot 5^{10}} \approx \frac{20 \ \pi^{9}}{1077}. \tag{15} \end{align}

From equation (13) the factor is \begin{align} \frac{2^{18} \ (10)!(20)!(30)!}{10^{62}} = 6.138858832\cdots = 6 + .138858832\cdots . \end{align} Since $625/4501 = .1388580315\cdots$ then to a fair approximation it can be stated \begin{align} \prod_{k=1}^{40} \Gamma\left(\frac{k}{10}\right) \approx \left( 6 + \frac{625}{4501} \right) \pi^{18}. \tag{16} \end{align}

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What was the reasoning behind the rational factors of $\pi^{9n/20}$? Why did you think that the answer was a rational multiple of $\pi^{9n/20}$? –  robjohn Jun 7 at 12:58
    
@robjohn This problem is similar to problem I, of the same name, where most viewers missed the point. Several factors went into the design of this series of questions. Those being that the sine values be rationals, or have nice products, be "quick" to calculate, if knowing a few ideas, the products not be quickly found on a website, ie wiki or mathworld, and finally the problem, or a related one, be considered for a problem column of a journal, and also be a teaching experience. In aspect to reduction of the large factors to a nice fraction was just a side thought in the design –  Leucippus Jun 7 at 14:09

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