I created a function that describes the product of the inverse multiples of a factorial $$ m(x) = \frac{1}{x}!\cdot\frac{2}{x}!\cdot\frac{3}{x}!\cdots\frac{x-1}{x}!\cdot\frac{x}{x}!$$ for some reasons i thought this function might be useful, thats why i'm posting an incomplete explanation to the best of my knowledge but after some rigorous calculation, i was able to express $m(x)$ as a formula $$ m(x) = \frac{ {x}!\cdot(2\pi)^{\frac{x-1}{2}} }{ x^x\cdot\sqrt{x} }$$ it turns out that the graph of $m(x)$ is a very simple one, makes me suggest it would be easy to interpolate and create a super factorial approximation from it $$ {x}! = x^x\cdot\sqrt{x}\cdot\sqrt{2\pi}\cdot m(x)\cdot\sqrt{{2\pi}^{-x}}$$ and it resembles striling's approximation $${x}! \approx x^x\cdot\sqrt{x}\cdot\sqrt{2\pi}\cdot e^{-x} $$

graph of mother function

$$ m(1)=1, m(2)=\frac{1}{2}!, m(3)=\frac{4\pi}{3^{5/2}}, m(4)=\frac{3\pi^{3/2}}{2^{9/2}}, m(5)=\frac{96\pi^2}{5^{9/2}}, m(6)=\frac{45\pi^{5/2}}{3^{13/2}} , m(7)=\cdots $$ now my question, can you help with the approximation, does the function gives any more information

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    $\begingroup$ Congratulations and $\to +1$. Can you explain how you have been able to find this beautiful $m(x)$ ? $\endgroup$ Feb 1 at 14:11
  • $\begingroup$ Can you explain what $\frac{1}{2}! = ?$ $\endgroup$
    – Gregory
    Feb 1 at 15:15
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    $\begingroup$ @Gregory. Use the gamma function instead $$m(x)=\prod _{i=1}^x \Gamma \left(1+\frac{i}{x}\right)$$ $\endgroup$ Feb 2 at 5:21
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    $\begingroup$ The plot of $\log(m(x))$ seems more interesting (IMHO) $\endgroup$ Feb 2 at 5:36
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    $\begingroup$ I verified the identity numerically to several decimal places. Very cool. $\endgroup$ Feb 2 at 21:03

2 Answers 2


The Gauss multiplication formula states that $$ \prod\limits_{k = 0}^{n - 1} {\Gamma\! \left( {z + \frac{k}{n}} \right)} = (2\pi )^{(n - 1)/2} n^{1/2 - nz} \Gamma (nz) $$ for $n\geq 1$ and for all complex $z$ for which both sides are defined. Taking $z=1$ and using $r!=\Gamma(1+r)$ gives $$ \prod\limits_{k = 1}^n {\left( {\frac{k}{n}} \right)!} = \prod\limits_{k = 0}^{n - 1} {\left( {\frac{k}{n}} \right)!} = \frac{{(2\pi )^{(n - 1)/2} n!}}{{n^n \sqrt n }}. $$ Thus your result is a special case of the Gauss multiplication formula. An asymptotic expansion for $m$ coming from the Stirling series is $$ m(x) \sim \left( {\frac{{\sqrt {2\pi } }}{e}} \right)^x \left( {1 + \frac{1}{{12x}} + \frac{1}{{288x^2 }} - \frac{{139}}{{51840x^3 }} - \cdots } \right) $$ as $x\to +\infty$.

  • $\begingroup$ Thanks for this sir, I didn't know I could also get $m$ directly from Gauss multiplication formula $$m(x) \approx e^{(0.5-\gamma)x}$$ But my question is about getting an asymptotic series of $m$ without meddling with Stirling series, just for the sake of getting a super factorial approximation $\endgroup$ Feb 3 at 13:42
  • $\begingroup$ Asymptotic expansions are unique. You may work out some other method but the resulting large-$x$ asymptotics will always be the same. Your $m(x)$ is more complicated than $x!$ (despite its graph may look simpler) and therefore there is no reason to analyse $x!$ through it. $\endgroup$
    – Gary
    Feb 3 at 13:47
  • $\begingroup$ okay, maybe you're right $\endgroup$ Feb 3 at 13:48

Here is the proof of how I got $m(x)$ and since we're till speaking on approximation i also want to share this to you (degrees) @claude leibovici $$ \sin{5} \approx \frac{1}{6+\sqrt{3}+\sqrt{14}}$$ The factorial function above is actually extended to the gamma function, but i am using factorial notation for simplicity $ \Gamma(x) = (x-1)!$ and i was able to write out the $m(x)$ function using help from euler reflextion formula check here $$ {x}!(-x)! = \frac{\pi x }{\sin{\pi x }}$$ I reduced each term in the factorial of $m(x)$ so that i would always get it's negative face by using ${x}!=x(x-1)!$ and ignore product of unity there $$ \begin{array} \\ m(1) = 1\\ m(2) = \frac{1}{2}!\\ m(3) = \frac{1}{3}!\cdot\frac{2}{3}!\\ m(3) = \frac{1}{3}!\cdot\frac{-1}{3}!\cdot\frac{2}{3}\\ m(4) = \frac{1}{4}!\cdot\frac{2}{4}!\cdot\frac{3}{4}!\\ m(4) = \frac{1}{4}!\cdot\frac{-1}{4}!\cdot\frac{2}{4}!\cdot\frac{3}{4}\\ m(5) = \frac{1}{5}!\cdot\frac{2}{5}!\cdot\frac{3}{5}!\cdot\frac{4}{5}!\\ m(5) = \frac{1}{5}!\cdot\frac{-1}{5}!\cdot\frac{2}{5}!\cdot\frac{-2}{5}!\cdot\frac{3}{5}\cdot\frac{4}{5}\\ m(6) = \frac{1}{6}!\cdot\frac{2}{6}!\cdot\frac{3}{6}!\cdot\frac{4}{6}!\cdot\frac{5}{6}!\\ m(6) = \frac{1}{6}!\cdot\frac{-1}{6}!\cdot\frac{2}{6}!\cdot\frac{-2}{6}!\cdot\frac{3}{6}!\cdot\frac{4}{6}\cdot\frac{5}{6}\\ m(x) = \frac{1}{x}!\cdot\frac{2}{x}!\cdot\frac{3}{x}!\cdot\frac{4}{x}\cdots\frac{x-1}{x}!\\ m(x) = \frac{1}{x}!\cdot\frac{-1}{x}!\cdot\frac{2}{x}!\cdot\frac{-2}{x}!\cdot\frac{3}{x}!\cdot\frac{-3}{x}!\cdots\frac{n}{x}!\cdot\frac{-n}{x}!\cdots\cdots\frac{x-n}{x}\cdots\frac{x-3}{x}\cdot\frac{x-2}{x}\cdot\frac{x-1}{x}\cdot f(n)\\ \end{array} $$ above, $n$ represents a function of the variable $x$ and $f(n)$ is a function of $n$ $$n= \begin{cases} \frac{x}{2}-1, &\text{if $x$ is even}\\ \frac{x+1}{2}-1, &\text{if $x$ is odd} \end{cases} $$

$$f(n)= \begin{cases} 1, &\text{if $x$ is odd}\\ \frac{1}{2}!, &\text{if $x$ is even} \end{cases} $$

$\frac{1}{2}! = m(2)$ and from the $m(x)$ formula it is directly equal to $\frac{\sqrt{\pi}}{2}$

$m(x)$ is variant, so i'll solve each part of it separately and apply some trigonometry identity to simplify $$\frac{1}{x}!\cdot\frac{-1}{x}!\cdot\frac{2}{x}!\cdot\frac{-2}{x}!\cdot\frac{3}{x}!\cdot\frac{-3}{x}!\cdots\frac{n}{x}!\cdot\frac{-n}{x}! = \frac{\frac{\pi}{x} }{\sin{\frac{\pi}{x} }}\cdot\frac{ \frac{2\pi}{x} }{\sin{\frac{2\pi}{x} }}\cdot\frac{\frac{3\pi}{x} }{\sin{\frac{3\pi}{x} }}\cdots\frac{\frac{n\pi}{x} }{\sin{\frac{n\pi}{x} }}$$ $$\frac{1}{x}!\cdot\frac{-1}{x}!\cdot\frac{2}{x}!\cdot\frac{-2}{x}!\cdot\frac{3}{x}!\cdot\frac{-3}{x}!\cdots\frac{n}{x}!\cdot\frac{-n}{x}! = \frac{ {n}!\cdot \pi^{n}}{ x^n \cdot (\sin{\frac{\pi}{x}}\cdot\sin{\frac{2\pi}{x}}\cdot\sin{\frac{3\pi}{x}}\cdots\sin{\frac{n\pi}{x}}) }$$ $$\frac{1}{x}!\cdot\frac{-1}{x}!\cdot\frac{2}{x}!\cdot\frac{-2}{x}!\cdot\frac{3}{x}!\cdot\frac{-3}{x}!\cdots\frac{n}{x}!\cdot\frac{-n}{x}! = \frac{{n}!\cdot \pi^n\cdot \sqrt{2^{x-1}}}{x^n\cdot \sqrt{x}}$$ $$\frac{x-n}{x}\cdots\frac{x-3}{x}\cdot\frac{x-2}{x}\cdot\frac{x-1}{x}\cdot f(n) = { (x-n)\cdots(x-3)(x-2)(x-1) }\cdot \frac{f(n)}{x^n}$$ $${x}! = x(x-1)(x-2)(x-3)\cdots(x-n)\cdot(x-n-1)!$$ $$(x-1)(x-2)(x-3)\cdots(x-n) = \frac{{x}!}{x(x-n-1)!}$$ $$\frac{x-n}{x}\cdots\frac{x-3}{x}\cdot\frac{x-2}{x}\cdot\frac{x-1}{x}\cdot f(n) = \frac{ {x}!f(n)}{x(x-n-1)!x^n}$$ $$m(x) = \frac{{n}!\cdot\pi^n\cdot\sqrt{2^{x-1}}\cdot{x}!\cdot f(n)}{x^{2n}\cdot\sqrt{x}\cdot x\cdot (x-n-1)!}$$ if we simplify further to remove the $n$ and $f(n)$, we would arrive at the formula i wrote in the question

  • $\begingroup$ I'm not able to follow how you simplified the product of sines, is there a reference for that trig identity? $\endgroup$ Feb 2 at 20:45
  • $\begingroup$ @JairTaylor I don't think so, I have a separate proof for it though because I worked it out by induction $\endgroup$ Feb 2 at 21:09
  • $\begingroup$ I see, can you provide a proof of that also then? (either on here or you could ask another question and answer it yourself for example) $\endgroup$ Feb 2 at 21:18
  • $\begingroup$ @JairTaylor i wrote it there above that $n$ is a function of $x$ and it depends on whether $x$ is even or odd, you can check the formula yourself and confirm it's correct..... i think it would be better if you ask the question then ill answer it $\endgroup$ Feb 2 at 21:42
  • $\begingroup$ Oh I missed that, thanks for clarifying. I may ask later $\endgroup$ Feb 2 at 22:12

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