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I know how to calculate the value of $\left(\dfrac{1}{2}\right)!$ using the gamma function, but I don't know how to find the value of $\left(\dfrac{1}{3}\right)!$ or $\left(\dfrac{1}{6}\right)!$ using the gamma function.

V.G
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Sergio Garcia
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    The value of these factorials is also given by the Gamma function at some point. Note that they may not have a closed form , unlike $(\frac 12)!$. In fact, how did you find $(\frac 12)!$? – Sarvesh Ravichandran Iyer Aug 29 '20 at 16:58
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    The $\Gamma(4/3)$ or $\Gamma(7/6)$ is a closed form. There is no more simple form. – openspace Aug 29 '20 at 17:01
  • The general rule is that $x!=\Gamma (x+1)$. If you have a problem about evaluating this expression, you should add more details to your Question. – hardmath Aug 29 '20 at 17:04
  • @openspace It is a subjective matter what counts as a closed form. Your $\Gamma(4/3)$ is a shorthand for $$\int_0^\infty x^{1/3}e^{-x}dx$$which many wouldn't count as a closed form. – Arthur Aug 29 '20 at 17:30
  • @Arthur that's some kind of definition. So it's the same – openspace Aug 29 '20 at 17:34

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$$ \Gamma\left(\frac{1}{3}\right)=\frac{2^{7/9}\pi^{2/3}}{3^{1/12}{\rm agm}(2,\sqrt{2+\sqrt{3}})^{1/3}} $$ where $\rm agm$ is the arithmetic-geometric mean. One can then deduce that $$ \Gamma\left(\frac{1}{6}\right)=\frac{2^{14/9}3^{1/3}\pi^{5/6}}{{\rm agm}(1+\sqrt{3},\sqrt{8})^{2/3}} $$ See : Upper bound on integral: $\int_1^\infty \frac{dx}{\sqrt{x^3-1}} < 4$.

To get the values of $\Gamma\left(\frac{4}{3}\right)$ and $\Gamma\left(\frac{7}{6}\right)$, you can use the formula $$\Gamma\left(n+\frac{1}{p}\right)=\Gamma\left(\frac{1}{p}\right)\frac{(pn-p+1)!^{(p)}}{p^n}$$

Tuvasbien
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According to Wikipedia, "It has been proved that $\Gamma(n+r)$ is a transcendental number and algebraically independent of $\pi$ for any integer $n$ and each of the fractions $r=\frac16,\frac14,\frac13,\frac23,\frac34,\frac56$," so it isn't possible to do what you ask.

saulspatz
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  • Just because it's transcendental and algebraically independent of $\pi$, that doesn't really prove anything. After all, the same is strongly suspected to hold for $e$, for instance. – Arthur Aug 29 '20 at 17:36
  • @Arthur Sure. I meant "It isn't possible in the current state of knowledge." – saulspatz Aug 29 '20 at 17:38
  • That's not as impossible to answer, but still a difficult thing for a single person to claim to know. Without knowing who you are, I still sincerely doubt you've read all the relevant literature and attended all relevant conferences to conclude that we currently do not know. – Arthur Aug 29 '20 at 17:43
  • @You're being awfully literal. Anything else I say will be impolite. – saulspatz Aug 29 '20 at 17:45
  • I'm being a bit pedantic here, yes, but it's rarely wrong to slightly moderate statements of the "we do not know" variety. – Arthur Aug 29 '20 at 17:46