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It can be shown that using the definition of the Gamma function as: $$\Gamma(t) = \int_0^\infty x^{t-1} e^{-x} dx $$ that $$\Gamma(\tfrac{1}{2}) = \sqrt{\pi}$$ or slightly abusing notation, that $(-\frac{1}{2})! = \sqrt{\pi}$. Is there an intuitive explanation to this?

I want to make clear that I am not per se interested in a proof of this fact (most often these are clever technical manipulations) but in insight into this phenomenon.

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  • $\begingroup$ I do not have the answer, but I was wondering how does one define the factorial of a negative non-integer value? $\endgroup$ – Yellow Skies Aug 21 '15 at 8:26
  • $\begingroup$ @YellowSkies: wikiwand.com/en/Gamma_function $\endgroup$ – Dair Aug 21 '15 at 8:27
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    $\begingroup$ I'm definitely not sure, but this seems strongly related to this question: math.stackexchange.com/questions/3444/… $\endgroup$ – Dair Aug 21 '15 at 8:29
  • $\begingroup$ @Dair, the second answer there does seem very related to this question, although still not quite what I had in mind. $\endgroup$ – Krijn Aug 21 '15 at 8:33
  • $\begingroup$ See for example this thread. $\endgroup$ – Raymond Manzoni Aug 21 '15 at 8:44
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Consider the area of the surface of the $n-$Ball with radius $1$ . It is given by:

$$ A_{n}=2\frac{\pi^{n/2}}{\Gamma(n/2)} $$

Our intuition tells us that for $n=1$ the surface "area" (or to be mathematically more precise the, Hausdorff measure as @Michael Galuza pointed out correctly)should be 2, because it consist of two points. To make this consistent with the above formula we have to demand that

$$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$

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  • $\begingroup$ More strictly, "area" is Hausdorff measure. $\endgroup$ – Michael Galuza Aug 21 '15 at 8:41
  • $\begingroup$ @MichaelGaluza I think you are right, but forgive i'm just a physicist with nearly no knowledge about measure theory. $\endgroup$ – tired Aug 21 '15 at 8:43
  • $\begingroup$ I'm a physicist too;) I just thought that persons who like rigor can make sound about this “area” of two points $\endgroup$ – Michael Galuza Aug 21 '15 at 8:47
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    $\begingroup$ Alternatively, take $n=3$; then $A_n = 4\pi$ from which it follows that $\Gamma\left(\frac32\right) = \frac12 \sqrt\pi$. Then apply the fact that $\Gamma(x+1) = x\Gamma(x)$, which for $x=\frac12$ says that $\Gamma\left(\frac32\right) = \frac12 \Gamma\left(\frac12\right)$. $\endgroup$ – David K Aug 21 '15 at 10:10
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Is there an intuitive explanation to this ?

Yes. There is an umbilical connection between $\bigg(\dfrac1n\bigg){\large!}$ or $\Gamma\bigg(\dfrac1n\bigg)$, and geometric shapes
of the form $X^n+Y^n=R^n\iff x^n+y^n=1\iff y=\sqrt[n]{1-x^n}$, whose area is
$\displaystyle\int_0^1\sqrt[n]{1-x^n}~dx$, which is nothing else than the beta function in disguise. Ultimately, it's
all related to Newton's binomial theorem. The latter expands the power of a sum into a
sum of powers, with the help of binomial coefficients, or beta functions, which are then
expressed in terms of factorials, or $\Gamma$ functions.

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$\color{red}{\text{beta-gamma function relation}}$

Notice, we know from beta function that $$\int_{0}^{\pi/2}\sin^m(\theta) \cos^n(\theta)d\theta$$ $$=\frac{\Gamma\left(\frac{m+1}{2}\right)\Gamma\left(\frac{n+1}{2}\right)}{2\Gamma\left(\frac{m+n+2}{2}\right)}$$ Substituting $m=n=0$, we get $$\int_{0}^{\pi/2}(\sin(\theta))^{0} (\cos^n(\theta))^{0}d\theta=\frac{\Gamma\left(\frac{0+1}{2}\right)\Gamma\left(\frac{0+1}{2}\right)}{2\Gamma\left(\frac{0+0+2}{2}\right)}$$

$$\int_{0}^{\pi/2}d\theta=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(1\right)}$$

$$[\theta]_{0}^{\pi/2}=\frac{\left(\Gamma\left(\frac{1}{2}\right)\right)^2}{2\times 1}$$ $$\frac{\pi}{2}=\frac{\left(\Gamma\left(\frac{1}{2}\right)\right)^2}{2}$$ $$\left(\Gamma\left(\frac{1}{2}\right)\right)^2=\pi$$ $$\Gamma\left(\frac{1}{2}\right)=\sqrt \pi$$

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  • $\begingroup$ Could some explain what is wrong in the answer? $\endgroup$ – Harish Chandra Rajpoot Aug 21 '15 at 10:18
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    $\begingroup$ Maybe because it was asked explicitly for an intuitive explanation? (i didn't downvote) $\endgroup$ – tired Aug 21 '15 at 11:07

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