4

According to Wikipedia, factorial only is defined for non-negative integers. How come Spotlight, the Windows calculator and the Google search engine come up with $\sqrt\pi$ if you try to solve $(-0.5)!$ ?

Michael
  • 221
  • 1
  • 8

2 Answers2

2

Hint: $\Gamma(n)=(n-1)!= \int_0^\infty x^{n-1}e^{-x}\, \mathrm{d}x$ so $(-0.5)!=(0.5-1)!=\Gamma(0.5)=\int_0^\infty x^{-0.5}e^{-x}\, \mathrm{d}x=\int_0^\infty \frac{1}{\sqrt x}e^{-x}\, \mathrm{d}x$ now you should just prove the above value equals to $\sqrt \pi$. By the way, it's better to use the term "evaluate" or "calculate" instead of "solve", because we can't actually "solve" a number!

CODE
  • 4,921
  • To be honest, I wouldn't use the integral representation of the gamma function to evaluate it at $z=0.5$. – Simply Beautiful Art Oct 07 '16 at 19:49
  • @SimplyBeautifulArt I would, to be honest. It gives an intuitive look into it. When you perform the right substitutions, you get the area under the curve of e^-x². Which is essentially a scaled version of the bell curve. After that, we can take a detour to find the volume of the graph e^(-x²-y²), which is essentially a 3D bell curve. If evaluated with rectangular coordinates, you get the square of the aforementioned area, i.e. Γ(1/2)². If evaluated with polar coordinates, however, we get π. Since the area of the scaled bell curve must be positive, this proves that Γ(1/2) has to be √(π). QED – Math Machine Nov 12 '21 at 18:32
2

The ordinary factorial function can be extended, in an essentially unique way, to a function defined everywhere except at negative integers. This extension preserves the important $f(x+1) = (x+1)f(x)$ relation that characterizes the factorial. (See Extension of factorial to non-integer values for details.)

This extended function is defined at $-\frac12$, and its value there is $\sqrt\pi$.

MJD
  • 65,394
  • 39
  • 298
  • 580
  • 1
    It can be extended in infinite ways with functions that evaluate to $n!$ for natural $n$. The commonest/most useful one is the $\Gamma$ function. See also question 1537 here. – vonbrand Feb 02 '14 at 15:54
  • The $\Gamma$ function is the only such extension that is log-convex, but I thought OP might not care about that, so I left it out, since it's mentioned in the page I linked. – MJD Feb 02 '14 at 20:06
  • The log-convex property is important for the factorial because it essentially means it's the least "wavy" version of the factorial. x! is approximately x^x as x → ∞. More precisely, it's approximately x^xe^-x√(2πx). If you evaluate the graph of x!/(that asymptotic expression) as x → ∞, you'd basically just see the graph y=1. However, if you were to do the same, but using literally any other definition of the factorial, you'd see a wavy graph, periodically bouncing above and below the line y=1 ad infinitum. In other words, this definition of x! has the least residue. – Math Machine Nov 12 '21 at 18:26