How to find the factorial of a fraction?

Question

From what I know, the factorial function is defined as follows:

$$n! = n(n-1)(n-2) \cdots(3)(2)(1)$$

And $0! = 1$. However, this page seems to be saying that you can take the factorial of a fraction, like, for instance, $\frac{1}{2}!$, which they claim is equal to $\frac{1}{2}\sqrt\pi$ due to something called the gamma function. Moreover, they start getting the factorial of negative numbers, like $-\frac{1}{2}! = \sqrt{\pi}$

How is this possible? What is the definition of the factorial of a fraction? What about negative numbers?

I tried researching it on Wikipedia and such, but there doesn't seem to be a clear-cut answer.

Answer 1

The gamma function is defined by the following integral, which converges for real $s>0$: $$\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}dt.$$

The function can also be extended into the complex plane, if you're familiar with that subject. I'll assume not and just let $s$ be real.

This function is like the factorial in the when $s$ is a positive integer, say $s=n$, it satisfies $\Gamma(n)=(n-1)!$. It generalizes the factorial in the sense that it is the factorial for positive integer arguments, and is also well-defined for positive rational (and even real) numbers. This is what it means to take a "rational factorial," but I would hesitate to call it that. Many functions have those two properties, and $\Gamma$ is chosen out of all of them because it is the most useful in other applications. Rather than the notation used in that article you refer to, it would be more accurate for you to say that "the gamma function takes these values for these arguments." Gamma is not a function that intends to generalize factorials; rather, generalizing factorials came along as something of an accident following the definition. Its true purpose is deeper.

As for why $\Gamma(1/2)=\sqrt{\pi}$, this comes out of an interesting property of the $\Gamma$ function: some of them are here http://en.wikipedia.org/wiki/Gamma_function#Properties. The property you are interested in is the reflection formula: $$\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}.$$ Set $z=1/2$ in the formula to get the desired identity.

If you want to learn more about the gamma function, the hard way is to learn a lot more math, in particular real and complex analysis. An easier way is to read this excellent set of notes: http://www.sosmath.com/calculus/improper/gamma/gamma.html.

Answer 2

The gamma function, shown with a Greek capital gamma $\Gamma$, is a function that extends the factorial function to all real numbers, except to the negative integers and zero, for which it is not defined. $\Gamma(x)$ is related to the factorial in that it is equal to $(x-1)!$. The function is defined as

$$\Gamma(z) = \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}$$

Simply use this to compute factorials for any number. A handy way of calculating for real fractions with even denominators is:

$$\Gamma(\tfrac12 + n) = {(2n)! \over 4^n n!} \sqrt{\pi}$$

Where n is an integer. But keep in mind that the gamma function is actually the factorial of 1 less than the number than it evaluates, so if you want $\frac{3}{2}!$ use n = 2 instead of 1.

Or, you could just put the fraction into Google Calculator, which uses the gamma function to evaluate factorials of any number.

For some more examples of the gamma function's values, see here.

(If you don't understand this, don't worry, because I don't either, and the Wikipedia article on the function seems to lack a clear-cut definition of it or how it relates to $\sqrt{\pi}$.)

Answer 3

try $$(n+t)! = n!+(n+1! \times t)$$ where t is any real number between 0 and 1

Answer 4

Others have mentioned that the answers on the page you linked come from the Gamma function, which has the property that Γ(n)=(n-1)! for all nonnegative integers n. It stands to reason, then, that to find the value of x! for any x, you can use Γ(x+1).

But you also asked specifically about negative numbers, and there's an important caveat there. This definition produces well-defined values for all real and even complex numbers except negative integers.

This is easily seen by simply extending the definition of factorial backward, no gamma function required. The problem is, you wind up dividing by 0:

Given 3! = 6.

Then:
   2! = 3!/3 = 2
   1! = 2!/2 = 1
   0! = 1!/1 = 1
  (-1)! = 0!/0 = oops

You can see that this sequence also invalidates any other negative integer factorials, since (-2)! would be equal to (-1)!/-1 and (-1)! is undefined. Since (-2)! is undefined, so is (-3)!, and so on.

The Gamma function happily produces values for negative reals, but if you plot it, you'll see that it zooms out to infinity at the negative integers. But even if we were willing to extend the range of our function to allow infinity as a value, as silly as that is, Γ(-n) would still be undefined, because they aren't even well-behaved infinities! For every integer n, the limit of Γ(x) as x → -n can be either +∞ or -∞, depending on which direction you approach n from. Specifically, the limit is positive above even integers and below odd ones, negative above odd and below even.

For instance, Γ(-1.9) is about 5; Γ(-1.99) is about 50; Γ(-1.999) is about 500; and so on. The closer you get to x=-2 coming from the positive direction, the greater Γ(x) gets, increasing without bound, implying that Γ(-2) = (-3)! is +∞. But on the other side of x=-2, Γ(-2.1) is about -5, Γ(-2.01) is about -50, and Γ(-2.001) is about -500; the closer you get to x=-2 coming from the negative side, the lower Γ(x) gets, decreasing without bound, implying that Γ(-2) = (-3)! = -∞.