1
$\begingroup$

I had always understood that the factorial n! was defined as

$$\prod_{k=1}^nk$$

However, this leaves several questions: Why does 0! exist?* By extension, why can't you take the factorial of a negative number? And why can you take the factorial of a decimal? Finally, what about imaginary numbers? Is it possible to take the factorial of an imaginary or complex number?

*Related is why it equals 1; my question isn't about the empty product, but rather why the formula allows for 0! to exist. (That is: I understand that $0!=\frac{1!}1$, but given the above formula, why does this not contradict itself?)

$\endgroup$
  • $\begingroup$ There is Gamma function that generalizes this concept $\endgroup$ – Navaro Jun 24 '16 at 3:10
  • 4
    $\begingroup$ $(n-1)!=\frac{n!}{n}$. Setting $n=1$ gives a value for $0!$ as $\frac{1!}{1}$ with no ambiguity. $\endgroup$ – JMoravitz Jun 24 '16 at 4:21
  • $\begingroup$ I rationalize 0! = 1 because 3! resembles the number of ways to order three items where there is only 1 way to order 0 items (0!), and that is to not order them at all. Similarly this doesn't make sense when you have negative numbers, decimals, or imaginary numbers. While 0! may be hard to visualize from the definition, it's proven otherwise and serves as a foundation to the definition. $\endgroup$ – Tommyixi Jun 24 '16 at 19:07
2
$\begingroup$

Why does $0!$ exist?

The ultimate reason that $0!$ is allowed to exist is because mathematicians define it to exist. We simply state $0!=1$ and continue from there. There are reasons why we do this, which others have expounded upon elsewhere, but ultimately there is nothing stopping us from defining whatever we like. You only see the useful definitions because those are the only ones that survive.

If you insist, however, the formula you are using "allows for a value to exist", which I interpret to mean "makes mathematical sense", if we interpret it in a certain way. I assume the main issue you are having is that, as far as you are concerned, the number on the top of the $\prod$ symbol should be greater than or equal to the number on the bottom. We can get around this as follows.

For all $n_{1},n_{2}\in\mathbb{Z}$, write $$I[n]=\{k\in\mathbb{Z}:1\leq k\leq n\}.$$ For $n<1$ this definition gives $I[n]=\varnothing$. We may now reasonably define the notation you have used, for all $n\in\mathbb{Z}$, by $$\prod_{k=1}^{n}k := \prod_{k\in I[1,n]}k.$$ In particular, if $n=0$ then there are no terms in the product, that is, we are computing $$\prod_{k\in\varnothing}k.$$ By convention we take the value of this to be $1,$ but that's not my point. My point is merely to demonstrate that the notation can be interpreted in a logical way.

Why can't you take the factorial of a negative number?

You can, in a sense, take factorials of negative numbers, but not of negative integers. One way to look at this is to notice that, using the definitions as above, we would have $$\prod_{k=1}^{-1}k = \prod_{k\in\varnothing}k = \prod_{k=1}^{0}k.$$ This would give $0!=(-1)!$, and the same for all the other negative integers. This is clearly very boring and should therefore be avoided unless very convenient, but it turns out there are better reasons to avoid this sort of thing. The best reasons involve the Gamma function $\Gamma$, but here's a naive reason why $(-1)!$ should be left undefined.

We give an alternative (in my opinion, better) definition of "factorial". First, we define $0!=1$ (or $1!=1$ if you prefer). Then, for each $n\in\mathbb{N}$ we define $n!$ inductively by $n!=n\cdot (n-1)!$. Clearly this definition agrees with your definition so far. Now, we would like to extend this definition so that we can give a value to $(-1)!$ and still keep this formula. But we would then have $$1 = 0! = 0\cdot(-1)! = 0,$$ a contradiction. So we cannot assign a value to $(-1)!$ and keep this (very useful) formula. In a sense, we can't have our cake and eat it too.

Taking factorials of other numbers

So we can't take factorials of negative integers. If you know enough analysis, you can look into the Gamma function, which, taken to its furthest, extends the idea of "factorial" to any complex number that is not a negative integer. The key points about the Gamma function, the facts that make it special, are as follows:

  • that it satisfies the very useful functional equation $\Gamma(z)=(z-1)\cdot\Gamma(z-1)$, a slight variant on the one mentioned above, which gives $\Gamma(n)=(n-1)!$ for $n\in\mathbb{N}$;
  • it's quite "smooth", in a slightly technical sense (except at the nonpositive integers, which correspond to negative integer factorials);
  • it starts at "the right place" (it satisfies the "initial condition" $\Gamma(1)=1$).
$\endgroup$
  • 2
    $\begingroup$ This was exactly the sort of thing I was looking for. Thank you so much. :) $\endgroup$ – user345895 Jun 26 '16 at 2:25
1
$\begingroup$

For zero is just conventional to set the empty multiplication to 1 as we set the empty sum to 0. To define the factorial function with the help of the gamma function. https://en.wikipedia.org/wiki/Gamma_function Basically this is the only function that agrees with the factorial in the natural numbers, and behaves nicely.

$\endgroup$
  • 2
    $\begingroup$ Probably not useful to OP, but the gamma function is not the only analytic function that agrees with the factorial function on the integers. Can you explain why it's unique given certain "nice" restrictions? $\endgroup$ – Andres Mejia Jun 24 '16 at 3:26
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem $\endgroup$ – Ian Jun 24 '16 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy