# College Math Teaching

## May 16, 2011

### Why is 0! = 1?

Filed under: advanced mathematics, calculus, integrals, mathematics education, media, uniqueness of solution — collegemathteaching @ 2:39 pm

This is in response to the following article which says, in part:

I do want to bring up one interesting case study I came across that points in favor of the “math is invented” side of the debate. My friends over at the popular blog Ask a Mathematician, Ask a Physicist did a great post a while ago addressing one of their readers’ questions: What is 0^0?

The reason this question is a head-scratcher is that our rules about how exponents work seem to yield two contradictory answers. On the one hand, we have a rule that zero raised to any power equals zero. But on the other hand, we have a rule that anything raised to the power of zero equals one. So which is it? Does 0^0 = 0 or does 0^0 = 1? […]

Indeed, the Mathematician at AAMAAP confirms, mathematicians in practice act as if 0^0 = 1. But why? Because it’s more convenient, basically. If we let 0^0=0, there are certain important theorems, like the Binomial Theorem, that would need to be rewritten in more complicated and clunky ways.

I provided an answer to the $0^0$ question in another post. But the binomial theorem (which indeed starts with $(a + b)^0$ uses the binomial coefficients; and for each $n$ we eventually have to do with $n$ choose $k$ notation ${n \choose k}$ which involves a $0!$ in the denominator when $k = n$ or when $k = 0$. And yes, the formulas don’t work unless $0! = 1$.

But is that the only reason that $0! = 1$?

Hardly. As before, we’ll use the properties of analysis to see why this should be true.
Note: I can recommend this internet article for those who are unfamiliar with some of the topics that we are going to discuss; in fact, that article is better than what I am about to write. 🙂

So, can the factorial function $n! = n(n-1)(n-2)....(3)(2)(1)$ be extended so as to be continuous on at least the non-negative part of the real line? The answer is “yes”; the usual extension is the so called gamma function $\Gamma (x) = \int^{\infty}_{0} t^{x-1} e^{-t} dt$

It is an easy exercise in improper integration, integration by parts and mathematical induction to see that $\Gamma (n + 1) = n!$ for all $n \in {1, 2, 3, ....}$ and that $\Gamma (1) = 1$; hence it makes sense to set $0! =1$.

It is a more advanced exercise to see that the gamma has continuous derivatives of all orders defined for all positive reals (hint: differentiate under the integral sign) and that, in fact, the gamma function is real analytic (has a Taylor series defined on some open interval at every point ).

Strictly speaking, the gamma function is NOT the only possible way to extend the factorial function to an analytic function (example: $\Gamma (x) + sin(\pi x) = \Gamma (x)$ on the non-negative integers ). But if we add the following restrictions:
1. $f(1) = 1$
2. $f(x+1) = xf(x)$ (a property the factorial function has) and
3. $f$ is logarithmically convex (e. g., $ln(f(x)$ is convex in $x$ then
the Bohr-Mollerup theorem shows that the gamma function is the only extension.

Now of course, the key here is the recursive property (2); this forces the $0! =1$ definition.