Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

By definition, the double factorial $(-1)!! = 1$. How can this be rationalized?

share|cite|improve this question
I couldn't find any specific answers to this on, but worked out an answer for myself. Therefore, I've answered the question myself. – James Womack Jan 6 '13 at 21:43
up vote 8 down vote accepted

This is a "double factorial":

The product of all odd integers up to some odd positive integer n is often called the double factorial of n (even though it only involves about half the factors of the ordinary factorial, and its value is therefore closer to the square root of the factorial). It is denoted by n!!

From the link above, we have that for odd $n$ there is a $k\in \mathbb{Z}$ such that $n = 2k-1$, so $$n!! = (2k-1)!! = \dfrac{(2k)!}{2^k k!}.$$

$$\text{At}\;k = 0,\;\;\;(2\cdot 0 - 1)!! = (-1)!! = \frac{0!}{2^0 0!} = \frac{1}{1\cdot 1} = 1.$$

Recall that $0! = 1$, by definition (as representing the "empty product").

For even $n = 2k\,$ for $k \in \mathbb{Z}$: $$n\,!\,! = (2k)\,!\,! \;= \;2^k\, k\,!$$

Note: in both the odd and even case, $k$ is usually taken to be $k \ge 1$.

share|cite|improve this answer
+1 for extending the answer to cover the even number double factorials, too. – James Womack Jan 7 '13 at 9:58

Looking at the expression for a double factorial in terms of ordinary factorials, $$(2k-1)!! = \frac{(2k)!}{2^{k} k!}$$ and setting $k=0$, $$(0-1)!! = \frac{0!}{2^0 0!} = 1.$$ This maintains consistency with the convention that $0! = 1$.

share|cite|improve this answer

One answer is to maintain that $$ n!!=\frac{(n+2)!!}{n+2} $$ for all $n$. In this case, we must have $$ (-1)!!=\frac{(-1+2)!!}{(-1+2)}=\frac{1!!}{1}=1. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.