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

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  • $\begingroup$ I couldn't find any specific answers to this on math.stackexchange.com, but worked out an answer for myself. Therefore, I've answered the question myself. $\endgroup$ – James Womack Jan 6 '13 at 21:43
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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 $k \ge 1$.

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  • $\begingroup$ +1 for extending the answer to cover the even number double factorials, too. $\endgroup$ – James Womack Jan 7 '13 at 9:58
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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 $0! = 1$.

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

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