Negative factorials and the creation of math functions

Question

I know that factorials are, by definition, positive integers which means you can not have n! where n is negative. My question is can you create a factorial specifically for negative integers? I thought of this idea as a sort of opposite factorial. I used an upside down exclamation mark to denote a negative factorial. It functions as such: A negative factorial can only be performed on negative integers. The negative factorial of an even number is positive and the negative factorial of an odd number is negative (assuming the number is negative to begin with). If you take a negative factorial of a positive number you get the same answer as a factorial of negative number.

n¡=+(0,2,4,6,8…) or –(1,3,5,7,9…)

│n│¡=error

I do not know of any useful applications for this or if this is even possible at all. That's why I want to know if it's possible to make up a function such as a negative factorial.

Answer 1

Your idea is a special case of what is known as rising factorials and falling factorials. The rising factorial is defined as $$ x^{(n)} = x(x+1)\cdots(x+n-1), $$ and the falling factorial is $$ x_{(n)} = x(x-1)\cdots(x-n+1), $$ where $n$ is a non-negative integer and $x$ is an arbitrary (complex) number. These notations are used in combinatorics and definitions of special functions. See the wiki article on the Pochhammer symbol for more info.

In particular, your "negative factorial" can be written as $$(-n)^{(n)} = (-1)^n n_{(n)} = (-1)^n n!$$

Answer 2

Sure, you could define $(-n)¡=(-n)\cdot(-n+1)\cdot(-n+2)\dots\cdot(-2)\cdot(-1)$, when $n$ is a nonnegative integer. Note that you can alternatively write this as $(-1)^n n!$, so you don't really need to introduce a new notation for it. I don't know of any applications of this concept, though. Generally, because you can just write it in terms of the ordinary factorial, people wouldn't mention it explicitly even if it did come up naturally somewhere.

Answer 3

The gamma function could possibly be used in this scenario, for negative $x$, as it is defined as follows, for $x << \infty$: $$ \Gamma(x+1)=x\Gamma(x) $$

Proof:

The gamma function is generally defined as the integral;

$$ \Gamma(x)=\int_0^\infty t^{x-1} e^{-t} dt $$

We can then get:

$$ \Gamma(x)=\lim_{a \to \infty}[-(a^{x-1}e^{-a} + (x-1)\Gamma(x-1))] $$

Obviously, for a small $x$, then we can get the following formula:

$$ \Gamma(x) = (x-1)\Gamma(x-1) $$

, or

$$ \Gamma(x+1) = x\Gamma(x) $$

This is one of the infamous results from Ramanujan's work on the gamma function and its properties.