I know that you can find the factorial of positive integers where n!= n(n-1)...2 x 1. However, what if you want to find the factorial of a negative integer or a decimal? I tried to do it on my calculator and it gave an answer however, I wasn't able to understand how they calculator got the answers.

I did some research and came across the gamma function which supposedly allowed you to solve such questions. However, I found it very hard to understand and still don't see the purpose of finding the factorial of a negative integers or decimals.

Help would be appreciated.

Thank you :)

  • $\begingroup$ The factorial is an integer function, but look up something called the gamma function ($\Gamma$.) It is the extension of factorial to the complex numbers (including the reals.) $\endgroup$ – Thomas Andrews Apr 19 '15 at 23:13
  • $\begingroup$ It might be worth mentioning that the gamma function (with its argument shifted by $1$) isn't the only possible extension of the factorial. It is, however, by far the most often used one. $\endgroup$ – Andreas Blass Apr 20 '15 at 1:00

The factorial function is extended by the $\Gamma$ function. The relation is $$(n-1)! = \Gamma(n) = \int_0^\infty t^{n-1} e^{-t}\, dt$$ This can be analytically continued as a meromorphic function in the complex plane. Ref: John Conway's book on Complex Analysis.


My favorite use of the factorial function for non-integer arguments is the formula for the volume of an $n$-dimensional ball of radius $r$: $$ \frac{\pi^{n/2}r^n}{(n/2)!}. $$

  • $\begingroup$ I'm just going to assume this is strictly about odd $n$, and you're not actually suggesting non-integral $n$, because the latter situation just broke my brain. $\endgroup$ – Kevin Apr 20 '15 at 2:36
  • $\begingroup$ @Kevin Right; the dimension $n$ is an integer. $\endgroup$ – Andreas Blass Apr 20 '15 at 14:20

As already noted by @ncmathsadist, the Gamma Function $$\Gamma(z)=(z-1)!$$ can be extended to a meromorphic function defined on the complex plane without the non-positive integers.

Here's an instructive example how to work with negative factorials presented in section $3.6$ of $A=B$ by M. Petkovsek, H. Wilf and D. Zeilberger.

Challenge: Find a closed expression for

\begin{align*} f(n)=\sum_{k=0}^{2n}t_k=\sum_{k}(-1)^k\binom{2n}{k}\binom{2k}{k}\binom{4n-2k}{2n-k} \end{align*}

We start with checking if the ratio $\frac{t_{k+1}}{t_k}$ gives rise to a known hypergeometric series. Indeed, with \begin{align*} \frac{t_{k+1}}{t_k}&=\frac{(-1)^{k+1}\binom{2n}{k+1}\binom{2k+2}{k+1}\binom{4n-2k-2}{2n-k-1}} {(-1)^k\binom{2n}{k}\binom{2k}{k}\binom{4n-2k}{2n-k}} =\frac{(k+\frac{1}{2})(k-2n)^2}{(k+1)^2(k-2n+\frac{1}{2})} \end{align*} we derive \begin{align*} f(n)=\binom{4n}{2n} {}_{3}F_{2}\left(-2n,-2n,\frac{1}{2};1,-2n+\frac{1}{2};1\right) \end{align*}

It turns out, that this hypergeometric series matches Dixon's identity and

we obtain \begin{align*} f(n)=\binom{4n}{2n}\frac{(-n)!(-2n-\frac{1}{2})!(n-\frac{1}{2})}{(-2n)!n!(-n-\frac{1}{2})!(-\frac{1}{2})!}\tag{1} \end{align*}

At first glance this expression is rather distressing, since it contains factorials of negative integers which are precisely the values, where the gamma function is not defined!

The clou: We have a ratio of two factorials at negative integers and if we can take an appropriate limit, the singularities will cancel leaving a pleasant limiting ratio. As the authors point out, this situation happens fairly frequently when using this approach.


We start analysing the ratio \begin{align*} \frac{(-n)!}{(-2n)!}\tag{2} \end{align*}

Let's assume, that $n$ is near a positive integer, but is not equal to a positive integer. Then we can use the reflection formula for the $\Gamma$-function \begin{align*} \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z} \end{align*} once more in the equivalent form \begin{align*} (-z)!=\frac{\pi}{(z-1)!\sin \pi z} \end{align*}

So, when $n$ is near a positive integer, the expression (2) becomes \begin{align*} \frac{(-n)!}{(-2n)!}=\frac{\pi}{(\sin n\pi)(n-1)!}\frac{(\sin 2n\pi)(2n-1)!}{\pi}=\frac{2(2n-1)!\cos n\pi}{(n-1)!} \end{align*} and we observe, if $n$ approaches a positive integer \begin{align*} \frac{(-n)!}{(-2n)!}\longrightarrow(-1)^n\frac{(2n)!}{n!} \end{align*} The expression (1) becomes \begin{align*} f(n)=(-1)^n\binom{4n}{2n}\binom{2n}{n}\frac{(-2n-\frac{1}{2})!(n-\frac{1}{2})}{(-n-\frac{1}{2})!(-\frac{1}{2})!} \end{align*} Similarly, we find \begin{align*} \frac{(-2n-\frac{1}{2})!}{(-n-\frac{1}{2})!}=\frac{(-1)^n(n-\frac{1}{2})!}{(2n-\frac{1}{2})!} \end{align*} and we obtain \begin{align*} f(n)=\binom{4n}{2n}\binom{2n}{n}\frac{(n-\frac{1}{2})!^2}{(2n-\frac{1}{2})!(-\frac{1}{2})!}\tag{3} \end{align*}

For every positive integer $m$, \begin{align*} (m-\frac{1}{2})!&=(m-\frac{1}{2})(m-\frac{3}{2})\cdots(\frac{1}{2})(-\frac{1}{2})!\\ &=\frac{(2m-1)(2m-3)\cdots 1}{2^m}(-\frac{1}{2})!\\ &=\frac{(2m)!}{4^mm!}(-\frac{1}{2})! \end{align*}

This way we can simplify the expression (3) to $f(n)=\binom{2n}{n}^2$ and we have shown the identity

\begin{align*} \sum_{k}(-1)^k\binom{2n}{k}\binom{2k}{k}\binom{4n-2k}{2n-k}=\binom{2n}{n}^2 \end{align*}


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