# What is the definition of $2.5!$? (2.5 factorial)

I was messing around with my TI-84 Plus Silver Edition calculator and discovered that it will actually give me values when taking the factorial of any number $n/2$ where $n$ is any integer greater than $-2$. Why does this happen? I thought factorials were only defined for positive integers and $0$, so what is my calculator doing to get the answer $3.32335097$ when I enter $2.5!$? Is there actually a definition of $2.5!$ or is my calculator just being weird? How is the factorial function implemented?

I understand the binary implications of $2.5$, so that could possibly have something to do with it. I get a domain error when trying to take the factorial of $-1, 2.3, e$, and any number that is not of the form $n/2$ where $n$ is any integer greater than $-2$.

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There is an extension of the factorial to "most" numbers (including complex numbers) called the Gamma Function, $\Gamma(z) = \int_0^\infty e^{-t} t^{z-1}dt$. It satisfies $\Gamma(n+1) = n!$, and more generally, that $\Gamma(z+1) = z\Gamma(z)$ for any number $z$.

It is a curious fact that $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$. Using this, together with the fact that $\Gamma(z+1) = z\Gamma(z)$, we get

\begin{align*}2.5! &= \Gamma(3.5) \\\ &= 2.5\cdot\Gamma(2.5) \\\ &= 2.5\cdot 1.5\cdot \Gamma(1.5) \\\ &= 2.5\cdot 1.5\cdot .5\cdot \Gamma(.5)\\\ &= 2.5\cdot 1.5\cdot .5\cdot \sqrt{\pi} \\\ &= 3.32335097... \end{align*}

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The other answers have already given nice explanations, so I shall instead be teaching a different lesson: as always, when confronted with seemingly peculiar behavior in your device, taking a look at your fine device's fine manual is almost always a profitable first step. As it turns out, looking at page 58 of your fine calculator's fine manual mentions the following:

So yes, do read the fine manual. ;)

As a further addition: there is what is called Gauss's duplication formula, which in factorial form goes as

$$\left(n+\frac12\right)!=\frac{\sqrt{\pi}(2n+2)!}{4^{n+1}(n+1)!}$$

This allows you to express factorials of semi-integers in terms of factorials and $\left(-\frac12\right)!=\sqrt \pi$.

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The factorial of a positive real number $s$ is defined as $$s! = \int_0^\infty x^s \exp(-x) \mathrm{d} x$$ It is easy to verify, integrating by parts, that the definition satisfies the recurrence equation of the factorial: $$\begin{eqnarray} s! &=& \int_0^\infty x^s \exp(-x) \mathrm{d} x = \int_0^\infty x^s \mathrm{d} (-\exp(-x) ) \\ &=& \left.\left(- x^s \exp(-x) \right) \right|_{x \downarrow 0}^{x \uparrow \infty} + s \int_0^\infty x^{s-1} \exp(-x) \mathrm{d} x = s (s-1)! \end{eqnarray}$$ The limit $\lim_{x \to +\infty} x^s \exp(-x) = \lim_{x \to +\infty} \frac{x^s}{\exp(x)} = 0$, since exponential function grows faster than any polynomial, and $\lim_{x \downarrow 0} x^s \exp(-x) \leqslant \lim_{x \downarrow 0} x^s = 0$, since $s$ was assumed positive.