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Is this the only product representation of the factorial function?

$$ {n!} =\prod_{k=1}^{n} k $$

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That's the definition. You could manipulate the right hand side of the equation in many ways still getting the same result , but another representation of it. It's like saying: is $1=0+1$ the only representation of $1$? Well $1=e^0$ is another one and there are many many others. It depends on how you define it. – sigmatau Jul 25 '13 at 0:31
$ \displaystyle n! = \frac{1}{z} \prod_{k=1}^{\infty} \frac{(1+\frac{1}{k})^{n+1}}{1+ \frac{n+1}{k}}$ – Random Variable Jul 25 '13 at 0:47
@RandomVariable What does z represent? Is it a constant or complex? – zerosofthezeta Jul 25 '13 at 1:02
It should be (n+1). But the product converges for all complex numbers except the negative integers and zero. – Random Variable Jul 25 '13 at 1:10
up vote 2 down vote accepted

There is the Weierstrass infinite product definition of Gamma function

$$\Gamma(z) = \frac{e^{-\gamma z}}{z}\prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)^{-1} e^{\frac{z}{k}}$$

which leads to $$n! = \Gamma(n+1) = \frac{e^{-\gamma (n+1)}}{n+1}\prod_{k=1}^{\infty}\frac{k}{k+n+1}e^{\frac{n+1}{k}}$$

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This is another one, for $n>2$

$${n!} =\prod_{k=2}^{n} k$$

And another one not so stupid:

$$n!= \int_0^\infty x^ne^{-x}dx\qquad n\in\mathbb{N}$$

That's the gamma function, that generalizes the factorial to complex numbers, when $n$ is allowed to be complex. I don't know more. But basically that's the definition for factorial... so every manipulation that you can think about and it's acceptable will be a different representation.

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The k=2 one is trivial! :) but I see what you are saying – zerosofthezeta Jul 25 '13 at 0:37
@euclid I know, it was supposed to be trivial just to show a simple change :) – MyUserIsThis Jul 25 '13 at 0:39

Euler probably did something like the following to extend the factorial function to values other than the positive integers.

Let $x$ be a positive integer.

Then $ \displaystyle x! = \frac{(x+n)!}{(x+1)(x+2) \cdots (x+n)} = \frac{(x+n)(x+n-1) \cdots (n+1) n!}{(x+1)(x+2) \cdots (x+n)}$

$ \displaystyle= \frac{(n+1)(n+2) \ldots (n+x)}{n^{x}}\frac{n! \ n^{x}}{(x+1)(x+2) \cdots (x+n)}$

And $ \displaystyle \lim_{n \to \infty} x! = x! = \lim_{n \to \infty} \frac{(n+1)(n+2) \ldots (n+x)}{n^{x}} \cdot \lim_{n \to \infty}\frac{n! \ n^{x}}{(x+1)(x+2) \cdots (x+n)}$

$ \displaystyle =\lim_{n \to \infty} \frac{n! \ n^{x}}{(x+1)(x+2) \cdots (x+n)} = \lim_{n \to \infty} \frac{n^{x}}{\left( 1 + x \right) \left( 1+ \frac{x}{2} \right) \cdots \left( 1 + \frac{x}{n} \right)}$

$ \displaystyle = \lim_{n \to \infty} \frac{\prod_{k=1}^{n-1} (1 + \frac{1}{k} )^{x}}{\left( 1 + x \right) \left( 1+ \frac{x}{2} \right) \cdots \left( 1 + \frac{x}{n} \right)} =\prod_{k=1}^{\infty} \frac{(1+\frac{1}{k})^{x}}{1+ \frac{x}{k}}$

Now let $\Gamma(x) = (x-1)!$ (which for some reason is how the gamma function is defined for positive integers).

Then $\Gamma(x+1) = x \Gamma(x)$ and $ \displaystyle \Gamma(x) = \frac{1}{x} \prod_{k=1}^{\infty} \frac{(1+\frac{1}{k})^{x}}{1+ \frac{x}{k}}$

But that infinite product converges not only when $x$ is a positive integer, but also when $x$ is any complex number excluding zero and the negative integers.

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Sorry I can't type in LATEX but does the last equation need the 1/x? Doesn't the x from xΓ(x) cancel the 1/x in the last equation leaving only the product? – zerosofthezeta Jul 25 '13 at 3:59
For a positive integer $x$, $\Gamma(x+1) = x! = x (x-1)! = x \Gamma(x)$. So $x! = \Gamma(x+1) = x \Gamma(x) = \prod_{k=1}^{\infty} \frac{(1+\frac{1}{k})^{x}}{1+ \frac{x}{k}}$. Then divide both sides by $x$. – Random Variable Jul 25 '13 at 4:55

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