Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is this the only product representation of the factorial function?

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

share|improve this question
    
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. –  Amire Jul 25 '13 at 0:31
1  
$ \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
1  
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

3 Answers 3

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

share|improve this answer

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.

share|improve this answer
    
The k=2 one is trivial! :) but I see what you are saying –  zerosofthezeta Jul 25 '13 at 0:37
1  
@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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.