# Is there a polynomial equation for $f(n) = n!$ and if so what is it?

And I am not necessarily talking about $f(n) = n(n-1)(n-2)...(3)(2)(1)$ in its factored form; Well it could be that but then I would like a general way of expansion. Thanks in advance!

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There is no polynomial $P(n)$ such that $n!=P(n)$ for all $n$: the factorial function grows faster than anypolynomial. – André Nicolas Jul 21 '14 at 23:57
@AndréNicolas Please post this as an answer. – Fly by Night Jul 22 '14 at 0:00
@FlybyNight : I don't think he bothers anymore. – Patrick Da Silva Jul 22 '14 at 0:06
I thought of it as not an answer, since proof was not given. – André Nicolas Jul 22 '14 at 0:14
@AndréNicolas Don't give me that. You've got 8 comment-upvotes, which doesn't give you anything. While Patrick, who posted a nice answer, has one single up-vote. Most users, and especially OPs, scroll until they see an answer and then stop. Come on, please, just play the game. – Fly by Night Jul 22 '14 at 0:17

You can always write a polynomial $P(x)$ in the following form : $$P(x) = \sum_{i=0}^m a_i x^i = x^m \left( \sum_{i=0}^m a_i x^{i-m} \right).$$ When you plug in an integer $n$, you get $$\frac{P(n)}{n^m} = \sum_{i=0}^m a_i n^{i-m} \to a_m \neq 0.$$ Therefore, for any polynomial $P$ of degree $m$, we have $$\frac{P(n)}{n!} = \frac{P(n)}{n^m} \frac{n^m}{n!} \to a_m \lim_{n \to \infty} \frac{n^m}{n!}.$$ You can easily check that the last limit is zero for all positive integers $m$. If $n!$ would be a polynomial, $P(n)/n!$ would be identically $1$ for some choice of polynomial $P$, a contradiction.

Hope that helps,

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Interesting proof, thanks for the response. – AmourK Jul 22 '14 at 0:18

There is no polynomial $P(n)$ such that $n!=P(n)$ for all $n$: the factorial function grows faster than any polynomial.

One way of showing this is to note that $n!\gt 2^n$ if $n\ge 4$. Then we can use L'Hospital's Rule repeatedly to show that if $P(x)$ is any polynomial, then $\lim_{x\to\infty} \frac{P(x)}{2^x}=0$.

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I didn't know that. Thanks. – Jacob Krall Jul 22 '14 at 3:58

The factorial can't be expressed as a polynomial. Polynomials as really quite simple objects and if you could write the factorial as a polynomial then you would most certainly have heard about it.

On really cool thing that you can do though is to extend the factorial to a function on the complex plane. This function, called the Gamma Function, has the property that $\Gamma(n)=(n-1)!$ for all positive integers $n$. However, $\Gamma$ is defined on all complex numbers except the non positive integers.

$$\Gamma(z) := \int_0^{\infty}x^{z-1}\mathrm{e}^{-x}~\mathrm{d}x$$

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I'll have to look into this gamma function, it seems interesting, thanks. – AmourK Jul 22 '14 at 0:21