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Is there a proof that the factorial function $!:\mathbb N\to\mathbb N$ is nonelementary?

If it were equal to an elementary function (call it $P(n)$), then it would extend the factorial function to the real and complex numbers. This sounds like the Gamma function, but we have $\dfrac{\Gamma(n+1)}{\Gamma(n)}=n$ for all real $n$. It's entirely possible that $\dfrac{P(n)}{P(n-1)}$ isn't $n$, but rather something like $n+\sin(\pi n)$ which is only equal to $n$ at the integers. (Also, I've never found a proof that Gamma is nonelementary, either. I do know that the incomplete Gamma function is nonelementary, due to differential Galois theory.)

Also, the fact that $\pi$ appears in limits involving factorials isn't a proof, by the way. For example, the fact that: $$\lim_{n\to\infty}\frac{(n!)^2(n+1)^{2n^2+n}}{n^{2n^2+3n+1}}=2\pi,$$ which comes from Stirling's approximation, doesn't prove that it can't be elementary; we also have: $$\lim_{n\to\infty}n(-1)^{1/n}-n=i\pi$$ so it's possible for elementary functions to have $\pi$ as a limiting value.

So, is there any proof that the factorial function is nonelementary?

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1 Answer 1

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We will use the following facts:

(i) The extension, to a larger domain, of a non-elementary function is also non-elementary;

(ii) The derivative of an elementary function is also elementary;

(iii) The product of finitely many elementary functions is also elementary;

(iv) The product of an elementary function times a non-elementary function is non-elementary.

Claim 1: $\Gamma(x)$ is a non-elementary function.

Proof. Assume the contrary. By (i) $n!$ must be elementary, and by (ii) so is $\Gamma'(x)=\Gamma(x)\psi^{(0)}(x)$, which by (iii) implies the same for $\psi^{(0)}(x)$ and all of its derivatives. But we have $$\psi^{(n)}(x)=(-1)^{n+1}\ n!\ \zeta(n+1,x),$$ where $\zeta(a,s)$ is the non-elementary Hurwitz zeta function, so combining (iii) and (iv) yields that $\psi^{(n)}(x)$ is a non-elementary function, contradiction. $ \ \ \ \text{QED} $

Claim 2: $n!$ is a non-elementary function.

Proof. The Riemann zeta function satisfies $$\begin{align} 2\ \pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)&= \int_0 ^ \infty \left(\vartheta(0,it) -1\right)t^{s/2-1}dt,\end{align}$$ where $\vartheta(z,q)$ is the non-elementary Jacobi theta function. So let $s=2n$ to obtain $$\begin{align}2\pi^{-n}\Gamma(n)\zeta(2n) &= \int_0 ^ \infty \left(\vartheta(0,it) -1\right)t^{n-1}dt \\ 2\ \pi^{-n} (n-1)! \frac{ (-1)^{n+1} B_{2n}(2\pi)^{2n}}{2(2n)!} &= \int_0 ^ \infty \left(\vartheta(0,it) -1\right)t^{n-1}dt \\ - (-\pi)^n 2^{2n} B_{2n}\frac{(n-1)!}{(2n)!}&=\int_0 ^ \infty \left(\vartheta(0,it) -1\right)t^{n-1}dt.\end{align}$$ Now, by (i) the Bernoulli numbers are elementary, due to being a restriction of the Bernoulli polynomials, which are elementary. But the RHS is non-elementary by (ii), therefore by (iii) the ratio of factorials is non-elementary as well, and the claim follows combining this and (iii). $ \ \ \ \text{QED} $


Of course Claim 1 directly follows from Claim 2 by (i), but I wanted to give two different and independent proofs.

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  • $\begingroup$ In fact there's no need of seeking for a contradiction, to prove Claim 2. $\endgroup$ Commented Aug 13, 2015 at 19:13
  • $\begingroup$ In fact, the gamma function doesn't even satisfy an algebraic differential equation -- see Expanded concept of elementary function?. $\endgroup$ Commented Aug 13, 2015 at 19:32
  • $\begingroup$ @Dave: Nice, thank you. And +1, by the way. $\endgroup$ Commented Aug 13, 2015 at 19:37
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    $\begingroup$ This does not quite make sense to me: each individual Bernoulli polynomial $B_n(x)$ is of course elementary, but why is $B_n = B_n(0)$ elementary as a function of $n$? If it is, then what's a closed form expression for it? We can't use a series that involves binomial coefficients since I'm pretty sure those aren't elementary (they seem strictly more complicated than the factorial function, right?).I don't think we can even say that IF the factorial is elementary, then $B_n$ is elementary. $\endgroup$
    – Davey
    Commented May 14, 2020 at 4:03
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    $\begingroup$ @Davey Bernoulli numbers as function of the order are essentially, zeta function. $\endgroup$
    – Anixx
    Commented Mar 29, 2021 at 15:47

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