Can there be only one extension to the factorial?

Question

Usually, when someone says something like $\left(\frac12\right)!$, they are probably referring to the Gamma function, which extends the factorial to any value of $x$.

The usual definition of the factorial is $x!=1\times2\times3\times\dots x$, but for $x\notin\mathbb{N}$, the Gamma function results in $x!=\int_0^\infty t^xe^{-t}dt$.

However, back a while ago, someone mentioned that there may be more than one way to define the factorial for non-integer arguments, and so, I wished to disprove that statement with some assumptions about the factorial function.

the factorial

is a $C^\infty$ function for $x\in\mathbb{C}$ except at $\mathbb{Z}_{<0}$ because of singularities, which we will see later.
is a monotone increasing function that is concave up for $x>1$.
satisfies the relation $x!=x(x-1)!$
and lastly $1!=1$

From $3$ and $4$, one can define $x!$ for $x\in\mathbb{N}$, and we can see that for negative integer arguments, the factorial is undefined. We can also see that $0!=1$.

Since we assumed $2$, we should be able to sketch the factorial for $x>1$, using our points found from $3,4$ as guidelines.

At the same time, when sketching the graph, we remember $1$, so there can be no jumps or gaps from one value of $x$ to the next.

Then we reapply $3$, correcting values for $x\in\mathbb{R}$, since all values of $x$ must satisfy this relationship.

Again, because of $1$, we must re-correct our graph, since having $3$ makes the derivative of $x!$ for $x\in\mathbb N$ undefined.

So, because of $1$ and $3$, I realized that there can only be one way to define the factorial for $x\in\mathbb R$.

Is my reasoning correct? And can there be only one extension to the factorial?

Oh, and here is a 'link' to how I almost differentiated the factorial only with a few assumptions, like that it is even possible to differentiate.

Putting that in mind, it could be possible to define the factorial with Taylor's theorem?

The factorial $x!$ can't be defined for all $x\in\mathbb{C}$. At least, all $x\in\mathbb{Z}_{<0}$ must be singularities. — Batominovski, Jul 03 '16 at 12:38
@Batominovski Well, yeah, I do mention that. Whats your point? Oh, should I state that in my bullet points? — Simply Beautiful Art, Jul 03 '16 at 12:39
There is a theorem of this type (but the condition is that the function is "log convex"). http://en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem — GEdgar, Jul 03 '16 at 12:45
@GEdgar Quite interesting, it is somewhat difficult to consume all at once, so I'll let it sit in for a while before I ask anything about it. — Simply Beautiful Art, Jul 03 '16 at 12:53
Maybe you could make a program that plots n!, then interpolate (by segments, splines or whatever) and apply 3. Then see if updating the plot converges to the gamma function. It's not a proof, though. — jinawee, Jul 03 '16 at 13:06
@jinawee I don't actually know how to do coding. If anyone else makes a code that can, that'd be awesome! (and yeah, it surely isn't anything proof worthy, but it'd be interesting) — Simply Beautiful Art, Jul 03 '16 at 13:10

Batominovski · Accepted Answer · 2016-07-04T08:56:33.657

8

First, for a fixed $c\in\mathbb{C}$, let $$F_c(z):=\Gamma(z+1)\cdot\big(1+c\,\sin(2\pi z)\big)$$ for all $z\in\mathbb{C}\setminus \mathbb{Z}_{<0}$, which defines an analytic function $F_c:\mathbb{C}\setminus\mathbb{Z}_{<0}\to\mathbb{C}$ such that $$F_c(z)=z\cdot F_c(z-1)$$ for all $z\in\mathbb{C}\setminus\mathbb{Z}_{\leq 0}$ and that $F_c(0)=F_c(1)=1$ (whence $F_c(n)=n!$ for every $n\in\mathbb{Z}_{\geq 0}$). Excluding the essential singularity at $\infty$, the negative integers are the only singularities of $F_c$, which are simple poles.

Here are some results I checked with Mathematica. If $c$ is a positive real number less than $0.022752$, then $F_c'(z)>0$ for all $z>1$ and $F_c''(z)>0$ for all $z>-1$, making $F_c$ monotonically increasing on $(1,\infty)$ and convex on $(-1,\infty)$. It also appears that, with $0<c<0.022752$, $F_c$ is convex on $(-2n,-2n+1)$ and concave on $(-2n-1,-2n)$ for every $n=1,2,\ldots$. (I have checked this with various values of $c$ and with $n\leq 30$.) It would be great if someone can find an actual proof. Hence, it seems to me that the conditions 1-4 do not give a unique factorial function.

edited Jul 04 '16 at 08:56

answered Jul 03 '16 at 17:21

Batominovski

49,629

What if from the definition of the factorial, it were findable that $\frac{d}{dx}\ln(x!)=G(x)$, where $G(x)$ satisfies the functional equation $G(x)=\frac1x+G(x-1)$, but from your function, I get $\frac{d}{dx}\ln(x!)=\frac{d}{dx}\ln(F_c(x))=\int_0^1\frac{t^x-1}{t-1}dt+C+\frac{2\pi c\cos(2\pi x)}{1+c\sin(2\pi x)}$, which doesn't satisfy that functional equation? $C$ is a constant. And I derive $G(x)$ only from $3$ (and maybe $1$). Derivation is in link near end of my post. (+1 for good try though) – Simply Beautiful Art Jul 04 '16 at 14:42
Haven't thought about this much, but $G_c(x)=\psi(x+1)+c,\sin(2\pi x)$ for some small $c$ may work. Here, $\psi$ is the digamma function. – Batominovski Jul 04 '16 at 15:02
But then that will alter your $F_c(x)$ to $F_c(x)=?$ The problem is that $\frac{d}{dx}F_c(x)$ isn't purely multiplication, like $x!$, which would result in our neat functional equations. And besides, $\int G_c(x)=\Gamma-\frac{c}{2\pi}\cos(2\pi x)$, or something like that, and that's probably not going to satisfy my $3,4$. More generally, and you may already know this, but if we define $$G^k(x):=\frac{d^k}{dx^k}\ln(x!)$$we find it satisfies$$G^k(x)=G^k(x-1)+\frac{(-1)^{k-1}(k-1)!}{x^k}$$ – Simply Beautiful Art Jul 04 '16 at 15:07
I thought you only wanted the functional equation $G(x)=\frac{1}{x}+G(x-1)$ to hold. I don't know the answer if you want $F$ to satisfy all functional equations $G^k(x)=G^k(x-1)+\frac{(-1)^{k-1}(k-1)!}{x^k}$ for all $k=1,2,\ldots$, where $G^k(x):=\frac{\text{d}^k}{\text{d}x^k},\ln\big(F(x)\big)$, and that $F$ satisfies 1-4. However, I have a feeling that with these very strong constraints, $F(x)=\Gamma(x+1)$ is the only possible solution. – Batominovski Jul 04 '16 at 15:52
No no, they just come naturally from differentiating $\ln(x!)$, using $x!=1\cdot2\cdot3\dots x$, which was bullet $3$. It requires knowing how to differentiate $\sum_{n=1}^x\ln(n)$, but I assure you that when done $k$ times, the result is some $G^k(x)$ that satisfies those functional equations. Did you see my link on how to differentiate the factorial without using the gamma function? Methods there work for this to any extent I do believe. – Simply Beautiful Art Jul 04 '16 at 16:03
http://math.stackexchange.com/questions/1840297/calculating-sum-n-1x-fracrnnk-with-integrals and http://math.stackexchange.com/questions/1633014/is-there-a-way-to-evaluate-the-derivative-of-x-without-using-gamma-function/1844903#1844903 – Simply Beautiful Art Jul 04 '16 at 16:09
@SimpleArt I got some time to think about your question again. It turns out that $F_c$ satisfies the functional equation $G_c^k(x)=G_c^k(x-1)+\frac{(-1)^{k-1}(k-1)!}{x^k}$ too, where $G_c^k(x):=\frac{\text{d}^k}{\text{d}x^k},F_c(x)$. – Batominovski Jul 04 '16 at 22:58
You mean $G_c^k(x):=\frac{d^k}{dx}^k\ln(F_c(x))$? – Simply Beautiful Art Jul 05 '16 at 22:27
@SimpleArt Yep, that is what I meant. In fact, you got a typo to :D. What I meant is this: $$G^k_c(x):=\frac{\text{d}^k}{\text{d}x^k},\ln\big(F_c(x)\big),.$$ – Batominovski Jul 06 '16 at 11:44

score 5 · Answer 2 · edited Apr 13 '17 at 12:19

5

The Bohr-Mollerup Theorem states that the Gamma function is the only log-convex function which satisfies $\Gamma(1)=1$ and $x\Gamma(x)=\Gamma(x+1)$. There are a continuum of functions that are not log-convex that satisfy the other two constraints.

This answer details some of the important points of this.

edited Apr 13 '17 at 12:19

Community

1

answered Jul 03 '16 at 21:20

robjohn

345,667

Can there be only one extension to the factorial?

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