# Analytically continuing Euler's partition function

Author's note: This question might be a little hopeless, but maybe someone has some form of good feedback. It's a long one because I tried to be very thorough. I tried to explained all the odds and ends I had at approaching this.

It is a preliminary fact when studying modern analytic number theory that the reader has seen Euler's partition function. The partition function $p$, as review, can be expressed as

$$\prod_{i=1}^\infty \frac{1}{1- x^{i}} = \sum_{k=0}^\infty p(k)x^k$$

Essentially, I want to analytically continue $p$ to some domain in $\mathbb{C}$.

I'll be clear now as to what this question is not asking. It is a basic graduate exercise in complex analysis to construct a holomorphic entire function $f(z)$ such that $f(n) = p(n)$ for $n \in \mathbb{N}$. I am not asking for such a simple analytic continuation of the partition function. I am asking for something a bit more nuanced. To explain this, I'll require Euler's Pentagonal Theorem.

To paraphase and for the readers' convenience: let us enforce the fact that $p(k) = 0$ for $k \in \mathbb{Z}$ and $k \le -1$. Then, if $g_i = \frac{1}{2}(3i^2 - i)$ it follows

$$\sum_{i=0}^\infty (-1)^i p(k-g_i) = 1_{k=0}$$

(Where $1_{k=0}$ equals $1$ if $k=0$ and $0$ otherwise.)

This can be rewritten for $k \neq 0$:

$$\sum_{i=1}^\infty (-1)^{i+1} p(k - g_i) = p(k)$$

I want to abuse this relationship in complex analysis. And by abuse, I mean it in a similar sense to how Euler would talk of the 'half-factorial' without actually rigorously describing the 'half-factorial'. I want to do something similar to the invention of the Gamma function (there are trillions of functions that interpolate the factorial, but only one nice one that satisfies $z(z-1)! = z!$).

Here comes the actual question. I want to find a function $\mathcal{P}$ that interpolates the values of $p :\mathbb{Z} \to \mathbb{Z}$, but it also satisfies the above functional equation. Essentially, I don't just want to interpolate the partition function, I want it to satisfy a nice equation. That equation being

$$\mathcal{P}(z) = \sum_{i=1}^\infty (-1)^{i+1} \mathcal{P}(z-g_i)$$

This is simply a rewrite of Euler's pentagonal identity for the partition function, except we're letting the variable be complex.

To make the question more concrete, we have to work around a few facts. First things first, we have to remove $0$ from the domain of $\mathcal{P}$ because $\sum_{i=1}^\infty (-1)^{i+1} \mathcal{P}(z-g_i)$ is a holomorphic function that equals $\mathcal{P}$ almost everywhere, except when $z = 0$—this sum equals $0$ but $\mathcal{P}(0) = 1$. So it's hopeless for this function to be holomorphic at zero. Because of this, we must weaken the solution so that $\mathcal{P}$ is not analytic at $0$.

Working from this we can't have $\mathcal{P}$ being analytic at $z=g_i$ as well, because then undefined values of $\mathcal{P}$ enter our sum ($z - g_i = 0$ at some point). Then continuing from this, just as well if $z= g_i + g_j$, we will get undefined values in our sum $z - g_i = g_j$. Similarly for $z = g_i + g_j + g_k$, so on and so forth. So $\mathcal{P}$ can't be holomorphic on $\mathbb{N}$ at all. So to be safe, we're just going to cut $\mathbb{R}$ entirely. This might seem like we've just lost the meat of the question, but I don't quite see it like that.

We're going to only talk about $\mathcal{P}$ in the upper half plane $\mathbb{H}$ with a continuous extension to $\mathbb{R}$. Because of this, we don't get a nice interpolation of the partition function, we get something more wonky.

The sum

$$\mathcal{P}(z) = \sum_{i=1}^\infty (-1)^{i+1}\mathcal{P}(z-g_i)$$

cannot converge for $z \in \mathbb{R}$, excepting when $z \in \mathbb{N}$ where it converges trivially. This way, we can't pull the limit through the sum and produce a contradiction (in contrast to if the function was holomorphic in a neighborhood of $\mathbb{R}$). Namely, if it did converge, we could pull the limit through and

$$1 = \mathcal{P}(0) = \lim_{z\to 0 } \sum_{i=1}^\infty (-1)^{i+1}\mathcal{P}(z-g_i) = \sum_{i=1}^\infty (-1)^{i+1}\lim_{z \to 0}\mathcal{P}(z-g_i) = \sum_{i=1}^\infty (-1)^{i+1}0=0$$

This would clearly muff everything up. Which better explains why $\mathcal{P}$ can't be holomorphic in a neighborhood of $\mathbb{R}$.

So now I can ask my question. It looks a lot like a boundary value ODE problem in complex analysis, except it's discrete. We write $\mathcal{P}\Big{|}_{\mathbb{Z}}$ to mean $\{\lim_{z \to k}\mathcal{P}(z)\,\Big{|}\, k \in \mathbb{Z}\}$

Is there a holomorphic function $\mathcal{P} : \mathbb{H} \to \mathbb{C}$ such that

$$\mathcal{P}\Big{|}_{\mathbb{Z}} = p$$

and, most importantly for $z \in \mathbb{H}$

$$\mathcal{P}(z) = \sum_{i=1}^\infty (-1)^{i+1} \mathcal{P}(z-g_i)$$

The best I could do at solving this was assume $\mathcal{P}$ belonged to a Hilbert space on $\mathbb{H}$. Using the fact the operator $T\mathcal{P}(z) = \mathcal{P}(z-1)$ acts on the Hilbert space, I tried to design the Hilbert space so that the operator $U = \sum_{i}(-1)^{i+1} T^{g_i}$ also acted on the Hilbert space . Then all we have to do is find an eigenfunction of $U$ with eigenvalue $1$ (or prove such an eigenfunction even exists). But I couldn't seem to get anywhere with this. Most importantly I couldn't really figure out what the Hilbert space "looked like".

I firstly assumed, for convenience, $\int_{-\infty}^\infty |\mathcal{P}(x+iy)|\,dx < \infty$ (seeing as the convergence of the sum slightly suggests that). But then the Fourier transfrom $\hat{\mathcal{P}}(\xi)$ satisfies $\hat{\mathcal{P}}(\xi) = f(\xi)\hat{\mathcal{P}}(\xi)$ where $f(\xi) = \sum_k (-1)^{k+1}e^{2\pi i g_k \xi}$. Therefore the only $L^1$ solution of $U\mathcal{P} = \mathcal{P}$ is the zero function.

So instead I thought more relaxed and assumed $\mathcal{P}$ is $L^2$ as $\Re(z) = x \to - \infty$ and grew however as $\Re(z) = x \to \infty$. But from there I wasn't sure what inner product to use to make manipulations convenient. (Haven't worked too much with Hilbert spaces in a while, a little rusty to be honest.)

So for this, I thought for convenience $\mathcal{P}$ was a Laplace transform. Then we can write

$$\mathcal{P}(z) = \int_{-\infty}^\infty e^{tz}\,d\mu$$

and

$$U\mathcal{P}(z) = \int_{-\infty}^{\infty} e^{tz} (\sum_{i=1}^\infty (-1)^{i+1}e^{- g_i t})\,d\mu = \int_{-\infty}^{\infty} f(-t) e^{tz}\,d\mu$$

and consequently $f(t)= 1$ which implies $\mathcal{P}(z)$ cannot be a ''sum'' of exponentials.

From here, I realized, without a base of solutions to the fixed point equation $U\mathcal{P} = \mathcal{P}$ I hadn't even thought about the initial conditions $\mathcal{P}\Big{|}_\mathbb{Z} = p$. It's like being an undergrad all over again: the first question is asking you to prove the existence of a solution to some differential equation, you try for thirty minutes; you get no where; then you move on to the next question and they ask you to construct a solution to the same differential equation while satisfying certain initial conditions. All in all, I'm at a loss as to how to approach this.

To motivate why I care about the construction of $\mathcal{P}$ is a little hard. I've always been very interested in analytically continuing sequences while preserving the recursion of the sequence. I've done so many times, to the weirdest recursions I could think up, but this case is a bit more troublesome.

I thought it would be cool to have an example of a nice number theoretic function that has an extension to the complex plane. The partition function is a nice example because it satisfies a recursion that is easy to sub complex numbers into. But, the more I looked at it, the more restrictions I had to place on $\mathcal{P}$. The more my toolset dissipated. The more I realized I was out of my depth. I was originally hoping for a nice entire function bounded in a certain way, but when doing math, it's very often you have to settle for something less than what you wished for.