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

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


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

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.

Any comments, partial answers, answers, suggestions, or edits are greatly welcomed. Thanks in advance to anyone who gives this a second thought.

Regards, James.

  • $\begingroup$ You might want to take a look at Justin Hsu's Stanford senior thesis: mathematics.stanford.edu/wp-content/uploads/2013/08/… $\endgroup$
    – Igor Rivin
    Jan 12, 2018 at 3:03
  • $\begingroup$ @IgorRivin I don't mean to be dense, but from what I can tell by reading the first few pages and a few minutes of perusing is that this is an overview of Rademacher's formula for the partition function with some modular function stuff stuffed in between. Is there a specific part of the paper I should be interested in which pertains to my question? Thanks, James. $\endgroup$
    – user78249
    Jan 12, 2018 at 3:27
  • 1
    $\begingroup$ See mathoverflow.net/questions/366733. $\endgroup$ Jul 28, 2020 at 23:26


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