Consider the famous formula of Rademacher (actually Hardy, Ramanujan, and Rademacher):

$$p(n) = \frac{1}{\pi \sqrt{2}} \sum_{k=1}^{\infty} \sqrt{k}\ A_k(n)\ F_k'(n)$$ $$A_k(n) = \sum_{0 \le m < k, \gcd(m, k) = 1} e^{i\pi\left(s(m, k) - 2nm/k\right)}$$ $$F_k(x) = \frac{1}{\sqrt{x - \frac{1}{24}}} \sinh\left(\frac{\pi}{k} \sqrt{\frac{2}{3}\left(x - \frac{1}{24}\right)}\right)$$

and $s(m, k)$ is the Dedekind sum given by

$$s(m, k) = \sum_{n=1}^{k} \left(\left(\frac{n}{k}\right)\right)\left(\left(\frac{mn}{k}\right)\right)$$

where $((x))$ is the sawtooth function

$$((x)) = \begin{cases} x - \lfloor x \rfloor - \frac{1}{2}, &\mbox{if } x \in \mathbb{R} \setminus \mathbb{Z}\\ 0, &\mbox{if }x \in \mathbb{Z} \end{cases}$$.

(see http://en.wikipedia.org/wiki/Partition_%28number_theory%29#Partition_function)

$p(n)$ is the $n$th partition number. But here's the interesting thing: this formula also actually seems to work not just for natural, but real and complex values of $n$ as well! So we could perhaps think of Rademacher's formula as giving an extension of the partition-number function to real and complex indices, much as how the gamma function extends the factorial. Albeit, however, it is complex-valued at real indices.

But the question I have is, where does this converge, when the range of $n$ is expanded from $\mathbb{N}$ to $\mathbb{C}$? It seems to converge okay for real values of $n$, and also those for complex $n$ (perhaps should now be called $z$?) with negative imaginary part. But what about with positive imaginary part? The formula is slow, and numerical experiments with $n = 2i$ don't seem to help. It looks like it diverges, but I'm not totally sure. This is a very complicated series formula, and I'm not sure where one would even begin to analyze it to determine the region of convergence.


A good place to start would be to look at the book titled "The Theory of Partitions" by George E. Andrews (Cambridge Univ. Press). Chapter 5 is devoted to a derivation of the Hardy-Ramanujan-Rademacher formula using the circle method. Just go through his arguments where he obtains an integral representation for F_k'(n) and check to see the region in the complex n-plane where the integral is convergent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.