I am going to explain The Little Circle Method; with the pivotal example. I'm not going to do the rigor, or any of the hard math. I'm just going to give the story board; of how this behaves. This is a very deep subject. I am not a master, or someone who is exceptionally good at these types of questions. But they appear, and I know a fair amount of literature.

We will call the function \(\#n\) as the number of partitions of the number \(n \in \mathbb{N}\). In which:

\[

\begin{align}

\# 0 &= 1\\

\# 1 &= 1\\

\# 2 &= 2\\

\# 3 &= 3\\

\# 4 &= 5\\

&\vdots \\

\#n &= \#\{A \in \{\mathbb{N}\cup \{0\}\}^n\,|\,\sum_{j=1}^n A[j] = n\}\\

\end{align}

\]

The function \(\#n\) is known as the partition function; or the additive partition function. The following result; low and behold; dates to Euler:

\[

\sum_{n=0}^\infty \frac{x^n}{n!} \#n = \prod_{j=1}^\infty \frac{1}{1-x^j}\\

\]

Euler proved his own results about this; which culminate in The Pentagonal Number Theorem. Which honestly, is such a bananas kind of result. You're missing out if you haven't seen it. We won't need that; we just need the above expression.

The goal that Ramanujan, and later, Hardy, set out; was to construct an analytic expression for \(\# n\). The truth is, no analytic expression exists which satisfied any of their constraints. Largely because; they weren't integrable in the manner they hoped. Instead, the best they could do, was find an asymptotic solution.

Begin by calling:

\[

F(z) = \prod_{j=1}^\infty \frac{1}{1-z^j}\\

\]

At every rational point \(z^k = 1\), we have a pole. And therefore, for \(|z| = 1\) we have a dense amount of singularities. There is no beating this. There is no making it better.

Additionally, we have the Euler formula (And For Fuck's Sake, this is why you study Euler!), which is written:

\[

F(z) = \sum_{n=0}^\infty \#n \frac{z^n}{n!}\\

\]

You are trying to guess something similar that Ramanujan was trying to guess. He wanted to write \(\# n \approx P(n)\), and he wanted arbitrary accuracy. (Please recall, I'm just giving a rough history overview, and relating your problem in broad motions).

[Enter LITTLE CIRCLE]

Let us take \(\gamma_\rho(t) : [0,2\pi] \to \rho e^{2 \pi i t}\). Where \(0 < \rho < 1\). Then, we have a formula for \(\#n\) which is written as:

\[

\#n = \frac{1}{2\pi i} \int_{\gamma_\rho} \frac{F(z)}{z^{n+1}}\,dz\\

\]

Where now, this result holds by Cauchy's Integral theorem, and there's really no magic involved.

\[

\frac{1}{2\pi i} \int_{\gamma_\rho} \frac{\sum_{n=0}^\infty c_n z^n}{z^{n+1}}\,dz = c_n\\

\]

Which is just Residue magic...

The idea of the Little Circle method, is to let \(\rho \to 1\). But as we let \(\rho \to 1\), we produce singularities at all the values \(z^k = 1\). And we want to avoid these values. So we make cuts in the perfect circle of \(\gamma_\rho\). We make "little circles", in the contour \(\gamma_\rho\).

What this eventually does (I'm not going to explain it, because it takes 40 pages of hard math to get here, I'm just giving you an idea), is that we get:

\[

\#n = \frac{1}{2\pi i}\int_{\tau_\rho} \frac{F(z)}{z^{n+1}}\,dz + \sum_{k=1}^M H_k(n)\\

\]

The arc \(\tau_\rho\), is a broken circle, with small circles deleted. And the sum is up to \(M\) terms, where \(M = M(\rho)\). And as \(\rho \to 1\), the small circles get smaller, and \(M\to \infty\). Where finally we end up with Ramanujan's first REAL HANDS DOWN UNBELIEVABLE RESULT, that \(\#n\) can be approximated with an exponential series. Because \(H_k(n)\) devolves to an \(O(e^{a_k\sqrt{n}})\) kind of expression.

You are trying to do something similar, Caleb. Whether you see it or not. You are. And I wish you luck. I'm not trying to discourage you. I just want you to know that, studying the asymptotic boundary values of:

\[

G(z)= \sum_{n=0}^\infty z^{n^2}\\

\]

And trying to guess the asymptotics of \(\lambda(n)\) from this--requires appreciating history of these results.

Nothing but love, Caleb.

Just giving my two cents. I'm excited for everything you post. You're going places if you keep working like this! I mean that sincerely! I might seem like a no-one, but I've met a lot of "famous mathematicians"--and I can smell it on you. Keep hitting the books and doing the raw demonstrations you are doing

Sincerely, James

PS:

The first German mathematician to run through this; discovered a faster Exponential square root series than Ramanujan-Hardy did. He also pretty much gave the "reduce to its elements" Little Circle, analysis. So, if you want the more modern, better explanation, study Hans Rademacher. Dude just stumbled upon a better Little Circle analysis. And the Germans do it better