[MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n))

[Caleb]

Hey, Caleb!

You're right; I don't cover shit like this til' close to the end; and there I void all possible contours like this.

If you take:

\[
z \mapsto e^{1/z}\\
\]

You will notice very similar behaviour; and that is where I cap the lambert reflection formula.

Where you are seeing here is more like:

\[
z \mapsto e^{1/\sqrt{z}}\\
\]

Which has a branching singularity at \(0\) (coincidentally similarly at \(z = \infty\)). The rough draft I sent you is intended solely for singularities which aren't branching. Your graph above is clearly a branching singularity.

That's why I don't want to comment on this new question too much; but it deeply relates to "residue at infinity" as I'm talking about. It's just a branching residue rather than a meromorphic, or essential residue. I'm focusing on \(z \mapsto 1/z^k\) or \(z \mapsto e^{1/z}\), I'm entirely ignoring \(z \mapsto e^{1/\sqrt{z}}\). Not because I can't speak on it, but because I'm less confident, and more unsure what happens here.

Awesome job still!! I love these graphs! Keep em coming!

Sincere Regards, Jame

I'm not sure what your definition of branching singularity is, but I'm not sure the object I have shown fits into that categorization. In particular, it only has a single branch, defined everywhere, unlike \( e^{\frac{1}{\sqrt{z}} \). In particular, if you don't consider \(e^\frac{1}{z}\) to have a branching singularity, then what do you thinking about \(e^\frac{1}{z^2}\). This has the same type of behaviour as this function. In particular, consider a contour of this shape
   
This has the same type of behaviour as the function in the original post, and you can't apply Cauchy's integral theorem to compute it. The behaviour gets more extreme when we increase the power, for instance, at \( e^{\frac{1}{z^10}} \) it looks like this
   
Actually, with these type of contours, you can do something like Cauchy's theorem, but weaker. In particular, there is actually a Cauchy's theorem that I devoloped a while ago that applies to open circles, instead of closed curves. Take \( f(z) = \sum_{n=-\infty}^\infty a_n x^n \). In the case where you take a closed curve (a closed circle in particular), you cancel out every term except \( a_{-1} \) (this is the Cauchy integral formula). But, if you take a semi circle, you cancel out every other term (except a_{-1}). A third of a circle cancels out every 3rd term, and a fourth of a circle cancels out every 4th, etc. You can use this to obtain a divergent series representation for the integral over these badly defined contours, but this series representation has generally seemed useless to me since the series is divergent.

I've found some old notes where I derived the precise formula for half circles (its actually easy to compute for other ones as well, but I haven't gone through the algebra yet), I took \(f(z)=\sum_{n=1}^\infty \frac{a_n}{z^n}\) and then we obtain \(\int_C f(z)dz = a_1\pi i-2ir\sum_{n=1}^\infty \frac{(-1)^n a_{2n}}{(2n-1)r^{2n}}\) if \(C\) is a half-circle from \(\theta=-\frac{\pi}{2}\) to \(\theta = \frac{\pi}{2}\) with radius \(r\). Thus, we may compute the residue as \(\lim_{r \to 0} a_1\pi i-2ir\sum_{n=1}^\infty \frac{(-1)^n a_{2n}}{(2n-1)r^{2n}}\), which allows us to obtain that, in general, the residue picked up by the contour for \( \frac{1}{z^2} t^{\frac{1}{z^2}}\) is equal to \(- \frac{\sqrt{\pi}}{\sqrt{-\ln(t)}}\).