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

[Caleb]

I agree with james.

The residue at oo is linked to those at 0.

However in case our function is more complicated or we do not have a reflection formula ...
Im not sure.

Maybe something like

f(x) + h(f(1/x)) = g(x)

Can be used and considered a reflection kinda.

regards

tommy1729

Hey, Tommy;

I sent Caleb the PDF, not you yet; but "reflection formula" is more a term I'm using loosely. It's not literally \(f(x) = f(1/x)\) or something like this. It's more so we can use \(f(x)\) to construct \(f(1/x)\) and vice versa. Which is why I refer to it as a reflection formula. Essentially; we can reconstruct \(f(x)\) about \(x =0\) using the taylor coefficients of \(f(x)\) about \(x =\infty\)--I just like using "reflection formula" as a term; because it relates to Lambert series; and generalizes the simple reflection formula you see. I'm not actually just saying \(f(1/x) = h(f(x))\) or something!

I haven't read all the way through the paper yet, but I'm guessing that the type of residues I'm considering here aren't quite covered under the cased you consider. These aren't your usual residue at ∞, they are more like parts or sections of a residue. Here's what it looks like if you map from z -> 1/z, so that the residues at ∞ get moved to zero. Here, for example, is what a countour looks like that picks up the residue
   
You can't use any simple for the the Cauchy integral formula here, because you can't close the contour. Its a bit interesting, because your integrating over a function that is analytic inside the whole region, except on only a single point at 0, and there's no way around that point, i.e. every contour needs to include 0 to be closed and integrate over the right region.