(03/02/2023, 07:21 PM)Caleb Wrote: Nice work you have here-- this gives us a unique way to connect these series outside their natural boundary into a function inside the natural boundary!
Also, based on your work, we get the 'Cauchy-like' integral formula
\[ F(x) = a_0 + \int_C \frac{F(z)}{z-\frac{1}{x}}dz\]
Where the contour C is taken outside the natural boundary and the pole at \(z = \frac{1}{x}\) needs to outside the contour integration. Actually, we could do that, or we could only pick up the pole at \(z = \frac{1}{x}\) is pick up no other residue (but then the constant term is a bit more complciated I think). Either way, we now have an easy way to use contours only inside of the natural boundary to compute values on the inside!
Yes!
This was the kind of idea I smelled when I first saw it. Though it definitely turned out different than I originally expected. I imagine something similar will happen in more general instances.
If we let:
\[
F(z) : \mathbb{D}_{|z| < 2}/U \to \mathbb{C}\\
\]
Where there is a wall of singularities along \(|z| = 1\); but other wise everything is holomorphic, we should have something very very similar. I'm going to see if this continues to happen--or if something close to this happens. I do not think your function \(f\) is too particularly special; so something similar should pop out. I'm mostly worried if the wall of singularities needs to be simple poles or not.
I'm going to start by testing this hypothesis on:
\[
f_{\theta}(x) = \sum_{n=0}^\infty \frac{x^n}{1+e^{i\theta n} x^n} \frac{1}{2^n}\\
\]
This will scramble the poles around; but other than that, should keep the same structure as your function--when \(\theta = 0\). If a similar result follows; then we know that this result is true up to permutation of the locations of the wall of singularities.
The next step would be to take:
\[
f_\theta^K(x) = \sum_{n=0}^\infty \frac{x^n}{(1+e^{i\theta n} x^n)^{k_n}} \frac{1}{2^n}\\
\]
for a sequence of natural numbers \(K = \{k_n\}_{n=0}^\infty\). This would allow us to look at when we have second order, third order, etc... poles. This will probably create more problems, but I don't think it'll be too unreasonable.
