Okay, I want to add a stronger result to my last result.
Assume \(|x| \neq 1\). Call the contour \(C\) the parameterization of \(|z| =R\) for \(R\neq 1\). Then for all \(|x| < R\) and \(|x|\neq 1\), we have that:
\[
f(x) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-x}\,dz\\
\]
At this point I have only shown this result for \(|x| < \text{min}(R,1)\). I want to show it for \(|x| < R\).
This is actually pretty simple...
Let's take the summand \(\frac{z^n}{(1+z^n)(z-x)}\)... The value of all the residues of this summand are:
\[
\frac{x^n}{1+x^n} + \frac{1}{n}\sum_{q^n = -1}\frac{q}{q-x}\\
\]
The sum:
\[
\sum_{q^n = -1}\frac{q}{q-x} = \sum_{j = 0}^{n-1} \frac{\sqrt[n]{-1}e^{2\pi i \frac{kj}{n}}}{\sqrt[n]{-1}e^{2\pi i \frac{kj}{n}}-x}\\
\]
So that:
\[
\sum_{j=0}^{n-1} \sum_{m=0}^\infty \sqrt[n]{-1}^{-m} e^{-2\pi i \frac{kjm}{n}}x^m=0\\
\]
Because:
\[
\sum_{j=0}^{n-1} e^{-2\pi i \frac{kjm}{n}} = 0\\
\]
And that, all we end up with is that:
\[
\int_C \frac{z^n}{1+z^n} \frac{dz}{z-x} = \frac{x^n}{1+x^n}\\
\]
So that the wall of singularities disappear even better than I originally thought.
My conclusion is that Caleb's function \(f(x)\) is holomorphic for \(\mathbb{C}/U\) where \(U\) is the unit circle. And that for all disks \(|z| = R\) where \(R \neq 1\)-- we have Cauchy's integral Formula:
For all \(|x| < R\) and \(|x| \neq 1\) and \(R \neq 1\)--where \(C\) is the parameterization of \(|z| =R\):
\[
f(x) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-x}\,dz\\
\]
For all \(|x| < R\)...
Fuck this function is cool as fuck!
This becomes not a question of "analytic continuation"--but whether it satisfies Cauchy's integral formula... This has to do with functional analysis (where advanced interpretations of Cauchy's integral formula exist). I'm super excited by your work, Caleb!
Assume \(|x| \neq 1\). Call the contour \(C\) the parameterization of \(|z| =R\) for \(R\neq 1\). Then for all \(|x| < R\) and \(|x|\neq 1\), we have that:
\[
f(x) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-x}\,dz\\
\]
At this point I have only shown this result for \(|x| < \text{min}(R,1)\). I want to show it for \(|x| < R\).
This is actually pretty simple...
Let's take the summand \(\frac{z^n}{(1+z^n)(z-x)}\)... The value of all the residues of this summand are:
\[
\frac{x^n}{1+x^n} + \frac{1}{n}\sum_{q^n = -1}\frac{q}{q-x}\\
\]
The sum:
\[
\sum_{q^n = -1}\frac{q}{q-x} = \sum_{j = 0}^{n-1} \frac{\sqrt[n]{-1}e^{2\pi i \frac{kj}{n}}}{\sqrt[n]{-1}e^{2\pi i \frac{kj}{n}}-x}\\
\]
So that:
\[
\sum_{j=0}^{n-1} \sum_{m=0}^\infty \sqrt[n]{-1}^{-m} e^{-2\pi i \frac{kjm}{n}}x^m=0\\
\]
Because:
\[
\sum_{j=0}^{n-1} e^{-2\pi i \frac{kjm}{n}} = 0\\
\]
And that, all we end up with is that:
\[
\int_C \frac{z^n}{1+z^n} \frac{dz}{z-x} = \frac{x^n}{1+x^n}\\
\]
So that the wall of singularities disappear even better than I originally thought.
My conclusion is that Caleb's function \(f(x)\) is holomorphic for \(\mathbb{C}/U\) where \(U\) is the unit circle. And that for all disks \(|z| = R\) where \(R \neq 1\)-- we have Cauchy's integral Formula:
For all \(|x| < R\) and \(|x| \neq 1\) and \(R \neq 1\)--where \(C\) is the parameterization of \(|z| =R\):
\[
f(x) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-x}\,dz\\
\]
For all \(|x| < R\)...
Fuck this function is cool as fuck!
This becomes not a question of "analytic continuation"--but whether it satisfies Cauchy's integral formula... This has to do with functional analysis (where advanced interpretations of Cauchy's integral formula exist). I'm super excited by your work, Caleb!
