THANK YOU, CALEB!
I AM AWARE OF EVERYTHING IN YOUR POST. IT MAY BE CONTINUOUS AT A BOUNDARY POINT! IT MAY EVEN BE COMPLEX DIFFERENTIABLE!
BUT!!!!!!!!!!!!!!!!!!!!!! IT IS NOT COMPLEX DIFFERENTIABLE ON AN OPEN DOMAIN!
Funny you mention \(1/n^2\). Let me take the function:
\[
f(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{(n+1)^2}\\
\]
Then MY FORMULA STILL WORKS!!!!!
My formula doesn't care that \(1/2^n\) goes to zero fast; All I care is that \(G(n) \to 0\) and:
\[
\sum_{n=0}^\infty |G(n)| < \infty\\
\]
Then:
\[
f_G(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} G(n)\\
\]
Satisfies the reflection formula. Let's make it even more complex. Let's say \(m(n) : \mathbb{N} \to \mathbb{N}\), so instead of \(n \mapsto n\), we have \(n \mapsto m(n)\). So we write:
\[
f_G^m(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^{m(n)}} G(n)\\
\]
THIS SATISFIES THE SAME REFLECTION FORMULA!
The fact you are noticing values in which this object is finite/differentiable along \(|z| = 1\), is entirely expected. At every value \(|z| = 1\) such that \(z \neq e^{\pi i/ n}e^{2\pi i \frac{k}{n}}\) for all \(n \ge 1\) and \(0 \le k < n\)-- this object is differentiable.
The point, and maybe, I'll chop it up to you not understanding some subtleties. Just because \(f(z)\) is complex differentiable at \(z_0\), does not mean it is holomorphic at \(z_0\). For \(f(z)\) to be holomorphic at \(z_0\), we need \(f(z)\) to be complex differentiable for \(|z-z_0| < \delta\) for some \(\delta> 0\). This is a SUPER SUBTLE DIFFERENCE!
What you are detailing is that \(f(z_0)\) and \(f'(z_0)\) exist! Which, all the power to you! That's correct as fuck! But HOLOMORPHY, is a stricter requirement. So when I say it is NOWHERE HOLOMORPHIC on \(|z| = 1\), I am correct. BUT IT ALLOWS, for \(f(z_0)\) and \(f'(z_0)\) to still exist!
I am entirely on board with you, Caleb! No where in this post am I trying to disagree. I am just clarifying terminology. This function is NOWHERE HOLOMORPHIC for \(|z| = 1\). It can be pointwise differentiable! BUT NOT HOLOMORPHIC! At least, assuming that there are dense poles about \(z^n = -1\)....
I AM AWARE OF EVERYTHING IN YOUR POST. IT MAY BE CONTINUOUS AT A BOUNDARY POINT! IT MAY EVEN BE COMPLEX DIFFERENTIABLE!
BUT!!!!!!!!!!!!!!!!!!!!!! IT IS NOT COMPLEX DIFFERENTIABLE ON AN OPEN DOMAIN!
Funny you mention \(1/n^2\). Let me take the function:
\[
f(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{(n+1)^2}\\
\]
Then MY FORMULA STILL WORKS!!!!!
My formula doesn't care that \(1/2^n\) goes to zero fast; All I care is that \(G(n) \to 0\) and:
\[
\sum_{n=0}^\infty |G(n)| < \infty\\
\]
Then:
\[
f_G(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} G(n)\\
\]
Satisfies the reflection formula. Let's make it even more complex. Let's say \(m(n) : \mathbb{N} \to \mathbb{N}\), so instead of \(n \mapsto n\), we have \(n \mapsto m(n)\). So we write:
\[
f_G^m(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^{m(n)}} G(n)\\
\]
THIS SATISFIES THE SAME REFLECTION FORMULA!
The fact you are noticing values in which this object is finite/differentiable along \(|z| = 1\), is entirely expected. At every value \(|z| = 1\) such that \(z \neq e^{\pi i/ n}e^{2\pi i \frac{k}{n}}\) for all \(n \ge 1\) and \(0 \le k < n\)-- this object is differentiable.
The point, and maybe, I'll chop it up to you not understanding some subtleties. Just because \(f(z)\) is complex differentiable at \(z_0\), does not mean it is holomorphic at \(z_0\). For \(f(z)\) to be holomorphic at \(z_0\), we need \(f(z)\) to be complex differentiable for \(|z-z_0| < \delta\) for some \(\delta> 0\). This is a SUPER SUBTLE DIFFERENCE!
What you are detailing is that \(f(z_0)\) and \(f'(z_0)\) exist! Which, all the power to you! That's correct as fuck! But HOLOMORPHY, is a stricter requirement. So when I say it is NOWHERE HOLOMORPHIC on \(|z| = 1\), I am correct. BUT IT ALLOWS, for \(f(z_0)\) and \(f'(z_0)\) to still exist!
I am entirely on board with you, Caleb! No where in this post am I trying to disagree. I am just clarifying terminology. This function is NOWHERE HOLOMORPHIC for \(|z| = 1\). It can be pointwise differentiable! BUT NOT HOLOMORPHIC! At least, assuming that there are dense poles about \(z^n = -1\)....