We are not matching the derivatives on the boundary!!!!!!!!!!!! I know it might look like that. But that's not what we want. I mean, \(1/z^2\) isn't continuous at \(0\)...
We are talking about a function \(L(z) : \mathbb{C}/\mathcal{U} \to \mathbb{C}\) where \(\mathcal{U}\) is the unit circle. It is discontinuous on this circle. That's not what's in question. The graphs you showed should behave like that. Also, the reason that your graph looks different--I have done a super precision version. You can't get these with MatLab, or other graphing software. I did the maximal amount of poles my computer could take. And I graphed it raw. Which means it'll look slightly different than the result you graphed. Because these can be computer intensive results to calculate. And something like Desmos/Matlab may default to a lower precision, to save CPU exhaustion. I'm making my CPU run marathons to graph this. Additionally; I am using mike3's colour schema--which is a tad different than a standard colour schema.
You have turned me on to these results; and perhaps I haven't made it clear that I am proving things you are trying to prove.
Let's define:
\[
c_m^{k} = \frac{1}{(m-1)!}\sum_{d \vert\ k} (-1)^{d+1} \frac{\prod_{i=0}^{m-2}(d+i)}{2^{k/d}}\\
\]
Where \(k \in \mathbb{Z}\); and when we take \(k=dn\), the value \(d\) is negative if \(k\) is negative. This can be written more clearly that \(k/d\) is always positive. This is how we take this divisor sum.
Then,
\[
\begin{align*}
\sum_{k=1}^\infty c_m^{k} z^k &= \sum_{n=1}^\infty \frac{z^n}{(1+z^n)^m}\frac{1}{2^n}\,\,\text{while}\,\,|z| < 1\\
-\sum_{k=1}^\infty c_m^{-k} z^{-k} &= \sum_{n=1}^\infty \frac{z^n}{(1+z^n)^m}\frac{1}{2^n}\,\,\text{while}\,\,|z| > 1\\
\end{align*}
\]
This is a perfectly, one-off, defined arithmetic function. It satisfies your curious functional relationship:
\[
\sum_{k=0}^\infty f(k)z^k = -\sum_{k=1}^\infty f(-k)z^{-k}\\
\]
It does so without referencing Ramanujan. While still; being written as an analytic expression.
This is a very deep result, Caleb. I am calling the function:
\[
c(z) = \sum_{n=0}^\infty \frac{z^n}{1+z^n} \frac{1}{2^n}\\
\]
The Caleb function. And it's a very good jumping off point, especially because the coefficients are geometrically converging. In my head I think of m'th order Caleb functions; which are just:
\[
c^m(z) = \sum_{n=0}^\infty \frac{z^n}{(1+z^n)^m} \frac{1}{2^n}\\
\]
There is so much more going on here; that if we restrict:
\[
L(z) = \infty\,\,\text{only when}\,\,z^n = -1\,\,\text{for some}\,\, n\\
\]
Then \(L(z)\) suffers this same reflection formula. We just have to express the residues at \(z^n = -1\) using a superposition of Lambert functions...
I plan to write this much more detailed. I apologize if I'm not making any sense. I'm prioritizing writing the paper now. But I have it figured out; such that we can generalize the behaviour of Caleb's function as \(z \mapsto 1/z\) to any function \(L(z) : \mathbb{C}/\mathcal{U} \to \mathbb{C}\)... At this point in time I am confident with my result; depending on two conditions.
1.) The function \(L(z)\) solely has singularities when \(|z| = 1\). These singularities are poles, just that, poles--not essential singularities/branching points/logarithmic singularities. If \(L(q) = \infty\) then, \(L(q+h) = O(1/h^m)\).
2.) The singularities can only occur at \(z =q\) where: \(|q| = 1\), and \(q^n = -1\). This is for any \(n > 0\).
Given these two conditions, we have the answer to the question you we're hinting at:
\[
-\sum_{k=1}^\infty L^{(-k)}(0)z^{k} = L(1/z) - L(\infty)\\
\]
While also satisfying:
\[
\sum_{k=1}^\infty L^{(k)}(0)z^k = L(z) - L(0)\\
\]
We can freely swap these variables; and every function with a dense amount of singularities on \(|z| =1\)--where the only singularities are at \(z^n =-1\)..... then there's your reflection formula.
Sincere regards, James
BONUS!
Here is the function:
\[
F(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{2^n}\\
\]
Which satisfies this reflection formula...
Graphed in a super hi res manner:
We are talking about a function \(L(z) : \mathbb{C}/\mathcal{U} \to \mathbb{C}\) where \(\mathcal{U}\) is the unit circle. It is discontinuous on this circle. That's not what's in question. The graphs you showed should behave like that. Also, the reason that your graph looks different--I have done a super precision version. You can't get these with MatLab, or other graphing software. I did the maximal amount of poles my computer could take. And I graphed it raw. Which means it'll look slightly different than the result you graphed. Because these can be computer intensive results to calculate. And something like Desmos/Matlab may default to a lower precision, to save CPU exhaustion. I'm making my CPU run marathons to graph this. Additionally; I am using mike3's colour schema--which is a tad different than a standard colour schema.
You have turned me on to these results; and perhaps I haven't made it clear that I am proving things you are trying to prove.
Let's define:
\[
c_m^{k} = \frac{1}{(m-1)!}\sum_{d \vert\ k} (-1)^{d+1} \frac{\prod_{i=0}^{m-2}(d+i)}{2^{k/d}}\\
\]
Where \(k \in \mathbb{Z}\); and when we take \(k=dn\), the value \(d\) is negative if \(k\) is negative. This can be written more clearly that \(k/d\) is always positive. This is how we take this divisor sum.
Then,
\[
\begin{align*}
\sum_{k=1}^\infty c_m^{k} z^k &= \sum_{n=1}^\infty \frac{z^n}{(1+z^n)^m}\frac{1}{2^n}\,\,\text{while}\,\,|z| < 1\\
-\sum_{k=1}^\infty c_m^{-k} z^{-k} &= \sum_{n=1}^\infty \frac{z^n}{(1+z^n)^m}\frac{1}{2^n}\,\,\text{while}\,\,|z| > 1\\
\end{align*}
\]
This is a perfectly, one-off, defined arithmetic function. It satisfies your curious functional relationship:
\[
\sum_{k=0}^\infty f(k)z^k = -\sum_{k=1}^\infty f(-k)z^{-k}\\
\]
It does so without referencing Ramanujan. While still; being written as an analytic expression.
This is a very deep result, Caleb. I am calling the function:
\[
c(z) = \sum_{n=0}^\infty \frac{z^n}{1+z^n} \frac{1}{2^n}\\
\]
The Caleb function. And it's a very good jumping off point, especially because the coefficients are geometrically converging. In my head I think of m'th order Caleb functions; which are just:
\[
c^m(z) = \sum_{n=0}^\infty \frac{z^n}{(1+z^n)^m} \frac{1}{2^n}\\
\]
There is so much more going on here; that if we restrict:
\[
L(z) = \infty\,\,\text{only when}\,\,z^n = -1\,\,\text{for some}\,\, n\\
\]
Then \(L(z)\) suffers this same reflection formula. We just have to express the residues at \(z^n = -1\) using a superposition of Lambert functions...
I plan to write this much more detailed. I apologize if I'm not making any sense. I'm prioritizing writing the paper now. But I have it figured out; such that we can generalize the behaviour of Caleb's function as \(z \mapsto 1/z\) to any function \(L(z) : \mathbb{C}/\mathcal{U} \to \mathbb{C}\)... At this point in time I am confident with my result; depending on two conditions.
1.) The function \(L(z)\) solely has singularities when \(|z| = 1\). These singularities are poles, just that, poles--not essential singularities/branching points/logarithmic singularities. If \(L(q) = \infty\) then, \(L(q+h) = O(1/h^m)\).
2.) The singularities can only occur at \(z =q\) where: \(|q| = 1\), and \(q^n = -1\). This is for any \(n > 0\).
Given these two conditions, we have the answer to the question you we're hinting at:
\[
-\sum_{k=1}^\infty L^{(-k)}(0)z^{k} = L(1/z) - L(\infty)\\
\]
While also satisfying:
\[
\sum_{k=1}^\infty L^{(k)}(0)z^k = L(z) - L(0)\\
\]
We can freely swap these variables; and every function with a dense amount of singularities on \(|z| =1\)--where the only singularities are at \(z^n =-1\)..... then there's your reflection formula.
Sincere regards, James
BONUS!
Here is the function:
\[
F(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{2^n}\\
\]
Which satisfies this reflection formula...
Graphed in a super hi res manner: