Okay........
I have figured out the formula for:
\[
f_m(\theta ; x) = \sum_{k=0}^\infty B^m_k(\theta) x^k\\
\]
And the formula for:
\[
F^m(\theta;x) = \sum_{k=1}^{\infty} F_{-k}^m(\theta) x^k
\]
The values \(B^m_k\) are only slightly different than \(F_{-k}^m(\theta)\)...
I need to sleep on this a bit, just to make sure everything is okay
My code is fucking perfect though!
EDIT:
Okay, so the way I see it is that:
\[
B^m_k(\theta) = e^{i\theta k} \sum_{d|k} h_m(d)e^{-i\theta k/d} \frac{(-1)^{d+1}}{2^{k/d}}\\
\]
Where:
\[
h_m(d) = \frac{1}{m-1!}\prod_{i=1}^{m-1} (d+i-1)\\
\]
Then for \(F_{-k}^m\) we have the exact same formula; but we have to take a slightly different divisor sum. It is technically the exact same formula; but we have to use \(-k\) and we are taking divisors \(d\) which are negative; so that:
\[
F_{-k}^m = e^{-i\theta k} \sum_{d|-k} h_m(d)e^{i\theta k/d} \frac{(-1)^{d+1}}{2^{-k/d}}\\
\]
Or if you prefer to write this with positive divisors:
\[
F_{-k}^m = -e^{-i \theta k} \sum_{d|k} h_m(-d)e^{-i\theta k/d} \frac{(-1)^{d+1}}{2^{k/d}}
\]
So that actually; TECHNICALLY:
\[
F_{-k}^m(\theta) = -B_{-k}^m(\theta)
\]
But the term \(h_m(-d)\) produces a lot of chaos in this formula; because it is zero for \(d < m-1\). So I believe your intuition is definitely correct. I think we will get that:
\[
\sum_{k=0}^\infty B_k^m(\theta) x^k = -\sum_{k=1}^\infty F_{-k}^m(\theta) x^{-k}\\
\]
Or something close to this........
Alright; I'm off to study the holy grail now. Let \(\mathcal{M} = \{m_n\}_{n=0}^\infty\) be a sequence of natural numbers \(m_n \in \mathbb{N}\) for \(m_n \ge 0\). And look at the generalized Caleb function:
\[
f_{\mathcal{M}}(\theta ; x) = \sum_{n=0}^\infty \frac{z^n}{\left(1+e^{i\theta n}z^n\right)^{m_n}} \frac{1}{2^n}\\
\]
If it works for this function; I think it'll work for every function. And by what I'm seeing so far; I see no reason this shouldn't produce the exact same result. I think it will be important to introduce arbitrary sequences \(a_n\) now, rather than \(1/2^n\)...
WAIT!
I think it's just:
\[
B^{\mathcal{M}}_k(\theta) = e^{i\theta k} \sum_{d|k} h_{m(k/d)}(d)e^{-i\theta k/d} \frac{(-1)^{d+1}}{2^{k/d}}\\
\]
I'll double check this though. The same formula should pop out for \(F_{-k}^{\mathcal{M}}\)....
UPDATE!!!!!!!!!
Okay, lets let \(m(n) : \mathbb{N} \to \mathbb{N}\). Let's let \(a_n : \mathbb{N} \to \mathbb{C}\), but additionally let's assume that:
\[
\sum_{n=0}^\infty |a_n| < \infty\\
\]
Let's define the generalized Lambert function as:
\[
L(\theta; z) = \sum_{n=0}^\infty a_n \frac{z^n}{\left(1+e^{i\theta n}z^n\right)^{m(n)}}\\
\]
Let:
\[
\ell_k = \frac{L^{(k)}(\theta ; 0)}{k!} = e^{i\theta k} \sum_{d | k} e^{-i\theta k/d}a_{k/d} h_{m(k/d)}(d)(-1)^{d+1}
\]
Then:
\[
L(\theta ; z) = \sum_{k=0}^\infty \ell_k z^k\\
\]
Let:
\[
P_k = \frac{1}{2\pi i} \int_{|z| = 2} L(\theta ; z) z^{k-1}\,dz\\
\]
for \(k > 0\). Then:
\[
P_k = e^{-i\theta k} \sum_{d | k} a_{k/d} e^{-i\theta k/d}h_{m(k/d)}(-d)(-1)^{d} = \ell_{-k}\\
\]
This is about how general I can make it. I will now work on showing that your identity that we can reconstruct \(L\) using only the coefficients \(P_k\)--as the coefficients of a Laurent series. We'll see how it goes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
This is super cool; because we don't need Ramanujan at all. But it's doing everything Ramanujan does! AWesome!
I have figured out the formula for:
\[
f_m(\theta ; x) = \sum_{k=0}^\infty B^m_k(\theta) x^k\\
\]
And the formula for:
\[
F^m(\theta;x) = \sum_{k=1}^{\infty} F_{-k}^m(\theta) x^k
\]
The values \(B^m_k\) are only slightly different than \(F_{-k}^m(\theta)\)...
I need to sleep on this a bit, just to make sure everything is okay
My code is fucking perfect though!
EDIT:
Okay, so the way I see it is that:
\[
B^m_k(\theta) = e^{i\theta k} \sum_{d|k} h_m(d)e^{-i\theta k/d} \frac{(-1)^{d+1}}{2^{k/d}}\\
\]
Where:
\[
h_m(d) = \frac{1}{m-1!}\prod_{i=1}^{m-1} (d+i-1)\\
\]
Then for \(F_{-k}^m\) we have the exact same formula; but we have to take a slightly different divisor sum. It is technically the exact same formula; but we have to use \(-k\) and we are taking divisors \(d\) which are negative; so that:
\[
F_{-k}^m = e^{-i\theta k} \sum_{d|-k} h_m(d)e^{i\theta k/d} \frac{(-1)^{d+1}}{2^{-k/d}}\\
\]
Or if you prefer to write this with positive divisors:
\[
F_{-k}^m = -e^{-i \theta k} \sum_{d|k} h_m(-d)e^{-i\theta k/d} \frac{(-1)^{d+1}}{2^{k/d}}
\]
So that actually; TECHNICALLY:
\[
F_{-k}^m(\theta) = -B_{-k}^m(\theta)
\]
But the term \(h_m(-d)\) produces a lot of chaos in this formula; because it is zero for \(d < m-1\). So I believe your intuition is definitely correct. I think we will get that:
\[
\sum_{k=0}^\infty B_k^m(\theta) x^k = -\sum_{k=1}^\infty F_{-k}^m(\theta) x^{-k}\\
\]
Or something close to this........
Alright; I'm off to study the holy grail now. Let \(\mathcal{M} = \{m_n\}_{n=0}^\infty\) be a sequence of natural numbers \(m_n \in \mathbb{N}\) for \(m_n \ge 0\). And look at the generalized Caleb function:
\[
f_{\mathcal{M}}(\theta ; x) = \sum_{n=0}^\infty \frac{z^n}{\left(1+e^{i\theta n}z^n\right)^{m_n}} \frac{1}{2^n}\\
\]
If it works for this function; I think it'll work for every function. And by what I'm seeing so far; I see no reason this shouldn't produce the exact same result. I think it will be important to introduce arbitrary sequences \(a_n\) now, rather than \(1/2^n\)...
WAIT!
I think it's just:
\[
B^{\mathcal{M}}_k(\theta) = e^{i\theta k} \sum_{d|k} h_{m(k/d)}(d)e^{-i\theta k/d} \frac{(-1)^{d+1}}{2^{k/d}}\\
\]
I'll double check this though. The same formula should pop out for \(F_{-k}^{\mathcal{M}}\)....
UPDATE!!!!!!!!!
Okay, lets let \(m(n) : \mathbb{N} \to \mathbb{N}\). Let's let \(a_n : \mathbb{N} \to \mathbb{C}\), but additionally let's assume that:
\[
\sum_{n=0}^\infty |a_n| < \infty\\
\]
Let's define the generalized Lambert function as:
\[
L(\theta; z) = \sum_{n=0}^\infty a_n \frac{z^n}{\left(1+e^{i\theta n}z^n\right)^{m(n)}}\\
\]
Let:
\[
\ell_k = \frac{L^{(k)}(\theta ; 0)}{k!} = e^{i\theta k} \sum_{d | k} e^{-i\theta k/d}a_{k/d} h_{m(k/d)}(d)(-1)^{d+1}
\]
Then:
\[
L(\theta ; z) = \sum_{k=0}^\infty \ell_k z^k\\
\]
Let:
\[
P_k = \frac{1}{2\pi i} \int_{|z| = 2} L(\theta ; z) z^{k-1}\,dz\\
\]
for \(k > 0\). Then:
\[
P_k = e^{-i\theta k} \sum_{d | k} a_{k/d} e^{-i\theta k/d}h_{m(k/d)}(-d)(-1)^{d} = \ell_{-k}\\
\]
This is about how general I can make it. I will now work on showing that your identity that we can reconstruct \(L\) using only the coefficients \(P_k\)--as the coefficients of a Laurent series. We'll see how it goes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
This is super cool; because we don't need Ramanujan at all. But it's doing everything Ramanujan does! AWesome!
