I am going to add my hand at an entire function here.
Let:
\[
f(z,q) = \sum_{n=0}^\infty \frac{z^n}{1+q^nz^n}\frac{1}{2^n}\\
\]
Then:
\[
\frac{\partial}{\partial z}\Big{|}_{z=0} f(z,q) = 1/2\\
\]
Let:
\[
J(q) = 2F_{-1}(q) = \frac{1}{\pi i} \int_{|z| =2} f(z,q)\,dz\\
\]
This function looks modular as fuck.
We get that:
\[
f(0,q)-f(J(q)z,q) = \sum_{k=1}^\infty F_{-k}(q)z^k\\
\]
Oh and I am taking the modular assumption that \(q = e^{i\theta}\). I should've metioned that, lol.
Holy, fuck!
I've been sitting on this formula for a while.
But let's let \(b_{kn} = (-1)^{n/k+1}\) if \(k=nd\) and \(b_{00} = 1/\)--otherwise it's zero. Then Caleb's original function is written as:
\[
f(x) = \sum_{k=0}^\infty\sum_{n=0}^\infty \frac{b_{kn}}{2^n}x^k\\
\]
The thing is; we can reduce this summation.
We only have non zero values when \(k = nd\) for some value \(d \in \mathbb{N}\); where we then get:
\[
f(x) = 1/2 + \sum_{k=1}^\infty x^k \sum_{d \mid k} \frac{(-1)^{d+1}}{2^{k/d}}\\
\]
...........
When we write:
\[
f^{(k)}_\theta(0) =e^{i\theta k} k! \sum_{d \mid k} e^{i\theta k/d}\frac{(-1)^{d+1}}{2^{k/d}}
\]
We have instantly added a summative/ to /multiplicative relationship \(\sum_{d \mid k}\) and added the multiplicative/ to /summative relationship \(e^{i\theta d}\). Rather than multiplying exponentials; we've taken factors and put them in the exponentials....
I'm not saying this is modular. But this is the shit modular relationships are built from.......
I HAVE DEFINITELY MADE SOME NUMERICAL MISTAKES! I'M TRYING TO CARVE A PATH IS ALLL!!!!
EDITTTTTT:::::
\[
\frac{f^{(k)}_\theta(0)}{k!} = e^{i\theta k}\sum_{d \mid k} e^{i\theta k/d}\frac{(-1)^{d+1}}{2^{k/d}}
\]
This is the perfect formula. Only trouble is at \(k=0\); where we have to just sub in \(1/2\). But if you know how technical divisor sums are this is fine.
The function:
\[
J(e^{i\theta}) =-2\sum_{d \mid -1} e^{-i\theta/d}\frac{(-1)^{d+1}}{2^{-1/d}}
\]
Despite this expression being meaningless.......
We have given meaning to it; by writing:
\[
2F_{-1}^{\theta} = J(e^{i\theta})\\
\]
Or....
\[
J(e^{i\theta}) = \frac{1}{\pi i} \int_{|z|=2} f(z,e^{i\theta})\,dz\\
\]
The function \(J\) takes the unit circle to itself. But also takes the upper half plane to itself. There is a boundary on the unit circle; but it's different. It looks fucking modular!
Let:
\[
f(z,q) = \sum_{n=0}^\infty \frac{z^n}{1+q^nz^n}\frac{1}{2^n}\\
\]
Then:
\[
\frac{\partial}{\partial z}\Big{|}_{z=0} f(z,q) = 1/2\\
\]
Let:
\[
J(q) = 2F_{-1}(q) = \frac{1}{\pi i} \int_{|z| =2} f(z,q)\,dz\\
\]
This function looks modular as fuck.
We get that:
\[
f(0,q)-f(J(q)z,q) = \sum_{k=1}^\infty F_{-k}(q)z^k\\
\]
Oh and I am taking the modular assumption that \(q = e^{i\theta}\). I should've metioned that, lol.
Holy, fuck!
I've been sitting on this formula for a while.
But let's let \(b_{kn} = (-1)^{n/k+1}\) if \(k=nd\) and \(b_{00} = 1/\)--otherwise it's zero. Then Caleb's original function is written as:
\[
f(x) = \sum_{k=0}^\infty\sum_{n=0}^\infty \frac{b_{kn}}{2^n}x^k\\
\]
The thing is; we can reduce this summation.
We only have non zero values when \(k = nd\) for some value \(d \in \mathbb{N}\); where we then get:
\[
f(x) = 1/2 + \sum_{k=1}^\infty x^k \sum_{d \mid k} \frac{(-1)^{d+1}}{2^{k/d}}\\
\]
...........
When we write:
\[
f^{(k)}_\theta(0) =e^{i\theta k} k! \sum_{d \mid k} e^{i\theta k/d}\frac{(-1)^{d+1}}{2^{k/d}}
\]
We have instantly added a summative/ to /multiplicative relationship \(\sum_{d \mid k}\) and added the multiplicative/ to /summative relationship \(e^{i\theta d}\). Rather than multiplying exponentials; we've taken factors and put them in the exponentials....
I'm not saying this is modular. But this is the shit modular relationships are built from.......
I HAVE DEFINITELY MADE SOME NUMERICAL MISTAKES! I'M TRYING TO CARVE A PATH IS ALLL!!!!
EDITTTTTT:::::
\[
\frac{f^{(k)}_\theta(0)}{k!} = e^{i\theta k}\sum_{d \mid k} e^{i\theta k/d}\frac{(-1)^{d+1}}{2^{k/d}}
\]
This is the perfect formula. Only trouble is at \(k=0\); where we have to just sub in \(1/2\). But if you know how technical divisor sums are this is fine.
The function:
\[
J(e^{i\theta}) =-2\sum_{d \mid -1} e^{-i\theta/d}\frac{(-1)^{d+1}}{2^{-1/d}}
\]
Despite this expression being meaningless.......
We have given meaning to it; by writing:
\[
2F_{-1}^{\theta} = J(e^{i\theta})\\
\]
Or....
\[
J(e^{i\theta}) = \frac{1}{\pi i} \int_{|z|=2} f(z,e^{i\theta})\,dz\\
\]
The function \(J\) takes the unit circle to itself. But also takes the upper half plane to itself. There is a boundary on the unit circle; but it's different. It looks fucking modular!
