Okay, let's keep the same notations as before, but write:
\[
F_k^\theta = \frac{1}{2\pi i} \int_{|z|=2} f_\theta(z) \frac{dz}{z^{k+1}}\\
\]
The value:
\[
F_0^\theta = \frac{1}{2} + e^{i\theta}\sum_{n=1}^\infty \frac{e^{-i\theta n}}{2^n} = \frac{1}{2} + \frac{1}{2}\frac{1}{1-\frac{e^{-i\theta}}{2}}\\
\]
This is observed because, for \(n \ge 1\):
\[
\frac{1}{2\pi i} \int_{|z|=2} \frac{z^n}{1+e^{i\theta n} z^n} \frac{1}{z}\,dz = e^{i\theta (1-n)}\\
\]
Which is just by linear substitution.
We know that:
\[
F_k^\theta = 0\\
\]
For \(k \ge 1\) just as well. (Multiply the left hand side by \(e^{i\theta(k+1-n)}\) and we're reduced to the same integral as before). The real trouble happens for \(k \le -1\). If we recall our value \(b_{kn}\); then:
\[
\frac{1}{2\pi i} \int_{|z|=2} \frac{z^n}{1+e^{i\theta n} z^n} \frac{1}{z^{k+1}}\,dz =\frac{e^{i\theta(k+1-n)}}{2\pi i} \int_{|z|=2} \frac{e^{i\theta n}z^n}{1+e^{i\theta n} z^n} \frac{1}{e^{i\theta (k+1)}z^{k+1}}\,dz = -e^{i\theta(k+1-n)} b_{kn}\\
\]
................
Now let's sum across \(n\) with the \(1/2^n\) weight; recalling that \(k \le -1\)...
\[
F_k^\theta = -\sum_{n=0}^\infty e^{i\theta(k+1-n)} \frac{b_{-kn}}{2^n}\\
\]
This equals:
\[
-e^{i\theta(k+1)} \sum_{n=0}^\infty e^{-i\theta n} \frac{b_{-kn}}{2^n}
\]
This is definitely not so obviously related to \(-f_\theta^{(-k)}(0)/(-k)!\). This is some kind of spectral decomposition...
I'm really not sure...
But now when we write:
\[
\sum_{k\in\mathbb{Z}} F_k^{\theta}x^k = F_0^\theta -\sum_{k=1}^\infty e^{i\theta(1-k)} x^{-k}\sum_{n=0}^\infty e^{-i\theta n} \frac{b_{-kn}}{2^n}\\
\]
This still looks A LOT like \(f(1/x)\)--but there's some kind of fourier magic I can't parse....
The real magic is that:
\[
\sum_{k=1}^\infty F_{-k}^{\theta} x^k \approx f_\theta(0) - f_\theta(x)\\
\]
So the same result seems to be popping out. I'm just not sure why yet. But I think I see why... We have to monitor the order of \(\theta\) though... this is really weird. But it's looking very very similar.
.................................................................................................................
Oh..... I'm a dumb ass it's a rotation.... This does sum to some thing close to \(f_\theta(x)\); but we apply a rotation in \(x\) before the sums line up. I'll explain in a bit. Need to double check everything....
Okay!
For example if you take \(\theta = \pi/2\), then:
\[
\sum_{k\in\mathbb{Z}}^\infty F_k^{\pi/2} x^k = C - f_{\pi/2}(-1/x)\\
\]
So the rotation is only one rotation. I'm not sure of the general rule. But i'll update as I figure it out. I'm not sure why \(i^4 = 1\) only induces a negative... So this is definitely a brain scratcher.... I thought it was going to be some \(if_{\pi/2}(-i/x)\) or something--but it appears to be much different...
The best I can say for the moment, is that:
\[
|F_k^\theta| = \left| \frac{f_\theta^{(-k)}(0)}{(-k)!}\right|\\
\]
And that these are rational rotations at worse. And they are some what related to the initial \(\theta\) value...
FUCK I'm an idiot.
Take the value:
\[
F_{-1}^{\theta}/f^{(1)}_\theta(0) = \zeta_\theta\\
\]
Then:
\[
\sum_{k\in \mathbb{Z}} F_k^{\theta} x^k = C - f_\theta(\zeta_\theta/x)\\
\]
EASY FUCKING BREEZY!!!
This actually makes perfect sense too!
All the pari code lines up, this is the answer.
So you're formula still holds, Caleb. You just need to insert a rotation depending on \(\theta\). That's it! Makes perfect sense too...
\[
F_k^\theta = \frac{1}{2\pi i} \int_{|z|=2} f_\theta(z) \frac{dz}{z^{k+1}}\\
\]
The value:
\[
F_0^\theta = \frac{1}{2} + e^{i\theta}\sum_{n=1}^\infty \frac{e^{-i\theta n}}{2^n} = \frac{1}{2} + \frac{1}{2}\frac{1}{1-\frac{e^{-i\theta}}{2}}\\
\]
This is observed because, for \(n \ge 1\):
\[
\frac{1}{2\pi i} \int_{|z|=2} \frac{z^n}{1+e^{i\theta n} z^n} \frac{1}{z}\,dz = e^{i\theta (1-n)}\\
\]
Which is just by linear substitution.
We know that:
\[
F_k^\theta = 0\\
\]
For \(k \ge 1\) just as well. (Multiply the left hand side by \(e^{i\theta(k+1-n)}\) and we're reduced to the same integral as before). The real trouble happens for \(k \le -1\). If we recall our value \(b_{kn}\); then:
\[
\frac{1}{2\pi i} \int_{|z|=2} \frac{z^n}{1+e^{i\theta n} z^n} \frac{1}{z^{k+1}}\,dz =\frac{e^{i\theta(k+1-n)}}{2\pi i} \int_{|z|=2} \frac{e^{i\theta n}z^n}{1+e^{i\theta n} z^n} \frac{1}{e^{i\theta (k+1)}z^{k+1}}\,dz = -e^{i\theta(k+1-n)} b_{kn}\\
\]
................
Now let's sum across \(n\) with the \(1/2^n\) weight; recalling that \(k \le -1\)...
\[
F_k^\theta = -\sum_{n=0}^\infty e^{i\theta(k+1-n)} \frac{b_{-kn}}{2^n}\\
\]
This equals:
\[
-e^{i\theta(k+1)} \sum_{n=0}^\infty e^{-i\theta n} \frac{b_{-kn}}{2^n}
\]
This is definitely not so obviously related to \(-f_\theta^{(-k)}(0)/(-k)!\). This is some kind of spectral decomposition...
I'm really not sure...
But now when we write:
\[
\sum_{k\in\mathbb{Z}} F_k^{\theta}x^k = F_0^\theta -\sum_{k=1}^\infty e^{i\theta(1-k)} x^{-k}\sum_{n=0}^\infty e^{-i\theta n} \frac{b_{-kn}}{2^n}\\
\]
This still looks A LOT like \(f(1/x)\)--but there's some kind of fourier magic I can't parse....
The real magic is that:
\[
\sum_{k=1}^\infty F_{-k}^{\theta} x^k \approx f_\theta(0) - f_\theta(x)\\
\]
So the same result seems to be popping out. I'm just not sure why yet. But I think I see why... We have to monitor the order of \(\theta\) though... this is really weird. But it's looking very very similar.
.................................................................................................................
Oh..... I'm a dumb ass it's a rotation.... This does sum to some thing close to \(f_\theta(x)\); but we apply a rotation in \(x\) before the sums line up. I'll explain in a bit. Need to double check everything....
Okay!
For example if you take \(\theta = \pi/2\), then:
\[
\sum_{k\in\mathbb{Z}}^\infty F_k^{\pi/2} x^k = C - f_{\pi/2}(-1/x)\\
\]
So the rotation is only one rotation. I'm not sure of the general rule. But i'll update as I figure it out. I'm not sure why \(i^4 = 1\) only induces a negative... So this is definitely a brain scratcher.... I thought it was going to be some \(if_{\pi/2}(-i/x)\) or something--but it appears to be much different...
The best I can say for the moment, is that:
\[
|F_k^\theta| = \left| \frac{f_\theta^{(-k)}(0)}{(-k)!}\right|\\
\]
And that these are rational rotations at worse. And they are some what related to the initial \(\theta\) value...
FUCK I'm an idiot.
Take the value:
\[
F_{-1}^{\theta}/f^{(1)}_\theta(0) = \zeta_\theta\\
\]
Then:
\[
\sum_{k\in \mathbb{Z}} F_k^{\theta} x^k = C - f_\theta(\zeta_\theta/x)\\
\]
EASY FUCKING BREEZY!!!
This actually makes perfect sense too!
All the pari code lines up, this is the answer.
So you're formula still holds, Caleb. You just need to insert a rotation depending on \(\theta\). That's it! Makes perfect sense too...
