From \(t =0\) to \(t = 2 \pi\)... Then \(J(e^{it})\) graphs as:
This looks just like a very simple modular kind of function. We only need a single frequency to get it.
I suspect this is because it's only simple poles on the wall. If we have arbitrary order poles; we will get more complication.
Please remember the entire point is that:
\[
f(0,e^{it}) - f(J(e^{it})x,e^{it}) = \sum_{k=1}^{\infty} F_{-k}^{t}x^k\\
\]
Recalling that:
\[
F_{-k}^t = \frac{1}{2\pi i} \int_{|z| =2} f(z,e^{it})z^{k-1}\,dz\\
\]
To remind everyone:
\[
f(x,e^{it}) = \sum_{n=0}^\infty \frac{x^n}{1+e^{itn}x^n} \frac{1}{2^n}\\\
\]
This looks just like a very simple modular kind of function. We only need a single frequency to get it.
I suspect this is because it's only simple poles on the wall. If we have arbitrary order poles; we will get more complication.
Please remember the entire point is that:
\[
f(0,e^{it}) - f(J(e^{it})x,e^{it}) = \sum_{k=1}^{\infty} F_{-k}^{t}x^k\\
\]
Recalling that:
\[
F_{-k}^t = \frac{1}{2\pi i} \int_{|z| =2} f(z,e^{it})z^{k-1}\,dz\\
\]
To remind everyone:
\[
f(x,e^{it}) = \sum_{n=0}^\infty \frac{x^n}{1+e^{itn}x^n} \frac{1}{2^n}\\\
\]
