I'm hesitant to post these pictures, as people might not understand. We are taking \(z^{16} =-1\) sample points. And we are writing:
\[
H(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{2^n}\\
\]
And,
\[
G(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^{n^2}} \frac{1}{2^n}\\
\]
I opted for higher res pictures of these objects; just because they encompass more chaos. But they are graphed over the same box domain: \(|\Re(z)| < 2\) and \(|\Im(z)|<2\).
Here is \(H\):
Here is \(G\):
These functions have a reflection formula... And that's my main point....
I'm going to take \(z^{40} = -1\) amount of sample points; and make a much bigger graph, with even higher res; but just for \(H(z)\). This will take 5-6 hours at best...
In the mean time, here is \(z^{40} =-1\) and a hi res graph of Caleb's original function...
\[
H(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{2^n}\\
\]
And,
\[
G(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^{n^2}} \frac{1}{2^n}\\
\]
I opted for higher res pictures of these objects; just because they encompass more chaos. But they are graphed over the same box domain: \(|\Re(z)| < 2\) and \(|\Im(z)|<2\).
Here is \(H\):
Here is \(G\):
These functions have a reflection formula... And that's my main point....
I'm going to take \(z^{40} = -1\) amount of sample points; and make a much bigger graph, with even higher res; but just for \(H(z)\). This will take 5-6 hours at best...
In the mean time, here is \(z^{40} =-1\) and a hi res graph of Caleb's original function...