04/12/2023, 12:15 AM
ALRIGHT!
So I no longer need the assumption \(\psi(z + 2\pi i) = \psi(z)\) in my code base. This was a fairly artificial condition, which would serve us fine for the real line; but would show too many errors as we go out. The correct version of \(\psi(z)\) is holomorphic for \(\Im(z) \neq 0\), where it is still analytic on \(\Im(z) = 0\) upto some singularities. Which we can expect to occur around \(0, 1 , e, e^e, e^{e^e},...\)--just as a normal theta function appears...
We can also expect a good amount of branching near these singularities; but still, we have that \(\psi(z)\) is holomorphic for \(\Im(z)>0\) and \(\Im(z) < 0\)--and it's related as \(\overline{\psi(\overline{z})} = \psi(z)\). So that this object is conjugate similar. The value on the real line is still escaping me, and near the real line my code kind of wonks out, but it still says something.
This version of \(\psi\) looks a lot like the first graphs I posted earlier today; but now the numbers line up much better. The earlier graphs only had 1 to 2 digit precision--where as the periodic version had 100 digit precision. This graphs the same thing as the first graphs, but graphs it with 100 digit precision.
So I've managed to write a fairly efficient algorithm to graph the function \(\psi(z+2\pi) = \psi(z) + 2\pi i k\), and find \(k\) in a quick and efficient way. It seems to hover near \(k=0\), but it can jump around a bit and can bite you in the ass if you aren't looking.
This function can take something like:
\[
\psi(e^{100+i100}) = \frac{e^{\psi(100+i100)}}{1+e^{-100-i100}}
\]
And confirm it to 100 digits; and does so without requiring periodicity (it finds the value \(k\)).
Here is \(|\Re(z)| < 5\) and \(|\Im(z)| < 5\). It looks a lot like my original graphs from earlier today, but the numbers are much much much more exact!
So I no longer need the assumption \(\psi(z + 2\pi i) = \psi(z)\) in my code base. This was a fairly artificial condition, which would serve us fine for the real line; but would show too many errors as we go out. The correct version of \(\psi(z)\) is holomorphic for \(\Im(z) \neq 0\), where it is still analytic on \(\Im(z) = 0\) upto some singularities. Which we can expect to occur around \(0, 1 , e, e^e, e^{e^e},...\)--just as a normal theta function appears...
We can also expect a good amount of branching near these singularities; but still, we have that \(\psi(z)\) is holomorphic for \(\Im(z)>0\) and \(\Im(z) < 0\)--and it's related as \(\overline{\psi(\overline{z})} = \psi(z)\). So that this object is conjugate similar. The value on the real line is still escaping me, and near the real line my code kind of wonks out, but it still says something.
This version of \(\psi\) looks a lot like the first graphs I posted earlier today; but now the numbers line up much better. The earlier graphs only had 1 to 2 digit precision--where as the periodic version had 100 digit precision. This graphs the same thing as the first graphs, but graphs it with 100 digit precision.
So I've managed to write a fairly efficient algorithm to graph the function \(\psi(z+2\pi) = \psi(z) + 2\pi i k\), and find \(k\) in a quick and efficient way. It seems to hover near \(k=0\), but it can jump around a bit and can bite you in the ass if you aren't looking.
This function can take something like:
\[
\psi(e^{100+i100}) = \frac{e^{\psi(100+i100)}}{1+e^{-100-i100}}
\]
And confirm it to 100 digits; and does so without requiring periodicity (it finds the value \(k\)).
Here is \(|\Re(z)| < 5\) and \(|\Im(z)| < 5\). It looks a lot like my original graphs from earlier today, but the numbers are much much much more exact!
