Well, I thought I'd do some level sets while I'm wondering what to do with this.

So here is:

\[

\beta_{1,1}(s) = \Omega_{j=1}^\infty \frac{e^z}{1+e^{j-s}}\,\bullet z\\

\]

Which satisfies \(\beta_{1,1}(s+1) = \frac{e^{\beta_{1,1}(s)}}{1+e^{-s}}\). And next to it we have the function:

\[

\Psi(e^z) = L \Psi(z)\\

\]

And we are drawing the level set:

\[

|\beta'(z)| = |\Psi'(z)|\\

\]

We draw this over \(-10 \le \Re(s) \le 5\) and \(|\Im(s)| \le 7.5\). This should be a pretty important set of values; and makes a pretty fucking cool graph to be honest:

I'm going to work on drawing the lines better; but it's pretty tough to do it without sacrificing run time; mike3's program as a base is already pretty slow. But I'll keep cracking at it, lol. It would really help Ember, exactly what you'd like me to draw. I don't have to draw level sets; I can superimpose parametric plots atop; and do strange things with this. Essentially this adds a tool to mike3's program to "draw yellow lines" atop the graph. Having trouble getting the lines perfect, lmfao; but it seems to be working fairly well.

I thought another interesting graph would be to take:

\[

F(z) = \beta(\log(\Psi(z))/L)\\

\]

And graph it next to the level set \(|F'(z)| = 1\). Which essentially equates to where the beta schroder function looks like the normal schroder function; and draws a path accordingly in the complex plane. I'm going to make a few of these graphs as I try to figure out how to draw the lines properly; lmao

.

Nonetheless this graph pretty clearly shows the inversion that happens that makes the beta function real valued; I fucked up the level set lines though

-- Over a 20x20 box centered at 0.

I've zoomed in some more from here; Which is over a 3x3 box and I used a different line drawing test condition. looks like a funny smiley face, lmaoooo

OKAY! LAST ONE. Here is:

\[

|\beta'(z)| = 1\\

\]

This creates a continuous unbreaking line... Similarly so does any \(|\beta'(z)| = \ell(z)\). So \(|\beta'(z)| = |\chi'(z)|\) is a jordan curve on \(\widehat{\mathbb{C}}\)-- mapping the level curve \(|\beta'(z)| = |\text{tet}_{K}'(z)|\), is a unique solution. Just a different way of saying the Riemann mappings.

If we think of \(|\beta'(z)| = |\text{tet}_K'(z)|\) as a global result; then we expect that \(\beta(z) = A + e^{i\theta}\text{tet}_K'(z)\). This creates a polar matrix (which is in the neighborhood of how Sheldon does it).

This relates to the chi star; but I'm still not sure what you want me to graph. Sorry, if I'm being dense as fuck.

OH! AND ALSO I FIXED MY STUPID LEVEL SET DRAWING FOR THE FIRST PICTURE!

You'd draw the chi star in my circumstance, by mapping the line:

\[

|\beta'(z)| = 1\\

\]

To the line:

\[

|\text{tet}_{K}'(z)| = 1

\]

Where if you wrote an analytic function which maps these lines to each other; you have a function:

\[

\tau(\beta(z)) = \text{tet}_{K}(z)

\]

This is because this is still an analytic curve. And the equating of two analytic curves means the functions are the same. We can do this about \(\text{tet}_K(-1) = 0\) probably most effectively. Because we have a whole half plane of well behaved beta to deal with.