I'm going to write a modified version of MakeGraph by mike3
#21
(02/26/2023, 05:10 PM)tommy1729 Wrote:
(02/26/2023, 04:43 PM)JmsNxn Wrote: As the great irrationalist Pythoras said:

There are no such things as quadratics. For everything is irrational. Even within perfect geometry; everything is irrational.

I just made this up, but this is a lot of his cults philosophy around \(\sqrt{2}\)

\(x^{2-\delta}\) is it getting so close to a quadratic... Just like the Pythagorean search for the physical manifestation of a perfect right triangle........ Sad

In nature everything is irrational whenever it can and "whenever it can " usually means involving conservation of energy and quantization.
And that is why we will never understand it.

Tom Raes

Lmao, much nicer, Tommy! Very Pythagorean... Only Geometry is Rational to Athenians, Pythagoreans knew that even Geometry was irrational Smile
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#22
(02/26/2023, 05:20 PM)JmsNxn Wrote:
(02/26/2023, 05:10 PM)tommy1729 Wrote:
(02/26/2023, 04:43 PM)JmsNxn Wrote: As the great irrationalist Pythoras said:

There are no such things as quadratics. For everything is irrational. Even within perfect geometry; everything is irrational.

I just made this up, but this is a lot of his cults philosophy around \(\sqrt{2}\)

\(x^{2-\delta}\) is it getting so close to a quadratic... Just like the Pythagorean search for the physical manifestation of a perfect right triangle........ Sad

In nature everything is irrational whenever it can and "whenever it can " usually means involving conservation of energy and quantization.
And that is why we will never understand it.

Tom Raes

Lmao, much nicer, Tommy! Very Pythagorean... Only Geometry is Rational to Athenians, Pythagoreans knew that even Geometry was irrational Smile

Yeah but they did not know if ( 2 + sin(pi/17) + sin(pi/29) )^( - 2) was constructible by compass or even origami.
And neither if it could be expressed by radicals.

But today almost everyone knows it and can prove it  Wink


regards

tommy1729
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#23
I've come to make a request. Big Grin Although this request is not very relevant to this post.

Can the beta method make a Pictures similar to the Chi-star function?
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#24
(03/31/2023, 02:10 PM)Ember Edison Wrote: I've come to make a request. Big Grin Although this request is not very relevant to this post.

Can the beta method make a Pictures similar to the Chi-star function?

Just to be clear; we are asking for an additional input to define a level set? And the specific level set is related to this first order expansion?

I can definitely code in level sets; I'd probably write this as:

Code:
if(abs(func(z)- level(z)) == 0, Pixel(z) = YELLOW); //not what the code actually looks like but basically this

Where you would be forced to define func(z), as usual, but then additionally define a level set function level(z); which I'll take to be zero with no input. I'll see if I can find a quick way to code in a manner of guessing the chi-star level set... but I'm not sure... I can definitely get level sets for this though!!!!!

I'll see what I can do! This seems interesting so I'll give it a shot! First up is level sets; then finding the chi-star level set in terms of beta!
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#25
Alright; so I'm fifty percent there with Ember Edison's request. I'm having trouble finding the exact "thickness of the line." But to begin, I'll start slow.

Let's let

Code:
func(z) = sin(z)
level(z) = 1

If we allow \(f(z) = \sin(z)\). Then we are going to do two things simultaneously. We are going to graph \(f(z)\) for \(-\pi \le \Re(z) \le \pi\) and \(-\pi \le \Im(z) \le \pi\). But!!!! We are going to draw a yellow line where \(|f'(z)| = 1\). This yellow line looks exactly like \(|\cos(z)| = 1\). The manner I draw the YELLOW PIXELS is \(\left||\cos(z)| - 1\right| \le 0.001\)--which is proving to work, but not be that graphically descriptive.

   

Obviously we would like something much cleaner looking; but for the moment this is what I have; and it only adds an O(1) time frame. This is always my point as a programmer; adding a level set is negligible in my code. I'm having trouble drawing the lines at the moment; but I have the interjection. I just have to clean up the "graphing" code. We can take something simpler if we'd like.

Let \(f(z) = e^z\) and we will graph over the same domain \(|\Re(z)| < \pi\) and \(|\Im(z)| < \pi\). But we want a yellow line where \(|f'(z)| = 1\). This is precisely a straight vertical line at 0. The manner my code will draw this line is ugly; but it's an O(1) addition to the original code. So it's successful for now.

   

I know this looks super unimpressive!!!!!!!!!!!! But it's the behind the scenes code which is an addition to the above code which is important. I have an idea of how to make the line look better in both cases; but still; I'm not some fucking graphics programmer.

If \(||f'(z)| - |L(z)|| \le 0.001\) PRINT(YELLOW). Where we define \(f(z) =\) func(z), and \(L(z) =\) level(z).

That's the thesis of the code, so to speak. Any suggestions are greatly appreciated.

I'll end with \(f(z) = 0.5z^2\) and \(|f'(z)|= 1\), which draws a yellow circle at \(|z| = 1\). I drew this circle a little thick; I'm still trying to fine tune the graphics without taking too much program time:

   

I'm not talking about tetration yet; but I imagine the manner of drawing the Chi-stars requires these graphing protocols. So I don't know how to write the Chi-stars yet; but I know I'll need level set code for Mike3's program.

Curves are drawn as \(|f'(z)| = |L(z)|\). I just need to write a beta function \(|\beta'(z)| = |\lambda|\) like the chi-star..... Still not sure yet. Wish me luck, lmao...

Regards, James
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#26
Just to go a bit further... Assume that \(f(z) = e^z\) and \(L(z) = \sqrt{z}\). Then if we graph \(f\) over \(|\Re(z)| < 1\) and \(|\Im(z)| < 1\)--where the line \(||f'(z)| - |L(z)|| < 0.001\) means the pixel is YELLOW!

We get this graph... I'm still fucking up the thickness... But it's O(1) in code at least, lmao:

   
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#27
Alright; so I've managed to code in the Schroder function to my satisfaction, and the inverse schroder function. I've always wanted to do this; as I've only done it inside the Shell-Thron region. I'm going to get it all working more detailed; but for the moment; here's a little preview of the chistar function:

   

You can see the singularity at \(0\) and \(1\); and they continue at each \(\exp^{\circ n}(0)\).

Here's the same function graphed parametrically:

   

You can kind of see the beginning of the star like shape that Sheldon saw.

Now, I'm pretty confused as to how you would like to overlay this over the beta method?

I've been fiddling and I've written:

\[
\chi(\beta_{1,1}(x))\\
\]

And this is again a parametric plot in the complex plane. We get a weird looking star like shape that is more prominent.

   

We can make a closer idea by taking:

\[
g_\beta(X) = \beta_{1,1}(\log(\chi(X))/L)
\]

Which satisfies:

\[
g_\beta(e^X) = \beta_{1,1}(\log(L \chi(X))/L) =\frac{ e^{g_\beta(X)}}{1+e^{-X\log(L \chi(X))/L}}\\
\]

So it's pretty close to Sheldon's \(\Psi^{-1}(\chi(X)) = X\).

   

But I'm still not sure precisely what you'd like to draw..?

I can try to draw anything you'd like; I'd just like some more context. Still not even sure what sheldon is trying to draw!
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#28
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  Tongue .

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  Rolleyes -- 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!  Wink  Wink  Wink

   

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.
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#29
It looks great, and it shows that the beta method and the Kneser's method weren't really that different before.
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