04/02/2023, 08:56 AM
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
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
Let's let
Code:
func(z) = sin(z)
level(z) = 1If 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