Talk:Polynomial lemniscate

Mandelbrot set question[edit]

The Mandelbrot picture was wrong, Mandelbrot level curve is connected. This was square root 2 curve, not level 2 curve. Mandelbrot level curve divides plane in 2 components because Mandelbrot set is connected, and contains in the bounded components all the zeros of all the Mandelbrot polynomials, which are centers of hyperbolic components of Mandelbrot set. I have removed the very offensive to intelligence picture. — Preceding unsigned comment added by 78.30.183.44 (talk) 11:37, 23 May 2021 (UTC)[reply]

Here are the real Mandelbrot curves https://www.desmos.com/calculator/bkhv7u4hif — Preceding unsigned comment added by 78.30.183.44 (talk) 11:38, 23 May 2021 (UTC)[reply]

If Abs(z)=1 the one gets circle of radius 1 centered at the origin . Mandelbrot set has a point c=-2 so it is out of that circle. IMHO proper definition should be Abs(z)=EscapeRadius where EscapeRadius>=2. --Adam majewski 19:59, 17 June 2007 (UTC)[reply]

Yes, c=-2 is outside the unit circle, but we are talking about the limit of a a sequence of sets, not just the unit circle. The limit of this sequence of sets is the boundary of the Mandelbrot set. Why not try generating some of the sets in this sequence and see for yourself? I've got a java applet lying around that shows this sequence: let me know if you'd like a link to it. Cheers, Doctormatt 22:08, 17 June 2007 (UTC)[reply]

Yes, I would like a link to your applet . Cheers --Adam majewski 14:24, 18 June 2007 (UTC)[reply]

Here it is: [1]. Let me know if you have any suggestions (some browsers work better with this applet than others, but it seems to work with everything I've tried. You might need to click on the image before you can use any key commands). Doctormatt 06:21, 19 June 2007 (UTC)[reply]

Thx. It works.
1. How can you explain ( I dont understand it):
- algoritm for drawing Mandelbrot set is : if Abs(Zn)>2 then it is outside
- algorithm for drawing lemniscates ( which approaches the boundary of the Mandelbrot set) is : if Abs(Pn(z))=1 then it belongs to lemniscate.
Where Pn(z)=Zn.
2. Mandelbrot lemniscate can be named level curves. See algorithm by : LCM
3. Can you look at wiki page about external ray. It needs an expert in math ( see talk page). I'm trying to do without results. (:-))

I saw your other works. Great.
Cheers. --Adam majewski 17:33, 19 June 2007 (UTC)[reply]

Let's see.

1. I'm not sure of the proof, but my guts and experiments tell me that the 1 in Abs(Pn(z))=1 is arbitrary: any finite number should work. As n goes to infinity, the lemniscates defined by Abs(Pn(z))=k all bunch up together at the boundary of the Mandelbrot set. Certainly if Abs(Pn(z))=k as n goes to infinity, for any finite k, then we would expect z to be either in the interior or on the boundary of the Mandelbrot set. Note that the 2 in Abs(Zn)>2 is arbitrary, too: 2 is just the most efficient choice (from an algorithm point of view), but you could use anything larger: the point is whether or not Zn diverges to infinity or not. I'll think more about this and see if I can make this all more solid.

2. Sure. Abs(Pn(z)) can be expressed as a real-valued function of (x,y) where z=x+iy, so Abs(Pn(z))=k defines a level curve of that function.

3. It's not really my area of expertise (if I have one), but I'll see what I can do.

Cheers, Doctormatt 07:02, 21 June 2007 (UTC)[reply]