## Bugs #20

No posts for the last two days. I have no excuses, I simply did not make the time. I will try to make up with multiple posts today.

The next few pictures are success out-zooms of Bugs #16 to get an idea of what the neighborhood looks like.

Today’s image is a zoom out by a factor of two.

For a quick recap, this formula is bugR, defined back in Bugs #14 , with parameter a = 2.5.

## Bugs #18

Yesterday’s image was a deep zoom into a “quiet” area. It appears that an infinite grid with ever finer lines is everywhere. At every intersection, a deep zoom holds something interesting. Here is a distorted minibrot.

With z^2+c and a little algebra, you can show that if |z| > 2 + |c| then z escapes. Yesterday I showed a similar argument for bugI() which concludes that if |x| > a + 1 + |c| then z =x + yi escapes. So for both of these formulas, there are easily identified large areas where the orbit is guaranteed to escape. Also we saw that this type of simple algebra argument breaks down for bugR(). That does not prove that no such area exists, only that if it does, it will be difficult to find. Let’s dig deeper

Since this is a blog, not a book, I should recap frequently. We are iterating

$bugR(z^2+c) \quad where \\ bugR(x+yi) = u(x)+yi, \\ u=a*sin(x/a), a>0$

Recall that $|u(x)|<= a \text{ and } u(a*n\pi)=0$

If a>.5 (an easy assumption, it has been true for every picture so far), u(p)=-.5 has a solution. In fact many solutions when you find one, $p+a*n\pi$ is also a solution for any integer n.

Look at $c=c_r+c_i i = p+a*m\pi + \pm \sqrt{p+.25+a*n\pi} i$. where n and m are arbitrary integers.

These points lie on a grid. Equal space horizontally, decreasing spacing vertically because of the square root. But extending to infinity in all directions. Where ever you are in the plane you are surrounded by these points. The following will show that none of these points escape, in fact every one is a 2-cycle.

Lets iterate bugR(z^2+c)

$\begin{matrix} z_0 & = & 0 \\ z_1 & = & bugR(c)\\ & = & -.5 & + & c_i i \\ z_2 & = & u(.25 - c_i^2+c_r) & + & (2*(-.5)*c_i + c_i)i \\ & = & u(.25 - (p +.25+an\pi)+p+am\pi) & + & (-c_i+c_i)i \\ & = & u(a*(m-n)\pi) & + & 0i \\ & = & 0 \end{matrix}$

So every one of these points generates a two cycle.

It is embarrassing to admit how much time I spent on this. My math skills are rusty.