# IntroductionEdit

Julia set drawn by inverse iteration of critical orbit ( in case of Siegel disc )
Periodic external rays of dynamic plane made with backward iteration

Iteration in mathematics refer to the process of iterating a function i.e. applying a function repeatedly, using the output from one iteration as the input to the next.[1] Iteration of apparently simple functions can produce complex behaviours and difficult problems.

One can make inverse ( backward iteration) :

• of repeller for drawing Julia set ( IIM/J)[2]
• of circle outside Jlia set (radius=ER) for drawing level curves of escape time ( which tend to Julia set)[3]
• of circle inside Julia set (radius=AR) for drawing level curves of attract time ( which tend to Julia set)
• of critical orbit ( in Siegel disc case) for drawing Julia set ( probably only in case of Goldem Mean )
• for drawing external ray

Repellor for forward iteration is attractor for backward iteration

# TestEdit

One can iterate ad infinitum. Test tells when one can stop

• bailout test for forward iteration

# Target set or trapEdit

Target set is used in test. When zn is inside target set then one can stop the iterations.

# PlanesEdit

## Dynamic plane $f_0$ for c=0Edit

Lets take c=0, then one can call dynamical plane $f_0$ plane.

Here dynamical plane can be divided into :

• Fatou set
• Julia set = $\{ z : |z| = 1 \}$

Fatou set consist from 2 subsets :

• interior of Julia set = basin of attraction of finite attractor = $\{ z : |z| < 1 \}$
• exterior of Julia set = basin of attraction of infinity = $\{ z : |z| > 1 \}$

### Forward iterationEdit

The 10 first powers of a complex number inside the unit circle
Exponential spirals

$z = r e^{i \theta} \,$

where $r = | z | \,$

so

$f_0(z) = z^2 = (r e^{i \theta})^2 = r^2 e^{i 2 \theta}\,$

and forward iteration :[4]

$f^n_0(z) = r^{2^n} e^{i 2^n \theta}\,$

Forward iteration gives forward orbit = list of points {z0, z1, z2, z3... , zn} which lays on exponential spirals.[5] [6]

#### Escape testEdit

If distance between point z of exterior of Julia set and Julia set is :

$distance(z, J_c) = 2^{-n}$

then point escapes ( measured using bailout test and escape time )

$|z_n| > ER$

after :

• $n$ steps in non-parabolic case
• $2^n$ steps in parabolic case [7]

See here for the precision needed for escape test

### Backward iterationEdit

Every angle α ∈ R/Z measured in turns has :

• one image = 2α mod 1 under doubling map
• "two preimages under the doubling map: α/2 and (α + 1)/2." [8]. Inverse of doubling map is multivalued function.

Note that difference between these 2 preimages

$\frac{\alpha}{2} - \frac{\alpha +1}{2} = \frac{1}{2}$

is half a turn = 180 degrees = Pi radians.

 $\alpha$ $d^1(\alpha)$ $d^{-1}(\alpha)$ $\frac{1}{2}$ $\frac{1}{1}$ $\left \{ \frac{1}{4} , \frac{3}{4} \right \}$ $\frac{1}{3}$ $\frac{2}{3}$ $\left \{ \frac{1}{6} , \frac{4}{6} \right \}$ $\frac{1}{4}$ $\frac{1}{2}$ $\left \{ \frac{1}{8} , \frac{5}{8} \right \}$ $\frac{1}{5}$ $\frac{2}{5}$ $\left \{ \frac{1}{10} , \frac{6}{10} \right \}$ $\frac{1}{6}$ $\frac{1}{3}$ $\left \{ \frac{1}{12} , \frac{7}{12} \right \}$ $\frac{1}{7}$ $\frac{2}{7}$ $\left \{ \frac{1}{14} , \frac{4}{7} \right \}$

On complex dynamical plane backward iteration using quadratic polynomial $f_c$

$f_c(z) = z^2 + c$

gives backward orbit = binary tree of preimages :

$z \,$

$-\sqrt{z-c} , +\sqrt{z-c} \,$

$-\sqrt{-\sqrt{z-c} -c} , +\sqrt{-\sqrt{z-c} -c}, -\sqrt{+\sqrt{z-c} -c}, +\sqrt{+\sqrt{z-c} -c} \,$

One can't choose good path in such tree without extra informations.

Not that preimages show rotational symmetry ( 180 degrees)

## Dynamic plane for $f_c$Edit

### Level curves of escape timeEdit

Preimages of circle under fc

### Julia set by IIM/JEdit

In escape time one computes forward iteration of point z.

In IIM/J one computes:

• repelling fixed point[9] of complex quadratic polynomial $Z_0=\beta_c \,$
• preimages of $Z_0\,$ by inverse iterations

$Z_{n-1} = \sqrt{Z_n - C}$

Because square root is multivalued function then each $Z_{n}\,$ has two preimages $Z_{n-1}\,$. Thus inverse iteration creates binary tree.

#### Root of treeEdit

• repelling fixed point[10] of complex quadratic polynomial $Z_0=\beta_c \,$
• - beta
• other repelling periodic points ( cut points of filled Julia set ). It will be important especially in case of the parabolic Julia set.

"... preimages of the repelling fixed point beta. These form a tree like

                                               beta
beta                                            -beta
beta                         -beta                    x                     y


So every point is computed at last twice when you start the tree with beta. If you start with -beta, you will get the same points with half the number of computations.

Something similar applies to the preimages of the critical orbit. If z is in the critical orbit, one of its two preimages will be there as well, so you should draw -z and the tree of its preimages to avoid getting the same points twice." (Wolf Jung )

#### Variants of IIMEdit

• random choose one of two roots IIM ( up to chosen level max). Random walk through the tree. Simplest to code and fast, but inefficient. Start from it.
• both roots with the same probability
• more often one then other root
• draw all roots ( up to chosen level max)[11]
• using recurrence
• using stack ( faster ?)
• draw some rare paths in binary tree = MIIM. This is modification of drawing all roots. Stop using some rare paths.
• using hits table ( while hit(pixel(iX,iY)) < HitMax )[12]
• using derivative ( while derivative(z) < DerivativeMax)[13]

Examples of code :

Compare it with:

# ReferencesEdit

1. wikipedia : Iteration
2. Inverse Iteration Algorithms for Julia Sets by Mark McClure