Editing Julia set (section)

===Using DEM/J===
<gallery caption="Images of Julia sets for {{tmath|1=f_c(z) = z^2 + c}}" widths="180px" heights="120px">
File:Demj.jpg|{{tmath|1=c=-0.74543+0.11301*i}}
File:Julia dem.png|{{tmath|1=c= -0.75+0.11*i}}
File:Julia dem c=-0.1+0.651.png|{{tmath|1=c=-0.1+0.651*i}}
File:Juliasetsdkjulia1.jpg|Julia set drawn by distance estimation, the iteration is of the form {{tmath|1=1-z^2+z^5/(2 + 4z)+c}}
File:Juliasetsdkland1.jpg|Three-dimensional rendering of Julia set using distance estimation
</gallery>

As a Julia set is infinitely thin we cannot draw it effectively by backwards iteration from the pixels. It will appear fragmented because of the impracticality of examining infinitely many startpoints. Since the iteration count changes vigorously near the Julia set, a partial solution is to imply the outline of the set from  the nearest color contours, but the set will tend to look muddy.

A better way to draw the Julia set in black and white is to estimate the distance of pixels (DEM) from the set and to color every pixel whose center is close to the set. The formula for the distance estimation is derived from the formula for the potential function ''φ''(''z''). When the equipotential lines for ''φ''(''z'') lie close, the number <math>|\varphi'(z)|</math> is large, and conversely, therefore the equipotential lines for the function <math>\delta(z) = \varphi(z)/|\varphi'(z)|</math> should lie approximately regularly. It has been proven that the value found by this formula (up to a constant factor) converges towards the true distance for z converging towards the Julia set.<ref name="Peitgen1986" />

We assume that ''f''(''z'') is rational, that is, <math>f(z) = p(z)/q(z)</math> where ''p''(''z'') and ''q''(''z'') are complex polynomials of degrees ''m'' and ''n'', respectively, and we have to find the derivative of the above expressions for ''φ''(''z''). And as it is only <math>z_k</math> that varies, we must calculate the derivative <math>z'_k</math> of <math>z_k</math> with respect to ''z''. But as <math>z_k = f(f(\cdots f(z)))</math> (the ''k''-fold composition), <math>z'_k</math> is the product of the numbers <math>f'(z_k)</math>, and this sequence can be calculated recursively by <math>z'_{k+1} = f'(z_k)z'_k</math>, starting with <math>z'_0 = 1</math> (''before'' the calculation of the next iteration <math>z_{k+1} = f(z_k)</math>).

For iteration towards ∞ (more precisely when {{math|1=''m'' ≥ ''n'' + 2}}, so that ∞ is a super-attracting fixed point), we have

:<math>|\varphi'(z)| = \lim_{k\to\infty} \frac{|z'_k|}{|z_k|d^k},</math>

({{math|1=''d'' = ''m'' − ''n''}}) and consequently:

:<math>\delta(z) = \varphi(z)/|\varphi'(z)| = \lim_{k\to\infty} \log|z_k||z_k|/|z'_k|. \, </math>

For iteration towards a finite attracting cycle (that is not super-attracting) containing the point {{tmath|1=z^*}} and having order ''r'', we have

:<math>|\varphi'(z)| = \lim_{k\to\infty} |z'_{kr}|/(|z_{kr} - z^*|^{2}\alpha^{k}), \, </math>

and consequently:

:<math>\delta(z) = \varphi(z)/|\varphi'(z)| = \lim_{k\to\infty} |z_{kr} - z^*|/|z'_{kr}|. \, </math>

For a super-attracting cycle, the formula is:

:<math>\delta(z) = \lim_{k\to\infty} \log|z_{kr} - z^*|^2/|z'_{kr}|. \, </math>

We calculate this number when the iteration stops. Note that the distance estimation is independent of the attraction of the cycle. This means that it has meaning for transcendental functions of "degree infinity" (e.g. sin(''z'') and tan(''z'')).

Besides drawing of the boundary, the distance function can be introduced as a 3rd dimension to create a solid fractal landscape.