Video 4, chapter 20

Julia Sets, One Rule and Many Starting Points

Which starting points stay bounded when the rule z squared plus c is repeated with c fixed, how the filled Julia set and the Julia set answer it, and how an escape radius proves that an orbit escapes.

  • Video 4, chapter 20
  • A map of mathematical worlds

Fix a complex number cc, choose a starting point z0z_0 in the complex plane, and replace the current number zz by z2+cz^2 + c again and again. Some starting points produce numbers that stay within a fixed distance of 00 forever, while others run off to infinity, and the edge between the two kinds can be extremely intricate. For a fixed cc, which starting points stay bounded? This page answers the question exactly for c=0c = 0, defines the filled Julia set and the Julia set for every cc, gives the test that proves an orbit escapes, and explains why a computer picture cannot prove the opposite.

Key terms

Julia set

Definition Fix a complex number cc and the rule fc(z)=z2+cf_c(z) = z^2 + c. Starting from a point z0z_0 of the complex plane, the rule gives z1=fc(z0)z_1 = f_c(z_0), then z2=fc(z1)z_2 = f_c(z_1), and so on, and this sequence is the orbit of z0z_0. The orbit is bounded when some number BB satisfies ∣zn∣≤B\lvert z_n \rvert \le B for every nn. The filled Julia set KcK_c is the set of starting points whose orbits are bounded. The Julia set JcJ_c is the boundary of KcK_c, the set of points pp such that every disk centered at pp, however small, contains points of KcK_c and points outside KcK_c. The filled Julia set contains its boundary, so the points of JcJ_c have bounded orbits, yet arbitrarily close to each of them there are starting points that escape. So the answer to the question of which starting points stay bounded is the set KcK_c, and the intricate part of any picture of it is its edge JcJ_c. For c=0c = 0 the filled Julia set is the closed unit disk and the Julia set is the unit circle, while for other values of cc the Julia set can be far more intricate, although the rule still has only a squaring and an addition. For the rules z2+cz^2 + c, a classical theorem says that JcJ_c is either in one piece or a dust of separate points, and The Mandelbrot Set, a Map of Julia Worlds explains which. That two-way choice is a theorem about this quadratic family, and it does not hold for every rule. Julia sets are named after Gaston Julia, who studied repeated rules given by polynomials and fractions of polynomials in a long memoir of 1918, as Pierre Fatou did in the same years.

In plain words One rule, many starting points. The filled Julia set collects the starting points whose orbits never run off, and the Julia set is the edge between them and the starting points that escape. Near that edge, a tiny change in the starting point can decide between staying and escaping.

Example With c=0c = 0 the rule only squares, and squaring squares the modulus, so

∣zn∣=∣z0∣2n.\lvert z_n \rvert = \lvert z_0 \rvert^{2^n}.

If ∣z0∣<1\lvert z_0 \rvert < 1 the moduli shrink toward 00, if ∣z0∣=1\lvert z_0 \rvert = 1 they all equal 11, and if ∣z0∣>1\lvert z_0 \rvert > 1 they grow without bound. So K0K_0 is the closed unit disk of all zz with ∣z∣≤1\lvert z \rvert \le 1, and J0J_0 is its edge, the unit circle of all zz with ∣z∣=1\lvert z \rvert = 1. Every disk around a point of the circle contains points just inside, which stay bounded, and points just outside, which escape. The start 0.990.99 stays bounded and the start 1.011.01 escapes, although they differ by only 0.020.02.

Quadratic iteration

Definition Quadratic iteration is the repeated application of a rule fc(z)=z2+cf_c(z) = z^2 + c, where cc is a fixed complex number called the parameter. Each step first squares the current number, which squares its modulus and doubles its argument, and then adds cc, which slides the result by the arrow from 00 to cc. The same cc is used at every step and for every starting point. All the rules z2+cz^2 + c together form the quadratic family. This family covers every quadratic rule z↦az2+bz+dz \mapsto az^2 + bz + d with a≠0a \neq 0, where the arrow ↦\mapsto reads “is replaced by”. Relabeling the plane by w=az+b2w = az + \tfrac{b}{2} changes such a rule into w↦w2+cw \mapsto w^2 + c with c=ad+b2−b24c = ad + \tfrac{b}{2} - \tfrac{b^2}{4}, so studying the rules z2+cz^2 + c studies all quadratic rules at once.

In plain words Square, then slide, and repeat. The parameter cc chooses the rule, and once chosen it never changes.

Example Take c=ic = i and the start z0=1+iz_0 = 1 + i. Squaring gives (1+i)2=2i(1 + i)^2 = 2i, with modulus 22 and argument 9090 degrees, and adding ii gives z1=3iz_1 = 3i. The next step gives

z2=(3i)2+i=−9+i.\begin{aligned} z_2 &= (3i)^2 + i \\[0.4em] &= -9 + i. \end{aligned}

The distances from 00 are 2\sqrt{2}, then 33, then 82≈9.06\sqrt{82} \approx 9.06. The escape radius card below shows that this orbit escapes, and that z1z_1 already proves it.

Orbit

Definition The orbit of a starting point z0z_0 under a rule ff is the sequence z0,z1,z2,…z_0, z_1, z_2, \dots in which zn+1=f(zn)z_{n+1} = f(z_n) for every n≥0n \ge 0. It is bounded when all its points lie within some fixed distance of 00, and it escapes when ∣zn∣\lvert z_n \rvert grows beyond every bound and tends to infinity. For a rule z2+cz^2 + c every orbit does one or the other, by the escape radius card below. A fixed point is a start with f(z0)=z0f(z_0) = z_0, and an orbit is a cycle of length kk when kk is the smallest positive whole number with zk=z0z_k = z_0. Bounded does not mean settled. A bounded orbit may approach a point, run around a cycle, or wander forever without repeating, as some starts on the unit circle do for c=0c = 0.

In plain words An orbit is the trail a starting point leaves when the rule is applied over and over. Staying bounded only means that the trail never runs off to infinity. It does not have to settle down.

Example For c=0c = 0, take the start with modulus 12\tfrac12 and argument 3030 degrees, which is 14(3+i)\tfrac14(\sqrt{3} + i). Its orbit has moduli 12\tfrac12, 14\tfrac14, 116\tfrac1{16} and 1256\tfrac1{256} and arguments 3030, 6060, 120120 and 240240 degrees, so it spirals in toward 00. The start ii goes to −1-1, then to 11, and stays at the fixed point 11. The start −12+32i-\tfrac12 + \tfrac{\sqrt{3}}{2}i lies on the unit circle with argument 120120 degrees. Squaring sends it to argument 240240 degrees and then to 480480 degrees, which is 120120 degrees again, so its orbit is a cycle of length 22. All three orbits are bounded, and only the first approaches 00.

Filled Julia set

Definition For a fixed complex number cc, the filled Julia set KcK_c is the set of starting points z0z_0 whose orbits under z2+cz^2 + c are bounded. By the escape radius card below, a start belongs to KcK_c exactly when every point of its orbit satisfies ∣zn∣≤max⁡(2,∣c∣)\lvert z_n \rvert \le \max(2, \lvert c \rvert), so KcK_c lies inside that closed disk. The set KcK_c contains its boundary, and that boundary is the Julia set JcJ_c. When KcK_c has an inside, made of points with a whole small disk around them in KcK_c, the Julia set is only its edge. When KcK_c has no inside, the two sets are equal. That happens for c=−2c = -2, where K−2K_{-2} is the segment of the real axis from −2-2 to 22. Every start in that segment stays in it, since z2z^2 is then between 00 and 44, and every other start escapes. Membership in KcK_c is a statement about every step of an orbit, so no finite computation proves it by running the rule, although an exact argument, such as finding a cycle, can.

In plain words The filled Julia set is everything that stays. Its edge is the Julia set, and when the filled set is too thin to have an inside, the edge is all there is.

Example Take c=−1c = -1. The start 00 gives 0,−1,0,−1,…0, -1, 0, -1, \dots and the start 11 gives 1,0,−1,0,…1, 0, -1, 0, \dots, so both lie in K−1K_{-1}. The golden ratio φ=1+52≈1.618\varphi = \tfrac{1 + \sqrt{5}}{2} \approx 1.618 satisfies φ2−1=φ\varphi^2 - 1 = \varphi, so it is a fixed point and lies in K−1K_{-1} too, but it sits on the edge. The nearby start 1.71.7 goes to 1.891.89 and then to 2.57212.5721, beyond the escape radius 22, while the start 1.61.6 drifts down through 1.561.56, 1.43361.4336 and 1.05521.0552 and is then drawn toward the cycle 0,−10, -1. On the real axis, K−1K_{-1} is exactly the segment from −φ-\varphi to φ\varphi. The rule sends that segment into its part from −1-1 to φ\varphi, while any real start farther from 00 than φ\varphi escapes, because from then on each step moves it farther out by at least as much as the step before. In the whole plane, K−1K_{-1} is made of rounded pieces that touch at single points, a shape known as the basilica.

Escape radius

Definition For the rule z2+cz^2 + c, the number

R=max⁡(2,∣c∣),R = \max(2, \lvert c \rvert),

the larger of 22 and ∣c∣\lvert c \rvert, is an escape radius. If some point of an orbit satisfies ∣zn∣>R\lvert z_n \rvert > R, strictly, then the orbit escapes, and from then on every point is farther from 00 than the one before. The reason is that adding cc can bring a point at most ∣c∣\lvert c \rvert closer to 00. For a point zz with r=∣z∣>Rr = \lvert z \rvert > R this gives

∣z2+c∣≥∣z∣2−∣c∣>r2−r=r(r−1),\begin{aligned} \lvert z^2 + c \rvert &\ge \lvert z \rvert^2 - \lvert c \rvert \\[0.4em] &> r^2 - r = r(r - 1), \end{aligned}

since ∣c∣≤R<r\lvert c \rvert \le R < r. As r>2r > 2, the factor r−1r - 1 is greater than 11, so the next point is farther out than rr and again beyond RR. If λ=r−1\lambda = r - 1 for the first point beyond RR, every later step multiplies the distance from 00 by more than λ\lambda, so the distances grow at least as fast as the powers of λ\lambda, which grow without bound. The strict inequality matters. For c=−2c = -2 the radius is 22, and the start 22 is a fixed point, since 22−2=22^2 - 2 = 2, so a point exactly at distance RR need not escape.

In plain words Once a point is far enough from 00, squaring beats the fixed slide by cc, and the orbit runs off faster and faster. Far enough means beyond both 22 and the length of cc. A computer can therefore prove that a start escapes by finding one point of its orbit beyond the radius, but a start that has not crossed after many steps may still cross later.

Example For c=−1c = -1 the escape radius is 22. The start 22 lies exactly on the radius, which decides nothing, but its next point is 3>23 > 2, so the orbit escapes, as 3,8,633, 8, 63 confirm. The start ii goes to −2-2, on the radius, and then to 33, beyond it. The start 1+i1 + i goes to (1+i)2−1=−1+2i(1 + i)^2 - 1 = -1 + 2i, whose modulus 5\sqrt{5} already exceeds 22. A computer picture of KcK_c runs the starting point of each pixel for a budget of NN steps, colors the starts that cross the radius by the step at which they cross, and gives the rest a separate color that means only that no escape was seen within NN steps. The colors are a choice of palette applied to those step counts, not a property of the set, so the legend and the budget belong with every picture.

Check yourself

  1. Take c=0c = 0. For each of the starts 0.9i0.9i, 1.11.1 and 35+45i\tfrac35 + \tfrac45 i, decide whether it lies in the filled Julia set K0K_0 and whether it lies in the Julia set J0J_0.
Show the answer

For c=0c = 0 each step squares the modulus, so only the modulus matters. The start 0.9i0.9i has modulus 0.9<10.9 < 1, so it lies in K0K_0, but it is inside the disk rather than on its edge, so it is not in J0J_0. The start 1.11.1 has modulus greater than 11, so it is in neither set, and its orbit 1.211.21, 1.46411.4641, 2.1436…2.1436\dots crosses the escape radius 22 at the third step. The start 35+45i\tfrac35 + \tfrac45 i has modulus

925+1625=1,\sqrt{\tfrac{9}{25} + \tfrac{16}{25}} = 1,

so it lies on the unit circle and belongs to both K0K_0 and J0J_0.

  1. Change a number. Someone uses the escape radius 22 for every rule z2+cz^2 + c. Test that choice with c=−3c = -3 and the start z0=1+132≈2.303z_0 = \tfrac{1 + \sqrt{13}}{2} \approx 2.303.
Show the answer

Squaring the start and subtracting 33 gives

z02−3=14+2134−3=1+132=z0,\begin{aligned} z_0^2 - 3 &= \frac{14 + 2\sqrt{13}}{4} - 3 \\[0.4em] &= \frac{1 + \sqrt{13}}{2} = z_0, \end{aligned}

so the start is a fixed point. Its orbit stays at about 2.3032.303 forever and is bounded, although its modulus is greater than 22. The radius 22 would wrongly mark it as escaping. The correct radius is max⁡(2,∣−3∣)=3\max(2, \lvert -3 \rvert) = 3, and since 2.303<32.303 < 3 the correct test makes no claim about this start. The radius 22 alone can fail when ∣c∣>2\lvert c \rvert > 2, as it does here, because the slide by cc can carry a point that lies beyond 22 back toward 00. Here squaring takes 2.3032.303 out to about 5.3035.303, and adding −3-3 brings it back to 2.3032.303. Taking the radius at least ∣c∣\lvert c \rvert rules this out, since a point beyond the radius is then farther from 00 than the length of the slide.

  1. Change the budget. A picture of K0K_0 runs each start for 55 steps with escape radius 22. What does the pixel at the start 1.011.01 show, and is that right? How many steps would reveal the truth?
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For c=0c = 0 the moduli are ∣zn∣=1.012n\lvert z_n \rvert = 1.01^{2^n}. The first five steps give about 1.02011.0201, 1.04061.0406, 1.08291.0829, 1.17261.1726 and 1.37491.3749, none beyond 22, so the pixel gets the color for no escape. That is wrong, because 1.011.01 has modulus greater than 11 and is not in K0K_0. Step 66 gives about 1.89051.8905 and step 77 about 3.57383.5738, which is beyond 22, so a budget of at least 77 steps reveals the escape. Starts closer to the circle need more steps, and 1.0011.001 first crosses the radius at step 1010. No finite budget catches every escaping start near the circle, which is why the color for no escape means only that no escape has been seen yet.

How this lesson connects