whereeach coefficient is either 0 or 1. Another way to put this: the set of all possible sums
for b = 1/2is a line segment.
What is this set for other values of “base” b, then?Let’s stick to|b| < 1for now, so that the series converges. Nothing interesting happens for realb between 1/2 and 1; the segment grows longer, to length b/(1-b). When b is between 0 and 1, we get Cantor sets, with the classical middle-third set being the case b = 1/3.
There is no need to consider negative b, because of a symmetry between b and -b. Indeed, up to scaling and translation, thecoefficients can be taken from {-1, 1} instead of {0, 1}. Then it’s obvious that changing the sign of b is the same as flipping half of coefficients the other way― does not change the set of possible sums.
Let’s look at purely imaginary b, then. Here is b = 0.6i
Why so rectangular? The real part is the sumof
over even k, and the imaginary part is the sum over odd k. Each of theseyields a Cantor type set as long as
. Since the odd- and even-numbered coefficients are independent of each other, we get the product of two Cantor sets. Which changes into a rectangle when
:
(I didn’t think a full-size picture of a solid rectangle was necessary here.)
This is already interesting: the phase transition from dust to solid (connected, and even with interior) happens at different values in the real and imaginary directions: 1/2 versus
. What will happen forother complex values? Using complex conjugation and the symmetry between b and -b, we reduce the problem to the quarter-disk in the first quadrant. Which still leaves a room for a lot of things to happen…
b = 0.6 + 0.3i
b = 0.7 + 0.2i
b = 0.4 + 0.3i
b = 0.2 + 0.7i
It’s clear that for |b| < 1/2 we get a totally disconnected set ― it is covered by 2 copies of itself scaled by the factor of |b|, so its Hausdorff dimension is less than 1when |b| is less than 1/2. Also, theargument of b is responsible for rotation of the scaled copies, and it looks like rotation favors disconnectivity… but then again, the pieces may link up again after being scaled-rotated a few times, so the story is not a simple one.
The set of basesb for which the complex Cantor set is connected is a Manderbrot-like setintroduced byBarnsley and Harrington in1985. It has the symmetries of a rectangle, and features a prominent hole centered at 0 (discussed above). Butit actually has infinitely many holes , with“exotic” holes being tiny islands of disconnectedness, surrounded by connected sets. This was proved in 2014 byCalegari, Koch, Walker, so I refer to Danny Calegari’s post for an explanation and more pictures (much better looking than mine).
Besides “disconnectedto connected”, there is another phase transition: empty interior to nonempty interior. Hare and Sidorov proved that the complex Cantor set has nonempty interior when
; their path to the proof involved a MathOverflow question The Minkowski sum of two curves which is of its own interest.
The pictures were made with a straightforward python script, using expansions of length 20:
import matplotlib.pyplot as plt import numpy as np import itertools n = 20 b = 0.6 + 0.3j c = np.array(list(itertools.product([0, 1], repeat=n))) w = np.array([b**k for k in range(n)]).reshape(1, -1) z = np.sum(c*w, axis=1) plt.plot(np.real(z), np.imag(z), '.', ms=4) plt.axis('equal') plt.show()Since we are looking at partial sums anyway, it’s not necessary to limit ourselves to |b| being less than 1. Replacing b by 1/b only scales the picture, so the place to look for new kinds of pictures is the unit circle. Let’s try a 7th root of unity:
b = exp(pi i / 7)
The set abovelooks sparse becausemany points overlap. Let’s change b to something non-algebraic:
b = exp(i)
What’s with the cusps along the perimeter?