 In these notes I will use mathematical notation similar to the syntax of Mathematica.

11. Counting in the Mandelbrot set: an excursion into number theory.

I this lecture I will discuss some coutning problems associated with periodic points in the Mandelbrot set. We saw in Lecture 7 that a point c of period n must be a root of the polynomial G[n, c, 0] defined by

G[0, c_, z_] := z; G[n_, c_, z_] := G[n-1, c, z]^2 + c.

This polynomial is easily seen to degree 2^(n-1). This raises the follwoing questions.

• How do we know that G[n, c, 0] has 2^(n-1) distinct roots?
• How many of these roots actually have period n, rather than some lower period?
• How many of the period n points are inside cardioids and how many are insdie buds?
• How many of them are real?

Each of these questions requires some elementry number theory to answer.

Why are there 2^(n-1) distinct roots?

Here are the first few values of G[n, c, 0], given by the code

Do[Print["G[",n,", c, 0] = ",InputForm[Expand[G[n,c,0]]]], {n, 2, 4}]

The output is

G[2, c, 0] = c + c^2

G[3, c, 0] = c + c^2 + 2*c^3 + c^4

G[4, c, 0] = c + c^2 + 2*c^3 + 5*c^4 + 6*c^5 + 6*c^6 + 4*c^7 + c^8

It is clear from th definition of G that the degree of this polynomial doubles each time we increase n by 1, so G[n, c, 0] as degree 2^(n-1).

The Fundamental Theorem of Algebra says that any polynomial of degree d of the form

p[c] = a + a*c + ... a[d-1]*c^(d-1) + c^d

an be factored as

p[c] = (c - r)(c - r)...(c - r[d]),

and the complex numbers r through r[d] are its roots. In genral these do not have to be distinct. For example we could have p[c] = (c - r)^d, in which ase the roots would all be the same.

There is a way to use the derivative of p at a root r to see if that root is repeated. Suppose that r is a repeated root, i.e. that

p[c] = (c - r)^2*q[c]

where q[c] is a polynomial of degree d-2. Then the derivative is

p'[c] = 2(c-r)*q[c] + (c - r)^2*q[c]

= (c-r)*(2*q[c] + (c-r)*q[c]).

Since this derivative is divisible by (c-r), it must vanish when c = r.  Hence if r is a repeated root, the deriviative of p (as well as p itself) must be zero when c = r. This means that

If the derivative of p does not vanish at a root r, then that root only occurs once.                                  (11.1)

We will use this fact to prove that G[n, c, 0] has no repeated roots by showing that its derivative is nonzero at each of its roots. By definition, G[n, c, 0] = G[n-1, c, 0]^2 + c, so its derivative is

D[G[n, c, 0], c] = 2*G[n-1, c, 0]*D[G[n - 1, c, 0], c] + 1.                     (11.2)

We want to show that this is nonzero whenever c is a root r of G[n, c, 0].

Now suppose that the root r is a whole number. (This is almost never the case, but nevermind.) The polynomials appearing in (11.2) all have integer coefficients, so the expression on the right is an odd number and hence nonzero. This means that

An integer root of G[n, c, 0] cannot be a repeated root.                                                             (11.3)

This statement as it stands is not worth very much, because the only integers that are ever roots of a G[n, c, 0] are 0 and -1, but there is a way to leverage this idea into something more useful. The key concept is the following.

Definition 11.4. A complex number z is an algebraic integer if it is the root of a polynomial of the form

p[z] = a + a*z + ... a[d-1]*z^(d-1) + z^d

here the coefficients a though a[d-1] are all integers.

Algebraic integers are discussed in many textbooks on number theory. They have many of the same formal properties as ordinary integers. The sum, difference and product of any two algebraic integers is again an algebraic integer, and one can distinguish between those that are even (i.e. two times another algebraic integer) and those which are odd.

Algebraic integers are relevant here because each root of G[n, c, 0] is one by definition, since G[n, c, 0] is a polynomials of the form shown in 11.4. This means we can conclude that the derivative in (11.2) is nonzero because it is an aglebraic integer not divisible by 2. Every root of G[n, c, 0] s an algebraic integr, so none of them is a repeated root.

Why does this argument not apply to preperiodic points?

How many period n points are there?

We now know that G[n, c, 0] has exactly 2^(n-1) roots, but not all of them have period n. For example, the 8 roots of G[4, c, 0] include the 2 roots of G[2, c, 0], so there are only 6 points of period 4. More generally the 2^(n-1) roots of G[n, c, 0] include all points with periods dividing n. If we define Period[n] to be the number of points with period n, then we have

1 = Period

2 = Period + Period

4 = Period + Period

8 = Period + Period + Period

16 = Period + Period

32 = Period + Period + Period + Period

etc.

It is possible to solve these equations for Period[n] in terms the values of Period[d] for numbers d dividing n. Mathematica has a built in function Divisors[n] which gives a list of all numbers d that divide a positive integer n in ascending order. The equations above can be written as

2^(n-1) == Sum[Period[Divisors[n][[i]]], {i, 1, Length[Divisors[n]]}]             (11.4)

This notation is cumbersome so we shorten it by defining

DivisorSum[f_, n_] :=

Sum[f[Divisors[n][[i]]],{i,1,Length[Divisors[n]]}]

so that (11.4) can be rewritten as

2^(n-1) = DivisorSum[Period, n]

There is a method for solving equations of this type called the Moebius inversion formula. The Möbius function MoebiusMu[n] is defined to be (-1)^k if n is a product of k distinct primes, and 0 if n is divisible by the square of any prime number. Now suppose that two functions f and g are related by the formula

f[n] = DivisorSum[g, n],

in other words f[n] is the sum over all divisors d of n of the quantities g[d]. Before we can state the inversion formula, we need another definition, namely

MoebiusDivisorSum[f_, n_] :=

Sum[MoebiusMu[n/Divisors[n][[i]]]*f[Divisors[n][[i]]],{i,1,Length[Divisors[n]]}]

Then the inversion formula says that

g[n] = MoebiusDivisorSum[f_, n_] ,                                         (11.5)

in other words g[n] is a similar sum of the quantities MoebiusMu[n/d]*f[n]. Applying this to the case f[n_] := 2^(n-1) and g[n_] := Period[n], we can rplace the defintion of Period[n] above with that

h[n_] := 2^(n-1)

Period[n_] := MoebiusDivisorSum[h, n],                                          (11.6)

in other words, Period[n] is the sum over all divisors d of n of the quantities MoebiusMu[n/d]*2^(d -1).

For example for a prime number p, (11.6) gives

Period[p] = 2^(p-1) - 1

Period[p^2] = 2^(p^2-1) - 2^(p-1),

and if q is another prime number we get

Period[p*q] = 2^(p*q-1) - 2^(p-1) - 2^(q-1) + 1.

How many period n points are in buds?

Now that we know the number Period[n] of points of period n, we can ask how many are inside cardioids and how many are inside buds.

For this we need the Euler totient function EulerPhi[n], which is defined to be the number of integers between 0 and n which have no common divisors with n. . EulerPhi[n] is also the number of fractions between 0 and 1 with denominator n that have been reduced to lowest terms, so it is the number of period n buds attached to the main cardioid. Similarly, a period m bud has EulerPhi[n] buds attached to it each having period m*n.

Let Buds[n] be the number of period n buds attached to the main cardioid either directly (as primary buds) or indirectly via other buds.

Now suppose we want to determine Buds, the numebr of period 10 buds attached directly or indirectly to the main cardioid. There are

• EulerPhi = 4 primary buds of period 10,
• EulerPhi = 1 secondary bud of period 10 attached to each of the Buds = 4 buds of period 5, and
• EulerPhi = 4 secondary buds of period 10 attached to the one bud of period 2.

It is known that the sum over all divisors d of n of the quantities EulerPhi[d] is equal to n

How many period n points are real?

This page was last revised on February 26, 1998.