Bifurcation in Complex Quadrat ic Polynomial and Some Folk Theorems Involving the Geometry of Bulbs of the Mandelbrot Set
Monzu Ara Begum
A Thesis
in
The Department
of
Mat hemat ics and St at ist ics
Presented in Partial F'uffilment of the Requirements
for the Degree of Master of Science at
Concordia University
Montreal, Quebec, Canada
May 2001
@Monzu Ara Begum, 2001 muisitions and Acquisitions et BiMiiriic Services senrices bibliographiques
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The auîhor retains ownership of the L'auteur conserve la propriété du copyright in this thesis. Neither the droit d'auteur qui protège cette thèse. thesis nor substantial extracts fiom it Ni La îhèse ni des extraits substantiels may be printed or otherwise de celle-ci ne doivent être imprimés reproduced without the author's ou autrement reproduits sans son permission. autorisation. Abstract
Bifurcation in Complex Quadratic Polynomial and Some Folk Theorems
Involving the Geometry of Bulbs of the Mandelbrot Set
Monzu Ara Begum
The Mandelbrot set M is a subset of the parameter plane for iteration of the complex quadratic polynomial Q,(z) = z2 +c. M consists of those c values for which the orbit of O is bounded. This set features a basic cardioid shape from which hang numerous
'bulbs' or 'decorations'. Each of these bulbs is a large disk that is directly attached to the main cardioid together with numerous other smaller bulbs and a prominent
'antenna'. In this thesis we study the geometry of bulbs and some 'folk theorems' about the geometry of bulbs involving spokes of the antenna. Acknowledgment s
1 would like to express my gratitude to my supervisor, professor Pawel G6ra of Con- cordia University, for his guidance of this thesis. Without his help and valuable suggestion it was impossible to mite this thesis. 1 am also grateful to the Depart- ment of Mathematics and Statistics of Concordia University for financial support during my Master's program. Contents
1 Introduction 1
1.1 General Introduction ...... 1
1.2 Scopeofthethesis ...... 2
2 Preliminaries 3
2.1 Fixed points. periodic points and their nature ...... 3
2.2 Fatou sets and Julia sets ...... 5
2.3 Some properties of the Julia and the Fatou sets ...... 9
3 The Mandelbrot Set and Bifurcation in Complex Quadratic Polyno-
mial 14
3.1 Complex Quadratic polynomial. Filled Julia set and the Mandelbrot set 15
3.2 Role of the critical orbit ...... 19
3.3 Bifurcation in Q, (z) and visible bulbs of the Mandelbrot set ..... 19
3.4 Algorithm for the Mandelbrot set ...... 25
3.5 Periods of the bulb ...... 26 3.6 Rotation numbers ...... 29
4 Geometry of bulbs in the Mandelbrot set. Farey tree and Fibonacci
sequence 32
4.1 Introduction ...... 32
4.2 Farey addition. Farey tree. Fibonacci sequence ...... 33
4.3 Angle doubling mod 1 ...... 39
4.4 Itinerary and Binary Expansion: ...... 42
4.5 External Rays ...... 44
4.6 Idea of Limbs ...... 47
4.7 Rays landing on the p/q bulb ...... 47
4.8 An algorithm for computing the angles of rays landing at c(p/q).... 50
4.9 The size of Limbs and the Farey Tree ...... 54
4.10 The Fibonacci sequence ...... 58
4.1 1 Conclusion ...... 60
Bibliography 61 Chapter 1
Introduction
1.1 General Introduction
In the study of complex dynarnical systerns the Mandelbrot set is one of the most intricate and beautiful objects in Mathematics. The geometry in the Mandelbrot set is quite complicated. In the last fifteen years many mathematicians such as
Paul Blanchard [BL], Pau Atela [A],Bodil Branner [BR],R.L. Devaney [D2],E.Lau and D. Sclileicher Yoccoz [LS]and recently R.L.Devaney [Dl] and R.L.Devaney and
M-Morenno-Rocha [DM] studied the geometry of the bulbs and antennae in the
Mandelbrot set. In this thesis we study the Mandelbrot set and geometry of bulbs.
We follow [Dl], [DM]and [02]. The study of the geometry of the Mandelbrot set needs basic concepts on iteration theory, Fatou set and Julia sets. First we review these basic concepts. Our main work is on 'some folk theorems' about the geometry Figure 1.1
- 1.2 Scope of the thesis
In Chapter 2 we revien- some basic concepts of Discrete Dynamical systerns, CompIes
Analysis, Fatou and Julia sets for a cornplex polynomial.
In Chapter 3 the definition and construction of the Mandelbrot set and some prop-
erties of the Mandelbrot set are given. We aIso discuss period doubling bifurcation.
We present our niain work on the geometry of bulbs in Chapter 4. In this Chapter
Ive discuss the geomerry of the bulbs, geometry of the aiitennas and esternal rays.
The main tool of the discussior~is the doubling fiiiictioii aiid rotation map. We state
md prove some folk theorcms in this Chapter. Chapter 2
Preliminaries
In this chapter we review some basic concepts of complex analysis, discrete dynamical systems. Throughout this chapter we follow [BE],[D3] and [KF].
2.1 Fixed points, periodic points and their nature
Definition 2.1 Let @, = @ U {m) and P be a polynomzal of degree n 3 1. A point zo E 2s sazd to be a fized point of P if P(ro)= zo. Moreover, zo 2s
(2) attracting zf 1 Pf(zo)l< 1;
(ii) super-attractzng 2j Pf(ro)= 0;
(222) repellzng if 1 Pt(ro ) 1 > 1 ;
(zv) neutml if 1 P1(zo)1 = 1; Definition 2.2 Let P be a polynomial of degree n > I and a E @,. The sequence
is called the orbit of zo under P. The point zo is a peBodic point of period k if
Pk(zo)= zo and Pq(zo) # zo for 1 5 q < k. If zo is a periodic point of period k, then the sequence {zo,P(zo), P2(zo), P3(4.. . Pk-l(~o)) is a k-cgcle.
Lemma 2.1 A point zo is an attractzng fied point of a polynomial P if and only if
3 a nezghborhood U of zo such that
(i) P(U) C U;
Definition 2.3 The basin of attraction ofzo 2s the set of al1 points whose orbit tends to q-, and it is denoted by B(zo),that is
B(zO)= {Z : Pn(z)+ 20, as n + m).
Example 2.1 Let P : & + @, be defined by P(z)= z2. Let U = {z : Izl < r <
1) be a neighborhood of z = O. Then
Inductively, we get Pn(U)= {z : [zl 5 rZn). SO Pn(U)= {O).
Definition 2.4 Suppose zo is a periodic point of period k for the polpomial P. Let
A = (Pk)'(zO)Then the k cycle {zo,P(zo), P2(zo), P3(z0)? ..., Ph-'(z0)}is
4 (2) attrncting if IXI < 1;
(iz) super-attractzng if X = 0:
(iii) repellzng zj 1 XI > 1;
(zv) neutral if IXI = 1;
2.2 Fatou sets and Julia sets
To study the Mandelbrot set we need the concept of Julia set. The Julia set is related to the Fatou set. In this subsection we define Fatou and Julia sets of a polynomial
P in terms of equicontinuity of the family {Pn),>o.- Since equicontinuity is closely related to normal families of analytic functions, using the notion of normality we can get more information about the Fatou and Julia sets. Hence, we discuss normality.
Definition 2.5 Let (X,d) and (Xi, dl) be two rnetric spaces. A family 7 of maps of
(X,d) into (Xi, dl) is equicontinuous ut xo if and only if for every E > O there exists
6 > O such that for al1 x E X and for al1 f E 3
The family 3 is eqzGicontinuous on a subset Xo of X if it is equicontinuous at each point xo of Xo.
Theorem 2.1 [BE]Let 3 be any family of maps , each mapping (X, d) into (Xi,di).
Then there is a maximal open subset of X on which 3 is equicontinuous. In particular,
5 if P maps a metric space (X,d) into itself, then there as a maximal open subset of A' on which the family of iterates Pn zs equicontinous.
Definition 2.6 Let P be a complex polynomzal of degree n 2 2. Then the Fatou set
F(P) of P is the maximal open subset of @, on which {Pn),>o- is equicontinous and the Julza set J(P) of P is the complement of the Fatou set in &. In other words,
Fatou set is the stable set and Julia set is the unstable set. We see the Fatou and the
Julia sets in the following example.
Definition 2.7 A family 3 of maps from (Xi,di) into (X2,d2) is said to be a normal or a normal farnily in Xi if every infinite sequence of functions from F contains a subsequence which converges locally unijormly on every compact subset of ATi.
Example 2.2 Let P : @, -+ @, be defined by P(z) = z2. Here fixed points of
P are 0,l and oo. P'(z) = 22 and IP'(Z)~.=~= O. Thus O is a super-attracting fixed point of P. Also IP'(z) l,=' = 2 > 1 and thus 1 is a repelling fixed point of P.
If we consider z E @, such that lzl < 1, Le., the open unit disc. Shen the family
{P"(Z)},~~is a normal family (Example 2.1). So the unit disc is a subset of the Fatou set. Similarly if lzl > 1, then the family {P"(Z))~>~- is also a normal family. We sketch the proof: Let Izl > R > 1. Then 1z21 > R2 and 1z2"1 > R2". Thus IPn(z)l -+ oo.
Thus {z : lzl > 1) is also a component of Fatou set. If we consider lzl = 1 then we have complicated situation and the family {Pn(z))n>o- is not a normal family. So from the definition of Fatou and Julia set, the set {z : Ir1 < 1} U {z : (21 > 1) is the
Fatou set for P and the Julia set for P is the unit circie.
6 Definition 2.8 Two maps F, G : @, + @, are conjugate to each other zf there e&ts a homeomorphism h : & -;, @, such that h O F = G o h. It is clear that h O Fn = GnO h. Thus, h maps orbits of F to those of G. Szmzlarly h- takes orbits of
G to those of F. Hence, the conjvgacy h gzves a one-to-one cowespondence between orbits of F and G.
Theorem 2.2 (Ascoli-Arzela)[BE]Let D be a subdomain of the complex sphere and let 3 be a family of continuous maps of D into the complex sphere. Then 3 zs equicontinuous in D if and only .if it is a normal family in D.
Theorem 2.3 (Montel)[KF/ Let {F,) be a family of complex analytic functions on an open dornain U. If {Fn}is not a normal family, then for al1 w E @ wzth at most one exceptzon we have F,(z) = w for some z E U and some n.
Example 2.3 Let Q,(z) = r2 +c. Let us study the dynamics of Qc for the following cases:
(i) QC(r)+ O if [z1 < !j + JG;
(ii) QF(z) -t oo if lzl > $ + Jf + ICI;
Let R = ? + and IzI < R The zeros of Q,(z) are I\/c and I\/cl < R
(shown below). Since R = 5 + \/f - ICI, we have R-R2=Icl- SinceR>l/2wehaveR-R2cR2andIcl So the zeros of Q,(z) are in Izl < R. Thus in lzl 3 R, the function does not have zeros and is holomorphic. Now on the circle C(0,R) we have Now, by the maximum principle < in the domain Izl < R. This implies 1 Qc(r) 1 < R 1 2 1 2. Applied inductively Hence, Again, let R = $ + \/a + Ici and lzl > R. The zeros of QC(z) are ffi and 1 fil < R. So the zeros of Q, (r)are in 1 zl < R and By the minimum principle I,*I Q( > in the domain 1.~1 > R. This implies IQc(2)l > Rlgl2. But > 1. Applied inductively, IQ;(z)l ~lftl~"+ 00. Hence, 2.3 Some properties of the Julia and the Fatou sets Definition 2.9 Let P be a poiynomial of degree n and E c cC,.Then E 2s (9 forward invariant if P(E) = E. (22) backward invariant if P-'(E) = E. (iii) completely invanant if P(E) = P-'(E) = E. If P is surjectzue then the concepts of backward invariance and complete invariance coincide. Proposition 2.1 Attmcting fized points of a pofynomial P lie in the Fatou set of P and repelling jixed points of P 12e in the Julia set. Proof. Let P(zO)= z0 and zo be an attracting fùred point of P, that is, 1 P' (zo)1 = X < 1. We need to show that zo E F(P). Let U = {z : lP1(z)I 5 X + c < i}n D(zo,r), where D(zo,r) is an open disk with radius r. Now IPn(z)- zo 1 + O for z E U,because It does not depend on z, Le., Pn(z)+ zo uniformly on U. Shus Li is a subset of the Fatou set for P and consequently zo E F(P). Now, let zo be a repelling Gxed point of P. We need to show that zo E J(P). It is clear that Pn + oo does not hold uniformly on U,because Pn (z0)= 20, n = 1,2, - - - . It is enough to show that I(Pn)'(zo)lis not bounded. Now Proposition 2.2 Let zo be an attracting fied point of P. Then the basin of attraction of ro, B(zo)= {z : Pn(z)+ zo, as n -t m), zs in F(P). Proof. In the Proposition [2.1] we have proved that the family {Pn) is normal on a neighborhood U, U C B(zo),of ro. Let z E B(zo). Then, for some k 3 1, Pk(z)E U. Let V c U be a neighborhood of Pk(r).Then P-~((Vis a neighborhood of z and P-*(V) c B(zo). Then the family {Pn}on PWk(V)is normal, because Pn + zo uniformly on P-&(v). Thus, z E F(P). This proves that the basin of attraction of zo is in F(P). m Proposition 2.3 Let P be an analytic and suppose that zo is a repelling periodic point for P. Then the family of iterates of P at ro zs not normal at so. Proof. Let z0 be a Lued point of P. Assume that {Pn)is normal on a neighbor- hood U of zo. Since Pn(zo)= zo for all n, it follows that Pn(z)does not converge to oo on LI. Thus assume some sequence of {Pn}has a subsequence {Pni)which converges uniformly to G on U. Hence 1 (Pn')'(zo)1 -t IGf(zo)1. But 1 (Pn')'(zo)1 -+ W. This contradiction establishes the result for repelling kedpoint. i Proposition 2.4 If P zs a polynomial, then J(P) is compact. Proof. We know by the definition of the Fatou set, F(P) = @ \ J(P). The set z E @ such that there is an open set V nrith z E V and {Pn)normal on V is an open set. So J(P) has open complement and is closed. Now we need to show that J(P) is bounded. Since P is a polynomial of degree n 2 2, so we rnay find r such that IP(z)[2 21.4 if lzl > r, which implies that IPk(z)l > 2"r if Ir1 > r. Thus Pk(z)+ rn uniformly on the open set V = {z : Izl > r). Now by definition {pk}is normal on V so that V c C\J(P) * VC1 (C\J(P))" = J(P). + VC= {z : Izl 5 r) 3 J(P).So J(P)is bounded and J(P) is compact. a Proposition 2.5 J(P) as non-empty. Proof. Suppose that J(P)= 0. Then for every r > O , the family {Pn}is normal on the open disc B(0,r) with center the origin and radius r. Since the closed disc B(0,r) is compact and it may be covered by a finite number of open sets on mhich Pn is normal. Since P is a polynomial, taking T large enough ensures that B(0,r) contains a point z for which IPn(z)1 -t ao and also contains a fixed point w of P with Pn(w) = w for al1 n. Thus, we see that it is impossible for any subsequence of {Pn)to converge uniformly either to a bounded function or to infinity on any compact subset of B(0,r) which contains both z and w, which is a contradiction for the normality of {Pn).i Proposition 2.6 Let P be a polynomial and w E J(P)and let U be any neighborhood of W. Then W = Uzl Pn(U) zs the whole of @, except possibly for a single point . Any such exceptional poznt is not in J(P) and is independent of w and U. Proof. We know that for the Julia set J(P),the family {Pn)is not normal at W. So by the Montel's Theorem [2.3] we have @ = W,except possibly for a single point. Now we need to show that exceptional point is not in J(P). Suppose v 4 W. If P(z) = u, then since P(W) c W,it follows that z @ W. As C\W consists of at most one point, then 2 = v. Hence P is a polynomial of degree n such that the only solution of P(z) - v = O is u, which implies that P(z) - v = c(z - v)~for some constant c. If t is sufficiently close to v, then convergence being uniform on {z : Iz - vl < 12~1-A).Thus {Pn) is normal at v, so the exceptional point u 4 J(P). Clearly v only depends on the polynomial P. i Corollary 2.1 [KE'] The following holds for al1 z E @ wzth at most one exception (2) If U is an open set intersecting J(P) then P-"(2) zntersects U for injnitely many value of n. (iz) If z E J(P) then J(P) zs the closure of Uz, Fn(z)- Corollary 2.2 If P is a polynomial, then J(P) has empty interior. Proof. Suppose J(P) contains an open set LI. Then J(P) > Pn(U)for al1 n and we know that the Jufia set is both forward and backward invariant. So we cari write J(P) > Ur=f=,Pn(U). Now by Proposition [2.6] J(P) is a11 of @ except possibiy for one point. Which is a contradiction, since by Proposition [2.4] J(P) is bounded. Hence, J(P) has empty interior. i Chapter 3 The Mandelbrot Set and Bifurcation in Complex Quadrat ic Polynomial In the study of Complex Dynamical system, the study of the Mandelbrot set is one of the most intricate and beautiful subjects in Mathematics. To study the Mandelbrot set, first we need to study the dynamics of complex quadratic polynomial. In this chapter we study the filled Julia set, the Mandelbrot set and some basic properties of them. At the end of this chapter we also discuss periods of the bulbs and rotation numbers. Throughout this chapter we follow [BE], [02], [D3]. 3.1 Complex Quadratic polynomial, Filled Julia set and the Mandelbrot set Definition 3.1 Consider the Quadratic compfex function Q, : cC, -+ @, defined by Q,(z) = z2 + C, where c is a complex parameter. In complex dynamics critical points play an important rofe. Note that O is the only critical point for Qc(z) . Proposition 3.1 Let P(z) = az2 + bz + d with a # O and b, d E @. Then P 2s conjugate to QC(z)= z2 + c for some c E @. Proof. We need to find a homeomorphism H : cC, + û& such that Let H (z) = az + b/2, where c = ad + b/2 - b2/4. Clearly H is a homeomorphism. Now H O P(z)= H(P(z))= ~(az~+ bt + d) = a(az2+ bz + d) + b/2. = a2z2+ abz + ad + b/2. Similarly QC H(z) = QC(H(z))= Q&z + b/2) = (az + l~/2)~+ C. = (az + b/2)* + ad + b/2 - b2/4 = a2z2+ abz + ad + b/2. Hence P is conjugate to Qc. rn Proposition 3.2 For any complex number X # 0, there is a c = c(X) for which the quadratic map P(z) = Xz(1 - z) is analytically conjugate to QC(z) with respect to a map of the form h(z) = az + b. Proof. Let QC(t)= z2 + c and P(z) = Xt(1 - I). By Proposition [3.1] a = -A, b = A, and d = 0 and H(z) = az + b/2 = -Xz + A/2, where Now Qc o H = H o P. Then, Q,(-Xz + X/2) = (-Xz + X/2)* + c = X2z2 - X2z + X2/4 + X/2 - X2/4 = h2z2 - X2z + X/2. Similarly, H(Xz - Xz2) = -X(Xz - Xz2) + ,412 = ~~z~- X2z + A/2. So P is conjugate to Q,. u Definition 3.2 (Filled Julia set) The filled Julia set, Kc of the complex quadratic polynomial Q, is the set of points whose orbits do not tend to m. That is Kc = {z E @ : QZ(z) is bounded ). Remark 3.1 in Chapter 2 we saw that if t E J(Qc) then the orbit of Q:(z) is bounded. So the Julia set is a subset of the filled Julia set. The filled Julia set also contains any attracting periodic orbit and its basin of attraction, if there is such an orbit. Both Kc and its complement are clearly completely invariant under Q,. Definition 3.3 (Mandelbrot set) The Mandelbrot set is a subset of the parameter plane which consists of those c values for which the orbit of 0 under the complex 16 quadratic functzon Q,(z) = z2 + c is bounded ( c is complex parameter). That zs, M = {c : Q:(O) is bounded ). Example 3.1 i E M and 2i $ M. Since c = i then the critical orbit of Qiis which is bounded. Otherwise the critical orbit for is : which is unbounded, since IQ&(O)I becomes bigger than max (2, (cl). Proposition 3.3 [D3] Let P(z) be an analytic funetion satisfying P(0) = O and Pf(0)= X with O < 1 XI < 1. Then there zs a neighborhood U of O and an analytic map H : U -t @ such that Po H(z) = H O L(z), where L(z) = Xz. Proposition 3.4 [D3] Let [al < 1. De/ine Ta(z)= E. Then Ta is analytic for Ir1 < lai-'. Moreouer T'' = T-. for Izl < 1 and Ta : D + D. Theorem 3.1 Let P be a polynornial and suppose that zo zs an attracting periodic point of P. Then, there is a cn'tical value whzch lies in the basin of attraction of zo. Proof. We prove this only for an attracting fixed point. Suppose zo is an attract- ing fked point. Then, by Proposition [3.3] there is a neighborhood U of zo and an analytic homeomorphism H : U + D(open unit disc) which linearizes P. Let V be an open set containing U and such that P : V + U is onto. We need to prove that either P has a critical point in V or else P has an analytic inverse P-' : U + V. Let us assume that P does not have an analytic inverse on U. Since P is analytic and surjective on V, it follows that P must not be one-to-one. So there exist 21, z;? E V such that P(q)= P(4= q. Suppose that H(q) = a and let Ta: D + D be as given in the previous proposition. Now, consider the circles Cr of radius r < 1 centered at O in D. TL' pulls this fam- ily of circles back to nested sequence of simple closed curves surrounding 0. Since P : V + U is analytic and P1(zi)# O for each i. It follows that for r small, P-'(H-' 0 TL' (Cr))is a pair of families of nested circles, one family centered at zl and the other centered at 22. Here P-' denotes the inverse image, not the in- verse map. Now, as r increses there is a smallest r, for which these two families first intersect. Let p be a point common to both simple closed curves of the form P-' (H-' O TL' (Cr.)).Shen, p is easily seen to be a critical point for P. Thus, we may continue to define P-' on successively larger domains of attraction of zo until we either meet a critical point or else exhaust the irnmediate attractive basin of zo by constructing Pdk : U LI C for al1 positive k. Since for any k, P-"U) cannot cover al1 of Cl indeed P-'(U) omits the basin of attraction of oo which con- tains (~11~1> R) for some sufficiently large R. However, the family of maps P-* is not normal on U,as zo is a repelling fixed point for each. So by Montel's Theorem we must have that UEoP-k covers @ minus at rnost one point. This contradiction establishes the result. i 3.2 Role of the critical orbit From the definition of the Mandelbrot set M we know that M consists of those c values for which the orbit of the critical point O under Qc is bounded. Theorem [3.l] says that the immediate basin of attraction of an attracting cycle contains a critical point. Since O is the only critical point of Q,, O is in the basin of attraction of an attracting cycle. That is, orbit of O under Qc converges to the cycle. So the Mandelbrot set M contains al1 c values for which Q, has an attracting cycle. The complement of the Mandelbrot set consists of those c values for which Q:(O) + W. 3.3 Bifurcation in Q,(z) and visible bulbs of the Mandelbrot set We will find values of c for which Qc has an attracting fixed point. Suppose cr and be two fxed points of QJz) = r2 + c in @ . So Qc(a) = a and Q,(P) = 0, that is, a and are the roots of z2 - z + c = O. Thus So Q:(a)+Q:(P) = 2(a+p) = 2. This shows that not both cr and B can be attracting. It follows that Q, can have at most one attracting fixed point, and the condition that one of the fixed points, Say cr is attracting is 214 = IQ:(a)l < 1, and c = a - a*. So the set of c we are seeking is just the image of the disc {a : la1 < $1 under the map Figure 3.1 Xow we want to find all c values for which Q, has an attracting 2-cycle. The equation for fixed points of Qz is Q:(z) = z. Thus, we have to solve Sirice cu and 0 are fised by Q,, so tliey are also fixed by Q:(z). Ths Q,(z) - z clil-ides -I=,/L--:,C-L) atid discriniiiianr of z2+ z + cf 1 = O is A = 1 - 4c+ 1). R,oots of QC are 2 For ail attracting 2 cycle we have Q,-(u) = v, Q&) = u,u # v, togethcr with 1 (Q;)'(U) 1 < 1, 1 (Q:)'(v) 1 < 1, so u and v are attracting fised points of Q:. Now the condition for attracting 2-cycle is I(Q:(u))Qé(v))J c 1. This implies Hence 11 + cl < f. So the set c we are looking for is the disc {c : 11 + cl < i). This Fiyre 3.2 set. of c 1-ziliics for \rliir.li Q, lias ail att.r;tr.t.irig3- c.yc.lc arc t.hc 1)iilIx iri t.Iict t.011 aiiiI 1~ot.toiiiof tlic carciioici. \Ire c-an contiiiuing this prowss to gct the al1 visihlc biilbs the Maiidelbrot set.. I\é shall non- describc eli~!change iii d'-ilarnics as c niows fr-oiii tlie cardioid to the disc tlirough the value -$. Wlieii c is in the cardioid Q, lias ail at.tractiiig fised point n aiid a repelling 2-cycle {u!v). -As c + - $: tlie poi1it.s a. 11 and v converges to a coilimon value -:- aiid wlieii c = -?, n, u and u coincide at a neutral fised point of Q,(z). Since fked points of z2 - z - c = O are o; = 1-"y- alid fl = L~J1T4;-2 AS c now moves into the dix, tliese points separate again, this thne with a being a repelliiig fised point and {u,u) being an attracting 2 - cycle: thus dunng this process the attracting nature of cu has beeii transferred to pair {u;Y). This process is repeated: as parameter c moves further along negative part of x- mis, {u, v) losing its attracting nature to a 4- cycle, then to an 8- cycle, and so on. This is known as period doubling and one can easily verify it on a computer by esainining the lirnit of Q:(O) for various values of c. 1+-,/m Proposition 3.5 Let QC(t)= z2 + c, ç, = Q:(O) and a = q,0 = 2 Then (i) ifc<-2, thenc, 22+n[c+21 whennz 2; (ii) if -2 5 c < O, then Qc maps [-,O, O] into itself; (iii) if O 5 c 5 f, then Qc maps [O,a] into itself; (iv) if c > f, then ç, 3 n(c - -&t Moreover, the intersection of the Mandelbrot set with the real line zs 1-2, f]. Proof. (2) Suppose c < -2 and n 2 2 . Let n = 2 then So it is true for n = 2 . Assume that it is true for k. Then, Now we need to show that it is true for k + 1 . We have C~+L= C: + c 2 4+4klc+ 21 + k21c + 212 + c > 2 + (k + l)lc+ 21, since (4k - (k+ 1) - l)lc + 21 + c + 2 + Ic + 21 + k21c + 212 > O- The last inequality follows from k 2 2 and t + Itl 2 O. So by mathematical induction it is true for any n > 2, i.e., ç, 3 2 + nlc + 21, Vn 2 2. (ii) Suppose -2 5 c < O and Qc : [-Pt fl] + @ by Q.(z) = z2 + c, where @= 2 which is equivalent to D2 + c = 0. Let z E [-O, ,BI + Izl 5 @ =+ z2 5 02. Thus Q,(z) = z2 + c 5 a2 + c = 0. Hence QC(t)5 ,O. Now p= If 2 and -2 5 c < O. This irnplies 1 < @ 5 2. Thus, z E [-O, O] 3 1 < z2 $ 4. Now -2 5 c < 0, 1 < z2 5 4, -2 5 -P < -1, which implies that z2 + c 2 -0, Vz E [-P, a]. SO -P 5 QC(z).Combining this two result nre get 1-JI-4c' (iii) Suppose O 5 c 5 a and Q, : [O, a] -t Q3 by Q,(z) = r2+c, where a = . Thus, O 5 c 5 ! + O 5 a 5 and hence, 1 1 1 z~[O,a]*O Thus, O 5 z2 + c $ i. Hence, QC(z)E [O, a]. (iv) Suppose c > !. Let S(n): c, 3 n(c - 2). So 1 S(l): = QJO) = 2 (C - 4)- Thus S(n)is true for n = 1. Suppose S(k) is true then ck 2 k(c - f). We have since k2t2- kt + 114 2 O for dlt. So S(n)is true for k + 1. By mathematical induction Qc(n) is true Vn 2 1. Hence c, 2 n(c - f). From the definition of the Mandelbrot set we conclude that M î) R = [-2, f]. i Theorem 3.2 The Mandelbrot set M is contained in the closed disk of radius 2. i. e., A!! C {c: ICI 5 2). Proof. Suppose that Ici > 2. Then, if Izl 2 Ici we have IQc(z)I = lz2 + cl = 14 jz + tl 2 14 (14- b) 2 lzl(lcl - 1) = l+, where r = ICI - 1 > 1. NOWQ,(O) = c, IQ:(o)l = [QC(c)l= 12 +cl 2 Iclr. So it is true for k = 2. Suppose that it is true for k, i.e, QE(0) 2 Iclrk-' . We need to show that QOC1(0) 2 Iclrk Now IQ:+~(o)I = IQ~Q:(O)I2 IQCW lr 3 lclrk. So by mathematical induction IQa (O)1 2 Iclrk for al1 k. Since r > 1 so IQ~(o)I> lclrk -t m as n -t m. Consequently c: 4 kr. i 3.4 Algorithm for the Mandelbrot set The above discussion in Section 3.3 gives us an algorithm for computing the Mandel- brot set. We consider a rectangular grid in the c-plane. As we know, the Mandelbrot set M is contained in the circle of radius 2 centered at the origin, so we will always assume that our grid contained inside the square centered at the origin wbose sides have length 4. For each point c in this grid, we compute the corresponding orbit of O under Qc and ask whether or not this orbit tends to infinity. If the orbit does not escape, then our original point is in M, so we will color the original point black. If the orbit escapes then we leave the original point white. More precisely we have the following algorithm: Choose a maximum number of iterations N. For each point c in a grid, compute the first N points on the orbit of O under Q,. If IQ:(0)[ > 2 for some i $ N, then stop iterating and color c white. If IQ:(0)1 5 2 for some i 5 N, then color c black. 3.5 Periods of the bulb The bulbs which are directly attached to the main cardioid are called the primary bulbs. In this section we mil1 only discuss the primary bulbs. From the discussion in Section 3.3 we see that for each c values such that Qc has an attracting k-cycle and we have a bulb which is a part of the Mandelbrot set M. So we can assign a number for each bulb of the Mandelbrot set. This number is called the period of the corre- sponding bulb. For example, the biggest bulb in the left of the cardioid has period 2, because this bulb corresponds to al1 c values for which Q, has an attracting 2 cycle. Another example, we see is the topmost bulb on the cardioid has period 3, because this bulb corresponding to al1 c values for which Q, has an attracting 3 cycle. Again consider the period 5 bulbs. Because these bulbs correspond to al1 c values for which Figure 3.4 3ow ive can also read off the period of the primary bulbs by simply counting the spokes of the antenna attaclied to this decoration [D']. For a rough proof consider the follon-ing example. Note that each bulb features a large antenna that contairis a ju11ci;ion point from ~vhicha number of spokes enlanate. It is a fact that the number of tliose spokes is esactly the period of the period of the biilb. When counting these spokes, it is importaiit to corint the main spoke ( the spoke that is attaclied directly to the biilb), ttiis spoke is callcd the principal spolie. Cotisider ci bulb n-hicli lias attracting 3 cycle. Tiim we sc!c that this bulb has 3 spokcs of anteiina attaclicct to of antcnria attached to this bulb and ont! spokc is directly attached to this brilb. Xow counting clocliwise direction froni the rtiuiri spoke we have 4 spokes of antenna. So this bulb has period 4. The following figure (figure 3.5) shows the antena arid spokes of period 3 and period 4 bulbs. Period 3 bulb Period 4 bulb Figure 3.5 Another esample, consider a bulb which has attracting 5 cycle. Then, Ive see thar tliis bulb has a spokes of antenna attached to this bulb and one spoke is Elirectly attached to this bulb. Now counting clockwise direction forni the niain spolce ive have 5 spokes of anteniia. So this bulb has period 5. Similady we cari fiiid periods of other bulbs. In the follou-iiig figure (figure 3.6) WC! Iiwe clisplayed several of the ~~riniirrydecoratioiis Period 7 bulb Period 5 bulb 3.6 Rotation numbers In Section 3.3, u-e saw that for each primary bulb we have a number q which is the period of the bulb. In this section we will associate a number p/q rvith each bulb where q is the period of the bulb. The rational number p/q is called a rotation number for the bulb. Ué have to determine the number p. There are some niethods involving Julia set J, to determine the nuniber p. In titis thesis we n-ant to determine tlic riumber p without cotiiputing the Julia set. The niettiod involves oiily looking at the antenna attaclieci to t lie bulb. To fitid p, note that tlicrc are CI spokcs eiiiaiiating froiii the junction poiiit in tlic iriûin iititciina attacheci to t!~(bIIIL~IJ- Lt~(-i~t.(! t,!lf! S~OLWS~of th(^*. SOI- I~~IISI.IJIL!I)S t.liis S~LO~~.VSI;S!IOI;(! is !CN:;LI.(![I J oi~sr ILCJIL~~S Ir.Fr 1 IL^ sol Xow tliis is n(J t iilwi~striie if rvc measure the 1crigt.ti irsi riz t,li(i iisiinl Eiicliclrari (listiince. Rather wc siioilld use a distance tliat assigns il stiort.er lciigtli to the spolie closer to the tnain spolie. Consider the period 3 bulb which has 3 spokes of antenna ernanating from the junction point and one spoke (principal spoke) directly attached to this bulb. We see that the shortest spoke is located 113 of a tiirn in the counterclockwise direction from the principal spoke. So this bulb is called l/3 bulb. 1/3 bulb 3/7 bulb Figure 3.7 Consider the period 7 bulb which lias i spokes of nnteiitia ernanating from the jiiiic- tion point and one spoke (principal spoke) directly attaciicd to this bulb.We see tliat the shortest spoke located 3/7 of a turn in the coiinter cioclcwise direction frotii the 11/12 bulb 5/12 bulb Figure 3.8 Similarly we can find 11/12 bulb and 5/12 bulb and so on. The following figure shows that various type of bulbs: Figure 3.9 31 Chapter 4 Geometry of bulbs in the Mandelbrot set, Farey tree and Fibonacci sequence 4.1 Introduction So far we have discussed the Mandeibrot set in Chapter 3. It turns out that the Mandelbrot set M consists of c values for which the orbit of O under the quadratic map Qc(z) = z2 + c is bounded. In the Mandelbrot set, the bulbs attached directly to the main cardioid are called the primary bulbs. Each of these bulbs has a rotation number p/q. Recently R.L. Devaney and M-Moreno Rocha [DMl, R.L.Devaney [02] studied the geometry of primary bulbs in the Mandelbrot set. In this chapter we 4.2 Farey addit ion, Farey tree, Fibonacci sequence The Mandelbrot set featirres a basic cardioid shczpe from which hang nurnerous " bulbs" or 'decorations" (Figure 4.1). Each of these bulbs is a large disk that is directly attached to the main cardioid, together wit h numerous other smaller deco- rations and a prominent " antenna" . Figure 4.1 In Chapter 3. we showveci that each of these large (%sks turns contains c values for mhicli Q, admits an attracting cycle with period q and rotation nttnher p/q. Tliat is, the attracting cycle of Q, tends to rotate about a central fised point, turning or1 average p'q bnlb reroliitioiis at each iteration. For tliis reasoii tliis l~iilbis callcd the p/q bulb. Each of the c values in this bulb has essentially the same dynamical behavior . In Chapter 3 we showed that we can recognize the p/q bulb from the geometry of the bulb itself. That is we can read off dynamical information from the geornetric information contained in the Mandelbrot set. In fact we can recognize the plq bulb by locating the "smallest" spoke in the antenna and determining its location relative to the main spoke. This is our first folk theorem that we want to discuss. Now consider 112 bulb and topmost 1/3 bulb. In between these two bulbs there are infinitely many bulbs directly attached to the main cardioid. Now we want to find the largest bulb of al1 other bulbs between 112 and 113 bulb. To find this we need the following definition. Definition 4.1 (Forey addition) Let pi/ql ,pz/qz E [O, 11. V plq2 - prql = =t1 then pl/ql and p2/q2 are called Farey neighbors. The Farey addition is defined bg i.e., we sirnply add the corresponding fractions just the way we always wanted tu add them, namely by addzng the numerators and denominators. Thus, Our second folk theorem that we want to discuss will state that 2/5 bulb is the largest bulb of the bulbs between 112 bulb and 113 bulb. That is the largest bulb between Figure 4.2 Similarly the largest bulb of bulbs between 113 and 7/5 is 318 bulb. Sirnilady the largest bulb of bitlbs bettveen 318 and 2/5 is 3/13 bulb. Continuing in this way we get a special sequence n-hose numerators and denominators correspond to the Fibonacci sequence. We n-il1 also discuss t his connect ion. Lemrna 4.1 Eue~yrational nurnber-p/q E (O, l),(p, q) = 1, can be exp-essed uniqzdy as a Fareg sum Proof. Arniiigc (1 points aroiiiid the circle atiti iiiiirk ii clistingiiiscd first point. Move counterclockwise around the circle, each time marking the point which is p points beyond the one previously marked. Continue as long as no two adjacent points are marked. Let ql be the number of marked points. By rotational symmetry, the next marked point would be adjacent to the first one, so that for some integer pl, which satisfies pl < p since q > ql. If Ive set Euclidean aigorithm. It remains to prove uniqueness. Suppose pll/ql' and p2'/q2' are a second pair with p/q as the Farey sum; without loss of generality we assume their order is the same as that of the non-primed pair of fractions. Then and plqi - ql'l = plpl - pl'l. Since p and q coprime, and Iql - ql1l < q, the Last eq is impossible unless q, - qlt = pl - = 0. . Definition 4.2 (The Farey Tree) The Farey tree is a tree which contains al1 of the rationals between O and 1. Let us understand this notion with an example: let 011 and 111 be a pair of neighbors (rationals). Using the Farey addition we get a Farey child which is the rational between 0/1 and 1/1 and whose denominators is the smallest [JF]. We Say that in this case 011 and 1/1 are Farey parent of the Farey child 1/2. Thus we get Now if we consider 011, and 112 as Farey parent we get a new Farey child 113 and if we consider 112 and 1/1 as Farey parent we get another new Farey child 213. In this stage we get In this next stage we get and so on. From lemma 4.1 we know that the Farey tree contains al1 rationals between O and 1. Recall from Chapter 3 that the visible bulbs in M correspond to c values for which Q, has an attracting cycle of some given period. For example, the main central cardioid in iM consists of c values for which Q, has an attracting hed point. This can be seen by solving for the fked points of Q,(z) = z2 + c that are attracting: IQC(z) ( = l2tl < 1. Solviog these equations simultaneously we see that the boundary of this region is given by c = z - z2, where z = That is, parametrizes the boundary of the cardioid with O 5 0 5 1. At c(B), Q,(B) has a fixed point that is neutral and the derivative of Qc(B) at this point is e2"'0. For each rational value of 0, there is a bulb tangent to the main cardioid at c(0). For c values in the bulb attached to the cardioid at c(p/q), Qc has an attracting cycle of period q. We cal1 this bulb the p/q bulb attached to the main cardioid and denoted it by BPI,. It is known that as c passes from the main cardioid through c(p/q), and into hl,, Qc undergoes a p/q bifurcation. By this we mean when c lies in the main cardioid near c(p/q), Q, has an attracting hed point with a nearby repelling cycle of period q. At c(p/q) the attracting fixed point and repelling cycle merge to produce a neutral fked 27ri 9 point with derivative e . When c lies in BPI,,Qc now has an attracting cycle of period q and a repelling fuced point. When c = c(p/q), the local(1inearized) dynamics are given by rotation through angle 2n(p/q). As a consequence, for nearby c E BPI,, the attracting cycle rotates about the repelling fixed point by jumping approximately 2~(p/q)radians at each iteration. The doubling map and the rotation map is the main tool of the discussion of the geometry of bulbs. We discuss the doubling map in the next section and the rotation map in Section 4.7. 4.3 Angle doubling mod 1 Definition 4.3 The doubling function, which is defined on the circle conszdered as the reals rnodtllo one, zs gzven by D(0) = 28 mod 1, i-e., Lemma 4.2 The angle 0 is periodic under doubléng function D af and only if 6 is a rational of the form plq wzth q odd. Proof. Let 9 = p/q, then D(p/q) = 2p/g, or D@/q)= (2p - q)/q. Similarl- we have Dn(9)= 2"8 - integer. Assume that 8 is eventually periodic then we have Since there is only finite number of possible numerators 0 is eventually periodic. Thus integer, - integer, 8 = 2" - p 7 which means that 0 is a rational number. Thus every eventually periodic point is rational. Now we have to prove that periodic points are k/odd. Assume that 0 = oddleven then 8 is not periodic. Because odd odd - euen12 or eue@ euen12 and al1 further images have denominator euen/2. Thus 0 cannot be periodic. Now if 0 = k/q with q odd then 0 is periodic. Now we want to prove that every image Dn(k/q)is of the form p/q , where O 5 p 5 q and we also proved that 0 is periodic. Let and let (n, j) be the srnallest such pair. Then (i) if j = O then O is periodic. (ii) if j > O then 8 is not periodic. If j > O we can consider Dn-l(k/q) and DjWL(klq). Since the pair (n,j) is minimal so Rn-'(O) and D-'(O)must be different and be on different sides of 112. Let Dn-'(8) = pl/q and @-'(O) = p2/q. Then This implies q is even. Which is a contradiction. Thus the only possibilities is (i) and 8 is periodic with q odd. i Remark 4.1 The rationals with odd denominator are periodic and the rational with even denominator are al1 strictly preperiodic- The following are some examples of periodic, pre periodic and eventually jixed points. 40 Example 4.1 The D-orbit of 213 is which has period 2. Similarly the D- orbit of 1/3 is which has period 2. Example 4.2 The rational 317 has penod 3 under doubling : Similarly 117 and 217 have period 3 under doubling: Example 4.3 The rational 2/5 has period 4: Similarly 1/15, 3/15 and 7/15 bave period 4: and and Example 4.4 The rationd 9/31 has period 5 : Similarly the rational 1/31 has period 5: and 21/31 has period 5: Example 4.5 Consider 1/10 ,where q is even.The rationals with even denominator are eventually periodic but not periodic. The D-orbit of 1/10 is 1 12431 ----r..., 1055555 So 1/10 lies on an eventual 3-cycle. Example 4.6 The rationai with even denominator 1/4 is eventually fixed : 4.4 Itinerary and Binary Expansion: Using the doubling function on the circle we can get the binary expansion of 8 where O 5 8 < 1 by noting the itinerary of @ in the circle relative to D. To define the itinerary, we denote the upper semicircle O 5 0 < !j by Io and the Iower sernicircle 5 9 < 1 by Il. Given O, we attach an infinite string of B(9) as follows: where O when Dj(0) E Io, Sj = 1 when Dj(f9)E Il That is ,we simply watch the orbit of O in the circle under doubling and assign O or 1 to the itinerary whenever Dj(0) lands in the arc Io or Il. Lemma 4.3 The ztznerary B(0) is the binary expansion of 0. Proof. Let where Oi are either O or 1, be the binary expansion of 0 E (O, 1). Apply the doubling function D then Here el is the integer part which is either O or 1. Now and Here O2 is the integer part which is again O or 1. We proceed in this way and we get B(0) = 818283- --a is the itinerary of 8. ) Example 4.7 If 8 = 2/3, then O E Il, while D(B) E Io and D2(8) = O. Hence B(2/3) is the repeating sequence m,which of course is the binary expansion of 213. - Similarly, B(3/7) = 011, while B(2/5)= 0110, B(9/31) = O1OO1 and B(10/31) = O1O1O 4.5 External Rays In order to make precise the folk theorems mentioned in the introduction, we recall some beautiful results of Douady and Hubbard [DHZ] concerning the esternal rays of the Mandelbrot set. Let E = {z : IzI > 1) denote the exterior of the unit circle in the plane. .4ccording to Douady and Hubbard, there is a unique analytic isomorphism @ that maps E to the exterior of the Mandelbrot set. The mapping O takes w to oo and positive mals to positive reals. The mapping is the uniformization of the exterior of the Mandelbrot, or the exterior Riemann map. The importance of O stems from the fact that the image under 44 the external ray with external angle O0 to be the a- image of 9 = e0. It is known that an external ray whose angle Bo is rational "lands" on M. That is li&+ @(~e'"'~~1 exists and is a unique point on the boundary of M. This c value is called the landing point of the ray with angle &. The ray with angle O lies on the real axis and lands on Ad at the cusp of the main cardioid, namely c = 1/4. Also the ray with angle 1/2 lies on the negative reaI axis and lands on M at the tip of the "tail" of M, which can be shown to be c = -2. Consider now the interior of M. The interior consists of infinitely many simple connected regions. A bulb of M is a component of the interior of kf in which each c-value corresponds to a quadratic function that admits an attract ing cycIe. The period of this cycle is constant over each bulb. In many cases, a bulb is attached to a component of lower period at a unique point called the root point of the cornponent. Theorem 4.1 [DH2] Suppose a bulb B consists of c-values for which the quadratic map has an attractzng q- cycle. Then the root point of this bulb is the landing point of exactly 2 rays, and the angles of each of these rays have period q under doubling. Example 4.8 The angles of the external rays of M determine the ordering of the bulbs in M. The large bulb directly attâched to the left of the main cardioid is the 1/2 bulb, so two rays with period 2 under doubling must land there. Now the only angles with period 2 under doubling are 113 and 2/3, so these are the angles of rays that land at the root point of B1p . In this case root point of BLI2= -3/4. Example 4.10 Consider the ?/5 blilh.Tli(: 2/5 httlb lies between the 1/13 and 1/2 bulbs. Hence the rays that land at c(2/5) must have period 5 under doubling and lie betrveen 2/7 and 1/3.The only angles that have this property are 9/31 and 10/31,so these rays must land at ~(215).we have secn in the ahove figure. Figure 4.3 4.6 Idea of Limbs The idea of external rays allow us to measure the "largeness" or "smallness" of por- tions of the Mandelbrot set. Suppose we have two rays with angles 0- and O+ that both land at a point c, in the boundary of M. Let M' be the component of Ad which touches the main cardiod at c,. Then, by the isomorphism a, al1 rays with angles between 0- and O+ must approach the component of M* \ {c.) cut off by 8- and O+. Thus, it is natural to measure the size of this portion M* of ILI by the lengtb of the interval [O-, O+]. The root point of the plq bulb of M divides M into two sets. The component M* containing the p/q bulb is called the p/q lirnb. We can then measure the size of the p/q limb if we know the external rays that land on the root point of the p/q bulb. 4.7 Rays landing on the p/q bulb In order to make the notion of 'large' or 'small' precise in the statement of the folk theorems, we need a way to determine the angles of the rays landing at the root point of BPI,. We denote the angles of these two rays in binary by l*(p/q), where 1- (p/q) < l+(p/q). We cd1 1-(p/q) the lower angle of BPI,and l+(p/q) the upper angle. As we will see , lI(p/q) is a string of q digits (O or 1) and so lI(p/q) denotes the infinite repeating sequence whose basic block is l*(p/q). Douady and Hubbard [DHl] have a geometric method involving Julia sets to determine these angles. Our method is more combinatorial and resembles algonthms due to Aleta [A]: La Vaurs [LV], and Lau and Schleicher[LS]. To describe this algorithm, let &/,(O) denote rotation of the unit circle through p/q turns, Le., We consider the itineraries of points in the circle under R using two different partitions of the circle. The lower partition of the circle is definecl as follows. Let The boundary point O belongs to 1; and -p/q = 1 - p/q belongs to Io. We then define s- @/q)to be the itinerary of p/q under hl,relative to this partition. CVe cal1 the basic repeating block of this itinerary, s-(plq), the lower itinerary of plq. That is where sj is either O or 1 and the digits O when R~~,@/q)E Io, Sj = 1 when Ri~;(p/q)E IF We also define the upper partition I,f and 1: as follows: The upper itinerary of p/q, s+@/q)is then the repeating block of the itinerary of p/q relative to this partition. Note that 1; and IF differ from Io and 1; only at the end points. - Example 4.11 Consider s- (l/3) = 001 , since and the orbit 113 + 2/3 + 1 -t 113 - - - lies in Io, Io, Ir, respectively. Similarly - s,(1/3) = 010, since 1: = [O, 2/3), 1: = [2/3, l), and the orbit is 1/3 + 213 +O + 113---. This orbit starts in I:, hops to I:, and then returns to IO before cycling and the orbit lies in I$, I:, I,f respectively. - Similarly ~(213)= 101 , since Io = (0,1/3], 1; = (1/3,1] and the orbit - 2/3 + 1/3 + 1 + 213- - - lies in Ir, Io, Ir, respectively. Sirnilarly s+(2/3) = 110' since Io' = [O, 1/31, IT = [1/3, l), and the orbit 2/3 + 113 + O -t 213 - - - lies in Ir, I:, Iz respectively. Example 4.12 Consider 315 then the lower itinerary of 315 is s-(315) = 10101, since 1, = (0:2/51, Ir = (2/5,1], 49 and the orbit is 3/5 -t 115 + 415 + 215 + lies in Ir,Ir, Ir, Io, Ir, respectively. Similarly the upper itinerary of 315 is s+(3/5) = 101 10, since and the orbit is 315 + 1/5 + 415 -+ 215 + lies in IF, I,f, I:, I:, Il, respectively. Example 4.13 Consider s-(215) = 01001 since the orbit is 215 + 415 + 115 + 3/5 -t 215 + - - and &- = (0,3151, I; = (3/5,1]. and the orbit lies in Io, IL, I;, Iz and 1; Similarly s+(2/5) = 01010, since the orbit is 215 + 4/5 + 115 -+ 3/5 + 215 + - and This orbit starts in I,f, hops to Ir, and then returns to 1: before cycling and the orbit lies in I:, Ir,I:, IF, I:, respectively. 4.8 An algorithm for computing the angles of rays landing at c(p/q). Theorem 4.2 [DHl,DM] The rays lI@/q) landing ut the root point c(p/q) of the p/q bu16 are given by s- (plq) and s+(p/g). Example 4.14 Let p/q = 114. Shen the lower external angle 1-(114) = O001 and the upper external angle 1+(1/4) = 0010, so that s-(1/4) = 1/15 = O001 in binary 50 and s+ (114) = 21 15 = O010 in binary. Since from the binary expansion we have Similarly we have - 1 - 2 1-(114) = ~(114)= O001 = - and 1+(1/4) = s+(1/4) = O010 = - 15 15- Example 4.15 Consider p/q = 315 , then the lower angle 1-(315) = 10101 and the upper angle 1+(3/5) = 10101, so that s- (315) = 21/31 = 10101 in binary and s+(3/5) = 22/31 = 10110 in binary. From binary expansion we have Similarly Thus and Remark 4.2 Note that sI(p/q) dger only in their last two digits provided q 2 2. Indeed we may write The reason for this is that the upper and lower itineraries are the same except at R:;:(~/~)= -p/q and ~z;i(p/q)= O, which jorm the endpoints of the two partitions of the circle. We now define the szze of the p/q Limb to be the length of interval [s-(p/q), s+(p/q)]. That zs, the size of the p/q limb is given by the number of extemal rays that approach thzs lzmb. We rnay cornpute the szze of these bulbs explzcitly by uszng the fact that s*(p/q) difler only in the last two digits. Theorem 4.3 The size of the p/q limb zs 1/(29 - 1). That zs, s+(plq) s-@/q) = - 2q - 1 Proof. We write the binary expansion of the difference in the form Example 4.16 Consider 1/3 limb then, 1 ~+(1/3)- S-(1/3) = 2/7 - 1/7 = 1/7 = - 23 - 1- Similarly the size of the 2/5 limb is 1/31, since 1 ~+(2/5)- S-(2/5) = 10/31- 9/31 = 1/31 = - 25 - 1- As we see in the following Figure , the visual size of the bulbs does indeed correspond to the size as defined above. 53 Figure 4.4 4.9 The size of Limbs and the Farey Tree In this section Ive relate the size of a p/q limb to the size of the limbs corresponding to the Farey parents of p/p. The following proposition relates the upper and lower itineraries of p/q and its Farey parents. Proposition 4.1 Suppose a//3 and y/d are the Farey parents ofp/q and that O < a/P < y/& < 1. Then the lower itineranj s-(p/q) consists of the first q digits of the upper angle s+(rr/4) of the srnaller parent, and the upper itineraq s+(p/q) consists of the first q digits of the lower angle s-(716) O/ the larger- parent. Proof. We consider only s+(p/q);the proof for s- (plq) is similar. From Defiiiitioii of Farey addition we have ph - qy = =tl, so ive have Consider the orbits of p/q and 7/6 relative to the respective rotations R,,/,and kls.Since y/b rotates faster than plq, the distance between these orbits advances by 1/96 at each iteration. We thus have It follows that R:/,(~/~)lies within 116 units of ~:/,(.//6) provided j < q - 1. Since points on the orbit of y/b under ~:/~(y/6)under Ryldeltalie exactly 1/6 units apart on the circle, it follows that the first q - 1 entries in the itineraries of p/q and y/6 are the same, provided we choose the lower itinerary for y/6 and the upper itinerary for p/q. The reason for this is that the orbit of y/6 lies ahead of that of p/q in the counterclockwise direction, but by no more than 1/6 units. Choosing the upper itinerary for p/q and the lower for y/b forces the corresponding digits to be the same. When j = q - 1, we have R;;,' (p/q) = O and Hence Therefore the qth digit in s+(p/q) is O and so is the qth digit of y/6, as long as ~/6# 1. i Example 4.17 Consider 113 and 112 are the Farey parents of 2/5 and O < a < 0 < 1 and the upper itinery of 1/3 is So by Proposition 4.1 we have s-(2/5) = 01001, since q = 5. Similarly the lower angle of 1/2 is So by Proposition 4.1 we have since q = 5 If one of the Farey parents is O or 1, we must modie Proposition 4.1 Proposition 4.2 Suppose that O zs a Farey parent of plq. Then the q digits zn the lower itinerary of plq are s-@/q) = 0.-U1. If 1 is a Farey parent of p/q then s+ (p/q)= 1 - . -10. Proof. For s-(p/q), we first consider 0/1 is a Farey parent of p/q then from the definition of Farey addition we have p/q = and po - qa = 1 and pd - qy = -1, where a = O and @ = 1, so we must have p = 1. Thus, s-(p/q) is given by the itinerary of l/q unit under counterclockwise rotation by l/q units. We therefore have It follows that the first q - 1 digits of s- (l/q)are 0, and the last digit is 1. If a Farey parent is 1/1, the proof is similar. For s+ (p/q),we note that, since 111 is a Farey parent, we must have p = q - 1, since 7 = 1 and 6 = 1. Thus, s+(p/q) is 56 given by the itinerary of (q - l)/q unit under counterclockwise rotation by (q - 1)/q units. We therefore have It follows that the first q - 1 digits of s+((q - l)/q)are 1, and the last digit is O. Thus s+((q- l)/q) = 1 o-*-. 10. . We now complete the proof of one of the folk theorerns mentioned in the intro- duction. Theorem 4.4 Suppose y/& 5 1. Then the sire of the p/q limb zs larger than the szze of any other lzmb between the a/B and y/& lzmbs. Proof. Assume first that neither parent is O or 1. Proposition 4.1 and 4.2 ensure that s- (plq) and s+(cY/~)agree in their first q digits. Using these binary representa- Similarly This implies that the arc of rays between the plq limb and either of its parents' limbs has length no longer than 1/29. From Theorem 4.3, we know that As this quantity is larger than 1/29, it follows that the p/q limb attracts the largest number of rays between its two parents. If one of the parents of l/q is 011, then the size of the l/q bulb is again 1/(2q - 1): while the gap between 011 and s- (plq) = 0 -..O1 is also 1/(2q - 1). But then any limb between the 1/q limb and the cusp of the cardioid must have size strictly smaller than 1/(29 - l), again showing that the l/q limb is the largest. Now if one of the parents of 1/q is 1/1 ,then the size of the l/q bulb is again 1/(29 - l),while the gap between 111 and s+(p/q) = 1 ---IO is also 1/(29 - 1). But then any limb between the l/p limb and the cusp of the cardioid must have size strictly smaller than 1/(2q - l),again showing that the l/q limb is the largest. m Example 4.18 Consider one of the 113 is 0/1 , then the size of the 113 bulb is 1 ~+(1/3)- S-(113) = 217 - 117 = -- 117. 23 - I Again if we consider 114 limb between 113 and the cusp of the cardioid must have strictly smaller than m,since s+(1/4) - s-(114) = 2/15 - 1/15 = 1/15 = - 1) < 1/(2~- 1) = 117. Thus 113 is the largest limb. 4.10 The Fibonacci sequence Theorem 4.4 shows that the Fibonacci sequence appears in the Mandelbrot set. As we have seen in Figure 4.2 the largest bulb between the 112 and 1/3 bulb is the 2/5 58 main cardiod. The foliowing figure stiow that the corresponding bulbs forming the Fibonacci sequence. The Fibonacci sequence. Figure 4.5 4.11 Conclusion The technique of measuring the size of certain portions of the Mandelbrot set by the length of interval of rays that land on the portion provides justification for other folk theorems involving the size of the Mandelbrot set M. For example this technique is used to identify the p/q bulb using the length of the spokes in its antenna. Once we know these rays we can easily compute the lengths of the various spokes. We can also use the two rays separating the principal spoke from the rest of the antenna to determine a list of the q rays that land on the junction point. Then we can determine that the shortest is located p/q turns in the counterclockwise direction from the principal spoke. Bibliography [A] P. Atela. Bifurcations of Dynamzc Rays in Complex Polynomials of Degree Two Ergod. Th. and Dynam. Sys. 12(1991), 401-423. [BE] A. F. Beardon. Iteration of Rational Functions, Springer-verleg, 1991. [BL] P. Blanchard. 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