THE AREA OF THE MANDELBROT SET AND ZAGIER S CONJECTURE

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1 #A57 INTEGERS 8 (08) THE AREA OF THE MANDELBROT SET AND ZAGIER S CONJECTURE Patrick F. Bray Department of Mathematics, Rowan University, Glassboro, New Jersey brayp4@students.rowan.edu Hieu D. Nguyen Department of Mathematics, Rowan University, Glassboro, New Jersey nguyen@rowan.edu Received: 9/6/7, Accepted: 6/8/8, Published: 6/9/8 Abstract We prove Zagier s conjecture regarding the -adic valuation of the coe cients b m that appear in Ewing and Schober s series formula for the area of the Mandelbrot set in the case where m mod 4.. Introduction The Mandelbrot set M is defined as the set of complex numbers c C for which the sequence {z n } defined by the recursion z n z n + c () with initial value z 0 0 remains bounded for all n 0. Douady and Hubbard [5] proved that M is connected and Shishikura [5] proved that M has fractal boundary of Hausdor dimension. However, it is unknown whether the boundary of M has positive Lebesgue measure, although Julia sets with positive area are known to exist (Bu and Chéritat []). A natural problem is to determine the area of the Mandelbrot set. Towards this end, Ewing and Schober [8] derived a series formula for the area of M by considering its complement, M, inside the Riemann sphere C C [ {}, i.e. M C M. It is known that M is simply connected with mapping radius ([5]). In other words, there exists an analytic homeomorphism (z) z + X b m z m () m0

2 INTEGERS: 8 (08) which maps the domain {z : < z apple } C onto M. It follows from the classic result of Gronwall [0] that the area of the Mandelbrot set M C M is given by A " # X m b m. () m Determining the exact value of A remains an open problem. Numerical approximations based on pixel-counting due to Förstemann [9] give an estimate of A based on a resolution of almost 88 trillion pixels. On the other hand, approximating the series () using the first 5 million terms of b m gives an upper bound of A apple.6888 as computed in [] (cf. [4, 6]). The discrepancy in the two approximations indicates that either the series () converges very slowly or else the pixel-counting method fails to include area due to the fractal boundary of M. Thus, it is important to investigate the coe cients b m in (), whose arithmetic properties have been studied in depth, first by Jungreis [], then independently by Levin [, ], Bielefeld, Fisher, and Haeseler [], Ewing and Schober [7, 8], and more recently by Shimauchi [4]. There are many approaches to calculating b m (see [, 7, 8]). In this paper, we shall focus on the following formula for b m. Theorem (Ewing-Schober [8]). Suppose m apple n+. Define the set of n-tuples J {j (j,..., j n ) : ( n )j ( )j n + ( )j n m + }; and given any j J, set Then j (k) : (k) : b m where C jk ( (k)) is the binomial coe m n k+ k j k j... j k. m X J cient ny C jk ( (k)) (4) k C jk ( (k)) ( )( ) ( (j k )). (5) j k! Using formula (4) to compute b m is impractical as it requires determining the set of tuples J, which is computationally hard. However, since it is known that each b m is rational and has denominator equal to a power of, it is then useful to find a formula for its -adic valuation, which we define next. Definition. Let n be a non-negative integer. We define (a) (n) to be the -adic valuation of n. (b) s(n) (called the sum-of-digits function) to be the sum of the binary digits of n.

3 INTEGERS: 8 (08) Don Zagier, as described in [], formulated several conjectures based on an empirical study of the coe cients b m. One conjecture gives an upper bound formula for (b m ) that is valid for all m with equality holding if and only if m is odd. A proof of this conjecture for m odd was given by Levin [] and a proof of the full conjecture was given by Shimauchi [5]. Theorem (Levin [], Shimauchi [5]). Let m be a non-negative integer. Then (b m ) apple (m + ) s((m + )) (6) Moreover, equality holds precisely when m is odd. Zagier also conjectured a formula for b m in the case where m mod 4, namely where j k (b m ) (m + ) (m) and a partial formula when m 4 mod 8: j m 5 k (b m ) 7 j k s (m + ) + (m), (7) ( 0, if m mod 4;, otherwise, jm 5 k s + (m 0 ), (8) 7 where m 4m 0 with m 0 odd and (m 0 ) is given by the incomplete table with two entries unknown: m 0 mod (m 0 ) ?? More generally, Zagier conjectured some periodic patterns when m n m 0 and m 0 ( n + )k + l with k 0 and l odd, apple l apple n+ : where (b m ) n+ k s( n+ k) + (m 0 ), (9) 8 >< l, if l n+, k odd; (m 0 ) l, if l >: n+, k even; l +, if n+ apple l apple n+ 5. In this paper we prove Zagier s formula for the case m mod 4. Theorem 4. Suppose m mod 4. Then formula (7) holds for b m.

4 INTEGERS: 8 (08) 4 Our proof relies on determining those tuples j max J that maximize V (j) : ( Q n k C j k ( (k))), i.e., V (j) < V (j max ) for all j J. In particular, we show for m mod 4 that this largest -adic valuation V (j max ) is achieved by exactly one tuple j max or else by exactly three tuples j max, j 0 max, j 00 max in the special case where m mod 4. To prove that V (j) < V (j max ) for all j J, we derive lemmas to compare the values of V (j) for di erent types of tuples. For example, if m 8, then it holds that V ((,, 0, )) < V ((0, 5,, )) < V ((0, 0,, 0)), where j max (0, 0,, 0). We refer to the chain of tuples (,, 0, )! (0, 5,, )! j max as a set of tuple transformations. As a result of our comparison lemmas (derived in Sections and ), we have the result (b m ) + V (j max ). (0) This follows from the fact that the -adic valuation of the sum of any number of fractions (whose denominators are powers of and whose numerators are odd) is equal to the largest -adic valuation of all the fractions, assuming that there are an odd number of fractions with the same largest -adic valuation. It remains to calculate V (j max ) in each case, which then establishes Zagier s conjecture.. Tuple Transformations We begin with preliminary definitions and lemmas. Definition 5. Given j J, define j(k) : (k) : n k+ (k) m n j n j n k+ j k and B(k) ( n k+ )( n k+ ) ( (j k ) n k+ ). Lemma. We have (B(k)) j k for apple k apple n (m). Proof. First, we establish that ( (k)) (m) for apple k apple n (m). This follows from ( ) (m n j n j n k+ j k ) (m ( n j n j n k+ j k )) (m),

5 INTEGERS: 8 (08) 5 which holds since ( n j n j n k+ j k ) n k + > (m). Then by definition we have B(k) ( d n k+ )( d n k+ ) ( (j k )d n k+ ). Taking the -adic valuation of both sides and expanding the right-hand side gives (B(k)) ( ( n k+ )( n k+ ) ( (j k ) n k+ )) ( ) + ( n k+ ) + ( n k+ )+ + ( (j k ) n k+ ). Since n k + > (m), ( p n k+ ) for all integers p. Thus, as desired. Lemma. We have (B(k)) j k (C jk ( (k)) (n k + )j k s(j k ) () for apple k apple n (m). Proof. It is clear from Definition 5 that and thus C jk ( (k)) ( )( ) ( (j k )) j k! ( n k+ )( n k+ ) ( (j k ) n k+ ) j k(n k+) j k! B(k) j k(n k+) j k! (C jk ( (k)) B(k) j k(n k+) j k! ( (B(k) ( j k(n k+) j k!)) (n k + )j k + j k s(j k ) (B(k)) (n k + )j k s(j k ) since we have from Lemma that (B(k)) j k for apple k apple n (m). We now consider the case where k > n (m). Define c(x, y) to be the number of carries performed when summing two non-negative integers x and y in binary. It is a well known result that c(x, y) s(x) + s(y) s(x + y).

6 INTEGERS: 8 (08) 6 Lemma. Let j J. Then for k > n (m), we have 8 >< c(j k, (k) ), (k) < 0; (C jk ( (k))), 0 apple (k) apple j k ; >: c(j k, (k) j k ), (k) > j k. () Proof. First, we demonstrate that (k) is an integer when k > n (m). By definition, we have (k) m n k+ k j k j... j k. Since (m) n k +, it follows that m is divisible by n k+ m. Thus, is n k+ an integer, and since the remaining terms are all integers, (k) must be an integer as well. If (k) < 0, we have ( )... ( jk + ) (C jk ( (k))) j k! j k s(j k ) (( j k + )... ( ) ) j k s(j k ) ( (( j k + )!) ( )) s(j k ) s( ) + s( + j k ) c(j k, ). On the other hand, if 0 apple (k), then C jk ( (k)) 0, and therefore (C jk ). Lastly, if (k) > j k, then we have! (C jk ( )) ( j k )!j k! as desired. s( ) ( j k ) + s( j k ) j k + s(j k ) s(j k ) + s( j k ) s( ) c(j k, (k) j k ) Definition 6. For convenience, define 8 >< c(j k, (k) ), (k) < 0; (m, k) : (k), 0 apple (k) apple j k ; >: c(j k, (k) j k ), (k) > j k, () and for any tuple j J, define v(m, j) n (m) X k [(n k + )j k s(j k )] (4)

7 INTEGERS: 8 (08) 7 and V (m, j) : V (j)! ny C jk ( (k)). (5) In the case where m mod 4, and thus (m), we shall simply write k v(j) : v(m, j) [(n k + )j k s(j k )]. (6) k The next lemma follows immediately from Definition 6 and Lemmas and. Lemma 4. We have V (j) v(m, j) + and in particular if m mod 4, then kn (m)+ (k) V (j) v(j) + (n). (7) We now consider tuple transformations that allow us to compare v(m, j) for di erent types of tuples. Lemma 5. Suppose (m). Let j be a J-tuple and i < n (m) be an integer such that j i 6 0. Define the tuple j 0 (j, 0..., jn) 0 by 8 j k, k 6 i, i +, n; >< jk 0 j i r, k i; j i+ + p, k i + ; >: j n + q, k n, where r is the largest power of less than j i, and p and q satisfy with q < n i. Then ( n i )p + q ( n i+ )r (8) v(m, j) < v(m, j 0 ). Proof. It is clear that p and q exist by Euclid s Division Theorem. Then since j k jk 0 for all k 6 i, i +, n, the corresponding terms will cancel when we compute the di erence v(j 0 ) v(j). If i < n, then v(m, j 0 ) v(m, j) (n i)p (n i + )r + s(j i ) s(j i r) + s(j i+ ) s(j i+ + p) (n i)p (n i + )r + s(p) > n i n i + p dlog (p)e 0

8 INTEGERS: 8 (08) 8 since r < (p + )/ and p. The remaining case, i n, can be easily proven by similar means. Observe that we can apply Lemma 5 repeatedly to transform any tuple j J containing a non-zero element j i, apple i apple n (m), to a tuple j 0 J with j 0 i 0. Thus, any tuple j J can be transformed to a tuple j 0, where all elements j 0 i 0 except for i n (m), with v(j) < v(j 0 ). We will make use of this fact later on. Lemma 6. Let j be a J-tuple where j n >, and j 0 be the tuple such that 8 j k, apple k apple n (m) ; >< jk 0 j n (m) + p, k n (m); 0, n (m) < k < n; >: P n kn (m)+ (n k+ )j k ( (m)+ )p, k n, where p is chosen to be as largest as possible so that j 0 n < (m)+. Then v(m, j) < v(m, j 0 ). Proof. We have that v(m, j 0 ) v(m, j) (n (m) + )(j n (m) + p) s(j n (m) + p) (n (m) + )j n (m) + s(j n (m) ) (n (m) + )p + s(j n (m) ) s(j n (m) + p) (n (m) + )p + c(j n (m), p) s(p) (n (m) + )p s(p) > 0. In particular, when m mod 4, Lemma 6 allows us to transform a tuple j J, whose elements are all zero except for j n and j n >, to a tuple j 0 J, whose elements are also all zero but with jn 0 apple, so that v(j) < v(j 0 ).. Zagier s Conjecture In this section we prove Zagier s conjecture for the case where m mod 4, which we assume throughout this section. In order to accomplish this, we first derive additional lemmas that allow us to compare V (j) for the tuple transformations described in the previous section.

9 INTEGERS: 8 (08) 9 Lemma 7. If m + 0 mod, then V (j) < V (j 0 ) for all j 6 j 0, where j 0 (0, 0,..., m+, 0). Proof. By Lemmas 5 and 6, we can transform j to a tuple j 0 so that ji 0 0 for all i < n since (m). Moreover, jn 0 (m + )/ and jn 0 0 since m + 0 mod. It follows that V (j) [(n k + )j k s(j k )] c(j n, (n) ) k apple [(n k + )j k s(j k )]] v(j) k < v(j 0 ) V (j 0 ) since c(j 0 n, 0 (n) ) 0 due to Lemma 4. Lemma 8. If m + mod, then V (j) < V (j 0 ) for all j 6 j 0, where j 0 (0, 0,..., m, ). Proof. We have V (j) < V (j 0 ) by the same reasoning as in the previous lemma. Lemma 9. If m + mod and m mod 8, then V (j) < V (j 0 ) for all j 6 j 0, where j 0 (0, 0,..., m, ). Proof. Again, such a tuple j 0 exists because of Lemmas 5 and 6. We first determine the binary representation of (n). Since (n) j n ( + j + + j n ) m m m m 0 6 and m mod 8 by assumption, it follows that (n) has binary representation b n b 00. It follows that c(, (n) ) 0 and thus V (j 0 ) v(j 0 ) by

10 INTEGERS: 8 (08) 0 Lemma 4. Moreover, we have V (j 0 ) [(n k + )j k s(j k )] c(j n, (n) ) k (m ) (m ) m s c(, (n) ) (m ) s. It remains to be shown that V (j) < V (j 0 ) for all j 6 j 0. This follows from This proves the lemma. V (j) [(n k + )j k s(j k )] c(j n, (n) ) k apple [(n k + )j k s(j k )]] v(j) k < v(j 0 ) V (j 0 ). In order to handle the case m + mod and m 6 mod 8 (or equivalently m mod 4), we will need the following lemma. First, we define the following three special tuples, which exist for this case: j 0 (0, 0,..., m, ) j 00 (0, 0,..., m j 000 (0, 0,...,, m, 5), ). Lemma 0. Suppose m + mod and m 6 mod 8. Then for all j / {j 0, j 00, j 000 }, we have V (j) < V (j 000 ). Proof. Since j 000(n) is odd and j 000 n, we have c(j 000 n, (n) ) 0 and thus

11 INTEGERS: 8 (08) V (j) v(j). Moreover, we have V (j 000 ) [(n k + )j 000 k k s(j 000 k )] c(jn 000, (n) ) (m 7) m 7 s() + s (m ) m s (m ) m s. Thus, it su ces to show that v(j) < v(j 000 ) since this will imply V (j) apple v(j) < v(j 0 ) V (j 0 ). Note that for any tuple j containing an element j i 6 0 with apple i apple n, we have v(j) < v(g) for some tuple g with g i 0 for all apple i apple n and g n k for some k. To construct such a tuple g, we simply apply the tuple transformation in Lemma 5 repeatedly. We now consider cases. First, if g j 000, then the theorem holds trivially. If g n >, we proceed in two steps. Let 7(g n ) p + q where q <, and let g 0 be such that Then we have g 0 i 0 for apple i apple n g 0 n g 0 n g n + p g 0 n g n + q. v(g 0 ) v(g) + (g n + p) s(g n+ + p) g n + g n + s(g n ) p g n dlog (p)e + p dlog 7 (p)e + 7 > 0. Then applying Lemma 6 to g 0 completes the proof for this case. If g n 0, then we proceed as follows. Let m g n + p. Note that because g / {j 0, j 00, j 000 }, we have p. Thus, v(j 000 ) v(g) (g n + p) s(g n + p) g n + s(g n ) > 0. This completes the proof. p s(p)

12 INTEGERS: 8 (08) Lemma. If m + mod and m 46 mod 48, then V (j) < V (j 000 ) for all j 6 j 0, where j 000 (0, 0,...,, m 7, ). Proof. In light of Lemma 0, it su ces to prove that V (j 0 ) < V (j 000 ) and V (j 00 ) < V (j 000 ). We first consider j 0. We have j 0(n) m/ n j n j j n m/ (m )/ (4 m)/6, which implies (n) (m 0)/6 has binary expansion b n... b 0. Thus, c(j 0 n, (n) ) > 0 since j 0 n. It follows that V (j 0 ) [(n k + )jk 0 s(jk)] 0 c(jn, 0 (n) ) k < [(n k + )jk 0 s(jk)] 0 k (m ) (m ) (m ) V (j 000 ). m s m s m s As for j 00, we have c(j 00 n, (n) ) > 0 since j 00 n 5 and j 00(n) m/ n j n j j n m/ (m 4)/ (6 m)/6, which implies (n) (m )/6 has binary expansion b n... b 00. It follows

13 INTEGERS: 8 (08) that V (j 00 ) [(n k + )j 00 k < [(n k + )j 00 k (m 4) (m ) (m ) V (j 000 ). k s(j 00 k s(j 00 k )] c(jn, 00 (n) ) k )] m 4 s m s m s This completes the proof. Lemma. If m + mod and m mod 48, then for all j / {j 0, j 00, j 000 }. Moreover, V (j 0 ) V (j 00 ) V (j 000 ) V (j) < V (j 0 ) (m ) (m ) s Proof. Again, in light of Lemma 0, it su ces to prove that V (j 0 ) V (j 00 ) V (j 000 ). Write m 48q + for q N and so that the elements of j 0, j 00, and j 000 take the form 8 0, apple i apple n ; >< ji 0, i n ; 6q + 5 i n ; >: i n, (9) 8 0, apple i apple n ; >< ji 00 0, i n ; 6q + 7 i n ; >: i n, (0) and j 000 i 8 0, apple i apple n ; >< 0, i n ; 6q + 6 i n ; >: 5 i n.. ()

14 INTEGERS: 8 (08) 4 Then It is straightforward to show that Similarly, we have and j (n) (8q + ) < 0, j 0(n) (8q + ) < 0, j 00(n) (8q + ) < 0. V (j 0 ) j 0 n s(j 0 n ) + j 0 n s(j 0 n ) c(j n, j 0(n) ) () s() + (6q + 5) s(6q + 5) c(, 8q + ) q + s(q) s(5) q + 0 s(q). V (j 00 ) j 00 n s(j 00 n ) + j 00 n s(j 00 n ) c(j 00 n, j 00(n) ) (0) s(0) + (6q + 7) s(6q + 7) c(, 8q + ) q + 4 s(q) s(7) c(, 8q + ) q + 0 s(q) V (j 000 ) j 000 n s(j 000 n ) + j 000 n s(j 000 n ) c(j 000 n, j 000(n) ) (0) s(0) + (6q + 6) s(6q + 6) c(5, 8q) q + s(q) s(6) c(5, 8q) q + 0 s(q). Thus, V (j 0 ) V (j 00 ) V (j 000 ). The following theorem summarizes the form of the maximum tuple j max for the case m mod 4. Theorem 7. Suppose m mod 4. The maximum tuple j max occurs in the following form:. If m + 0 mod, then j max (0,..., 0, p, 0) where p (m + )/.. if m + mod, then j max (0,..., 0, p, ) where p m/.. If m + mod and (a) If m mod 8, then j max (0,..., 0, p, ) where p (m )/. (b) If m 46 mod 48, then j max (0,...,, p, ) where p (m )/.

15 INTEGERS: 8 (08) 5 (c) If m mod 48, then j max {(0,...,, (m 7)/, ), (0,..., 0, (m )/, ), (0,..., 0, (m 4)/, 5)}. We now have all the necessary ingredients to prove Zagier s conjecture. Proof of Theorem 4 (Zagier s Conjecture): We divide the proof into the following cases:. m + 0 mod.. m + mod.. m + mod and (a) m mod 8. (b) m 46 mod 48. (c) m mod 48. Case (): Write m + p for some positive integer p. Since m mod 4, it follows that p mod 4 and so p mod 4. Now, recall that j max (0,..., 0, j n, j n ) (0,..., 0, p, 0), we have (n) ( + p)/. Then using the relation we have c(j n, (n) ) s ( j n ) + s( (n) ) s(j n (n) ), (b,m ) (m) + [(n k + )j k s(j k )] s(j n ) s( (n) ) k + s(j n (n) ) + j n s(j n ) + p s(p) + p s(p) + bpc s(bpc) (m) + (m + ) s (m + ). Case (): Write m + p + for some positive integer p. Since m mod 4, it follows that p mod 4 and so p mod 4. Since in this case j max

16 INTEGERS: 8 (08) 6 (0,..., 0, j n, j n ) (0,..., 0, p, ), we have (n) p/. It follows that (b,m ) (m) + kn + s(j n (n) ) [(n k + )j k s(j k )] s(j n ) s( (n) ) + j n s(j n ) s(j n ) s(p/ ) + s(j n + p/ ) + p s(p) s((p )/) + s(p/) p s(p ) + p s(p) + p s(p) + bp + /c s(bp + /c) (m) + (m + ) s (m + ). Case ()-(a): Write m + p + for some positive integer p. Since m mod 8, it follows that p + mod 8 and so p mod 8. Thus, p has binary representation b r... b 0. Since j max (0,..., 0, j n, j n ) (0,..., 0, p, ), we have (n) ( p)/. It follows that (b,m ) (m) + kn + s(j n (n) ) [(n k + )j k s(j k )] s(j n ) s( (n) ) + j n s(j n ) s(j n ) s((p )/ ) + s(j n + (p )/ ) + p s(p) s() s((p )/) + s((p + )/) p s(p) s(p ) + s(p + ) p s(p) + s(4) + (p + ) s(p + ) + bp + 4/c s(bp + 4/c) (m) + (m + ) s (m + ). Case ()-(b): Write m + p + for some positive integer p. Since m 46 mod 48, it follows that p+ 46 mod 48 and so p 5 mod 48. Thus, p has binary representation b r... b 5. Since j max (0,..., 0, j n, j n ) (0,..., 0,, p

17 INTEGERS: 8 (08) 7, ), we have (n) ( p)/. It follows that (b,m ) (m) + kn + s(j n (n) ) [(n k + )j k s(j k )] s(j n ) s( (n) ) + j n s(j n ) + j n s(j n ) s(j n ) s((p )/ ) + s(j n + (p )/ ) + s() + (p ) s(p ) s() s((p )/) + s((p )/) p s(p ) s(p ) + s(p ) p (s(p) s()) (s(p) s()) + (s(p) s()) p s(p) p + s(p + ) 0 + bp + 4/c s(bp + 4/c) (m) + (m + ) s (m + ). Here, (m) 0 since m m 0, where m 0 mod. Case ()-(c): Write m + p + for some positive integer p. Since m mod 48, it follows that p + mod 48 and so p 7 mod 48. In this case j max (0,..., 0, j n, j n, j n ) (0,..., 0,, p, ) and thus the same argument applies as in Case ()-(b). This completes the proof of Zagier s conjecture. References [] B. Bielefeld, Y. Fisher and F. Haeseler, Computing the Laurent series of the map : C D! C M, Adv. Appl. Math. 4 (99), 5 8. [] D. Bittner, L. Cheong, D. Gates, and H. D. Nguyen, New approximations for the area of the Mandelbrot set, Involve 0 (07), No. 4, [] X. Bu and A. Chéritat, Quadratic Julia sets with positive area, Ann. of Math. 76 (0), No., [4] Y. Chen, T. Kawahira, H. Li, J. Yuan, Family of invariant Cantor sets as orbits of di erential equations. II. Julia sets, Intern. J. Bifur. Chaos Appl. Sci. Engrg. (0), No., [5] A. Douady and J. Hubbard, Iteration des polynomes quadratiques complexes, C.R. Math. Acad. Sci. Paris 94 (98), 6. [6] J. Ewing, Can we see the Mandelbrot set?, College Math. J. 6 (990), No.,

18 INTEGERS: 8 (08) 8 [7] J. Ewing and G. Schober, On the coe cients of the mapping to the exterior of the Mandelbrot set, Michigan Math. J. 7 (990), 5 0. [8] J. Ewing and G. Schober, The area of the Mandelbrot set, Numer. Math. 6 (99), [9] T. Förstemann, Numerical estimation of the area of the Mandelbrot set, preprint, 0, area.pdf [0] T. H. Gronwall, Some remarks on conformal representation, Ann. of Math. 6 (94-5), [] I. Jungreis, The uniformization of the complement of the Mandelbrot set, Duke Math. J. 5 (985), [] G. M. Levin, Theory of iterations of polynomial families in the complex plane, Teor. Funktsiĭ Funktsional. Anal. i Prilozhen 5 (989), (Russian), English translation in J. Soviet Math. 5 (990), 5-5. [] G. M. Levin, On the Laurent coe cients of the Riemann map for the complement of the Mandelbrot set, arxiv: [4] H. Shimauchi, A remark on Zagier s observation of the Mandelbrot set, Osaka J. Math. 5 (05), [5] M. Shishikura, The Hausdor dimension of the boundary of the Mandelbrot set and Julia sets, Ann. of Math. 47 (998), 5 67.