# On Connectivity of Fatou Components concerning a Family of Rational Maps

Authors: Junyang Gao; Gang Liu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

I. N. Baker established the existence of Fatou component with any given finite connectivity by the method of quasi-conformal surgery. M. Shishikura suggested giving an explicit rational map which has a Fatou component with finite connectivity greater than 2. In this paper, considering a family of rational mapsRz,t that A. F. Beardon proposed, we prove that Rz,t has Fatou components with connectivities 3 and 5 for any t∈0,1/12. Furthermore, there exists t∈0,1/12 such that Rz,t has Fatou components with connectivity nine.


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## Body

## 1. Introduction and Main Results

By Sullivan’s theorem [1], each Fatou component of a rational map is eventually periodic. Moreover, for periodic Fatou components, there are only four possibilities: attracting basin, parabolic basin, Siegel disk, and Herman ring. Attracting basins and parabolic basins are either simply connected or infinitely connected, a Siegel disk is simply connected, and a Herman ring is doubly connected. However, for nonperiodic Fatou components, the corresponding connectivity may be bigger than two.For any givenn∈ℤ+, Baker et al. [2] proved that there exists a rational map R which has a Fatou component with connectivity n by the method of quasiconformal surgery. M. Shishikura suggested giving an explicit example such that it has a Fatou component with finite connectivity greater than two. Beardon [3] investigated the family of rational maps as follows:
(1)R(z,t)=z2(1+t12z3)(1-tz)3(1-t4z),(z∈ℂ¯,t∈ℝ).He proved the following result.Theorem A.
For sufficiently smallt>0, there exists a Fatou component D of R(z,t) with connectivity three or four.At the same time, he claimed that one may be able to compute the connectivity ofD by further discussion. Qiao and Gao [4] verified that D has connectivity three for t∈(0,10-4). Moreover, for any given positive integer n, two different families of rational maps were constructed such that one of them has a Fatou component with connectivity n (see [4, 5]). However, the degree of rational maps satisfies with those conditions are increased as the number n increases. As the first step to study the problem of connectivity number of Fatou components in rational maps space with fixed degree, we just investigate the connectivity of any other Fatou component of R(z,t) as the real parameter t varies. In fact, we have the following results.Theorem 1.
Suppose thatR(z,t) is defined as in (1); then we have the following.(1)
For anyt∈(0,1/12], there exist two Fatou components Dt and Dt~ of R(z,t) with connectivities three and five, respectively. Moreover, R:Dt~→Dt is an unbranched covering.(2)
There existst∈(0,1/12] such that R(z,t) has one Fatou component with connectivity nine.Remark 2.
In order to draw the graphs of Julia sets and Fatou components of such rational maps in complex plane, we consider its conformal conjugate. Letg(z)=1/z, and put R~(z,t)=g∘R∘g-1(z,t)=(z(z-t)3(z-t4))/(z3+t12). It is easy to see that ∞ is a superattracting fixed point of R~(z,t) for any t∈(0,1/12]. By Figure 1, R(z,t) has Fatou components with connectivities three, five, and nine for some t∈(0,1/12] since R(z,t) and R~(z,t) have the same dynamical properties. Furthermore, by Figure 2, we know that R(z,0.004355) has Fatou components with connectivities eight and fourteen, and we conjecture that for any large integral N∈ℕ, there exists t0∈(0,1/12) such that R(z,t0) has a Fatou component with its connectivity bigger than N.The Julia setJ(R~(z,1/20)) and three Fatou components of R~(z,1/20) with connectivities 3, 5, and 9, respectively.
(a)
(b)
(c)
(d)Two Fatou components ofR~(z,0.004355) with connectivities 8 and 14, respectively.
(a)
(b)


## 2. Preliminary Lemmas

For the fundamental concepts and classical results of iteration theory of rational maps, see [2, 3, 6, 7]. In order to prove Theorem 1, we need the following four lemmas. Except for Lemma 3, the others are certain modifications of results which have been verified in [3].Lemma 3 (see [8, Proposition 2.5]).
Letf be a rational map of a degree larger than one, and let {Ui}i=1p be its one (super)attractive or parabolic cycle of periodic Fatou components. If one of Ui(i=1,…,p) is not simply connected, then ∪i=1pUi contains at least two different critical values for critical points in itself.In what follows,t∈(0,1/12] in R(z,t). It is easy to see that z=0 (resp., z=∞) is a superattracting (resp., repelling) fixed point of R(z,t). Let D00 be the Fatou component that contains z=0.Lemma 4.
The nonzero critical points ofR(z,t) lie outside the circle {z:|z|=3}.Proof.
R
(
z
,
t
) has exactly eight critical points, that is, 0,1/t,1/t and ξj(j=1,2,3,4,5), which are the solutions of the equation R′(z,t)=0. Obviously, z=1/t lies outside the circle {|z|=3}. By a calculation, each ξj(j=1,2,3,4,5) satisfies the following equation:
(2)2+tzH(z,t)=0,
where H(z,t)=t16z4-2t12(1+2t3)z3+5t11z2-2t4z+1-t3. Putting
(3)M=sup⁡{|H(z,t)|:|z|≤3,t∈(0,112]},
we can deduce that
(4)M≤81t16+54t12(1+2t3)+45t11+6t4+t3+1≤2.
By (2), the nonzero critical points of R(z,t) lie outside the circle {z:|z|=3}.Lemma 5.
D
00 is simply connected and {z:|z|≤1/2}⊂D00⊂{z:|z|≤3/2}.Proof.
Forz∈K={z:0<|z|≤3/2}, we have
(5)|R(z,t)z2|≤1+t12|z|3(1-t|z|)3(1-t4|z|)≤1+(3/2)3t12(1-(3/2)t)3(1-(3/2)t4)<32,|R(z,t)z2|≥1-t12|z|3(1+t|z|)3(1+t4|z|)≥1-(3/2)3t12(1+(3/2)t)3(1+(3/2)t4)>23.
If|z|≤1/2, then |R(z)|≤(3/2)|z|2≤(3/4)|z|; we can deduce that {z:|z|≤1/2}⊂D00.
Suppose thatD00 meets the circle {z:|z|=3/2}; take a point ω∈D00∩{z:|z|=3/2} and join ω to the origin by a curve σ⊂D00. It is easy to see that Rn→0 uniformly on σ; then there exists a unique positive integer k, such that Rk(σ) meets the circle {z:|z|=3/2}, but Rn(σ) does not meet {z:|z|=3/2} for n>k. Let ζ be a point where Rk(σ) meets {z:|z|=3/2}; we have
(6)|ζ|>|R(ζ)|>23|ζ|2=|ζ|.
It is a contradiction and thus D00⊂{z:|z|≤3/2}. Obviously, D00 contains only one critical point z=0 by Lemma 4; then D00 is simply connected by Lemma 3.Lemma 6.
R
-
1
(
D
00
) only consists of two Fatou components, that is, D00 and D01, which contains a triply connected domain Ω. Here, Ω={z:1/2t4-1/t≤|z|≤3/2t4+1/t,|z-1/t4|≥1/2t4}.Proof.
Takez∈Ω; by a simple calculation, we have
(7)t|z|-1≥12t3-2>800,12-t3<t4|z|<32+t3,|1-t4z|≥12.
It is easy to see that1/(t|z|-1)<1.002/t|z|; we have
(8)|z2(1+t12z3)(1-tz)3|≤|z|2(1+t12|z|3)(t|z|-1)3<(1.002)3|z|2(1+t12|z|3)(t|z|)3<3t.
Moreover, R(Ω)⊂{z:|z|≤6t}⊂{z:|z|≤1/2}. By Lemma 5, we have R(Ω)⊂D00 and Ω⊊D00, and there exists at least one Fatou component D01(≠D00) of R-1(D00). Noting that there are three (resp., two) zeros of R(z,t) in Ω (resp., D00), this implies that R-1(D00)=D00∪D01 and D01⊃Ω.


## 3. Proofs

LetD⊂ℂ be a bounded Fatou component of R(z,t). We denote the connectivity of D by n(D) and the unbounded component of ℂ¯∖D by Out(D). Let Int(D)=ℂ¯∖Out(D) and d|D:=deg⁡(R:D→R(D)). Moreover, we say that a component of R-n(D)(n∈ℕ) is a component D~ such that Rn(D~)=D. We say that D surrounds a point z(∈ℂ) (or a domain U(⊂ℂ)) if it satisfies z∈Int(D) (or U⊂Int(D)) and denote by D↺z (or D↺U). Denote the number of zeros and poles of R(z,t) in the interior of Jordan curve γ⊂ℂ by N(R,γ) and P(R,γ). In order to prove Theorem 1, we need the following propositions. Considering the connectivity of D01 in Lemma 6, we have the following result.Proposition 7.
D
01 is a triply-connected domain.Proof.
By a simple calculation, we have(9)R′(z,t)=zQ(z,t)(1-tz)4(1-t4z)2.
Here
(10)Q(z,t)=t17z5-2t13(1+2t3)z4+5t12z3-2t5z2+t(1-t3)z+2.
Note that
(11)Q(-2t+5t2,t)=-t3+35t6-122t9+524t12-910t15-1775t18+7750t21-8750t24+3125t27<0,Q(-2t+6t2,t)=42t6-144t9+648t12-1560t15-1800t18+12960t21-18144t24+7776t27>0,
we can deduce that there exists a point c0∈(-2/t+5t2,-2/t+6t2) such that Q(c0,t)=0 (c0 is a critical point of R(z,t)). Note that (-2/t+5t2,-2/t+6t2)⊂(-2/t,-1/t) and
(12)R(-xt,t)=x2(1-t9x9)t2(1+x)3(1+t3x),
and for any x∈[1,2], we have |R(-x/t,t)|>3/2; thus |R(c0,t)|>3/2. By Lemma 5, it follows that c0∉D01. Moreover, three critical points 0, 1/t (twice) lie outside of D01. Hence, D01 contains at most four critical points. By Lemma 6, since the triply-connected domain Ω⊂D01 and 1/t4∈J(R(z,t))∩{|z-1/t4|<1/2t4}, then χ(D1)≤-1. Applying the Riemann-Hurwitz formula to the threefold covering map R:D1→D00, we have
(13)χ(D1)+∑(kj-1)=3χ(D00)=3,
so ∑‍(kj-1)≥4. It follows that ∑‍(kj-1)=4 and χ(D1)=-1, and thus n(D01)=3.By Lemma5, D00¯ is bounded. Since ∞ is a repelling fixed point, then each component in the preimage of D00 is bounded. In fact, we have the following result.Proposition 8.
Each Fatou component ofR(z,t) is bounded.Proof.
By Proposition7, D01 contains four critical points and they tend to z=0 under Rn(z,t)(n→∞). Note that R(1/t,t)=∞; then the dynamics of R(z,t) are decided by the forward orbit of the critical point c0 in Proposition 7. If lim⁡n→∞Rn(c0)≠0, there exists at most one cycle of periodic components which is distinct from D00 by Sullivan’s theorem. Assume that this cycle exists, denoted by U1,…,Up; then
(14)F(R(z,t))=⋃n=0∞R-n(D00)∪(⋃n=0∞⋃i=1pR-n(Ui)).
Otherwise, F(R(z,t))=∪n=0∞R-n(D00).
Below we will prove that each component ofF(R(z,t)) is bounded. From the above analysis, let D be any component of F(R(z,t)); there exists n∈ℕ such that Rn(D)=D00 or Rn(D)=U1 (if it exists). In order to show that D is bounded, we need to prove that U1¯ is bounded. Suppose that U1,…,Up exist, and note that ∪n=0+∞Rn(c0)⊂ℝ∪{∞}; we have U1,…,Up which are (super)attracting or parabolic components by the Sullivan theorem, and thus c0∈∪i=1pUi. Without loss of generality, c0∈U1. By Lemma 3, Ui(i=1,…,p) is simply connected; then U1¯ is contained in the bounded component of ℂ.Remark 9.
By Proposition8, z=∞ and its preimages are buried points; here a point z0 in Julia set is called buried point if it is not on the boundary of any Fatou component.By Lemma6 and Propositions 7 and 8, let M1 be the component of ℂ¯∖D01 which contains 1/t4. Clearly, M1≠{1/t4}, and in the following, we denote M=Int(D01)∖(D01∪M1); then 1/t∈M. Below we consider the connectivity of R-1(D01).Proposition 10.
R
-
1
(
D
01
) consists of three Fatou components with connectivities 5, 3, and 3, respectively.Proof.
We claim thatR-1(D01) consists of three Fatou components. On one hand, since R(Out(D01))=R(M1)=Out(D00) and D01⊂Out(D00), there exists at least one component of R-1(D01) in Out(D01) and M1, respectively. On the other hand, since R(z,t) is monotone increasing from 0 to +∞ for z∈(0,1/t) and D01⊃Ω by Lemma 6, there exists a unique component D2 of R-1(D01) with D2∩(0,1/t)≠∅ and D2⊂M. We claim that
(15)D2↺1t.
Assume that (15) is true (in what follows, we will return to the proof of this fact later in the proof), and by the definition of interior at the beginning of this section, 1/t∈Int(D2). Since 1/t is a critical point with multiplities 2 and 1/t4,∞∉Int(D2), then R(z,t) is a 3-fold map from Int(D2) to some neighborhood U of ∞. Furthermore, we can easily deduce that D2 is the unique component of R-1(D01) in Int(D2) (otherwise, R(z,t):Int(D2)→U is at least a 4-fold map; it is a contradiction). Hence, d|D2=3 owing to R(z,t):D2→D01 which is a proper map. Obviously, we have d|R(z,t)=5, so the number of components of R-1(D01) in Out(D01) and M1 is exactly one, respectively, denoted by D1 and D3. Clearly, D1, D2, and D3 are mutually disjoint preimage components of R-1(D01) and d|D1=d|D3=1.
Below we prove (15). In fact, if D2 does not surround 1/t, we distinguish two cases to discuss and get a contradiction.(i)
IfD2 does not surround D00, it is easy to see that R(Int(D2))=Int(D01) and D00⊂Int(D01), and since D2 has no pole, then R-1(D00)∩Int(D2)≠∅, but it is a contradiction to Lemma 6.(ii)
IfD2↺D00, note that R(z,t)<0 for z∈(1/t,1/t4) and lim⁡z→1/t+R(z,t)=-∞, and we can deduce that there exists at least one component (denoted by D2^) of R-1(D01) with D2^∩(1/t,1/t4)≠∅. Obviously, D2^⊂Out(D01) and D2^∩D2=∅ since D2 does not surround 1/t. If D2^↺D00, then D2^↺1/t, and by a similar discussion as used in the case of D2, it is easy to see that d|D2^=3, but it is a contradiction to the fact that both M and M1 contain some connected components of R-1(D01). If D2^ does not surround D00, we also get a contradiction by a similar discussion of case (i). Hence, we get D2↺1/t.
Next we will acquire the connectivity ofDi(i=1,2,3). Obviously, c0∉D1∪D3, and by the Riemann-Hurwitz formula, n(D1)=n(D3)=3. Furthermore, we claim that the “free” critical point c0 in Proposition 7 is not contained in D2, and thus n(D2)=5. In order to prove that c0∉D2, we turn to show the stronger result as follows:
(16)D2¯∩ℝ-=∅.
Otherwise, assume that x∈D2¯∩ℝ-. Note that t is a real parameter and D2∩ℝ+≠∅; then D2 is symmetric with respect to real axis ℝ. We choose a Jordan curve γ in the interior of D2 such that γ is very close to ∂Out(D2) and symmetric with respect to ℝ, and we take a point z0 in γ∩{z∣z∈ℂ,Im⁡z≥0} such that z0 is one of the nearest points from x; denote the arc of γ from z0¯ to z0 in counterclockwise direction by Γ. Moreover, we can also choose a Jordan arc η between x and z0 in γ∩{z∣z∈ℂ,Im⁡z≥0} such that η∩Γ={z0}. Set η~={z∣z¯∈η} and take Γ~=Γ∪η∪η~; then Γ~ is a Jordan curve in D2¯ and Γ~∩∂D2⊂{x}. Since N(R,Γ~)=2, P(R,Γ~)=3, and argR(z,t) changes by -2π as z goes around Γ~ by argument principle, but argR(z,t) changes by -6π since d|D2=3; it is a contradiction. Hence, we get (16).By a similar argument as the one used in (15), we can deduce that D3↺1/t4 and D1↺D01. In fact, it is decided by the “similarity” of the Julia sets J(R(z,t)). By the definition of M1 and Lemma 6, for any x0∈M1∩ℝ, x0>1/2t4. Let x1∈ℝ- be the largest point of ∂D01 and let x2∈ℝ+ be the largest point of ∂D00; by Lemma 5, R(x1,t)=x2<2. Furthermore, by Lemma 6 and (16), the unique “free” critical point c0 in Proposition 7 satisfies c0∈(x1,0). Note that R(z,t) is monotone increasing in (x1,c0) and monotone decreasing in (c0,0); by a calculation, we can easily deduce that |R(c0,t)|<1/t2 for any x∈(-1/t4,0). It is easy to see that R(z,t) is monotone increasing in (0,1/t) from 0 to +∞; then the equation R(z,t)=x0 has only one real root in (-1/t4,1/t4) since R(z,t)<0 for z∈(1/t,1/t4). Since D2⊂M and M∩ℝ⊂(-1/t4,1/t4), we can deduce that the number of bounded components of ℂ¯∖D2 which intersects with ℝ is two (see Figure 3).Figure 3
Schematic diagram of∪i=02R-i(D00).For any componentD in the preimage of Di(i=1,2,3), it is easy to see that n(D)≥3 by the Riemann-Hurwitz formula. Furthermore, the connectivity n(D) is decided by the number of critical points in D and local degree d|D. It is easy to see that there exists at most one Fatou component which contains the free critical point c0; the following Proposition 13 shows that even if there is no critical point in D, its local degree may be larger than one. Therefore, we cannot give a complete description of connectivity of this family of rational maps. In order to get Proposition 13, we first consider the number of poles in Int(D).Proposition 11.
No preimage component ofD2 or D3 surrounds 0, 1/t or 1/t4.Proof.
We argue by contradiction and induction. LetD be a preimage component of D2 or D3; we distinguish the following three cases to discuss.
(i) Suppose thatD↺0; it is easy to see that D↺D00 since 0∈D00 and D≠D00.
(ii) Suppose thatD↺1/t; then D⊂M2 or D⊂Out(D2). If D⊂M2, since R(z,t)>0 for any z∈D∩(0,1/t)(≠∅) and R(z,t)<0 for any z∈D∩(1/t,1/t4)(≠∅), R(D)↺D00 by symmetry. If D⊂Out(D2), then D↺D00 or there exist two points z1∈(0,1/t)∩D and z2∈(1/t,1/t4)∩D. Since R(z1,t)>0 and R(z2,t)<0, then R(D) surrounds D00 by symmetry.
(iii) Suppose thatD↺1/t4; then either D or R(D) surrounds D00 by the similar proof of (i) and (ii).
In all, ifD surrounds any of 0, 1/t or 1/t4, then either D or R(D) surrounds D00. Note that if R(D) surrounds D00, we have D⊊R-1(D2) and D⊊R-1(D3) since D2∩ℝ-=∅ and D3⊂M1. To get the conclusion in this proposition, it suffices to prove that no preimage component of D2 or D3 surrounds D00.
Below we prove that no preimage component ofD2 surrounds D00 by induction.
LetD be a component of R-1(D2). Obviously, we have D∩D00=∅. Suppose that D↺D00; we will get contradictions by discussion.
(iv) Suppose thatD⊂Out(D01), and since (-∞,-1/t4)∩D≠∅, we get R(D)∩ℝ-1≠∅ which contradicts with D2∩ℝ-1=∅.
(v) Suppose thatD⊂M, and choosing a Jordan curve γ in D such that γ↺D00, it is easy to know that N(R,γ)=2 and P(R,γ)=0 or 3. Since R(D)=D2 and D2 does not surround D00, the contradiction can be deduced by argument principle.
Hence, any componentD of R-1(D2) cannot surround D00. Assume that any component of R-n(D2) cannot surround D00. Again, let D be a component of R-(n+1)(D2). Suppose that D↺D00; we still distinguish two cases to discuss.
(vi) Suppose thatD⊂Out(D01), and since R(z,t)<0 for any z∈D∩(-∞,-1/t4)(≠∅) and R(z,t)>0 for any z∈D∩(1/t4,+∞)(≠∅), then R(D) surrounds D00 by symmetry which contradicts with the assumption.
(vii) Suppose thatD⊂M, and since R(D) is a component of R-n(D2) and R(D)≠D, then R(D) cannot surround D00 by assumption. By a similar analysis as used in the case (v), we also deduce a contradiction.
Therefore, we get that no preimage component ofD2 surrounds D00. By a similar discussion as the above used in D2, any preimage component of D3 cannot surround D00.However, the conclusion in Proposition11 cannot fit for D1. For simplicity, the symbol (1×n)(n∈ℕ) is 1⋯1︸n and D(1×0)i=Di for i=1,2,3.Proposition 12.
Given thatn∈[1,+∞), only one component of R-n(D1) surrounds 0, 1/t, and 1/t4. Moreover, only one component of R-n(D1) surrounds 1/t but does not surround 0 and 1/t4. In addition, only one component of R-n(D1) surrounds 1/t4 but does not surround 0 and 1/t.Proof.
By induction and a similar discussion as the one used in Proposition10, it is easy to get the following conclusion. For any integer n∈ℕ, R-1(D(1×n)) consists of three components: one is contained in Out(D(1×n)), denoted by D(1×(n+1)); one is contained in M2, denoted by D(1×n)2; one is contained in M1, denoted by D(1×n)3. Moreover, D(1×(n+1))↺D(1×n) and D(1×n)i↺D(1×(n+1))i(i=2,3). By a similar discussion as used in Proposition 11, we can deduce that any preimage component of D(1×m)2 or D(1×m)3 (m=1,…,n-1) cannot surround 0 or 1/t. Since Rn(D(1×n)i)=Di(i=1,2,3), then D(1×(n+1)), D(1×n)2 and D(1×n)3 are satisfied with conditions of this proposition in turn as follows:
(17)D00⟵D01⟵{D2D1D3⟵{D12D11D13⟵{D112D111D113⟵⋯.By Proposition10, we can deduce that n(D(1×n)1)=n(D(1×n)3)=3, n(D(1×n)2)=5 for any n∈[1,+∞). However, for the other components in the preimage of Di(i=1,2,3), we have the following result.Proposition 13.
Given thatn∈ℕ, let D be a component of preimage of D(1×n)2 or D(1×n)3, and if c0∈Out(D), then d|D=1 or d|D=2.Proof.
By Propositions11 and 12, Int(D) contains no pole of R(z,t), and thus R(Int(D))=IntR(D) and R is a proper map in Int(D). We distinguish three cases to discuss the local degree of D as follows.(1)
Ifc0∈Out(D), then d|Int(D)=1 by the Riemann-Hurwitz formula. Obviously, d|D=1.(2)
Ifc0∈D, then d|D≥2. By the Riemann-Hurwitz formula, d|Int(D)=2, so d|D=2.(3)
Ifc0∈Int(D)∖D, then d|Int(D)=2 by the Riemann-Hurwitz formula. Furthermore, we can deduce that d|D=2. Otherwise, d|D=1. Note that Int(D) has no poles of R(z,t), and for any bounded components of ℂ¯∖R(D), the forward components of it in Int(D) are corresponding to all bounded components of ℂ¯∖D. Hence, R-1(R(D))∩Int(D)=D; it is a contradiction to d|Int(D)=2.Proof of Theorem1.
(1) It is an immediate result of Propositions7 and 10.
(2) We show thatR(z,1/20) has a Fatou component with connectivity 9.
Putt0=1/20, and by Proposition 7, the “free” critical point c0 satisfies c0∈(-2/t0+5t02,-2/t0+6t02). Set R(x)=R(x,t0), r1(x)=x2, r2(x)=1+t012x3, r3(x)=(1-t0x)3, r4(x)=1-t04x (x∈ℝ); then R(x)=(r1(x)r2(x))/(r3(x)r4(x)). Put I1=[-2/t0+5t02,-2/t0+6t02], and for any x1∈I1, we have
(18)r1(-2/t0+6t02)r2(-2/t0+5t02)r3(-2/t0+5t02)r4(-2/t0+5t02)≤R(x1)≤r1(-2/t0+5t02)r2(-2/t0+6t02)r3(-2/t0+6t02)r4(-2/t0+6t02).
Furthermore, we can deduce that
(19)R(x1)∈[59.237,59.252]:=I2,∀x1∈I1.
If necessary, we enlarge or reduce the length of corresponding interval, by a similar discussion as the one used in inequality 5, as follows:
(20)R(x2)∈[-465.125,-464.357]:=I3,∀x2∈I2;R(x3)∈[15.065,15.188]:=I4,∀x3∈I3;R(x4)∈[15108,16564]:=I5,∀x4∈I4;R(x5)∈[-0.714,-0.445]:=I6,∀x5∈I5;R(x6)∈[0.178,0.478]:=I7,∀x6∈I6.
For example,(21)r1(59.252)r2(59.252)r3(59.237)r4(59.252)≤R(x2)≤r1(59.237)r2(59.237)r3(59.252)r4(59.237).
By Lemmas5 and 6, I7∪I6⊂D00 and I5⊂D01. Note that I4=[15.065,15.188]; then I4⊂D2. Let D be the Fatou component that contains c0, and since R(c0)>0, then R(D)∩ℝ-=∅ by Proposition 11, and thus c0∈Out(R(D)). As R(z,t0)=-1/(t0)4 has an approximate root 22.9777(>1/t0), we have R(D)⊂Out(D2). By Proposition 12, we can deduce that R2(D) cannot surround D. By Proposition 13, d|D=2 and dRn(D)=1(n=1,2), and thus n(R(D))=n(R2(D))=5, n(D)=9 by the Riemann-Hurwitz formula.



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*Source: 621312-2014-03-23.xml*