Fractal property of generalized M-set with rational number exponent

Dynamic systems described by f"c(z)=z^2+c is called Mandelbrot set (M-set), which is important for fractal and chaos theories due to its simple expression and complex structure. f"c(z)=z^k+c is called generalized M set (k-M set). This paper proposes a new theory to compute the higher and lower bounds of generalized M set while exponent k is rational, and proves relevant properties, such as that generalized M set could cover whole complex number plane when k 1), and that k-M set can be divided into |p-q| isomorphic parts.

Fractal property of generalized M-set with rational number exponent 1

. Introduction
The dynamic system f c (z) = z 2 + c is called Mandelbrot set (M-set) [1] as Mandelbrot first proposed it in 1982. Later, Gujar and Dhurandhar extended M-set into generalized M-set f c (z) = z k + c with different exponent k. When exponent is k, it is called k-M set. Thus we can see that original M-set is 2-M set.
k-M set is important for chaos and fractal theories. Many scholars have conducted researches on k-M sets. Huang studied periodic orbit of k-M set when k is a positive integer [2]. Gujar and Dhurandhar explored the structure of k-M set where k is a real number [3,4]. Shirriff researched characteristics of k-M set where k is a complex number [5].
After year 2000, more progresses have been made with advancement of computer techniques, especially computer graphic techniques. Wang explored escape time of M-set [6]. Noah analyzed the radius of M-set [7]. Pastor and Ashish analyzed chaotic features of M-set [8][9]. Fractal is widely used with computer techniques [10][11][12]14] when M-set continues to attract researchers' attentions [13].
However, there are still some problems unsolved. For example, the structure of k-M set is unknown when k is not an integer.
This paper presents our following results in research of k-M set where k is a rational number: (i) Boundary of k-M set falls between circle with radius R ¼ ðk À 1Þ Á k À k kÀ1 and circle with radius R ¼ 2 1 kÀ1 when k > 1. (ii) A k-M set could contain whole complex number plane when k < 1. (iii) If k = p/q, p and q are irreducible integers, (GCD(p, q) = 1, k > 1); then the symmetry property of k-M set is decided by p and q.
The remainder of the paper is organized as follows. We analyze the boundary of k-M set with k > 0 in Section 2. Then, we analyze the boundary of k-M set with k < 0 in Section 3. Then, we analyze symmetry property of k-M Set with exponent k when k > 1 in Section 4. Finally, Section 5 summarizes the main results of the paper.
2. Boundary of k-M set with exponent k > 0 Falconer [14] gives the following extended M Set definition: A definition of k-M set can be generalized from above M set definition: Lemma 1. Let 0 < k < 1, all positive real numbers are in k-M set.
Proof. When 0 < k < 1 and In the following paragraphs, these three cases are analyzed.
The following proof is based on mathematical induction. First, let i = 1, we have Then, assume when i 6 j, we have Hence, let i = j + 1, we have Then, lim Then we know f jþ1 xþ1 ð0Þ is not divergent from case (i). Hence, f jþ1 x ð0Þ is not divergent. Thus, x is in k-M set.
. Thus it is convergent by case (i). Hence, x = 1 is in k-M set.
From cases (i), (ii), (iii), Lemma 1 is proved. h Lemma 1 shows that all positive real numbers are in k-M set when 0 < k < 1 and k is a rational number. The following Lemma 2 and Inference 1 extend the result to whole complex number plane.

Lemma 2.
For any complex number z and natural number n = 1,2,. . ., jf n z ð0Þj 6 jf n c ð0Þj when jzj ¼ c; for any k-M set.
Proof. Considering any point z in the complex plane, we assume |z| = c.
Then, jf i z ð0Þj 6 jf i c ð0Þj can be proved by mathematical induction.
Then, assume when i 6 n, we have jf n z ð0Þj 6 jf n c ð0Þj: ð7Þ From Eqs. (6)-(8), Lemma 2 is proved. h The following Inference 1 extends Lemma 1 to complex number plane.
Inference 1. Let 0 < k < 1 and k is a rational number, all points in complex number plane are in k-M set.
Proof. We assume z as any complex number and |z| = c. Lemma 3. Let k > 1 and k is a rational number. k-M set is bounded and the minimum absolute value of points at boundary is Proof. The minimum absolute value c of points at boundary must satisfy equation We obtain Eq. (11) using Eqs. (9) and (10).
Then, from f 2 It means that c is in k-M set. Then, from Lemma 2, we know that any complex number z is in k-M set when |z| < c.Then we prove that c is the lower bound value, which means that there is at least one point z that is not in k-M set with |z| > c.Let c Ã ¼ c þ x 1 > c and there exists a positive integer n that makes f n c Ã ð0Þ ¼ x þ x 2 > x:To use Eq. (11) and kx kÀ1 ¼ 1, we prove it as follows.Let i = n + 1, f nþ1 To conclude Eqs. (12)- (14), we know when i ! 1, m ! 1, x þ x 1 Á m ! 1, f i c Ã ð0Þ ! 1.It means c ⁄ is not in k-M set, hence, c is the minimum boundary point.Lemma 3 is proved. h Furthermore, considering process of proof in Lemma 3, we affirm that point z at boundary with the maximum absolute value c must satisfy z 0 ¼ ze In other words, function f maps z to z 0 with same module and different modular phase angle. Thus, it means that we can calculate z from the equation. So we reach Lemma 4.
Then we solve c Á e We assume that k=p/q is irreducible, it makes e 2npi k ¼ e 2qnpi p . So when p is even, let n=p/2, we have e 2npi k ¼ e qpi . Then, let q is odd, we have e qpi =-1 and |c|=2 1 kÀ1 . On opposite side, when p is odd, we know there exist u and v make pu+qv=1. So when we set n=vn 1 , we find Eq. (17).
In other words, when n 1 ¼ pÀ1 2 or pþ1 2 as well as n ¼ v pAE1 2 , we find Eq. (19) Then, we prove that a point z with jzj > 2 1 kÀ1 is not in k-M set. When we assume jzj ¼ 2 1 kÀ1 þ (e > 0), we use mathematical induction to prove it. Let i = 0, we get Eq. (20).
It means z is out of k-M set.
To conclude, Lemma 4 is proved. h Then we gain Theorem 1 from Lemmas 1-4 and Inference 1. Theorem 1 is boundary conclusion of k-M set with exponent k > 0.  Proof. We know lim . So Inference 2 is proved. h 4. Symmetry characteristics of k-M set with exponent k (k>1) As we know, fractals of k-M set with integer exponent n, which is called n-M, show their symmetry by n > 2. But there is no proved conclusion about k-M set when exponent k is not integer. In this section, we will prove the symmetry and other fractal characteristics of k-M set when k is positive.
Firstly, we know that k-M set are axial symmetrical about real axis because z k þ C ¼ Z k þ C ¼ Z k þ C for any z and c in complex plane. Then, when rational number k = p/q (p and q are irreducible), we reach Theorem 2.
Theorem 2. Number of isomorphic subset clusters in k-M set is p-q.
Proof. At first, we explain that when we say k-M set has p-q isomorphic subset clusters, if and only if, we prove all points z in one subset clusters of k-M set, there are p-q-1 other points z i in other p-q-1 subset clusters with same modular in k-M set which are iterated to same modular and phase angle except origin in complex plane. In other words, if jf p z ð0Þj ¼ jf p z 1 ð0Þj and f p z ð0Þ has same different phase angle to f p z 1 ð0Þ, we call z and z i are isomorphic. Isomorphic clusters are make up of all isomorphic points.
For each point z ¼ r Á e hi in k-M set, if there exist z 1 ¼ r Á e h 1 i makes jf p z ð0Þj ¼ jf p z1 ð0Þj when p>0, it means that we find an isomorphic point of z. To solve this equation, we first solve equation jz k þ zj ¼ jz k 1 þ z 1 j. To predigest it, we get Eq. (27).
It means that our assuming is true. Furthermore, we can calculate the number of isomorphic points of z. To use k = p/q in Eq. (31), we get z ¼ z 1 Á e 2qnp pÀq . It means that z and z i are isomorphic when n = 1 $ p-q. Based on all above, Theorem 2 is proved. h Then, we have to say that there are some minor errors in some thesis that draws fractal figures of k-M set by using computer. The errors are caused by calculation. When we calculate a point z in complex variable function, we also change the point z to complex exponent form that is used to operate. But in calculation, computer will translate it into form a + bi to operate. Then some values would be changed. The difference between the results calculated by these two kinds once is e 2mph , and the results of following mappings will be wrong. For example, when a computer calculates e ð ap b ÞiÁ b a , the result is correct. However, when it calculates e ð ap b Þi À Áb a , the result is always wrong. Then, the resulting mapping will be always false. For example, we use k = 3.5 to validate it. In this case, p = 7, q = 2. So we get minimum and maximum radius are 5 7 Á 2 7 À Á2 5 and 2 2 5 . We validate our opinion by use Fig. 1. In Fig. 1a-c, we create some classic k-M set with p = 7, 9, 11 and q = 2. We can see generation of k-M set by k increase by compare Fig. 1a to b and c. But just in another example in Fig. 1d, e and f, we trust there are 7 À 2 = 5 isomorphic parts in fractal figure of 3.5-M set. Then we use a point to execute an experiment in Fig. 1. As we seen, a point z ¼ e  Fig. 1d-f, which is drawn by classic method, it is not suitable. In fact, center point in 'd' and 'e' are convergence, but 'f' are divergence. It validates our conclusion.
In fact, when we calculate these points in k-M set, we find they are all 2-periodic points. For example, let z ¼ e 1:2pi % -0.8090-0.5878i, z 3:5 þ z ¼ zðz 2:5 þ 1Þ ¼ zðz 3pi þ 1Þ ¼ 0, then the other points make z 3:5 þ z ¼ 0. This is also a validation of our idea. Additionally, if there are some isomorphic subsets with different period m and n overlapped, it means these parts have both characteristics both of periods m and n. The conclusion is absurd when m and n are irreducible. So we conjecture that all periods are disjunctive in k-M set. This is equivalent to that radii of all periodic areas are less than p pÀq when we assume a periodic area is an isomorphic circle. Conjecture 1. Assuming a periodic area is a isomorphic circle, radii of periodic areas are less than p pÀq . Admittedly, our idea considers the complex plane as composed by plane-layers with phase angle range [2np,2np + 2p). When considering complex plane that does not consist of plane-layers with phase angle more than 2p, we should subtract some boundary points from Theorem 2.

Conclusion and future work
We analyzed k-M set when k is rational number in this paper. We first computed bounds of k-M set and proved its symmetry is determined by k = p/q when p and q are irreducible. Then, we proved k-M set contains whole complex plane when k 6 1. In other words, k-M set has no bound when k 6 1. Next, except for some classic figures of k-M set, we found that k-M set can be divided into p-q isomorphic parts. Of course, these parts cannot overlap with each other. Finally, we give a conjecture to find bound of periodic subsets.
Our future work can be conducted in two steps. At first, we will find a new computational algorithm in complex plane. This algorithm is used to find correct results of exponent calculation in complex plane. We will design the algorithm to avoid phase angle errors by considering intermediate results in computing process. Secondly, we will improve our design. We can assume complex plane is constructed by plane layers of phase angle ranges of [2np, 2np + 2p) where n = 1,2. . . A mapping will be executed from one plane layer to another. All plane-layers form the whole perfect complex plane.