1. Introduction
Let
be a monic complex polynomial of degree
. We denote by
the
j-th iterate of
, that is,
The
filled-in Julia set of
is defined as
and the
Julia set of the function
is defined to be the boundary of the set
, i.e.,
(see, e.g., [
1]).
In this work, we are interested in a modified version of the “classical” filled-in Julia set
and the Julia set
of functions in the
quadratic family . We observe that the Mandelbrot set
is the fractal defined as
We point out that there is a more “workable” way of considering the Mandelbrot set (we refer to [
2], Theorem 14.14) for a proof of the usually referred fundamental theorem of the Mandelbrot set):
Some other recent results related to the Mandelbrot set can be found for example in [
3,
4,
5,
6,
7,
8,
9,
10].
In 2019, Mork et al. [
11] constructed filled-in Julia sets for a
lacunary function , where
is the sequence of
centered k-gonal numbers and
k is any positive integer (for more facts and history of lacunary functions see, e.g., [
12,
13,
14]).
In 2021, Mork et al. [
15] followed up on the aforementioned article and considered a generalization of the filled-in Julia sets and their corresponding Mandelbrot sets by composing the lacunary function
with a fixed Möbius transformation
(with
, where 𝔻 denotes the the unit disc) at each iteration step. More precisely
Very recently, Mork and Ulness [
16] continued the previous line of research by dealing with the so-called
j-averaged Mandelbrot set which is a set generated by iterating a function obtained by composing the function
and the Möbius transformation
, where
. Thus,
The name “
j-averaged” is used here since the points of the resulting fractal are colored according to the total number of members of the following sequence of iterations
, that escaped from the circle with radius 2 (the concrete algorithm for coloring of points of this fractal you can find in Appendix 1 of [
16]), see
Figure 1,
Mork and Ulness ([
16] Theorem 1) proved that the
j-averaged Mandelbrot set for the Möbius transformation
with
has threefold rotational symmetry and dihedral mirror symmetry. Additionally, they raised a conjecture (see [
16], Conjecture 2)) concerning the coefficients of these iterations. Before stating their conjecture, we introduce some basic notations.
Let
be a non-zero complex number. Define the function
by
, with
. Therefore,
Observe that the
n-th iteration of
H at
is a function of
, say
, which satisfies the relations:
The sequence
of the
Catalan numbers, which is called the sequence A000108 in the
OEIS [
18], is often defined with the help of the central binomial coefficient
by
thus, its first terms are in
Table 1.
Table 1.
Values of for n from 0 to 14.
Table 1.
Values of for n from 0 to 14.
n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
| 1 | 1 | 2 | 5 | 14 | 42 | 132 | 429 | 1430 | 4862 | 16,796 | 58,786 | 208,012 | 742,900 | 2,674,440 |
which can lead us to the following recurrence relation (it was first discovered by Euler in 1761; for more facts, see [
19])
with the initial condition
. Sometimes the sequence
is defined on the basis of the generating function
, as the following holds (for
)
The aim of this paper is to obtain a (quantitative) result for the coefficients of the power series of which implies the Mork–Ulness’ conjecture (qualitative version). More precisely,
Theorem 1. For all , we havewhere is the n-th Catalan number. Remark 1. We remark that Mork and Ulness [16] posed a slightly different conjecture. In fact, we can express their question by defining and asand They also asserted that these functions should converge in the whole unit disk (or the punctured one for ). However, this is not true (this is expected because of the exponential nature of Catalan numbers). For example, the simple bound , which comes from the fact that , implies that (some other bounds can be found in ([19] Chapter 2) and [20]) and so if then (for ) yielding the divergence of . In order to compute the radius of convergence, say r, of , one can write this function as , where Thus, and, by using (which comes from the Stirling formula ), we obtain Therefore, is the disk of convergence of (observe that ).
2. Auxiliary Results
Before proceeding further, we shall present some useful tools related to the previous sequences.
Our the first ingredient provides a useful form to the Laurent series of .
Lemma 1. For any , there exists a power series such that
Proof. By definition in (
1),
satisfies the following recurrence relation
with
(since
). Now, by defining
and using the previous recurrence, we obtain
and so
with
. We claim that
for some rational function
, where
n is any positive integer. Indeed, we can proceed by induction on
n. For
, we can take
. Suppose (by induction hypothesis) that
, for some formal power series
, then, by (
3), we have
where
can be chosen by satisfying the recurrence
with
. The inductive process is finished. Observe that, since
, then it suffices to prove that
The proof is also by induction on
n (more precisely, a double induction). For the basis cases, we have
and
where we used that for
, one has
(in general, it holds that
). Suppose that (
4) is valid for all
. Then,
where we used
, since
.
Now, we use the previous fact
This completes the induction proof of (
4).
Therefore, since
, we can write
and so
This completes the proof. □
Remark 2. Note that, by using Lemma 1, we can writewhere , i.e., is 1
if n is odd and 0 if n is even. In particular, is an analytic function in some neighborhood of , when n is even, and for n odd, has a simple pole at origin (with residue equal to 1
). Remark 3. Another viewpoint of Lemma 1 (and consequently, of Remark 2) is that the k-th derivative of as , for any or . This fact can also be proved by a harder (but maybe theoretically useful) combination of induction, the generalized Chain Rule (Faà di Bruno’s formula) and the fact that all odd order derivatives of vanish (for fixed λ) at . This last assertion follows from Cauchy’s integral formula. Indeed, we havewhere is the circle , for and . Now, we can use the partial fraction decomposition to deduce thatfor computable constants A and B. Hence, again by the Cauchy integral formula, we havewhere , for all z. Thus, is equal to zero as claimed. Now we show the important connection of the sequence
to the Catalan numbers. For the simplicity of notation, we use the following notation in the rest of the text:
Lemma 2. Let be the Catalan sequence. We have
- (i)
If is defined by the recurrence,with , then , for all . - (ii)
If is defined by the recurrence,with , then , for all .
Proof. Let us recall that Catalan numbers satisfy the Segner recurrence relation (see, e.g., [
19], p. 117)
with
.
(i). We shall proceed by induction on
k. For
, one has
(by definition). Suppose
, for all
. Then,
which completes the proof (where we used (
6)).
(ii). Again by induction on
k, the basis case
follows by definition. Assume now that
, for all
. Then, by the recurrence for
together with the induction hypothesis, we obtain
which finishes the proof (where we used again (
6)). □
The next lemma gives a helpful recurrence for
, depending on the parity of
n. The proof follows by induction together with (
6) (we leave the details to the readers).
Lemma 3. Let be the Catalan sequence. Then,
and ,
for all (with ).
Now, we are ready to deal with the proof.