# A Numerical Investigation on the Structure of the Zeros of Euler Polynomials

Authors: C. S. Ryoo; J. Y. Kang
Journal: Discrete Dynamics in Nature and Society
 (2015)

---

## Abstract

Using numerical investigation, we observe the behavior of complex roots of the Euler polynomialsE
n
(
x
). By means of numerical experiments, we demonstrate a remarkably regular structure of the complex roots of the Euler polynomials E
n
(
x
). Finally, we show the Julia set of Newton iteration function R
(
x
)
=
x
-
E
n
(
x
)
/
E
n
′
(
x
).


---

## Body

## 1. Introduction

The computing environment would make more and more rapid progress and there has been increasing interest in solving mathematical problems with the aid of computers. By using software, mathematicians can explore concepts much more easily than in the past. The ability to create and manipulate figures on the computer screen enables mathematicians to quickly visualize and produce many problems, examine properties of the figures, look for patterns, and make conjectures. This capability is especially exciting because these steps are essential for most mathematicians to truly understand even basic concept. Numerical experiments of Bernoulli polynomials, Euler polynomials, and Genocchi polynomials have been the subject of extensive study in recent year and much progress has been made both mathematically and computationally. Mathematicians have studied different kinds of the Euler, Bernoulli, Tangent, and Genocchi numbers and polynomials. Recently, many authors have studied in the area of theq-analogues of these numbers and polynomials (see [1–20]). Using computer, a realistic study for the zeros of Euler polynomials E
n
(
x
) is very interesting. The main purpose of this paper is to observe an interesting phenomenon of “scattering” of the zeros of the Euler polynomials E
n
(
x
) in complex plane. Throughout this paper, we always make use of the following notations: N
=
{
1,2
,
3
,
…
} denotes the set of natural numbers, N
0
=
{
0,1
,
2,3
,
…
} denotes the set of nonnegative integers, Z denotes the set of integers, R denotes the set of real numbers, and C denotes the set of complex numbers.The classical Euler numbersE
n and Euler polynomials E
n
(
x
) are usually defined by the following generating functions: (1)
2
e
t
+
1
=
∑
n
=
0
∞
E
n
t
n
n
!
,
(2)
2
e
t
+
1
e
x
t
=
∑
n
=
0
∞
E
n
x
t
n
n
!
,where we use the technique method notation by replacing E
(
x
)
n by E
n
(
x
) symbolically.Clearly,E
n
=
E
n
(
0
). These polynomials and numbers play important roles in many different areas of mathematics such as combinatorics, number theory, special function, and analysis, and numerous interesting results for them have been explored (see [2, 3, 7–9, 13, 15, 18–20]). The following elementary properties of Euler polynomials E
n
(
x
) are readily derived from (1) and (2). We, therefore, choose to omit the details involved. For more studies and results in this subject we may see [2, 3, 8, 13, 15, 17–20].Theorem 1.
Forn
∈
N
0, one has (3)
E
n
x
=
∑
l
=
0
n
n
l
E
l
x
n
-
l
.By Theorem1, after some elementary calculations, we get (4)
∫
a
b
E
x
d
x
=
∑
l
=
0
n
n
l
E
l
∫
a
b
x
n
-
l
d
x
=
∑
l
=
0
n
n
l
E
l
x
n
-
l
+
1
n
-
l
+
1
a
b
=
1
n
+
1
∑
l
=
0
n
+
1
n
+
1
l
E
l
x
n
-
l
+
1
a
b
.By Theorem 1, we have (5)
∫
a
b
E
n
x
d
x
=
E
n
+
1
b
-
E
n
+
1
a
n
+
1
.Since E
n
(
0
)
=
E
n, by (5), we have the following theorem.Theorem 2.
Forn
∈
N, one has (6)
E
n
x
=
E
n
+
n
∫
0
x
E
n
-
1
t
d
t
.Then, it is easy to deduce thatE
k
(
x
) are polynomials of degree k. Here is the list of the first Euler polynomials: (7)
E
0
x
=
1
,
E
1
x
=
x
-
1
2
,
E
2
x
=
x
2
-
x
,
E
3
x
=
x
3
-
3
x
2
2
+
1
4
,
E
4
x
=
x
4
-
2
x
3
+
x
,
E
5
x
=
x
5
-
5
x
4
2
+
5
x
2
2
-
1
2
,
E
6
x
=
x
6
-
3
x
5
+
5
x
3
-
3
x
,
E
7
x
=
x
7
-
7
x
6
2
+
35
x
4
4
-
21
x
2
2
+
17
8
,
E
8
x
=
x
8
-
4
x
7
+
14
x
5
-
28
x
3
+
17
x
,
E
9
x
=
x
9
-
9
x
8
2
+
21
x
6
-
63
x
4
+
153
x
2
2
-
31
2
,
E
10
x
=
x
10
-
5
x
9
+
30
x
7
-
126
x
5
+
255
x
3
-
155
x
.


## 2. The Phenomenon of Scattering of Zeros

In this section, an interesting phenomenon of scattering of zeros ofE
n
(
x
) is observed. By (2), we obtain (8)
∑
n
=
0
∞
E
n
1
-
x
-
1
n
t
n
n
!
=
2
e
-
t
+
1
e
1
-
x
-
t
=
2
e
t
+
1
e
x
t
=
∑
n
=
0
∞
E
n
x
t
n
n
!
.Hence we have the following theorem.Theorem 3 (see [13, 15]).
Forn
∈
N
0, one has (9)
E
n
x
=
-
1
n
E
n
1
-
x
.By (9), we also have the following theorem.Theorem 4.
Forn
∈
N
0, if n
≡
1
(
m
o
d
2
), then E
n
1
/
2
=
0; if n
≡
0
(
m
o
d
2
), then E
n
(
0
)
=
E
n
(
1
)
=
0.In [13, 15], we made a series of the following conjectures.Conjecture 5.
Prove thatE
n
(
x
), x
∈
C, has R
e
(
x
)
=
1
/
2 reflection symmetry in addition to the usual I
m
(
x
)
=
0 reflection symmetry analytic complex functions.Conjecture 6.
Prove thatE
n
(
x
)
=
0 has n distinct solutions.We find a counterexample of Conjecture6. Let n
=
5. Then there are five numbers, x
i (i
=
1,2
,
3,4
,
5), such that E
5
(
x
i
)
=
0. That is, we obtain x
1
=
1
/
2, x
2
=
(
1
/
2
)
(
1
-
5
), x
3
=
(
1
/
2
)
(
1
-
5
), x
4
=
(
1
/
2
)
(
1
+
5
), and x
5
=
(
1
/
2
)
(
1
+
5
). Hence, Conjecture 6 is not true for all n. Using computers, many more values of n have been checked. It still remains unknown if the conjecture fails or holds for any value n
≠
5.Sincen is the degree of the polynomial E
n
(
x
), the number of real zeros R
E
n
(
x
) lying on the real plane I
m
(
x
)
=
0 is then R
E
n
(
x
)
=
n
-
C
E
n
(
x
), where C
E
n
(
x
) denotes complex zeros. See Table 1 for tabulated values of R
E
n
(
x
) and C
E
n
(
x
).Table 1
Numbers of real and complex zeros ofE
n
(
x
).
Degreen
Real zeros
Complex zeros
41
13
28
42
10
32
43
11
32
44
12
32
45
13
32
46
14
32
47
15
32
48
12
36
49
13
36
50
14
36
51
15
36
52
16
36
53
13
40
54
14
40
55
15
40
56
16
40
57
17
40
58
14
44
59
15
44
60
16
44
61
17
44
62
18
44
63
15
48
64
16
48
65
17
48
66
18
48
67
19
48
68
20
48
69
17
52
70
18
52
71
19
52
72
20
52Conjecture 7.
Prove that the number of complex zerosC
E
n
(
x
) of E
n
(
x
)
,
I
m
(
x
)
≠
0 is (10)
C
E
n
x
=
4
n
-
α
5
,
α
=
n
+
20
21
,where  denotes taking the integer part.Subsequently, much theoretical and computational work has been done, extending and testing these conjectures, with particular attention paid recently to certain refined conjectures for the analytic continued Euler polynomials (see [15]). By means of numerical experiments, we observe that E
n
(
x
), x
∈
C, has R
e
(
x
)
=
1
/
2 reflection symmetry in addition to the usual I
m
(
x
)
=
0 reflection symmetry analytic complex functions (Figures 1 and 2). The obvious corollary is that the zeros of E
n
(
x
) will also inherit these symmetries: (11)
If
E
n
x
0
=
0
,
then
E
n
1
-
x
0
=
E
n
x
0
∗
=
E
n
1
-
x
0
∗
=
0
,where ∗ denotes complex conjugation (see Figures 1 and 2). By Theorem 4 and Conjecture 5, the center of the structure of zeros is 1
/
2.Figure 1
Zeros ofE
n
(
x
).
(a)
(b)Figure 2
Stacks of zeros ofE
n
(
x
) for 1
≤
n
≤
70.
(a)
(b)
(c)
(d)The data concerning the numerical verification of Conjectures6 and 7 are contained in Tables 1 and 2. See Table 1 for tabulated values of R
E
n
(
x
) and C
E
n
(
x
). First, we investigate the beautiful zeros of the E
n
(
x
) by using a computer. We plot the zeros of E
n
(
x
) for n
=
70 and x
∈
C (Figure 1). In Figure 1(a), the zeros of E
70
(
x
) structure are presented. In Figure 1(b), the real zeros of E
n
(
x
) structure are presented for 1
≤
n
≤
70.Table 2
Approximate solutions ofE
n
(
x
)
=
0, x
∈
R.
Degreen
x
1
0.5
2
0, 1
3
−0.366025404, 0.5, 1.36602540
4
−0.61803399, 0, 1, 1.61803399
5
−0.6180, −0.6180, 0.5, 1.618, 1.618
6
0, 1
7
−0.4977314, 0.5, 1.49773
8
−0.93233, 0, 1, 1.9323
9
−1.21973, −0.50008, 0.5, 1.50008, 2.21973
10
−1.36521, −1.01498, 0, 1, 2.01498, 2.36521
11
−0.499998, 0.5, 1.5Stacks of zeros ofE
n
(
x
) for 1
≤
n
≤
70, forming a 3D structure, are presented (Figure 2). In Figure 2(a), we plot stacks of zeros of E
n
(
x
) for 1
≤
n
≤
70. In Figure 2(b), we draw x and y axes but no z axis in three dimensions. In Figure 2(c), we draw y and z axes but no x axis in three dimensions. In Figure 2(d), we draw x and z axes but no y axis in three dimensions.Our numerical results for the numbers of real and complex zeros ofE
n
(
x
) are displayed (Table 1).We observe a remarkably regular structure of the complex roots of Euler polynomials.Next, we calculated an approximate solution satisfyingE
n
(
x
), x
∈
R. The results are given in Table 2.From (2), we have (12)
∑
n
=
0
∞
E
n
x
+
1
+
E
n
x
t
n
n
!
=
-
2
∑
n
=
0
∞
-
1
n
+
1
e
n
+
x
+
1
t
+
2
∑
n
=
0
∞
-
1
n
e
n
+
x
t
=
2
e
x
t
=
∑
n
=
0
∞
2
x
n
t
n
n
!
.Comparing the coefficient of t
n
/
n
! on both sides of (12), we get the following theorem.Theorem 8.
For any positive integern, one has (13)
E
n
x
+
1
+
E
n
x
=
2
x
n
.By (13), we have the following corollary.Corollary 9.
Forn
∈
N, one has (14)
E
n
=
-
E
n
1
.


## 3. Julia Set of the Euler Polynomials

In this section, we will present a series of diagrams showing the Julia set of theE
n
(
x
) and its related Mandelbrot set. Computations of the Julia and Mandelbrot sets of the E
n
(
x
) and observations of their properties are made. Let C denote the set of complex numbers and C
∞
=
C
∪
{
∞
}. We generate graphic images using the software Mathematica. We define and construct orbits of points under the action of a complex function. Let f
:
D
→
D be a complex function, with D being a subset of C. The iterates of f are the functions f
,
f
∘
f
,
f
∘
f
∘
f
,
…, which are denoted by f
1
,
f
2
,
f
3
,
…. If x
∈
C, then the orbit of x under f is the sequence (
x
,
f
(
x
)
,
f
(
f
(
x
)
)
,
…
).Definition 10 (see [21]).
The orbit of the pointx under the action of the function f is said to be bounded if there exists M
∈
R such that |
f
n
(
x
)
|
≤
M for all n
∈
N. If the orbit is not bounded, it is said to be unbounded.Plotting the orbit of length 3 of the point1
+
0.5
i under the action of the function E
3
(
x
), we get the results shown in Figure 3.Figure 3
Unbounded orbit ofE
3
(
x
).When plotting the orbit of a point, it is a good idea to start by plotting a few points, in order to obtain an idea if the orbit is bounded or unbounded. We see that the orbit ofE
3
(
x
) is unbounded.Definition 11 (see [21]).
Letf
:
D
→
D be a complex function, with D being a subset of C. The point z is said to be a fixed point of f if f
(
z
)
=
z. One also says that ∞ is a fixed point of f if |
f
(
z
)
|
→
∞ as |
z
|
→
∞.Definition 12 (see [21]).
A fixed pointp of f
(
x
) is an attractor if |
f
′
(
p
)
|
<
1, a repeller if |
f
′
(
p
)
|
>
1, and a neutral fixed point if |
f
′
(
p
)
|
=
1.As stated above, it can be proved that ifα is an attracting fixed point of f, then there exists a neighbourhood D of α such that if b
∈
D the orbit of b converges to α. If α is a repelling periodic point of f, then there is a neighbourhood N of α such that if b
∈
N there are points in the orbit of b which are not in N. In the case of polynomials of degree greater than 0 and some rational functions, ∞ is also called an attracting fixed point, as, for each such function, f, there exists R
>
0 such that if |
x
|
>
R then f
n
(
x
)
→
∞ as n
→
∞.Theorem 13 (fixed points ofE
n
(
x
)).
The Euler polynomialsE
3
(
x
)
+
0.2
+
0.5
i have one attractor fixed point (see Figure 4) at(15)
α
=
0.347610414
+
0.283937229
i
.
Our numerical results for fixed point ofE
n
(
x
) are displayed (Table 3). The results are obtained by Mathematica software.Table 3
Numbers of attractor, repeller, and neutral fixed points ofE
n
(
x
).
Degreen
Attractor
Repeller
Neutral
2
0
1
1
3
1
2
0
4
0
1
3
5
0
5
0
6
0
6
0
7
0
7
0
8
0
8
0
9
0
9
0
10
0
10
0Figure 4
Attractor fixed point ofE
3
(
x
)
+
0.2
+
0.5
i.Conjecture 14.
Forn
>
3, the Euler polynomials E
n
(
x
) have no attractor fixed point except for infinity.Definition 15 (basins of attraction).
Letf be a complex function with attracting fixed point α. The basin of attraction of α under f is defined to be the set (16)
z
∈
C
∣
f
n
z
⟶
α
as
n
⟶
∞
.Definition 16 (basins of attraction of infinity).
If infinity is an attracting fixed point off, then the basin of attraction of infinity is defined to be the set (17)
x
∈
C
∣
f
n
x
⟶
α
as
n
⟶
∞
.
Letf
:
C
→
C be a polynomial in x of degree >
2 where x
∈
C.Definition 17 (Julia set).
The filled Julia set off is the set (18)
K
f
=
x
∈
C
∣
the  orbit  of
x
under
f
is  bounded
.The Julia set of f, J
f, is the boundary of the filled Julia set of f.Consider the family of functions{
E
3
(
c
)
:
x
→
E
3
(
x
)
+
c
∣
c
∈
C
}, where c is a parameter. Julia sets can be divided into 2 classes. They are either connected or totally disconnected. Roughly speaking, a set which is connected is all in one piece (no breaks) while a set which is totally disconnected is like a cloud of dust particles, with its only connected components being points. If |
c
|
<
2 then both types of Julia sets occur and both types are generally aesthetically pleasing. If |
c
|
>
2 then the Julia set of E
3
(
c
) is totally disconnected. This fact, together with the largeness of c, produces Julia sets which are rather dispersed and spread out and so not very pleasing. We show (a) in Figure 5(a) the Julia set of E
3
(
-
0.7280
+
0.0900
i
) which is connected; (b) in Figure 5(b) the Julia set of E
3
(
0.0340
+
0.6250
i
) which is totally disconnected; (c) in Figure 5(c) the Julia set of E
3
(
0.4010
+
0.0900
i
) which is totally disconnected.Figure 5
Julia setsE
3
(
c
).
(a)
(b)
(c)Examples of Julia sets for the family of{
E
5
(
c
)
:
x
→
E
5
(
x
)
+
c
∣
c
∈
C
}are shown in Figure 6.Figure 6
Julia setsE
5
(
c
).Newton tells us to consider the dynamical system(19)
C
∞
:
R
x
=
x
-
S
x
S
′
x
.
R is called the Newton iteration function of S. It can be shown that the fixed points of R are zeros of S and that all fixed points of R are attracting. R may also have one or more attracting cycles. To obtain the Julia set of R, plot, in white, the set J of points whose orbits do not converge, and plot the set U of remaining points whose orbits converge in a different coloring. The boundary of the set J or of the set U is the Julia set of R.Consider the Euler polynomialE
4
(
x
) for x
∈
C. There are four distinct complex numbers, a
i (i
=
1,2
,
3,4), such that E
4
(
a
i
)
=
0. Newton’s method provides a means to compute them. We obtain a
1
=
0, a
2
=
2, a
3
=
(
1
/
2
)
(
1
-
5
), and a
4
=
(
1
/
2
)
(
1
+
5
). Then Newton tells us to consider the dynamical system (20)
C
∞
:
R
x
=
x
-
E
4
x
E
4
′
x
.The general expectation is that a typical orbit {
R
n
(
x
)
}, which starts from an initial “guess” x
0
∈
C, will converge to one of the roots of E
4
(
x
). If we choose x
0 close enough to a
i then it is readily proved that (21)
lim
n
→
∞
⁡
R
n
x
0
=
a
i
,
for
i
=
1,2
,
3,4
.Given a pointx
0 in the plane, we wish to find out if the orbit of x
0 under the action of R
(
x
) does or does not converge to one of the roots of the equation, and if so, which one.WhenR
(
x
) is applied to x
0, the orbit of x
0 under the action of R is calculated until the absolute value of the last 2 iterations differs by an amount less than 1
0
-
10 or until 30 iterations have been carried out. The output is the last orbit value calculated. We construct a function which assigns one of 4 colors to each point in the plane, according to the outcome of R. We assign the red, yellow, green, and blue to x
0 if its orbit converges to 0,1
,
(
1
/
2
)
(
1
-
5
)
,
(
1
/
2
)
(
1
+
5
), respectively. We assign the black to it if the orbit does not converge to any of the points (Figure 7). The viewing windows are {
(
x
,
y
)
:
-
6
≤
x
≤
6
,
-
6
≤
y
≤
6
}. In Figure 7, the red region represents part of the basin of attraction of a
1
=
0.Figure 7
Orbit ofx
0 under the action of R for E
4
(
x
).As the Newton iteration function ofE
4
(
x
) is of degree 4, we plot, in white, the set of points whose orbits do not converge. The Julia set will be the boundary of the set of black points (see Figure 8). The viewing windows are {
(
x
,
y
)
:
-
5.5
≤
x
≤
5.5
,
-
5.5
≤
y
≤
5.5
}.Figure 8
Julia set ofR
(
x
) for E
4
(
x
).A zoom-in on a version of Figure8, in which the contours and contour lines are colored, is shown in Figure 9. The viewing windows are {
(
x
,
y
)
:
-
2.5
≤
x
≤
3.5
,
-
2.5
≤
y
≤
2.5
}.Figure 9
Julia set ofR
(
x
) for E
4
(
x
).This color palette explains the coloring of Figure9. For example, points which escape after 1 to 30 iterations are colored red to green. Points which do not escape after 30 iterations and so are taken to be in the filled Julia set are colored black (Figure 10).Figure 10
Palette for escaping points.The abovementioned rapid change can also be illustrated by applying the three-dimensional structure to the escape-time function (Figure11). The viewing windows are {
(
x
,
y
)
:
-
5.5
≤
x
≤
5.5
,
-
5.5
≤
y
≤
5.5
}. When R
(
x
) is applied to x
0, the orbit of x
0 under the action of R is calculated until the absolute value of the last 2 iterations differs by an amount less than 1
0
-
10 or until 30 iterations have been carried out. The output is the last orbit value calculated.Figure 11
3D structure of Julia set ofR
(
x
) for E
4
(
x
).



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*Source: 174173-2015-05-31.xml*