# Analytic Continuation of Euler Polynomials and the Euler Zeta Function

Authors: C. S. Ryoo
Journal: Discrete Dynamics in Nature and Society
 (2014)

---

## Abstract

We study that the Euler numbersE
n and Euler polynomials E
n
z are analytically continued to E
s and E
(
s
,
w
). We investigate the new concept of dynamics of the zeros of analytica continued polynomials. Finally, we observe an interesting phenomenon of “scattering” of the zeros of E
(
s
,
w
).


---

## Body

## 1. Introduction

Throughout this paper,ℤ, ℝ, and ℂ will denote the ring of integers, the field of real numbers, and the complex numbers, respectively. Recently, many mathematicians have studied different kinds of the Euler, Bernoulli, and Genocchi numbers and polynomials (see [1–19]). 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 years and much progress has been made both mathematically and computationally. Using computer, a realistic study for Euler polynomials E
n
(
x
) is very interesting. It is the aim of this paper to observe an interesting phenomenon of “scattering” of the zeros of the Euler polynomials E
n
(
x
) in complex plane. First, we introduce the Euler numbers and Euler polynomials. As a well-known definition, the Euler numbers E
n are defined by
(1)
2
e
t
+
1
=
∑
n
=
0
∞
‍
E
n
t
n
n
!
.
Here is the list of the first Euler numbers:
(2)
E
0
=
1
,
E
1
=
-
1
2
,
E
2
=
0
,
E
3
=
1
4
,
E
4
=
0
,
E
5
=
-
1
2
,
E
6
=
0
,
E
7
=
17
8
,
E
8
=
0
,
E
9
=
-
31
2
,
E
10
=
0
,
E
11
=
691
4
,
E
12
=
0
,
E
13
=
-
5461
2
,
E
14
=
0
,
E
15
=
929569
16
,
E
16
=
0
,
E
17
=
-
3202291
2
,
E
18
=
0
,
E
19
=
221930581
4
,
E
20
=
0
,
E
21
=
-
4722116521
2
,
E
22
=
0
,
E
23
=
968383680827
8
,
E
24
=
0
,
E
25
=
-
14717667114151
2
,
E
26
=
0
,
E
27
=
2093660879252671
4
,
E
28
=
0
.The Euler polynomialsE
n
(
x
) are defined by the generating function:
(3)
F
(
x
,
t
)
=
∑
n
=
0
∞
‍
E
n
(
x
)
t
n
n
!
=
(
2
e
t
+
1
)
e
x
t
,
where we use the technique method notation by replacing E
(
x
)
n by E
n
(
x
) symbolically.Because(4)
∂
F
∂
x
(
x
,
t
)
=
t
F
(
x
,
t
)
=
∑
n
=
0
∞
‍
d
E
n
d
x
(
x
)
t
n
n
!
,
an important relation follows:
(5)
d
E
k
d
x
(
x
)
=
k
E
k
-
1
(
x
)
.
Then, it is easy to deduce that E
k
(
x
) are polynomials of degree k. Here is the list of the first Euler polynomials:
(6)
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
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. Generating Euler Polynomials and Numbers

Since(7)
∑
n
=
0
∞
‍
E
n
(
x
)
t
n
n
!
=
2
e
t
+
1
e
x
t
=
e
E
t
e
x
t
=
e
(
E
+
x
)
t
=
∑
n
=
0
∞
‍
(
E
+
x
)
n
t
n
n
!
,
we have the following theorem.Theorem 1.
Forn, one has
(8)
E
n
(
x
)
=
∑
k
=
0
n
‍
(
n
k
)
E
k
x
n
-
k
.Definition 2.
Fors
∈
ℂ with Re
(
s
)
>
0, define the Euler zeta function by
(9)
ζ
E
(
s
)
=
2
∑
n
=
1
∞
‍
(
-
1
)
n
n
s
;
see [7–10].Notice that the Euler zeta function can be analytically continued to the whole complex plane, and these zeta functions have the values of the Euler numbers at negative integers. That is, Euler numbers are related to the Euler zeta function as(10)
ζ
E
(
-
n
)
=
E
n
.Definition 3.
We define the Hurwitz zeta functionζ
E
(
s
,
x
) for s
∈
ℂ with Re
(
s
)
>
0 and x
∈
ℝ with 0
≤
x
<
1 by
(11)
ζ
E
(
s
,
x
)
=
2
∑
n
=
0
∞
‍
(
-
1
)
n
(
n
+
x
)
s
;
see [3, 7–10, 12].Euler polynomials are related to the Hurwitz zeta function as(12)
ζ
E
(
-
n
,
x
)
=
E
n
(
x
)
.We now consider the functionE
(
s
) as the analytic continuation of Euler numbers. From the above analytic continuation of Euler numbers, we consider
(13)
E
n
⟼
E
(
s
)
,
ζ
E
(
-
n
)
=
E
n
⟼
ζ
E
(
-
s
)
=
E
(
s
)
.All the Euler numbersE
n agree with E
(
n
), the analytic continuation of Euler numbers evaluated at n (see Figure 1),
(14)
E
n
=
E
(
n
)
for
n
≥
1
,
(15)
except
E
(
0
)
=
-
1
,
but
E
0
=
1
.Figure 1
The curveE
(
s
) runs through the points of all E
n except E
0.In fact, we can expressE
′
(
s
) in terms of ζ
E
′
(
s
), the derivative of ζ
E
(
s
),
(16)
E
(
s
)
=
ζ
E
(
-
s
)
,
E
′
(
s
)
=
-
ζ
E
′
(
-
s
)
,
E
′
(
2
n
+
1
)
=
-
ζ
E
′
(
-
2
n
-
1
)
for
n
∈
ℕ
∪
{
0
}
.From relation (16), we can define the other analytic continued half of Euler numbers as
(17)
E
(
s
)
=
ζ
E
(
-
s
)
,
E
(
-
s
)
=
ζ
E
(
s
)
⟹
E
(
-
n
)
=
ζ
E
(
n
)
,
n
∈
ℕ
.
By (17), we have
(18)
lim
⁡
n
→
∞
E
-
n
=
ζ
E
(
n
)
=
-
2
.The curveE
(
s
) runs through the points E
-
n
=
E
(
-
n
) and grows ~−2 asymptotically as -
n
→
∞(see Figure 2).Figure 2
The curveE
(
s
) runs through the points E
-
n.


## 3.  Analytic Continuation of Euler Polynomials

Looking back at (1) and (3), we can see that the sign convention of E
0 was actually arbitrary. Equation (15) suggests that consistent definition of Euler numbers should really have been
(19)
∑
n
=
0
∞
‍
(
-
1
)
n
+
1
E
n
t
n
n
!
=
2
e
t
+
1
,
|
t
|
<
π
,
n
=
0,1
,
2
,
…
∈
ℕ
∪
{
0
}
which only changes the sign in the conventional definition of the only nonzero even Euler numbers, E
0, from E
0
=
1 to E
0
=
E
(
0
)
=
-
1.By using Cauchy product, we have(20)
∑
n
=
0
∞
‍
(
-
1
)
n
+
1
E
n
(
x
)
t
n
n
!
=
(
2
e
t
+
1
)
e
x
t
=
(
∑
k
=
0
∞
‍
(
-
1
)
k
+
1
E
n
t
k
k
!
)
(
∑
m
=
0
∞
‍
x
m
t
m
m
!
)
=
∑
n
=
0
∞
‍
(
∑
k
=
0
n
‍
(
n
k
)
(
-
1
)
k
+
1
E
k
x
n
-
k
)
t
n
n
!
.
For consistency with the redefinition of E
n
=
E
(
n
) in (19), Euler polynomials should be analogously redefined as
(21)
E
n
(
x
)
=
∑
n
=
0
n
‍
(
-
1
)
n
+
k
(
n
k
)
E
k
x
n
-
k
.
The analytic continuation can be then obtained as
(22)
n
⟼
s
∈
ℝ
,
x
⟼
w
∈
ℂ
,
E
k
⟼
E
(
k
+
s
-
[
s
]
)
=
ζ
E
(
-
(
k
+
(
s
-
[
s
]
)
)
)
,
(
n
k
)
⟼
Γ
(
1
+
s
)
Γ
(
1
+
k
+
(
s
-
[
s
]
)
)
Γ
(
1
+
[
s
]
-
k
)
⟹
E
n
(
x
)
⟼
E
(
s
,
w
)
=
∑
k
=
-
1
[
s
]
‍
(
-
1
)
k
+
[
s
]
Γ
(
1
+
s
)
E
(
k
+
s
-
[
s
]
)
w
[
s
]
-
k
Γ
(
1
+
k
+
(
s
-
[
s
]
)
)
Γ
(
1
+
[
s
]
-
k
)
=
∑
k
=
0
[
s
]
+
1
‍
(
-
1
)
k
-
1
+
[
s
]
Γ
(
1
+
s
)
E
(
(
k
-
1
)
+
s
-
[
s
]
)
w
[
s
]
+
1
-
k
Γ
(
k
+
(
s
-
[
s
]
)
)
Γ
(
2
+
[
s
]
-
k
)
,
where [
s
] gives the integer part of s, and so s
-
[
s
] gives the fractional part.By (22), we obtain analytic continuation of Euler polynomials:
(23)
E
0
(
w
)
=
-
1
,
E
1
(
w
)
=
E
(
1
,
w
)
=
-
0.5
+
w
,
E
2
(
w
)
=
E
(
2
,
w
)
=
w
-
w
2
,
E
(
2.2
,
w
)
≈
0.0799175
+
0.866594
w
-
1.19765154
w
2
+
0.116134
w
3
,
E
(
2.4
,
w
)
≈
0.147412
+
0.693349
w
-
1.36116
w
2
+
0.279503
w
3
,
E
(
2.6
,
w
)
≈
0.199873
+
0.485705
w
-
1.47547
w
2
+
0.486915
w
3
,
E
(
2.8
,
w
)
≈
0.234806
+
0.251271
w
-
1.52609
w
2
+
0.731015
w
3
,
E
3
(
w
)
=
E
(
3
,
w
)
=
0.25
-
1.5
w
2
+
w
3
.By using (23), we plot the deformation of the curve E
(
2
,
w
) into the curve of E
(
3
,
w
) via the real analytic continuation E
(
s
,
w
), 2
≤
s
≤
3, w
∈
ℝ (see Figure 3).Figure 3
The curve ofE
(
s
,
w
), 2
≤
s
≤
3, -
0.5
≤
w
≤
0.2.Next, we investigate the beautiful zeros of theE
(
s
,
w
) by using a computer. We plot the zeros of E
(
s
,
w
) for s
=
9,9.6,9.8,10 and w
∈
ℂ (Figure 4).Zeros ofE
(
s
,
w
) for s
=
9,9.6,9.8,10.
(a)
(b)
(c)
(d)In Figure4(a), we choose s
=
9. In Figure 4(b), we choose s
=
9.6. In Figure 4(c), we choose s
=
9.8. In Figure 4(d), we choose s
=
10.Since(24)
∑
n
=
0
∞
‍
E
n
(
1
-
x
)
(
-
1
)
n
t
n
n
!
=
F
(
1
-
x
,
-
t
)
=
2
e
-
t
+
1
e
(
1
-
x
)
(
-
t
)
=
2
e
t
+
1
e
x
t
=
F
(
x
,
t
)
=
∑
n
=
0
∞
‍
E
n
(
x
)
t
n
n
!
,
we obtain
(25)
E
n
(
x
)
=
(
-
1
)
n
E
n
(
1
-
x
)
.Hence, we have the following theorem.Theorem 4.
Ifn
≡
1
(
mod
⁡
2
), then E
n
(
1
/
2
)
=
0, for n
∈
ℕ.The question is, what happens with the reflexive symmetry (25) when one considers Euler polynomials? Prove that E
n
(
x
), x
∈
ℂ, has Re
(
x
)
=
1
/
2 reflection symmetry in addition to the usual Im
⁡
(
x
)
=
0 reflection symmetry analytic complex functions. However, we observe that E
(
s
,
w
), w
∈
ℂ, does not have Re
(
w
)
=
1
/
2 reflection symmetry analytic complex functions (Figure 4).Our numerical results for approximate solutions of real zeros ofE
(
s
,
w
) are displayed. We observe a remarkably regular structure of the complex roots of Euler polynomials. We hope to verify a remarkably regular structure of the complex roots of Euler polynomials (Table 1). Next, we calculated an approximate solution satisfying E
(
s
,
w
), w
∈
ℝ. The results are given in Table 2.Table 1
Numbers of real and complex zeros ofE
(
s
,
w
).
s
Real zeros
Complex zeros
1.5
2
0
2.5
3
0
3.5
4
0
4.5
5
0
5.5
2
4
6.5
3
4
7.5
4
4
8.5
5
4
9
5
4
9.6
6
4
9.8
6
4
10
6
4Table 2
Approximate  solutions ofE
(
s
,
w
)
=
0, w
∈
ℝ.
s
w
6
0.0000
,
1.000
6.5
-
0.24986
,
0.749547
,
6.94675
7
-
0.49773
,
0.49999
,
1.49773
7.5
-
0.731622
,
0.249968
,
1.23584
,
7.86641
8
-
0.93231
,
-
0.0000242346
,
1.00002
,
1.93232
8.5
-
1.09269
,
-
0.250159
,
0.750199
,
1.64301
,
8.78539
9
-
1.21928
,
-
0.501115
,
0.500956
,
1.49919
,
2.22008
9.6
-
1.32375
,
-
0.810502
,
0.206935
,
1.19507
,
1.90363
,
7.44777
9.8
-
1.34314
,
-
0.922069
,
0.112954
,
1.08884
,
1.90735
,
4.56809
10
-
1.34708
,
-
1.0482
,
0.0238982
,
0.97898
,
2.03661
,
2.35464Euler polynomialsE
n
(
w
) are polynomials of degree n. Thus, E
n
(
w
) has n zeros and E
n
+
1
(
w
) has n
+
1 zeros. When discrete n is analytically continued to continuous parameter s, it naturally leads to the following question: how does E
(
s
,
w
), the analytic continuation of E
n
(
w
), pick up an additional zero as s increases continuously by one?This introduces the exciting concept of the dynamics of the zeros of analytic continued polynomials, the idea of looking at how the zeros move about in thew complex plane as we vary the parameter s. For more studies and results in this subject you may see [11, 14–16].



---

*Source: 568129-2014-04-03.xml*