# Dynamics of the Zeros of Analytic Continued(h,q)-Euler Polynomials

Authors: C. S. Ryoo
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

In this paper, we study that the(h,q)-Euler numbers En,q(h) and (h,q)-Euler polynomials En,q(h)(x) are analytic continued to Eq(h)(s) and Eq(h)(s,w). We investigate the new concept of dynamics of the zeros of analytic continued polynomials related to solution of Bernoulli equation. Finally, we observe an interesting phenomenon of “scattering” of the zeros of Eq(h)(s,w).


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## Body

## 1. Introduction

By using software, many 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. Recently, 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. Mathematicians have studied different kinds of the Euler, Bernoulli, Tangent, and Genocchi numbers and polynomials. Numerical experiments of Bernoulli polynomials, Euler polynomials, Genocchi polynomials, and Tangent polynomials have been the subject of extensive study in recent year and much progress has been made both mathematically and computationally (see [1–18]). Throughout this paper, we always make use of the following notations: N denotes the set of natural numbers, N0 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. Let q be a complex number with |q|<1 and h∈Z. Bernoulli equation is one of the well known nonlinear differential equations of the first order. It is written as
(1)dydt+p(x)y=g(x)ym(manyrealnumber),
where p(x) and g(x) are continuous functions. For m=0 and m=1 the equation is linear, and otherwise it is nonlinear. When m=2, the Bernoulli equation has the solution which is the function of exponential generating function of the Euler numbers. Simsek [18] introduced the (h,q)-Euler numbers En,q(h) and polynomials En,q(h)(x). He gave recurrence identities (h,q)-Euler polynomials and the alternating sums of powers of consecutive (h,q)-integers. In [13], we described the beautiful zeros of the (h,q)-Euler polynomials En,q(h)(x) using a numerical investigation. Also we investigated distribution and structure of the zeros of the (h,q)-Euler polynomials En,q(h)(x) by using computer.Let us define the(h,q)-Euler numbers En,q(h) and polynomials En,q(h)(x) as follows:
(2)Fq(h)(t)=2qhet+1=∑n=0∞‍En,q(h)tnn!,(3)Fq(h)(x,t)=2qhet+1ext=∑n=0∞‍En,q(h)(x)tnn!.
Observe that if q→1, then En,q(h)(x)=En(x) and En,q(h)=En, where En(x) and En denote the Euler polynomials and the numbers, respectively (see [2, 5, 8, 16, 17]).Thus(h,q)-Euler numbers En,q(h) are defined by means of the generating function
(4)Fqht=∑n=0∞‍En,q(h)tnn!=2∑m=0∞‍-1mqhmemt.
As is well known, when m=2 a special Bernoulli equation
(5)dydt+y=12y2
has the solution
(6)y=2qhet+1=∑n=0∞‍En,q,(h)tnn!.
That is, the Bernoulli equation has the solution which is the function of exponential generating function of the (h,q)-Euler numbers. Thus, a realistic study for the analytic continued polynomials Eq(h)(s,w) is very interesting by using computer. It is the aim of this paper to observe an interesting phenomenon of “scattering” of the zeros of the analytic continued polynomials Eq(h)(s,w) in complex plane.By using computer, the(h,q)-Euler numbers En,q,(h) can be determined explicitly. A few of them are
(7)E0,q(h)=21+qh,E1,q(h)=-2qh1+qh2,E2,q(h)=-2qh1+qh2+4q2h1+qh3,E3,q(h)=-2qh1+qh2+12q2h1+qh3-12q3h1+qh4.Theorem 1.
Forn∈N0, we have
(8)En,q,(h)(x)=∑k=0n‍nkEk,q(h)xn-k.By Theorem1, after some elementary calculations, we have
(9)∫ab‍En,qhxdx=∑l=0n‍nlEl,q(h)∫ab‍xn-ldx=1n+1∑l=0n+1‍n+1lEl,q(h)xn-l+1ab=En+1,q(h)(b)-En+1,q(h)(a)n+1.SinceEn,q,(h)(0)=En,q(h), by (9), we obtain
(10)En,qh(x)=En,qh+n∫0x‍En-1,qhtdtforn∈N.Then, it is easy to deduce thatEn,q(h)(x) are polynomials of degree n. Here is the list of the first (h,q)-Euler’s polynomials:
(11)E0,q(h)(x)=21+qh,E1,q(h)(x)=-2qh1+qh2+2x1+qh,E2,q(h)(x)=-2qh1+qh2+4q2h1+qh3-4qhx1+qh2+2x21+qh,E3,qhx=-2qh1+qh2+12q2h1+qh3-12q3h1+qh4+12q2hx1+qh3-6qhx1+qh2-6qhx21+qh2+2x31+qh.


## 2. Analytic Continuation of(h,q)-Euler Numbers En,q(h)

In this section, we introduce the(h,q)-Euler zeta function and Hurwitz (h,q)-Euler zeta function. By (h,q)-Euler zeta function, we consider the function Eq(h)(s) as the analytic continuation of (h,q)-Euler numbers. For more studies in this subject, you may see [2–5, 7–9, 12, 13, 18].From (4), we note that
(12)dkdtkFq(h)(t)t=0=2∑m=0∞‍-1mqhmmk=Ek,q(h),(k∈N).
By using the above equation, we are now ready to define (h,q)-Euler zeta functions.Definition 2.
Lets∈C with Re(s)>0. Consider
(13)ζE,q(h)(s)=2∑n=1∞‍-1nqhnns.
Observe thatζE,q(h)(s) is a meromorphic function on C. Clearly, lim⁡q→1ζE,q(h)(s)=ζE(s) (see [3, 4, 7–9, 12, 13, 18]). Notice that the (h,q)-Euler zeta function can be analytically continued to the whole complex plane, and these zeta functions have the values of the (h,q)-Euler numbers at negative integers.Theorem 3.
Fork∈N, we have
(14)ζE,q(h)(-k)=Ek,q(h).Observe thatζE,q(h)(s) function interpolates Ek,q(h) numbers at nonnegative integers.By using (3), we note that
(15)dkdtkFq(h)(x,t)t=0=2∑m=0∞‍-1mqhmx+mk=Ek,q(h)(x),(k∈N),(16)ddtk∑n=0∞‍En,q,(h)(x)tnn!t=0=Ek,qh(x),fork∈N.
By (16), we are now ready to define the Hurwitz-type (h,q)-Euler zeta functions.Definition 4.
Lets∈C with Re(s)>0. Consider
(17)ζE,q(h)(s,x)=2∑n=0∞‍-1nqhnn+xs.
Note thatζE,q(h)(s,x) is a meromorphic function on C (see [3, 4, 7–9, 12, 13, 18]). Relation between ζE,q(h)(s,x) and Ek,q(h)(x) is given by the following theorem.Theorem 5.
Fork∈N, we have
(18)ζE,q(h)(-k,x)=Ek,q(h)(x).We now consider the functionEq(h)(s) as the analytic continuation of (h,q)-Euler numbers. From the above analytic continuation of (h,q)-Euler numbers, we consider
(19)En,q(h)⟼Eq(h)(s),ζE,q(h)(-n)=En,q(h)⟼ζE,q(h)(-s)=Eq(h)(s).In Figure1(a), we choose q=-1/2 and h=3. In Figure 1(b), we choose q=1/2 and h=3.The curveEqh(s) runs through the points of all En,q(h) except E0,q(h).
(a)
(b)All the(h,q)-Euler numbers En,q(h) agree with Eq(h)(n), the analytic continuation of (h,q)-Euler numbers evaluated at n (see Figure 1),
(20)En,q(h)=Eq(h)(n)forn≥1exceptEqh(0)=-2qh1+qh,butE0,q(h)=21+qh.In fact, we can expressEqh′(s) in terms of ζE,qh′(s), the derivative of ζE(h)(s), as follows:
(21)Eq(h)(s)=ζE,q(h)(-s),Eqh′s=-ζE,qh′-s,Eqh′2n+1=-ζE,qh′-2n-1forn∈N0.
From the relation (21), we can define the other analytic continued half of (h,q)-Euler numbers
(22)Eqhs=ζE,q(h)(-s),Eq(h)(-s)=ζE,q(h)(s)⟹Eq(h)(-n)=ζE,q(h)(n),n∈N.
By (22), we have
(23)lim⁡n→∞E-n,q(h)=ζE,q(h)(n)=-2qh.
The curve Eq(h)(s) runs through the points E-n,q(h)=Eq(h)(-n) and grows ~-2qh asymptotically as -n→∞ (see Figure 2).The curveEqh(s) runs through the points E-n,q(h).
(a)
(b)In Figure2(a), we choose q=-1/2 and h=3. In Figure 2(b), we choose q=1/2 and h=3.


## 3. Analytic Continuation of Euler PolynomialsEn,q(h)(x)

In this section, we observe the analytic continued(h,q)-Euler polynomials. Looking back at (13) and (22), for consistency with the definition of En,q(h)(x)=Eq(h)(n,x), (h,q)-Euler polynomials should be analogously redefined as
(24)Eq(h)(0,x)=-qhE0,q(h)(x),Eq(h)(n,x)=∑l=0n‍nlEl,q(h)xn-l.
Let Γ(s) be the gamma function. The analytic continuation can be then obtained as
(25)n⟼s∈R,x⟼w∈C,E0,q(h)⟼Eq(h)(0)=-1qhζE,q(h)(0),Ek,q(h)⟼Eq(h)(k+s-[s])=ζE,q(h)(-(k+(s-[s]))),nk⟼Γ1+sΓ1+k+s-sΓ1+s-k⟹En,q(h)(w)⟼Eq(h)(s,w)=∑k=-1[s]‍Γ1+sEqhk+s-sw[s]-kΓ1+k+s-sΓ1+s-k=∑k=0[s]+1‍Γ(1+s)Eq(h)((k-1)+s-[s])w[s]+1-kΓ(k+(s-[s]))Γ(2+[s]-k),
where [s] gives the integer part of s, and so s-[s] gives the fractional part.By (25), we obtain analytic continuation of (h,q)-Euler polynomials for q=-1/2 and h=3 as follows:
(26)Eq(h)(0,w)≈2.28571,Eq(h)(1,w)≈0.326531+2.28571w,Eq(h)(2,w)≈0.41982+0.65306w+0.28571w2,Eqh2.2,w≈0.45027+0.74682w+0.38478w2+0.02373w3,Eqh2.4,w≈0.48668+0.85139w+0.50111w2+0.06112w3,Eqh2.6,w≈0.53037+0.96937w+0.63694w2+0.11503w3,Eqh2.8,w≈0.58293+1.10403w+0.79519w2+0.18867w3,Eqh2.9,w≈0.61316+1.178859w+0.88385w2+0.23401w3,Eqh3,w≈0.64639+1.25947w+0.97959w2+2.28571w3.
By using (26), we plot the deformation of the curve Eq(h)(2,w) into the curve of Eq(h)(3,w) via the real analytic continuation Eq(h)(s,w), 2≤s≤3, w∈R (see Figure 3).Figure 3
The curve ofEqh(s,w), 2≤s≤3, -0.1≤w≤0.1.Next, we investigate the beautiful zeros of theEq(h)(n,w) by using a computer. We plot the zeros of Eq(h)(n,w) for n∈N, q=1/2, h=3, and w∈C (Figure 4). In Figure 4(b), we draw x and y axes but no z axis in three dimensions. In Figure 4(c), we draw y and z axes but no x axis in three dimensions. In Figure 4(d), we draw x and z axes but no y axis in three dimensions.Stacks of zeros ofEq(h)(n,w) for 1≤n≤30.
(a)
(b)
(c)
(d)In Figure4, we observe that Eq(h)(n,w), w∈C, has Im⁡(w)=0 reflection symmetry analytic complex functions (Figure 4). The obvious corollary is that the zeros of Eq(h)(n,w) will also inherit these symmetries:
(27)ifEqh(n,w0)=0,thenEq(h)(n,w0*)=0,
where * denotes complex conjugation.Finally, we investigate the beautiful zeros of theEq(h)(s,w) by using a computer. We plot the zeros of Eq(h)(s,w) for s=28,20.8,20.9,29, q=-1/2, h=3, and w∈C (Figure 5).Zeros ofEq(h)(s,w) for s=28,28.8,28.9,29.
(a)
(b)
(c)
(d)In Figure5(a), we choose s=28. In Figure 5(b), we choose s=28.8. In Figure 5(c), we choose s=28.9. In Figure 5(d), we choose s=29.Since(28)∑n=0∞‍En,q-1h1-x-1ntnn!=2q-he-t+1e(1-x)(-t)=qh2qhet+1ext=∑n=0∞‍qhEn,qhxtnn!,
we obtain
(29)qhEn,q(h)(x)=-1nEn,q-1(h)(1-x).Observe thatEn(x), x∈C, has Re(x)=1/2 reflection symmetry in addition to the usual Im⁡(x)=0 reflection symmetry analytic complex functions (see [14]). The question is, what happens with the reflexive symmetry (29), when one considers (h,q)-Euler polynomials? Prove that Eq(h)(n,w), w∈C, has no Re(w)=1/2 reflection symmetry analytic complex functions (Figure 4). However, we observe that Eq(h)(s,w), w∈C, has Im⁡(x)=0 reflection symmetry analytic complex functions (Figure 5).Stacks of zeros ofEq(h)(s,w) for s=n+1/2, h=3, 1≤n≤30 from a 3-D structure are presented (Figure 6).Stacks of zeros ofEq(h)(s,w) for 1≤n≤30.
(a)
(b)
(c)
(d)In Figure6(b), we draw y and z axes but no x axis in three dimensions. In Figure 6(c), we draw x and y axes but no z axis in three dimensions. In Figure 6(d), we draw x and z axes but no y axis in three dimensions.Our numerical results for approximate solutions of real zeros ofEq(h)(s,w), q=-1/2, h=3 are displayed. We observe a remarkably regular structure of the complex roots of (h,q)-Euler polynomials. We hope to verify a remarkably regular structure of the complex roots of (h,q)-Euler polynomials (Table 1).Table 1
Numbers of real and complex zeros ofEq(h)(s,w).
s
Real zeros
Complex zeros
1.5
2
0
2.5
1
2
3.5
2
2
4.5
1
4
5.5
2
4
6.5
1
6
7.5
2
6
8
0
8
8.5
1
8
9
1
8
9.5
2
8
10
0
10
10.5
1
10
11
1
10
11.5
2
10Next, we calculated an approximate solution satisfyingEq(h)(s,w), q=-1/2, h=3, w∈R. The results are given in Table 2.Table 2
Approximate solutions ofEq(h)(s,w)=0,h=3,w∈R.
s
w
6
×
6.5
-
9.47111
7
-
1.06648
7.5
-
10.6998
,
-
1.8568
8
×
8.5
-
11.9289
9
-
1.34687
9.5
-
13.1581
,
-
2.12315
10
×
10.5
-
14.3874
11
-
1.62471
11.5
-
15.6169
,
-
2.3908In Figure7, we plot the real zeros of the (h,q)-Euler polynomials Eq(h)(s,w) for s=n+1/2, q=-1/2, h=3, and w∈C (Figure 7). In Figure 7(a), we choose s=n+1/2. In Figure 7(b), we choose s=n. We want to find a formula that best fits a given set of data points. The least squares method is used to fit polynomials or a set of functions to a given set of data points. Using the least squares method, we can find a and b such that x=a+bn is the least squares fit to the data given in Table 2. The graph of the data points is shown in Figure 7. We obtain x=-0.0818486-0.138376n for n=1,3,5,…. We also obtain x=-1.48453-1.22923n for n=1,3,5,… and x=-0.850454-0.134332n for n=1.5,2.5,3.5,4.5,…. The real zero x~-∞ asymptotically as n→∞.Real zeros ofEq(h)(s,w).
(a)
(b)The(h,q)-Euler polynomials Eq(h)(n,w) are polynomials of degree n. Thus, Eq(h)(n,w) has n zeros and Eq(h)(n+1,w) has n+1 zeros. When discrete n is analytic continued to continuous parameter s, it naturally leads to the following question.How doesEq(h)(s,w), the analytic continuation of Eq(h)(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 Euler polynomials, the idea of looking at how the zeros move about in thew complex plane as we vary the parameter s.To have a physical picture of the motion of the zeros in the complexw plane, imagine that each time as s increases gradually and continuously by one, an additional real zero flies in from positive infinity along the real positive axis, gradually slowing down as if “it is flying through a viscous medium.”For more studies and results in this subject, you may see [6, 11–15].



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