A Research on a Certain Family of Numbers and Polynomials Related to Stirling Numbers, Central Factorial Numbers, and Euler Numbers

Abstract

Recently, many mathematicians have studied different kinds of the Euler, Bernoulli, and Genocchi numbers and polynomials. In this paper, we give another definition of polynomials 𝑈_𝑛(𝑥). We observe an interesting phenomenon of “scattering” of the zeros of the polynomials 𝑈_𝑛(𝑥) in complex plane. We find out some identities and properties related to polynomials 𝑈_𝑛(𝑥). Finally, we also derive interesting relations between polynomials 𝑈_𝑛(𝑥), Stirling numbers, central factorial numbers, and Euler numbers.

1. Introduction

Recently, many mathematicians have studied in the areas of the Bernoulli numbers and polynomials, Euler numbers and polynomials, Genocchi numbers and polynomials, Stirling numbers, and central factorial numbers (see [1–17]). These numbers and polynomials possess many interesting properties and are arising in many areas of mathematics and physics. In this paper, we give another definition of polynomials 𝑈_𝑛(𝑥). We obtain some interesting identities and properties related to polynomials 𝑈_𝑛(𝑥). In order to study the polynomials 𝑈_𝑛(𝑥), we must understand the structure of the polynomials 𝑈_𝑛(𝑥). Therefore, using computer, a realistic study for the polynomials 𝑈_𝑛(𝑥) is very interesting. It is the aim of this paper to observe an interesting phenomenon of “scattering” of the zeros of the polynomials 𝑈_𝑛(𝑥) in complex plane.

The Stirling numbers of the first kind 𝑠(𝑛,𝑘) define that ^𝑛_𝑘=0𝑠(𝑛,𝑘)𝑥^𝑘=𝑥(𝑥−1)(𝑥−2)⋯(𝑥−𝑛+1). (1) The generating function of (1) is as follows:log(1+𝑥)^𝑘=𝑘!^∞_𝑛=𝑘𝑠()𝑥𝑛,𝑘^𝑛.𝑛! (2) From (1) and (2), we are aware of some properties of the Stirling numbers of the first kind 𝑠(𝑛,𝑘) as follows (see [1, 2, 15]): 𝑠(𝑛,𝑘)=𝑠(𝑛−1,𝑘−1)−(𝑛−1)𝑠(𝑛−1,𝑘), (3) with 𝑠()(),𝑠()𝑛,0=0𝑛∈ℕ𝑛,𝑛=1𝑛∈ℤ⁺,𝑠()=()=ℕ∪0𝑛,1−1^𝑛−1()!(),𝑠()().𝑛−1𝑛∈ℕ𝑛,𝑘=0𝑘>𝑛or𝑘<0 (4)

We usually define the central factorial numbers 𝑇(𝑛,𝑘) as the following expansion formula (see [1, 7]): ^𝑛_𝑘=0𝑇(𝑛,𝑘)𝑥𝑥−1²𝑥−2²⋯𝑥−(𝑘−1)²=𝑥^𝑛. (5) The generating function of (5) is as follows: 𝑒^𝑥+𝑒^−𝑥−2^𝑘=(2𝑘)!^∞_𝑛=𝑘𝑇()𝑥𝑛,𝑘^2𝑛()!.2𝑛 (6) By using (5) and (6), we are aware of some properties of the central factorial numbers 𝑇(𝑛,𝑘) as follows: 𝑇()𝑇()(),𝑇()(),𝑇()()0,0=1,𝑛,0=0𝑛∈ℕ𝑛,1=1𝑛∈ℕ𝑛,𝑘=𝑇𝑛−1,𝑘−1+𝑘²𝑇(𝑛−1,𝑘), (7) with 𝑇()=1𝑛,244^𝑛−1,𝑇()=9−1𝑛,3^𝑛−4360^𝑛+160().24𝑛∈ℕ (8) The Euler numbers 𝐸_𝑛 are defined by the generating function: 2𝑒^𝑡=+1^∞_𝑛=0𝐸_𝑛𝑡^𝑛.𝑛! (9) We introduce the Euler polynomials 𝐸_𝑛(𝑥) as follows: 2𝑒^𝑡𝑒+1^𝑥𝑡=^∞_𝑛=0𝐸_𝑛(𝑥)𝑡^𝑛.𝑛! (10)

Zhi-Hong Sun introduces the sequence {𝑈_𝑛} similar to Euler numbers as follows: 𝑈₀=1,𝑈_𝑛=−2^[𝑛/2]_𝑘=1𝑛𝑈2𝑘^𝑛−2𝑘(𝑛≥1), (11) where [𝑥] is the greatest integer not exceeding 𝑥 (see [4, 5]).

The outline of this paper is as follows. In Section 2, this paper is to define polynomials 𝑈_𝑛(𝑥). In Section 3, we describe the beautiful zeros of the polynomials 𝑈_𝑛(𝑥) using a numerical investigation. We investigate the roots of the polynomials 𝑈_𝑛(𝑥). Also we carried out computer experiments for demonstrating a remarkably regular structure of the complex roots of the polynomials 𝑈_𝑛(𝑥). In Section 4, we derive some special relations of polynomials 𝑈_𝑛(𝑥) and Euler numbers.

2. Some Properties Involving a Certain Family of Numbers and Polynomials

In this section, we introduce the polynomials 𝑈_𝑛(𝑥) and investigate some interesting properties and identities which are related to polynomials 𝑈_𝑛(𝑥). We also try to find relations between polynomials 𝑈_𝑛(𝑥), Stirling numbers 𝑠(𝑛,𝑘), and central factorial numbers 𝑇(𝑛,𝑘).

Definition 1. For 𝑥∈ℝ or ℂ, the polynomials 𝑈_𝑛(𝑥) are defined by ^∞_𝑛=0𝑈_𝑛(𝑥)𝑡^𝑛=1𝑛!𝑒^𝑡+𝑒^−𝑡−1^𝑥𝑒^𝑥𝑡. (12) From Definition 1, we have Theorem 2.

Theorem 2. Let 𝑛,𝑘,𝑙∈ℤ⁺ and 𝑛≥𝑘−𝑙−1. Then one obtains 𝑈_2𝑛(𝑥)=^2𝑛_𝑘=0^[𝑘/2]_𝑙=0()2𝑛2𝑙−1^𝑘−2𝑙×^𝑛−𝑙_{𝑗=𝑘−2𝑙}2𝑗!𝑇𝑠𝑥𝑗!𝑛−𝑙,𝑗𝑗,𝑘−2𝑙^𝑘,𝑈_2𝑛+1(𝑥)=^2𝑛+1_𝑘=1^[𝑘−1/2]_𝑙=0()2𝑛+12𝑙−1−1^{𝑘−(2𝑙+1)}×^𝑛−𝑙_{()𝑗=𝑘−2𝑙+1}2𝑗!𝑇𝑠()𝑥𝑗!𝑛−𝑙,𝑗𝑗,𝑘−2𝑙+1^𝑘, (13) where [𝑥] is the greatest integer not exceeding 𝑥 and ℤ⁺=ℕ∪{0}.

Proof. By Definition 1, (2), and (6), we have ^∞_𝑛=0𝑈_𝑛(𝑥)𝑡^𝑛=1𝑛!𝑒^𝑡+𝑒^−𝑡−1^𝑥𝑒^𝑡𝑥=^∞_𝑗=0𝑗()𝑥+𝑗−1−1^𝑗𝑒^𝑡+𝑒^−𝑡−2^𝑗𝑒^𝑡𝑥=^∞_𝑗=0(−1)^𝑗𝑗!𝑥+𝑗−12𝑗^∞_𝑛=𝑗𝑇𝑡𝑛,𝑗^2𝑛()!𝑒2𝑛^𝑡𝑥=^∞𝑛_𝑛=0𝑗=0(−1)^𝑗𝑗𝑥+𝑗−1𝑡2𝑗!𝑇𝑛,𝑗^2𝑛()!𝑒2𝑛^𝑡𝑥=^∞𝑛_𝑛=0^𝑗_𝑗=0𝑘=0(−1)^𝑘𝑠𝑗,𝑘𝑥^𝑘2𝑗!𝑇𝑡𝑗!𝑛,𝑗^2𝑛(2𝑛)!^∞_𝑛=0𝑥^𝑛𝑡^𝑛.𝑛! (14) From now on, we have to consider odd terms and even terms by using Cauchy product. So, we get to generating terms by dividing the odd terms and the even terms, respectively, ^∞_𝑛=0𝑈_𝑛(𝑥)𝑡^𝑛=𝑛!^∞_𝑛=0𝑈_2𝑛(𝑥)𝑡^2𝑛()!+2𝑛^∞_𝑛=0𝑈_2𝑛+1(𝑥)𝑡^2𝑛+1()!.2𝑛+1 (15) The following equation is the generating even terms: 𝑈_2𝑛(𝑥)=^𝑛_𝑙=02𝑛2𝑙^{𝑛−𝑙𝑗=0}2𝑗!𝑇𝑗!𝑛−𝑙,𝑗^𝑗_𝑘=0(−1)^𝑘𝑠𝑗,𝑘𝑥^𝑘+2𝑙=^2𝑛_𝑘=0^[𝑘/2]_𝑙=0)2𝑛2𝑙(−1^{𝑘−2𝑙𝑛−𝑙}_{𝑗=𝑘−2𝑙}2𝑗!𝑇𝑠𝑥𝑗!𝑛−𝑙,𝑗𝑗,𝑘−2𝑙^𝑘, (16) where [𝑥] is the greatest integer not exceeding 𝑥.
We also derive the generating odd terms: 𝑈_2𝑛+1(𝑥)=^𝑛_𝑙=02𝑛+12𝑙+1^{𝑛−𝑙𝑗=0}2𝑗!𝑇𝑗!𝑛−𝑙,𝑗^𝑗_𝑘=0(−1)^𝑘𝑠𝑗,𝑘𝑥^{𝑘+(2𝑙+1)}=^2𝑛+1_𝑘=1^{[(𝑘−1)/2]}_𝑙=0)2𝑛+12𝑙−1(−1^{𝑘−(2𝑙+1)}×^𝑛−𝑙_{()𝑗=𝑘−2𝑙+1}2𝑗!𝑇𝑠()𝑥𝑗!𝑛−𝑙,𝑗𝑗,𝑘−2𝑙+1^𝑘, (17) where [𝑥] is the greatest integer not exceeding 𝑥.
Thus, we complete the proof of Theorem 2.

Example 3. Let 𝑛=1,2,3,4,5,6,7,8. Then we can know the following polynomials: 𝑈₁(𝑥)𝑈=𝑥,₂(𝑥)=𝑥²𝑈−2𝑥,₃(𝑥)=𝑥³−6𝑥²,𝑈₄(𝑥)=𝑥⁴−12𝑥³+12𝑥²𝑈+10𝑥,₅(𝑥)=𝑥⁵−20𝑥⁴+60𝑥³+50𝑥²,𝑈₆(𝑥)=𝑥⁶−30𝑥⁵+180𝑥⁴+30𝑥³−300𝑥²𝑈−182𝑥,₇(𝑥)=𝑥⁷−42𝑥⁶+420𝑥⁵−490𝑥⁴−2100𝑥³−1274𝑥²,𝑈₈(𝑥)=𝑥⁸−56𝑥⁷+840𝑥⁶−2660𝑥⁵−6720𝑥⁴+3304𝑥³+13692𝑥²+6970𝑥,…. (18)

Corollary 4. Let 𝑛,𝑙∈ℤ⁺ and 𝑛≥𝑙. Then one has 𝑈_2𝑛(𝑥)=^𝑛_𝑙=02𝑛2𝑙^{𝑛−𝑙𝑗=0}2𝑗!𝑇𝑗!𝑛−𝑙,𝑗^𝑗_𝑘=0(−1)^𝑘𝑠𝑗,𝑘𝑥^𝑘+2𝑙,𝑈_2𝑛+1(𝑥)=^𝑛_𝑙=02𝑛+12𝑙+1^{𝑛−𝑙𝑗=0}2𝑗!𝑇×𝑗!𝑛−𝑙,𝑗_𝑗𝑘=0(−1)^𝑘𝑠𝑗,𝑘𝑥^{𝑘+(2𝑙+1)}. (19)

Proof. Corollary 4 shows identities of 𝑈_2𝑛(𝑥) and 𝑈_2𝑛+1(𝑥). The proof of Corollary 4 is contained in Theorem 2.

Setting 𝑥=1 in Corollary 4, we easily see the following corollary.

Corollary 5. Relation of  𝑈_𝑛(1) and central factorial numbers is shown.
Let 𝑛,𝑙,𝑗∈ℤ⁺ and 𝑛≥𝑙+𝑗. Then one has 𝑈_2𝑛(1)=^𝑛_𝑙=02𝑛2𝑙^{𝑛−𝑙𝑗=0}(−1)^𝑗,𝑈2𝑗!𝑇𝑛−𝑙,𝑗_2𝑛+1(1)=^𝑛_𝑙=02𝑛+12𝑙+1^{𝑛−𝑙𝑗=0}(−1)^𝑗2𝑗!𝑇𝑛−𝑙,𝑗. (20)

Proof. Consider the following: 𝑈_2𝑛(𝑥)=^𝑛_𝑙=02𝑛2𝑙^{𝑛−𝑙𝑗=0}2𝑗!𝑇𝑗!𝑛−𝑙,𝑗^𝑗_𝑘=0(−1)^𝑘𝑠𝑗,𝑘𝑥^𝑘+2𝑙=^𝑛_𝑙=02𝑛2𝑙^{𝑛−𝑙𝑗=0}2𝑗!𝑇)𝑗!𝑛−𝑙,𝑗(−1^𝑗2𝑗_𝑘=0(−1)^𝑘𝑠𝑗,𝑘𝑥^𝑘+2𝑙=^𝑛_𝑙=02𝑛2𝑙^{𝑛−𝑙𝑗=0}2𝑗!()𝑗!−1^−𝑗𝑇𝑛−𝑙,𝑗^𝑗_𝑘=0(−1)^𝑘−𝑗𝑠𝑗,𝑘𝑥^𝑘+2𝑙. (21) We find out the following equation from the above equation by substituting 𝑥=1: 𝑈_2𝑛(1)=^𝑛_𝑙=02𝑛2𝑙^{𝑛−𝑙𝑗=0}2𝑗!𝑇=𝑗!𝑛−𝑙,𝑗𝑗!_𝑛𝑙=02𝑛2𝑙^{𝑛−𝑙𝑗=0}(−1)^𝑗2𝑗!𝑇𝑛−𝑙,𝑗. (22) We also can find out 𝑈_2𝑛+1(1) by using similar method as above: 𝑈_2𝑛+1(𝑥)=^𝑛_𝑙=02𝑛+12𝑙+1^{𝑛−𝑙𝑗=0}2𝑗!𝑇×𝑗!𝑛−𝑙,𝑗_𝑗𝑘=0(−1)^𝑘𝑠𝑗,𝑘𝑥^𝑘+2𝑙=^𝑛_𝑙=02𝑛+12𝑙+1^{𝑛−𝑙𝑗=0}2𝑗!𝑇()𝑗!𝑛−𝑙,𝑗−1^2𝑗×^𝑗_𝑘=0(−1)^𝑘𝑠𝑗,𝑘𝑥^𝑘+2𝑙=^𝑛_𝑙=02𝑛+12𝑙+1^{𝑛−𝑙𝑗=0}2𝑗!()𝑗!−1^−𝑗𝑇×𝑛−𝑙,𝑗_𝑗𝑘=0(−1)^𝑘−𝑗𝑠𝑗,𝑘𝑥^𝑘+2𝑙. (23) If we substitute 𝑥=1, then 𝑈_2𝑛+1(1)=^𝑛_𝑙=02𝑛+12𝑙+1^{𝑛−𝑙𝑗=0}2𝑗!𝑇=𝑗!𝑛−𝑙,𝑗𝑗!_𝑛𝑙=02𝑛+12𝑙+1^{𝑛−𝑙𝑗=0}(−1)^𝑗2𝑗!𝑇𝑛−𝑙,𝑗. (24)

Let 𝑛,𝑘,𝑙∈ℤ⁺ and 𝑛≥𝑘−𝑙−1 and 𝑎(2𝑛,𝑘)=^[𝑘/2]_𝑙=0()2𝑛2𝑙−1^𝑘−2𝑙×^𝑛−𝑙_{𝑗=𝑘−2𝑙}2𝑗!𝑇𝑠,𝑏()=𝑗!𝑛−𝑙,𝑗𝑗,𝑘−2𝑙2𝑛+1,𝑘^{[(𝑘−1)/2]}_𝑙=0()2𝑛+12𝑙−1−1^{𝑘−(2𝑙+1)}×^𝑛−𝑙_{()𝑗=𝑘−2𝑙+1}2𝑗!𝑇𝑠(),𝑗!𝑛−𝑙,𝑗𝑗,𝑘−2𝑙+1 (25) where [𝑥] is the greatest integer not exceeding 𝑥.

Then we have 𝑈_2𝑛(𝑥)=^2𝑛_𝑘=0𝑎(2𝑛,𝑘)𝑥^𝑘,𝑈_2𝑛+1(𝑥)=^2𝑛+1_𝑘=1𝑏(2𝑛+1,𝑘)𝑥^𝑘. (26)

By using 𝑎(2𝑛,𝑘) and 𝑏(2𝑛+1,𝑘), we obtain Theorem 6.

Theorem 6. Let 𝑛,𝑘∈ℤ⁺. Then one gets 𝑈_2𝑛(1)=21𝑖𝑓𝑛=0_𝑛−1𝑘∑₌₀𝑎(2𝑛,2𝑘)−2^2𝑛𝑈+1𝑖𝑓𝑛≥𝑘+1,_2𝑛+1(1)=2^𝑛_𝑘=1𝑏(2𝑛+1,2𝑘)+2^2𝑛+1𝑈−1._2𝑛(1)=21𝑖𝑓𝑛=0_𝑛−1𝑘∑₌₀𝑎(2𝑛,2𝑘+1)+2^𝑛𝑈−1𝑖𝑓𝑛≥𝑘+1,_2𝑛+1(1)=2^2𝑛+1_𝑘=1𝑏(2𝑛+1,2𝑘−1)−2^2𝑛+1+1. (27)

Proof. From Definition 1, one easily obtains the following equation: ^∞_𝑛=0𝑈_𝑛(𝑥)𝑡^𝑛=𝑛!^∞_𝑛=0𝑈_2𝑛(𝑥)𝑡^2𝑛()!+2𝑛^∞_𝑛=0𝑈_2𝑛+1(𝑥)𝑡^2𝑛+1()!=𝑒2𝑛+1_𝑡𝑒^𝑡+𝑒^−𝑡−1^𝑥. (28) We also easily get the following equation by substituting 𝑥=−1: ^∞_𝑛=0𝑈_𝑛()𝑡−1^𝑛=𝑛!^∞_𝑛=0𝑈_2𝑛()𝑡−1^2𝑛()!+2𝑛_∞𝑛=0𝑈_2𝑛+1()𝑡−1^2𝑛+1()!=𝑒2𝑛+1_𝑡𝑒^𝑡+𝑒^−𝑡−1⁻¹=1+^∞_𝑛=0(−1)^𝑛2^𝑛𝑡−1^𝑛𝑛!=1+^∞_𝑛=0(−1)^𝑛2^𝑛𝑡−1^𝑛𝑛!=1+^∞_𝑛=12^2𝑛𝑡−1^2𝑛−2𝑛!_∞𝑛=02^2𝑛+1𝑡−1^2𝑛+1()!.2𝑛+1 (29) Thus, ^∞_𝑛=0𝑈_2𝑛()𝑡−1^2𝑛(2𝑛)!=1+^∞_𝑛=12^2𝑛𝑡−1^2𝑛,𝑈2𝑛!_2𝑛+1(−1)=1−2^2𝑛+1. (30) We have to consider odd terms and even terms in (26) and the previous equations, respectively.
Odd terms are in the following form: 𝑈_2𝑛+1(−1)=^2𝑛+1_𝑘=1𝑏(2𝑛+1,𝑘)(−1)^𝑘=^𝑛_𝑘=1𝑏(2𝑛+1,2𝑘)−^𝑛+1_𝑘=1𝑏(2𝑛+1,2𝑘−1)=1−2^2𝑛+1,𝑈_2𝑛+1(1)=2^𝑛_𝑘=1𝑏(2𝑛+1,2𝑘)+2^2𝑛+1−1=2^2𝑛+1_𝑘=1𝑏(2𝑛+1,2𝑘−1)−2^2𝑛+1+1, (31) and even terms are similar to the way of the proof process of the odd terms. Therefore, we omit the proof process of the even terms.

Theorem 7. Let 𝑛,𝑘∈ℤ⁺ and 𝑘+1≤𝑛. Then one has 𝑈_𝑛+1(1)−𝑈_𝑛(1)=^𝑛−1_𝑘=0𝑛𝑘1−2^𝑛−𝑘𝑈_𝑘+1(1)−𝑈_𝑘(1). (32)

Proof. From Definition 1, one easily obtains the following equation: ^∞_𝑛=0𝑈_𝑛(1)𝑡^𝑛=𝑒𝑛!𝑒^𝑡+𝑒^𝑡−𝑡.−1 (33) We also easily get the following by using differential, namely, ^∞_𝑛=1𝑈_𝑛(1)𝑡^𝑛−1()!=𝑛−1−𝑒^𝑡+2(𝑒^𝑡+𝑒^−𝑡−1)². (34) Thus, 𝑒^2𝑡−𝑒^𝑡+1^∞_𝑛=1𝑈_𝑛(1)𝑡^𝑛−1()!=𝑒𝑛−1_2𝑡−𝑒^𝑡+1−𝑒^𝑡+2(𝑒^𝑡+𝑒^−𝑡−1)²=2−𝑒^𝑡∞_𝑛=0𝑈_𝑛(1)𝑡^𝑛.𝑛! (35) Then, right-hand side is in the following form: 2−𝑒^𝑡∞_𝑛=0𝑈_𝑛(1)𝑡^𝑛𝑛!=2^∞_𝑛=0𝑈_𝑛(1)𝑡^𝑛−𝑛!^∞∞_𝑛=0𝑘=0𝑛𝑘𝑈_𝑘(1)𝑡^𝑛=𝑛!_∞𝑛=02𝑈_𝑛(1)−^𝑛_𝑘=0𝑛𝑘𝑈_𝑘(1)𝑡^𝑛,𝑛! (36) and left-hand side is in the following form: 𝑒^2𝑡−𝑒^𝑡+1^∞_𝑛=1𝑈_𝑛(1)𝑡^𝑛−1()!=𝑛−1_∞𝑛=02^𝑛𝑡^𝑛𝑛!^∞_𝑛=0𝑈_𝑛+1(1)𝑡^𝑛−𝑛!^∞∞_𝑛!𝑛=0𝑈_𝑛+1(1)𝑡^𝑛+𝑛!^∞_𝑛=0𝑈_𝑛+1(1)𝑡^𝑛=𝑛!_{∞𝑛=0𝑛𝑘=0}𝑛𝑘2^𝑛−𝑘−1𝑈_𝑘+1(1)+𝑈_𝑛+1(1)𝑡^𝑛.𝑛! (37) By using comparing coefficients of 𝑡^𝑛/𝑛! in the previous equations, we can represent the equation; that is, ^𝑛_𝑘=0𝑛𝑘2^𝑛−𝑘−1𝑈_𝑘+1(1)+𝑈_𝑛+1(1)=2𝑈_𝑛(1)−^𝑛_𝑘=0𝑛𝑘𝑈_𝑘(1). (38) By simple calculation, we get 𝑈_𝑛+1(1)−2𝑈_𝑛(1)=^𝑛_𝑘=0𝑛𝑘1−2^𝑛−𝑘𝑈_𝑘+1(1)−^𝑛_𝑘=0𝑛𝑘𝑈_𝑘(1)=^𝑛−1_𝑘=0𝑛𝑘1−2^𝑛−𝑘𝑈_𝑘+1(1)−𝑈_𝑘(1)−𝑈_𝑛(1). (39) Therefore, we proved that 𝑈_𝑛+1(1)−𝑈_𝑛(1)=^𝑛−1_𝑘=0𝑛𝑘1−2^𝑛−𝑘𝑈_𝑘+1(1)−𝑈_𝑘(1). (40)

3. Zeros of the Polynomials 𝑈_𝑛(𝑥)

In this section, we investigate the reflection symmetry of the zeros of the polynomials 𝑈_𝑛(𝑥).

We investigate the beautiful zeros of the polynomials 𝑈_𝑛(𝑥) by using a computer. We plot the zeros of the polynomials 𝑈_𝑛(𝑥) for 𝑛=20,30,40,50 and 𝑥∈ℂ (Figure 1). In Figure 1(a), we choose 𝑛=20. In Figure 1(b), we choose 𝑛=30. In Figure 1(c), we choose 𝑛=40. In Figure 1(d), we choose 𝑛=50.

(a)

(b)

(c)

(d)

Figure 1 

Zeros of 𝑈𝑛(𝑥) for 𝑛=20,30,40,50.

Throughout the numerical experiments, we can finally conclude that polynomials 𝑈_𝑛(𝑥) have no Re(𝑥)=0 reflection symmetry analytic complex functions. However, we observe that 𝑈_𝑛(𝑥) has Im(𝑥)=0 reflection symmetry (see Figures 1, 2, and 3). The obvious corollary is that the zeros of 𝑈_𝑛(𝑥) will also inherit these symmetries: if𝑈_𝑛(𝑥)=0,then𝑈_𝑛𝑥^∗=0, (41) where ∗ denotes complex conjugation (see Figure 1).

Figure 2 

Real zeros of 𝑈𝑛(𝑥) for 1≤𝑛≤50.

Figure 3 

Stacks of zeros of 𝑈𝑛(𝑥) for 1≤𝑛≤50.

Plots of real zeros of 𝑈_𝑛(𝑥) for 1≤𝑛≤50 structure are presented (Figure 2).

Our numerical results for approximate solutions of real zeros of the 𝑈_𝑛(𝑥) are displayed (Tables 1 and 2).

Table 1 

Numbers of real and complex zeros of 𝑈_𝑛(𝑥).

Table 2 

Approximate solutions of 𝑈_𝑛(𝑥)=𝑥∈ℝ.

We observe a remarkably regular structure of the complex roots of the polynomials 𝑈_𝑛(𝑥). We hope to verify a remarkably regular structure of the complex roots of the polynomials 𝑈_𝑛(𝑥) (Table 1).

Stacks of zeros of 𝑈_𝑛(𝑥) for 1≤𝑛≤50 from a 3-D structure are presented (Figure 3).

Next, we calculated an approximate solution satisfying 𝑈_𝑛(𝑥),𝑥∈ℝ. The results are given in Table 2.

Finally, we will consider the more general problems. Find the numbers of complex zeros 𝐶𝑈_𝑛(𝑥) of 𝑈_𝑛(𝑥),Im(𝑥)≠0. Since 𝑛 is the degree of the polynomial 𝑈_𝑛(𝑥), the number of real zeros 𝑅𝑈_𝑛^(𝑥) lying on the real plane Im(𝑥)=0 is then 𝑅𝑈_𝑛(𝑥)𝑈=𝑛−𝐶_𝑛(𝑥), where 𝐶𝑈_𝑛(𝑥) denotes complex zeros. See Table 1 for tabulated values of 𝑅𝑈_𝑛^(𝑥) and 𝐶𝑈_𝑛(𝑥). We plot the zeros of 𝑈_𝑛(𝑥), respectively (Figures 1–3). These figures give mathematicians an unbounded capacity to create visual mathematical investigations of the behavior of the roots of the 𝑈_𝑛(𝑥). Moreover, it is possible to create new mathematical ideas and analyze them in ways that generally are not possible by hand. The authors have no doubt that investigation along this line will lead to a new approach employing numerical methods in the field of research of 𝑈_𝑛(𝑥) to appear in mathematics and physics.

4. Some Relations of Polynomials 𝑈_𝑛(𝑥) and Euler Numbers

In this section, we find out interesting relations between polynomials 𝑈_𝑛(𝑥) and Euler numbers. We usually use the values of polynomials 𝑈_𝑛(𝑥) and Euler polynomials.

Theorem 8. Let 𝑛∈ℤ⁺. Then one has 23^𝑛𝑈_𝑛(1)=𝐸_𝑛23+(−1)^𝑛𝐸_𝑛,23^𝑛𝑈_𝑛(2)=𝐸23−𝐸^𝑛+2^𝑛−1𝐸_𝑛23+(−1)^𝑛𝐸_𝑛, (42) with the usual convention of replacing 𝐸^𝑛 by 𝐸_𝑛.

Proof. By setting 𝑥=1 in Definition 1 and using Euler numbers, then we have ^∞_𝑛=0𝑈_𝑛(1)𝑡^𝑛=𝑒𝑛!𝑒^𝑡+𝑒^𝑡−𝑡=1−11+𝑒^−3𝑡+11+𝑒^3𝑡𝑒^2𝑡=12^∞_𝑛=03^𝑛(−1)^𝑛𝐸_𝑛+𝐸_𝑛23𝑡^𝑛.𝑛! (43)
Hence, 23^𝑛𝑈_𝑛(1)=𝐸_𝑛23+(−1)^𝑛𝐸_𝑛. (44)
We also obtain the above equation by taking 𝑥=2 and using Euler numbers: ^∞_𝑛=0𝑈_𝑛(2)𝑡^𝑛=𝑒𝑛!𝑒^𝑡+𝑒^𝑡−𝑡−1²=11+𝑒^−3𝑡+11+𝑒^3𝑡𝑒^2𝑡×11+𝑒^−3𝑡+11+𝑒^3𝑡𝑒^2𝑡=12^∞_𝑛=03^𝑛(−1)^𝑛𝐸_𝑛+𝐸_𝑛23𝑡^𝑛×1𝑛!2^∞_𝑛=03^𝑛(−1)^𝑛𝐸_𝑛+𝐸_𝑛23𝑡^𝑛.𝑛! (45) By using Cauchy product in the above equation, we have ^∞_𝑛=0𝑈_𝑛(2)𝑡^𝑛=1𝑛!4^∞𝑛_𝑛=0𝑘=0𝑛𝑘3^𝑛(−1)^𝑘𝐸_𝑘+𝐸_𝑘23×(−1)^𝑛−𝑘𝐸_𝑛−𝑘+𝐸_𝑛−𝑘23𝑡^𝑛.𝑛! (46) By comparing the coefficient of both sides of 𝑡^𝑛/𝑛! and some calculations, we have the following equation: 23^𝑛𝑈_𝑛(2)=𝐸_𝑛23−𝐸^𝑛+2^𝑛−1𝐸_𝑛23+(−1)^𝑛𝐸_𝑛, (47) with the usual convention of replacing 𝐸^𝑛 by 𝐸_𝑛.
Therefore, we consummated the proof of Theorem 8.

From (26), we can suppose that 𝑈_𝑛(𝑥)=^2𝑛_𝑘=0𝑎(2𝑛,𝑘)𝑥^𝑘+^2𝑛+1_𝑘=1𝑏(2𝑛+1,𝑘)𝑥^𝑘=^𝑛_𝑘=0𝑐(𝑛,𝑘)𝑥^𝑘, (48) where 𝑐()=𝑎()𝑏()𝑛,𝑘2𝑛,𝑘if𝑛=odd,2𝑛+1,𝑘if𝑛=even. (49) Therefore, we have Theorem 9.

Theorem 9. Let 𝑛,𝑣₁,…,𝑣_𝑘,𝑘∈ℕ. Then one has ()=3𝑘!𝑐𝑛,𝑘_𝑛−𝑘𝑛!2^𝑘×_{𝑣1,…,𝑣𝑘𝑣∈ℕ1+⋯+𝑣𝑘=𝑛}𝐸_𝑣1−113−𝐸_𝑣1−123+2𝐸_𝑣1−1⋯𝐸_𝑣𝑘−113−𝐸_𝑣𝑘−123+2𝐸_𝑣𝑘−1×𝑣₁!⋯𝑣_𝑘!⁻¹. (50)

Proof. From Definition 1, we can differentiate the 𝑘-times as follows: ^∞_𝑛=𝑘𝑑^𝑘𝑑𝑥^𝑘𝑈_𝑛(𝑥)_𝑥=0𝑡^𝑛=𝑒𝑛!log^𝑡𝑒^𝑡+𝑒^−𝑡−1^𝑘. (51) Here we required that 𝑒log^𝑡𝑒^𝑡+𝑒^−𝑡=−1_∞𝑛=03^𝑛2𝐸_𝑛13−𝐸_𝑛23+2𝐸_𝑛𝑡^𝑛+1()!=𝑛+1_∞𝑛=13^𝑛−12𝐸_𝑛−113−𝐸_𝑛−123+2𝐸_𝑛−1𝑡^𝑛.𝑛! (52) Hence, we derive that ^∞_𝑛=𝑘𝑑^𝑘𝑑𝑥^𝑘𝑈_𝑛(𝑥)_𝑥=0𝑡^𝑛=𝑛!_∞𝑛=13^𝑛−12𝐸_𝑛−113−𝐸_𝑛−123+2𝐸_𝑛−1𝑡^𝑛𝑛!^𝑘=^∞_𝑛=𝑘3^𝑛−𝑘𝑛!2^𝑘×_{𝑣1,…,𝑣𝑘𝑣∈ℕ1+⋯+𝑣𝑘=𝑛}𝐸_𝑣1−113−𝐸_𝑣1−123+2𝐸_𝑣1−1⋯𝐸_𝑣𝑘−113−𝐸_𝑣𝑘−123+2𝐸_𝑣𝑘−1×𝑣₁!⋯𝑣_𝑘!⁻¹𝑡^𝑛.𝑛! (53) By comparing the coefficient of both sides of 𝑡^𝑛/𝑛!, we have the following equation: 𝑑^𝑘𝑑𝑥^𝑘𝑈_𝑛(𝑥)_𝑥=0=3^𝑛−𝑘𝑛!2^𝑘_{𝑣1,…,𝑣𝑘𝑣∈ℕ1+⋯+𝑣𝑘=𝑛}𝐸_𝑣1−113−𝐸_𝑣1−123+2𝐸_𝑣1−1⋯𝐸_𝑣𝑘−113−𝐸_𝑣𝑘−123+2𝐸_𝑣𝑘−1×𝑣₁!⋯𝑣_𝑘!⁻¹. (54) From (48), we easily see that 𝑑^𝑘𝑑𝑥^𝑘𝑈_𝑛(𝑥)_𝑥=0=𝑘!𝑐(𝑛,𝑘). (55)
Therefore, we clear off the proof of Theorem 9.

Theorem 10. Let 𝑛,𝑣₁,…,𝑣_𝑘,𝑘∈ℕ. Then one has 𝑘!𝑐(𝑛,𝑘)=𝑛!_{𝑣1,…,𝑣𝑘𝑣∈ℕ1+⋯+𝑣𝑘=𝑛}𝑈(𝑥)−3^𝑣1−1−𝑈(𝑥)−1^𝑣1−1⋯𝑈(𝑥)−3^𝑣𝑘−1−𝑈(𝑥)−1^𝑣𝑘−1×𝑣₁!⋯𝑣_𝑘!⁻¹. (56)

Proof. This proof can be proved by the similar method of Theorem 9.

We also can derive relation polynomials 𝑈_𝑛(𝑥) and Euler numbers as follows.

Theorem 11. Let 𝑛,𝑘∈ℤ⁺ and 𝑘+1≤𝑛. Then one derives ^{𝑛−1𝑘=0}𝑘()𝑛−1−1^{𝑛−𝑘−1}2^𝑛−𝑘−1𝑈_𝑘(1)=3^𝑛−12𝐸_𝑛−113−𝐸_𝑛−123+2𝐸_𝑛−1. (57)

Proof. By the proof of Theorem 9, we can differentiate polynomials 𝑈_𝑛(𝑥) as follows: 𝑑𝑒𝑑𝑡log^𝑡𝑒^𝑡+𝑒^−𝑡=−12𝑒^−𝑡−1𝑒^𝑡+𝑒^−𝑡=𝑒−1_𝑡𝑒^𝑡+𝑒^−𝑡−12𝑒^−2𝑡−𝑒^−𝑡=^∞_𝑛=0𝑈_𝑛(1)𝑡^𝑛2𝑛!^∞_𝑛=0(−2)^𝑛𝑡^𝑛−𝑛!^∞_𝑛=0(−1)^𝑛𝑡^𝑛=𝑛!_{∞𝑛𝑛=0𝑘=0}𝑛𝑘(−1)^𝑛−𝑘2^{𝑛−𝑘+1}−1𝑈_𝑘(1)𝑡^𝑛.𝑛! (58) By integrating from 0 to 𝑡, we deduce that 𝑒log^𝑡𝑒^𝑡+𝑒^−𝑡=−1_{∞𝑛𝑛=0𝑘=0}𝑛𝑘(−1)^𝑛−𝑘2^{𝑛−𝑘+1}−1𝑈_𝑘(1)𝑡^𝑛+1()!=𝑛+1_{∞𝑛=1𝑛−1𝑘=0}𝑘()𝑛−1−1^{𝑛−(𝑘+1)}2^𝑛−𝑘−1𝑈_𝑘(1)𝑡^𝑛.𝑛! (59)
We already knew from the proof process of Theorem 9 that 𝑒log^𝑡𝑒^𝑡+𝑒^−𝑡=−1_∞𝑛=13^𝑛−12𝐸_𝑛−113−𝐸_𝑛−123+2𝐸_𝑛−1𝑡^𝑛.𝑛! (60) By using the above equation, we can derive that ^{𝑛−1𝑘=0}𝑘()𝑛−1−1^{𝑛−𝑘−1}2^𝑛−𝑘−1𝑈_𝑘(1)=3^𝑛−12𝐸_𝑛−113−𝐸_𝑛−123+2𝐸_𝑛−1. (61)
Therefore, we completely demonstrated the proof of Theorem 11.

Acknowledgments

The authors express their gratitude to the referee for his/her valuable comments. This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korean Government (NRF-2013-Fostering Core Leaders of the Future Basic Science Program).