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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 14, Issue 1 (Jan 2016)

Issues

Differential equations associated with generalized Bell polynomials and their zeros

Seoung Cheon Ryoo
  • Corresponding author
  • Department of Mathematics, Hannam University, Daejeon 306-791, Korea (Republic of)
  • Email:
Published Online: 2016-10-29 | DOI: https://doi.org/10.1515/math-2016-0075

Abstract

In this paper, we study differential equations arising from the generating functions of the generalized Bell polynomials.We give explicit identities for the generalized Bell polynomials. Finally, we investigate the zeros of the generalized Bell polynomials by using numerical simulations.

KeyWords: Differential equations; Bell polynomials; Generalized Bell polynomials; Zeros

MSC 2010: 05A19; 11B83; 34A30; 65L99

1 Introduction

Recently, many mathematicians have worked in the are of the Bernoulli numbers, Euler numbers, Genocchi numbers, and tangent numbers (see [19]). The moments of the Poisson distribution are well-known to be connected to the combinatorics of the Bell and Stirling numbers. As is well known, the Bell numbers Bn are given by the generating function e(et1)=n=0Bntnn!.(1)

The Bell polynomials Bn(λ) are given by the generating function eλ(et1)=n=0Bn(λ)tnn!.(2)

The generalized Bell polynomials Bn(x, λ) are defined by the generating function F=F(t,x,λ)=n=0Bn(x,λ)tnn!=extλ(ett1)(see[10]).(3)

In particular the generalized Bell polynomials Bn(x, −λ) = Eλ[(Z+x−λ)n], λ, x ∈ ℝ, n ∈ ℕ where Z is a Poission random variable with parameter λ > 0 (see 10). The first few examples of generalized Bell polynomials are B0(x,λ)=1,B1(x,λ)=x,B2(x,λ)=x2λ,B3(x,λ)=x3λ3xλ,B4(x,λ)=x4λ4xλ,6x2λ+3λ2,B5(x,λ)=x5λ5xλ,10x2λ10x3λ+10λ2+15xλ2,B6(x,λ)=x6λ6xλ,15x2λ20x3λ+15x4λ+25xλ2+60λ2+45x2λ215λ3,B7(x,λ)=x7λ7xλ,21x2λ35x3λ+35x4λ+21x5λ+56λ2+175xλ2+210x2λ2+105x3λ2105λ3105xλ3.

From (2) and (3), we see that n=0Bn(x,λ)tnn!=e(x+λ)te(λ)(et1)=(k=0Bk(λ)tkk!)(m=0Bk(x+λ)mtmm!)=n=0(k=0n(nk)Bk(λ)(x+λ)nk)tnn!.(4)

Comparing the coefficients on both sides of (4), we obtain Bn(x,λ)=k=0n(nk)Bk(λ)(x+λ)nk(n0).(5)

Recently, many mathematicians have studied the differential equations arising from the generating functions of special polynomials (see [1113]). In this paper, we study differential equations arising from the generating functions of generalized Bell polynomials. We give explicit identities for the generalized Bell polynomials. In addition, we investigate the zeros of the generalized Bell polynomials with numerical methods. Finally, we observe an interesting phenomenon of ‘scattering’ of the zeros of generalized Bell polynomials.

2 Differential equations associated with generalized Bell polynomials

Differential equations arising from the generating functions of special polynomials are studied by many authors in order to give explicit identities for special polynomials (see [1113]). In this section, we study differential equations arising from the generating functions of generalized Bell polynomials.

Let F=F(t,x,λ)=extλ(ett1)=n=0Bn(x,λ)tnn!,λ,x,t.(6)

Then, by (6), we have F(1)=ddtF(t,x,λ)=ddt(extλ(ett1))=extλ(ett1)(xλ(et1))=(x+λ)F(t,x,λ)λF(t,x+1,λ),(7) F(2)=ddtF(1)=(x+λ)F(1)(t,x,λ)λF(1)(t,x+1,λ)=(x+λ)2F(t,x,λ)λ(2x+2λ+1)(t,x+1,λ)+λ2F(t,x+2,λ),(8) and F3=ddtF(2)=(x+λ)2F(t,x,λ)+(1)λ((x+λ)2+(2x+2λ+1)(x+1+λ))F(t,x+1,λ)+(1)2λ2(3x+3λ+3)F(t,x+2,λ)+(1)3λ3F(t,x+3,λ).

Continuing this process, we can guess that F(N)=(ddt)NF(t,x,λ)=i=0N(1)iai(N,x,λ)F(t,x+i,λ),(N=0,1,2,...).(9)

Taking the derivative with respect to t in (9), we have F(N+1)=dFNdt=∑!i=0N1iaiN,x,λF1t,x+i,λ=i=0N1iaiN,x,λx+i+λFt,x+i,λλFt,x+i+1,λ=i=0N1iaiN,x,λx+i+λFt,x+i,λ+i=0N1i+1λaiN,x,λFt,x+i+1,λ=i=0N1iaiN,x,λx+i+λFt,x+i,λ+i=1N+11iλai1N,x,λFt,x+i+1,λ.(10)

On the other hand, by replacing N by N + 1 in (9), we get F(N+1)=i=0N+1(1)iai(N+1,x)F(t,x+i,λ).(11)

Comparing the coefficients on both sides of (10) and (11), we obtain a0(N+1,x,λ)=(x+λ)a0(N,x,λ),aN+1(N+1,x,λ)=λaN(N,x,λ),(12) and ai(N+1,x,λ)=λai1(N,x,λ)+(x+i+λ)ai(N,x,λ),(1iN).(13)

In addition, by (9), we get F(t,x,λ)=F(0)(t,x,λ)=a0(0,x,λ)F(t,x,λ).(14)

By (14), we get a0(0,x,λ)=1.(15)

It is not difficult to show that (x+λ)F(t,x,λ)λF(t,x+1,λ)=F(1)(t,x,λ)=i=01(1)iai(1,x,λ)F(t,x+i,λ)=a0(1,x,λ)F(t,x,λ)a1(1,x,λ)F(t,x+1,λ).(16)

Thus, by (16), we also get a0(1,x,λ)=x+λ,a1(1,x,λ)=λ.(17)

From (12), we note that a0(N+1,x,λ)=(x+λ)a0(N,x,λ)==x+λNa0(1,x,λ)=x+λN+1,(18) and aN+1(N+1,x,λ)=λaN(N,x,λ)==λNa1(1,x,λ)=λN+1.(19)

For i = 1, 2, 3 in (13), we have a1(N+1,x,λ)=λk=0N(x+1+λ)ka0(Nk,x,λ),(20) a2(N+1,x,λ)=λk=0N1(x+2+λ)ka1(Nk,x,λ),(21) and a3(N+1,x,λ)=λk=0N2(x+3+λ)ka2(Nk,x,λ).(22)

Continuing this process, we can deduce that, for 1 ≤ iN, ai(N+1,x,λ)=k=0Ni+1(x+i+λ)kai1(Nk,x,λ).(23)

Here, we note that the matrix ai (j,x,λ)0≤i,jN+1 is given by (1x+λ(x+λ)2(x+λ)3(x+λ)N+10λ00λ2000λ30000λN+1)

Now, we give explicit expressions for ai (N + 1, x, λ). By (20), (21) and (22), we get a1(N+1,x,λ)=λK1=0N(x+1+λ)k1a0(Nk1,x,λ)=λk1=0N(x+1+λ)k1(x+λ)Nk1,a2(N+1,x,λ)=λK2=0N1(x+2+λ)k2a1(Nk2,x,λ)=λ2K2=0N1k1=0N1k2(x+2+λ)k2(x+1+λ)k1(x+λ)Nk2k11, and a3(N+1,x,λ)=λk3=0N2(x+2+λ)k3a2(Nk3,x,λ)=λ3k3=0N2k2=0N2k3k1=0N2K3k2(x+2+λ)k3(x+2+λ)k2(x+2+λ)k1(x+λ)Nk3k2k12.

Continuing this process, we have ai(N+1,x,λ)=λiki=0Ni+1ki1=0Ni+1kik1=0Ni+1kik2.(l=1i(x+l+λ)kl)(x+λ)Ni+1=l=1ikl.(24)

Therefore, by (24), we obtain the following theorem.

For N = 0, 1, 2, . . . , the differential equations F(N)=i=0N(1)iai(N,x,λ)F(t,x+i,λ)=(i=0N(1)iai(N,x,λ)eit)F(t,x,λ) have a solution F=F(t,x,λ)=extλ(ett1), where a0(N,x,λ)=(x+λ)N,aN(N,x,λ)=(λ)Nai(N,x,λ)=λiki=0Niki1=0Nikik1=0Nikik2(l=1i(x+l+λ)kl)(x+λ)Nil=1ikl,(1iN).

From (6), we note that F(N)=(ddt)NF(t,x,λ)=k=0Bk+N(x,λ)tkk!.(25)

From Theorem 2.1 and (25), we can derive the following equation: k=0Bk+N(x,λ)tkk!=F(N)=(i=0N(1)iai(N,x,λ)eit)F=i=0N(1)iai(N,x,λ)(l=0iltll!)(m=0Bm(x,λ)tmm!)=i=0N(1)iai(N,x,λ)(k=0m=0k(km)ikmBm(x,λ)tkk!)=k=0(i=0Nm=0k(km)ikm(1)iai(N,x,λ)Bm(x,λ)tkk!).(26)

By comparing the coefficients on both sides of (26), we obtain the following theorem.

For k, N = 0, 1, 2, . . . , we have Bk+N(x,λ)=i=0Nm=0k(km)ikm(1)iai(N,x,λ)Bm(x,λ),(27) where a0(N,x,λ)=(x+λ)N,aN(N,x,λ)=(λ)Nai(N,x,λ)=λiki=0Niki1=0Nikik1=0Nikik2(l=1i(x+l+λ)kl)(x+λ)Nil=1ikl,(1iN).

Let us take k = 0 in (27). Then, we have the following corollary.

For N = 0, 1, 2, . . . , we have BN(x,λ)=i=0N(1)iai(N,x,λ).

For N = 0, 1, 2, . . . , the functional equations F(N)=i=0N(1)iai(N,x,λ)F(t,x+i,λ)=(i=0N(1)iai(N,x,λ)eit)F(t,x,λ) have a solution F=F(t,x,λ)=extλ(ett1).

Here is a plot of the surface for this solution.

In Figure 1 (left), we choose −3 ≤ x ≤ 3, −1 ≤ t ≤ 1, and λ = −4. In Figure 1 (right), we choose −3 ≤ x ≤ 3, −1 ≤ t ≤ 1, and λ = 4.

Fig. 1

The surface for the solution F (t, x, λ)

3 Zeros of the generalized Bell polynomials

This section aims to demonstrate the benefit of using numerical investigation to support theoretical prediction and to discover new interesting pattern of the zeros of the generalized Bell polynomials Bn(x, λ). By using computer, the generalized Bell polynomials Bn(x, λ) can be determined explicitly. We display the shapes of the generalized Bell polynomials Bn(x, λ) and investigate the zeros of the generalized Bell polynomials Bn(x, λ). For n = 1, ..., 10, we can draw a plot of the generalized Bell polynomials Bn(x, λ), respectively. This shows the ten plots combined into one. We display the shape of Bn(x, λ), −10 ≤ x ≤ −10, λ = 4 (Figure 2).

Fig. 2

Zeros of Bn(x, λ)

We investigate the beautiful zeros of the generalized Bell polynomials Bn(x, λ) by using a computer. We plot the zeros of the Bn(x, λ) for n = 5, 10, 15, 20, λ = 4, and x ∈ ℂ (Figure 3).

Fig. 3

Zeros of Bn(x, λ)

In Figure 3 (top-left), we choose n = 5 and λ = 4. In Figure 3 (top-right), we choose n = 10 and λ = 4. In Figure 3 (bottom-left), we choose n = 15 and λ = 4 . In Figure 3 (bottom-right), we choose n = 20 and λ = 4. Prove that Bn(x, λ), x ∈ ℂ, has I m(x) = 0 reflection symmetry analytic complex functions (see Figure 3). Stacks of zeros of the generalized Bell polynomials Bn(x, λ) for 1 ≤ n ≤ 20, λ = 4 from a 3-D structure are presented (Figure 4).

Fig. 4

Stacks of zeros of Bn(x, λ), 1 ≤ n ≤ 20

Our numerical results for approximate solutions of real zeros of the generalized Bell polynomials Bn(x, λ) are displayed (Tables 1, 2).

Table 1

Numbers of real and complex zeros of Bn(x, 4)

Table 2

Approximate solutions of Bn(x, 4) = 0, x ∈ ℝ

Plot of real zeros of Bn(x, λ) for 1 ≤ n20 structure are presented (Figure 5).

Fig. 5

Real zeros of Bn(x, λ) for 1 ≤ n ≤ 20

We observe a remarkably regular structure of the complex roots of the generalized Bell polynomials Bn(x, λ). We hope to verify a remarkably regular structure of the complex roots of the generalized Bell polynomials Bn(x, λ) (Table 1). Next, we calculated an approximate solution satisfying Bn(x, λ) = 0, x ∈ ℂ. The results are given in Table 2.

Finally, we shall consider the more general problems. How many zeros does Bn(x, λ) have? Prove or disprove: Bn(x, λ) = 0 has n distinct solutions (see Table 2). Find the numbers of complex zeros CBn(x,λ) of Bn(x, λ), I m(x) ≠ 0. Since n is the degree of the polynomial Bn(x, λ), the number of real zeros RBn(x,λ) lying on the real line I m(x) = 0 is then RBn(x,λ) = nCBn(x,λ), where CBn(x,λ) denotes complex zeros. See Table 1 for tabulated values of RBn(x, λ) and CBn(x,λ). The author has no doubt that investigations along this line will lead to a new approach employing numerical method in the research field of the generalized Bell polynomials Bn(x, λ) to appear in mathematics and physics. The reader may refer to [14, 15] for the details.

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About the article

Received: 2016-08-02

Accepted: 2016-09-24

Published Online: 2016-10-29

Published in Print: 2016-01-01


Citation Information: Open Mathematics, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0075.

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© 2016 Ryoo, published by De Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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