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Some properties for 2-variable modified partially degenerate Hermite (MPDH) polynomials derived from differential equations and their zeros distributions

  • The 2-variable modified partially degenerate Hermite (MPDH) polynomials are the subject of our study in this paper. We found basic properties of these polynomials and obtained several types of differential equations related to MPDH polynomials. Based on the MPDH polynomials, we looked at the structures of the approximation roots for a particular polynomial and checked the values of the approximate roots. Further, we presented some conjectures for MPDH polynomials.

    Citation: Gyung Won Hwang, Cheon Seoung Ryoo, Jung Yoog Kang. Some properties for 2-variable modified partially degenerate Hermite (MPDH) polynomials derived from differential equations and their zeros distributions[J]. AIMS Mathematics, 2023, 8(12): 30591-30609. doi: 10.3934/math.20231564

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  • The 2-variable modified partially degenerate Hermite (MPDH) polynomials are the subject of our study in this paper. We found basic properties of these polynomials and obtained several types of differential equations related to MPDH polynomials. Based on the MPDH polynomials, we looked at the structures of the approximation roots for a particular polynomial and checked the values of the approximate roots. Further, we presented some conjectures for MPDH polynomials.



    A solution of the following second-order differential equation

    g2γg+2ng=0

    is the ordinary Hermite polynomials Hn(γ).

    Actually, for n{0,1,2,}, the following form

    d2H(γ)dγ22γdH(γ)dγ+2nH(γ)=0

    has a solution which is the generating function of the ordinary Hermite polynomials Hn(γ).

    Let Hn(γ,ν) be 2-variable Hermite polynomials given by the following generating function:

    n=0Hn(γ,ν)τnn!=eτ(γ+ντ)(see [9]). (1)

    In other words, this polynomial provides the solution of the following heat equation:

    νHn(γ,ν)=2γ2Hn(γ,ν). (2)

    It is clear that Hn(γ,0)=γn and Hn(2γ,1)=Hn(γ). Mathematicians who study polynomials have introduced and continuously developed new special polynomials that can be applied in fields such as combinatorics, numerical analysis, physics, and so on (see [1,2,3,4,5,7,12,14]). Carlitz pioneered the concept of degenerate polynomials, and since then several mathematicians have worked to extend well-known special polynomials such as Bernoulli ([6,15]), Euler [6] and tangent [13] to encompass their degenerate counterparts. With the discovery of new polynomials, they studied various properties and identities of polynomials, the structure of approximate roots of polynomials, differential equations and more (see [10,11]). Young [15] showed the symmetric properties, congruence and identities and the relationship with the Stirling number for degenerate Bernoulli polynomials. Ryoo [13] confirmed the structures of approximate roots, properties of approximated real and imaginary roots and the identity for degenerate tangent polynomials. Carlitz [6] has defined the degenerate Stirling, Bernoulli and Eulerian numbers and he studied some degenerate properties of these numbers. In [8], Hwang and Ryoo introduced the new generating function of the 2-variable degenerate Hermite polynomials Hn(γ,ν,ζ) as follows:

    n=0Hn(γ,ν,ζ)τnn!=(1+ζ)(γ+ντ)τζ. (3)

    We can see that (3) reduces to (1) when (1+ζ)τζeτ as ζ0. Hn(γ,ν,ζ) is the solution of the following partial differential equation:

    νHn(γ,ν,ζ)=ζlog(1+ζ)2γ2Hn(γ,ν,ζ),Hn(γ,0,ζ)=(ζlog(1+ζ))nγn. (4)

    Since ζlog(1+ζ)1 as ζ approaches to 0, it is clear that (4) becomes (2).

    The purpose of this paper is to construct new modified partially degenerate Hermite (MPDH) polynomials based on the results mentioned above and find some conjectures by looking at the properties of differential equations and approximate roots related to these polynomials.

    The structure of this paper is as follows. We find some properties of the newly defined MPDH polynomials in Section 2. Here, we also obtain the relationship between the Stirling numbers of the first kind and the MPDH polynomials. In Section 3, we derive the symmetric properties, which are important properties of MPDH polynomials, and Section 4 presents several differential equations with MPDH polynomials as solutions. Section 5 introduces the structures and data of approximation roots of MPDH polynomials. We try to understand higher-order MPDH polynomials by looking at various experimental results by substituting various values of variables into the MPDH polynomials. In Section 6, we present some conjectures as a result of the paper, as well as directions for further research.

    In this section, a new class of the MPDH polynomials are considered. Further, some properties of these polynomials are also obtained.

    We define the MPDH polynomials Hn(γ,ν,ζ) by means of the generating function

    n=0Hn(γ,ν,ζ)τnn!=eγτ(1+ζ)ντ2ζ. (5)

    Since (1+ζ)νt2ζeνt2 as ζ0, it is evident that (5) reduces to (1). Observe that degenerate Hermite polynomials Hn(γ,ν,ζ) and MPDH polynomials Hn(γ,ν,ζ) are completely different.

    Now, we recall that the ζ-analogue of the falling factorial sequences are as follows:

    (γ|ζ)0=1,(γ|ζ)n=γ(γζ)(γ2ζ)(γ(n1)ζ),(n1).

    Note that limζ1(γ|ζ)n=γ(γ1)(γ2)(γ(n1))=(γ)n,(n1). We remember that the classical Stirling numbers of the first kind S1(n,k) and the second kind S2(n,k) are defined by the relations (see [5,6,8,9,10,11,13,15])

    (γ)n=nk=0S1(n,k)γk and γn=nk=0S2(n,k)(γ)k,

    respectively. We also have

    n=mS2(n,m)τnn!=(eτ1)mm! and n=mS1(n,m)τnn!=(log(1+τ))mm!, (6)

    and we need the binomial theorem: For a variable v,

    (1+ζ)ντ2ζ=m=0(ντ2ζ)mζmm!=l=0(m=lS1(m,l)νlζmll!m!)τ2ll!. (7)

    Note that

    G(τ,γ,ν,ζ)=eγτ(1+ζ)ντ2ζ

    satisfies

    G(τ,γ,ν,ζ)τ(γ+2νlog(1+ζ)ζ)G(τ,γ,ν,ζ)=0.

    Substitute the series in (5) for G(τ,γ,ν,ζ) to get

    Hn+1(γ,ν,ζ)γHn(γ,ν,ζ)2νnlog(1+ζ)ζHn1(γ,ν,ζ)=0,n=1,2,. (8)

    This is the recurrence relation for MPDH polynomials. Another recurrence relation comes from

    G(τ,γ,ν,ζ)γτG(τ,γ,ν,ζ)=0.

    This implies

    Hn(γ,ν,ζ)γnHn1(γ,ν,ζ)=0,n=1,2,. (9)

    Eliminate Hn1(γ,ν,ζ) from (8) and (9) to obtain

    Hn+1(γ,ν,ζ)γHn(γ,ν,ζ)2νlog(1+ζ)ζHn(γ,ν,ζ)γ=0.

    Differentiate this equation and use (9) again to get

    nHn(γ,ν,ζ)γHn(γ,ν,ζ)γ2νlog(1+ζ)ζ2Hn(γ,ν,ζ)γ2=0,n=0,1,2,.

    Thus the MPDH polynomials Hn(γ,ν,ζ) in generating function (5) are the solution of differential equation

    2νlog(1+ζ)ζ2Hn(γ,ν,ζ)γ2+γHn(γ,ν,ζ)γnHn(γ,ν,ζ)=0,Hn(γ,0,ζ)=γn.

    As another application of the differential equation for Hn(γ,ν,ζ), we derive

    log(1+ζ)ζ2Hn(γ,ν,ζ)γ2Hn(γ,ν,ζ)ν=0,Hn(γ,0,ζ)=γn.

    The generating function (5) is useful for deriving several properties of the MPDH polynomials Hn(γ,v,ζ).

    The following basic properties of the MPDH polynomials Hn(γ,ν,ζ) are derived form (5). We, therefore, choose to omit the details involved.

    Since (8), we have

    n=0Hn(γ,ν,ζ)τnn!=eγτ(1+ζ)ντ2ζ=k=0γkτkk!m=0(ντ2ζ)mζmm!=k=0γkτkk!l=0m=lS1(m,l)νlζmll!m!τ2ll!=k=0l=0A(l,k),

    where

    A(l,k):=m=kS1(m,l)νlζmlγkτ2l+km!k!.

    Using the well-known identity

    k=0l=0A(l,k)=k=0[k2]l=0A(l,k2l),

    we obtain

    n=0Hn(γ,ν,ζ)τnn!=k=0τkk![k2]l=0(m=lS1(m,l)νlζmlγk2lk!m!(k2l)!).

    On comparing the coefficients of τnn!, we have the following theorem.

    Theorem 1. For any positive integer n, we have

    Hn(γ,ν,ζ)=[n2]k=0(m=kS1(m,k)νkζmkγn2kn!m!(n2k)!).

    By (5) and Theorem 1, we have the following corollary.

    Corollary 1. For any positive integer n, we have

    Hn(γ,ν,ζ)=[n2]k=0(log(1+ζ)ζ)kνkγn2kn!k!(n2k)!.

    Since limζ0log(1+ζ)ζ=1, we get

    Hn(γ,ν)=n![n2]k=0νkγn2kk!(n2k)!.

    The following basic properties of the MPDH polynomials Hn(γ,ν,ζ) are derived from (5). We, therefore, choose to omit the details involved.

    Theorem 2. For any positive integer n, we have

    (1)Hn(γ1+γ2,ν,ζ)=nl=0(nl)γl2Hnl(γ1,ν,ζ).(2)Hn(γ,ν1+ν2,ζ)=[n2]k=0m=kHn2k(γ,ν1,ζ)S1(m,k)νk2ζmkn!m!(n2k)!.(3)Hn(γ1+γ2,ν1+ν2,ζ)=nl=0(nl)Hl(γ1,ν1,ζ)Hnl(γ2,ν2,ζ).

    In this section, we give some new symmetric identities for the MPDH polynomials. We also get some explicit formulas and properties for the MPDH polynomials.

    Theorem 3. Let w1,w2>0 and w1w2. The following identity holds true:

    wm1Hm(w2γ,w22ν,ζ)=wm2Hm(w1γ,w21ν,ζ).

    Proof. Let w1,w2>0 and w1w2. We start with

    G(τ,ζ)=ew1w2γτ(1+ζ)w21w22ντ2ζ,

    then the expression for G(τ,μ) is symmetric in w1 and w2:

    G(τ,ζ)=n=0Hn(w1γ,w21ν,ζ)(w2τ)nn!=n=0wn2Hn(w1γ,w21ν,ζ)τnn!.

    On the similar lines we can obtain that

    G(τ,ζ)=n=0Hn(w2γ,w22ν,ζ)(w1τ)nn!=n=0wn1Hn(w2γ,w22ν,ζ)τnn!.

    Comparing the coefficients of τnn! in the last two equations, the expected result of Theorem 1 is achieved.

    For each integer k0, Sk(n)=0k+1k+2k++(n1)k is called the sums of powers of consecutive integers. A generalized falling factorial sum σk(n,μ) can be defined by the generating function (see [5,6,13])

    k=0σk(n,ζ)τkk!=(1+ζ)(n+1)τζ1(1+ζ)τζ1.

    Note that limζ0σk(n,ζ)=Sk(n). For μC, we defined the degenerate Bernoulli polynomials given by the generating function

    n=0βn(γ,ζ)τnn!=τ(1+ζ)τζ1(1+ζ)γτζ.

    When γ=0 and βn(ζ)=βn(0,ζ) are called the modified degenerate Bernoulli numbers, note that

    limζ0βn(ζ)=Bn,

    where Bn are called the Bernoulli numbers (see [6,7]).

    The first few of them are

    β0(ζ)=ζlog(1+ζ),β1(ζ)=12,β2(ζ)=log(1+ζ)6ζ,β3(ζ)=0,β4(ζ)=130(log(1+ζ)ζ)3,β5(ζ)=0,β6(ζ)=142(log(1+ζ)ζ)5.

    Again, we now use

    F(τ,μ)=w1w2τew1w2γτ(1+ζ)w21w22ντ2ζ((1+ζ)w1w2τζ1)((1+ζ)w1τζ1)((1+ζ)w2τζ1).

    From F(τ,μ), we get the following result:

    F(τ,μ)=w1w2τew1w2γτ(1+ζ)w21w22ντ2ζ((1+ζ)w1w2τζ1)((1+ζ)w1τζ1)((1+ζ)w2τζ1)=w1w2τ((1+ζ)w1τζ1)ew1w2γτ(1+ζ)w21w22ντ2ζ((1+ζ)w1w2τζ1)((1+ζ)w2τζ1)=w2n=0βn(ζ)(w1τ)nn!n=0Hn(w2γ,w22ν,ζ)(w1τ)nn!n=0σk(w11,ζ)(w2τ)nn!=n=0(ni=0im=0(ni)(im)wi1wn+1i2βm(ζ)Him(w2γ,w22ν,ζ)σni(w11,ζ))τnn!.

    In a similar fashion, we have

    F(τ,ζ)=w1w2τ((1+ζ)w2τζ1)ew1w2γτ(1+ζ)w21w22ντ2ζ((1+ζ)w1w2τζ1)((1+ζ)w1τζ1)=w1n=0βn(ζ)(w2τ)nn!n=0Hn(w1γ,w21ν,ζ)(w2τ)nn!n=0σk(w21,ζ)(w1τ)nn!=n=0(ni=0im=0(ni)(im)wi2wn+1i1βm(ζ)Him(w1γ,w21ν,ζ)σni(w21,ζ))τnn!.

    By comparing the coefficients of τnn! on the right-hand sides of the last two equations, we have the below theorem.

    Theorem 4. Let w1,w2>0 and w1w2, then the following identity holds true:

    ni=0im=0(ni)(im)wi1wn+1i2βm(μw1)Him(w2γ,w22ν,ζ)σni(w11,ζ)=ni=0im=0(ni)(im)wi2wn+1i1βm(ζ)Him(w1γ,w21ν,ζ)σni(w21,ζ).

    By taking the limit as μ0, we have the following corollary.

    Corollary 2. Let w1,w2>0 and w1w2, then the following identity holds true:

    ni=0im=0(ni)(im)wi1wn+1i2BmHim(w2γ,w22ν)Sni(w11)=ni=0im=0(ni)(im)wi2wn+1i1BmHim(w1γ,w21ν)Sni(w21).

    In this section, we construct the differential equations with coefficients ai(N,γ,ν,ζ) arising from the generating functions of the MPDH polynomials:

    (τ)NG(t,γ,ν,ζ)a0(N,γ,ν,ζ)G(τ,γ,ν,ζ)aN(N,γ,ν,ζ)τNG(τ,γ,ν,ζ)=0.

    By using the coefficients of this differential equation, we can derive explicit identities for the 2-variable MPDH polynomials Hn(γ,ν,ζ). Recall that

    G=G(τ,γ,ν,ζ)=eγτ(1+ζ)ντ2ζ=n=0Hn(γ,ν,ζ)τnn!,ζ,γ,ν,τC. (10)

    Then, by (10) we have

    G(1)=τG(τ,γ,ν,ζ)=τ(eγτ(1+ζ)ντ2ζ)=(γ+log(1+ζ)ζ2ντ)eγτ(1+ζ)ντ2ζ=(γ)G(τ,γ,ν,ζ)+(2νlog(1+ζ)ζ)τG(τ,γ,ν,ζ) (11)
    G(2)=τG(1)(τ,γ,ν,ζ)=(2νlog(1+ζ)ζ)G(τ,γ,ν,ζ)+((γ+2ντ)log(1+ζ)ζ)G(1)(τ,γ,ν,ζ)=(γ2+2νlog(1+ζ)ζ)G(τ,γ,ν,ζ)+(4γνlog(1+ζ)ζ)τG(τ,γ,ν,ζ)+(4ν2(log(1+ζ)ζ)2)τ2G(t,γ,ν,ζ). (12)

    Continuing this process as shown in (12), we can guess that

    G(N)=(τ)NG(τ,γ,ν,ζ)=Ni=0ai(N,γ,ν,ζ)τiG(τ,γ,ν,ζ),(N=0,1,2,). (13)

    Differentiating (13) with respect to τ, we have

    G(N+1)=G(N)τ=Ni=0ai(N,γ,ν,ζ)(i)τi1G(τ,γ,ν,ζ)+Ni=0ai(N,γ,ν,ζ)τiG(1)(τ,γ,ν,ζ)=Ni=0(i)ai(N,γ,ν,ζ)τi1G(τ,γ,ν,ζ)+Ni=0(γ)ai(N,γ,ν,ζ)τiG(τ,γ,ν,ζ)+Ni=0(2νlog(1+ζ)ζ)ai(N,γ,ν,ζ)τi+1G(τ,γ,ν,ζ)=N1i=0(i+1)ai+1(N,γ,ν,ζ)τiG(τ,γ,ν,ζ)+Ni=0(γ)ai(N,γ,ν,ζ)τiG(τ,γ,ν,ζ)+N+1i=1(2νlog(1+ζ)ζ)ai1(N,γ,ν,ζ)τiG(τ,γ,ν,ζ). (14)

    Now, replacing N by N+1 in (13), we find

    G(N+1)=N+1i=0ai(N+1,γ,ν,ζ)τiG(τ,γ,ν,ζ). (15)

    Comparing the coefficients on both sides of (14) and (15), we obtain

    a0(N+1,γ,ν,ζ)=a1(N,γ,ν,ζ)+(γ)a0(N,γ,ν,ζ). (16)

    For 1iN1, we obtain

    ai(N+1,γ,ν,ζ)=(i+1)ai+1(N,γ,ν,ζ)+(γ)ai(N,γ,ν,ζ)+(2νlog(1+ζ)ζ)ai1(N,γ,ν,ζ). (17)

    For i=N, we obtain

    aN(N+1,γ,ν,ζ)=(γ)aN(N,γ,ν,ζ)+(2νlog(1+ζ)ζ)aN1(N,γ,ν,ζ). (18)

    For i=N+1, we obtain

    aN+1(N+1,γ,ν,ζ)=(2νlog(1+ζ)ζ)aN(N,γ,ν,ζ). (19)

    In addition, by (13) we have

    G(τ,γ,ν,ζ)=G(0)(τ,γ,ν,ζ)=a0(0,γ,ν,ζ)G(τ,γ,ν,ζ). (20)

    By (20), we get

    a0(0,γ,ν,ζ)=1. (21)

    It is not difficult to show that

    (γ)G(τ,γ,ν,ζ)+(2νlog(1+ζ)ζ)τG(τ,γ,ν,ζ)=G(1)(τ,γ,ν,ζ)=2i=0ai(1,γ,ν,ζ)τiG(τ,γ,ν,ζ)=a0(1,γ,ν,ζ)G(τ,γ,ν,ζ)+a1(1,γ,ν,ζ)τG(τ,γ,ν,ζ). (22)

    Thus, by (10) and (22), we also get

    a0(1,γ,ν,ζ)=γ,a1(1,γ,ν,ζ)=2νlog(1+ζ)ζ. (23)

    From (16), we note that

    a0(N+1,γ,ν,ζ)=a1(N,γ,ν,ζ)+(γ)a0(N,γ,ν,ζ),a0(N,γ,ν,ζ)=a1(N1,γ,ν,ζ)+(γ)a0(N1,γ,ν,ζ),,a0(N+1,γ,ν,ζ)=N1j=0γja1(Nj,γ,ν,ζ)+γN+1. (24)

    For 1iN1, from (17) we note that

    ai(N+1,γ,ν,ζ)=(i+1)ai+1(N,γ,ν,ζ)+(γ)ai(N,γ,ν,ζ)+(2νlog(1+ζ)ζ)ai1(N,γ,ν,ζ),ai(N,γ,ν,ζ)=(i+1)ai+1(N1,γ,ν,ζ)+(γ)ai(N1,γ,ν,ζ)+(2νlog(1+ζ)ζ)ai1(N1,γ,ν,ζ),,ai(N+1,γ,ν,ζ)=(i+1)Nj=0γjai+1(Nj,γ,ν,ζ)+(2νlog(1+ζ)ζ)Nj=0γjai1(Nj,γ,ν,ζ). (25)

    From (18), we have

    aN(N+1,γ,ν,ζ)=(γ)aN(N,γ,ν,ζ)+(2νlog(1+ζ)ζ)aN1(N,γ,ν,ζ),aN1(N,γ,ν,ζ)=(γ)aN1(N1,γ,ν,ζ)+(2νlog(1+ζ)ζ)aN2(N1,γ,ν,ζ),,aN(N+1,γ,ν,ζ)=γN1j=0(2νlog(1+ζ)ζ)jaNi(Ni,γ,ν,ζ)+γ(2νlog(1+ζ)ζ)N. (26)

    Again, by (19) we have

    aN+1(N+1,γ,ν,ζ)=(2νlog(1+ζ)ζ)aN(N,γ,ν,ζ),aN(N,γ,ν,ζ)=(2νlog(1+ζ)ζ)aN1(N1,γ,ν,ζ),,aN+1(N+1,γ,ν,ζ)=(2νlog(1+ζ)ζ)N+1. (27)

    Note that here, the matrix ai(j,γ,ν,ζ)0iN+1,0jN+1 is given by

    (1γγ2+2νlog(1+ζ)ζ02νlog(1+ζ)ζ4γνlog(1+ζ)ζ00(2νlog(1+ζ)ζ)200000000000000(2νlog(1+ζ)ζ)N+1)

    Therefore, by (24)–(27), we obtain the following theorem.

    Theorem 5. For N=0,1,2,, the differential equation

    (τ)NG(τ,γ,ν,ζ)Ni=0ai(N,γ,ν,ζ)τiG(τ,γ,ν,ζ)=0

    has a solution of

    G=G(τ,γ,ν,ζ)=eγτ(1+ζ)ντ2ζ,

    where

    a0(N+1,γ,ν,ζ)=N1j=0γja1(Nj,γ,ν,ζ)+γN+1,aN(N+1,γ,ν,ζ)=γN1j=0(2νlog(1+ζ)ζ)jaNi(Ni,γ,ν,ζ)+γ(2νlog(1+ζ)ζ)N,aN+1(N+1,γ,ν,ζ)=(2νlog(1+ζ)ζ)N+1,ai(N+1,γ,ν,ζ)=(i+1)Nj=0γjai+1(Nj,γ,ν,ζ)+(2νlog(1+ζ)ζ)Nj=0γjai1(Nj,γ,ν,ζ),(1iN1).

    Here is a plot of the surface for this solution.

    In the left picture of Figure 1, we choose 3γ3,14τ14,ζ=1/3, and ν=2. In the right picture of Figure 1, we choose 1ν1,13τ13,ζ=1/3, and γ=5.

    Figure 1.  The surface for the solution G(τ,γ,ν,ζ).

    Making N-times derivative for (8) with respect to τ, we have

    (τ)NG(τ,γ,ν,ζ)=m=0Hm+N(γ,ν,ζ)tmm!. (28)

    By the Cauchy product and by multiplying the exponential series eγτ=m=0γmτmm! in both sides of (28), we get

    enlog(1+ζ)ζτ(τ)NG(τ,γ,ν,ζ)=(m=0(nlog(1+ζ)ζ)mτmm!)(m=0Hm+N(γ,ν,ζ)τmm!)=m=0(mk=0(mk)(nlog(1+ζ)ζ)mkHN+k(γ,ν,ζ))τmm!. (29)

    By the Leibniz rule and inverse relation, we have

    enlog(1+ζ)ζτ(τ)NG(τ,γ,ν,ζ)=Nk=0(Nk)(nlog(1+ζ)ζ)Nk(τ)k(enlog(1+ζ)ζtG(τ,γ,ν,ζ))=m=0(Nk=0(Nk)(nlog(1+ζ)ζ)NkHm+k(γnlog(1+ζ)ζ,ν,ζ))τmm!. (30)

    Hence, by (29) and (30) and comparing the coefficients of τmm!, we get the following theorem.

    Theorem 6. Let m,n,N be nonnegative integers, then

    mk=0(mk)(n)mk(log(1+ζ)ζ)mkHN+k(γ,ν,ζ)=Nk=0(Nk)nNk(log(1+ζ)ζ)NkHm+k(γnlog(1+ζ)ζ,ν,ζ). (31)

    If we take m=0 in (31), then we have the following corollary.

    Corollary 3. For N=0,1,2,, we have

    HN(γ,ν,ζ)=Nk=0(Nk)nNk(log(1+ζ)ζ)NkHk(γnlog(1+ζ)ζ,ν,ζ).

    By (26) and Theorem 6, we have

    a0(N,γ,ν,ζ)G(τ,γ,ν,ζ)+a1(N,γ,ν,ζ)τG(τ,γ,ν,ζ)++a2N1(N,γ,ν,ζ)τN1G(τ,γ,ν,ν)+a2N(N,γ,ν,ζ)τNG(τ,γ,ν,ζ)=m=0Hm+N(γ,ν,ζ)τmm!.

    Hence, we have the following theorem.

    Theorem 7. For N=0,1,2,, we get

    HN+m(γ,ν,μ)=mi=0Hmi(γ,ν,μ)ai(N,γ,ν,μ)m!(mi)!. (32)

    If we take m=0 in (32), then we have the below corollary.

    Corollary 4. For N=0,1,2,, we have

    HN(γ,ν,μ)=a0(N,γ,ν,μ)H0(γ,ν,μ)=a0(N,γ,ν,μ),

    where

    a0(0,γ,ν,μ)=1,a0(N+1,γ,ν,μ)=Ni=0γia1(Ni,γ,ν,μ)+γN+1.

    Hence, we have the following theorem.

    Theorem 8. For N=0,1,2, and 0m2N, we get

    [m2]k=0(N)kζkHN+m2k(γ,ν,ζ)m!(m2k)!k!=mi=0Hmi(γ,ν,ζ)ai(N,γ,ν,ζ)m!(mi)!. (33)

    If we take m=0 in (33), then we have the below corollary.

    Corollary 5. For N=0,1,2,, we have

    HN(γ,ν,ζ)=a0(N,γ,ν,ζ)H0(γ,ν,ζ)=a0(N,γ,ν,ζ),

    where

    a0(0,γ,ν,ζ)=1,a0(N+1,γ,ν,ζ)=N1j=0γja1(Nj,γ,ν,ζ)+γN+1.

    The first few of them are

    H0(γ,ν,ζ)=1,H1(γ,ν,ζ)=γ,H2(γ,ν,ζ)=γ2+2νlog(1+ζ)ζ,H3(γ,ν,ζ)=γ3+6γνlog(1+ζ)ζ,H4(γ,ν,ζ)=γ4+12γ2νlog(1+ζ)ζ+12ν2(log(1+ζ)ζ)2,H5(γ,ν,ζ)=γ5+20γ3νlog(1+ζ)ζ+60γν2(log(1+ζ)ζ)2,
    H6(γ,ν,ζ)=γ6+30γ4νlog(1+ζ)ζ+180γ2ν2(log(1+ζ)ζ)2+120ν3(log(1+ζ)ζ)3.

    This section shows the benefits of supporting theoretical prediction through numerical experiments and finding new interesting pattern of the zeros of the MPDH equations Hn(γ,ν,ζ)=0. By using a computer, the MPDH polynomials Hn(γ,ν,ζ) can be determined explicitly. We investigate the zeros of the MPDH equations Hn(γ,ν,ζ)=0. The zeros of the Hn(γ,ν,ζ)=0 for n=60,ν=5,5,5+i,5i,ζ=1/3 and γC are displayed in Figure 2.

    Figure 2.  Zeros of Hn(γ,ν,ζ)=0.

    In Figure 2(a), we choose n=50,ζ=1/3 and ν=5. In Figure 2(b), we choose n=50,ζ=1/3 and ν=5. In Figure 2(c), we choose n=50,ζ=1/3 and ν=5+i. In Figure 2(d), we choose n=50,ζ=1/3 and ν=5i.

    Stacks of zeros of the MPDH equations Hn(γ,ν,ζ)=0 for 1n50,ζ=1/3 from a 3D structure are presented in Figure 3.

    Figure 3.  Stacks of zeros of Hn(γ,ν,ζ)=0,1n60.

    In Figure 3(a), we choose ν=5. In Figure 3(b), we choose ν=5. In Figure 3(c), we choose ν=5+i. In Figure 3(d), we choose ν=5i.

    Our numerical results for the approximate solutions of real zeros of the MPDH equations Hn(γ,ν,ζ)=0 are displayed as Tables 1 and 2.

    Table 1.  Numbers of real and complex zeros of Hn(γ,ν,ζ)=0.
    ν=5,ζ=1/3 ν=5,ζ=1/3
    degree n real zeros complex zeros real zeros complex zeros
    1 1 0 1 0
    2 0 2 2 0
    3 1 2 3 0
    4 0 4 4 0
    5 1 4 5 0
    6 0 6 6 0
    7 1 6 7 0
    8 0 8 8 0
    9 1 8 9 0
    10 0 10 10 0
    11 1 10 11 0
    12 0 12 12 0

     | Show Table
    DownLoad: CSV
    Table 2.  Approximate solutions of Hn(γ,ν,ζ)=0,γR.
    degree n γ
    1 0
    2 -2.9378, 2.9378
    3 -5.0884, 0, 5.0884
    4 -6.8580, -2.1797, 2.1797, 6.8580
    5 -8.3931, -3.9825, 0, 3.9825, 8.3931
    6 -9.7659, -5.5500, -1.8117, 1.8117
    5.5500, 9.7659
    7 -11.018, -6.9530, -3.3914, 0, 3.3914
    6.9530, 11.018
    8 -12.176, -8.2330, -4.8077, -1.5837
    1.5837, 4.8077, 8.2330, 12.176

     | Show Table
    DownLoad: CSV

    We observed a remarkable regular structure of the complex roots of the MPDH equations Hn(γ,ν,ζ)=0, and also hope to verify the same kind of regular structure of the complex roots of the MPDH equations Hn(γ,ν,ζ)=0 (Table 1).

    Plot of real zeros of the MPDH equations Hn(γ,ν,ζ)=0 for 1n50,ζ=1/3 structure are presented in Figure 4.

    Figure 4.  Real zeros of Hn(γ,ν,ζ)=0 for 1n60.

    In Figure 4(a), we choose ν=5. In Figure 4(b), we choose ν=5. In Figure 4(c), we choose ν=5+i. In Figure 4(d), we choose ν=5i.

    Next, we calculated an approximate solution satisfying Hn(γ,ν,ζ)=0,γC. The results are given in Table 2. In Table 2, we choose ν=5 and ζ=1/3.

    In this article, we introduced the MPDH polynomials and got new symmetric identities for MPDH polynomials. We derived the symmetric property, one of the important properties of MPDH polynomials. We have shown several types of differential equations with Hm(γ,ν,ζ) as their solution. We also observed the symmetric properties of the roots of Hm(γ,ν,ζ)=0, which appeared differently as the values of the variables γ and ν changed. As a result, it was found that the distribution of the roots of Hm(γ,ν,ζ)=0 had a very regular pattern, and through numerical experiments we found the following conjectures are possible.

    Here, we use the notation as follows.

    (i) RHm(γ,ν,ζ): the number of real zeros of Hm(γ,ν,ζ)=0 lying on the real plane Im(γ)=0,

    (ii) CHm(γ,ν,ζ): the number of complex zeros of Hm(γ,ν,ζ)=0.

    (iii) C : the set of complex numbers.

    Since m is the degree of the polynomial Hm(γ,ν,ζ), we have RHm(γ,ν,ζ)=mCHm(γ,ν,ζ).

    We realized the regular pattern of the complex roots of the MPDH equations Hn(γ,ν,ζ)=0 related to ν and ζ. We made these conjectures. Proving or disproving the following conjectures will be our future task.

    Conjecture 1. Let m be an odd positive integer and b>0.

    RHm(γ,b,ζ)=1,CHm(γ,b,ζ)=2[m2].

    Conjecture 2. For b<0,

    RHm(γ,b,ζ)=m,CHm(γ,b,ζ)=0.

    Conjecture 3. Let m be odd positive integer and bC.

    Hm(0,b,ζ)=0.

    When we study more ν and ζ variables, it is still unsolved that the Conjectures 1 and 2 are true or false for all variables v and ζ.

    We observe that solutions of the MPDH equations Hm(γ,b,ζ)=0 have Re(γ)=0 reflection symmetry for bR. We guess that solutions of the MPDH equations Hm(γ,b,ζ)=0 do not have Re(γ)=b reflection symmetry for bCR (see Figures 24).

    Conjecture 4. The zeros of Hm(γ,a,ζ)=0, aR have Im(γ)=0 reflection symmetry analytic complex functions. The zeros of Hm(γ,a,ζ)=0,aCR do not have Im(γ)=0 reflection symmetry analytic complex functions.

    Finally, how many zeros does Hm(γ,ν,ζ)=0 have? We cannot determine whether Hm(γ,ν,ζ)=0 has m distinct solutions. We want to know the number of complex zeros CHm(γ,ν,ζ) of Hm(γ,ν,ζ)=0.

    Conjecture 5. For bC, Hm(γ,b,ζ)=0 has m distinct solutions.

    When we study more m variables, it is still unsolved whether the conjecture is true or false for all variables m (see Tables 1 and 2).

    The authors declare that they have not used Artificial Intelligence tools in the creation of this article.

    This work was supported by the Dong-A University research fund.

    The authors declare no conflict of interest.



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