Research article

Three identities and a determinantal formula for differences between Bernoulli polynomials and numbers

  • Received: 14 February 2023 Revised: 11 August 2023 Accepted: 30 August 2023 Published: 21 December 2023
  • In the paper, the authors simply review recent results of inequalities, monotonicity, signs of determinants, determinantal formulas, closed-form expressions, and identities of the Bernoulli numbers and polynomials, establish an identity involving the differences between the Bernoulli polynomials and the Bernoulli numbers, present two identities among the differences between the Bernoulli polynomials and the Bernoulli numbers in terms of a determinant and a partial Bell polynomial, and derive a determinantal formula of the differences between the Bernoulli polynomials and the Bernoulli numbers.

    Citation: Jian Cao, José Luis López-Bonilla, Feng Qi. Three identities and a determinantal formula for differences between Bernoulli polynomials and numbers[J]. Electronic Research Archive, 2024, 32(1): 224-240. doi: 10.3934/era.2024011

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  • In the paper, the authors simply review recent results of inequalities, monotonicity, signs of determinants, determinantal formulas, closed-form expressions, and identities of the Bernoulli numbers and polynomials, establish an identity involving the differences between the Bernoulli polynomials and the Bernoulli numbers, present two identities among the differences between the Bernoulli polynomials and the Bernoulli numbers in terms of a determinant and a partial Bell polynomial, and derive a determinantal formula of the differences between the Bernoulli polynomials and the Bernoulli numbers.



    Recall from [1, p. 804, Entry 23.1.1] that the Bernoulli numbers Bn can be generated by

    zez1=n=0Bnznn!=1z2+n=1B2nz2n(2n)!,|z|<2π. (1.1)

    Since the function xex11+x2 is even in xR, all the Bernoulli numbers B2n+1 for n1 are equal to 0. The first six non-zero Bernoulli numbers Bn are

    B0=1,B1=12,B2=16,B4=130,B6=142,B8=130.

    Recall from [2, Chapter 1] that the Bernoulli polynomials Bn(x) can be generated by

    zexzez1=n=0Bn(x)znn!,|z|<2π. (1.2)

    It is clear that Bn(0)=Bn. The first four Bernoulli polynomials Bn(x) are

    B0(x)=1,B1(x)=x12,B2(x)=x2x+16,B3(x)=x332x2+12x.

    The Bernoulli numbers Bn and the Bernoulli polynomials Bn(x) are classical and fundamental notions in both mathematical sciences and engineering sciences.

    We now give a simple review of recent developments of the Bernoulli numbers Bn and the Bernoulli polynomials Bn(x), including inequalities, monotonicity, determinantal expressions, signs of determinants, and identities related to the Bernoulli numbers Bn and the Bernoulli polynomials Bn(x).

    In [3], Alzer bounded the Bernoulli numbers Bn by the double inequality

    2(2n)!(2π)2n112α2n|B2n|2(2n)!(2π)2n112β2n (1.3)

    for n1, where α=0 and β=2+ln(16/π2)ln2=0.6491 are the best possible in the sense that they can not be replaced by any bigger and smaller constants respectively in the double inequality (1.3).

    In [4,5], Qi bounded the ratio B2n+2B2n by

    22n1122n+11(2n+1)(2n+2)π2<|B2n+2B2n|<22n122n+21(2n+1)(2n+2)π2. (1.4)

    The double inequality (1.4) was generalized and refined in [6,7]. This double inequality has had a number of non-self citations in over forty-eight articles or preprints published by other mathematicians.

    In [8], Y. Shuang et al. proved that the sequence |B2n+2B2n| for n0 and the sequences

    k=1[2(n+1)+k]k=1(2n+k)|B2n+2B2n|,n1 (1.5)

    for fixed 1 are increasing in n.

    In the papers [9,10], many determinantal expressions of the Bernoulli numbers Bn and the Bernoulli polynomials Bn(x) are reviewed and discovered. For example, the Bernoulli polynomials Bn(x) for n0 can be expressed in terms of the determinant of a Hessenberg matrix as

    Bn(x)=(1)n|110000x12(10)1000x213(20)12(21)000x314(30)13(31)000xn31n2(n30)1n3(n31)100xn21n1(n20)1n2(n21)12(n2n3)10xn11n(n10)1n1(n11)13(n1n3)12(n1n2)1xn1n+1(n0)1n(n1)14(nn3)13(nn2)12(nn1)| (1.6)

    and, consequently, the Bernoulli numbers Bn for n0 can be expressed as

    Bn=(1)n|110000012(10)1000013(20)12(21)000014(30)13(31)00001n2(n30)1n3(n31)10001n1(n20)1n2(n21)12(n2n3)1001n(n10)1n1(n11)13(n1n3)12(n1n2)101n+1(n0)1n(n1)14(nn3)13(nn2)12(nn1)|. (1.7)

    In [11], basing on the increasing property of the sequences in (1.5), among other things, Qi determined signs of certain Toeplitz–Hessenberg determinants whose elements involve the Bernoulli numbers B2n. For example, for n1 and α>56,

    (1)n|B2α0000B4B2α000B6B4B2000B2n4B2n6B2n8B2α0B2n2B2n4B2n6B4B2αB2nB2n2B2n4B6B4B2|<0 (1.8)

    and

    (1)n|B2B00000B4B2B0000B6B4B2000B2n4B2n6B2n8B2B00B2n2B2n4B2n6B4B2B0B2nB2n2B2n4B6B4B2|<0. (1.9)

    The rising factorial (α)k is defined [12] by

    (α)k=k1=0(α+)={α(α+1)(α+k1),k1;1,k=0.

    The central factorial numbers of the second kind T(n,k) for nk0 can be generated [13,14] by

    1k!(2sinhx2)k=n=kT(n,k)xnn!.

    In [15], considering the power series expansion

    (sinxx)α=1+m=1(1)m[2mk=1(α)kk!kj=1(1)j(kj)T(2m+j,j)(2m+jj)](2x)2m(2m)!

    for α<0, which was established in [16, Theorem 4.1], X.-Y. Chen et al. derived the closed-form expression

    B2n=22n122n112nk=1kj=1(1)j+1(kj)T(2n+j,j)(2n+jj),n1 (1.10)

    and two identities

    2nj=1(1)j(4n+22j)(22j11)(24n2j+11)B2jB4n2j+2=0,n1 (1.11)

    and

    n1j=1(2n2j)(122j122n2j1)B2jB2n2j=(22n1)B2n,n2. (1.12)

    There have been a simple review about closed-form formulas for the Bernoulli numbers and polynomials at the web sites https://math.stackexchange.com/a/4256911 (accessed on 5 February 2023), https://math.stackexchange.com/a/4256914 (accessed on 5 February 2023), and https://math.stackexchange.com/a/4656534 (accessed on 11 March 2023). For more recent developments of the Bernoulli numbers Bn and the Bernoulli polynomials Bn(x), please refer to the monograph [17], to the papers [18,19,20,21,22,23,24], and to the articles [25,26,27,28,29,30,31,32,33].

    Let

    Qn(x)=Bn(x)Bn,n0

    denote the differences between the Bernoulli polynomials Bn(x) and the Bernoulli numbers Bn. Subtracting (1.1) from (1.2) on both sides yields

    z(exz1)ez1=n=0[Bn(x)Bn]znn!=n=0Qn(x)znn!,|z|<2π. (2.1)

    It is easy to see that

    Q0(x)=B0(x)B0=0 (2.2)

    and Qn(0)=0 for n0. Accordingly, Eq (2.1) can be reformulated as

    exz1ez1=n=0Qn+1(x)(n+1)!zn,|z|<2π. (2.3)

    The values of Qn(x) for 1n4 are

    Q1(x)=x,Q2(x)=x2x,Q3(x)=x332x2+12x,Q4(x)=x42x3+x2.

    For α,βR such that αβ, (α,β)(0,1), and (α,β)(1,0), let

    Qα,β(t)={eαteβt1et,t0;βα,t=0.

    In the papers [34,35,36], the monotonicity and logarithmic convexity of Qα,β(t) were discussed and the following conclusions were acquired:

    1. the function Qα,β(t) is increasing on (0,) if and only if (βα)(1αβ)0 and (βα)(|αβ|αβ)0,

    2. the function Qα,β(t) is decreasing on (0,) if and only if (βα)(1αβ)0 and (βα)(|αβ|αβ)0,

    3. the function Qα,β(t) is increasing on (,0) if and only if (βα)(1αβ)0 and (βα)(2|αβ|αβ)0,

    4. the function Qα,β(t) is decreasing on (,0) if and only if (βα)(1αβ)0 and (βα)(2|αβ|αβ)0,

    5. the function Qα,β(t) is increasing on (,) if and only if (βα)(|αβ|αβ)0 and (βα)(2|αβ|αβ)0,

    6. the function Qα,β(t) is decreasing on (,) if and only if (βα)(|αβ|αβ)0 and (βα)(2|αβ|αβ)0,

    7. the function Qα,β(t) on (,) is logarithmically convex if βα>1 and logarithmically concave if 0<βα<1,

    8. if 1>βα>0, then Qα,β(t) is 3-log-convex on (0,) and 3-log-concave on (,0),

    9. if βα>1, then Qα,β(t) is 3-log-concave on (0,) and 3-log-convex on (,0).

    The monotonicity of Qα,β(t) on (0,) was used in [34,37,38] to present necessary and sufficient conditions for some functions involving ratios of the gamma and q-gamma functions to be logarithmically completely monotonic. The logarithmic convexity of Qα,β(t) on (0,) was employed in [36,39] to provide alternative proofs for Elezović-Giordano-Pečarić's theorem. For more detailed information, please refer to [40,41] and related references therein. The above texts are extracted and modified from [42, pp. 486–487].

    The generating function exz1ez1 in (2.3) can be reformulated as

    exz1ez1=e(1x)zez1ez=Q1x,1(z).

    Consequently, we deduce properties of the generating function Q1x,1(t)=ext1et1 in (2.3) as follows:

    1. the function Q1x,1(t) is increasing on (0,) if and only if x(x1)0 and x(|x|+x2)0,

    2. the function Q1x,1(t) is decreasing on (0,) if and only if x(x1)0 and x(|x|+x2)0,

    3. the function Q1x,1(t) is increasing on (,0) if and only if x(x1)0 and x(x|x|)0,

    4. the function Q1x,1(t) is decreasing on (,0) if and only if x(x1)0 and x(x|x|)0,

    5. the function Q1x,1(t) is increasing on (,) if and only if x(|x|+x2)0 and x(x|x|)0,

    6. the function Q1x,1(t) is decreasing on (,) if and only if x(|x|+x2)0 and x(x|x|)0,

    7. the function Q1x,1(t) on (,) is logarithmically convex if x>1 and logarithmically concave if 0<x<1,

    8. if 0<x<1, then the function Q1x,1(t) is 3-log-convex on (0,) and 3-log-concave on (,0),

    9. if x>1, then Q1x,1(t) is 3-log-concave on (0,) and 3-log-convex on (,0).

    What properties do the polynomials Qn(x)=Bn(x)Bn for n0, the differences between the Bernoulli polynomials Bn(x) and the Bernoulli numbers Bn, possess?

    In this section, we establish an identity involving the polynomials Qn(x)=Bn(x)Bn for n0, the differences between the Bernoulli polynomials Bn(x) and the Bernoulli numbers Bn.

    Theorem 3.1. For n1, we have

    nk=0(n+2k+1)Qk+1(1x)Qnk+1(x)xk=0. (3.1)

    Proof. The identity (3.1) can be reformulated as

    nk=0(n+2k+1)Qk+1(1x)Qn+2(k+1)(x)xk+1=0,n+1k=1(n+2k)Qk(1x)Qn+2k(x)xk=0,

    and

    n+2k=0(n+2k)Qk(1x)Qn+2k(x)xk=Q0(1x)Qn+2(x)+Qn+2(1x)Q0(x)xn+2=0,

    where we used the identity (2.2). The last equation means that the identity (3.1) is equivalent to

    An(x)=0,n3, (3.2)

    where

    An(x)=n1k=1Qk(1x)xkk!Qnk(x)(nk)!,n2.

    On both sides of the identity

    Bn(x+h)=nk=0(nk)Bk(x)hnk,n0,

    which is listed in [1, p. 804, Entry 23.1.7], taking x=0 yields

    Bn(h)=nk=0(nk)Bkhnk,n0.

    This implies that

    Qn(x)n!=n1k=0Bkk!xnk(nk)!,n0.

    Let

    Pk(x)=Qk(1x)xkk!=k1j=0Bjxjj!(kj)!andRn,k(x)=Qnk(x)(nk)!=nkj=1Bnkjxjj!(nkj)!

    for 1kn1 and n3. Therefore, we obtain

    An(x)=n1k=1Pk(x)Rn,k(x)

    with An(0)=0. This means that An(x) is a polynomial in x of degree n1. Hence, in order to verify the equality (3.2), it is sufficient to show

    A(q)n(0)=0,0qn1,n3.

    It is immediate that

    Rn,k(0)=0,R(m)n,k(0)={Bnkm(nkm)!,1mnk;0,mnk+1, (3.3)

    and

    P(m)k(0)={Bm(km)!,0mk1;0,mk. (3.4)

    Differentiating q2 times the polynomial An(x), taking the limit x0, and interchanging the order of repeated sums give

    A(q)n(0)=n1k=1q1j=0(qj)P(j)k(0)R(qj)n,k(0)=q1j=0(qj)n1k=1P(j)k(0)R(qj)n,k(0)={0,nq<1q1j=0(qj)j+nqk=j+1Bj(kj)!Bnk(qj)[nk(qj)]!,nq1={0,nq<1[q1j=0(qj)Bj][nq=1Bnq!(nq)!],nq1=0,

    where we used the derivatives in (3.3) and (3.4) and utilized the identity

    n1k=0(nk)Bk=0,n=2,3,, (3.5)

    which is collected in [43, p. 591, Entry 24.5.3] and [44, p. 206, (15.14)].

    Moreover, by the identity (3.5) again, it is easy to see that

    An(0)=n2j=0Bjj!(n1j)!=1(n1)!n2j=0(n1j)Bj=0,n3.

    The proof of the identity (3.1) is thus complete.

    In this section, we demonstrate two identities among Qn(x) and Qn(1x).

    The partial Bell polynomials Bn,k for nk0 are defined in [45, Definition 11.2] and [46, p. 134, Theorem A] by

    Bn,k(x1,x2,,xnk+1)=i0for1ink+1,nk+1i=1ii=n,nk+1i=1i=kn!nk+1i=1i!nk+1i=1(xii!)i.

    This kind of polynomials Bn,k are important in analytic combinatorics, analytic number theory, analysis, and other areas in mathematical sciences. In recent years, some novel conclusions and applications of partial Bell polynomials Bn,k have been discovered, carried out, reviewed, and surveyed in the papers [12,16,47,48,49,50,51,52,53,54], for example.

    Theorem 4.1. For n1, we have

    Qn(1x)=(1)n1n!x2n1|Q2(x)2!Q1(x)1!0000Q3(x)3!Q2(x)2!Q1(x)1!000Q4(x)4!Q3(x)3!Q2(x)2!000Qn2(x)(n2)!Qn3(x)(n3)!Qn4(x)(n4)!Q2(x)2!Q1(x)1!0Qn1(x)(n1)!Qn2(x)(n2)!Qn3(x)(n3)!Q3(x)3!Q2(x)2!Q1(x)1!Qn(x)n!Qn1(x)(n1)!Qn2(x)(n2)!Q4(x)4!Q3(x)3!Q2(x)2!|, (4.1)

    where the determinant of order 0 is regarded as 1 by convention.

    For n0, we have

    Qn+1(1x)=n+1xn+1nk=0(1)kk!xkBn,k(Q2(x)2,Q3(x)3,,Qnk+2(x)nk+2). (4.2)

    Proof. The Wronski formula reads that, if a00 and

    P(x)=a0+a1x+a2x2+ (4.3)

    is a formal series, then the coefficients of the reciprocal series

    1P(x)=b0+b1x+b2x2+ (4.4)

    are given by

    bn=(1)nan+10|a1a000000a2a1a00000a3a2a1a0000a4a3a2a1000an3an2an3an4a000an2an3an4an5a1a00an1an2an3an4a2a1a0anan1an2an3a3a2a1|,n0. (4.5)

    This can be found in [55, p. 17, Theorem 1.3], [11, Lemma 2.1 and Section 5], and [9, Lemma 2.4]. It is easy to see that the equalities (4.3) and (4.4) are equivalent to the identities a0b0=1 and nk=0akbnk=0 for n1. See [47,56,57,58].

    Let β be a fixed real number and let

    an=xn+β(n+1)!Qn+1(1x)andbn=1(n+1)!xβQn+1(x) (4.6)

    for n0. It is easy to verify that a0b0=1. The identity (3.1) in Theorem 3.1 is equivalent to the equality nk=0akbnk=0 for n1. Therefore, the sequences an and bn defined in (4.6) satisfy the relation (4.5). Interchanging the roles of an and bn and simplifying yield (4.1).

    On the other hand, if the sequences an and bn satisfy a0=b0=1 and meet the equalities (4.3) and (4.4), then

    bn=1n!nk=0(1)kk!Bn,k(1!a1,2!a2,,(nk+1)!ank+1). (4.7)

    See the papers [47,48,54,59]. When β=1 in (4.6), it follows that a0=b0=1 and nk=0akbnk=0 for n1. Interchanging the roles of an and bn in (4.7) and applying the sequences an and bn in (4.6) result in (4.2). The proof of Theorem 4.1 is complete.

    In this section, we derive a determinantal formula of the difference Qn(x) as follows.

    Theorem 5.1. For n1, the difference Qn(x) can be computed by

    Qn(x)=(1)n1nx|11000x212(10)100x2313(20)12(21)00x3414(30)13(31)00xn3n21n2(n30)1n3(n31)10xn2n11n1(n20)1n2(n21)12(n2n3)1xn1n1n(n10)1n1(n11)13(n1n3)12(n1n2)|. (5.1)

    Proof. The power series expansion (2.3) implies that

    Qn+1(x)n+1=limz0dndzn(exz1ez1)=limz0dnQ1x,1(z)dzn,n0.

    The generating function Q1x,1(z) can be rewritten as

    Q1x,1(z)=x(exz1)/(xz)(ez1)/z=xe1sxz1dse1sz1ds

    with

    limz0dkdzke1sz1ds=limz0e1sz1lnksds=e1lnkssds=1k+1

    and

    limz0dkdzke1sxz1ds=xklimz0e1sxz1lnksds=xke1lnkssds=xkk+1

    for k0.

    Let u(z) and v(z)0 be two differentiable functions, let U(n+1)×1(z) be an (n+1)×1 matrix whose elements are uk,1(z)=u(k1)(z) for 1kn+1, let V(n+1)×n(z) be an (n+1)×n matrix whose elements are

    vij(z)={(i1j1)v(ij)(z),ij00,ij<0

    for 1in+1 and 1jn, and let |W(n+1)×(n+1)(z)| denote the determinant of the (n+1)×(n+1) matrix

    W(n+1)×(n+1)(z)=(U(n+1)×1(z)V(n+1)×n(z)).

    Then the nth derivative of the ratio u(z)v(z) can be computed [60, p. 40, Exercise 5] by

    dndtn[u(z)v(z)]=(1)n|W(n+1)×(n+1)(z)|vn+1(z). (5.2)

    See also [61, Lemma 1], [11, Section 2], [9, p. 94, The first proof of Theorem 1.2], and [10, Lemma 1].

    Applying the formula (5.2) to the functions

    u(z)=e1sxz1dsandv(z)=e1sz1ds

    yields

    limz0dnQ1x,1(z)dzn=xlimz0dndzn(e1sxz1dse1sz1ds)=xlimz0dndzn[u(z)v(z)]=xlimz0(1)nvn+1(z)|u(z)v(z)00u(z)v(z)v(z)0u(z)v(z)(21)v(z)0u(n1)(z)v(n1)(z)(n11)v(n2)(z)v(z)u(n)(z)v(n)(z)(n1)v(n1)(z)(nn1)v(z)|=(1)nxvn+1(0)|u(0)v(0)000u(0)v(0)v(0)00u(0)v(0)(21)v(0)00u(n1)(0)v(n1)(0)(n11)v(n2)(0)(n1n2)v(0)v(0)u(n)(0)v(n)(0)(n1)v(n1)(0)(nn2)v(0)(nn1)v(0)|=(1)nx|110000x212(10)1000x2313(20)12(21)100x3414(30)13(31)12(32)00xn2n11n1(n20)1n2(n21)(n22)1n310xn1n1n(n10)1n1(n11)1n2(n12)12(n1n2)1xnn+11n+1(n0)1n(n1)1n1(n2)13(nn2)12(nn1)|.

    The determinantal formula (5.1) is thus proved.

    Remark 5.1. The formula (4.5) can also be proved by the formula (5.2). For details, please refer to [11, Section 5].

    Remark 5.2. For n1, the determinantal formula (5.1) can be reformulated as

    Qn(x)x=(1)n1n|11000x212(10)100x2313(20)12(21)00x3414(30)13(31)00xn3n21n2(n30)1n3(n31)10xn2n11n1(n20)1n2(n21)12(n2n3)1xn1n1n(n10)1n1(n11)13(n1n3)12(n1n2)|.

    Since

    limx0Qn(x)x=limx0Bn(x)Bnx=limx0Bn(x)=nlimx0Bn1(x)=nBn1,

    taking the limit x0 on both sides of the above determinantal formula gives

    nBn1=(1)n1n|11000012(10)100013(20)12(21)00014(30)13(31)0001n2(n30)1n3(n31)1001n1(n20)1n2(n21)12(n2n3)101n(n10)1n1(n11)13(n1n3)12(n1n2)|

    for n1. Consequently, we recover the determinantal formula (1.7).

    Remark 5.3. The determinantal formula (5.1) can be rearranged as

    Qn(x)=(1)n1n|x1000x2212(10)100x3313(20)12(21)00x4414(30)13(31)00xn2n21n2(n30)1n3(n31)10xn1n11n1(n20)1n2(n21)12(n2n3)1xnn1n(n10)1n1(n11)13(n1n3)12(n1n2)|. (5.3)

    Differentiating with respect to x on both sides of (5.3) and making use of the relation Bn(x)=nBn1(x), we recover the determinantal formula (1.6).

    Remark 5.4. In theory, the determinantal formula (5.1) in Theorem 5.1 can be obtained by algebraically subtracting the determinant (1.7) from the determinant (1.6).

    In this paper, about the Bernoulli numbers Bn and the Bernoulli polynomials Bn(x), we simply reviewed the inequalities (1.3) and (1.4), the increasing property of the sequence in (1.5), the determinantal formulas (1.6) and (1.7), the negativity of two determinants in (1.8) and (1.9), the closed-form formula (1.10), and the identities (1.11) and (1.12), established the identity (3.1) in Theorem 3.1 in which the differences Qn(x) between the Bernoulli polynomials Bn(x) and the Bernoulli numbers Bn are involved, presented two identities (4.1) and (4.2) among the differences Qn(x) in terms of a beautiful Hessenberg determinant and the partial Bell polynomials Bn,k in Theorem 4.1, and derived a determinantal formula (5.1) for the difference Qn(x) in Theorem 5.1.

    To the best of our authors' knowledge, the difference Qn(x) has been investigated in this paper for the first time in the mathematical community.

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

    The authors are grateful to three anonymous referees for their careful reading, valuable comments, and helpful suggestions to the original version of this paper.

    The authors declare there is no conflicts of interest.



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