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
zez−1=∞∑n=0Bnznn!=1−z2+∞∑n=1B2nz2n(2n)!,|z|<2π. | (1.1) |
Since the function xex−1−1+x2 is even in x∈R, all the Bernoulli numbers B2n+1 for n≥1 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
zexzez−1=∞∑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)=x−12,B2(x)=x2−x+16,B3(x)=x3−32x2+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)2n11−2α−2n≤|B2n|≤2(2n)!(2π)2n11−2β−2n | (1.3) |
for n≥1, where α=0 and β=2+ln(1−6/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
22n−1−122n+1−1(2n+1)(2n+2)π2<|B2n+2B2n|<22n−122n+2−1(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 n≥0 and the sequences
∏ℓk=1[2(n+1)+k]∏ℓk=1(2n+k)|B2n+2B2n|,n≥1 | (1.5) |
for fixed ℓ≥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 n≥0 can be expressed in terms of the determinant of a Hessenberg matrix as
Bn(x)=(−1)n|110⋯000x12(10)1⋯000x213(20)12(21)⋯000x314(30)13(31)⋯000⋮⋮⋮⋱⋮⋮⋮xn−31n−2(n−30)1n−3(n−31)⋯100xn−21n−1(n−20)1n−2(n−21)⋯12(n−2n−3)10xn−11n(n−10)1n−1(n−11)⋯13(n−1n−3)12(n−1n−2)1xn1n+1(n0)1n(n1)⋯14(nn−3)13(nn−2)12(nn−1)| | (1.6) |
and, consequently, the Bernoulli numbers Bn for n≥0 can be expressed as
Bn=(−1)n|110⋯000012(10)1⋯000013(20)12(21)⋯000014(30)13(31)⋯000⋮⋮⋮⋱⋮⋮⋮01n−2(n−30)1n−3(n−31)⋯10001n−1(n−20)1n−2(n−21)⋯12(n−2n−3)1001n(n−10)1n−1(n−11)⋯13(n−1n−3)12(n−1n−2)101n+1(n0)1n(n1)⋯14(nn−3)13(nn−2)12(nn−1)|. | (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 n≥1 and α>56,
(−1)n|B2−α0⋯000B4B2−α⋯000B6B4B2⋯000⋮⋮⋮⋱⋮⋮⋮B2n−4B2n−6B2n−8⋯B2−α0B2n−2B2n−4B2n−6⋯B4B2−αB2nB2n−2B2n−4⋯B6B4B2|<0 | (1.8) |
and
(−1)n|B2−B00⋯000B4B2−B0⋯000B6B4B2⋯000⋮⋮⋮⋱⋮⋮⋮B2n−4B2n−6B2n−8⋯B2−B00B2n−2B2n−4B2n−6⋯B4B2−B0B2nB2n−2B2n−4⋯B6B4B2|<0. | (1.9) |
The rising factorial (α)k is defined [12] by
(α)k=k−1∏ℓ=0(α+ℓ)={α(α+1)⋯(α+k−1),k≥1;1,k=0. |
The central factorial numbers of the second kind T(n,k) for n≥k≥0 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[2m∑k=1(−α)kk!k∑j=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=22n−122n−1−12n∑k=1k∑j=1(−1)j+1(kj)T(2n+j,j)(2n+jj),n≥1 | (1.10) |
and two identities
2n∑j=1(−1)j(4n+22j)(22j−1−1)(24n−2j+1−1)B2jB4n−2j+2=0,n≥1 | (1.11) |
and
n−1∑j=1(2n2j)(1−22j−1−22n−2j−1)B2jB2n−2j=(22n−1)B2n,n≥2. | (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,n≥0 |
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(exz−1)ez−1=∞∑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 n≥0. Accordingly, Eq (2.1) can be reformulated as
exz−1ez−1=∞∑n=0Qn+1(x)(n+1)!zn,|z|<2π. | (2.3) |
The values of Qn(x) for 1≤n≤4 are
Q1(x)=x,Q2(x)=x2−x,Q3(x)=x3−32x2+12x,Q4(x)=x4−2x3+x2. |
For α,β∈R such that α≠β, (α,β)≠(0,1), and (α,β)≠(1,0), let
Qα,β(t)={e−αt−e−βt1−e−t,t≠0;β−α,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 (β−α)(1−α−β)≥0 and (β−α)(|α−β|−α−β)≥0,
2. the function Qα,β(t) is decreasing on (0,∞) if and only if (β−α)(1−α−β)≤0 and (β−α)(|α−β|−α−β)≤0,
3. the function Qα,β(t) is increasing on (−∞,0) if and only if (β−α)(1−α−β)≥0 and (β−α)(2−|α−β|−α−β)≥0,
4. the function Qα,β(t) is decreasing on (−∞,0) if and only if (β−α)(1−α−β)≤0 and (β−α)(2−|α−β|−α−β)≤0,
5. the function Qα,β(t) is increasing on (−∞,∞) if and only if (β−α)(|α−β|−α−β)≥0 and (β−α)(2−|α−β|−α−β)≥0,
6. the function Qα,β(t) is decreasing on (−∞,∞) if and only if (β−α)(|α−β|−α−β)≤0 and (β−α)(2−|α−β|−α−β)≤0,
7. the function Qα,β(t) on (−∞,∞) is logarithmically convex if β−α>1 and logarithmically concave if 0<β−α<1,
8. if 1>β−α>0, then Qα,β(t) is 3-log-convex on (0,∞) and 3-log-concave on (−∞,0),
9. if β−α>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 exz−1ez−1 in (2.3) can be reformulated as
exz−1ez−1=e−(1−x)z−e−z1−e−z=Q1−x,1(z). |
Consequently, we deduce properties of the generating function Q1−x,1(t)=ext−1et−1 in (2.3) as follows:
1. the function Q1−x,1(t) is increasing on (0,∞) if and only if x(x−1)≥0 and x(|x|+x−2)≥0,
2. the function Q1−x,1(t) is decreasing on (0,∞) if and only if x(x−1)≤0 and x(|x|+x−2)≤0,
3. the function Q1−x,1(t) is increasing on (−∞,0) if and only if x(x−1)≥0 and x(x−|x|)≥0,
4. the function Q1−x,1(t) is decreasing on (−∞,0) if and only if x(x−1)≤0 and x(x−|x|)≤0,
5. the function Q1−x,1(t) is increasing on (−∞,∞) if and only if x(|x|+x−2)≥0 and x(x−|x|)≥0,
6. the function Q1−x,1(t) is decreasing on (−∞,∞) if and only if x(|x|+x−2)≤0 and x(x−|x|)≤0,
7. the function Q1−x,1(t) on (−∞,∞) is logarithmically convex if x>1 and logarithmically concave if 0<x<1,
8. if 0<x<1, then the function Q1−x,1(t) is 3-log-convex on (0,∞) and 3-log-concave on (−∞,0),
9. if x>1, then Q1−x,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 n≥0, 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 n≥0, the differences between the Bernoulli polynomials Bn(x) and the Bernoulli numbers Bn.
Theorem 3.1. For n≥1, we have
n∑k=0(n+2k+1)Qk+1(1x)Qn−k+1(x)xk=0. | (3.1) |
Proof. The identity (3.1) can be reformulated as
n∑k=0(n+2k+1)Qk+1(1x)Qn+2−(k+1)(x)xk+1=0,n+1∑k=1(n+2k)Qk(1x)Qn+2−k(x)xk=0, |
and
n+2∑k=0(n+2k)Qk(1x)Qn+2−k(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,n≥3, | (3.2) |
where
An(x)=n−1∑k=1Qk(1x)xkk!Qn−k(x)(n−k)!,n≥2. |
On both sides of the identity
Bn(x+h)=n∑k=0(nk)Bk(x)hn−k,n≥0, |
which is listed in [1, p. 804, Entry 23.1.7], taking x=0 yields
Bn(h)=n∑k=0(nk)Bkhn−k,n≥0. |
This implies that
Qn(x)n!=n−1∑k=0Bkk!xn−k(n−k)!,n≥0. |
Let
Pk(x)=Qk(1x)xkk!=k−1∑j=0Bjxjj!(k−j)!andRn,k(x)=Qn−k(x)(n−k)!=n−k∑j=1Bn−k−jxjj!(n−k−j)! |
for 1≤k≤n−1 and n≥3. Therefore, we obtain
An(x)=n−1∑k=1Pk(x)Rn,k(x) |
with An(0)=0. This means that An(x) is a polynomial in x of degree n−1. Hence, in order to verify the equality (3.2), it is sufficient to show
A(q)n(0)=0,0≤q≤n−1,n≥3. |
It is immediate that
Rn,k(0)=0,R(m)n,k(0)={Bn−k−m(n−k−m)!,1≤m≤n−k;0,m≥n−k+1, | (3.3) |
and
P(m)k(0)={Bm(k−m)!,0≤m≤k−1;0,m≥k. | (3.4) |
Differentiating q≥2 times the polynomial An(x), taking the limit x→0, and interchanging the order of repeated sums give
A(q)n(0)=n−1∑k=1q−1∑j=0(qj)P(j)k(0)R(q−j)n,k(0)=q−1∑j=0(qj)n−1∑k=1P(j)k(0)R(q−j)n,k(0)={0,n−q<1q−1∑j=0(qj)j+n−q∑k=j+1Bj(k−j)!Bn−k−(q−j)[n−k−(q−j)]!,n−q≥1={0,n−q<1[q−1∑j=0(qj)Bj][n−q∑ℓ=1Bn−q−ℓℓ!(n−q−ℓ)!],n−q≥1=0, |
where we used the derivatives in (3.3) and (3.4) and utilized the identity
n−1∑k=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
A′n(0)=n−2∑j=0Bjj!(n−1−j)!=1(n−1)!n−2∑j=0(n−1j)Bj=0,n≥3. |
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 n≥k≥0 are defined in [45, Definition 11.2] and [46, p. 134, Theorem A] by
Bn,k(x1,x2,…,xn−k+1)=∑ℓi≥0for1≤i≤n−k+1,∑n−k+1i=1iℓi=n,∑n−k+1i=1ℓi=kn!∏n−k+1i=1ℓi!n−k+1∏i=1(xii!)ℓi. |
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 n≥1, we have
Qn(1x)=(−1)n−1n!x2n−1|Q2(x)2!Q1(x)1!0⋯000Q3(x)3!Q2(x)2!Q1(x)1!⋯000Q4(x)4!Q3(x)3!Q2(x)2!⋯000⋮⋮⋮⋱⋮⋮⋮Qn−2(x)(n−2)!Qn−3(x)(n−3)!Qn−4(x)(n−4)!⋯Q2(x)2!Q1(x)1!0Qn−1(x)(n−1)!Qn−2(x)(n−2)!Qn−3(x)(n−3)!⋯Q3(x)3!Q2(x)2!Q1(x)1!Qn(x)n!Qn−1(x)(n−1)!Qn−2(x)(n−2)!⋯Q4(x)4!Q3(x)3!Q2(x)2!|, | (4.1) |
where the determinant of order 0 is regarded as 1 by convention.
For n≥0, we have
Qn+1(1x)=n+1xn+1n∑k=0(−1)kk!xkBn,k(Q2(x)2,Q3(x)3,…,Qn−k+2(x)n−k+2). | (4.2) |
Proof. The Wronski formula reads that, if a0≠0 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|a1a000⋯000a2a1a00⋯000a3a2a1a0⋯000a4a3a2a1⋯000⋮⋮⋮⋮⋱⋮⋮⋮an−3an−2an−3an−4⋯a000an−2an−3an−4an−5⋯a1a00an−1an−2an−3an−4⋯a2a1a0anan−1an−2an−3⋯a3a2a1|,n≥0. | (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=0akbn−k=0 for n≥1. 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 n≥0. It is easy to verify that a0b0=1. The identity (3.1) in Theorem 3.1 is equivalent to the equality ∑nk=0akbn−k=0 for n≥1. 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!n∑k=0(−1)kk!Bn,k(1!a1,2!a2,…,(n−k+1)!an−k+1). | (4.7) |
See the papers [47,48,54,59]. When β=1 in (4.6), it follows that a0=b0=1 and ∑nk=0akbn−k=0 for n≥1. 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 n≥1, the difference Qn(x) can be computed by
Qn(x)=(−1)n−1nx|110⋯00x212(10)1⋯00x2313(20)12(21)⋯00x3414(30)13(31)⋯00⋮⋮⋮⋱⋮⋮xn−3n−21n−2(n−30)1n−3(n−31)⋯10xn−2n−11n−1(n−20)1n−2(n−21)⋯12(n−2n−3)1xn−1n1n(n−10)1n−1(n−11)⋯13(n−1n−3)12(n−1n−2)|. | (5.1) |
Proof. The power series expansion (2.3) implies that
Qn+1(x)n+1=limz→0dndzn(exz−1ez−1)=limz→0dnQ1−x,1(z)dzn,n≥0. |
The generating function Q1−x,1(z) can be rewritten as
Q1−x,1(z)=x(exz−1)/(xz)(ez−1)/z=x∫e1sxz−1ds∫e1sz−1ds |
with
limz→0dkdzk∫e1sz−1ds=limz→0∫e1sz−1lnksds=∫e1lnkssds=1k+1 |
and
limz→0dkdzk∫e1sxz−1ds=xklimz→0∫e1sxz−1lnksds=xk∫e1lnkssds=xkk+1 |
for k≥0.
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(k−1)(z) for 1≤k≤n+1, let V(n+1)×n(z) be an (n+1)×n matrix whose elements are
vij(z)={(i−1j−1)v(i−j)(z),i−j≥00,i−j<0 |
for 1≤i≤n+1 and 1≤j≤n, 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)=∫e1sxz−1dsandv(z)=∫e1sz−1ds |
yields
limz→0dnQ1−x,1(z)dzn=xlimz→0dndzn(∫e1sxz−1ds∫e1sz−1ds)=xlimz→0dndzn[u(z)v(z)]=xlimz→0(−1)nvn+1(z)|u(z)v(z)0⋯0u′(z)v′(z)v(z)⋯0u″(z)v″(z)(21)v′(z)⋯0⋮⋮⋮⋱⋮u(n−1)(z)v(n−1)(z)(n−11)v(n−2)(z)⋯v(z)u(n)(z)v(n)(z)(n1)v(n−1)(z)⋯(nn−1)v′(z)|=(−1)nxvn+1(0)|u(0)v(0)0⋯00u′(0)v′(0)v(0)⋯00u″(0)v″(0)(21)v′(0)⋯00⋮⋮⋮⋱⋮⋮u(n−1)(0)v(n−1)(0)(n−11)v(n−2)(0)⋯(n−1n−2)v′(0)v(0)u(n)(0)v(n)(0)(n1)v(n−1)(0)⋯(nn−2)v″(0)(nn−1)v′(0)|=(−1)nx|1100⋯00x212(10)10⋯00x2313(20)12(21)1⋯00x3414(30)13(31)12(32)⋯00⋮⋮⋮⋮⋱⋮⋮xn−2n−11n−1(n−20)1n−2(n−21)(n−22)1n−3⋯10xn−1n1n(n−10)1n−1(n−11)1n−2(n−12)⋯12(n−1n−2)1xnn+11n+1(n0)1n(n1)1n−1(n2)⋯13(nn−2)12(nn−1)|. |
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 n≥1, the determinantal formula (5.1) can be reformulated as
Qn(x)x=(−1)n−1n|110⋯00x212(10)1⋯00x2313(20)12(21)⋯00x3414(30)13(31)⋯00⋮⋮⋮⋱⋮⋮xn−3n−21n−2(n−30)1n−3(n−31)⋯10xn−2n−11n−1(n−20)1n−2(n−21)⋯12(n−2n−3)1xn−1n1n(n−10)1n−1(n−11)⋯13(n−1n−3)12(n−1n−2)|. |
Since
limx→0Qn(x)x=limx→0Bn(x)−Bnx=limx→0B′n(x)=nlimx→0Bn−1(x)=nBn−1, |
taking the limit x→0 on both sides of the above determinantal formula gives
nBn−1=(−1)n−1n|110⋯00012(10)1⋯00013(20)12(21)⋯00014(30)13(31)⋯00⋮⋮⋮⋱⋮⋮01n−2(n−30)1n−3(n−31)⋯1001n−1(n−20)1n−2(n−21)⋯12(n−2n−3)101n(n−10)1n−1(n−11)⋯13(n−1n−3)12(n−1n−2)| |
for n≥1. Consequently, we recover the determinantal formula (1.7).
Remark 5.3. The determinantal formula (5.1) can be rearranged as
Qn(x)=(−1)n−1n|x10⋯00x2212(10)1⋯00x3313(20)12(21)⋯00x4414(30)13(31)⋯00⋮⋮⋮⋱⋮⋮xn−2n−21n−2(n−30)1n−3(n−31)⋯10xn−1n−11n−1(n−20)1n−2(n−21)⋯12(n−2n−3)1xnn1n(n−10)1n−1(n−11)⋯13(n−1n−3)12(n−1n−2)|. | (5.3) |
Differentiating with respect to x on both sides of (5.3) and making use of the relation B′n(x)=nBn−1(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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