# Some Formulae for the Product of Two Bernoulli and Euler Polynomials

Authors: D. S. Kim; D. V. Dolgy; T. Kim; S.-H. Rim
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate some formulae for the product of two Bernoulli and Euler polynomials arising from the Euler and Bernoulli basis polynomials.


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## Body

## 1. Introduction

As is well known, the Bernoulli polynomials are defined by the generating function as follows:(1.1)tet-1ext=eB(x)t=∑n=0∞Bn(x)tnn!
(see [1–21]), with the usual convention about replacing Bn(x) by Bn(x). In the special case, x=0, Bn(0)=Bn are called the nth Bernoulli numbers. The Euler polynomials are also defined by the generating function as follows:(1.2)2et+1ext=eE(x)t=∑n=0∞En(x)tnn!
(see [6–11]), with the usual convention about replacing En(x) by En(x). In the special case, x=0, En(0)=En are called the nth Euler numbers. From (1.1) and (1.2), we can derive the following recurrence relations for the Bernoulli and Euler numbers:(1.3)B0=1,Bn(1)-Bn=δ1,n,E0=1,En(1)+En=2δ0,n,
where δk,n is the Kronecker symbol. By (1.1) and (1.2), we get(1.4)Bn(x)=∑l=0nxn-l(nl)Bl,En(x)=∑l=0nxn-l(nl)El.
From (1.4), we can derive(1.5)ddxBn(x)=nBn-1(x),ddxEn(x)=nEn-1(x).
By (1.4) and (1.5), we get(1.6)∫01Bn(x)dx=δ0,nn+1=δ0,n,∫01En(x)dx=-2En+1n+1.
It is easy to show that(1.7)etx=1t(te(x+1)tet-1-textet-1)=1t∑n=0∞(Bn(x+1)-Bn(x))tnn!.
Thus, we have(1.8)xn=1n+1∑l=0n(n+1l)Bl(x)
(see [11–18]). By the definition of the Euler polynomials, we get(1.9)etx=12(2e(x+1)tet+1+2extet+1)=12∑n=0∞(En(x+1)+En(x))tnn!.
From (1.9), we have(1.10)xn=En(x)+12∑l=0n-1(nl)El(x)
(see [1–18]). By (1.8) and (1.10), we see that the set {E0(x),…,En(x)} and {B0(x),…,Bn(x)} are the basis for the space of polynomials of degree less than or equal to n with coefficients in ℚ (see [1–21]).Fromm,n∈ℤ+=ℕ⋃{0}, let Im,n=∫01Bm(x)xndx. Then, we note that(1.11)I0,n=∫01xndx=1n+1,Im,0=∫01Bm(x)dx=δ0,m.
Let us assume that m,n≥1. Then, we have(1.12)Im,n=Bm+1m+1-nm+1Im+1,n-1=Bm+1m+1+(-1)n(m+1)(m+2)Bm+2+(-1)2n(n-1)(m+1)(m+2)Im+2,n-2.Continuing this process, we get(1.13)Im,n=∑j=1n-1(-1)j-1n-j+1(∏l=1jn-l+1m+l)Bm+j+(-1)n-1n(n-1)⋯2(m+1)⋯(m+n-1)∫01Bm+n-1(x)xdx=∑j=1n(-1)j-1(n+1j)(n+1)(m+jm)Bm+j.
Let Jm,n=∫01Em(x)xndx for m,n∈ℤ+. Then, we have(1.14)J0,n=∫01xndx=1n+1,Jm,0=∫01Em(x)dx=-2m+1Em+1.
Assume that m,n≥1. Then, we get(1.15)Jm,n=∫01Em(x)xndx=-Em+1m+1-nm+1∫01Em+1(x)xn-1dx=-1m+1Em+1-nm+1Jm+1,n-1=-Em+1m+1+(-1)2n(m+1)(m+2)Em+2+(-1)2n(n-1)(m+1)(m+2)Jm+2,n-2.Continuing this process, we get(1.16)Jm,n=∑j=1n-1(-1)j1n-j+1(∏l=1jn-l+1m+l)Em+j+(-1)n-1n(n-1)⋯2(m+1)⋯(m+n-1)Jm+n-1,1,Jm+n-1,1=∫01Em+n-1(x)xdx=-Em+nm+n+(-1)22Em+n+1(m+n)(m+n+1).
By (1.16), we get(1.17)Jm,n=1n+1∑j=1n(-1)j(n+1j)(m+jm)Em+j+2(-1)n+1En+m+1(n+m+1)(n+mm).
From the properties of the Bernoulli and Euler basis for the space of the polynomials of degree less than or equal to n with coefficients in ℚ, we derive some identities for the product of two Bernoulli and Euler polynomials.


## 2. Some Identities for the Bernoulli and Euler Numbers

Let us consider the polynomialp(x)=∑k=0nBk(x)xn-k, with n∈ℕ. Then, we have(2.1)p(k)(x)=(n+1)!(n-k+1)!∑l=knBl-k(x)xn-l(k=0,1,2,…,n).
From the properties of the Bernoulli basis for the space of polynomials of degree less than or equal to n with coefficients in ℚ, p(x) is given by(2.2)p(x)=∑k=0nakBk(x).
Thus, by (2.2), we get(2.3)a0=∫01p(t)dt=∑k=0n∫01Bk(t)xn-kdt=∑k=0nIk,n-k=I0,n+∑k=1n-1Ik,n-k+In,0=1n+1+∑k=1n-1∑j=1n-k(-1)j-1(n-k+1j)(n-k+1)(k+jk)Bk+j+δ0,n.
By (2.1) and (2.2), we get(2.4)ak=(p(k-1)(1)-p(k-1)(0))=(n+1)!k!(n-k+2)!∑l=k-1n{Bl-k+1(1)-Bl-k+10n-l}=(n+2k)n+2(∑l=k-1n-1Bl-k+1+1).
Therefore, by (2.3) and (2.4), we obtain the following theorem.Theorem 2.1.
Forn∈ℕ, one has
(2.5)∑k=0nBk(x)xn-k=1n+1+∑l=1n-1∑j=1n-l(-1)j-1(n-l+1j)(n-l+1)(l+jl)Bl+j+1n+2∑k=1n(n+2k)(∑l=k-1n-1Bl-k+1+1)Bk(x).From the properties of the Euler basis for the space of polynomials of degree less than or equal ton with coefficients in ℚ, p(x) is given by(2.6)p(x)=∑k=0nbkEk(x).
By (2.1) and (2.6), we get(2.7)bk=12k!(p(k)(1)+p(k)(0))=12k!(n+1)!(n-k+1)!∑l=kn{Bl-k(1)+Bl-k0n-l}=(n+1k)2(∑l=knBl-k+1-δk,n+Bn-k)={(n+1k)2(∑l=knBl-k+1+Bn-k)ifk≠n,n+1ifk=n.
Therefore, by (2.6) and (2.7), we obtain the following theorem.Theorem 2.2.
Forn∈ℤ+, one has
(2.8)∑k=0nBk(x)xn-k=12∑k=0n-1(n+1k){∑l=knBl-k+1+Bn-k}Ek(x)+(n+1)En(x).Let us take polynomialp(x) with p(x)=∑k=0nEk(x)xn-k. Then, we have(2.9)p(k)(x)=(n+1)!(n-k+1)!∑l=knEl-k(x)xn-l(k=0,1,2,…,n).
By the basis set {B0(x),…,Bn(x)} for the space of polynomials of degree less than or equal to n with coefficients in ℚ, we see that p(x) is given by(2.10)p(x)=∑k=0nakBk(x).
From (2.10), we note that(2.11)a0=∫01p(t)dt=∑l=0n∫01El(t)xn-ldt=∑l=0nJl,n-l=J0,n+∑l=1n-1Jl,n-l+Jn,0=1n+1+∑l=1n-1Jl,n-l-2n+1En+1=(∑j=1n(-1)j(n+1j)n+1Ej+2(-1)n+1n+1En+1)+∑l=1n-1{∑j=1n-l(-1)j(n-l+1j)(n-l+1)(l+jl)El+j+2(-1)n-l+1(n+1)(nl)En+1}-2n+1En+1=∑l=0n{∑j=1n-l(-1)j(n-l+1j)(n-l+1)(l+jl)El+j+2(-1)n-l+1(n+1)(nl)En+1}.
Note that(2.12)∑l=0n(-1)l-1(n+1)(nl)=-∑l=0n(n-l)!l!(n+1)!(-1)l=-∑l=0nB(n-l+1,l+1)(-1)l=-1+(-1)nn+2,
where B(α,β) is the beta function.From (2.11) and (2.12), we have(2.13)a0=∑l=0n∑j=1n-l(-1)j(n-l+1j)(n-l+1)(l+jl)El+j+4(-1)n+1n+2En+1.
For k=1,2,…,n, by (2.9) and (2.10), we get(2.14)ak=1k!(p(k-1)(1)-p(k-1)(0))=(n+1)!k!(n-k+2)!∑l=k-1n{El-k+1(1)-El-k+10n-l}=1n+2(n+2k){-∑l=k-1nEl-k+1+2-En-k+1}.
Therefore, by (2.13) and (2.14), we obtain the following theorem.Theorem 2.3.
Forn∈ℤ+, one has
(2.15)∑k=0nEk(x)xn-k=∑k=0n-1∑j=1n-l(-1)j(n-l+1k)(n-l+1)(l+jl)El+j+4(-1)n+1n+2En+1+1n+2∑k=1n(n+2k){-∑l=k-1nEl-k+1+2-En-k+1}Bk(x).From the Euler basis{E0(x),E1(x),…,En(x)} for the space of polynomials of degree less than or equal to n with coefficients in ℚ, we note that p(x) can be written as follows:(2.16)p(x)=∑k=0nbkEk(x).
Thus, we have(2.17)bk=12k!(p(k)(1)+p(k)(0))=12k!(n+1)!(n-k+1)!∑l=kn(El-k(1)+El-k0n-l)=12(n+1k)(-∑l=kn-1El-k+2).
Therefore, by (2.16) and (2.17), we obtain the following theorem.Theorem 2.4.
Forn∈ℤ+, one has
(2.18)∑k=0nEk(x)xn-k=12∑k=0n-1(n+1k){-∑l=kn-1El-k+2}Ek(x)+(n+1)En(x).Let us consider the polynomialp(x)=∑k=0n(Bk(x)xn-k)/(k!(n-k)!)=∑k=0nakBk(x). Then, we have(2.19)a0=∫01p(t)dt=∑l=0n1l!(n-l)!Il,n-l=1n!I0,n+∑l=1n-1Il,n-ll!(n-l)!+In,0n!=1(n+1)!+1(n+1)!∑l=1n-1∑j=1n-l(-1)j-1(n+1l+j)Bl+j.
It is easy to show that(2.20)p(k)(x)=2k∑l=kn1(l-k)!(n-l)!Bl-k(x)xn-l.
For k=1,2,…,n, we have(2.21)ak=1k!(p(k-1)(1)-p(k-1)(0))=2k-1k!∑l=k-1n1(l-k+1)!(n-l)!(Bl-k+1(1)-Bl-k+10n-l)=2k-1k!{∑l=k-1nBl-k+1+δ1,l-k+1(l-k+1)!(n-l)!-Bn-k+1(n-k+1)!}=2k-1k!(∑l=k-1n-1Bl-k+1(l-k+1)!(n-l)!+1(n-k)!).
Therefore, by (2.19) and (2.21), we obtain the following theorem.Theorem 2.5.
Forn∈ℤ+, one has
(2.22)∑k=0nBk(x)k!(n-k)!xn-k=1(n+1)!+1(n+1)!∑l=1n-1∑j=1n-l(-1)j-1(n+1l+j)Bl+j+∑k=1n2k-1k!(∑l=k-1n-1Bl-k+1(l-k+1)!(n-l)!+1(n-k)!)Bk(x).Remark 2.6.
Ifp(x)=∑k=0nbkEk(x), by the same method, we get
(2.23)∑k=0nBk(x)xn-kk!(n-k)!=∑k=0n-12k-1k!{∑l=knBl-k(l-k)!(n-l)!+1(n-k-1)!+Bn-k(n-k)!}Ek(x)+2nn!En(x).Let us consider the polynomialp(x)=∑k=0n(Ek(x)xn-k)/(k!(n-k)!)=∑k=0nakBk(x). Then, we have(2.24)a0=∫01p(t)dt=∑l=0n1l!(n-l)!Jl,n-l=∑l=0n1l!(n-l)!{∑j=1n-l(-1)j(n-l+1j)(n-l+1)(l+jl)El+j+2(-1)n-l+1(n+1)(nl)En+1}=∑l=0n∑j=1n-l(-1)jEl+j(l+j)!(n+1-l-j)!+2En+1(-1)n+1(n+1)!∑l=0n(-1)l=∑l=0n∑j=1n-l(-1)jEl+j(l+j)!(n+1-l-j)!+2En+1(-1)n+1(n+1)!.
It is easy to show that(2.25)p(k)(x)=2k∑l=kn1(l-k)!(n-l)!El-k(x)xn-l.Fork=1,2,…,n, we have(2.26)ak=1k!(p(k-1)(1)-p(k-1)(0))=2k-1k!∑l=k-1n1(l-k+1)!(n-l)!(El-k+1(1)-El-k+10n-l)=2k-1k!{-∑l=k-1nEl-k+1(l-k+1)!(n-l)!+2(n-k+1)!-En-k+1(n-k+1)!}.
Therefore, by (2.24) and (2.26), we obtain the following theorem.Theorem 2.7.
Forn∈ℤ+, one has
(2.27)∑k=0n(nk)Ek(x)xn-k=∑l=0n∑j=1n-l(-1)j(n+1l+j)n+1El+j+2(-1)n+1(n+1)En+1+∑k=1n{-1n+1∑l=k-1n2k-1(n+1k)(n-k+1l-k+1)El-k+1+2(n+1)(n+1k)-2k-1n+1(n+1k)En-k+1}Bk(x).Let us consider the polynomialp(x)=∑k=0n(Ek(x)xn-k)/(k!(n-k)!)=∑k=0nbkEk(x). By the same method, we obtain the following identity:(2.28)∑k=0nEk(x)xn-kk!(n-k)!=∑k=0n2k-1k!{-∑l=knEl-k(l-k)!(n-l)!+2(n-k)!+En-k(n-k)!}Ek(x).
Let us take p(x)=∑k=1n-1(1/(k(n-k)))Bk(x)xn-k. Then, the kth derivative of p(x) is given by(2.29)p(k)(x)=Ck(xn-k+Bn-k(x))+(n-1)⋯(n-k)∑l=k+1n-1Bl-k(x)xn-l(n-l)(l-k),
where(2.30)Ck=∑j=1k(n-1)⋯(n-j+1)(n-j-1)⋯(n-k)n-k(k=1,2,…,n-1),C0=0.
Note that(2.31)p(n)(x)=(p(n-1)(x))′=Cn-1(x+B1(x))′=2Cn-1=2(n-1)!Hn-1,
where Hn-1=∑j=1n-1(1/j).By the properties of the Bernoulli basis for the space of polynomials of degree less than or equal ton with coefficients in ℚ, p(x) is given by(2.32)p(x)=∑k=0nakBk(x).
Thus, by (2.32), we get(2.33)a0=∑l=1n-11l(n-l)∫01Bl(t)xn-ldt=∑l=0n-11l(n-l)Il,n-l=∑l=0n-11l(n-l){∑j=1n-l(-1)j-1(n-l+1j)Bl+j(n-l+1)(l+jl)}=1n(n2-1)∑l=1n-1∑j=1n-l(-1)j-1(n+1j+l)(n-2l-1)Bl+j.From (2.29), (2.31), and (2.32), we note that(2.34)ak=1k!(p(k-1)(1)-p(k-1)(0))=Ck-1k!{Bn-k+1(1)-Bn-k+1}+{1-0n-k+1}+(nk)n∑l=kn-11(n-l)(l-k+1){Bl-k+1(1)-Bl-k+10n-l}=Ck-1k!(δ1,n-k+1+1)+(nk)n∑l=kn-1Bl-k+1+δ1,l-k+1(l-k+1)(n-l)=Ck-1k!(δ1,n-k+1+1)+(nk)n{∑l=kn-1Bl-k+1(l-k+1)(n-l)+1n-k}={Ck-1k!+(nk)n{∑l=kn-1Bl-k+1(l-k+1)(n-l)+1n-k}if1≤k≤n-1,2Hn-1nifk=n.
From (2.30), we have(2.35)Ck-1k!=1k!∑j=1k-1(n-1)!(n-k+1)(n-k)!(n-j)=(n!k!(n-k)!)(1n(n-k+1))∑j=1k-11n-j=(nk)1n(n-k+1)(∑j=1n-11j-∑j=1n-k1j)=(nk)n(n-k+1)(Hn-1-Hn-k).
Therefore, by (2.32), (2.34), and (2.35), we obtain the following theorem.Theorem 2.8.
Forn∈ℕ, one has
(2.36)∑k=1n-11k(n-k)Bk(x)xn-k=1n(n2-1)∑l=1n-1∑j=1n-l(-1)j-1(n+1l+j)Bl+j(n-2l-1)+1n∑k=1n-1(nk)×(Hn-1-Hn-kn-k+1+∑l=kn-1Bl-k+1(l-k+1)(n-l)+1n-k)×Bk(x)+2Hn-1Bn(x)n.We assume thatp(x)=∑k=1n-1(1/k(n-k))Bk(x)xn-k=∑k=0nbkEk(x). For k=0,1,2,…,n-1, we have(2.37)bk=12k!(p(k)(1)+p(k)(0))=Ck2k!{Bn-k(1)+1+Bn-k0n-k}+(n-1)⋯(n-k)2k!∑l=k+1n-11(l-k)(n-l){Bl-k(1)+Bl-k0n-l}={Ck2k!(2Bn-k+1)+(n-1k)2(∑l=k+1n-1Bl-k(l-k)(n-l)+1n-k-1)if0≤k≤n-2,Cn-12(n-1)!=12Hn-1ifk=n-1.
Finally,(2.38)bn=12n!(p(n)(1)+p(n)(0))=4Cn-12n!=2Cn-1n!=2Hn-1n.By the same method, we obtain the following identity:(2.39)∑k=1n-11k(n-k)Bk(x)xn-k=12∑k=0n-2{(nk)n(Hn-1-Hn-k-1)(2Bn-k+1)+(n-1k)(∑l=k+1n-1Bl-k(l-k)(n-l)+1n-k-1)}×Ek(x)+12Hn-1En-1(x)+2Hn-1nEn(x).
Let us take p(x)=∑k=1n-1(1/k(n-k))Ek(x)xn-k. Then, for k=0,1,2,…,n-1, we have(2.40)p(k)(x)=Ck(En-k(x)+xn-k)+(n-1)⋯(n-k)∑l=k+1n-1El-k(x)xn-l(l-k)(n-l),
where Ck=(∑j=1k(n-1)⋯(n-j+1)(n-j-1)⋯(n-k))/(n-k).Note that(2.41)p(n)(x)=(p(n-1)(x))′=(Cn-1(E1(x)+x))′=2Cn-1=2(n-1)!Hn-1.
By the properties of the Bernoulli basis for the space of polynomials of degree less than or equal to n with coefficients in ℚ, p(x) can be written as(2.42)p(x)=∑k=0nakBk(x).
Thus, by (2.42), we get(2.43)a0=∫01p(t)dt=∑l=1n-11l(n-l)∫01El(t)tn-ldt=∑l=1n-11l(n-l)Jl,n-l=1n(n2-1)∑l=1n-1∑j=1n-l(-1)j(n+1j+l)(n-2l-1)El+j+2(-1)n+1n(n2-1)En+1∑l=1n-1(-1)l(n-2l-1).It is easy to show that(2.44)∑l=1n-1(-1)l(n-1)(n-2l-1)=∑l=1n-1(-1)lB(l,n-l)=(4n(n+1)(n+2)-2(n-1)n2(n+1)2)En+1.
For k=1,2,…,n, one has(2.45)ak=1k!(p(k-1)(1)-p(k-1)(0))=Ck-1k!(En-k+1(1)+1-En-k+1-0n-k+1)+(n-1)⋯(n-k+1)k!∑l=kn-11(l-k+1)(n-l)(El-k+1(1)-El-k+10n-l)=1k!Ck-1(-2En-k+1+1)-(n-1)⋯(n-k+1)k!∑l=kn-1El-k+1(l-k+1)(n-l)=(nk)(Hn-1-Hn-k)n(n-k+1)(-2En-k+1+1)-(nk)n∑l=kn-1El-k+1(l-k+1)(n-l).
Therefore, by (2.42), (2.44), and (2.45), we obtain the following theorem.Theorem 2.9.
Forn∈ℕ, one has
(2.46)∑k=1nEk(x)xn-kk(n-k)=1n(n2-1)∑l=1n-1∑j=1n-l(-1)j(n+1l+j)El+j(n-2l-1)+2(-1)n+1n(n+1)(4n(n+1)(n+2)-2(n-1)n2(n+1)2)En+1+1n{∑k=1n(nk)(Hn-1-Hn-kn-k+1)(1-2En-k+1)-(nk)∑l=kn-1El-k+1(l-k+1)(n-l)}Bk(x).We may assume thatp(x)=∑k=0n(1/(k(n-k)))Ek(x)xn-k=∑k=0nbkEk(x). Then, we note that(2.47)bk=12k!(p(k)(1)+p(k)(0))(k=0,1,2,…,n-1).Thus, we have(2.48)bk=Ck2k!{En-k(1)+1+En-k+0n-k}+(n-1k)2∑l=k+1n-11(l-k)(n-l){El-k(1)+El-k0n-l}=Ck2k!+(n-1k)2∑l=k+1n-1-El-k(l-k)(n-l)=Ck2k!-(n-1k)2∑l=k+1n-1El-k(l-k)(n-l),bn=12n!(p(n)(1)+p(n)(0))=12n!4Cn-1=2Cn-1n!=2Hn-1n.By the same method, we obtain the following identity:(2.49)∑k=1n-11k(n-k)Ek(x)xn-k=∑k=0n-1{12n(nk)(Hn-1-Hn-k-1)-(n-1k)2∑l=k+1n-1El-k(l-k)(n-l)}×Ek(x)+2nHn-1En(x).



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