# Bernoulli Basis and the Product of Several Bernoulli Polynomials

Authors: Dae San Kim; Taekyun Kim
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

---

## Abstract

We develop methods for computing the product of several Bernoulli and Euler polynomials by using Bernoulli basis for the vector space of polynomials of degree less than or equal ton.


---

## Body

## 1. Introduction

It is well known that, thenth Bernoulli and Euler numbers are defined by
(1.1)∑l=0n‍(nl)Bl-Bn=δ1,n,∑l=0n‍(nl)El+En=2δ0,n,
where B0=E0=1 and δk,n is the Kronecker symbol (see [1–20]).The Bernoulli and Euler polynomials are also defined by(1.2)Bn(x)=∑l=0n‍(nl)Bn-lxl,En(x)=∑l=0n‍(nl)En-lxl.
Note that {B0(x),B1(1),…,Bn(x)} forms a basis for the space ℙn={p(x)∈ℚ[x]∣degp(x)≤n}.So, for a givenp(x)∈ℙn, we can write
(1.3)p(x)=∑k=0n‍akBk(x),
(see [8–12]) for uniquely determined ak∈ℚ.Further,(1.4)ak=1k!{p(k-1)(1)-p(k-1)(0)},wherep(k)(x)=dkp(x)dxk,a0=∫01‍p(t)dt,wherek=1,2,…,n.
Probably, {1,x,…,xn} is the most natural basis for the space ℙn. But {B0(x),B1(x),…,Bn(x)} is also a good basis for the space ℙn, for our purpose of arithmetical and combinatorial applications.What are common toBn(x), En(x), xn? A few proportion common to them are as follows: (i)
they are all monic polynomials of degreen with rational coefficients;(ii)
(Bn(x))′=nBn-1(x), (En(x))′=nEn-1(x), (xn)′=nxn-1;(iii)
∫‍Bn(x)dx=Bn+1(x)/(n+1)+c, ∫‍En(x)dx=En+1(x)/(n+1)+c, ∫‍xndx=xn+1/(n+1)+c.In [5, 6], Carlitz introduced the identities of the product of several Bernoulli polynomials:
(1.5)Bm(x)Bn(x)=∑r=0∞‍{(m2r)n+(n2r)m}B2rBm+n-2r(x)m+n-2r+(-1)m+1×m!n!(m+n)!Bm+n(m+n≥2),∫01‍Bm(x)Bn(x)Bp(x)Bq(x)dx=(-1)m+n+p+q∑r,s=0∞‍{(m2r)n+(n2r)m}{(p2s)q+(q2s)p}×(m+n-2r-1)!(p+q-2s-1)!(m+n+p+q-2r-2s)!BrBsBm+n+p+q-2r-2+(-1)m+pm!n!(m+n)!p!q!(p+q)!Bm+nBp+q.
In this paper, we will use (1.4) to derive the identities of the product of several Bernoulli and Euler polynomials.


## 2. The Product of Several Bernoulli and Euler Polynomials

Let us consider the following polynomials of degreen:
(2.1)p(x)=∑i1+⋯+ir+j1+⋯+js=n‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x),
where the sum runs over all nonnegative integers i1,…,ir,  j1,…js satisfying i1+⋯+ir+j1+⋯+js=n,  r+s=1,  r,s≥0.Thus, from (2.1), we have
(2.2)p(k)(x)=(n+r+s-1)(n+r+s-2)⋯(n+r+s-k)×∑i1+⋯+ir+j1+⋯+js=n-k∞‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x).
For k=1,2,…,n, by (1.4), we get
(2.3)ak=1k!{p(k-1)(1)-p(k-1)(0)}=(n+r+sk)n+r+s∑i1+⋯+ir+j1+⋯+js=n-k+1‍{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejs}=(n+r+sk)n+r+s{∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍(ra)(sc)(-1)c2s-c=(n+r+sk)n+r+s×∑i1+⋯+ir+j1+⋯+js=n+a+1-k-r∞‍Bi1⋯BiaEj1⋯Ejc=(n+r+sk)n+r+s-∑i1+⋯+ir+j1+⋯+js=n-k+1‍Bi1⋯BirEj1⋯Ejs∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍}.
From (2.3), we note that
(2.4)an=(n+r+sn)n+r+s∑i1+⋯+ir+j1+⋯+js=1‍{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejs}=(n+r+sn)n+r+s{(-12+1)r-(-12)s-(-12)(r+s)}=(n+r+sn)n+r+s(r+s)=(n+r+s-1n),an-1=1n+r+s(n+r+sn-1)×∑i1+⋯+ir+j1+⋯+js=2‍{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejs}=1n+r+s(n+r+sn-1){16r+1212(r+s2)-16r-(-12)(-12)(r+s2)}=0,a0=∫01‍p(t)dt=∑i1+⋯+ir+j1+⋯+js=n∞‍∑l1=0i1‍⋯∑lr=0ir‍∑p1=0j1‍⋯∑ps=0js‍(i1l1)⋯(irlr)(j1p1)⋯(jsps)×Bi1-l1⋯Bir-lrEj1-p1Ejs-psl1+⋯+lr+p1+⋯+ps+1.
Therefore, by (1.3), (2.1), (2.3), and (2.4), we obtain the following theorem.Theorem 2.1.
Forn∈ℕ with n≥2, we have
(2.5)∑i1+⋯+ir+j1+⋯+js=n‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)=1n+r+s∑k=1n-2‍(n+r+sk)×{∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍(ra)(sc)(-1)c2s-c∑i1+⋯+ia+j1+⋯+jc=n+a+1-k-r‍Bi1⋯BiaEj1⋯Ejc-∑i1+⋯+ir+j1+⋯+js=n-k+1∞‍Bi1⋯BirEk1⋯Ejs∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍}Bk(x)+(n+r+s-1n)Bn(x)+∑i1+⋯+ir+j1+⋯+js=n‍∑l1=0i1‍…∑lr=0ir‍∑p1=0j1‍⋯∑ps=0js‍(i1l1)⋯(irlr)(j1p1)⋯(jsps)×Bi1-l1⋯Bir-lrEj1-p1Ejs-psl1+⋯+lr+p1+⋯ps+1.Let us take the polynomialp(x) of degree n as follows:
(2.6)p(x)=∑i1+⋯+ir+j1+⋯+js+t=n‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)xt,
Then, from (2.6), we have
(2.7)p(k)(x)=(n+r+s)(n+r+s-1)⋯(n+r+s-k+1)×∑i1+⋯+ir+j1+⋯+js+t=n-k‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)xt,
By (1.4) and (2.7), we get, for k=1,2,…,n,
(2.8)ak=1k!{p(k-1)(1)-p(k-1)(0)}=1n+r+s+1(n+r+s+1k)×∑i1+⋯+ir+j1+⋯+js+t=n-k+1‍{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejs0t}=(n+r+s+1k)n+r+s+1{∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍(ra)(sc)(-1)c2s-c(n+r+s+1k)n+r+s+12×∑t=0n+a+1-k-r‍∑i1+⋯+ia+j1+⋯+jc=t‍Bi1⋯BiaEj1⋯Ejc(n+r+s+1k)n+r+s+12-∑i1+⋯+ir+j1+⋯+js=n-k+1‍Bi1⋯BirEj1⋯Ejs∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍},
Now, we look at an and an-1.
(2.9)an=(n+r+s+1n)n+r+s+1∑i1+⋯+ir+j1+⋯+js+t=1‍{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejs0t}=(n+r+s+1n)n+r+s+1{12(r+s)+1-(-12)(r+s)}=r+s+1n+r+s+1(n+r+s+1n)=(n+r+sn),an-1=(n+r+s+1n-1)n+r+s+1∑i1+⋯+ir+j1+⋯+js+t=2‍{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejs0t}=(n+r+s+1n-1)n+r+s+1{16r+1+1212(r+s2)+12(r+s)-16r-(-12)(-12)(r+s2)}=1n+r+s+1(n+r+s+1n-1)r+s+22=12(n+r+sn-1),
From (2.6), we note that
(2.10)a0=∫01‍p(t)dt=∑i1+⋯+ir+j1+⋯+js+t=n‍∑l1=0i1‍⋯∑lr=0ir‍∑p1=0j1‍⋯∑ps=0js‍(i1l1)⋯(irlr)(j1p1)⋯(jsps)×Bi1-l1⋯Bir-lrEj1-p1Ejs-ps1l1+⋯+lr+p1+⋯ps+t+1.
Therefore, by (1.3), (2.6), (2.8), (2.9), and (2.10), we obtain the following theorem.Theorem 2.2.
Forn∈ℕ with n≥2, one has
(2.11)∑i1+⋯+ir+j1+⋯+js+t=n‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)xt=1n+r+s+1∑k=1n-2‍(n+r+s+1k)×{∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍(ra)(sc)(-1)c2s-c∑t=0n+a+1-k-r‍∑i1+⋯+ia+j1+⋯+jc=t∞‍Bi1⋯BiaEj1⋯Ejc-∑i1+⋯+ir+j1+⋯+js=n-k+1‍Bi1⋯BirEj1⋯Ejs∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍}Bk(x)+12(n+r+sn-1)Bn-1(x)+(n+r+sn)Bn(x)+∑i1+⋯+ir+j1+⋯+js+t=n‍∑l1=0i1‍⋯∑lr=0ir‍∑p1=0j1‍⋯∑ps=0js‍{(i1l1)⋯(irlr)(j1p1)⋯(jsps)×Bi1-l1⋯Bir-lrEj1-p1⋯Ejs-ps×1l1+⋯+lr+p1+⋯ps+t+1(i1l1)}.Consider the following polynomial of degreen:
(2.12)p(x)=∑i1+⋯+ir+j1+⋯+js=n∞‍1i1!i2!⋯ir!j1!⋯js!Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x).
Then, from (2.12), one has
(2.13)p(k)(x)=(r+s)k∑i1+⋯+ir+j1+⋯+js=n-k‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)i1!i2!⋯ir!j1!⋯js!.
By (1.4) and (2.13), one gets, for k=1,2,…,n,
(2.14)ak=1k!{p(k-1)(1)-p(k-1)(0)}=(r+s)k-1k!∑i1+⋯+ir+j1+⋯+js+t=n-k+1‍{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejsi1!i2!⋯ir!j1!⋯js!}=(r+s)k-1k!{∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍(ra)(sc)(-1)c2s-c∑i1+⋯+ia+j1+⋯+jc=n+a+1-k-r‍Bi1⋯BiaEj1⋯Ejci1!i2!⋯ia!j1!⋯jc!(r+s)k-1k!22-∑i1+⋯+ir+j1+⋯+js=n-k+1‍1i1!i2!⋯ir!j1!⋯js!Bi1⋯BirEj1⋯Ejs∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍}.
Now look at an and an-1:
(2.15)an=(r+s)n-1n!∑i1+⋯+ir+j1+⋯+js=1‍{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejsi1!i2!⋯ir!j1!⋯js!}=(r+s)n-1n!{12(r+s)-(-12)(r+s)}=(r+s)nn!,an-1=(r+s)n-2(n-1)!∑i1+⋯+ir+j1+⋯+js=2‍{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejsi1!i2!⋯ir!j1!⋯js!}=(r+s)n-2(n-1)!{1216r+1212(r+s2)-1216r-(12)(-12)(r+s2)}=0.It is easy to show that(2.16)a0=∫01‍p(t)dt=∑i1+⋯+ir+j1+⋯+js=n‍1i1!⋯ir!j1!⋯js!×∑l1=0i1‍⋯∑lr=0ir‍∑p1=0j1‍⋯∑ps=0js‍{Bi1-l1⋯Bir-lrEj1-p1Ejs-psl1+⋯+lr+p1+⋯ps+1(i1l1)⋯(irlr)(j1p1)⋯(jsps)}.
Therefore, by (1.3), (2.14), and (2.15), one obtains the following theorem.Theorem 2.3.
Forn∈ℕ with n≥2, one has
(2.17)∑i1+⋯+ir+j1+⋯+js=n‍1i1!i2!⋯ir!j1!⋯js!Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)=∑k=1n-2‍(r+s)k-1k!{∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍(ra)(sc)(-1)c2s-c×∑i1+⋯+ia+j1+⋯+jc=n+a+1-k-r‍Bi1⋯BiaEj1⋯Ejci1!i2!⋯ia!j1!⋯jc!-∑i1+⋯+ir+j1+⋯+js=n-k+1‍1i1!i2!⋯ir!j1!⋯js!×Bi1⋯BirEj1⋯Ejs∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍}Bk(x)+(r+s)nn!Bn(x)+∑i1+⋯+ir+j1+⋯+js=n‍∑l1=0i1‍⋯∑lr=0ir‍∑p1=0j1‍⋯∑ps=0js‍(i1l1)⋯(irlr)(j1p1)⋯(jsps)×Bi1-l1⋯Bir-lrEj1-p1Ejs-psi1!i2!⋯ir!j1!⋯js!(l1+⋯+lr+p1+⋯ps+1).Take the polynomialp(x) of degree n as follows:
(2.18)p(x)=∑i1+⋯+ir+j1+⋯+js+t=n‍1i1!i2!⋯ir!j1!⋯js!t!Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)xt.
Then, from (2.18), one gets
(2.19)p(k)(x)=(r+s+1)k×∑i1+⋯+ir+j1+⋯+js+t=n-k‍1i1!i2!⋯ir!j1!⋯js!t!Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)xt.
By (1.4) and (2.19), one gets, for k=1,…,n,
(2.20)ak=1k!{p(k-1)(1)-p(k-1)(0)}=(r+s+1)k-1k!∑i1+⋯+ir+j1+⋯+js+t=n-k+1‍1i1!i2!⋯ir!j1!⋯js!t!×{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejs0t}=(r+s+1)k-1k!{∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍(ra)(sc)(-1)c2s-c∑t=0n+a+1-k-r‍1(n+a+1-k-r-t)!×∑i1+⋯+ia+j1+⋯+jc=t‍1i1!i2!⋯ia!j1!⋯jc!Bi1⋯BiaEj1⋯Ejc-∑i1+⋯+ir+j1+⋯+js=n-k+1‍Bi1⋯BirEj1⋯Ejs∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍}.Now look atan and an-1:
(2.21)an=(r+s+1)n-1n!∑i1+⋯+ir+j1+⋯+js+t=1‍1i1!i2!⋯ir!j1!⋯js!t!×{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejs0t}=(r+s+1)n-1n!{12(r+s)+1-(-12)(r+s)}=(r+s+1)n-1n!(r+s+1)=(r+s+1)nn!,an-1=(r+s+1)n-2(n-1)!∑i1+⋯+ir+j1+⋯+js+t=2‍1i1!i2!⋯ir!j1!⋯js!t!×{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)-Bi1⋯BirEj1⋯Ejs0t}=(r+s+1)n-2(n-1)!{1216r+12+1212(r+s2)+12(r+s)-1216r-(-12)(-12)(r+s2)}=(r+s+1)n-2(n-1)!r+s+12=(r+s+1)n-12(n-1)!.
From (2.18), one can derive the following identity:
(2.22)a0=∫01‍p(t)dt=∑i1+⋯+ir+j1+⋯+js+t=n‍1i1!⋯ir!j1!⋯js!t!∫01‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)xtdt=∑i1+⋯+ir+j1+⋯+js+t=n‍1i1!⋯ir!j1!⋯js!t!∑l1=0i1‍⋯∑lr=0ir‍∑p1=0j1‍⋯×∑ps=0js‍(i1l1)⋯(irlr)(j1p1)⋯(jsps)Bi1-l1⋯Bir-lrEj1-p1Ejs-ps1l1+⋯+lr+p1+⋯ps+t+1.
Therefore, by (1.3), (2.20), (2.21), and (2.22), one obtains the following theorem.Theorem 2.4.
Forn∈ℕ with n≥2, one has
(2.23)∑i1+⋯+ir+j1+⋯+js+t=n‍1i1!i2!⋯ir!j1!⋯js!t!Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)xt=∑k=1n-2‍(r+s+1)k-1k!{∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍(ra)(sc)(-1)c2s-c∑t=0n+a+1-k-r‍1(n+a+1-k-r-t)!×∑i1+⋯+ia+j1+⋯+jc=t‍1i1!i2!⋯ia!j1!⋯jc!Bi1⋯BiaEj1⋯Ejc-∑i1+⋯+ir+j1+⋯+js=n-k+1‍Bi1⋯BirEj1⋯Ejsi1!i2!⋯ir!j1!⋯js!∑0≤a≤r0≤c≤sk+r-n-1≤a≤r‍}Bk(x)+(r+s+1)n-12(n-1)!Bn-1(x)+(r+s+1)nn!∑i1+⋯+ir+j1+⋯+js+t=n‍∑l1=0i1‍⋯∑lr=0ir‍∑p1=0j1‍⋯∑ps=0js‍=×(i1l1)⋯(irlr)(j1p1)⋯(jsps)i1!⋯ir!j1!⋯js!t!Bi1-l1⋯Bir-lrEj1-p1Ejs-ps1l1+⋯+lr+p1+⋯ps+t+1.



---

*Source: 463659-2012-09-08.xml*