# Euler Basis, Identities, and Their Applications

Authors: D. S. Kim; T. Kim
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

LetVn={p(x)∈ℚ[x]|degp(x)≤n} be the (n+1)-dimensional vector space over ℚ. We show that {E0(x),E1(x),…,En(x)} is a good basis for the space Vn, for our purpose of arithmetical and combinatorial applications. Thus, if p(x)∈ℚ[x] is of degree n, then p(x)=∑l=0nblEl(x) for some uniquely determined bl∈ℚ. In this paper we develop method for computing bl from the information of p(x).


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## Body

## 1. Introduction

The Euler polynomials,En(x), are given by
(1.1)2et+1ext=eE(x)t=∑n=0∞‍En(x)tnn!,(see [1–20]) 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. The Bernoulli numbers are also defined by
(1.2)tet-1=eBt=∑n=0∞‍Bntnn!,
(see [1–20]) with the usual convention about replacing Bn by Bn. As is well known, the Bernoulli polynomials are given by
(1.3)Bn(x)=∑l=0n‍(nl)Blxn-l=∑l=0n‍(nl)Bn-lxl,
(see [9–15]) From (1.1), (1.2), and (1.3), we note that
(1.4)Bn(1)-Bn=δ1,n,En(1)+En=2δ0,n,
where δk,n is the kronecker symbol.Letm,n∈ℤ+ with m+n≥2. The formula
(1.5)Bm(x)Bn(x)=∑r‍{(m2r)n+(n2r)m}B2rBm+n-2r(x)m+n-2r+(-1)m+1m!n!Bm+n(m+n)!,
is proved in [4–6]. Let Vn={p(x)∈ℚ[x]∣degp(x)≤n} be the (n+1)-dimensional vector space over ℚ. Probably, {1,x,…,xn} is the most natural basis for this space. But {E0(x),E1(x),…,En(x)} is also a good basis for the space Vn, for our purpose of arithmetical and combinatorial applications. Thus, if p(x)∈ℚ[x] is of degree n, then
(1.6)p(x)=∑l=0n‍blEl(x),
for some uniquely determined bl∈ℚ. Further,
(1.7)bk=12k!{p(k)(1)+p(k)(0)}(k=0,1,2,…,n),
where p(k)(x)=dkp(x)/dxk. In this paper we develop methods for computing bl from the information of p(x). Apply these results to arithmetically and combinatorially interesting identities involving E0(x),E1(x),…,En(x),B0(x),…,Bn(x). Finally, we give some applications of those obtained identities.


## 2. Euler Basis, Identities, and Their Applications

Let us takep(x) the polynomial of degree n as follows:
(2.1)p(x)=∑k=0n‍Bk(x)Bn-k(x).
From (2.1), we have
(2.2)p(k)(x)=(n+1)!(n-k+1)!∑l=kn‍Bl-k(x)Bn-l(x).
By (1.7) and (2.2), we get
(2.3)bk=12k!{p(k)(1)+p(k)(0)}=12(n+1k)∑l=kn‍{(Bl-k+δ1,l-k)(Bn-l+δ1,n-l)+Bl-kBn-l},
Thus, we have
(2.4)bk=(n+1k)(∑l=kn‍Bl-kBn-l+Bn-k-1),(0≤k≤n-3),(2.5)bn-2=772n(n2-1),bn=n+1,bn-1=0.
By (1.6), (2.1), (2.3), and (2.4), we get
(2.6)∑k=0n‍Bk(x)Bn-k(x)=∑k=0n-3‍(n+1k)(∑l=kn‍Bl-kBn-l+Bn-k-1)Ek(x)+772n(n2-1)En-2(x)+(n+1)En(x).
Let us consider the following triple identities:
(2.7)p(x)=∑r+s+t=n‍Br(x)Bs(x)Bt(x)=∑k=0n‍bkEk(x),
where the sum runs over all r,s,t∈ℤ+ with r+s+t=n. Thus, by (2.7), we get
(2.8)p(k)(x)=(n+2)(n+1)n(n-1)⋯(n-k+3)∑r+s+t=n-k‍Br(x)Bs(x)Bt(x).
From (1.7) and (2.8), we have
(2.9)bk=12k!{p(k)(1)+p(k)(0)}=(n+2k)2∑r+s+t=n-k‍{Br(1)Bs(1)Bt(1)+BrBsBt}=(n+2k)2{2∑r+s+t=n-k‍BrBsBt+∑r+s+t=n-k‍δ1,rBsBt+∑r+s+t=n-k‍Brδ1,sBt+∑r+s+t=n-k‍BrBsδ1,t+∑r+s+t=n-k‍δ1,rδ1,sBt+∑r+s+t=n-k‍δ1,rBsδ1,t+∑r+s+t=n-k‍Brδ1,sδ1,t+∑r+s+t=n-k‍δ1,rδ1,sδ1,t}.
Therefore, by (2.7) and (2.9), we obtain the following theorem.Theorem 2.1.
Forr,s,t∈ℤ+, and n∈ℕ with n≥3, one has
(2.10)∑r+s+t=n‍Br(x)Bs(x)Bt(x)=12∑k=0n-2‍(n+2k){2∑r+s+t=n-k‍BrBsBt+3∑r+s=n-k-1‍BrBs+3Bn-k-2+δk,n-3}Ek(x)+(n+22)En(x).Let us take the polynomialp(x) as follows:
(2.11)p(x)=∑r+s+t=n‍Br(x)Bs(x)Et(x).
Then, by (2.11), we get
(2.12)p(k)(x)=(n+2)(n+1)n(n-1)⋯(n-k+3)∑r+s+t=n-k‍Br(x)Bs(x)Et(x).From (1.6), (1.7), and (2.12), we have
(2.13)bk=12k!{p(k)(1)+p(k)(0)}=(n+2k)2∑r+s+t=n-k‍{Br(1)Bs(1)Et(1)+BrBsEt}=(n+2k)2∑r+s+t=n-k‍{(Br+δ1,r)(Bs+δ1,s)(-Et+2δ0,t)+BrBsEt}=(n+2k)2{-∑r+s+t=n-k‍δ1,rBsEt-∑r+s+t=n-k‍Brδ1,sEt+2∑r+s+t=n-k‍BrBsδ0,t-∑r+s+t=n-k‍δ1,rδ1,sEt+2∑r+s+t=n-k‍δ1,rBsδ0,t+2∑r+s+t=n-k‍Brδ1,sδ0,t+2∑r+s+t=n-k‍δ1,rδ1,sδ0,t}.
Note that
(2.14)bn-1=(n+2n-1){-∑s+t=0‍BsEt-∑r+t=0‍BrEt+2∑r+s=1‍BrBs-0+2B0+2B0+2⋅0}=12(n+2n-1){-1-1+2(B1+B1)+2+2}=0.
Therefore, we obtain the following theorem.Theorem 2.2.
Forn∈ℕ with n≥2, one has
(2.15)∑r+s+t=n‍Br(x)Bs(x)Et(x)=12∑k=0n-2‍(n+2k){2∑r+s=n-k‍BrBs-2∑r+t=n-k-1‍BrEt-En-k-2+4Bn-k-1+2δk,n-2}Ek(x)+(n+22)En(x).Remark 2.3.
By the same method, we obtain the following identities.
(I)(2.16)∑r+s+t=n‍Br(x)Es(x)Et(x)=12∑k=0n-2‍(n+2k){2∑r+s+t=n-k‍BrEsEt+∑s+t=n-k-1‍EsEt-4∑r+s=n-k‍BrEs+4Bn-k-4En-k-1}Ek(x)+(n+22)En(x).
(II)(2.17)∑r+s+t=n‍Er(x)Es(x)Et(x)=3∑k=0n-2‍(n+2k){∑r+s=n-k‍ErEs-2En-k}Ek(x)+(n+22)En(x).
Let us consider the polynomial p(x) as follows:
(2.18)p(x)=∑r+s+t=n‍Br(x)Bs(x)xt.
Thus, by (2.18), we get
(2.19)p(k)(x)=(n+2)(n+1)n(n-1)⋯(n-k+3)∑r+s+t=n-k‍Br(x)Bs(x)xt.
From (1.6), (1.7), (2.18), and (2.19), we have
(2.20)bk=12k!{p(k)(1)+p(k)(0)}=(n+2k)2∑r+s+t=n-k‍{Br(1)Bs(1)+BrBs0t}=(n+2k)2∑r+s+t=n-k‍{(Br+δ1,r)(Bs+δ1,s)+BrBs0t}=(n+2k)2{∑r+s+t=n-k‍BrBs+∑r+s+t=n-k‍Brδ1,s+∑r+s+t=n-k‍δ1,rBs+∑r+s+t=n-k‍δ1,rδ1,s+∑r+s+t=n-k‍BrBs0t}.
Here we note that
(2.21)∑r+s+t=n-k‍BrBs=∑t=0n-k‍∑r+s=n-k-t‍BrBs=∑t=0n-k‍∑r+s=t‍BrBs∑r+s+t=n-k‍Brδ1,s={∑r=0n-k-1‍Br,ifk≤n-1,0,ifk=n,∑r+s+t=n-k‍Bsδ1,r={∑r=0n-k-1‍Br,ifk≤n-1,0,ifk=n,∑r+s+t=n-k‍δ1,rδ1,s={1,ifk≤n-2,0,ifk=n-1orn,∑r+s+t=n-k‍BrBs0t=∑r+s=n-k‍BrBs,∀k.
It is easy to show that
(2.22)bn-1=12(n+2n-1){∑r+s=0‍BrBs+2∑r+s=1‍BrBs+2B0}=12(n+2n-1){1+2(B1+B2)+2}=12(n+2n-1).
Therefore, by (1.6), (2.18), (2.20), and (2.22), we obtain the following theorem.Theorem 2.4.
Forn∈ℕ with n≥2, one has
(2.23)∑r+s+t=n‍Br(x)Bs(x)xt=12∑k=0n-2‍(n+2k){∑t=0n-k-1‍∑r+s=t‍BrBs+2∑r+s=n-k‍BrBs+2∑r=0n-k-1‍Br+1}Ek(x)+12(n+2n-1)En-1(x)+(n+2n)En(x).Remark 2.5.
By the same method, we can derive the following identities.
(I)(2.24)∑r+s+t=n‍Br(x)Es(x)xt=12∑k=0n-2‍(n+2k){-∑t=0n-k-1‍∑r+s=t‍BrEs-∑s=0n-k-1‍Es+2∑r=0n-k‍Br+2}Ek(x)+12(n+2n-1)En-1(x)+(n+2n)En(x).
(II)(2.25)∑r+s+t=n‍Er(x)Es(x)xt=12∑k=0n-2‍(n+2k){∑t=0n-k-1‍∑r+s=t‍ErEs+2∑r+s=n-k‍ErEs-4∑r=0n-k‍Er+4}Ek(x)+12(n+2n-1)En-1(x)+(n+22)En(x).Now we generalize the above consideration to the completely arbitrary case. Let(2.26)p(x)=∑i1+⋯+ir+j1+⋯+js=n‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x),
where the sum runs over all nonnegative integers i1,i2,…,ir,j1,…,js satisfying i1+i2+⋯+ir+j1+⋯+js=n. From (2.26), we note that
(2.27)p(k)(x)=(n+r+s-1)⋯(n+r+s-k)∑i1+⋯+ir+j1+⋯+js=n-k‍Bi1(x)⋯Bir(x)×Ej1(x)⋯Ejs(x).By (1.6), (1.7), (2.18), and (2.27), we get
(2.28)bk=12k!{p(k)(1)+p(k)(0)}=12(n+r+s-1k)∑i1+⋯+ir+j1+⋯+js=n-k‍{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)+Bi1⋯BirEj1⋯Ejs}=12(n+r+s-1k)×∑i1+⋯+ir+j1+⋯+js=n-k‍{(Bi1+δ1,i1)⋯(Bir+δ1,ir)×(-Ej1+2δ0,j1)⋯(-Ejs+2δ0,js)+Bi1⋯BirEj1⋯Ejs}=12(n+r+s-1k){∑0≤a≤r0≤c≤sa≥k+r-n‍(ra)(sc)(-1)c2s-c×∑i1+⋯+ia+j1+⋯+jc=n+a-k-r‍Bi1⋯BiaEj1⋯Ejc+∑i1+⋯+ir+j1+⋯+js=n-k‍Bi1⋯BirEj1⋯Ejs∑0≤a≤r0≤c≤sa≥k+r-n‍}.
Note that
(2.29)bn=12(n+r+s-1n){∑0≤c≤s‍(sc)(-1)c2s-c×∑i1+⋯+ir+j1+⋯+jc=0‍Bi1⋯BirEj1⋯Ejc+∑i1+⋯+ir+j1+⋯+js=0‍Bi1⋯BirEj1⋯Ejs}=12(n+r+s-1n)((2-1)s+1)=(n+r+s-1n),bn-1=12(n+r+s-1n-1){∑r-1≤a≤r0≤c≤s‍(ra)(sc)(-1)c2s-c×∑i1+⋯+ia+j1+⋯+jc=1+a-r‍Bi1⋯BiaEj1⋯Ejc+∑i1+⋯+ir+j1+⋯+js=1‍Bi1⋯BirEj1⋯Ejs∑r-1≤a≤r0≤c≤s‍}=12(n+r+s-1n-1){r(2-1)s+∑0≤c≤s‍(sc)(-1)c2s-c[-12(r+c)]-12(r+s)}=12(n+r+s-1n-1){r-12r+12s-12r-12s}=0.
Therefore, by (1.6), (2.28), and (2.29), we obtain the following theorem.Theorem 2.6.
Forn∈ℕ with n≥2, one has
(2.30)∑i1+⋯+ir+j1+⋯+js=n‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)=12∑k=0n-2‍(n+r+s-1k){∑0≤a≤r0≤c≤sa≥k+r-n‍(ra)(sc)(-1)c2s-c×∑i1+⋯+ia+j1+⋯+jc=n+a-k-r‍Bi1⋯BiaEj1⋯Ejc+∑i1+⋯+ir+j1+⋯+js=n-k‍Bi1⋯BirEj1⋯Ejs∑0≤a≤r0≤c≤sa≥k+r-n‍}Ek(x)+(n+r+s-1n)En(x).Let us consider the polynomialp(x) of degree n as
(2.31)p(x)=∑t+i1+⋯+ir+j1+⋯+js=n‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)xt.
Then, from (2.31), we have
(2.32)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 (1.7) and (2.32), we get
(2.33)bk=12k!{p(k)(1)+p(k)(0)}=12(n+r+sk)∑i1+⋯+ir+j1+⋯+js+t=n-k‍{Bi1(1)⋯Bir(1)Ej1(1)⋯Ejs(1)+Bi1⋯BirEj1⋯Ejs0t}=12(n+r+sk)×∑i1+⋯+ir+j1+⋯+js+t=n-k‍{(Bi1+δ1,i1)⋯(Bir+δ1,ir)×(-Ej0+2δ0,j1)⋯(-Ejs+2δ1,js)+Bi1⋯BirEj1⋯Ejs0t}
From (2.33), we can derive the following equation:
(2.34)bk=12(n+r+sk){∑0≤a≤r0≤c≤sa≥k+r-n‍(ra)(sc)(-1)c2s-c×∑t=0n+a-k-r‍∑i1+⋯+ia+j1+⋯+jc=t‍Bi1⋯BiaEj1⋯Ejc+∑i1+⋯+ir+j1+⋯+js=n-k‍Bi1⋯BirEj1⋯Ejs∑0≤a≤r0≤c≤sa≥k+r-n‍}.
Observe now that
(2.35)bn=12(n+r+sn){∑c=0s‍(sc)(-1)c2s-c×∑i1+⋯+ir+j1+⋯+jc=0‍Bi1⋯BirEj1⋯Ejc+∑i1+⋯+ir+j1+⋯+js=0‍Bi1⋯BirEj1⋯Ejs}=12(n+r+sn)[(2-1)s+1]=(n+r+sn),(2.36)bn-1=12(n+r+sn-1){∑r-1≤a≤r0≤c≤s‍(ra)(sc)(-1)c2s-c×∑t=01+a-r‍∑i1+⋯+ia+j1+⋯+jc=t‍Bi1⋯BiaEj1⋯Ejc+∑i1+⋯+ir+j1+⋯+js=1‍Bi1⋯BirEj1⋯Ejs∑r-1≤a≤r0≤c≤s‍}=12(n+r+sn-1){r+1-12r+12s-12r-12s}=12(n+r+sn-1).
Therefore, by (1.6), (2.31), (2.34), (2.35), and (2.36), we obtain the following theorem.Theorem 2.7.
Forn∈ℕ with n≥2, one has
(2.37)∑i1+⋯+ir+j1+⋯+js+t=n‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)xt=12∑k=0n-2‍(n+r+sk){∑0≤a≤r0≤c≤sa≥k+r-n‍(ra)(sc)(-1)c2s-c×∑t=0n+a-k-r‍∑i1+⋯+ia+j1+⋯+jc=t‍Bi1⋯BiaEj1⋯Ejc+∑i1+⋯+ir+j1+⋯+js=n-k‍Bi1⋯BirEj1⋯Ejs∑0≤a≤r0≤c≤sa≥k+r-n‍}Ek(x)+12(n+r+sn-1)En-1(x)+(n+r+sn)En(x).Let us consider the following polynomial of degreen. 						(2.38)p(x)=∑i1+⋯+ir+j1+⋯+js=n‍1i1!⋯ir!j1!⋯js!Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x).
Thus, by (2.38), we get
(2.39)p(k)(x)=(r+s)k∑i1+⋯+ir+j1+⋯+js=n-k‍1i1!⋯ir!j1!⋯js!×Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x).
From (1.7), we have
(2.40)bk=12k!{p(k)(1)+p(k)(0)}=(r+s)k2k!∑i1+⋯+ir+j1+⋯+js=n-k‍1i1!⋯ir!j1!⋯js!×{Bi1(1)⋯Bir(1)×Ej1(1)⋯Ejs(1)+Bi1⋯BirEj1⋯Ejs}=(r+s)k2k!∑i1+⋯+ir+j1+⋯+js=n-k‍1i1!⋯ir!j1!⋯js!×{(Bi1+δ1,i1)⋯(Bir+δ1,ir)×(-Ej1+2δ0,j1)⋯(-Ejs+2δ0,js)+Bi1⋯BirEj1⋯Ejs}.
Thus, by (2.40), we get
(2.41)bk=(r+s)k2k!{∑0≤a≤r0≤c≤sa≥k+r-n‍(ra)(sc)(-1)c2s-c×∑i1+⋯+ia+j1+⋯+jc=n+a-k-r‍Bi1⋯BiaEj1⋯Ejci1!⋯ia!j1!⋯jc!+∑i1+⋯+ir+j1+⋯+js=n-k‍Bi1⋯BirEj1⋯Ejsi1!⋯ir!j1!⋯js!∑0≤a≤r0≤c≤sa≥k+r-n‍}.
Now, we note that
(2.42)bn=(r+s)n2n!{∑i1+⋯+ir+j1+⋯+js=0‍Bi1⋯BirEj1⋯Ejsi1!⋯ir!j1!⋯js!∑c=0s‍(sc)(-1)c2s-c×∑i1+⋯+ir+j1+⋯+jc=0‍Bi1⋯BirEj1⋯Ejci1!⋯ir!j1!⋯jc!+∑i1+⋯+ir+j1+⋯+js=0‍Bi1⋯BirEj1⋯Ejsi1!⋯ir!j1!⋯js!}=(r+s)n2n![(2-1)s+1]=(r+s)nn!,bn-1=(r+s)n-12(n-1)!{∑r-1≤a≤r0≤c≤s‍(ra)(sc)(-1)c2s-c×∑i1+⋯+ia+j1+⋯+jc=1+a-r‍Bi1⋯BiaEj1⋯Ejci1!⋯ia!j1!⋯jc!+∑i1+⋯+ir+j1+⋯+js=1‍Bi1⋯BirEj1⋯Ejsi1!⋯ir!j1!⋯js!∑r-1≤a≤r0≤c≤s‍}=(r+s)n-12(n-1)!{r(2-1)s+∑c=0s‍(sc)(-1)c2s-c[-12(r+c)]-12(r+s)}=0.
Therefore, by (1.6), (2.38), (2.41), and (2.42), we obtain the following theorem.Theorem 2.8.
Forn∈ℕ with n≥2, one has
(2.43)∑i1+⋯+ir+j1+⋯+js=n‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)i1!⋯ir!j1!⋯js!=12∑k=0n-2‍(r+s)kk!{∑0≤a≤r0≤c≤sa≥k+r-n‍(ra)(sc)(-1)c2s-c×∑i1+⋯+ia+j1+⋯+jc=n+a-k-r‍Bi1⋯BiaEj1⋯Ejci1!⋯ia!j1!⋯jc!+∑i1+⋯+ir+j1+⋯+js=n-k‍Bi1⋯BirEj1⋯Ejsi1!⋯ir!j1!⋯js!∑0≤a≤r0≤c≤sa≥k+r-n‍}Ek(x)+(r+s)nn!En(x).By the same method, we can obtain the following identity:(2.44)∑i1+⋯+ir+j1+⋯+js+t=n‍Bi1(x)⋯Bir(x)Ej1(x)⋯Ejs(x)xti1!⋯ir!j1!⋯js!t!=12∑k=0n-2‍(r+s+1)kk!{∑0≤a≤r0≤c≤sa≥k+r-n‍(ra)(sc)(-1)c2s-c×∑t=0n+a-k-r‍1(n+a-k-r-t)!∑i1+⋯+ia+j1+⋯+jc=t‍Bi1⋯BiaEj1⋯Ejci1!⋯ia!j1!⋯jc!+∑i1+⋯+ir+j1+⋯+js=n-k‍Bi1⋯BirEj1⋯Ejsi1!⋯ir!j1!⋯js!∑0≤a≤r0≤c≤sa≥k+r-n‍}Ek(x)+(r+s+1)n-12(n-1)!En-1(x)+(r+s+1)nn!En(x).



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