Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2012, Article ID 343981, 15 pages
doi:10.1155/2012/343981
Research Article
Euler Basis, Identities, and Their Applications
D. S. Kim1 and T. Kim2
1
2
Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea
Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
Correspondence should be addressed to D. S. Kim, dskim@sogang.ac.kr
Received 11 June 2012; Accepted 9 August 2012
Academic Editor: Yilmaz Simsek
Copyright q 2012 D. S. Kim and T. Kim. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
Let Vn {px ∈ Qx| deg px ≤ n} be the n 1-dimensional vector space over Q. We show
and
that {E0 x, E1 x, . . . , En x} is a good basis for the space Vn , for our purpose of arithmetical
n
combinatorial applications. Thus, if px ∈ Qx is of degree n, then px b
E
x
for
l
l
l0
some uniquely determined bl ∈ Q. In this paper we develop method for computing bl from the
information of px.
1. Introduction
The Euler polynomials, En x, are given by
∞
tn
2
xt
Ext
,
e
e
E
x
n
n!
et 1
n0
1.1
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
∞
t
tn
Bt
,
e
B
n
n!
et − 1
n0
1.2
see 1–20 with the usual convention about replacing Bn by Bn . As is well known, the
Bernoulli polynomials are given by
Bn x n n n
n
Bl xn−l Bn−l xl ,
l
l
l0
l0
1.3
2
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see 9–15 From 1.1, 1.2, and 1.3, we note that
Bn 1 − Bn δ1,n ,
En 1 En 2δ0,n ,
1.4
where δk,n is the kronecker symbol.
Let m, n ∈ Z with m n ≥ 2. The formula
Bm xBn x m r
m!n!Bmn
B2r Bmn−2r x
n
−1m1
,
n
m
2r
2r
m n − 2r
m n!
1.5
is proved in 4–6. Let Vn {px ∈ Qx | deg px ≤ 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 px ∈ Qx is of degree n, then
px n
bl El x,
1.6
l0
for some uniquely determined bl ∈ Q. Further,
bk 1 k
p 1 pk 0
2k!
k 0, 1, 2, . . . , n,
1.7
where pk x dk px/dxk . In this paper we develop methods for computing bl
from the information of px. 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 take px the polynomial of degree n as follows:
px n
Bk xBn−k x.
2.1
k0
From 2.1, we have
pk x n
n 1! Bl−k xBn−l x.
n − k 1! lk
2.2
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3
By 1.7 and 2.2, we get
1 k
p 1 pk 0
2k!
n
1 n1 {Bl−k δ1,l−k Bn−l δ1,n−l Bl−k Bn−l },
k
2
lk
bk 2.3
Thus, we have
bk n
n1
Bl−k Bn−l Bn−k−1 ,
k
lk
bn−2 7 2
n n −1 ,
72
0 ≤ k ≤ n − 3,
2.4
bn−1 0.
2.5
bn n 1,
By 1.6, 2.1, 2.3, and 2.4, we get
n
Bk xBn−k x
k0
n−3 n
7 n1
Bl−k Bn−l Bn−k−1 Ek x n n2 − 1 En−2 x n 1En x.
k
72
k0
lk
2.6
Let us consider the following triple identities:
px Br xBs xBt x rstn
n
bk Ek x,
2.7
k0
where the sum runs over all r, s, t ∈ Z with r s t n. Thus, by 2.7, we get
pk x n 2n 1nn − 1 · · · n − k 3
Br xBs xBt x.
rstn−k
From 1.7 and 2.8, we have
1 k
p 1 pk 0
2k!
n2 k
{Br 1Bs 1Bt 1 Br Bs Bt }
2 rstn−k
n2 2
k
Br Bs Bt δ1,r Bs Bt Br δ1,s Bt
2
rstn−k
rstn−k
rstn−k
bk 2.8
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Br Bs δ1,t δ1,r δ1,s Bt δ1,r Bs δ1,t
rstn−k
rstn−k
Br δ1,s δ1,t rstn−k
rstn−k
δ1,r δ1,s δ1,t .
rstn−k
2.9
Therefore, by 2.7 and 2.9, we obtain the following theorem.
Theorem 2.1. For r, s, t ∈ Z , and n ∈ N with n ≥ 3, one has
Br xBs xBt x
rstn
n−2 1
n2
Br Bs Bt 3
Br Bs 3Bn−k−2 δk,n−3 Ek x
2
k
2 k0
rstn−k
rsn−k−1
2.10
n2
En x.
2
Let us take the polynomial px as follows:
px Br xBs xEt x.
2.11
rstn
Then, by 2.11, we get
pk x n 2n 1nn − 1 · · · n − k 3
Br xBs xEt x.
rstn−k
2.12
From 1.6, 1.7, and 2.12, we have
n2 1 k
k
p 1 p 0 k
bk {Br 1Bs 1Et 1 Br Bs Et }
2k!
2 rstn−k
n2 k
{Br δ1,r Bs δ1,s −Et 2δ0,t Br Bs Et }
2 rstn−k
n2 −
k
δ1,r Bs Et −
Br δ1,s Et 2
Br Bs δ0,t
2
rstn−k
rstn−k
rstn−k
−
δ1,r δ1,s Et 2
δ1,r Bs δ0,t 2
Br δ1,s δ0,t
rstn−k
2
rstn−k
rstn−k
δ1,r δ1,s δ0,t .
rstn−k
2.13
International Journal of Mathematics and Mathematical Sciences
Note that
bn−1
n2
Bs Et −
Br Et 2
Br Bs − 0 2B0 2B0 2 · 0
−
n−1
st0
rt0
rs1
1 n2
{−1 − 1 2B1 B1 2 2} 0.
2 n−1
5
2.14
Therefore, we obtain the following theorem.
Theorem 2.2. For n ∈ N with n ≥ 2, one has
Br xBs xEt x
rstn
n−2 1
n2
Br Bs − 2
Br Et − En−k−2 4Bn−k−1 2δk,n−2 Ek x
2
k
2 k0
rsn−k
rtn−k−1
n2
En x.
2
2.15
Remark 2.3. By the same method, we obtain the following identities.
I
Br xEs xEt x
rstn
n−2 1
n2
Br Es Et Es Et − 4
Br Es 4Bn−k − 4En−k−1 Ek x
2
k
2 k0
rstn−k
stn−k−1
rsn−k
n2
En x.
2
2.16
II
Er xEs xEt x
rstn
n−2 n2
n2
Er Es − 2En−k Ek x 3
En x.
k
2
k0
rsn−k
2.17
Let us consider the polynomial px as follows:
px Br xBs xxt .
rstn
2.18
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International Journal of Mathematics and Mathematical Sciences
Thus, by 2.18, we get
pk x n 2n 1nn − 1 · · · n − k 3
Br xBs xxt .
rstn−k
2.19
From 1.6, 1.7, 2.18, and 2.19, we have
n2 1 k
k
p 1 p 0 k
bk Br 1Bs 1 Br Bs 0t
2k!
2 rstn−k
n2 k
Br δ1,r Bs δ1,s Br Bs 0t
2 rstn−k
n2 k
t
Br Bs Br δ1,s δ1,r Bs δ1,r δ1,s Br Bs 0 .
2
rstn−k
rstn−k
rstn−k
rstn−k
rstn−k
2.20
Here we note that
Br Bs rstn−k
n−k
Br Bs t0 rsn−k−t
n−k Br Bs
t0 rst
⎧
n−k−1
⎪
⎨ B , if k ≤ n − 1,
r
Br δ1,s r0
⎪
⎩0,
rstn−k
if k n,
⎧
n−k−1
⎪
⎨ B , if k ≤ n − 1,
r
Bs δ1,r r0
⎪
⎩
rstn−k
0,
if k n,
1, if k ≤ n − 2,
δ1,r δ1,s 0, if k n − 1 or n,
rstn−k
Br Bs 0t Br Bs , ∀k.
rstn−k
2.21
rsn−k
It is easy to show that
bn−1
1 n2
Br Bs 2
Br Bs 2B0
2 n−1
rs0
rs1
1 n2
1 n2
.
{1 2B1 B2 2} 2 n−1
2 n−1
Therefore, by 1.6, 2.18, 2.20, and 2.22, we obtain the following theorem.
2.22
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7
Theorem 2.4. For n ∈ N with n ≥ 2, one has
Br xBs xxt
rstn
n−k−1
n−2 n−k−1
1
n2
Br Bs 2
Br Bs 2
Br 1 Ek x
k
2 k0
t0 rst
r0
rsn−k
2.23
1 n2
n2
En−1 x En x.
n
2 n−1
Remark 2.5. By the same method, we can derive the following identities.
I
Br xEs xxt
rstn
n−k−1
n−2 n−k−1
n−k
1
n2
−
Br Es −
Es 2 Br 2 Ek x
k
2 k0
t0 rst
s0
r0
2.24
1 n2
n2
En x.
En−1 x n
2 n−1
II
Er xEs xxt
rstn
n−k−1
n−2 n−k
1
n2
Er Es 2
Er Es − 4 Er 4 Ek x
k
2 k0
t0 rst
r0
rsn−k
2.25
1 n2
n2
En x.
En−1 x 2
2 n−1
Now we generalize the above consideration to the completely arbitrary case. Let
px Bi1 x · · · Bir xEj1 x · · · Ejs x,
i1 ···ir j1 ···js n
2.26
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
pk x n r s − 1 · · · n r s − k
i1 ···ir j1 ···js n−k
Bi1 x · · · Bir x × Ej1 x · · · Ejs x.
2.27
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International Journal of Mathematics and Mathematical Sciences
By 1.6, 1.7, 2.18, and 2.27, we get
1 k
p 1 pk 0
2k!
1 nrs−1
Bi1 1 · · · Bir 1Ej1 1 · · · Ejs 1 Bi1 · · · Bir Ej1 · · · Ejs
k
2
i1 ···ir j1 ···js n−k
1 nrs−1
k
2
×
{Bi1 δ1,i1 · · · Bir δ1,ir bk i1 ···ir j1 ···js n−k
× −Ej1 2δ0,j1 · · · −Ejs 2δ0,js Bi1 · · · Bir Ej1 · · · Ejs
⎧
⎪
⎪
⎪
⎪
1 nrs−1 ⎨ r
s
Bi1 · · · Bia Ej1 · · · Ejc
−1c 2s−c ×
⎪
k
c
2
⎪ 0≤a≤r a
i1 ···ia j1 ···jc na−k−r
⎪
⎪
⎩ 0≤c≤s
a≥kr−n
i1 ···ir j1 ···js n−k
Bi1 · · · Bir Ej1 · · · Ejs
⎫
⎪
⎪
⎪
⎪
⎬
.
⎪
⎪
⎪
⎪
⎭
2.28
Note that
⎧
1 nrs−1 ⎨ s
Bi1 · · · Bir Ej1 · · · Ejc
bn −1c 2s−c ×
⎩0≤c≤s c
n
2
i1 ···ir j1 ···jc 0
Bi1 · · · Bir Ej1 · · · Ejs
i1 ···ir j1 ···js 0
⎫
⎬
⎭
1 nrs−1 nrs−1
,
2 − 1s 1 n
n
2
⎧
⎪
1 nrs−1 ⎨ r
s
−1c 2s−c
⎪
a
n−1
c
2
⎩r−1≤a≤r
bn−1
0≤c≤s
×
i1 ···ia j1 ···jc 1a−r
Bi1 · · · Bia Ej1 · · · Ejc
Bi1 · · · Bir Ej1 · · · Ejs
i1 ···ir j1 ···js 1
⎫
⎪
⎬
⎪
⎭
International Journal of Mathematics and Mathematical Sciences
s
1 nrs−1
1
1
s
c s−c
− r c − r s
r2 − 1 −1 2
c
n−1
2
2
2
0≤c≤s
9
1 nrs−1
1
1
1
1
r − r s − r − s 0.
n−1
2
2
2
2
2
2.29
Therefore, by 1.6, 2.28, and 2.29, we obtain the following theorem.
Theorem 2.6. For n ∈ N with n ≥ 2, one has
Bi1 x · · · Bir xEj1 x · · · Ejs x
i1 ···ir j1 ···js n
⎧
⎪
⎪
⎪
⎪
n−2
1 n r s − 1 ⎨ r
s
−1c 2s−c
⎪
k
a
c
2 k0
⎪
0≤a≤r
⎪
⎪
⎩ 0≤c≤s
a≥kr−n
×
i1 ···ia j1 ···jc na−k−r
i1 ···ir j1 ···js n−k
Bi1 · · · Bia Ej1 · · · Ejc
Bi1 · · · Bir Ej1 · · · Ejs
2.30
⎫
⎪
⎪
⎪
⎪
⎬
Ek x
⎪
⎪
⎪
⎪
⎭
nrs−1
En x.
n
Let us consider the polynomial px of degree n as
px Bi1 x · · · Bir xEj1 x · · · Ejs xxt .
ti1 ···ir j1 ···js n
2.31
Then, from 2.31, we have
pk x n r sn r s − 1 · · · n r s − k 1
Bi1 x · · · Bir xEj1 x · · · Ejs xxt .
×
i1 ···ir j1 ···js tn−k
2.32
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International Journal of Mathematics and Mathematical Sciences
By 1.7 and 2.32, we get
bk 1 k
p 1 pk 0
2k!
1 nrs
k
2
i
1 ···ir j1 ···js tn−k
Bi1 1 · · · Bir 1Ej1 1 · · · Ejs 1 Bi1 · · · Bir Ej1 · · · Ejs 0t
1 nrs
k
2
×
i1 ···ir j1 ···js tn−k
{Bi1 δ1,i1 · · · Bir δ1,ir × −Ej0 2δ0,j1 · · · −Ejs 2δ1,js Bi1 · · · Bir Ej1 · · · Ejs 0t
2.33
From 2.33, we can derive the following equation:
⎧
⎪
⎪
⎪
⎪
na−k−r
1 nrs ⎨ r
s
bk Bi1 · · · Bia Ej1 · · · Ejc
−1c 2s−c ×
⎪
k
a
c
2
⎪
0≤a≤r
t0 i1 ···ia j1 ···jc t
⎪
⎪
⎩ 0≤c≤s
a≥kr−n
i1 ···ir j1 ···js n−k
Bi1 · · · Bir Ej1 · · · Ejs
⎫
⎪
⎪
⎪
⎪
⎬
.
⎪
⎪
⎪
⎪
⎭
2.34
Observe now that
⎧
s 1 n r s ⎨
s
bn Bi1 · · · Bir Ej1 · · · Ejc
−1c 2s−c ×
⎩ c0 c
n
2
i1 ···ir j1 ···jc 0
Bi1 · · · Bir Ej1 · · · Ejs
i1 ···ir j1 ···js 0
1 nrs nrs
,
2 − 1s 1 n
n
2
⎫
⎬
⎭
2.35
International Journal of Mathematics and Mathematical Sciences
11
⎧
⎪
1a−r
1 nrs ⎨ r
s
bn−1 Bi1 · · · Bia Ej1 · · · Ejc
−1c 2s−c ×
⎪
c
n−1
2
⎩r−1≤a≤r a
t0 i ···i j ···j t
1
0≤c≤s
Bi1 · · · Bir Ej1 · · · Ejs
i1 ···ir j1 ···js 1
a
1
c
⎫
⎪
⎬
⎪
⎭
1 nrs
1
1
1
1 nrs
1
r1− r s− r− s .
n−1
n−1
2
2
2
2
2
2
2.36
Therefore, by 1.6, 2.31, 2.34, 2.35, and 2.36, we obtain the following theorem.
Theorem 2.7. For n ∈ N with n ≥ 2, one has
Bi1 x · · · Bir xEj1 x · · · Ejs xxt
i1 ···ir j1 ···js tn
⎧
⎪
⎪
⎪
⎪
n−2 1
nrs ⎨ r
s
−1c 2s−c
⎪ 0≤a≤r a
k
c
2 k0
⎪
⎪
⎪
⎩ 0≤c≤s
a≥kr−n
×
na−k−r
t0
i1 ···ia j1 ···jc t
i1 ···ir j1 ···js n−k
Bi1 · · · Bia Ej1 · · · Ejc
Bi1 · · · Bir Ej1 · · · Ejs
2.37
⎫
⎪
⎪
⎪
⎪
⎬
Ek x
⎪
⎪
⎪
⎪
⎭
1 nrs
nrs
En x.
En−1 x n
n−1
2
Let us consider the following polynomial of degree n.
px 1
Bi x · · · Bir xEj1 x · · · Ejs x.
i ! · · · ir !j1 ! · · · js ! 1
i1 ···ir j1 ···js n 1
2.38
Thus, by 2.38, we get
pk x r sk
i1 ···ir j1
1
× Bi1 x · · · Bir xEj1 x · · · Ejs x.
i
!
·
·
·
i
!j
r 1 ! · · · js !
···j n−k 1
s
2.39
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International Journal of Mathematics and Mathematical Sciences
From 1.7, we have
bk 1 k
p 1 pk 0
2k!
1
r sk
2k! i ···i j ···j n−k i1 ! · · · ir !j1 ! · · · js !
1
r
1
s
× Bi1 1 · · · Bir 1 × Ej1 1 · · · Ejs 1 Bi1 · · · Bir Ej1 · · · Ejs
1
r sk
2k! i ···i j ···j n−k i1 ! · · · ir !j1 ! · · · js !
1
r
1
s
× Bi1 δ1,i1 · · · Bir δ1,ir × −Ej1 2δ0,j1 · · · −Ejs 2δ0,js Bi1 · · · Bir Ej1 · · · Ejs .
2.40
Thus, by 2.40, we get
⎧
⎪
⎪
⎪
k⎪
Bi1 · · · Bia Ej1 · · · Ejc
r s ⎨ r
s
bk −1c 2s−c ×
⎪
a
c
2k! ⎪
i ! · · · ia !j1 ! · · · jc !
0≤a≤r
i1 ···ia j1 ···jc na−k−r 1
⎪
⎪
⎩ 0≤c≤s
a≥kr−n
⎫
⎪
⎪
⎪
⎪
Bi1 · · · Bir Ej1 · · · Ejs ⎬
.
i ! · · · ir !j1 ! · · · js ! ⎪
⎪
i1 ···ir j1 ···js n−k 1
⎪
⎪
⎭
Now, we note that
⎧
s r sn ⎨
s
bn −1c 2s−c
2n! ⎩ c0 c
⎫
Bi1 · · · Bir Ej1 · · · Ejc
Bi1 · · · Bir Ej1 · · · Ejs ⎬
×
i ! · · · ir !j1 ! · · · jc !
i ! · · · ir !j1 ! · · · js ! ⎭
i1 ···ir j1 ···jc 0 1
i1 ···ir j1 ···js 0 1
r sn
r sn ,
2 − 1s 1 2n!
n!
2.41
International Journal of Mathematics and Mathematical Sciences
⎧
n−1 ⎪
r s ⎨ r
s
bn−1 −1c 2s−c
c
2n − 1! ⎪
⎩r−1≤a≤r a
13
0≤c≤s
⎫
⎪
Bi1 · · · Bia Ej1 · · · Ejc
Bi1 · · · Bir Ej1 · · · Ejs ⎬
×
i ! · · · ia !j1 ! · · · jc !
i ! · · · ir !j1 ! · · · js ! ⎪
⎭
i1 ···ia j1 ···jc 1a−r 1
i1 ···ir j1 ···js 1 1
s 1
1
r sn−1
s
s
c s−c
r2 − 1 − r c − r s 0.
−1 2
c
2n − 1!
2
2
c0
2.42
Therefore, by 1.6, 2.38, 2.41, and 2.42, we obtain the following theorem.
Theorem 2.8. For n ∈ N with n ≥ 2, one has
Bi1 x · · · Bir xEj1 x · · · Ejs x
i1 ! · · · ir !j1 ! · · · js !
i1 ···ir j1 ···js n
⎧
⎪
⎪
⎪
k⎪
n−2
Bi1 · · · Bia Ej1 · · · Ejc
1 r s ⎨ r
s
−1c 2s−c ×
⎪ 0≤a≤r a
c
2 k0 k! ⎪
i ! · · · ia !j1 ! · · · jc !
i1 ···ia j1 ···jc na−k−r 1
⎪
⎪
⎩ 0≤c≤s
a≥kr−n
⎫
⎪
⎪
⎪
⎪
Bi1 · · · Bir Ej1 · · · Ejs ⎬
Ek x
i ! · · · ir !j1 ! · · · js ! ⎪
⎪
i1 ···ir j1 ···js n−k 1
⎪
⎪
⎭
r sn
En x.
n!
2.43
By the same method, we can obtain the following identity:
Bi1 x · · · Bir xEj1 x · · · Ejs xxt
i1 ! · · · ir !j1 ! · · · js !t!
i1 ···ir j1 ···js tn
⎧
⎪
⎪
⎪
k⎪
n−2
1 r s 1 ⎨ r
s
−1c 2s−c
⎪
a
c
2 k0
k!
⎪ 0≤a≤r
⎪
⎪
⎩ 0≤c≤s
a≥kr−n
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International Journal of Mathematics and Mathematical Sciences
×
na−k−r
t0
Bi1 · · · Bia Ej1 · · · Ejc
1
i ! · · · ia !j1 ! · · · jc !
a
−
k
−
r
−
t!
n
i1 ···ia j1 ···jc t 1
⎫
⎪
⎪
⎪
⎪
Bi1 · · · Bir Ej1 · · · Ejs ⎬
E x
⎪ k
i ! · · · ir !j1 ! · · · js ! ⎪
i1 ···ir j1 ···js n−k 1
⎪
⎪
⎭
r s 1n−1
r s 1n
En−1 x En x.
2n − 1!
n!
2.44
Acknowledgments
This paper was supported by Basic Science Research Program through the National Research
Foundation of Korea NRF funded by the Ministry of Education, Science and Technology
2012R1A1A2003786.
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