Volume 2012, Article ID 343981,15pages doi:10.1155/2012/343981
Research Article
Euler Basis, Identities, and Their Applications
D. S. Kim
1and T. Kim
21Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea
2Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
Correspondence should be addressed to D. S. Kim,[email protected]
Received 11 June 2012; Accepted 9 August 2012
Academic Editor: Yilmaz Simsek
Copyrightq2012 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.
LetVn {px∈Qx|degpx≤n}be then1-dimensional vector space overQ. We show
that{E0x, E1x, . . . , Enx}is a good basis for the spaceVn, for our purpose of arithmetical and
combinatorial applications. Thus, ifpx ∈ Qxis of degree n, thenpx nl0blElx for
some uniquely determinedbl ∈Q. In this paper we develop method for computingblfrom the
information ofpx.
1. Introduction
The Euler polynomials,Enx, are given by
2
et1e
xteExt∞
n0
Enxtn
n!, 1.1
see1–20with the usual convention about replacingEnxbyEnx. In the special case, x 0, En0 Enare called thenth Euler numbers. The Bernoulli numbers are also defined by
t
et−1 e
Bt∞
n0
Bnt
n
n!, 1.2
see 1–20 with the usual convention about replacing Bn by Bn. As is well known, the
Bernoulli polynomials are given by
Bnx n
l0
n l
Blxn−l
n
l0
n l
see9–15From1.1,1.2, and1.3, we note that
Bn1−Bnδ1,n, En1 En2δ0,n, 1.4
whereδk,nis the kronecker symbol.
Letm, n∈Zwithmn≥2. The formula
BmxBnx
r
m
2r
n
n
2r
m
B2rBmn−2rx
mn−2r −1
m1m!n!Bmn
mn! , 1.5
is proved in 4–6. Let Vn {px ∈ Qx | degpx ≤ n} be the n1-dimensional vector space over Q. Probably, {1, x, . . . , xn} is the most natural basis for this space. But
{E0x, E1x, . . . , Enx}is also a good basis for the spaceVn, for our purpose of arithmetical
and combinatorial applications. Thus, ifpx∈Qxis of degreen, then
px n
l0
blElx, 1.6
for some uniquely determinedbl∈Q. Further,
bk 1
2k!
pk1 pk0 k0,1,2, . . . , n, 1.7
where pkx dkpx/dxk. In this paper we develop methods for computing bl
from the information of px. Apply these results to arithmetically and combinatorially interesting identities involvingE0x, E1x, . . . , Enx, B0x, . . . , Bnx. Finally, we give some applications of those obtained identities.
2. Euler Basis, Identities, and Their Applications
Let us takepxthe polynomial of degreenas follows:
px n
k0
BkxBn−kx. 2.1
From2.1, we have
pkx nn−1!
k1!
n
lk
By1.7and2.2, we get
bk 1
2k!
pk1 pk0
1
2
n1
k
n
lk
{Bl−kδ1,l−kBn−lδ1,n−l Bl−kBn−l},
2.3
Thus, we have
bk
n1
k
n
lk
Bl−kBn−lBn−k−1
, 0≤k≤n−3, 2.4
bn−2 7
72n
n2−1, bnn1, bn−10. 2.5
By1.6,2.1,2.3, and2.4, we get
n
k0
BkxBn−kx
n−3
k0
n1
k
n
lk
Bl−kBn−lBn−k−1
Ekx 7
72n
n2−1En−2x n1Enx.
2.6
Let us consider the following triple identities:
px
rstnBrxBsxBtx
n
k0
bkEkx, 2.7
where the sum runs over allr, s, t∈Zwithrstn. Thus, by2.7, we get
pkx n2n1nn−1· · ·n−k3
rstn−k
BrxBsxBtx. 2.8
From1.7and2.8, we have
bk 1
2k!
pk1 pk0
n2
k
2
rstn−k
{Br1Bs1Bt1 BrBsBt}
n2
k
2
2
rstn−k
BrBsBt
rstn−k
δ1,rBsBt
rstn−k
rstn−k
BrBsδ1,t
rstn−k
δ1,rδ1,sBt
rstn−k
δ1,rBsδ1,t
rstn−k
Brδ1,sδ1,t
rstn−k
δ1,rδ1,sδ1,t
.
2.9
Therefore, by2.7and2.9, we obtain the following theorem.
Theorem 2.1. Forr, s, t∈Z, andn∈Nwithn≥3, one has
rstn
BrxBsxBtx
1
2
n−2
k0
n2
k
2
rstn−k
BrBsBt3
rsn−k−1
BrBs3Bn−k−2δk,n−3
Ekx
n2
2
Enx.
2.10
Let us take the polynomialpxas follows:
px
rstn
BrxBsxEtx. 2.11
Then, by2.11, we get
pkx n2n1nn−1· · ·n−k3
rstn−k
BrxBsxEtx. 2.12
From1.6,1.7, and2.12, we have
bk 1
2k!
pk1 pk0
n2
k
2
rstn−k
{Br1Bs1Et1 BrBsEt}
n2
k
2
rstn−k
{Br δ1,rBsδ1,s−Et2δ0,t BrBsEt}
n2
k
2
−
rstn−k
δ1,rBsEt−
rstn−k
Brδ1,sEt2
rstn−k BrBsδ0,t
−
rstn−k
δ1,rδ1,sEt2
rstn−k
δ1,rBsδ0,t2
rstn−k
Brδ1,sδ0,t
2
rstn−k
δ1,rδ1,sδ0,t
.
Note that
bn−1
n2
n−1
−
st0
BsEt−
rt0
BrEt2
rs1
BrBs−02B02B02·0
1
2
n2
n−1
{−1−12B1B1 22}0.
2.14
Therefore, we obtain the following theorem.
Theorem 2.2. Forn∈Nwithn≥2, one has
rstn
BrxBsxEtx
1
2
n−2
k0
n2
k
2
rsn−k
BrBs−2
rtn−k−1
BrEt−En−k−24Bn−k−12δk,n−2
Ekx
n2
2
Enx.
2.15
Remark 2.3. By the same method, we obtain the following identities.
I
rstn
BrxEsxEtx
1
2
n−2
k0
n2
k
2
rstn−k
BrEsEt
stn−k−1
EsEt−4
rsn−k
BrEs4Bn−k−4En−k−1
Ekx
n2
2
Enx.
2.16
II
rstn
ErxEsxEtx
3
n−2
k0
n2
k rsn−kErEs−2En−k
Ekx
n2
2
Enx.
2.17
Let us consider the polynomialpxas follows:
px
rstn
BrxBsxxt.
Thus, by2.18, we get
pkx n2n1nn−1· · ·n−k3
rstn−k
BrxBsxxt. 2.19
From1.6,1.7,2.18, and2.19, we have
bk 1
2k!
pk1 pk0
n2
k
2
rstn−k
Br1Bs1 BrBs0t
n2
k
2
rstn−k
Brδ1,rBsδ1,s BrBs0t
n2
k
2
rstn−k
BrBs
rstn−k
Brδ1,s
rstn−k
δ1,rBs
rstn−k
δ1,rδ1,s
rstn−k
BrBs0t
.
2.20
Here we note that
rstn−k
BrBs
n−k
t0
rsn−k−t
BrBs
n−k
t0
rst
BrBs
rstn−k
Brδ1,s
⎧ ⎪ ⎨ ⎪ ⎩
n−k−1
r0 Br, if k≤n−1,
0, if kn,
rstn−k
Bsδ1,r
⎧ ⎪ ⎨ ⎪ ⎩
n−k−1
r0 Br, if k≤n−1,
0, if kn,
rstn−k
δ1,rδ1,s
1, ifk≤n−2,
0, ifkn−1 orn,
rstn−k
BrBs0t
rsn−k
BrBs, ∀k.
2.21
It is easy to show that
bn−1 1
2
n2
n−1 rs0BrBs2
rs1
BrBs2B0
1
2
n2
n−1
{12B1B2 2} 1
2
n2
n−1
.
2.22
Theorem 2.4. Forn∈Nwithn≥2, one has
rstn
BrxBsxxt
1
2
n−2
k0
n2
k
n−k−1
t0
rst
BrBs2
rsn−k
BrBs2
n−k−1
r0
Br 1
Ekx
1
2
n2
n−1
En−1x
n2
n
Enx.
2.23
Remark 2.5. By the same method, we can derive the following identities.
I
rstnBrxEsxx
t
1
2
n−2
k0
n2
k
−n−k−1
t0
rst
BrEs−
n−k−1
s0
Es2
n−k
r0
Br 2
Ekx
1
2
n2
n−1
En−1x
n2
n
Enx.
2.24
II
rstn
ErxEsxxt
1
2
n−2
k0
n2
k
n−k−1
t0
rst
ErEs2
rsn−k
ErEs−4
n−k
r0
Er4
Ekx
1
2
n2
n−1
En−1x
n2
2
Enx.
2.25
Now we generalize the above consideration to the completely arbitrary case. Let
px
i1···irj1···jsn
Bi1x· · ·BirxEj1x· · ·Ejsx, 2.26
where the sum runs over all nonnegative integersi1, i2, . . . , ir, j1, . . . , jssatisfyingi1i2· · ·
irj1· · ·jsn. From2.26, we note that
pkx nrs−1· · ·nrs−k
i1···irj1···jsn−k
Bi1x· · ·Birx×Ej1x· · ·Ejsx.
By1.6,1.7,2.18, and2.27, we get
bk 1
2k!
pk1 pk0
1
2
nrs−1
k
i1···irj1···jsn−k
Bi11· · ·Bir1Ej11· · ·Ejs1 Bi1· · ·BirEj1· · ·Ejs
1
2
nrs−1
k
×
i1···irj1···jsn−k
{Bi1δ1,i1· · ·Birδ1,ir
×−Ej12δ0,j1
· · ·−Ejs2δ0,js
Bi1· · ·BirEj1· · ·Ejs
1
2
nrs−1
k
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
0≤a≤r 0≤c≤s a≥kr−n
r a
s c
−1c2s−c ×
i1···iaj1···jcna−k−r
Bi1· · ·BiaEj1· · ·Ejc
i1···irj1···jsn−k
Bi1· · ·BirEj1· · ·Ejs
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ .
2.28
Note that
bn 1
2
nrs−1
n
⎧⎨
⎩
0≤c≤s
s c
−1c2s−c×
i1···irj1···jc0
Bi1· · ·BirEj1· · ·Ejc
i1···irj1···js0
Bi1· · ·BirEj1· · ·Ejs
⎫ ⎬ ⎭
1
2
nrs−1
n
2−1s1
nrs−1
n
,
bn−1 1
2
nrs−1
n−1
⎧⎪⎨
⎪ ⎩
r−1≤a≤r 0≤c≤s
r a
s c
−1c2s−c
×
i1···iaj1···jc1a−r
Bi1· · ·BiaEj1· · ·Ejc
i1···irj1···js1
Bi1· · ·BirEj1· · ·Ejs
1
2
nrs−1
n−1
r2−1s
0≤c≤s
s c
−1c2s−c
−1
2rc
−1
2rs
1
2
nrs−1
n−1
r−1
2r 1 2s−
1 2r−
1 2s
0.
2.29
Therefore, by1.6,2.28, and2.29, we obtain the following theorem.
Theorem 2.6. Forn∈Nwithn≥2, one has
i1···irj1···jsn
Bi1x· · ·BirxEj1x· · ·Ejsx
1
2
n−2
k0
nrs−1
k
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
0≤a≤r 0≤c≤s a≥kr−n
r a
s c
−1c2s−c
×
i1···iaj1···jcna−k−r
Bi1· · ·BiaEj1· · ·Ejc
i1···irj1···jsn−k
Bi1· · ·BirEj1· · ·Ejs
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭
Ekx
nrs−1
n
Enx.
2.30
Let us consider the polynomialpxof degreenas
px
ti1···irj1···jsn
Bi1x· · ·BirxEj1x· · ·Ejsxx
t.
2.31
Then, from2.31, we have
pkx nrsnrs−1· · ·nrs−k1
×
i1···irj1···jstn−k
Bi1x· · ·BirxEj1x· · ·Ejsxx
By1.7and2.32, we get
bk 1
2k!
pk1 pk0
1
2
nrs
k
i1···irj1···jstn−k
Bi11· · ·Bir1Ej11· · ·Ejs1 Bi1· · ·BirEj1· · ·Ejs0
t
1
2
nrs
k
×
i1···irj1···jstn−k
{Bi1δ1,i1· · ·Bir δ1,ir
×−Ej02δ0,j1
· · ·−Ejs2δ1,js
Bi1· · ·BirEj1· · ·Ejs0
t
2.33
From2.33, we can derive the following equation:
bk 1
2
nrs
k
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
0≤a≤r 0≤c≤s a≥kr−n
r a
s c
−1c2s−c×
na−k−r
t0
i1···iaj1···jct
Bi1· · ·BiaEj1· · ·Ejc
i1···irj1···jsn−k
Bi1· · ·BirEj1· · ·Ejs
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ .
2.34
Observe now that
bn 1
2
nrs
n
⎧⎨
⎩ s
c0
s c
−1c2s−c×
i1···irj1···jc0
Bi1· · ·BirEj1· · ·Ejc
i1···irj1···js0
Bi1· · ·BirEj1· · ·Ejs
⎫ ⎬ ⎭
1
2
nrs
n
2−1s1
nrs
n
,
bn−1 1
2
nrs
n−1
⎧⎪⎨
⎪ ⎩
r−1≤a≤r 0≤c≤s
r a
s c
−1c2s−c×
1a−r
t0
i1···iaj1···jct
Bi1· · ·BiaEj1· · ·Ejc
i1···irj1···js1
Bi1· · ·BirEj1· · ·Ejs
⎫ ⎪ ⎬ ⎪ ⎭
1
2
nrs
n−1
r1−1
2r 1 2s−
1 2r−
1 2s
1
2
nrs
n−1
.
2.36
Therefore, by1.6,2.31,2.34,2.35, and2.36, we obtain the following theorem.
Theorem 2.7. Forn∈Nwithn≥2, one has
i1···irj1···jstn
Bi1x· · ·BirxEj1x· · ·Ejsxx
t
1
2
n−2
k0
nrs
k
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
0≤a≤r 0≤c≤s a≥kr−n
r a
s c
−1c2s−c
×na−k−r
t0
i1···iaj1···jct
Bi1· · ·BiaEj1· · ·Ejc
i1···irj1···jsn−k
Bi1· · ·BirEj1· · ·Ejs
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭
Ekx
1
2
nrs
n−1
En−1x
nrs
n
Enx.
2.37
Let us consider the following polynomial of degreen.
px
i1···irj1···jsn
1
i1!· · ·ir!j1!· · ·js!Bi1x· · ·BirxEj1x· · ·Ejsx. 2.38
Thus, by2.38, we get
pkx rsk
i1···irj1···jsn−k
1
From1.7, we have
bk 1
2k!
pk1 pk0
rsk
2k!
i1···irj1···jsn−k
1
i1!· · ·ir!j1!· · ·js!
×Bi11· · ·Bir1×Ej11· · ·Ejs1 Bi1· · ·BirEj1· · ·Ejs
rsk
2k!
i1···irj1···jsn−k
1
i1!· · ·ir!j1!· · ·js!
×Bi1δ1,i1· · ·Birδ1,ir×
−Ej12δ0,j1
· · ·−Ejs2δ0,js
Bi1· · ·BirEj1· · ·Ejs
.
2.40
Thus, by2.40, we get
bk rs
k
2k!
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
0≤a≤r 0≤c≤s a≥kr−n
r a
s c
−1c2s−c×
i1···iaj1···jcna−k−r
Bi1· · ·BiaEj1· · ·Ejc
i1!· · ·ia!j1!· · ·jc!
i1···irj1···jsn−k
Bi1· · ·BirEj1· · ·Ejs
i1!· · ·ir!j1!· · ·js!
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ .
2.41
Now, we note that
bn rs
n
2n!
⎧ ⎨ ⎩ s
c0
s
c
−1c2s−c
×
i1···irj1···jc0
Bi1· · ·BirEj1· · ·Ejc
i1!· · ·ir!j1!· · ·jc!
i1···irj1···js0
Bi1· · ·BirEj1· · ·Ejs
i1!· · ·ir!j1!· · ·js!
⎫ ⎬ ⎭
rsn
2n!
2−1s1 rs
n
bn−1 rs n−1
2n−1!
⎧ ⎪ ⎨ ⎪ ⎩ r−1≤a≤r 0≤c≤s r a s c
−1c2s−c
×
i1···iaj1···jc1a−r
Bi1· · ·BiaEj1· · ·Ejc
i1!· · ·ia!j1!· · ·jc!
i1···irj1···js1
Bi1· · ·BirEj1· · ·Ejs
i1!· · ·ir!j1!· · ·js!
⎫ ⎪ ⎬ ⎪ ⎭
rsn−1
2n−1!
r2−1s
s c0 s c
−1c2s−c
−1
2rc
−1
2rs
0.
2.42
Therefore, by1.6,2.38,2.41, and2.42, we obtain the following theorem.
Theorem 2.8. Forn∈Nwithn≥2, one has
i1···irj1···jsn
Bi1x· · ·BirxEj1x· · ·Ejsx
i1!· · ·ir!j1!· · ·js!
1
2
n−2
k0
rsk
k! ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 0≤a≤r 0≤c≤s a≥kr−n r a s c
−1c2s−c×
i1···iaj1···jcna−k−r
Bi1· · ·BiaEj1· · ·Ejc
i1!· · ·ia!j1!· · ·jc!
i1···irj1···jsn−k
Bi1· · ·BirEj1· · ·Ejs
i1!· · ·ir!j1!· · ·js!
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ Ekx
rsn
n! Enx.
2.43
By the same method, we can obtain the following identity:
i1···irj1···jstn
Bi1x· · ·BirxEj1x· · ·Ejsxxt
i1!· · ·ir!j1!· · ·js!t!
1
2
n−2
k0
rs1k
k! ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 0≤a≤r 0≤c≤s a≥kr−n r a s c
×na−k−r t0
1
na−k−r−t!
i1···iaj1···jct
Bi1· · ·BiaEj1· · ·Ejc
i1!· · ·ia!j1!· · ·jc!
i1···irj1···jsn−k
Bi1· · ·BirEj1· · ·Ejs
i1!· · ·ir!j1!· · ·js!
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭
Ekx
rs1n−1
2n−1! En−1x
rs1n
n! Enx.
2.44
Acknowledgments
This paper was supported by Basic Science Research Program through the National Research Foundation of Korea NRFfunded by the Ministry of Education, Science and Technology 2012R1A1A2003786.
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