New Convolution Identities for Hypergeometric Bernoulli Polynomials

0.19 MB·

by  Hieu D. Nguyen

NEW CONVOLUTION IDENTITIES FOR
HYPERGEOMETRIC BERNOULLI POLYNOMIALS
HIEUD.NGUYENANDLONGG.CHEONG
Abstract. NewconvolutionidentitiesofhypergeometricBernoullipolynomialsarepresented. Twodifferent
4 approaches to proving these identities are discussed, corresponding to the two equivalent definitions of
1 hypergeometricBernoullipolynomialsasAppellsequences.
0
2

n
a
1. Introduction
J

3 It is well known that the Euler-Maclaurin Summation (EMS) formula given by
1
n n 1 ∞ B
] f(k)= f(x)dx+ [f(n)+f(0)]+ k f(k−1)(n)−f(k−1)(0) (1)
T Xk=0 Z0 2 Xk=2 k! h i
N
isextremelyusefulforapproximatingsums andintegralsandforderivingspecialformulas. Here, B arethe
n
h. Bernoulli numbers defined by the exponential generating function
at t ∞ tn
m et−1 = Bnn!
n=0
[ X
For example, if we set f(x) = xp in (1) and use the fact that B0 = 1 and B1 = −1/2, then we obtain the
1
classical sums of powers formula first discovered by Jacob Bernoulli:
v
0 n p
p!
7 kp =np+ B np+1−k
k
9 k!(p−k+1)!
k=1 k=0
2 X X
Consider next the special case of the EMS formula where n=1, which we shall write in the form
.
1
0 1 1 ∞ B
f(x)dx= [f(1)+f(0)]− k f(k−1)(1)−f(k−1)(0) (2)
4
2 k!
1 Z0 Xk=2 h i
v: If we again set f(x) = xn in (2), then we obtain the classic Bernoulli number identity first discovered by
i Euler:
X
n
n+1
r Bk =0
a k
k=0(cid:18) (cid:19)
X
It is natural to ask if other Bernoulli number identities can be obtained by substitution. For example, is
there a function f(x) which when substituted into (2) will yield the following quadratic identity?
n+1
n+1
B B =−(n+1)B −nB (3)
k n−k+1 n n+1
k
k=0(cid:18) (cid:19)
X
The answer,not surprisingly,is yes. The surprise howeveris the choice for f(x). It is clearthat f(x) should
involvethe Bernoullinumbers since (3) contains products of Bernoullinumbers. Therefore,a naturalchoice
forf(x)wouldbetosetitequaltoaBernoullipolynomial,sayB (x). However,thereaderwilldiscoverthat
n
substituting f(x)=B (x) into (2) yields the trivial identity. The correct answer is f(x)=(1−x)B (x).
n n
Date:11-14-2013.
2000 Mathematics Subject Classification. Primary11B68.
Keywords and phrases. hypergeometricBernoullipolynomial,Appellsequence, convolution,sumsofproducts.
1
TheBernoullipolynomialsB (x)giveanexampleofanAppellsequence. Assuch,therearetwoequivalent
n
definitions for B (x): one via the exponential generating function
n
text ∞ tn
= B (x) (4)
et−1 n n!
n=0
X
and the other as a polynomial sequence with the following properties:
B (x)=1,
0
B ′(x)=nB (x),
n n−1
(5)
1 1 if n=0
B (x)dx=δ ≡
n n
Z0 (0 if n6=0
where δ is the Kronecker delta function. In either approach the Bernoulli numbers can be computed as
n
the evaluation B =B (0). Beginning with the second approach, we shall demonstrate that this technique
n n
of substitution to obtain Bernoulli number identities can be extended to Appell sequences by generalizing
the EMS formula. This was achieved by recognizing that (2) can be derived from the following repeated
integration by parts formula
p
f(x)g(x)dx= (−1)kf(k)(x)g−(k+1)(x)+(−1)−(p+1) f(p+1)(x)g−(p+1)(x)dx (6)
Z k=0 Z
X
where f(k)(x) denotes the k-th derivative of f(x) and g−(k)(x) denotes the k-th integral (or anti-derivative)
of g(x). In particular, if we set g(x) = B (x) in (6), then g−(k)(x) = B /k! because of the derivative
0 k
property in (5). Then integrating over the interval [0,1] and using the fact that B =−1/2, B =0 and
1 2k+1
B (1)=(−1)kB (0) for k ≥1, we find that (6) reduces to (2) in the limit where p→∞.
k k
Let q (x) be an Appell sequence, i.e. q (x)= 1 and q′(x) =nq′ (x). Then setting g(x)= q (x) in (6)
n 0 n n−1 0
yields the following generalized EMS formula:
p
f(x)dx= (−1)kf(k)(x)g−(k+1)(x)+(−1)−(p+1) f(p+1)(x)g−(p+1)(x)dx (7)
Z k=0 Z
X
More interestingly, if we also set f(x) = p (x) to be another Appell sequence in (7), then we obtain the
n
following convolution identity (Theorem 1) for any pair of Appell sequences p (x) and q (x):
n n
n
n
(−1)k p (x)q (x)=c (8)
n−k k n
k
k=0 (cid:18) (cid:19)
X
where {c } are constants independent of x.
n
Equation (8) is our starting point for deriving new identities. We shall prove (8) in the next section and
apply it to a class of generalized Bernoulli polynomials known as hypergeometric Bernoulli polynomials,
defined by equation (12) (see also [2]), in order to obtain new convolution identities. For example, the
following sums of products formula holds for any two hypergeometric Bernoulli numbers B and B of
M,n N,n
order M and N, respectively, which we prove in Theorem 5:
n
n
(−1)k B B =
N,n−k−j M,k−i
j;i;k−i
" #
0≤i≤M−1 k=0 (cid:18) (cid:19)
X X
0≤j≤N−1
(i,j)6=(0,0)
N−1 M−1
M +N n n
(−1)M−1 δ + B +(−1)n−M−N B (9)
n−M−N N,n−M−j M,n−N−i
 M M;j N;i 
(cid:18) (cid:19) j=0 (cid:18) (cid:19) i=0 (cid:18) (cid:19)
X X
 2 
where a denotes the multinomial coefficient defined by
b;c
(cid:0) (cid:1) a a!
=
b;c b!c!(a−b−c)!
(cid:18) (cid:19)
In the special case where M =N =2, we obtain the identities
n
n 1
(−1)k B (x)B (x)= δ +nB +B
2,n−k 2,k n−2 2,n−1 2,n
k 2
k=0 (cid:18) (cid:19)
X
and
n
2n 1
B B = [δ +2nB −(2n−4)B ]
2,2n−2k 2,2k 2n−2 2,2n−1 2,2n
2k 4
k=0(cid:18) (cid:19)
X
whichweproveinCorollaries9and11,respectively. ThelatterformulageneralizesEuler’squadraticformula
for the classical Bernoulli numbers:
n
2n
B B =−(2n−1)B
2k 2n−2k 2n
2k
k=0(cid:18) (cid:19)
X
Lastly,insection3wedemonstratethatthesesameidentitiescanbederivedbytakingtheotherapproach
to hypergeometricBernoullipolynomials,namelybyconsideringtheirexponentialgeneratingfunctions,and
employingspecialpartialfractionexpansionformulasgiveninLemma13. Asaresult,weobtainthefollowing
identities (Theorem 15):
M+N−2 n−m
n M +N
m!a (−1)k B (x)B (x)=(−1)M δ
m N,n−m−k M,k n−M−N
m;k M
m=1 k=0 (cid:18) (cid:19) (cid:18) (cid:19)
X X
N−1 M−1
n n
+(−1)M B +(−1)n−N B (10)
N,n−m−M M,n−m−N
m;M m;N
m=0(cid:18) (cid:19) m=0(cid:18) (cid:19)
X X
and if N ≥M,
N−1 n−m
n n n
B (x )B (x )= B (z)− B (z) (11)
N,n−m−k 1 M,k 2 N,n−M M,n−N
m;k M N
m=M k=0 (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X X
wherez =x +x . Observethat(10)isequivalentto(9). Moreover,(11)istrivialwhenM =N. Toremedy
1 2
this, we employ a derivative expansion formula to establish a recurrence formula for sums of products of
hypergeometric Bernoulli polynomials. We then use this recurrence to obtain a general formula for these
sums of products, thereby generalizingKamano’sformula for hypergeometricBernoullinumbers givenin[4]
and Dilcher’s formula for Bernoulli polynomials given in [1]. An example is the identity
n
n 1
B (x )B (x )= [(N −n)B (z)+n(z−1)B (z)]
N,n−k 1 N,k 2 N,n N,n−1
k N
k=0(cid:18) (cid:19)
X
2. Hypergeometric Bernoulli Polynomials
Let N be a positive integer. Following the work of A. Hassen and the first author in [2] we define
hypergeometric Bernoulli polynomials B (x) of order N by the exponential generating function
N,n
(tN/N!)ext ∞ tn
F (x,t)≡ = B (x) (12)
N et−T (t) N,n n!
N−1
n=0
X
Here, T (t) = N tn is the N-th Taylor polynomial of et. In particular, when N = 1, we recover
N n=0 n!
the classical Bernoulli polynomials, i.e. B (x) = B (x). Hypergeometric Bernoulli polynomials play an
1,n n
P
important role in the study of hypergeometric zeta functions (see [3]).
3
It is known that hypergeometric Bernoulli polynomials can also be defined by the following properties:
(i)B (x)=1 (13)
N,0
(ii)B′ (x)=nB (x) (14)
N,n N,n−1
1 1
(iii) (1−x)N−1B (x)dx= δ (15)
N,n n
N
Z0
These properties were shown to be equivalent to definition (12) in [2]. In either definition, hypergeometric
Bernoulli numbersaredefinedanalogouslybyB =B (0). Moreover,wedefineB (x)=0forn<0.
N,m N,m N,n
We begin with an alternating convolution formula valid for Appell sequences, i.e. polynomial sequences
that satisfy conditions (13) and (14).
Theorem 1. Let p (x) and q (x) be two Appell sequences. Then
n n
n
n
(−1)k p (x)q (x)=c (16)
n−k k n
k
k=0 (cid:18) (cid:19)
X
where {c } are constants independent of x. Moreover, if p (x) = q (x), then c = 0 for every odd positive
n n n n
integer n.
Proof. Set f(x) = p (x) , g(x) = q (x), and p = n− 1 in (6). Then using property (14) for Appell
n−1 0
sequences and the fact that f(p+1) =0 (since f is a polynomial of degree p), we obtain
n−1 (n−1)
p (x)= (−1)k (k)p (x)q (x)
n−1 n−1−k k+1
(k+1)!
Z k=0
X
where n =n(n−1)(n−2)...(n−k+1) is the falling factorial. On the other hand, since p is an Appell
(k) n
sequence we have
p (x)−c
n n
p (x)=
n−1
n
Z
where c is the constant of integration. It follows from the substitution ℓ=k+1 that
n
n
n
p (x)−c =− (−1)ℓ p (x)q (x)
n n n−ℓ ℓ
ℓ
ℓ=1 (cid:18) (cid:19)
X
Rearranging terms, we get
n
n
p (x)+ (−1)ℓ p (x)q (x)=c
n n−ℓ ℓ n
ℓ
ℓ=1 (cid:18) (cid:19)
X
or equivalently,
n
n
(−1)ℓ p (x)q (x)=c
n−ℓ ℓ n
ℓ
ℓ=0 (cid:18) (cid:19)
X
If we now assume p (x) = q (x) and n is odd, then all the terms in the convolution sum cancel. Thus,
n n
c =0, which completes the proof. (cid:3)
n
We note that Theorem 1 can also be proven by differentiating the left-hand side of (16) or by using
exponential generating functions. More precisely, define
n
n
C (x)= (−1)k p (x)q (x) (17)
n n−k k
k
k=0 (cid:18) (cid:19)
X
4
Then using the fact that p′ (x)=(n−k)p (x) and q′(x)=kq (x), we have
n−k n−k−1 k k−1
n
n
C′(x)= (−1)k [(n−k)p (x)q (x)+kp (x)q (x)]
n k n−k−1 k n−k k−1
k=0 (cid:18) (cid:19)
X
n n
n−1 n−1
=n (−1)k p (x)q (x)+n (−1)k p (x)q (x)]
n−k−1 k n−k k−1
k k−1
k=0 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
X X
n−1 n−1
n−1 n−1
=n (−1)k p (x)q (x)−n (−1)k p (x)q (x)]
n−k−1 k n−k−1 k
k k
k=0 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
X X
=0
Thus,C(x)isconstant,whichprovesTheorem1. Alternatively,p (x)andq (x)haveexponentialgenerating
n n
functions of the form
∞ tn
F(x,t)=f(t)ext = p (x)
n
n!
n=0
X
∞ tn
G(x,t)=g(t)ext = q (x)
n
n!
n=0
X
It follows that
∞ n n tn ∞ tn
F(x,t)G(x,−t)= (−1)k p (x)q (x) = C (x)
n−k k n
k n! n!
n=0 k=0 (cid:18) (cid:19) ! n=0
X X X
But we also have F(x,t) = G(x,−t) = f(t)g(t), which is independent of x. Thus, C (x) = c is constant
n n
and Theorem 1 follows.
Let us now set p (x)=B (x) and q (x)=B in Theorem 1. This yields the following theorem.
n N,n n M,n
Theorem 2. Let B (x) and B (x) be two hypergeometric Bernoulli polynomials of orders M and N,
M,n N,n
respectively. Then
n
n
(−1)k B (x)B (x)=c (18)
N,n−k M,k n
k
k=0 (cid:18) (cid:19)
X
where {c } are constants independent of x. Moreover, if M =N, then c =0 for every odd positive integer
n n
n.
Since the left hand side of (2) is independent of x, we obtain the following corollary as a result.
Corollary 3. For any two real values a and b, we have
n n
n n
c = (−1)k B (b)B (b)= (−1)k B (a)B (a) (19)
n N,n−k M,k N,n−k M,k
k k
k=0 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
X X
Surprisingly,in the case ofclassicalBernoullinumbers,(19)becomes trivialwhenM =N =1 anda=0,
b=1 since it is known that B (1)=(−1)nB (0). If we denote B =B (0), then (19) becomes
n n n n
n n
n n
(−1)n (−1)k B B = (−1)k B B (20)
n−k k n−k k
k k
k=0 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
X X
When n is odd, the left and right hand sums in (20) are negatives of each other, which implies
n
n
(−1)k B B =0 (21)
n−k k
k
k=0 (cid:18) (cid:19)
X
However,we had already deduced this earlier since (21) is just a restatement of (18) with c =0 for n odd.
n
On the other hand, when n is even, then the left and right hand sides of (20) are identical and we are left
with a trivial identity.
5
Fortunately, (19) can be used to obtain new non-trivial identities for hypergeometric Bernoulli numbers
B with N >1. We demonstratethis nextwith the use ofthe followinglemma whichhelps us to evaluate
N,n
hypergeometric Bernoulli polynomials B (x) at x=1.
N,n
Lemma 4.
(n+1)(N) N n+N
B (1)= δ + B (22)
N,n+N n N,n+j
N! n+j
j=1(cid:18) (cid:19)
X
or equivalently,
(k−N +1)(N) N−1 k
B (1)= δ + B (23)
N,k k−N N,k−j
N! k−j
j=0 (cid:18) (cid:19)
X
where we define B =0 for k <0.
N,k
Proof. Set f(x)=(1−x)N−1, g(x)=B (x), and p=N −1. As f and g are Appell sequences, it follows
N,n
from (6) and the fact f(p+1) =0 (since f is a polynomial of degree p) that
N−1 (N −1)
(1−x)N−1B (x)dx= (k) (1−x)N−1−kB (x)
N,n (n+1)(k+1) N,n+k+1
Z k=0
X
where (N−1) and(n+1)(k+1) denote falling andrising factorials,respectively. Integrating this equation
(k)
over the interval [0,1] yields
1 (N −1)! N−1 (N −1)
(1−x)N−1B (x)dx= B (1)− (k) B (0)
N,n (n+1)N N,n+N (n+1)(k+1) N,n+k+1
Z0 k=0
X
Equating this answer with (14) in the definition of hypergeometric Bernoulli polynomials and solving for
B gives(22)asdesired. Formula(23)nowfollowseasilyfrom(22)bymakingachangeofvariableand
N,n+N
re-indexing. (cid:3)
We now makeuse of Lemma 4 to proveidentities involvingsums of products of hypergeometricBernoulli
numbers.
Theorem 5. For positive integers N and M, we have
n
n
(−1)k B B
N,n−k−j M,k−i
j;i;k−i
" #
0≤i≤M−1 k=0 (cid:18) (cid:19)
X X
0≤j≤N−1
(i,j)6=(0,0)
N−1 M−1
M +N n n
=(−1)M−1 δ + B +(−1)n−M−N B (24)
n−M−N N,n−M−j M,n−N−i
 M M;j N;i 
(cid:18) (cid:19) j=0 (cid:18) (cid:19) i=0 (cid:18) (cid:19)
X X
 
where
n n!
=
n ;n ;...;n n !n !...n !(n−n −n −...−n )!
(cid:18) 1 2 k(cid:19) 1 2 k 1 2 k
denote a multinomial coefficient.
Proof. We set a=0 and b=1 in (19) and use (23) to obtain
n n (n−k−N +1)(N) N−1 n−k (k−M +1)(M)
(−1)k δ + B (0) δ
n−k−N N,n−k−j k−M
k  N! n−k−j  M!
k=0 (cid:18) (cid:19) j=0 (cid:18) (cid:19) (cid:20)
X X
 M−1 n 
k n
+ B (0) = (−1)k B (0)B (0) (25)
M,k−i N,n−k M,k
k−i k
i=0 (cid:18) (cid:19) # k=0 (cid:18) (cid:19)
X X
6
Thenexpandingtheproductinsidetheleft-mostsummationaboveyieldsfourseparatesummationsthatcan
be rearrangedand simplified as follows:
N−1 M−1
M +N n n
(−1)M δ + B +(−1)n−M−N B
n−M−N N,n−M−j M,n−N−i
 M M;j N;i 
(cid:18) (cid:19) Xj=0 (cid:18) (cid:19) Xi=0 (cid:18) (cid:19) (26)
N−1M−1 n n 
n n
+ (−1)k B B = (−1)k B B
N,n−k−j M,k−i N,n−k M,k
j;i;k−i k
j=0 i=0 k=0 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
X X X X
Next, observethat the terms in the triple summation above for i=j =0 yield the same sum as that on the
right hand side, thus cancelling each other. Then solving for the remaining terms in the triple summation
yields (24) as desired. (cid:3)
We now consider special cases of Theorem 5. For example, the following corollary holds if M =1.
Corollary 6. For any positive integer N we have
N−1 n N−1
n n n
(−1)k B B =(N+1)δ +(−1)n−N−1 B + B (27)
k N,n−k−j n−N−1 n−N N,n−1−j
j, k N 1, j
j=1 k=0 (cid:18) (cid:19) (cid:18) (cid:19) j=0 (cid:18) (cid:19)
X X X
In addition, if we set N =2 in Corollary 6, then we obtain
Corollary 7. For n≥1 we have
n
n n
(−1)k B B =B +nB − B (28)
k 2,n−k 2,n 2,n−1 n−1
k 2
k=0 (cid:18) (cid:19)
X
or equivalently,
n
n n
B B =B − B (29)
k 2,n−k 2,n n−1
k 2
k=0(cid:18) (cid:19)
X
Proof. Observe that if N = 2 in (27), then j only takes on the value 1 for the left-hand side, in which case
n =n n−1 . Moreover,on the right-hand side we have n =n n−1 . This simplifies (28) as follows:
1, k k 1, j j
(cid:0) (cid:1) (cid:0)n (cid:1) n−1 (cid:0) n(cid:1) (cid:0) (cid:1) 1 n−1
(−1)kn B B =3δ +(−1)n−3 B + n B
k 2,n−k−1 n−3 n−2 2,n−1−j
k 2 j
k=0 (cid:18) (cid:19) (cid:18) (cid:19) j=0 (cid:18) (cid:19)
X X
Next, we replace n by n+1 and simplify to obtain
n+1
n (n+1)n
(n+1) (−1)k B B =3δ +(−1)n−2 B +(n+1)B +(n+1)nB
k 2,n−k n−2 n−1 2,n 2,n−1
k 2
k=0 (cid:18) (cid:19)
X
We then divide both sides by n+1 and use the fact that 3δ /(n+1)=δ on the right-hand side and
n−2 n−2
B =0 on the left-hand side for every even integer n>1 to obtain
n+1
n
n n
(−1)k B B =δ +(−1)n−2 B +B +nB
k 2,n−k n−2 n−1 2,n 2,n−1
k 2
k=0 (cid:18) (cid:19)
X
Equation(28) now followsfromthe fact that δ +(−1)n−2nB =−nB . To obtain(29), we use the
n−2 2 n−1 2 n−1
identity
n n
n n
(−1)k B B =nB + B B
k 2,n−k 2,n−1 k 2,n−k
k k
k=0 (cid:18) (cid:19) k=0(cid:18) (cid:19)
X X
which holds since B =−1/2 and B =0 for k ≥0. (cid:3)
1 2k+1
Observethatformula(29)generalizes(3)andallowsustocalculatethehypergeometricBernoullinumbers
B in terms ofthe classicalBernoullinumbers B . This is usefulsince the odd B areknownto vanish
2,n n 2n−1
except for B and thus (29) allows us to calculate B more efficiently.
1 2,n
Next, we discuss another interesting case of Theorem 5, namely when M =N. This is the content of the
following corollary.
7
Corollary 8. For any positive integer N we have
n
n
(−1)k B B
N,n−k−j N,k−i
j;i;k−i
" #
0≤i≤N−1 k=0 (cid:18) (cid:19)
X X
0≤j≤N−1
(i,j)6=(0,0)
N−1
2N n
=(−1)N−1 δ +(1+(−1)n) B (30)
n−2N N,n−N−j
 N N;j 
(cid:18) (cid:19) j=0 (cid:18) (cid:19)
X
 
Observe that if n is odd, then both sides of (30) vanish. This is clear for the right-hand side because of the
delta term and the sign alternation. As for the left-hand side, this follows from the fact that the k-th term
of the inner summation corresponding to (i,j) cancels with the (n−k)-th term corresponding to (j,i).
In particular, setting N =2 in Corollary 8 yields
Corollary 9. For any even integer n≥0, we have
n
n 1
(−1)k B (x)B (x)= δ +nB +B (31)
2,n−k 2,k n−2 2,n−1 2,n
k 2
k=0 (cid:18) (cid:19)
X
Proof. Assume n is even. Then substituting N =2 into (30) yields
n n
n n
(−1)k B B + (−1)k B B
2,n−k−1 2,k 2,n−k 2,k−1
1;0;k 0;1;k−1
k=0 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
X X
n 1
n 4 n
+ (−1)k B B =− δ −2 B
2,n−k−1 2,k−1 n−4 2,n−2−j
1;1;k−1 2 2;j
k=0 (cid:18) (cid:19) (cid:18) (cid:19) j=0(cid:18) (cid:19)
X X
which simplifies to
n−1 n−1
n−1 n−2
2n (−1)k B B +n(n−1) (−1)k B B
2,n−k−1 2,k 2,n−k−1 2,k−1
k k−1
kX=0 (cid:18) (cid:19) kX=1 (cid:18) (cid:19) (32)
1
n
=−6δ −2 B
n−4 2,n−2−j
2;j
j=0(cid:18) (cid:19)
X
where we have used the fact that B = 0 for k < 0. Next, observe that the first summation in (32) is an
2,k
alternating convolution, which vanishes since n−1 is odd. The second summation can also be rewritten as
an alternating convolution after re-indexing k. This leads to
n−2 1
n−2 n−2
n(n−1) (−1)k B B =6δ +n(n−1) B
2,n−k−2 2,k n−4 2,n−2−j
k j
k=0 (cid:18) (cid:19) j=0(cid:18) (cid:19)
X X
and gives a formula for c if we re-index n:
n
n 1
n 6 n
c = (−1)k B B = δ + B
n 2,n−k 2,k n−2 2,n−j
k (n+1)(n+2) j
k=0 (cid:18) (cid:19) j=0(cid:18) (cid:19)
X X
This completes the proof since 6δ /[(n+1)(n+2)]=δ /2. (cid:3)
n−2 n−2
As acorollary,we nowderiveidentities forsums ofproducts involvingonly the evenB andseparately
2,2n
for the odd B . Recall the following convolution formula due to Kamano [4]:
2,2n−1
Theorem 10 ([4]). Let N be a positive integer. Then
n
n 1
B B =− [nB +(n−N)B ] (33)
N,n−k N,k N,n−1 N,n
k N
k=0(cid:18) (cid:19)
X
8
It follows from averaging equations (31) and (33) above with N =2 that
n
n 1
B B = [δ +nB −(n−4)B ]
2,n−k 2,k n−2 2,n−1 2,n
k 4
k=0 mod 2(cid:18) (cid:19)
X
or equivalently,
Corollary 11.
n
2n 1
B B = [δ +2nB −(2n−4)B ] (34)
2,2n−2k 2,2k 2n−2 2,2n−1 2,2n
2k 4
k=0(cid:18) (cid:19)
X
Observe that (34) generalizes Euler’s quadratic formula for the classical Bernoulli numbers:
n
2n
B B =−(2n−1)B
2n−2k 2k 2n
2k
k=0(cid:18) (cid:19)
X
On the other hand, subtracting the same two equations above yields
n
n 1
B B =− [δ +3nB +nB ]
2,n−k 2,k n−2 2,n−1 2,n
k 4
k=1 mod 2(cid:18) (cid:19)
X
or equivalently,
Corollary 12.
n
2n 1
B B =− [δ +6nB +2nB ] (35)
2,2n−2k+1 2,2k−1 2n−2 2,2n−1 2,2n
2k−1 4
k=1(cid:18) (cid:19)
X
3. Exponential Partial Fraction Expansion
In this section we demonstrate that some of the identities obtained in the previous section can also be
obtained fromexponentialgenerating functions defined for hypergeometricBernoullipolynomials, whichwe
recall from (12):
(tN/N!)ext ∞ tn
F (x,t)≡ = B (x)
N et−T (t) N,n n!
N−1
n=0
X
This was achieved by taking advantage of elementary expansion formulas that hold for special partial frac-
tions, in particular those that we refer to as exponential partials fractions. We write out these formulas in
the following lemma without proof.
Lemma 13. Let A and B be two quantities. Then
1−AB A B
(1) =1+ + (36)
(et−A)(e−t−B) et−A e−t−B
A−B 1 1
(2) = − (37)
(et−A)(et−B) et−A et−B
Next, we express Lemma 13 in terms of the exponential generating function F (x,t).
N
Theorem 14. Let M and N be positive integers. Then
(1) [1−T (t)T (−t)]F (x,t)F (x,−t)
N−1 M−1 N M
tM+N tM tN
=(−1)M +(−1)M T (t)F (0,t)+ T (−t)F (0,−t) (38)
N−1 N M−1 M
M!N! M! N!
tM tN
(2) [T (t)−T (t)]F (x ,t)F (x ,t)= F (x,t)− F (x,t) (39)
N−1 M−1 N 1 M 2 N M
M! N!
where x=x +x .
1 2
9
Proof. For (1), set A=T (t) and B =T (−t). Then
N−1 M−1
[1−T (t)T (−t)]F (x,t)F (x,−t)
N−1 M−1 N M
1−AB
=(−1)M(tN/N!)(tM/M!)
(et−A)(e−t−B)
tM+N A B
=(−1)M 1+ + (40)
M!N! et−A e−t−B
(cid:18) (cid:19)
tM+N tM tN
=(−1)M +(−1)M T (t)F (0,t)+ T (−t)F (0,−t) (41)
N−1 N M−1 M
M!N! M! N!
For (2), set A=T (t) and B =T (t). Then
N−1 M−1
A−B
[T (t)−T (t)]F (x ,t)F (x ,t)=(tN/N!)(tM/M!)e(x1+x2)t
N−1 M−1 N 1 M 2 (et−A)(et−B)
tM+N 1 1
= e(x1+x2)t − (42)
M!N! et−A et−B
(cid:18) (cid:19)
tM tN
= F (x,t)− F (x,t) (43)
N M
M! N!
(cid:3)
Define a to be coefficients of the polynomial
m
M+N−2
1−T (t)T (−t)= a tm
N−1 M−1 m
m=1
X
We are ready to prove our first main result in this section.
Theorem 15. For positive integers M and N, we have
M+N−2 n−m
n M +N
(1) m!a (−1)k B (x)B (x)=(−1)M δ
m N,n−m−k M,k n−M−N
m;k M
m=1 k=0 (cid:18) (cid:19) (cid:18) (cid:19)
X X
N−1 M−1
n n
+(−1)M B +(−1)n−N B (44)
N,n−m−M M,n−m−N
m;M m;N
m=0(cid:18) (cid:19) m=0(cid:18) (cid:19)
X X
and if N ≥M,
N−1 n−m
n n n
(2) B (x )B (x )= B (x)− B (x) (45)
N,n−m−k 1 M,k 2 N,n−M M,n−N
m;k M N
m=M k=0 (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X X
Proof. To prove (44), we begin with the left-hand side of (38) which simplifies to
[1−T (t)T (−t)]F (x,t)F (x,−t) (46)
N−1 M−1 N M
M+N−2 ∞ n n tn
= a tm (−1)k B (x)B (x)
m N,n−k M,k
k n!
m=1 !n=0 k=0 (cid:18) (cid:19) !
X X X
M+N−2 ∞ n n tn+m
= a (−1)k B (x)B (x)
m N,n−k M,k
k n!
m=1 n=0k=0 (cid:18) (cid:19)
X XX
∞ M+N−2 n−m n tn
= m!a (−1)k B (x)B (x) (47)
m N,n−m−k M,k
m;k n!
n=0 m=1 k=0 (cid:18) (cid:19) !
X X X
10