The values of the high order Bernoulli polynomials at integers and the r-Stirling numbers

0.1 MB·

by  Miloud Mihoubi

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The values of the high order Bernoulli polynomials at integers
and the r-Stirling numbers
Miloud Mihoubi1 and Meriem Tiachachat2
USTHB, Faculty of Mathematics, RECITS Laboratory,
PB 32 El Alia 16111 Algiers, Algeria.
[email protected] [email protected] [email protected]
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Abstract. In this paper, we exploit the r-Stirling numbers of both kinds in order to give explicit
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formulae for the values of the high order Bernoulli numbers and polynomials of both kinds at integers.
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We give also some identities linked the r-Stirling numbers and binomial coefficients.
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J
Keywords. The r-Stirling numbers; the high order Bernoulli polynomials; binomial coefficients.
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2 MSC 2010: 11B68; 11B73; 11B83.

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T
N 1 Introduction
.
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t
a Thestudyof thehigher order Bernoulli polynomials of the bothkindshave extensively used in various
m
branches of mathematics and have extended in various directions. For details on the higher order
[ Bernoulli polynomials of the first kind B(α)(x) one can see [13, 14, 15, 17, 19], and, for the higher
n
1 order Bernoulli polynomials of the second kind b(α)(x) one can see [1, 3, 11, 12]. These polynomials
n
v
are defined by the generating function to be
8
5 tn t α
9 B(α)(x) = exp(xt), (1)
5 n n! exp(t)−1
1. nX≥0 tn (cid:18) t (cid:19)α
0 b(α)(x) = (1+t)x. (2)
4 n n! ln(1+t)
1 nX≥0 (cid:18) (cid:19)
:
v (α) (α) (1)
ThenumbersB := B (0)arethehighorderBernoullinumbersofthefirstkindandB := B (0)
i n n n n
X (α) (α)
are the Bernoulli numbers of the second kind. Also, the numbers b := b (0) are the high order
n n
r (1)
a Bernoulli numbers of the second kind and bn := bn (0) are the Bernoulli numbers of the second kind.
These numbers and polynomials are connected to the r-Stirling numbers of the first and second kind
n n
and introduced by Broder [2, 8, 9].
k r k r
Recall that the number n counts the number of permutations of the set [n] := {1,...,n} into k
(cid:2) (cid:3) (cid:8) (cid:9) k r
cycles such that the elements of the set [r] are in different cycles, and, the number n counts the
(cid:2) (cid:3) k r
number of partitions of the set [n] into k non-empty subsets such that the elements of the set [r] are
(cid:8) (cid:9)
in different subsets. These numbers are determined by the generating function to be
n+r tn 1 (−ln(1−t))k
= , (3)
k+r n! k! (1−t)r
n≥k(cid:20) (cid:21)r
X
n+r tn 1
= (exp(t)−1)kexp(rt). (4)
k+r n! k!
n≥k(cid:26) (cid:27)r
X
1This researchis supported by the PNR project 8/U160/3172..
2
By combining (1) and (4) we obtain
n+k −1 n+r+k
B(−k)(r)= , r,k ∈ N, (5)
n k k+r
(cid:18) (cid:19) (cid:26) (cid:27)r
and by combining (2) and (3) we get
n+k −1 n+r+k
b(−k)(−r)=(−1)n , r,k ∈ N. (6)
n k k+r
(cid:18) (cid:19) (cid:20) (cid:21)r
In this paper, we give formulas for the values of the high order Bernoulli polynomials at integers in
terms of the r-Stirlingnumbersof both kinds. In particular, we may prove that the Bernoulli numbers
of the first kind admit the following representations
2n −1 n (−1)n+j 2n n+j
b = (n+1)
n
n j +1 n+j j
(cid:18) (cid:19) j=0 (cid:18) (cid:19)(cid:20) (cid:21)
X
n (−1)j 2n n+j +1
= n
n+j n+j j+1
j=0 (cid:18) (cid:19)(cid:26) (cid:27)
X
and the Bernoulli numbers of the second kind admit the following representations
n (−1)n+j 2n n+j +1
B = n
n
n+j n+j j +1
j=0 (cid:18) (cid:19)(cid:20) (cid:21)
X
2n −1 n (−1)j 2n n+j
= (n+1) .
n j +1 n+j j
(cid:18) (cid:19) j=0 (cid:18) (cid:19)(cid:26) (cid:27)
X
As consequences, we give in the third section some identities linked r-Stirling numbers and binomial
coefficients. The mathematical tools used are the identities (5), (6) and the Melzak’s formula [6, 7]
given by
α+p p (−1)j p
f (α+x)= α f (−j +x)., (7)
n n
p α+j j
(cid:18) (cid:19)j=0 (cid:18) (cid:19)
X
where f is a polynomial of degree n ≤ p, x := x(x−1)···(x−k+1), k ≥ 1, and x := 1.
k k! 0
We use also the notation
(cid:0) (cid:1) (cid:0) (cid:1)
xn = x(x−1)···(x−n+1), n ≥ 1, x0 = 1 and xn = x(x+1)···(x+n−1), n ≥ 1, x0 = 1.
2 The values of the high order Bernoulli polynomials at integers
For such applications of (7), we consider the Bernoulli polynomials of both kinds. Indeed, the def-
(α) (α)
initions (1) and (2) show that B (0) and b (0) represent (potential) polynomials in α of degree
n n
(α) (α)
≤ n, see [4, Thm. B, p. 141]. So, the polynomials b (x) and B (x) are also polynomials in α of
n n
degree ≤ n. This help to give new formulas for the high order Bernoulli polynomials in terms of the
r-Stirling numbers. The following proposition gives formulas for the values of the high order Bernoulli
polynomials of both kinds at non-positive integers in terms of the r-Stirling numbers of the first kind.
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Proposition 1 Let α be a real number and p, q, r, n be non-negative integers with p ≥ n. We have
b(α)(−r) = (α+q) α+p+q p (−1)n+j p n+r+r+j+j+qq r,
n p α+q+j j n+j+q
(cid:2) (cid:3)
(cid:18) (cid:19)j=0 (cid:18) (cid:19) n
X
B(α)(−r) = (n+1−α+q) n+1−α+p+q p (cid:0) (−(cid:1)1)n+j p n+r+r+j+j+q+q+11 r+1.
n p n+1−α+q+j j (cid:2) n+j+q (cid:3)
(cid:18) (cid:19)j=0 (cid:18) (cid:19) n
X
(cid:0) (cid:1)
Proof. Setting x = −q and replace α by α+q in (7) to get
p
α+q+p p f(−j−q)
f (α) = (α+q) (−1)j . (8)
p j α+q+j
(cid:18) (cid:19)j=0 (cid:18) (cid:19)
X
(x)
By setting f (x) =b (−r) in (8) we obtain
n
p (−j−q)
α+p+q p b (−r)
b(α)(−r)= (α+q) (−1)j n .
n p j α+q+j
(cid:18) (cid:19)j=0 (cid:18) (cid:19)
X
On using (6), the last identity being the first identity of the proposition.
(α) (n+1−α)
Upon using the Carlitz’s identity B (x) = b (x−1), see [3, Eqs. (2.11), (2.12)], the second
n n
identity is equivalent to the first one. (cid:3)
For p = n and q = 0 in Proposition 1, we get the following corollary.
Corollary 2 Let α be a real number and r, n be non-negative integers. We have
α+n 2n −1 n (−1)n+j 2n n+r+j
b(α)(−r) = α ,
n n n α+j n+j r+j
(cid:18) (cid:19)(cid:18) (cid:19) j=0 (cid:18) (cid:19)(cid:20) (cid:21)r
X
2n−α+1 2n −1 n (−1)n+j 2n n+r+j +1
B(α)(−r) = (n+1−α) .
n n n n+1−α+j n+j r+j+1
(cid:18) (cid:19)(cid:18) (cid:19) j=0 (cid:18) (cid:19)(cid:20) (cid:21)r+1
X
In particular, for α= 1 the values of the classical Bernoulli polynomials at non-positive integers are
2n −1 n (−1)n+j 2n n+r+j
b (−r) : = b(1)(−r)= (n+1) ,
n n n j +1 n+j r+j
(cid:18) (cid:19) j=0 (cid:18) (cid:19)(cid:20) (cid:21)r
X
n (−1)n+j 2n n+r+j +1
B (−r) : = B(1)(−r)= n .
n n n+j n+j r+j +1
j=0 (cid:18) (cid:19)(cid:20) (cid:21)r+1
X
These representations show that the classical Bernoulli numbers admit the representations
2n −1 n (−1)n+j 2n n+j
b = (n+1) ,
n
n j +1 n+j j
(cid:18) (cid:19) j=0 (cid:18) (cid:19)(cid:20) (cid:21)
X
n (−1)n+j 2n n+j +1
B = n .
n
n+j n+j j +1
j=0 (cid:18) (cid:19)(cid:20) (cid:21)
X
Similarly to Proposition 1, the following proposition gives formulas for the values of the high order
Bernoulli polynomials of both kinds at non-negative integers in terms of the r-Stirling numbers of the
second kind.
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Proposition 3 Let α be a real number and p, q, r, n be non-negative integers with p ≥ n. We have
B(α)(r) = (α+q) α+q+p p (−1)j p n+r+r+q+q+jj r,
n p α+q+j j n+q+j
(cid:8) (cid:9)
(cid:18) (cid:19)j=0 (cid:18) (cid:19) n
X
b(α)(r) = (n+1−α+q) n+1−α+q+p p (cid:0) ((cid:1)−1)j p n+r+r+q+q+j+j+11 r+1.
n p n+1−α+q+j j (cid:8) n+q+j (cid:9)
(cid:18) (cid:19)j=0 (cid:18) (cid:19) n
X
(cid:0) (cid:1)
(x)
Proof. By setting f(x) = B (r) in (8) we obtain
n
p (−j−q)
α+q+p p B (r)
B(α)(r) = (α+q) (−1)j n .
n p j α+q+j
(cid:18) (cid:19)j=0 (cid:18) (cid:19)
X
On using (5), the last identity being the first identity of the proposition.
(α) (n+1−α)
Upon using the Carlitz’s identity b (x) = B (x+1), see [3, Eqs. (2.11), (2.12)], the second
n n
identity is equivalent to the first one. (cid:3)
For p = n and q = 0 in Proposition 3, we get the following corollary.
Corollary 4 Let α be a real number and r, n be non-negative integers. We have
α+n 2n −1 n (−1)j 2n n+r+j
B(α)(r) = α ,
n n n α+j n+j r+j
(cid:18) (cid:19)(cid:18) (cid:19) j=0 (cid:18) (cid:19)(cid:26) (cid:27)r
X
2n+1−α 2n −1 n (−1)j 2n n+r+j +1
b(α)(r) = (n+1−α) .
n n n n+1−α+j n+j r+j +1
(cid:18) (cid:19)(cid:18) (cid:19) j=0 (cid:18) (cid:19)(cid:26) (cid:27)r+1
X
In particular, for α= 1 the values of the classical Bernoulli polynomials at non-negative integers are
2n −1 n (−1)j 2n n+r+j
B (r) : = B(1)(r) = (n+1) ,
n n n j +1 n+j r+j
(cid:18) (cid:19) j=0 (cid:18) (cid:19)(cid:26) (cid:27)r
X
n (−1)j 2n n+r+j +1
b (r) : = b(1)(r)= n .
n n n+j n+j r+j +1
j=0 (cid:18) (cid:19)(cid:26) (cid:27)r+1
X
These representations show that the classical Bernoulli numbers admit the representations
2n −1 n (−1)j 2n n+j
B = (n+1) ,
n
n j +1 n+j j
(cid:18) (cid:19) j=0 (cid:18) (cid:19)(cid:26) (cid:27)
X
n (−1)j 2n n+j +1
b = n .
n
n+j n+j j +1
j=0 (cid:18) (cid:19)(cid:26) (cid:27)
X
Note that the above formula of Bernoulli numbers B is exactly the formula given in [10, Thm. 3.1].
n
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Remark 5 The Genocchi numbers (G ;n ≥ 0) given by G = 2 1−22n B , n ≥ 1, see [5, Propo-
n n 2n
sition 2.1], can be written via the above two expressions of B as
n
(cid:0) (cid:1)
2n (−1)j 4n 2n+j +1
G = 4n 1−22n ,
n
2n+j 2n+j j +1
j=0 (cid:18) (cid:19)(cid:20) (cid:21)
(cid:0) (cid:1)X
4n −1 2n (−1)j 4n 2n+j
G = 2(2n+1) 1−22n .
n
2n j +1 2n+j j
(cid:18) (cid:19) j=0 (cid:18) (cid:19)(cid:26) (cid:27)
(cid:0) (cid:1) X
Furthermore, the link between the Euler and Bernoulli polynomials via the identity E (2x) =
n−1
2 (B (2x)−2nB (x)), see [16, p. 88], shows that the values of the Euler polynomials at even integers
n n n
can be written on using the above expressions of B (−r) and B (r) as
n n
2 2
E (−2r) = (B (−2r)−2nB (−r)) and E (2r) = (B (2r)−2nB (r)).
n−1 n n n−1 n n
n n
n
Remark 6 It is known that B = (−1)j j! n . Similarly of the proof of this identity, we have
n j+1 j
j=0
P (cid:8) (cid:9)
tn ln(1+exp(t)−1) j! 1
B (r) = exp(rt)= (−1)j (exp(t)−1)jexp(rt)
n
n! exp(t)−1 j +1 j!
n≥0 j≥0 (cid:18) (cid:19)
X X
which gives
n
j! n+r
B (r)= (−1)j .
n
j+1 j +r
j=0 (cid:26) (cid:27)r
X
3 Identities linked r-Stirling numbers and binomial coefficients
The above Propositions can be used to deduce relations between the r-Stirling numbers and binomial
coefficients as it is shown by the following two corollaries..
Corollary 7 Let r, n, p, q, k be non-negative integers with p ≥ n. We have
p
j +q n+k+q+j n+k+p+q+1 n+r+j +q
j
(−1)
q k p−j r+j +q
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:20) (cid:21)r
X
n
n+k+q n+k j +k
= (−1)n−j (r−1)n−j
q j +k k
(cid:18) (cid:19)j=0 (cid:18) (cid:19)(cid:26) (cid:27)
X
and
p
j+q n+k+q+j n+k+p+q+1 n+r+j +q
j
(−1)
q k p−j r+j +q
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:26) (cid:27)r
X
n
n+k+q n+k j +k
= (−1)j (r−1)n−j.
q j +k k
(cid:18) (cid:19)j=0 (cid:18) (cid:19)(cid:20) (cid:21)
X
6
Proof. For α = n+k+1 in the first identities of Propositions 1 and 3, the left hand side of the first
identity of this corollary is equal to n+k b(n+k+1)(−r) and the left hand side of the second identity
k n
is equal to n+k B(n+k+1)(r) and use the fact that:
k n (cid:0) (cid:1)
(cid:0)B(n+(cid:1)k+1)(x) = b(−k)(x−1)
n n
dn ln(1+t) k
= (1+t)x−1
dtn t
!
(cid:18) (cid:19) t=0
n! dn+k
= (ln(1+t))k(1+t)x−1
(n+k)!dtn+k t=0
(cid:16) (cid:17)(cid:12)
= n! n+k n+k dj (ln(1+t))k dn(cid:12)(cid:12)+k−j (1+t)x−1
(n+k)! j dtj dtn+k−j (cid:12)
j=0(cid:18) (cid:19) (cid:12)
X (cid:12)t=0
n (cid:12)
= 1 (−1)j n+k j+k (x−1)n−j (cid:12)(cid:12)
n+k j +k k
k j=0 (cid:18) (cid:19)(cid:20) (cid:21)
X
and (cid:0) (cid:1)
b(n+k+1)(x) = B(−k)(x+1)
n n
dn exp(t)−1 k
= exp((x+1)t)
dtn t
!
(cid:18) (cid:19) t=0
n! n+k n+k dj dn+k−j
= (exp(t)−1)k (exp((x+1)t))
(n+k)! j dtj dtn+k−j (cid:12)
j=0(cid:18) (cid:19) (cid:12)
X (cid:12)t=0
n (cid:12)
= 1 n+k j +k (x+1)n−j. (cid:12)(cid:12)
n+k j +k k
k j=0(cid:18) (cid:19)(cid:26) (cid:27)
X
(cid:3)
(cid:0) (cid:1)
Below, we present some particular cases of Corollary 7.
Example 1 For r = 1 in Corollary 7 we get
p
j +q n+k+q+j n+k+p+q+1 j +n+q+1
j
(−1)
q k p−j j+q+1
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:20) (cid:21)
X
n+k+q n+k
= ,
q k
(cid:18) (cid:19)(cid:26) (cid:27)
p
j +q n+k+q+j n+k+p+q+1 j+n+q+1
n−j
(−1)
q k p−j j +q+1
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:26) (cid:27)
X
n+k+q n+k
= .
q k
(cid:18) (cid:19)(cid:20) (cid:21)
Example 2 For k = 0 in Corollary 7 we get
p
j +q n+p+q+1 n+r+j +q n+q
(−1)n−j = (r−1)n,
q p−j r+j+q q
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:20) (cid:21)r (cid:18) (cid:19)
X
p
j +q n+p+q+1 n+r+j +q n+q
(−1)j = (r−1)n.
q p−j r+j +q q
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:26) (cid:27)r (cid:18) (cid:19)
X
7
Example 3 For n = 0 in Corollary 7 we get
p
j +q k+q+j k+q+p+1 k+q
(−1)j = .
q k p−j q
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
X
Corollary 8 Let r, n, p, q, k be non-negative integers with p ≥ n. We have
p
n+p+q+k+1 q+j q+k+1+j n+r+q+k+1+j
j
(−1)
p−j j k r+q+k+1+j
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:20) (cid:21)r
X
n+p+q+k+1 n+k+r
n
= (−1)
n+k k+r
(cid:18) (cid:19)(cid:20) (cid:21)r
and
p
n+p+q+k+1 q+j q+k+1+j n+r+q+k+1+j
j
(−1)
p−j j k r+q+k+1+j
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:26) (cid:27)r
X
n+p+q+k+1 n+k+r
= .
n+k k+r
(cid:18) (cid:19)(cid:26) (cid:27)r
Proof. Choice α = −k with k+1 ≤ q in the first identities of Propositions 1 and 3, the left hand side
of the first identity of this corollary is equal to n+k n+p+q b(−k)(−r) and the left hand side of the
k n+k n
second identity is equal to n+k n+p+q B(−k)(r) and use the identities given by (5) and (6). After
k n+k n (cid:0) (cid:1)(cid:0) (cid:1)
that, replace q by q+k+1. (cid:3)
(cid:0) (cid:1)(cid:0) (cid:1)
Example 4 For k = 0 in Corollary 8 we get
p
n+p+q+1 q+j n+r+q+1+j n+p+q+1
(−1)j = rn,
p−j j r+q+1+j n
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:20) (cid:21)r (cid:18) (cid:19)
X
p
n+p+q+1 q+j n+r+q+1+j n+p+q+1
(−1)j = rn.
p−j j r+q+1+j n
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:26) (cid:27)r (cid:18) (cid:19)
X
Example 5 For n = 0 in Corollary 8 we get
p
p+q+k+1 q+j q+k+1+j p+q+k+1
(−1)j = .
p−j q k k
j=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
X
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