aa r X i v : . [ m a t h . N T ] D ec Three pearls of Bernoulli numbers
Abdelmoum´ene Z´ekiri , ∗ and Farid Bencherif , † Faculty of Mathematics, USTHB, Algiers, Algeria Faculty of Mathematics, USTHB, Algiers, Algeria
Abstract
The Bernoulli numbers are fascinating and ubiquitous numbers; they occur in several domainsof Mathematics like Number theory ( FLT), Group theory, Calculus and even in Physics. SinceBernoulli’s work, they are yet studied to understand their deep nature [9], [6] and particularly tofind relationships between them. In this paper, we give, firstly, a short response [15] to a problemstated, in 1971, by Carlitz [4] and studied by many authors like Prodinger [10]; the second pearlis an answer to a question raised, in 2008, by Tom Apostol [1].The third pearl is another proof ofa relationship already given in 2011, by the authors [14]
Keywords:
Bernoulli numbers, Bernoulli polynomials.
Mathematics Subject Classification:
The aim of this work is to give original proofs of three relationships involving Bernoulli numbers. Inthe fisrt section, we give a short proof to a problem stated, in 1971, by Carlitz [4] and studied by manyauthors like Prodinger [10]. In the second section, we give a response to a question raised by Apostolin 2008 in his relevant paper [1]. In the third section, we expose a different proof of a relationshipalready given by us in 2011 [15].
In Mathematics Magazine, Vol. 44, No. 2 (Mar., 1971), pp. 105-114+101, Carlitz states the followingproblem : define { B n } by means of B = 1 and for n > n X k =0 (cid:18) nk (cid:19) B k = B n show that for arbitrary m, n > − m m X k =0 (cid:18) mk (cid:19) B n + k = ( − n n X k =0 (cid:18) nk (cid:19) B m + k This identity was firstly proved by Shanon [11] in 1971, by Gessel [7] in 2003, by Wu, Sun and Pan [13]in 2004, by Vassilev-Missana [12] in 2005, by Chen and Sun [5] in 2009, by Gould and J. Quaintance[8] in 2014 and by Prodinger [10] in 2014. The Prodinger’s proof is very short and uses a two variablesformal series. In fact, one can see that Carlitz’s problem can be easily deduced from the followingrelationship already proved in 2012 by Bench´erif and Garici in 2012 [3] :( − m m + q X k =0 (cid:18) m + qk (cid:19)(cid:18) n + q + kq (cid:19) B n + k − ( − n + q n + q X k =0 (cid:18) n + qk (cid:19)(cid:18) m + q + kq (cid:19) B m + k = 0 . Hereafter, we give a proof different from that was given by Prodinger. ∗ A. Zekiri : [email protected] † F. Bencherif : [email protected] roof. We consider the linear functional L defined on Q [ x ] by L ( x n ) = B n for n ≥
0, which gives L (cid:18) x + 12 (cid:19) n +1 ! = B n +1 (cid:18) (cid:19) = 0 , see [1], p.182, then the polynomial defined by : P ( x ) = ( − m + q x n + q (1 + x ) m + q − ( − n x m + q (1 + x ) n + q satisfies P (cid:18) −
12 + x (cid:19) + ( − q P (cid:18) − − x (cid:19) = 0and P ( q ) ( −
12 + x ) + P ( q ) ( − − x ) = 0Now, with use of the equality : L (cid:18) x + 12 (cid:19) n +1 ! = B n +1 (cid:18) (cid:19) = 0 . And as P ( q ) is an even polynomial , we get : L (cid:16) P ( q ) ( x ) (cid:17) = 0Thus1 q ! P ( q ) ( x ) = ( − m m + q X k =0 (cid:18) m + qk (cid:19)(cid:18) n + q + kq (cid:19) x n + k − ( − n + q n + q X k =0 (cid:18) n + qk (cid:19)(cid:18) m + q + kq (cid:19) x m + k = 0and finally:( − m m + q X k =0 (cid:18) m + qk (cid:19)(cid:18) n + q + kq (cid:19) B n + k − ( − n + q n + q X k =0 (cid:18) n + qk (cid:19)(cid:18) m + q + kq (cid:19) B m + k = 0which yields the identity wanted by Carlitz, by taking q = 0 . In his relevant paper published in 2008, Tom Apostol writes : we leave it as a challenge to the readerto find another proof of (42) as a direct consequence of (3) without the use of integration . In Apostol’spaper, (42) denotes the relationship: n X k =0 (cid:18) nk (cid:19) B k ( n + 2 − k ) = B n +1 n + 1 , n ≥ six definitions of the Bernoulli numbers he recalls to show his relationship andwhich is : B = 1 , n − X k =0 (cid:18) nk (cid:19) B k = 0 for n ≥ integration method to deduce his (42)- numbered relation from the Bernoullinumbers’s definition that he has chosen. To take up the challenge he has launched, we expose a proofwithout use of integration method . 2 roof. ( Answer to Apostol’s problem)Let’s define the sequence ( u n ) by: u n := n X k =0 (cid:18) n + 1 k (cid:19) B k We can see that u n = 0 for n ≥
1. Writing: (cid:18) nk (cid:19) n + 2 − k = 1 n + 1 (cid:18) n + 1 k (cid:19) − n + 1)( n + 2) (cid:18) n + 2 k (cid:19) we get : n X k =0 (cid:18) nk (cid:19) B k n + 2 − k = 1 n + 1 n X k =0 (cid:18) n + 1 k (cid:19) B k − n + 1)( n + 2) n X k =0 (cid:18) n + 2 k (cid:19) B k which yields : n X k =0 (cid:18) nk (cid:19) B k n + 2 − k = 1 n + 1 u n − n + 1)( n + 2) (cid:18) u n +1 − (cid:18) n + 2 n + 1 (cid:19) B n +1 (cid:19) As if n ≥
1, we have u n = u n +1 = 0 and as (cid:0) n +2 n +1 (cid:1) = n + 2, we get : − n + 1)( n + 2) (cid:18) − (cid:18) n + 2 n + 1 (cid:19) B n +1 (cid:19) = B n +1 n + 1 . which gives the asked relation : n X k =0 (cid:18) nk (cid:19) B k ( n + 2 − k ) = B n +1 n + 1 , n ≥ integration method . In our paper [15], we proved the following relationship: n + q X k =0 (cid:18) n + qk (cid:19) q Y j =1 ( n + k + j ) B n + k = 0where q is an odd number. For this, we showed that the two well-suited polynomials : H n ( x ) = 12 x n + q ( x − n + q and K n ( x ) = n + q X k =0 ε n + k n + q + k + 1 (cid:18) n + qk (cid:19) B n + q + k +1 ( x ) − B n + q + k +1 Are equal, where B n ( x ) and B n are respectively the Bernoulli polynomials and the Bernoulli numbersdefined by the generating function: xe x − e xz = + ∞ X n =0 B n ( x ) x n n !knowing that B n = B n (0) = B n (1), n ≥ B n +1 = 0, n ≥
1, see [1] , relations (11), (12) (13) and(15). Furthermore, we shall use the well-known equalities: B n ( x + 1) − B n ( x ) = nx n − , B ′ n ( x ) = nB n − ( x ) , Z x +1 x B n ( t ) dt = x n , n ≥ n ≥
1, see e.g. [1], relations (14), (27) and (30).Now, to give another proof of the relation already proved in [15], we consider the polynomials: P n ( x ) := 12 x n +1 ( x − n +1 K n ( x ) := n +1 X k =0 (cid:18) n + 1 k (cid:19) − ( − n +1 − k B n +1+ k ( x ) H n ( x ) := 12 ( n + 1) x n ( x − n (2 x − Q -space vector Q [ x ] defined by f ( P ( x )) = R x +1 x P ( t ) dt .First of all, let’s prove the Theorem 4.1.
The two polynomials K n ( x ) and H n ( x ) are equal, i.e. K n ( x ) = H n ( x ) .Proof. Z x +1 x P ′ n ( t ) dt = P n ( x + 1) − P n ( x )= 12 x n +1 ( x + 1) n +1 − x n +1 ( x − n +1 = 12 x n +1 (cid:0) ( x + 1) n +1 − ( x − n +1 (cid:1) = x n +1 n +1 X k =0 (cid:18) n + 1 k (cid:19) (1 − ( − n +1 − k x k = n +1 X k =0 (cid:18) n + 1 k (cid:19) − ( − n +1 − k x n +1+ k = n +1 X k =0 (cid:18) n + 1 k (cid:19) − ( − n +1 − k Z x +1 x B n +1+ k ( t ) dt = Z x +1 x n +1 X k =0 (cid:18) n + 1 k (cid:19) − ( − n +1 − k B n +1+ k ( t ) dt ! Thus, we can see that f ( P ′ n ( t )) = f n +1 X k =0 (cid:18) n + 1 k (cid:19) − ( − n +1 − k B n +1+ k ( t ) ! As f is bijective, we get : P ′ n ( t ) = n +1 X k =0 (cid:18) n + 1 k (cid:19) − ( − n +1 − k B n +1+ k ( t )i.e. P ′ n ( t ) = K n ( t )Let’s compute P ′ n ( x ) : P ′ n ( x ) = 12 (cid:0) x − x ) n +1 (cid:1) ′ = 12 ( n + 1)( x − x ) n (2 x − n + 1) x n ( x − n (2 x − H n ( x ) 4.e. K n ( x ) = H n ( x ) . Theorem 4.2.
The following identity holds: n + q X k =0 (cid:18) n + qk (cid:19) q Y j =1 ( n + k + j ) B n + k = 0 Proof.
To get this, let’s replace n by n + q − q ≥ q odd. Then we compute the coefficient of x q in the equality: K n + q − ( x ) = H n + q − ( x ). The coefficient of x q in the polynomial K n ( x ) K n ( x ) := n + q X k =0 (cid:18) n + qk (cid:19) − ( − n + q − k B n +1+ k ( x )is : C q := [ x ] q B n + q − ( x ) , so that C q has the value C q = n + q X k =0 (cid:18) n + qk (cid:19) q Y j =1 ( n + k + j ) − ( − n + q − k B n +1+ k On the other hand, the coefficient of x q in H n + q − ( x ) is : (cid:26) n ≥ − q +1 q if n = 0As we have : 1 + ( − m B m = (cid:26) B m if m = 10 if m = 1We get the aimed relationship. Remark 4.3.
I would like to dedicate this modest contribution to the memory of Tom Mike Apostolwho passed away on 8 May of this year 2016.(APO2)
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