A closed form for the generalized Bernoulli polynomials via Faà di Bruno's formula
aa r X i v : . [ m a t h . G M ] M a y A closed form for the generalized Bernoullipolynomials via Fa`a di Bruno’s formula
Sumit Kumar JhaInternational Institute of Information TechnologyHyderabad-500 032, India [email protected]
Abstract
We derive a closed form for the generalized Bernoulli polynomial of order a in termsof Bell polynomials and Stirling numbers of the second kind using the Fa`a di Bruno’sformula. Definition 1.
The generalized Bernoulli polynomial of order a ( a ∈ N ), denoted by B an ( x ),can be defined by the following exponential generating function (cid:18) te t − (cid:19) a e − xt = ∞ X k =0 B ak ( − x ) k ! t k , (1)where | t | < π .Two closed forms for B an ( x ) are given in [2] and [3]. We prove the following formula. Theorem 2.
We have B an ( − x ) = n X k =0 ( − k k ! B n,k ( λ , · · · , λ n − k +1 ) (2) where λ m = m X l =0 S ( l + a, a ) x m − l (cid:18) ml (cid:19)(cid:18) l + aa (cid:19) − ( m = 1 , , · · · , n − k + 1) , and B n,k ( λ , λ , . . . , λ n − k +1 ) are the Bell polynomials [1, p. 206] defined by B n,k ( λ , λ , . . . , λ n − k +1 ) = X ≤ i ≤ n,ℓ i ∈ N P ni =1 iℓ i = n P ni =1 ℓ i = k n ! Q n − k +1 i =1 ℓ i ! n − k +1 Y i =1 (cid:16) λ i i ! (cid:17) ℓ i , (3)1 nd S ( n, k ) are the Stirling numbers of the second kind for n ≥ k ≥ which can be generatedby ( e x − k k ! = ∞ X n = k S ( n, k ) x n n ! , k ∈ N . (4) Proof.
Let f ( t ) = 1 /t and g ( t ) = (cid:16) e t − t (cid:17) a e xt . Using Fa`a di Bruno’s formula [1, p. 134] d n dx n f ( g ( t )) = n X k =0 f ( k ) ( g ( t )) · B n,k (cid:0) g ′ ( t ) , g ′′ ( t ) , . . . , g ( n − k +1) ( t ) (cid:1) , and the expansion g ( t ) = (cid:18) e t − t (cid:19) a e xt = a ! t a ∞ X l = a S ( l, a ) t l l ! ∞ X k =0 x k t k k != a ! t a ∞ X l =0 S ( l + a, a ) t l + a ( l + a )! ∞ X k =0 x k t k k ! = a ! ∞ X l =0 S ( l + a, a ) t l ( l + a )! ∞ X k =0 x k t k k != a ! ∞ X m =0 t m X k + l = m S ( l + a, a ) x k ( l + a )! k != a ! ∞ X m =0 t m m X l =0 S ( l + a, a ) x m − l ( l + a )! ( m − l )!= a ! ∞ X m =0 t m m ! m X l =0 S ( l + a, a ) x m − l m !( l + a )! ( m − l )!= ∞ X m =0 t m m ! m X l =0 S ( l + a, a ) x m − l m ! ( l + a )! l ! a ! ( m − l )! l != ∞ X m =0 t m m ! m X l =0 S ( l + a, a ) x m − l (cid:18) ml (cid:19)(cid:18) l + aa (cid:19) − = ∞ X k =0 g ( k ) (0) t k k ! . we conclude our formula (2). Remark . Letting x = 0 and a = 1 in equation (2) gives us B n = n X k =0 ( − k k ! B n,k (cid:18) , , · · · , n − k + 1 (cid:19) which was obtained in [4]. Here B n denote the Bernoulli numbers. Remark . Similar method yields a closed form for the Bernoulli polynomials B n ( x ) obtainedin [5]. 2 valuation of B n,k ( λ , · · · , λ n − k +1 ) We have ∞ X n = k B n,k ( λ , . . . , λ n − k +1 ) t n n ! = 1 k ! ∞ X j =1 λ j t j j ! ! k = 1 k ! ( g ( t ) − k = 1 k ! k X r =0 ( − k − r (cid:18) kr (cid:19) g ( t ) r = 1 k ! k X r =0 ( − k − r (cid:18) kr (cid:19) ∞ X n =0 t n n ! n X l =0 S ( l + ar, ar )( rx ) n − l (cid:18) nl (cid:19)(cid:18) l + arar (cid:19) − . Thus B n,k ( λ , · · · , λ n − k +1 ) = 1 k ! k X r =0 ( − k − r (cid:18) kr (cid:19) n X l =0 S ( l + ar, ar )( rx ) n − l (cid:18) nl (cid:19)(cid:18) l + arar (cid:19) − . We can use the above to conclude from equation (2) that B an ( − x ) = n X k =0 k X r =0 ( − r (cid:18) kr (cid:19) n X l =0 S ( l + ar, ar )( rx ) n − l (cid:18) nl (cid:19)(cid:18) l + arar (cid:19) − = n X r =0 n X k = r ( − r (cid:18) kr (cid:19) n X l =0 S ( l + ar, ar )( rx ) n − l (cid:18) nl (cid:19)(cid:18) l + arar (cid:19) − = n X r =0 n X k = r (cid:18) kr (cid:19) ( − r n X l =0 S ( l + ar, ar )( rx ) n − l (cid:18) nl (cid:19)(cid:18) l + arar (cid:19) − = n X r =0 (cid:18) n + 1 r + 1 (cid:19) ( − r n X l =0 S ( l + ar, ar )( rx ) n − l (cid:18) nl (cid:19)(cid:18) l + arar (cid:19) − . References [1] L. Comtet,
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