Int. Journa of Math. Anaysis, Vo. 7, 2013, no. 5, 229-240 Discrete Bernoui s Formua and its Appications Arising from Generaized Difference Operator G. Britto Antony Xavier 1 Department of Mathematics, Sacred Heart Coege Tirupattur - 635601, Veore District, Tami Nadu, S.India shcbritto@yahoo.co.in H. Nasira Begum Department of Mathematics, Sacred Heart Coege Tirupattur - 635601, Veore District, Tami Nadu, S.India Abstract In this paper, the authors derive discrete Bernoui s formua of the form, [u(k)v(k)] = ( 1) t u (t) (k)v t+1 (k + t) t=0 for the rea vaued rea variabe functions u(k) and v(k) using the generaized differences, u (t) (k) =Δ t u(k) and its inverses, v t(k) =Δ t v(k) and obtain severa formuae on finite and infinite series as appication of the discrete Bernoui s formua in number theory. Suitabe exampes are provided to iustrate the main resuts. Mathematics Subject Cassification: 39A70, 47B39, 39A10 Keywords: Generaized difference operator, Discrete Bernoui s Formua, Partia sums 1 Introduction The theory and some appications of the generaized difference operator Δ defined as Δ u(k) =u(k + ) u(k), k [0, ) and (0, ) had been deveoped in [2]. The generaized versions of Leibnitz, Binomia, Montmorte s Theorems, Newton s formua aong with formuae for the sums and partia sums of the n th powers of an arithmetic progression, the sums and partia 1 Research Supported by University Grants Commission, SERO, Hyderabad, India.
230 G. Britto Antony Xavier and H. Nasira Begum sums of the products of n consecutive terms of an arithmetic progression and the sums and partia sums of an arithmetic-geometric progression using the generaized difference operators of the first, second and n th kinds and their inverses were obtained in ([2], [7], [8], [9]). Generaized Bernoui poynomias B n (k, ±) with appications using Δ ± and its inverses ([4, 5]), quaitative and rotatory properties of certain cass of generaized difference equations ([3, 6]) are some of the appications of difference equations invoving Δ. In this paper, we derive the discrete version of the Bernoui s formua and some appications in number theory which are not obtained earier. A the resuts are iustrated with suitabe exampes. 2 Preiminaries Before stating and proving our resuts, we present some notations, basic definitions and preiminary resuts which wi be used in the subsequent discussions. Let > 0, k [0, ), j = k [k/], where [k/] denotes the integer part of k/, N (j) ={j, + j, 2 + j, }, N 1 (j) =N(j) and c j is a constant for a k N (j). For m N(1), we denote Δ m u(k) k ( (m 1)+j = 1 ( Δ Δ 1 u(k) k ) j k ) +j k (m 1)+j,Δ 1 u(k) k j = u 1(k) = u(k) ( u(j), Δ 1 u(k) k ) j k +j = u 2(k) = u 1 (k) u 1 ( + j), and so on. Definition 2.1 [2] For a rea vaued function u(k), the generaized difference operator Δ and its inverse are respectivey defined as Δ u(k) =u(k + ) u(k), k [0, ), (0, ), (1) if Δ v(k) =u(k), then v(k) = u(k)+c j. (2) Definition 2.2 [11] For k, n (0, ), the -factoria function is defined by k (n) = n Γ( k +1) (3) +1 n), where Γ is the Euer gamma function and k (n) 1 = k (n). Remark 2.3 When n N(1), (3) and its Δ -difference become Γ( k k (n) n 1 = (k t), and Δ k (n) t=0 =(n)k (n 1). (4) Theorem 2.4 [2] Let u(k), k [0, ) be rea vaued function. Then for k [, ), [k/] u(k) k j = u(k r). (5)
Discrete Bernoui s formua 231 Lemma 2.5 [2] For n N(0), and for n N(2), 1 k (n) k j k (n) k j = k(n+1) = 1 (n +1) (n 1)(j ) (n 1) j(n+1) (n +1) 1 (n 1)(k ) (n 1) In particuar, when =1, k =(r 1) N(1), j =0, then (6) becomes (6). (7) k (n) r 1 0 = (r 1)(n+1). (8) (n +1) Lemma 2.6 Let n N(0) and r N(3). Then, k (n) r 1 0 =(r 2) (n) +(r 3) (n) + + (1) (n) = (r 1)(n+1). (9) (n +1) Proof. The proof foows by taking =1,u(k) =k (n) in (5) and (6). Theorem 2.7 Let m N(1), (0, ) and k [m, ). Then, Δ m Proof. Taking ( [k/] u(k) k (m 1)+j = r=m (r 1) (m 1) u(k r). (10) (m 1)! on (5), and appying (5) for u(k r), we get [k/] u(k) k j ) k +j = u(k r) = [k/] [k r/] s=1 u(k r s). From the notation given this section and ordering the terms u(k r), we find Δ 2 [k/] u(k) k +j = (r 1) (1) u(k r). (11) 1! Again, taking on (11), by (5) for u(k r), we arrive Δ 3 [k/] u(k) k 2+j = [ (r 2) (1) 1! which yieds by (9), r=3 + (r 3)(1) 1! + + (1)(1) ] u(k r), 1! Δ 3 [k/] u(k) k 2+j = r=3 (r 1) (2) u(k r). (12) 2! Now, (10) wi be obtained by continuing this process and using (9).
232 G. Britto Antony Xavier and H. Nasira Begum Theorem 2.8 Let k [0, ) and im k u(k) =0. Then, u(k) k = u(k + r). (13) Proof. From (5), and expressing its terms in reverse order, we find r=0 u(k) k j +Δ 1 u(k) k = u(j)+u(+j)+ +u(k )+u(k)+u(k+)+ +u( ) which is same as u(k) j = u(j + r). (14) r=0 Now, (13) foows by given condition u( ) = 0 and then repacing j by k. Lemma 2.9 [2] Let Sr n s are the stiring numbers of the second kind. Then n k n = Sr n n r k (r). (15) Lemma 2.10 Let u(k) and v(k) be two rea vaued functions. Then [u(k)v(k)] = u(k) Proof. From (1), we find v(k) [ v(k + )Δ u(k)]. (16) Δ [u(k)w(k)] = w(k + )Δ u(k)+u(k)δ w(k). (17) Appying (4) in (17), we obtain [u(k)δ w(k)] = u(k)w(k) [w(k + )Δ u(k)]. (18) The proof foows from the reation v(k) =Δ w(k) and (18). 3 Main Resuts and Appications In this section, we present the discrte version of Bernoui s Formua and deveop reative resuts of it in number theory. Theorem 3.1 (Discrete Bernoui s Formua) Let u(k) and v(k) be two rea vaued functions, u (t) (k) =Δ t u(k), v t (k) =Δ t v(k) for t =1, 2, and u (0) (k) =u(k). Then [u(k)v(k)] = ( 1) t u (t) (k)v t+1 (k + t). (19) t=0
Discrete Bernoui s formua 233 Proof. Taking v(k) =v 1 (k) and Δ u(k) =u (1) (k) in (16), we have [u(k)v(k)] = u(k)v 1 (k) [v 1 (k + )u (1) (k)] (20) Appying the equation (16) again and again in (20) we get proof. Theorem 3.2 Let m N(1) and t be any integer. Then, [k/] r=m where F (n) m F (n) (r 1) (m 1) (m 1)! 1 (k) = (k) =Δ m [k (n) a tk ]. (k r) (n) Proof. From (16), we find [k (n) a t(k r) = F (n) m (k) F (n) m ((m 1) + j), (21) [ F (n) m 1(k) F (n) m 1((m 2) + j) ], m =2, 3,... and a tk ]=k (n) Appying (10) in (22), we get [k/] (k r) (n) a tk (a t 1) Δ 1 { a t(k+) a t 1 (n)k(n 1) a t(k r) = F (n) 1 (k) F (n) 1 (j), F (n) 1 (k) = }. (22) [k (n) a tk ]. Again operating on both sides and appying (10), we obtain [k/] where F (n) 2 (k) = (r 1)(k r) (n) F (n) a t(k r) = F (n) 2 (k) F (n) 2 ( + j), 1 (k). By continuing this process, we get the proof. Coroary 3.3 Let k [2, ), a 1and j = k [k/]. Then, [k/] where F (n) 2 (k) = (r 1)(k r) (n) n r=0 ( ) r n (r) r!(a 1) a k r = F (n) 2 (k) F (n) 2 ( + j), (23) (r + 1)!k (n r) a k+r r!j(n r) (a 1) r+1 Proof. The proof foows by taking m = 2 and t = 1 in (21). Exampe 3.4 Putting n =3in (23), we obtain [k/] (r 1)(k r) (3) a k r = F (3) 2 (k) F (3) a j+r k (1) (a 1) r. 2 ( + j), (24)
234 G. Britto Antony Xavier and H. Nasira Begum where F (3) 2 (k) = k(3) a k (a 1) 2 6k(2) a k+ (a 1) 3 + 182 k (1) a k+2 (a 1) 4 243 a k+3 (a 1) 5 (j) (3) a j (a 1) 3(j)(2) a +j (a 1) 2 + 62 (j) (1) a 2+j 63 a 3+j k (1) (a 1) 3 (a 1) 4. In particuar, when k =16.4,=2.4,a=2and j =2in (24), we get [ 16.4] 2.4 (r 1)(16.4 2.4r) (3) 2.42 16.4 2.4r = F (3) 2 (16.4) F (3) 2 (4.4) = 2596232.003. Coroary 3.5 If a 1, then [k/] here F (n) 2 (k) = (r 1)(k r) (n) n r=0 ( ) r n (r) r!(a 1) a (k r) = F (n) 2 (k) F (n) 2 ( + j), (25) (r + 1)!k (n r) a (k+r) r!j(n r) (a 1) r+1 a (j+r) k (1) (a 1) r. Proof. The proof foows by taking m = 2 and t = 1 in (21). Exampe 3.6 Taking n =2in (25), we get [k/] (r 1)(k r) (2) a (k r) = F (2) 2 (k) F (2) a k a (k+) 2 ( + j), (26) where F (2) 2 (k) = k(2) (a 1) 4k(1) + 62 a (k+2) 2 (a 1) 3 (a 1) 4 (j) (2) a j (a 1) 2(j)(1) a (j+) + 22 a (j+2) k (1) (a 1) 2 (a 1) 3. In particuar, when k =28.9,=10.9,a=2and j =7.1 in (26), we obtain [ 28.9] 10.9 (r 1)(28.9 10.9r) (2) 10.92 (28.9 10.9r) = F (2) 2 (28.9) F (2) 2 (18) = 0.19666586. Coroary 3.7 Let p, q N(1), q p +2 and k (q) [k/] (k +(s r))(p) = F (k r) (q) 1 (k) F 1 (j), F 1 (k) = 0. Then, (k + s)(p) k (q) (27) Proof. The vaue of F 1 (k) can be obtained by substituting u(k) =(k + s) (p) and v(k) = 1 in (16) and using (6), (7). The proof foows by (10). k (q)
Discrete Bernoui s formua 235 Exampe 3.8 Taking p =2, q =4in (27), and by (6), (7),(16), we get [k/] (k +(s r))(2) = F (k r) (4) 1 (k) F 1 (j), (28) (k + s)(2) (k + s)(1) 1 where F 1 (k) =. 3(k ) (3) 3(k ) (2) 3(k ) (1) In particuar, when k =60.5,=20.5,s=2and j =19.5 in (28), we have [ 60.5] 20.5 (60.5+(2 r)20.5)(2) 20.5 = F (60.5 20.5r) (4) 1 (60.5) F 1 (19.5) = 0.15478462. 20.5 Theorem 3.9 If q p +3, then [k/] (k +(s r))(p) (r 1) = F (k r) (q) 2 (k) F 2 ( + j), (29) where F 2 (k) = [F 1 (k) F 1 (j)] and F 1 (k) = (k + s)(p). k (q) Proof. Cosed form expression of F 1 (k) and hence F 2 (k) can be obtained by taking u(k) =(k + s) (p) and v(k) = 1 in (16) and using (6) and (7). The k (q) proof foows from (10). Exampe 3.10 In (29), by taking p =3and q =6, we have [k/] (k +(s r))(3) (r 1) = F (k r) (6) 2 (k) F 2 ( + j), (30) (k + s)(3) (k + s)(2) 3(k + s)(1) 1 F 2 (k) = + + + 20 2 (k 2) (4) 10 2 (k 2) (3) 20 2 (k 2) (2) 5 2 (k 2) (1) (j + s) (3) 3(j + s)(2) (j + s)(1) 1 k (1) + + + + 5(j ) (5) 20(j ) (4) 10(j ) (3) 20(j ) (2). 2 In particuar, when k =9,=2,s=2and j =1in (30), we obtain [ 9] 2 (9 + (2 r)2)(3) (r 1) 2 = F (9 2r) (6) 2 (9) F 2 (3) = 0.780952381. 2 Coroary 3.11 If q p +2, then (k +(s 1) + r)(p) = (k + s)(p) (k +(r 1)) (q) (31) k (q) k
236 G. Britto Antony Xavier and H. Nasira Begum Proof. The proof foows by (13) and the cosed form expression of F 1 (k) can be obtained from (16). Exampe 3.12 Substituting p =2and q =4in (31), we obtain (k +(s 1) + r) (2) (k +(r 1)) (4) = (k + s)(2) + 3(k ) (3) (k + s)(1) + 3(k ) (2) In particuar, when k =11, =2, s =3in (32), we get (11 + (3 1)2 + 2r)(2) 2 =0.197260195. (11 + 2(r 1)) (4) 2 1 3(k ) (1). (32) Coroary 3.13 Let p, q N(1), q p +3. Then, (k +(s 2) + r)(p) (r 1) = F (k +(r 2)) (q) 2 (k), (33) where F 2 (k) = F 1 (k) and F 1 (k) = u(k). Proof. The proof foows by (13) and the cosed form expression of F 1 (k) and hence F 2 (k) can be obtained by taking u(k) =(k + s) (p), v(k) = 1 in (16) k (q) and using (6) and (7). Exampe 3.14 Put p =3and q =6in (33), we get (k +(s 2) + r)(3) (r 1) = F (k +(r 2)) (6) 2 (k), (34) (k + s)(3) (k + s)(2) 3(k + s)(1) F 2 (k) = + + + 20 2 (k 2) (4) 10 2 (k 2) (3) 20 2 (k 2) (2) (j + s) (3) 3(j + s)(2) (j + s)(1) 1 k (1) + + + + 5(j ) (5) 20(j ) (4) 10(j ) (3) 20(j ) (2). 2 In particuar, when k =5, =2, s =2and j =1in (34), we get (r 1) = 0.02865598197. (5 + (2 2)2 + 2r)(3) 2 (5 + (r 2)2) (6) 2 Theorem 3.15 Let (0, ), a 1, t =1and k>. Then, [k/] where F n 1 (k) = (k n a k ). 1 5 2 (k 2) (1) (k r) n a k r = F1 n (k) F 1 n (j), (35)
Discrete Bernoui s formua 237 Proof. From (15), we have where (k n a k )= n Sr n n r [k (r) a k ], (36) [k (r) a k ] wi be obtained from (22), by taking n = r and t = 1. The proof foows from (10). The foowing coroary iustrates Theorem 3.15, when n = 3. Coroary 3.16 Let k [, ), (0, ) and a 1. Then, F 3 1 (k) = [k/] (k r) 3 a k r = F 3 1 (k) F 3 1 (j), (37) a k [ (3) k (a +3k (2) + 2 k (1) ] a k+ [ 3 +3k (2) 1) (a 1) 2 +6 2 k (1) ] + 62 a k+2 [ ] (1) 6 3 a k+3 + k (a 1) 3 (a 1). (38) 4 Exampe 3.17 In (37), by taking k =35.5, =20.5, a =2and j =15, we obtain [ 35.5] 20.5 (35.5 20.5r) 3 2 35.5 20.5r = F2 3 (35.5) F 2 3 (15) = 110592000, where F2 3 (35.5) is obtained by substituting k =35.5 in (38). Theorem 3.18 Let k [2, ), (0, ) and a 1. Then, where F n 2 [k/] (k) =Δ 1 (r 1)(k r) n a k r = F2 n (k) F 2 n ( + j), (39) [F n 1 (k) F n 1 (j)] and F n 1 (k) =Δ 1 (k n a k ). Proof. The proof foows from (4), (10) and (37). The foowing coroary iustrates Theorem 3.18, when n = 3. Coroary 3.19 If k [2, ) and a 1, then [k/] (r 1)(k r) 3 a k r = F2 3 (k) F 2 3 ( + j), (40)
238 G. Britto Antony Xavier and H. Nasira Begum where F 3 2 (k) = k(1) a j a k (a 1) 2 [k(3) +3k (2) + 2 k (1) ] 2ak+ (a 1) 3 [2 +3k (2) +6k (1) ] (a 1) [(j)(3) +3(j) (2) + 2 (j) (1) ]+ k(1) a +j (a 1) 2 [2 +3(j) (2) +6(j) (1) ] (41) + 182 a k+2 [ + (a k(1) 1) 4 ] 6k(1) a 2+j [ (a +(j)(1) 1) 3 ]+ 62 k (1) a 3+j 243 a k+3 (a 1) 4 (a 1) 5 and F2 3 (k) is given in (38). Exampe 3.20 Putting k =9, =2, a =2and j =1in (40), we get [ 9] 2 (r 1)(9 2r) 3 2 9 2r = F2 3 (9) F 2 3 (3) = 4438. Theorem 3.21 If is a positive rea number, a 1, t = 1 and k>, then [k/] (k r) n a (k r) = F1 n (k) F 1 n (j), (42) where Proof. From (15), we have F n 1 (k n a k )= (k) =Δ 1 (k n a k ). (43) n Sr n n r [k (r) a k ]. (44) The proof foows from (10), (16) and (44). The foowing coroary iustrates Theorem 3.21, when n = 3. Coroary 3.22 If k [, ), (0, ) and a 1, then F 3 1 (k) = [k/] (k r) 3 a (k r) = F1 3 (k) F 1 3 (j), (45) a k [ (3) k (a +3k (2) + 2 k (1) ] a (k+) [ 3 +3k (2) 1) (a 1) 2 +6 2 k (1) ] + 62 a (k+2) [ ] (1) 6 3 a (k+3) + k (a 1) 3 (a 1). (46) 4 Exampe 3.23 In (45), by substituting k =62.1, =12.1, a =2and j =1.6, we obtain [ 62.1] 12.1 (62.1 12.1r) 3 2 62.1 12.1r = F1 3 (62.1) F 1 3 (1.6) = 1.544689685. where F1 3 (62.1) is obtained by putting k =62.1.
Discrete Bernoui s formua 239 Theorem 3.24 If k [2, ), t = 1 and a 1, then where F n 2 [k/] (r 1)(k r) n a (k r) = F n 2 (k) F n 2 ( + j), (47) (k) =Δ 1 F1 n(k) and F 1 n (k) is given in (43). Proof. The proof foows from (10) and (42). Coroary 3.25 If k [2, ) and a 1, then [k/] (r 1)(k r) 3 a (k r) = F2 3 (k) F 2 3 ( + j), (48) F2 3 (k) = [k(3) +3k (2) + 2 k (1) ] 2[2 +3k (2) a k (a 1) 2 a (k+) (a 1) 3 +6k (1) ] k(1) [(j) (3) +3(j) (2) + 2 (j) (1) ] + k(1) [ 2 +3(j) (2) +6(j) (1) ] (49) a j (a 1) a (+j) (a 1) 2 + 182 [ + k (1) ] a (k+2) (a 1) 6k(1) [ +(j) (1) ] 4 a (2+j) (a 1) + 62 k (1) a (3+j) 243 a (k+3) 3 (a 1) 4 (a 1) 5 and F1 3 (k) is given in (46). Proof. The proof foows by taking n = 3 in (47). Exampe 3.26 Putting k =9, =2, a =2and j =1in (40), we get 4 (r 1)(9 2r) 3 2 9 2r = F2 3 (9) F 2 3 (3) = 12.15625, where F2 3 (9) is obtained by substituting k =9in (49). References [1] R.P Agarwa, Difference Equations and Inequaities, Marce Dekker, New York, 2000. [2] M.Maria Susai Manue, G.Britto Antony Xavier and E.Thandapani, Theory of Generaized Difference Operator and Its Appications, Far East Journa of Mathematica Sciences, 20(2) (2006), 163-171. [3] M.Maria Susai Manue, G.Britto Antony Xavier and E.Thandapani, Quaitative Properties of Soutions of Certain Cass of Difference Equations, Far East Journa of Mathematica Sciences, 23(3) (2006), 295-304.
240 G. Britto Antony Xavier and H. Nasira Begum [4] M.Maria Susai Manue, G.Britto Antony Xavier and E.Thandapani, Generaized Bernoui Poynomias Through Weighted Pochhammer Symbos, Far East Journa of Appied Mathematics, 26(3) (2007), 321-333. [5] M. Maria Susai Manue, A. George Maria Sevam and G. Britto Antony Xavier, On the soutions and appications of some cass of generaized difference equations, Far East Journa of Appied Mathematics 28(2)(2007), 223-241. [6] M.Maria Susai Manue and G.Britto Antony Xavier, Recessive, Dominant and Spira Behaviours of Soutions of Certain Cass of Generaized Difference Equations, Internationa Journa of Differentia Equations and Appications, 10(4) (2007), 423-433. [7] M.Maria Susai Manue, G.Britto Antony Xavier and V.Chandrasekar, Generaized Difference Operator of the Second Kind and Its Appication to Number Theory, Internationa Journa of Pure and Appied Mathematics, 47(1) (2008), 127-140. [8] M.Maria Susai Manue, G.Britto Antony Xavier and V.Chandrasekar, Theory and Appication of the Generaized Difference Operator of the n th Kind (Part - I), Demonstratio Mathematica, 45(1) (2012), 95-106. [9] M.Maria Susai Manue, G.Britto Antony Xavier and V.Chandrasekar, Some Appications of the Generaized Difference Operator of the n th Kind, Far East Journa of Appied Mathematics, 66(2) (2012), 107-126. [10] Ronad E.Mickens, Difference Equations, Van Nostrand Reinhod Company, New York, 1990. [11] Ruta.C.Ferreira and Defim F.M.Jorres,Fractiona h-difference equations arising from the Cacuus of variations, Journa of Appicabe Anaysis and Discrete Mathematics, (2011). Received: September, 2012