... M´exico,Circuito Exterior, C.U. Coyoac´an 04 5 10, M´exico D.F.merino@matem.unam.mxSubmitted: Nov 21 , 20 07 ; Accepted: Jul 11, 20 08 ; Published: Jul 21 , 20 08 Mathematics Subject Classifications: 05 A19AbstractIf ... thatn 0 Jn +2 (−1)( 2) ntnn! = 1(cos(t) + sin(t)) 2 . (5)Taking x = 2 and y = −1 in Equation (3), we get the following identities1 +n≥1( 2) nTn (2, −1)tnn! = n 0 (−1)(n 2 )tnn! 2 = 1(cos(t) ... following:Theorem 2. For n ≥ 0, Jn +2 (−1) = Tn (2, −1).Proof. By taking y = −1 in Equation (2) we getn 0 Jn+1(−1)( 2) ntnn! = n 0 (−1)(n+1 2 )tnn!n 0 (−1)(n 2 )tnn! = F (t)H(t).Clearly,...