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Some Remarkable Infinite Product Identities Involving Fibonacci

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Some Remarkable Infinite Product Identities Involving Fibonacci Some remarkable infinite product identities involving Fibonacci and Lucas numbers∗ Kunle Adegoke † Department of Physics and Engineering Physics, Obafemi Awolowo University, Ile-Ife, Nigeria Abstract By applying the classic telescoping summation formula and its vari- ants to identities involving inverse hyperbolic tangent functions hav- ing inverse powers of the golden ratio as arguments and employing subtle properties of the Fibonacci and Lucas numbers, we derive interesting general infinite product identities involving these num- bers. 1 Introduction The following striking (in Frontczak’s description [1]) infinite product iden- tity involving Fibonacci numbers, ∞ F +1 2k+2 =3 , F2k+2 1 Yk=1 − originally obtained by Melham and Shannon [2], is an n =1= q case of a more general identity (to be derived in this present paper): arXiv:1702.08321v2 [math.NT] 29 Apr 2017 ∞ F + F q F L (2n−1)(2k+2q) 2q(2n−1) = (2n−1)(2k−1) (2n−1)2k . F(2n−1)(2k+2q) F2q(2n−1) L(2n−1)(2k−1)F(2n−1)2k Yk=1 − Yk=1 ∗AMS Classification Numbers : 11B39, 11Y60 †[email protected], [email protected] 1 + The above identity is valid for q N0, n Z , where we use N0 to denote the set of natural numbers including∈ zero.∈ Here the Fibonacci numbers, Fn, and Lucas numbers, Ln, are defined, for n N , as usual, through the recurrence relations F = F + F , with ∈ 0 n n−1 n−2 F0 =0, F1 =1 and Ln = Ln−1 + Ln−2, with L0 =2, L1 =1. Our goal in this paper is to derive several general infinite product identi- ties, including the one given above. Our method consists of applying the following general telescoping summation property (equation (2.1) of [3], also (2.17) of [4]) N m m [f(k) f(k + m)] = f(k) f(k + N), for N m 1 , − − ≥ ≥ k=1 k=1 k=1 X X X (1.1) and its alternating version, obtained by replacing f(k) with ( 1)k−1f(k) in (1.1), namely, − N ( 1)k−1 f(k)+( 1)m−1f(k + m) − − k=1 X m m (1.2) = ( 1)k−1f(k)+( 1)N−1 ( 1)k−1f(k + N) − − − Xk=1 Xk=1 to some inverse hyperbolic tangent identities with the inverse powers of the golden mean as arguments of the inverse hyperbolic tangent functions. If f(N) 0 as N , then we have, from (1.1) and (1.2), the useful identities→ → ∞ ∞ m [f(k) f(k + m)] = f(k) (1.3) − Xk=1 Xk=1 and ∞ m ( 1)k−1 f(k)+( 1)m−1f(k + m) = ( 1)k−1f(k) . (1.4) − − − k=1 k=1 X X In handling infinite products, the following identity will often come in handy m ⌈m/2⌉ ⌊m/2⌋ f(k)= f(2k 1) f(2k) , (1.5) − Yk=1 kY=1 kY=1 2 where u is the greatest integer less than or equal to u and u is the smallest⌊ integer⌋ greater than or equal to u. ⌈ ⌉ Specifically, we have 2q q f(k)= f(2k 1)f(2k) − kY=1 kY=1 and 2q−1 q q−1 f(k)= f(2k 1) f(2k) . − kY=1 Yk=1 Yk=1 We shall adopt the following empty product convention: 0 f(k)=1 . kY=1 The golden ratio, having the numerical value of (√5 + 1)/2, denoted throughout this paper by φ, will appear frequently. We shall often make use of the following algebraic properties of φ φ2 = 1+ φ , (1.6a) √5 = 2φ 1 , (1.6b) − φ 1 = 1/φ , (1.6c) − n φ = φFn + Fn−1 , (1.6d) φ−n = ( 1)n( φF + F ) , (1.6e) − − n n+1 and φn = φn−1 + φn−2 . (1.6f) n −n Using Binet’s formula, Fn√5= φ ( φ) , n Z and identities (1.6d) and (1.6e), it is straightforward to establish− − the followin∈ g useful identities n −n φ φ = Fn√5, n even (1.7a) n − −n φ + φ = Ln, n even (1.7b) n −n φ + φ = Fn√5, n odd (1.7c) φn φ−n = L , n odd (1.7d) − n 3 From (1.7) we also have the following useful results φn +1 F √5+2 = n , n odd, (1.8a) φn 1 L − n φn +1 F √5 = n , n even . (1.8b) φn 1 L 2 − n − We shall make repeated use of the following identities connecting Fi- bonacci and Lucas numbers: F2n = FnLn , (1.9a) L 2( 1)n =5F 2 , (1.9b) 2n − − n 5F 2 L2 = 4( 1)(n+1) , (1.9c) n − n − L + 2( 1)n = L2 . (1.9d) 2n − n Identities (1.9) or their variations can be found in [5, 6, 7]. Finally, the relevant inverse hyperbolic function identities that we shall require are the following x + y tanh−1 x + tanh−1 y = tanh−1 , xy< 1 (1.10a) 1+ xy x y tanh−1 x tanh−1 y = tanh−1 − , xy> 1 , (1.10b) − 1 xy − − x 1 y + x tanh−1 = log , x < y , (1.10c) y 2 y x | | | | − y x π tanh−1 = tanh−1 i , x < y . (1.10d) x y − 2 | | | | Infinite product identities involving Fibonacci numbers and Lucas num- bers were also derived in [1, 2, 8] and references therein. 4 2 Infinite product identities 1 2pk 2.1 Identities resulting from taking f(k) = tanh− (φ− ) in (1.3) Theorem 2.1. For q, n Z+ the following infinite product identities hold: ∈ ∞ L + L q F q−1 L (2n−1)(2k+2q−1) (2n−1)(2q−1) = √5 (2k−1)(2n−1) 2k(2n−1) , L(2n−1)(2k+2q−1) L(2n−1)(2q−1) L(2k−1)(2n−1) F2k(2n−1) k=1 − k=1 k=1 Y Y Y (2.1) 2q−1 ∞ F + F 1 L 2n(2k+2q−1) 2n(2q−1) = 2nk , (2.2) F2n(2k+2q−1) F2n(2q−1) (√5)2q−1 F2nk kY=1 − kY=1 ∞ 2q F4n(k+q) + F4nq 1 L2nk = q (2.3) F4n(k+q) F4nq 5 F2nk kY=1 − kY=1 and ∞ F + F q F L (2n−1)(2k+2q) 2q(2n−1) = (2n−1)(2k−1) (2n−1)2k . (2.4) F(2n−1)(2k+2q) F2q(2n−1) L(2n−1)(2k−1)F(2n−1)2k Yk=1 − Yk=1 Proof. Choosing f(k) = tanh−1(φ−2pk) in (1.3) and using (1.10b) and (1.7), we obtain ∞ m −1 Lpm −1 1 tanh = tanh 2pk , p m odd (2.5) Lp(2k+m) φ Xk=1 Xk=1 and ∞ m −1 Fpm −1 1 tanh = tanh 2pk , p m even . (2.6) Fp(2k+m) φ Xk=1 Xk=1 Setting p =2n 1 and m =2q 1 in (2.5) and converting the infinite sum identity to an infinite− product identity,− employing also the identities (1.8), 5 (1.9b), (1.9d) and (1.5), we have ∞ L(2n−1)(2k+2q−1) + L(2n−1)(2q−1) L(2n−1)(2k+2q−1) L(2n−1)(2q−1) kY=1 − 2q−1 F √5 = 2k(2n−1) L2k(2n−1) 2 kY=1 − 2q−1 F =(√5)2q−1 2k(2n−1) L 2 k=1 2k(2n−1) Y − (2.7) q F q−1 F =(√5)2q−1 2(2k−1)(2n−1) 4k(2n−1) L2(2k−1)(2n−1) 2 L4k(2n−1) 2 kY=1 − Yk=1 − q F L q−1 F L √ 2q−1 (2k−1)(2n−1) (2k−1)(2n−1) 2k(2n−1) 2k(2n−1) =( 5) 2 2 L(2k−1)(2n−1) 5F2k(2n−1) kY=1 kY=1 q F q−1 L = √5 (2k−1)(2n−1) 2k(2n−1) , L(2k−1)(2n−1) F2k(2n−1) kY=1 kY=1 which proves (2.1). The possibilities for the choice of m and p in (2.6) are (i) p even, m odd, (ii) p even, m even and (iii) p odd, m even. Converting the infinite sums in (2.6) to infinite products and setting p = 2n and m = 2q 1 proves − identity (2.2); while the identity (2.3) is proved by setting p = 2n and m =2q. Finally, identity (2.4) is proved by setting p =2n 1 and m =2q in the infinite product identity resulting from (2.6). − 1 p(2k 1) 2.2 Identities resulting from taking f(k) = tanh− (φ− − ) in (1.3) Theorem 2.2. For q, n Z+ the following infinite product identities hold: ∈ ∞ q F4n(2k+q−1) + F4nq 1 L2n(2k−1) = q , (2.8) F4n(2k+q−1) F4nq √5 F2n(2k−1) Yk=1 − Yk=1 ∞ q F + F q F (4n−2)(2k+q−1) 2q(2n−1) = √5 (2n−1)(2k−1) , (2.9) F(4n−2)(2k+q−1) F2q(2n−1) L(2n−1)(2k−1) kY=1 − Yk=1 6 ∞ L + √5F 2q F √5+2 (2k−1+2q)(2n−1) 2q(2n−1) = (2k−1)(2n−1) (2.10) L(2k−1+2q)(2n−1) √5F2q(2n−1) L(2k−1)(2n−1) kY=1 − kY=1 and ∞ √5F + L 2q−1 F √5+2 2(2n−1)(k+q−1) (2n−1)(2q−1) = (2k−1)(2n−1) . (2.11) √5F2(2n−1)(k+q−1) L(2n−1)(2q−1) L(2k−1)(2n−1) kY=1 − kY=1 Proof. Choosing f(k) = tanh−1(φ−p(2k−1)) in (1.3) and proceeding as in the previous section we obtain the following infinite sum identities: ∞ m −1 Fmp −1 1 tanh = tanh p(2k−1) , p even , F2kp+mp−p φ Xk=1 Xk=1 ∞ m −1 Fmp√5 −1 1 tanh = tanh p(2k−1) , p odd, m even , "L2kp+mp−p # φ Xk=1 Xk=1 and ∞ m −1 Lmp −1 1 tanh = tanh p(2k−1) , p m odd .
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