Some Remarkable Infinite Product Identities Involving Fibonacci

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arXiv:1702.08321v2 [math.NT] 29 Apr 2017 oegnrliett t edrvdi hspeetpaper): present this in derived be (to identity general more h following The Introduction 1 rgnlyotie yMla n hno 2,i an is [2], Shannon and Melham by obtained originally iyivligFbncinumbers, Fibonacci involving tity oermral nnt rdc identities product infinite remarkable Some ∗ † M lsicto ubr 13,11Y60 [email protected] 11B39, [email protected], : Numbers Classification AMS novn ioac n ua numbers Lucas and Fibonacci involving utepoete fteFbnciadLcsnmes eder we involving numbers, identities Lucas bers. product and infinite emplo Fibonacci general and the arguments interesting as of ratio properties golden subtle the of powers inverse fu ing tangent hyperbolic it inverse involving and identities formula to summation ants telescoping classic the applying By eateto hsc n niern Physics, Engineering and Physics of Department k Y ∞ =1 bfm wlw nvriy l-f,Nigeria Ile-Ife, University, Awolowo Obafemi F F (2 (2 striking n n − − 1)(2 1)(2 k k +2 +2 i rnca’ ecito 1)ifiiepoutiden- product infinite [1]) description Frontczak’s (in q q ) ) − + ul Adegoke Kunle k Y F F ∞ =1 2 2 q q (2 (2 F F Abstract n n 2 2 − − k k +2 +2 1) 1) 1 − = 1 + 1 k Y =1 q 3 = F L (2 (2 , † n n − − 1)(2 1)(2 k k − − 1) 1) n L F = 1 = cin hav- nctions (2 (2 hs num- these n n − − 1)2 1)2 vari- s k k q ying . aeo a of case ive ∗ + 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 identities, 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 numbers 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 . √5F2kp+mp−p φ Xk=1 Xk=1 Identities (2.8) — (2.11) are established by converting each inﬁnite sum identity to an inﬁnite product identity with the appropriate choice of m and p in each case.

In closing this section we observe that q =1= n in (2.10) and (2.11) yield the following special evaluations:

∞ L + √5 ∞ √5F +1 2k+1 = φ4 , 2k = φ3 . L2k+1 √5 √5F2k 1 kY=1 − kY=1 − 1 2pk 2.3 Identities resulting from taking f(k) = tanh− (φ− ) in (1.4) Theorem 2.3. The following inﬁnite product identities hold for q, n Z+: ∈ ∞ k−1 q F4nq+4nk +( 1) F4nq L2n(2k−1)F4nk − k = , (2.12) F4nq+4nk +( 1) F4nq F2n(2k−1)L4nk kY=1 − kY=1 7 ∞ k−1 q F(2n−1)(2q+2k) +( 1) F(2n−1)2q q F(2n−1)(2k−1)F(2n−1)2k − k =5 , F(2n−1)(2q+2k) +( 1) F(2n−1)2q L(2n−1)(2k−1)L(2n−1)2k k=1 − k=1 Y Y (2.13) ∞ k−1 q q−1 L4nk+4nq−2n +( 1) L4nq−2n 1 L2n(2k−1) F4nk − k = (2.14) L4nk+4nq−2n +( 1) L4nq−2n √5 F2n(2k−1) L4nk Yk=1 − kY=1 kY=1 and

∞ F +( 1)k−1F (2n−1)(2q+2k−1) − (2n−1)(2q−1) F +( 1)kF k=1 (2n−1)(2q+2k−1) (2n−1)(2q−1) Y − (2.15) q F q−1 F =(√5)2q−1 (2n−1)(2k−1) (2n−1)2k . L(2n−1)(2k−1) L(2n−1)2k Yk=1 kY=1 Proof. Taking f(k) = tanh−1(φ−2pk) in (1.4), setting m =2q and making use of the identities (1.10b) and (1.7a) we have

∞ 2q k−1 −1 F2qp k−1 −1 1 ( 1) tanh = ( 1) tanh 2pk , − F2kp+2qp − φ Xk=1 Xk=1 from which follows

∞ k−1 2q 2pk k−1 F2pq+2pk +( 1) F2pq φ +( 1) − k = 2pk − k F2pq+2pk +( 1) F2pq φ +( 1) k=1 − k=1 − Y Yq q φ2p(2k−1) +1 φ4pk 1 = − φ2p(2k−1) 1 φ4pk +1 k=1 − k=1 Yq Y (2.16) F2p(2k−1)√5 L 2 = 4pk − L2p(2k−1) 2 F √5 k=1 − 4pk Y q F F =5q 2p(2k−1) 2pk . L2p(2k−1) 2 L2pk kY=1 − The ﬁnal evaluation of the denominator of the product term in the last line of identity (2.16) depends on the parity of p. Setting p =2n in (2.16), making use of (1.9b) proves (2.12) while setting p = 2n 1 in (2.16), − utilizing (1.9d) proves (2.13).

8 Taking f(k) = tanh−1(φ−2pk) in (1.4), setting m =2q 1 and making use of the identities (1.10a) and (1.7) we have −

∞ 2q−1 k−1 −1 L2qp−p k−1 −1 1 ( 1) tanh = ( 1) tanh 2pk , p even , − L2kp+2qp−p − φ Xk=1 Xk=1 and

∞ 2q−1 k−1 −1 F2qp−p k−1 −1 1 ( 1) tanh = ( 1) tanh 2pk , p odd , − F2kp+2qp−p − φ Xk=1 Xk=1 from which identities (2.14) and (2.15) follow immediately.

1 p(2k 1) 2.4 Identities resulting from taking f(k) = tanh− (φ− − ) in (1.4) Theorem 2.4. The following inﬁnite product identities hold for q, n Z+: ∈ ∞ k−1 q F8nk+8nq−4n +( 1) F8nq L2n(4k−3)F2n(4k−1) − k = , (2.17) F8nk+8nq−4n +( 1) F8nq F2n(4k−3)L2n(4k−1) kY=1 − Yk=1 ∞ F +( 1)k−1F q F L (4n−2)(2q+2k−1) − (2n−1)4q = (2n−1)(4k−3) (2n−1)(4k−1) , F +( 1)kF L F k=1 (4n−2)(2q+2k−1) − (2n−1)4q k=1 (2n−1)(4k−3) (2n−1)(4k−1) Y Y (2.18) ∞ L +( 1)k−1√5F (2n−1)(2q+2k−1) − (2n−1)2q k L(2n−1)(2q+2k−1) +( 1) √5F(2n−1)2q kY=1 − (2.19) q (F √5+2)L = (2n−1)(4k−3) (2n−1)(4k−1) , (F(2n−1)(4k−1)√5+2)L(2n−1)(4k−3) Yk=1 ∞ k−1 q q−1 L8n(k+q−1) +( 1) L4n(2q−1) 1 L2n(4k−3) F2n(4k−1) − k = , L8n(k+q−1) +( 1) L4n(2q−1) √5 F2n(4k−3) L2n(4k−1) k=1 − k=1 k=1 Y Y Y (2.20) ∞ k−1 L(8n−4)(k+q−1) +( 1) L(4n−2)(2q−1) − k L(8n−4)(k+q−1) +( 1) L(4n−2)(2q−1) k=1 Y − (2.21) q F q−1 L = √5 (2n−1)(4k−3) (2n−1)(4k−1) L(2n−1)(4k−3) F(2n−1)(4k−1) kY=1 kY=1 9 and ∞ L +( 1)k−1√5F (4n−2)(q+k−1) − (2n−1)(2q−1) k√ k=1 L(4n−2)(q+k−1) +( 1) 5F(2n−1)(2q−1) Y − (2.22) q (F √5+2) q−1 L = (2n−1)(4k−3) (2n−1)(4k−1) . L(2n−1)(4k−3) (F(2n−1)(4k−1)√5+2) kY=1 Yk=1 Proof. Taking f(k) = tanh−1(φ−p(2k−1)) in (1.4) and setting m = 2q we obtain the following summation identities

∞ 2q k−1 −1 F2qp k−1 −1 1 ( 1) tanh = ( 1) tanh 2pk−p , p even , − F2kp+2qp−p − φ k=1 k=1 X X (2.23) and

∞ 2q k−1 −1 F2qp√5 k−1 −1 1 ( 1) tanh = ( 1) tanh 2pk−p , p odd . − L2kp+2qp−p − φ k=1 " # k=1 X X (2.24) Identities (2.17) and (2.18) follow from (2.23) while the identity (2.19) follows directly from (2.24).

Taking f(k) = tanh−1(φ−p(2k−1)) in (1.4) and setting m =2q 1 we obtain − the following summation identities

2q−1 ∞ L 1 ( 1)k−1 tanh−1 p(2q−1) = ( 1)k−1 tanh−1 , p even − L − φ2pk−p k=1 2p(k+q−1) k=1 X X (2.25) and

∞ 2q−1 k−1 −1 Fp(2q−1)√5 k−1 −1 1 ( 1) tanh = ( 1) tanh 2pk−p , p odd . − L2p(k+q−1) − φ k=1 " # k=1 X X (2.26) Identities (2.20) and (2.21) follow from evaluating the summation identity (2.25) at p = 4n and p = 4n 2n, respectively. Identity (2.22) is obtained by converting the summation− identity (2.26) to an inﬁnite product identity at p =2n 1. −

10 From (2.19) and (2.22) at q =1= n come the special evaluations

∞ L +( 1)k−1√5 ∞ L +( 1)k−1√5 2k+1 − = φ2 , 2k − = φ3 . k k L2k+1 +( 1) √5 L2k +( 1) √5 kY=1 − kY=1 − References

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