The k-fibonacci Dual Quaternions

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1 International Journal of Mathematical Analysis Vol. 12, 2018, no. 8, HIKARI Ltd, The k-fibonacci Dual Quaternions Fügen Torunbalcı Aydın Yildiz Technical University Faculty of Chemical Metallurgical Engineering Department of Mathematical Engineering Davutpasa Campus, Esenler, Istanbul, Turkey Copyright c 2018 Fügen Torunbalcı Aydın. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, reproduction in any medium, provided the original work is properly cited. Abstract In this paper, k-fibonacci dual quaternions are defined. Also, some algebraic properties of k-fibonacci dual quaternions which are connected with k-fibonacci numbers Fibonacci numbers are investigated. Furthermore, d Ocagne s identity, the Honsberger identity, Binet s formula, Cassini s identity Catalan s identity for these quaternions are given. Mathematics Subject Classification: 11R52; 11B37; 11B39; 20G20 Keywords: Fibonacci number, k-fibonacci number, k-fibonacci quaternion, dual quaternion, k-fibonacci dual quaternion 1. Introduction The quaternions constitute an extension of complex numbers into a fourdimensional space can be considered as four-dimensional vectors, in the same way that complex numbers are considered as two-dimensional vectors. Quaternions were first described by Irish mathematician Hamilton in Hamilton [10] introduced a set of quaternions which can be represented as H = { q = q 0 + i q 1 + j q 2 + k q 3 q 0, q 1, q 2, q 3 R } (1.1) i 2 = j 2 = k 2 = 1, i j = j i = k, j k = k j = i, k i = i k = j. (1.2)

2 364 Fügen Torunbalcı Aydın Several authors worked on different quaternions their generalizations. ([1],[12]-[15],[17],[20]). Now, let s talk about the work done on Fibonacci quaternion dual Fibonacci quaternion: Horadam [11] defined complex Fibonacci Lucas quaternions as follows Q n = F n + i F n+1 + j F n+2 + k F n+3 K n = L n + i L n+1 + j L n+2 + k L n+3 i 2 = j 2 = k 2 = 1, i j = j i = k, j k = k j = i, k i = i k = j. In 2012, Halıcı [8] gave generating functions Binet s formulas for Fibonacci Lucas quaternions. In 2013, Halıcı [9] defined complex Fibonacci quaternions as follows: H F C ={R n = C n + e 1 C n+1 + e 2 C n+2 + e 3 C n+3 C n = F n + i F n+1, i 2 = 1} e 2 1 = e 2 2 = e 2 3 = e 1 e 2 e 3 = 1, e 1 e 2 = e 2 e 1 = e 3, e 2 e 3 = e 3 e 2 = e 1, e 3 e 1 = e 1 e 3 = e 2, n 1. Majernik [16] defined dual quaternions as follows: { } Q = a + b i + c j + d k a, b, c, d R, i 2 = j 2 = k 2 = i j k = 0, H D =. i j = j i = j k = k j = k i = i k = 0 For more details on dual quaternions, see [3]. In 2015, Yüce Torunbalcı Aydın [22] defined dual Fibonacci quaternions as follows: H D = { Q n = F n + i F n+1 + j F n+2 + k F n+3 F n, n th Fibonacci number}, i 2 = j 2 = k 2 = i j k = 0, i j = j i = j k = k j = k i = i k = 0. Ramirez[19] defined the the k-fibonacci the k-lucas quaternions as follows: D ={F + i F +1 + j F +2 + k F +3 F, n th k-fibonacci number}, P ={L + i L +1 + j L +2 + k L +3 L, n th k-lucas number} i, j, k satisfy the multiplication rules (1.2).

3 The k-fibonacci dual quaternions 365 In 2015, Polatlı Kızılateş Kesim [18] defined split k-fibonacci split k-lucas quaternions (M ) (N ) respectively as follows: M ={F + i F +1 + j F +2 + k F +3 F, n th k-fibonacci number} i, j, k are split quaternionic units which satisy the multiplication rules i 2 = 1, j 2 = k 2 = i j k = 1, i j = j i = k, j k = k j = i, k i = i k = j. In 2017, Catarino Vasco [2] defined dual k-pell Quaternions Octanions ( R ) as follows: R ={R + ε R +1 R = P e 0 + P +1 e 1 + P +2 e 2 + P +3 e 3, P, n th k-pell number} e 0 = 1, e i e j = e j e i, (e i ) 2 = 1, (i, j = 1, 2, 3). In this paper, the k-fibonacci dual quaternions the k-lucas dual quaternions will be defined respectively, as follows DF = {DF =F + i F +1 + j F +2 + k F +3 F, n th k-fibonacci number} DL = {DL =L + i L +1 + j L +2 + k L +3 L, n th k-lucas number} (1.3) (1.4) i 2 = j 2 = k 2 = i j k = 0, i j = j i = j k = k j = k i = i k = 0. (1.5) The aim of this work is to present in a unified manner a variety of algebraic properties of both the k-fibonacci dual quaternions as well as the k- Fibonacci quaternions. In accordance with these definitions, we given some algebraic properties Binet s formula for k-fibonacci dual quaternions. Moreover, some sums formulas some identities such as d ocagnes, Honsberger, Cassini s Catalan s identities for these quaternions are given. 2. The k-fibonacci Dual Quaternion The k-fibonacci sequence {F } n N [19] is defined as F k,0 = 0, F k,1 = 1 F +1 = k F + F 1, n 1 or {F } n N = { 0, 1, k, k 2 + 1, k k, k k 2 + 1,...} Here, k is a positive real number. (2.1)

4 366 Fügen Torunbalcı Aydın In this section, firstly the k-fibonacci dual quaternions will be defined. The k-fibonacci dual quaternions are defined by using the k-fibonacci numbers as follows DF = { DF = F + i F +1 + j F +2 + k F +3 F, n th (2.2) k-fibonacci number}, i 2 = j 2 = k 2 = i j k = 0, i j = j i = j k = k j = k i = i k = 0. Let DF 1 DF 2 be two k-fibonacci dual quaternions such that DF 1 = F + i F +1 + j F +2 + k F +3 (2.3) DF 2 = G + i G +1 + j G +2 + k G +3 (2.4) Then, the addition subtraction of two k-fibonacci dual quaternions are defined in the obvious way, DF 1 ± DF 2 = (F + i F +1 + j F +2 + k F +3 ) ±(G + i G +1 + j G +2 + k G +3 ) = (F ± G ) + i (F +1 ± G +1 ) + j (F +2 ± G +2 ) +k (F +3 ± G +3 ). (2.5) Multiplication of two k-fibonacci dual quaternions is defined by DF 1 DF 2 = (F + i F +1 + j F +2 + k F +3 ) (G + i G +1 + j G +2 + k G +3 ) = (F G ) + i (F G +1 + F +1 G ) +j (F G +2 + F +2 G ) +k (F G +3 + F +3 G ) = (F G ) + F (i G +1 + j G +2 + k G +3 ) +(i F +1 + j F +2 + k F +3 )G. The scaler the vector parts of the k-fibonacci dual quaternion denoted by (2.6) DF are S DF = F V DF = i F +1 + j F +2 + k F +3. (2.7) Thus, the k-fibonacci dual quaternion DF is given by DF = S DF + V DF. Then, relation (2.6) is defined by DF 1 DF 2 =S DF 1 S DF 2 + S DF 1 V DF 2 + S DF 2 V DF 1. (2.8) The conjugate of k-fibonacci dual quaternion DF is denoted by DF it is DF = F i F +1 j F +2 k F +3. (2.9) The norm of the k-fibonacci dual quaternion DF is defined as follows DF 2 = DF DF = (F ) 2. (2.10)

5 The k-fibonacci dual quaternions 367 In the following theorem, some properties related to the k-fibonacci dual quaternions are given. Theorem 2.1. Let F DF be the n th terms of k-fibonacci sequence (F ) the k-fibonacci dual quaternion (DF ), respectively. In this case, for n 1 we can give the following relations: DF +2 = k DF +1 + DF, (2.11) (DF ) 2 = 2 F DF DF DF = 2 F DF (F ) 2, (2.12) DF i DF +1 j DF +2 k DF +3 = F. (2.13) Proof. (2.11): By the equation (2.2) we get, DF + k DF +1 = (F + i F +1 + j F +2 + k F +3 ) +k (F +1 + i F +2 + j F +3 + k F +4 ) = (F + k F +1 ) + i (F +1 + k F +2 ) +j (F +2 + k F +3 ) + k (F +3 + k F +4 ) = F +2 + i F +3 + j F +4 + k F +5 = DF +2. (2.12): By the equation (2.2) we get, (DF ) 2 = F i (F F +1 ) + 2 j (F F +2 ) + 2 k (F F +3 ) = 2 F (F + i F +1 + j F +2 + k F +3 ) F 2 = 2F DF F 2. (2.13): By the equation (2.2) we get, DF i DF +1 j DF +2 k DF +3 = (F + i F +1 + j F +2 + k F +3 ) i (F +1 + i F +2 + j F +3 + k F +4 ) j (F +2 + i F +3 + j F +4 + k F +5 ) k (F +3 + i F +4 + j F +5 + k F +6 ) = F. Theorem 2.2. For m n + 1 the d Ocagne s identity for the k-fibonacci dual quaternions DF k,m DF is given by DF k,m DF +1 DF k,m+1 DF =( 1) n F k,m n ( DF k,1 + j + k k ) =( 1) n F k,m n ( DF k,1 + DF k, 1 1 ). (2.14)

6 368 Fügen Torunbalcı Aydın Proof. (2.14): By using (2.2) DF k,m DF +1 DF k,m+1 DF = (F k,m F +1 F k,m+1 F ) +i [ (F k,m F +2 F k,m+1 F +1 ) +((F k,m+1 F +1 F k,m+2 F ) ] +j [ (F k,m F +3 F k,m+1 F +2 ) +(F k,m+2 F +1 F k,m+3 F ) ] +k [ (F F +4 F k,m+1 F +3 ) +(F k,m+3 F +1 F k,m+4 F ) ] = ( 1) n F k,m n [ 1 + k i + (k 2 + 2) j + (k k)k ] = ( 1) n F k,m n ( DF k,1 + j + k k ) = ( 1) n F k,m n ( DF k,1 + DF k, 1 1 ). Here, d Ocagne s identity of k-fibonacci number F k,m F +1 F k,m+1 F = ( 1) n F k,m n in [5] was used. Theorem 2.3. For n, m 0 the Honsberger identity for the k-fibonacci dual quaternions DF DF k,m is given by DF +1 DF k,m + DF DF k,m 1 =2 DF +m+1 F +m+1. (2.15) Proof. (2.15): By using (2.6) (2.11) DF DF k,m = F F k,m + i (F F k,m+1 + F +1 F k,m ) +j (F F k,m+2 + F +2 F k,m ) +k (F F k,m+3 + F +3 F k,m ), DF +1 DF k,m+1 = F +1 F k,m+1 +i (F +1 F k,m+2 + F +2 F k,m+1 ) +j (F +1 F k,m+3 + F +3 F k,m+1 ) +k (F +1 F k,m+4 + F +4 F k,m+1 ). Finally, adding by two sides to the side, we obtain DF DF k,m + DF +1 DF k,m+1 = (F F k,m + F +1 F k,m+1 ) +i [F F k,m+1 + F +1 F k,m + F +1 F k,m+2 + F +2 F k,m+1 ] +j [F F k,m+2 + F +2 F k,m + H +1 F k,m+3 + F +3 F k,m+1 ] +k [F F k,m+3 + F +3 F k,m + F +1 F k,m+4 + F +4 F k,m+1 ] = 2 [ F +m+1 + i F +m+2 + j F +m+3 + k F +m+4 ] F +m+1 = 2 DF +m+1 F +m+1. Here, the Honsberger identity of k-fibonacci number F +1 F k,m +F F k,m 1 = F +m in [4] was used. Theorem 2.4. Let DF DL be n-th terms of k-fibonacci dual quaternion (DF ) k-lucas dual quaternion (DL ), respectively. The following relation is satisfied DF +1 + DF 1 = DL. (2.16)

7 The k-fibonacci dual quaternions 369 Proof. (2.16): From equation (2.2) identity F +1 + F 1 = L n 1 [19] between k-fibonacci number k-lucas number, it follows that DF +1 + DF 1 = (F +1 + F 1 ) + i (F +2 + F ) + j (F +3 + F +1 ) +k (F +4 + F +2 ) = (L + i L +1 + j L +2 + kl +3 ) = DL. Theorem 2.5. Let DF be conjugation of the k-fibonacci dual quaternion (DF ). In this case, we can give the following relations between these quaternions: DF + DF = 2 F, (2.17) DF DF + DF 1 DF 1 = F 2 + F 2 1 = F k,2n 1, (2.18) 19(DF ) 2 + (DF n 1 ) 2 = 2 DF k,2n 1 F k,2n 1, (2.19) DF 2 +1 DF 2 1 = k ( 2 DF k,2n F k,2n ). (2.20) Proof. (2.17): By using (2.9), we get DF + DF = (F + i H n+1 + j F +2 + k F +3 ) +(F i F +1 j F +2 k F +3 ) = 2 F. (2.18): By using (2.10), we get DF DF + DF 1 DF 1 = F 2 + F 2 1 = F k,2n 1. the identity of k-fibonacci number F 2 + F 2 +1 = F k,2n+1, n 0 [19] was used. (2.19): By using (2.12), we get DF 2 + DF 2 1 = 2 (DF F + DF 1 F 1 ) (F 2 + F 2 1 ) = 2 [ (F 2 + F 2 1 ) + i (F F +1 + F 1 F ) +j (F F +2 + F 1 F +1 ) +k (F F +3 + F 1 F +2 )] (F 2 + F 2 1 ) = 2 (F k,2n 1 + i F k,2n + j F k,2n+1 + k F k,2n+2 ) F k,2n 1 = 2 DF k,2n 1 F k,2n 1. the identity of k-fibonacci number F 2 + F 2 +1 = F k,2n+1, n 0 [19] was used. (2.20): By using (2.12), we get DF 2 +1 DF 2 1 = 2 (DF +1 F +1 DF 1 F 1 ) (F F 2 1 ) = 2 [ (F 2 +1 F 2 1 ) + i (F +1 F +2 F 1 F ) +j (F +1 F +3 F 1 F +1 ) +k (F +1 F +4 F 1 F +2 )] (F 2 +1 F 2 1 ) = 2 k (F k,2n + i F k,2n+1 + j F k,2n+2 + k F k,2n+3 ) F k,2n = k [ 2 DF k,2n F k,2n ].

8 370 Fügen Torunbalcı Aydın the identity of k-fibonacci number F+1 2 F 1 2 = k F k,2n, [4] was used. Theorem 2.6. Let DF be the k-fibonacci dual quaternion. Then, we have the following identities DF k,s = 1 k (DF +1 + DF DF k,1 DF k,0 ), (2.21) DF k,2s 1 = 1 k (DF k,2n DF k,0 ), (2.22) DF k,2s = 1 k (DF k,2n+1 DF k,1 ). (2.23) Proof. (2.21): Since n i=1 F k,i = 1 k (F +1 + F 1), [?], we get DF k,s = F k,s + i F k,s+1 + j F k,s+2 + k = 1 k { [(F +1 + F 1) + i [F +2 + F +1 k 1] + j [F +3 + F +2 (k 2 + 1) k] + k [F +4 + F +3 (k 3 + 2k) (k 2 + 1)] } = 1 k {(F +1 + i F +2 + j F +3 + k F +4 ) + (F + i F +1 + j F +2 + k F +3 ) F k,s+3 [1 + i (k + 1) + j (k 2 + k + 1) + k (k 3 + k 2 + 2k + 1)] } = 1 k {DF +1 + DF [ F k,1 + i (F k,2 + F k,1 ) + j (F k,3 + F k,2 ) + k (F k,4 + F k,3 ) ] } = 1 k (DF +1 + DF DF k,1 DF k,0 ). (2.22): Using n F k,2i 1 = 1 F k k,2n i=1 F k,2i = 1 (F k 2n+1 1) [4], we get DF k,2s 1 = 1 { (F k k,2n) + i (F k,2n+1 1) + j (F k,2n+2 k) i=1 +k [ F k,2n+3 (k 2 + 1) ]} = 1 k {[ F k,2n + i F k,2n+1 + j F k,2n+2 + k F k,2n+3 ] [ i + k j + (k 2 + 1) k ]} = 1 k {DF k,2n (F k,0 + i F k,1 + j F k,2 + k F k,3 )} = 1 k (DF k,2n DF k,0 ).

9 The k-fibonacci dual quaternions 371 (2.23): Using n F k,2i = 1(F k k,2n+1 1) [?], we obtain i=1 DF k,2s = 1{(F k k,2n+1 1) + i [F k,2n+2 k] + j [F 2n+3 (k 2 + 1)] +k [F k,2n+4 (k 3 + 2k)]} = 1 k { (F k,2n+1 + i F k,2n+2 + j F k,2n+3 + k F k,2n+4 ) [1 + k i + (k 2 + 1) j + (k 3 + 2k) k ] } = 1 k (DF k,2n+1 DF k,1 ). Theorem 2.7. (Binet s Formula). Let DF be the k-fibonacci dual quaternion. For n 1, Binet s formula for these quaternions is as follows: 1 ( DF = ˆα α n ˆβ ) β n (2.24) k2 + 4 ˆα = 1 + i (k β) + j ( (k 2 + 1) k β ) + k ( (k k) (k 2 + 1)β ), ˆβ = 1 + i (α k) + j [ k α (k 2 + 1) ] + k [ (k 2 + 1) α (k k) ]. Proof. The characteristic equation of recurrence relation DF +2 = k DF +1 + DF is t 2 k t 1 = 0. The roots of this equation are α = k + k β = k k α + β = k, α β = k 2 + 4, αβ = 1. Using recurrence relation initial values DF k,0 = (0, 1, k, k 2 + 1), DF k,1 = (1, k, k 2 + 1, k k), the Binet formula for DF is DF = A α n + B β n = 1 k2 + 4 [ ˆα α n ˆβ β n ], A = DF k,1 β DF k,0, B = α DF k,0 DF k,1 α β α β ˆα = 1 + i α + j α 2 + k α 3, ˆβ = 1 + i β + j β 2 + k β 3. Theorem 2.8. (Cassini s Identity). Let DF be the k-fibonacci dual quaternion. For n 1, Cassini s identity for DF is as follows: DF 1 DF +1 (DF ) 2 = ( 1) n (DF k,1 + j + k k) = ( 1) n (DF k,1 + DF k, 1 1). (2.25)

10 372 Fügen Torunbalcı Aydın Proof. (2.25): By using (2.2) (2.5), we get DF 1 DF +1 (DF ) 2 = (F 1 + i F + j F +1 + k F +2 ) (F +1 + i F +2 + j F +3 + k F +4 ) [F + i F +1 + j F +2 + k F +3 ] 2 = [F 1 F +1 F 2 ] +i [F 1 F +2 + F F +2 2 F F +1 ] +j [F 1 F +3 2 F F +2 + F 2 +1 ] +k [F 1 F +4 + F +1 F +2 2 F F +3 ] = ( 1) n (DF k,1 + j + k k) = ( 1) n (DF k,1 + DF k, 1 1). Here, the identity of the k-fibonacci number F 1 F n+1 F 2 = ( 1)n [4] was used. Theorem 2.9. (Catalan s Identity). Let DF be the k-fibonacci dual quaternion. For n 1, Catalan s identity for DF is as follows: DF +r 1 DF +r+1 (DF +r ) 2 = ( 1) n+r (DF k,1 + DF k, 1 1). (2.26) Proof. (2.26): By using (2.16) (2.17) we get DF +r 1 DF +r+1 (DF +r ) 2 = (F +r 1 F +r+2 F +r F +r+1 ) +j (F +r 1 F +r+3 + F 2 +r+1 2 F +rf +r+2 ) +k (F +r 1 F +r+4 + F +r+2 F +r+1 2 F +r F +r+3 ) = ( 1) n+r (1 + k i + (k 2 + 2) j + (k k)k ) = ( 1) n+r [ 1 + k i + (k 2 + 1) j + (k k)k + j + k k ] = ( 1) n+r (DF k,1 + DF k, 1 1). we use Catalan s identity of the k-fibonacci number F +r 1 F +r+1 Fn+r 2 = ( 1) n+r [4]. 3. Conclusion In this study, a number of new results on k-fibonacci dual quaternions were derived. Quaternions have great importance as they are used in quantum physics, applied mathematics, quantum mechanics, Lie groups, kinematics differential equations.this study fills the gap in the literature by providing the k-fibonacci dual quaternion using definitions of the k-fibonacci number dual Fibonacci quaternion [22]. References [1] M. Akyiğit, H. H. Kösal M. Tosun, Split Fibonacci Quaternions, Adv. Appl.Clifford Algebras, 23 (2013), no. 3, [2] P. Catarino, P. Vasco, On Dual k-pell Quaternions Octonions, Mediterr. J. Math., 14 (2017), no. 2, [3] Z. Ercan, S. Yüce, On properties of the dual quaternions, Eur. J. Pure Appl. Math., 4 (2011), no. 2, [4] S. Falcon, A. Plaza, On the Fibonacci k-numbers, Chaos Solitons Fractals, 32 (2007), no. 5,

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