Dual Bicomplex Fibonacci Numbers with Fibonacci

Journal of Informatics and Mathematical Sciences Vol. 10, Nos. 1 & 2, pp. 161–172, 2018 ISSN 0975-5748 (online); 0974-875X (print) Published by RGN Publications

http://www.rgnpublications.com

http://dx.doi.org/10.26713/jims.v10i1-2.575

Research Article

Dual Bicomplex Fibonacci Numbers with Fibonacci and Lucas Numbers Faik Babadag˘ Department of Mathematics, Art & Science Faculty, Kırıkkale University, Ankara Yolu 7. Km, 71450 Yah¸sihan/Kırıkkale, Turkey [email protected]

Abstract. Recently, the authors give some results about Fibonacci and Lucas numbers. In this present paper, our object introduce a detailed study of a new generation of dual bicomplex Fibonacci numbers. We define new dual vector which is called dual Fibonacci vector. We give properties of dual Fibonacci vector to expert in geometry and then we introduce some formulas, facts and properties about dual bicomplex Fibonacci numbers and variety of geometric and algebraic properties which are not generally known. Keywords. Dual bicomplex Fibonacci numbers; Dual Fibonacci numbers; Dual Lucas numbers; Dual Fibonacci vector MSC. 11B39; 15A24 Received: February 22, 2018

Accepted: March 3, 2018

˘ This is an open access article distributed under the Creative Commons Attribution Copyright © 2018 Faik Babadag. License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction The defnition of dual numbers was first given by William Kingdon Clifford (1845-1879) [1] and some properties of that were studied in 1873 as a tool for his geometrical investigations. After him Eduard Study (1862-1930) used dual numbers and dual vectors in his research on line geometry and kinematics. He devoted special attention to the representation of oriented lines by dual unit vectors and defined the famous mapping. The set of oriented lines in a Euclidean three-dimensional space. E 3 is one to one correspondence with the points of a dual space D 3

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Dual Bicomplex Fibonacci Numbers with Fibonacci and Lucas Numbers: F. Babadag˘

of triples of dual numbers [1]. Dual numbers form two dimensional commutative associative algebra over the real numbers. Also the algebra of dual numbers is a ring. Dual numbers are used as an appliance for expressing and analyzing the physical properties of rigid bodies. They are computationally efficient approach of representing rigid transforms like translation and rotation. The remainder of the paper is organized as follows. In Section 2 we recall to some needed basic and fundamental concepts of dual numbers and dual bicomplex numbers. In Section 3 it is given a detailed study of a new generation of dual bicomplex Fibonacci numbers and in Section 4 it is given new a dual vector which is called dual Fibonacci vector and properties of this vector to expert in geometry. Finally, the conclusions and future works will be presented.

2. Preliminary This section is devoted to some basic and fundamental concepts of dual numbers and dual bicomplex numbers. A dual number is a number of the form e x = x + ε x∗ , where x and x∗ are real numbers called the components of e x and ε = (0, 1) arbitrary dual unit satisfy, the relation

ε2 = 0. Let us consider D = {e x|e x = x + ε x∗ , x, x∗ ∈ R}. Addition of two elements in D and scalar

multiplication of an element in D by a real number are defined in the usual way. D is a vector space with respect to addition and scalar multiplication. The product e x⊗ e y is the element in

D obtained by multiplying x + ε x∗ and y + ε y∗ as if they were polynomials and then using the relation ε2 = 0 to simplify the result. e x⊗ e y=e xe y = ( x + ε x∗ ) · ( y + ε y∗ ) = x y + ε( x y∗ + x∗ y).

Then the dual number x + ε x∗ divided by the dual number y + ε y∗ provided y 6= 0 can be defined as e x x + ε x∗ x x∗ y − x y∗ = + ε . = e y y + ε y∗ y y2 A dual bicomplex number is defined in the form [2, 3], e = X + εX ∗ X = ( x0 + x i 1 + x2 i 2 + x3 i 3 ) + ε( x0∗ + x1∗ i 1 + x2∗ i 2 + x3∗ i 3 ) = ( x0 + ε x0∗ ) + ( x1 + ε x1∗ ) i 1 + ( x2 + ε x2∗ ) i 2 + ( x3 + ε x3∗ ) i 3 =e x0 + e x1 i 1 + e x2 i 2 + e x3 i 3 ,

where e x0 , e x1 , e x2 , e x3 ∈ D and i 1 , i 2 , i 3 are the imaginary units, ε is dual unit. The dual bicomplex e is constructed from eight real parameters as x0 , x1 , x2 , x3 , ε x∗ , ε x∗ , ε x∗ , ε x∗ . number X 0

1

2

3

e = X + ε X ∗ and Y e = Y + εY ∗ be two dual bicomplex number, then the addition and Let X

subtraction are given by e ∓Y e = ( X ∓ Y ) + ε( X ∗ ∓ Y ∗ ) X

and multiplication is e ·Y e = X Y + ε(Y X ∗ + Y ∗ X ). X

Journal of Informatics and Mathematical Sciences, Vol. 10, No. 1 & 2, pp. 161–172, 2018

163


e are denoted by X e . There are different The conjugates of the dual bicomplex number X

conjugations of dual bicomplex number according to the imaginary units i 1 , i 2 and i 3 as follows:  e = ( x0 + ε x∗ ) − ( x1 + ε x∗ ) i 1 + ( x2 + ε x∗ ) i 2 − ( x3 + ε x∗ ) i 3  (a) X  0 1 2 3   ∗ ∗ ∗ ∗ e (b) X = ( x0 + ε x0 ) + ( x1 + ε x1 ) i 1 − ( x2 + ε x2 ) i 2 − ( x3 + ε x3 ) i 3 )    e = ( x0 + ε x∗ ) − ( x1 + ε x∗ ) i 1 − ( x2 + ε x∗ ) i 2 + ( x3 + ε x∗ ) i 3 .  (c) X 0 1 2 3

(1)

The Fibonacci and Lucas sequence are given as

F0 = 0, F1 = 1, F n+2 = F n + F n+1 and

L 0 = 2, L 1 = 1, L n = L n−1 + L n−2 . For Fibonacci and Lucas numbers, there are many study related on Fibonacci and Lucas numbers. Dunlap [4], Vajda [5] and Hoggatt [6] defined the properties of Fibonacci and Lucas numbers and gave the relation between them. The generalized Fibonacci sequences are described by Horadam [7, 8]

Q n = F n + i 1 F n+1 + i 2 F n+2 + i 3 F n+3 and

K n = L n + i 1 L n+1 + i 2 L n+2 + i 3 L n+3 respectively, where F n is Fibonacci number and L n is Lucas number. Also, i 1 , i 2 and i 3 are the imaginary units. Swamy [9] gave the relations of generalized Fibonacci quaternions. Iyer studied Fibonacci quaternions in [10] and obtained some other relations about Fibonacci and Lucas quaternions. In [11], Halıcı expressed the generating function and Binet formulas for ˘ Köksal and Tosun [12] defined the split Fibonacci and split Lucas these quaternions. Akyigit, quaternions. They also gave Binet formulas and Cassini identities for these quaternions. In this paper, our object give a detailed study of a new generation of dual bicomplex Fibonacci numbers. We define new a dual vector which is called dual Fibonacci vector. We give properties of this vector to expert in geometry.

3. Dual Fibonacci Bicomplex Numbers with Fibonacci and Lucas Number Components The n th dual Fibonacci numbers and the n th dual Lucas numbers are given by en = F n + εF n+1 , F e n = L n + εL n+1 , L

respectively, a dual bicomplex Fibonacci number and a dual bicomplex Lucas number can be given with Fibonacci and Lucas number components as following formulae, e xn = (F n + εF n+1 ) + i 1 (F n+1 + εF n+2 ) + i 2 (F n+2 + εF n+3 ) + i 3 (F n+3 + εF n+4 ), e n = (L n + εL n+1 ) + i 1 (L n+1 + εL n+2 ) + i 2 (L n+2 + εL n+3 ) + i 3 (L n+3 + εL n+4 ). k

)


(2)

164


Also, i 1 , i 2 and i 3 are the imaginary units and ε is the dual unit which satisfy the following rules: (

i 21 = −1, i 22 = −1, i 23 = 1, ε2 = 0

)

(3)

i 1 i 2 = i 2 i 1 = i 3 : i 1 i 3 = i 3 i 1 = −i 2 : i 2 i 3 = i 3 i 2 = −i 1.

The addition, subtraction and multiplication with scalar of dual bicomplex Fibonacci numbers are given by e xn ∓ e yn = ( xn ∓ yn ) + ε( xn+1 + yn+1 ) , λe xn = λ xn + λε xn+1 .

From equation (1), (2) and (3), the conjugates of dual bicomplex Fibonacci numbers are defined by e x n according to the imaginary units i 1 , i 2 and i 3 . We can be written as

x n = (F n + εF n+1 ) − (F n+1 + εF n+2 ) i 1 + (F n+2 + εF n+3 ) i 2 − (F n+3 + εF n+4 ) i 3 (a) e en − i 1 F en+1 + i 2 F en+2 − i 3 F en+3 , =F

            

x n = (F n + εF n+1 ) + (F n+1 + εF n+2 ) i 1 − (F n+2 + εF n+3 ) i 2 − (F n+3 + εF n+4 ) i 3 (b) e  en + i 1 F en+1 − i 2 F en+2 − i 3 F en+3 , =F      (c) e x n = (F n + εF n+1 ) − (F n+1 + εF n+2 ) i 1 − (F n+2 + εF n+3 ) i 2 + (F n+3 + εF n+4 ) i 3       e e e e = F n − i 1 F n+1 − i 2 F n+2 + i 3 F n+3 .

(4)

Theorem 1. Let e xn be dual bicomplex Fibonacci number. Then, Fibonacci number is the 2th 0 linear recurrence sequence. After that, we introduce the sets X D and X D as fallows

en , F en+1 , F en+2 , F en+3 ); F en is nth dual Fibonacci number} X D = {e xn | e xn = (F

and 0 XD

½ · ¸ ¾ e z1 −e z2 en | γ en = = γ ; (e z1 , e z2 ) are dual complex Fibonacci number . e z2 e z1

0 Then there is an isomorphism X D and X D , we can write · ¸ en + i 1 F en+1 −F en+2 − i 1 F en+3 F e e e e en = e e xn = (F n , F n+1 , F n+2 , F n+3 ) → γ . en+3 en + i 1 F en+1 F n+2 + i 1 F F

Therefore, we can give ½ · ¸ · ¸ ¾ 1 0 0 −1 0 en | γ en = e XD = γ z1 +e z2 ; (e z1 , e z2 ) are dual complex Fibonacci number 0 1 1 0 and en = F nU1 + F n+1U2 + F n+2U3 + F n+3U4 , γ

where [13], · ¸ · 1 0 i1 U1 = , U2 = 0 1 0

¸ · ¸ · ¸ 0 0 −1 0 −i1 , U3 = , U4 = . i1 1 0 i1 0

0 en ) 6= 0, the matrix γ en has the inverse and it is in X D Since det(γ .


165


e n , dual bicomplex Fibonacci and Lucas numbers. Then, for Theorem 2. Consider e xn and k

n ≥ 1, according to the imaginary units i 1 , i 1 and i 3 , the following relations hold:  (a) e xn + e xn+1 = e xn+2      e e e  xn − e xn+1 i 1 ) + i 2 ( e xn+2 − e xn+3 i 1 ) = −5F n+3 + 2 j L n+3  (b) X n = ( e (c)

e n = (e en+3 X xn + e xn+1 i 1 ) − i 2 ( e xn+2 + e xn+3 i 1 ) = (−3 + 2 i 1 )F

e n = (e e n+3 . (d) X xn − e xn+1 i 1 ) − i 2 ( e xn+2 − e xn+3 i 1 ) = 3L

      

Proof. By using equations (2) and (4), we will obtain the proof (a), (b), (c), (d). (a): By putting equation F n + F n+1 = F n+2 , according to the imaginary unit i 1 , we obtain e xn + e xn+1 = (F n + F n+1 ) + i 1 (F n+1 + F n+2 ) + i 2 (F n+2 + F n+3 ) + i 3 (F n+3 + F n+4 ) ε((F n+1 + F n+2 ) + i 1 (F n+2 + F n+3 ) + i 2 (F n+3 + F n+4 ) + i 3 (F n+4 + F n+5 ))

=e xn+2 .

(b): By putting equations F n+1 + F n−1 = L n and L n − L n+4 = 5F n+2 , according to the imaginary unit i 1 , we obtain e n = − [(F n + εF n+1 ) + (F n+2 + εF n+3 ) − (F n+4 + εF n+5 ) − (F n+6 + εF n+7 )] X + 2 j ((F n+2 + εF n+3 ) + (F n+4 + εF n+5 )) ¡ ¢ en + F en+2 − F en+4 − F en+6 + 2 j (F en+2 + F en+4 ) =− F ¡ ¢ e n+2 + L e n+4 + 2 j L e n+3 =− L en+3 + 2 j L e n+3 . = −5F

(d): By putting equations the identity of Fibonacci numbers and F n+4 + F n = 3F n+3 , according to the imaginary unit i 2 , we obtain e n = [(F n + εF n+1 ) − (F n+2 + εF n+3 ) + (F n+4 + εF n+5 ) − (F n+6 + εF n+7 )] X + 2 i 2 (F n+1 + +εF n+2 ) + (F n+5 + εF n+6 ) en+3 . = (−3 + 6 i )F

(d) By putting equations the identity of Fibonacci numbers and F n+1 + F n−1 = L n , L n + L n+4 = 3L n+2 , according to the imaginary unit i 3 , we obtain e n = [(F n + εF n+1 ) + (F n+2 + εF n+3 ) + (F n+4 + εF n+5 ) + (F n+6 + εF n+7 )] X e n+3 . = 3L

Lemma 3.1. Let e xn be dual bicomplex Fibonacci number. Then, we can give different conjugations according to i 1 , i 2 and i 3 the following relations,  en + i 2 F en+2 )  (a) e xn + e x n = 2((F n + εF n+1 ) + i 2 (F n+2 + εF n+3 )) = 2(F   e e (b) e xn + e x n = 2((F n + εF n+1 ) + i 1 (F n+1 + εF n+2 )) = 2(F n + i 1 F n+1 )   en + i 3 F en+3 )  (d) e xn + e x n = 2((F n + εF n+1 ) + i 3 (F n+3 + εF n+4 )) = 2(F

Proof. The proofs of (a), (b), (c), according to the imaginary units i 1 , i 2 and i 3 , are clear.


(5)

166


e n and X e n be dual bicomplex Fibonacci numbers and different conjugations Theorem 3. Let X

of dual bicomplex Fibonacci number according to the imaginary units i 1 , i 2 and i 3 . Then, we give the following relations between these numbers (a)

e n = −L e 2n+3 + 2 i 2 F e2n+3 en X X

(b)

en X e n = (−1 + 2 i 1 )F e2n+3 X

(c)

en X e n = 3F e2n+3 + 2 i 3 [(−1)n−1 ] X

Proof. From equations (2), (3) and (4), according to the imaginary units i 1 , i 2 and i 3 . (a): Using the identities of Fibonacci and Lucas numbers and equations F n F m + F n+1 F m+1 =

F n+m+1 , F n2 + F n2+1 = F2n+1 , according to the imaginary units i 1 , we see that en X en = F en2 + F e2 − F e2 − F e2 + 2 i 2 (F en F en+2 + F en+1 F en+3 ) X n+1 n+2 n+3 = F n2 + F n2+1 − F n2+2 − F n2+3 + 2ε([F n F n+1 + F n+1 F n+2 ] − [F n+2 F n+3 + F n+3 F n+4 ]) + 2 i 2 ((F n + εF n+1 )(F n+2 + εF n+3 ) + (F n+1 + εF n+2 )(F n+3 + εF n+4 )) = (F2n+1 − F2n+5 ) + ε(F2n+2 − F2n+6 ) + 2 i 2 (F2n+3 + εF2n+4 ) e 2n+3 + 2 i 2 F e2n+3 . = −L

(b): By using the equations F n+2 F n−1 = F n2+1 − F n2 , F n F n+1 = F n2+1 + (−1)n , F n F m + F n+1 F m+1 =

F n+m+1 and the identities of Fibonacci numbers and Lucas numbers, according to the imaginary units i 2 . We prove en X en = F en2 − F e2 + F e2 − F e2 + 2 i 1 (F en F en+1 + F en+2 F en+3 ) X n+1 n+2 n+3 e2n+3 e2n+3 + +2 i 1 F = −F e2n+3 . = (−1 + +2 i 1 )F

(c): By using the equations F n2 + F n2+1 = F2n+1 , F n F n+1 − F n+2 F n−1 = (−1)n−1 and the identities of Fibonacci numbers and Lucas numbers, according to the imaginary units i 1 . We prove en X en = F en2 + F e2 + F e2 + F e2 + 2 i 3 (F en F en+3 − F en+1 F en+2 ) X n+1 n+2 n+3 e2n+3 + 2 i 3 [(−1)n−1 ] . = 3F

Theorem 4. From equations (2), (3) and (4), let us consider dual bicomplex Fibonacci number e xn . Then, we give the following relations of this number.

(a)

en X en − X e n−1 X e n−1 = −L e 2n+2 + 2 i 2 F e2n+2 X

(b)

e n2 = 2 X n F n − 2L 2n+3 + 2 i 2 ( X n F n+2 − L n+1 ) X

Proof. (a): By using F n2 + F n2+1 = F2n+1 , F n+1 + F n−1 = L n , L n + L n+4 = 5F n and the identities of Fibonacci and Lucas numbers, we see that e n−1 X e n−1 = F e2 + F en2 − F e2 − F e2 + 2 i 2 (F en−1 F en+1 + F en F en+2 ) X n−1 n+1 n+2 = F n2−1 + F n2 − F n2+1 − F n2+2 + 2ε([F n−1 F n + F n F n+1 ] − [F n+1 F n+2 + F n+2 F n+3 ]) + 2 i 2 ((F n−1 + εF n )(F n+1 + εF n+2 ) + (F n + εF n+1 )(F n+2 + εF n+3 ))



167

e 2n+1 + 2 i 2 F e2n+1 , = −L en X en − X e n−1 X e n−1 = F en2 + F e2 − F e2 − F e2 + 2 i 2 (F en F en+2 + F en+1 F en+3 ) X n+1 n+2 n+3 e2 + F en2 − F e2 − F e2 + 2 i 2 (F en−1 F en+1 + F en F en+2 )) − (F n−1 n+1 n+2 e 2n+2 + 2 i 2 F e2n+2 . = −L

(b): By using equation (5), we have

X n2 = X n X n £ ¤ = X n 2(F n + i 2 F n+2 ) − X n∗ = X n (2F n + 2 i 2 F n+2 ) − X n X n∗ e 2n+3 + 2 i 2 F e2n+3 . = 2 X n (F n + i 2 F n+2 ) − L

In the the same manner, according to the imaginary units i 1 , i 3 , we can write X n2 . The proof is completed.

4. Dual Fibonacci Vector with the Dual Bicomplex Fibonacci Numbers Now, we describe a dual Fibonacci vector with vectorial part of dual bicomplex Fibonacci numbers asfollow  − e  → en + i 2 F en+1 + i 3 F en+2  V n = i1F   en , F en+1 , F en+2 ) = (F → − − e e en , F en+1 , F en+2 ) and → em , F em+1 , F em+2 ) are dual Fibonacci vectors and where V n = (F V m = (F en = F n + εF n+1 is n th dual Fibonacci number. F → − → − e e Definition 4.1. Consider dual Fibonacci vector, V n and V m , then dot product of these vectors

are defined by ¡→ ¢ → − − → − → − − → − → − → − e → e 〈 V n , V m 〉 = 〈 F n , F m 〉 + ε 〈 F n+1 , F m 〉 + 〈 F n , F m+1 〉 . → − − e → e Theorem 5. Let V n , V m be dual Fibonacci vectors, The dot product of these vectors is given as → − − e → e em+n+1 + (1 + ε)F n+2 F m+2 + εL m+n+3 . 〈V n, V m〉 = F

Proof. By using equations F n+1 F m+1 + F n+2 F m+2 = F n+m+3 , F n+1 + F n−1 = L n , we obtain → − − e → e en , F en+1 , F en+2 ), (F em , F em+1 , F em+2 )〉 〈 V n , V m 〉 = 〈(F = F n F m + F n+1 F m+1 + F n+2 F m+2 + ε(F n+1 F m + F n+2 F m+1 + F n+3 F m+2 + F n F m+1 + F n+1 F m+2 + F n+2 F m+3 ) = F n+2 F m+2 + F m+n+1 + ε(2F m+n+2 + F m+n+4 + F n+2 F m+2 ) em+n+1 + (1 + ε)F n+2 F m+2 + εL m+n+3 . =F

Definition 4.2. The permanent of a matrix A n×n is defined as p( A ) or per( A ) and n X Y a iσ(i) . per( A ) = σ∈S n i =1


168


Here S n indicates the symmetric group and permutation σ. The definition of the permanent of a matrix A n×n differs from that of the determinant of a matrix A n×n . Since the signatures of the permutations are not taken into account [14]. → − − → Definition 4.3 (The cross product of two vector). For vectors T = ( t 1 , t 2 , t 3 ) and W = ( y1 , y2 , y3 )

obtained by vectorial part of tessarine, the cross product is defined by using the permanent as following [14]. ¯ ¯ i −i2 → − − → ¯¯ 1 T ∧ W = ¯ t1 t2 ¯ y y2 1

i3 t3 y3

¯ ¯ ¯ ¯ = − i 1 [ t 2 y3 + t 3 y2 ] − i 2 [ t 1 y3 + t 3 y1 ] + i 3 [ t 1 y2 + t 2 y1 ] . ¯ ¯

→ − → − e e Theorem 6. Let V n , V m be dual Fibonacci vectors, then by using Definition 4.2 and

Definition 4.3, the cross product of these vectors is defined by → − → − → − → − → − → − → − → − e e V n ∧ V m = F n ∧ F m + ε( F n ∧ F m+1 + F n+1 ∧ F m ) . Proof. Using the equations L n+4 + L n = 3L n+2 , 5F m F n = L n+m − (−1)m L n−m , we get ¯ ¯ ¯ i1 ¯ − i i 2 3 ¯ ¯ → − → − ¯ F n ∧ F m = ¯ F n F n+1 F n+2 ¯¯ ¯ F ¯ m F m+1 F m+2

(6)

(7)

1 = {− i 1 (2L n+m+3 + (−1)m (L n−m−1 − L n−m+1 )) 5 − i 2 (2L n+m+2 − (−1)m (L n−m−2 + L n−m+2 )) + i 3 (2L n+m+1 + (−1)m (L n−m−1 − L n−m+1 ))}

1 = {− i 1 (2L n+m+3 − (−1)m L n−m ) − i 2 (2L n+m+2 − 3(−1)m L n−m ) 5 + i 3 (2L n+m+1 − (−1)m L n−m )} ¯ ¯ ¯ i1 −i2 i 3 ¯¯ ¯ → − → − F n+1 F n+2 ¯¯ F n ∧ F m+1 = ¯¯ F n ¯ F ¯ m+1 F m+2 F m+3

(8)

1 = {− i 1 (2L n+m+4 − (−1)m L n−m−1 ) − i 2 (2L n+m+3 + 3(−1)m L n−m−1 ) 5 + i 3 (2L n+m+2 + (−1)m L n−m−1 ).} and ¯ ¯ i1 −i2 i3 ¯ → − → − ¯ F n+1 × F m = ¯ F n+1 F n+2 F n+3 ¯ F F m+1 F m+2 m

¯ ¯ ¯ ¯ ¯ ¯

(9)

1 = {− i 1 (2L n+m+4 + (−1)m (L n−m+1 ) − i 2 (2L n+m+3 − 3(−1)m (L n−m+1 ) 5 + i 3 (2L n+m+2 − (−1)m (L n−m+1 ).} Substituting the equations (7), (8) and (9) in equation (6), using the identities of Fibonacci and



169

Lucas numbers, we have 1 e n+m+3 + (−1)m L n−m (1 + ε) + 2εL n+m+4 ) = {− i 1 (2L 5 e n+m+2 − 3(−1)m L n−m (1 + ε) + 2εL n+m+3 ) − i 2 (L e n+m+1 + (−1)m L n−m (1 + ε) + 2εL n+m+2 ).} + i 3 (L

(10)

→ − → − e e Lemma 4.4. Consider dual Fibonacci vector V n . Then, V n is a dual unit Fibonacci vector if

and only if L 2n+2 = 5−(−21)

n+1

.

Proof. Using the equations 5F m F n = L n+m − (−1)m L n−m , we have °→ e °1 °− V n ° 2 = F n2 + F n2+1 + F n2+2 = F n F n + F n+1 F n+1 + F n+2 F n+2 ¢ 1¡ = L 2n + L 2n+2 + L 2n+4 + (−1)n+1 5 ¢ 1¡ = 2L 2n+2 + (−1)n+1 . 5 °→ n+1 − e ° Since ° V n ° = 1, we obtain L 2n+2 = 5−(−21) . → − → − e e Theorem 7. Get two units of dual Fibonacci Vector, V n and V m . Then the dual bicomplex

number product of these vectors is given as £→ ¡→ ¢¤ → − → − − → − − → − → − → − e e V n × V m = 〈 F n , F m 〉 + ε 〈 F n+1 , F m 〉 + 〈 F n , F m+1 〉 → − → − → − → − → − → − + F n Λ F m + ε( F n Λ F m+1 + F n+1 Λ F m ).

(11)

Proof. Substituting the equations (6) and (10) in equation (11), using the identities of dual Fibonacci numbers and dual Lucas numbers, we obtain e m+n+4 + F em+n+1 − (−1)m+2 L n−m (1 + ε) =L

1 e n+m+3 + (−1)m L n−m (1 + ε) + 2εL n+m+4 ) + {− i 1 (2L 5 e n+m+2 − 3(−1)m L n−m (1 + ε) + 2εL n+m+3 ) − i 2 (L e n+m+1 + (−1)m L n−m (1 + ε) + 2εL n+m+2 )}. + i 3 (L → − − − Definition 4.5 (Study Mapping). Let us consider a dual unit vector V = → u + ε→ u ∗ . Then consider → − − → − − the equation of oriented line which is related to V , → γ =− u ×→ u ∗ + w→ u , where t ∈ I ⊂ R is a

parameter. → − → − − Proof. Denote a dual unit vector V , then the oriented line related to V is in direction of → u as

Study mapping, any two point M and X are on line [15]. Let us consider origin which is point

O , then we can write −−→ −−→ −−→ OX = OM + M X


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−−→ −−→ → OX = OM + t− u. − − Let a point X be on the line vectors → u and → u ∗ , then we can write −−→ − → − u ∗ = OX × → u.

(12)

By using equation (12), we obtain −−→ − → − − − u ×→ u∗ =→ u × (OX × → u) −−→ − −−→ → − − = 〈→ u ,→ u 〉OX − 〈→ u , OX 〉− u.

In equation (12), if we write the last equation, we obtain −−→ → −−→ − − − OM = − u ×→ u ∗ + (〈→ u , OX 〉 + t)→ u. −−→ − Let us consider 〈→ u , OX 〉 + t = w then if a new parameter is γ, we can write the result as: → − − − − γ =→ u ×→ u ∗ + w→ u. (13)

Figure 1. DStudy Mapping.

→ − → − e e Corollary 4.6. Let us denote dual Fibonacci bicomplex unit vector V o n . A unit vector V o n is → − → − → − → − e e terminal point V n = F n = F n + ε F n+1 on the dual unit sphere. Then the equation of oriented → − e line which is according to F n is given as → − − − − γ =→ u ×→ u ∗ + w→ u → − → − γ = 2 i 1 (L 2n+4 + (−1)n ) + i 2 (2L 2n+3 − 3(−1)n ) + i 3 (2L 2n+4 − 3(−1)n ) + w F n ,

where w ∈ I ⊂ R is a parameter.



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Proof. We can obtain the equation of oriented line by using equation (13), which is related to → − e F n as following: → − → − → − → − γ = F n × F n+1 + w F n . By using Definition 4.3 and Definition 4.4, the identities of Fibonacci numbers and equation 5F m F n = L n+m − (−1)m L n−m , we obtain → − → − → − − u ×→ u ∗ = F n × F m+1 ¯ ¯ −i1 −i2 i3 ¯ ¯ = ¯ F n F n+1 F n+2 ¯F n+1 F n+2 F m+3

¯ ¯ ¯ ¯ ¯ ¯

= 2 i 1 (L 2n+4 + (−1)n ) − i 2 (2L 2n+3 − 3(−1)n ) + i 3 (2L 2n+4 − 3(−1)n ).

In equation (13), if we write last equation. This completes the procession.

5. Conclusion bicomplex numbers that are emerged as a generalization of the best known dual bicomplex numbers. We gave the dual vector which is called dual Fibonacci vector for these dual bicomplex numbers. By using dual Fibonacci vector, we get some results of this vector to expert in geometry such as the cross product of two vector and study mapping.

Competing Interests The author declares that he has no competing interests.

Authors’ Contributions The author wrote, read and approved the final manuscript.

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