Golden Binomials and Carlitz Characteristic Polynomials
aa r X i v : . [ m a t h . QA ] D ec Golden Binomials and Carlitz CharacteristicPolynomials
Oktay K Pashaev and Merve ¨OzvatanDepartment of MathematicsIzmir Institute of TechnologyUrla-Izmir, 35430, TurkeyDecember 22, 2020
Abstract
The golden binomials, introduced in the golden quantum calculus,have expansion determined by Fibonomial coefficients and the set ofsimple zeros given by powers of Golden ratio. We show that thesegolden binomials are equivalent to Carlitz characteristic polynomialsof certain matrices of binomial coefficients. It is shown that traceinvariants for powers of these matrices are determined by Fibonaccidivisors, quantum calculus of which was developed very recently.
The golden quantum calculus, based on the Binet formula for Fibonaccinumbers F n as q -numbers, was introduced in [1]. In this calculus, the finite-difference q -derivative operator is determined by two Golden ratio bases ϕ and ϕ ′ , while the golden binomial expansion, by Fibonomial coefficients. Thecoefficients are expressed in terms of Fibonacci numbers, while zeros of thesebinomials are given by powers of Golden ratio ϕ and ϕ ′ . It was observed thatsimilar polynomials were introduced by Carlitz in 1965 from different reason,as characteristic polynomials of certain matrices of binomial coefficients [3].The goal of the present paper is to show equivalence of Carlitz characterisitcpolynomials with golden binomials. In addition, the proof and interpretation1f main formulas for trace of powers of the matrix A n +1 in terms of Fibonaccidivisors and corresponding quantum calculus, developed recently in [4] wouldbe given. The binomial coefficients defined by " nk F = [ n ] F ![ n − k ] F ![ k ] F ! = F n ! F n − k ! F k ! , (1)with n and k being non-negative integers, n ≥ k , are called the Fibonomials.Using the addition formula for Fibonacci numbers [1], F n + m = ϕ n F m + ϕ ′ m F n (2)we have following expression F n = F n − k + k = − ϕ ! k F n − k + ϕ n − k F k . (3)By using ϕ n = ϕF n + F n − , ϕ ′ n = ϕ ′ F n + F n − , (4)it can be rewritten as follows F n = F n − k − F k + F n − k F k +1 = F n − k F k − + F n − k +1 F k . (5)With the above definition (1) it gives recursion formula for Fibonomials intwo forms, " nk F = ( − ϕ ) k [ n − F ![ k ] F ![ n − k − F ! + ϕ n − k [ n − F ![ n − k ] F ![ k − F != − ϕ ! k " n − k F + ϕ n − k " n − k − F (6)= ϕ k " n − k F + − ϕ ! n − k " n − k − F . (7)These formulas, for 1 ≤ k ≤ n −
1, determine the Golden Pascal triangle forFibonomials [1]. 2 .2 Golden Binomial
The Golden Binomial is defined as [1],( x + y ) nF = ( x + ϕ n − y )( x + ϕ n − ϕ ′ y ) ... ( x + ϕϕ ′ n − y )( x + ϕ ′ n − y ) (8)or due to ϕϕ ′ = − x + y ) nF = ( x + ϕ n − y )( x − ϕ n − y ) ... ( x + ( − n − ϕ − n +1 y ) . (9)It has n-zeros at powers of the Golden ratio xy = − ϕ n − , xy = − ϕ n − , ..., xy = − ϕ − n +1 . For Golden binomial the following expansion in terms of Fibonomials is valid[1] ( x + y ) nF = n X k =0 " nk F ( − k ( k − x n − k y k = n X k =0 F n ! F n − k ! F k ! ( − k ( k − x n − k y k . (10)The proof is easy by induction and using recursion formulas (6), (7) . Interms of Golden binomials we introduce the Golden polynomials P n ( x ) = ( x − a ) nF F n ! , (11)where n = 1 , , ... , and P ( x ) = 1. These polynomials satisfy relations D xF P n ( x ) = P n − ( x ) , (12)where the Golden derivative is defined as D xF P n ( x ) = P n ( ϕx ) − P n ( ϕ ′ x )( ϕ − ϕ ′ ) x . (13)For even and odd polynomials we have different products P n ( x ) = 1 F n ! n Y k =1 ( x − ( − n + k ϕ k − a )( x + ( − n + k ϕ − k +1 a ) , (14)3 n +1 ( x ) = ( x − ( − n a ) F n +1 ! n Y k =1 ( x − ( − n + k ϕ k a )( x − ( − n + k ϕ − k a ) . (15)By using (4) it is easy to find ϕ k + 1 ϕ k = F k + 2 F k − , (16) ϕ k +1 − ϕ k +1 = F k +1 + 2 F k . (17)Then we can rewrite our polynomials in terms of Fibonacci numbers P n ( x ) = 1 F n ! n Y k =1 ( x − ( − n + k ( F k − + 2 F k − ) xa − a ) , (18) P n +1 ( x ) = ( x − ( − n a ) F n +1 ! n Y k =1 ( x − ( − n + k ( F k + 2 F k − ) xa + a ) . (19)The first few odd polynomials are P ( x ) = ( x − a ) , (20) P ( x ) = 12 ( x + a )( x − xa + a ) , (21) P ( x ) = 12 · · x − a )( x + 3 xa + a )( x − xa + a ) , (22) P ( x ) = 12 · · · ·
13 ( x + a )( x − xa + a )( x + 7 xa + a )( x − xa + a ) , (23)and the even ones P ( x ) = ( x − xa − a ) , (24) P ( x ) = 12 · x + xa − a )( x − xa − a ) , (25) P ( x ) = 12 · · · x − xa − a )( x + 4 xa − a )( x − xa − a ) . (26)4 .3 Golden analytic function By golden binomials in complex domain, the golden analytic function can bederived, which is complex valued function of complex argument, not analyticin usual sense [2]. The complex golden binomial is defined as( x + iy ) nF = ( x + iϕ n − y )( x − iϕ n − y ) ... ( x + i ( − n − ϕ − n y ) (27)= n X k =0 " nk F ( − k ( k − x n − k i k y k . (28)It can be generated by the golden translation E iyD xF F x n = ( x + iy ) nF , where E xF = ∞ X n =0 ( − n ( n − x n F n ! . The binomials determine the golden analytic function f ( z, F ) = E iyD xF F f ( x ) = ∞ X n =0 a n ( x + iy ) nF F n ! , satisfying the golden ¯ ∂ F equation12 ( D xF + iD y − F ) f ( z ; F ) = 0 , (29)where D x − F = ( − x ddx D xF . For u ( x, y ) = Cos F ( yD xF ) f ( x ) and v ( x, y ) = Sin F ( yD xF ) f ( x ), the golden Cauchy-Riemann equations are D xF u ( x, y ) = D y − F v ( x, y ) , D y − F u ( x, y ) = − D xF v ( x, y ) , (30)and the golden-Laplace equation is( D xF ) u ( x, y ) + ( D y − F ) u ( x, y ) = 0 . (31) The golden binomial ( x − a ) nF can be also generated by the golden translation E − aD xF F x n = ( x − a ) nF . (32)5n particular case a = 1 we have( x − mF = ( x − ϕ m − )( x + ϕ m − ) ... ( x − ( − m − ϕ − m +1 ) . (33)First few binomials are( x − F = x − , (34)( x − F = ( x − ϕ )( x − ϕ ′ ) , (35)( x − F = ( x − ϕ )( x + 1)( x − ϕ ′ ) , (36)( x − F = ( x − ϕ )( x + ϕ )( x + ϕ ′ )( x − ϕ ′ ) , (37)and corresponding zeros m = 1 ⇒ x = 1 (38) m = 2 ⇒ x = ϕ, x = ϕ ′ (39) m = 3 ⇒ x = ϕ , x = − , x = ϕ ′ (40) m = 4 ⇒ x = ϕ , x = − ϕ, x = − ϕ ′ , x = ϕ ′ . (41)For arbitrary even and odd n we have following zeros of Golden binomials n = 2 k ⇒ ( x − kF : ϕ n − , ϕ ′ n − , − ϕ n − , − ϕ ′ n − , ..., ± ϕ, ± ϕ ′ ; (42) n = 2 k + 1 ⇒ ( x − k +1 F : ϕ n − , ϕ ′ n − , − ϕ n − , − ϕ ′ n − , ..., ± . (43) In Section 2 we have introduced the Golden binomials. Now we are goingto relate these binomials with characteristic equations for some matrices,constructed from binomial coefficients by Carlitz [3].
Definition 3.0.1
We define an ( n + 1) × ( n + 1) matrix A n +1 with binomialcoefficients, A n +1 = " rn − s ! , (44) where r, s = 0 , , , ..., n. Here, nk ! = ( n !( n − k )! k ! , if k ≤ n; , k > n. (45)6irst few matrices are, n = 0 ⇒ r = s = 0 ⇒ A = " ! = (1) n = 1 ⇒ r, s = 0 , ⇒ A = " r − s ! = (cid:16) (cid:17) (cid:16) (cid:17)(cid:16) (cid:17) (cid:16) (cid:17) = ! n = 2 ⇒ r, s = 0 , , ⇒ A = " r − s ! = (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) = Continuing, the general matrix A n +1 of order ( n + 1) can be written as, A n +1 = . . . . . . . . . . . . . . . ( n +1) × ( n +1) , where the lower triangular matrix is build from Pascal’s triangle. We noticethat trace of first few matrices A n +1 gives Fibonacci numbers. As would beshown, it is valid for any n (Theorem (4 . .
5) equation (60)) .
Definition 3.0.2
Characteristic polynomial of matrix A n +1 is determinedby, Q n +1 ( x ) = det ( xI − A n +1 ) . (46)First few polynomials explicitly are n = 0 : Q ( x ) = x − ,n = 1 : Q ( x ) = det( xI − A ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − − x − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = x − x − ,n = 2 : Q ( x ) = det( xI − A ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − x − − − − x − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = x − x − x + 1 ,n = 3 : 7 ( x ) = det( xI − A ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − x − − − x − − − − − x − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − x + 3 x + 6 x − x − . Corresponding eigenvalues are represented by powers of ϕ and ϕ ′ ; n = 0 ⇒ x = 1, n = 1 ⇒ x = ϕ, x = ϕ ′ , n = 2 ⇒ x = ϕ , x = − , x = ϕ ′ , n = 3 ⇒ x = ϕ , x = − ϕ, x = − ϕ ′ , x = ϕ ′ .Comparing zeros of first few characteristic polynomials, with zeros ofGolden Binomial (33), we notice that they coincide. According to this, wehave following conjecture. Conjecture:
The characteristic equation (46) of matrix A n +1 coincideswith Golden Binomial; Q n +1 ( x ) = det( xI − A n +1 ) = ( x − n +1 F . (47)To prove this conjecture, firstly we represent Golden binomials in the productform. Proposition 3.0.3
The Golden binomial can be written as a product, ( x − n +1 F = n Y j =0 (cid:16) x − ϕ j ϕ ′ n − j (cid:17) . (48) Proof 3.0.4
Starting from Golden binomial in product representation ( x + y ) nF ≡ n − Y j =0 (cid:16) x − ( − j − ϕ n − ϕ − j y (cid:17) (49) by using ϕ − j = ϕ ! j = − ϕ ! j = ϕ ′ j , (50) after substitution y = − we have ( x − nF ≡ n − Y j =0 (cid:16) x − ( − j ϕ n − ϕ ′ j (cid:17) . y shifting n → n + 1 , ( x − n +1 F = n Y j =0 (cid:16) x − ( − j ϕ n ϕ ′ j (cid:17) = n Y j =0 x − ( − j ϕ n ( − j ϕ j ϕ j ! = n Y j =0 x − ϕ n − ϕ ! j ϕ j = n Y j =0 (cid:16) x − ϕ n − j ϕ ′ j (cid:17) and substituting j = n − m we get, ( x − n +1 F = n Y m =0 (cid:16) x − ϕ m ϕ ′ n − m (cid:17) . The formula shows explicitly that zeros of Golden binomial in (42) and (43) are given by powers of ϕ and ϕ ′ . Corollary 3.0.5
The eigenvalues of matrix A n +1 are the numbers, ϕ n , ϕ n − ϕ ′ , ϕ n − ϕ ′ , . . . , ϕ ϕ ′ n − , ϕ ′ n . (51)As it was shown by Carlitz [3], this product formula is just characteristicequation (46) for matrix A n +1 . Since zeros of two polynomials det( xI − A n +1 )and ( x − n +1 F coincide, then the conjecture is correct and we have followingtheorem. Theorem 3.0.6
Characteristic equation for combinatorial matrix A n +1 isgiven by Golden binomial: Q n +1 ( x ) = det ( xI − A n +1 ) = ( x − n +1 F . (52) A n +1 and Fibonacci Divisors Proposition 4.0.1
Arbitrary n th power of A matrix is written in terms ofFibonacci numbers, A n = F n − F n F n F n +1 ! . (53)9 roof 4.0.2 Proof will be done by induction. For n = 1 , A = ! = F F F F ! , and for n = 2 , A = ! = F F F F ! . Suppose for n = k , A k = F k − F k F k F k +1 ! , then A k +12 = A k A = F k − F k F k F k +1 ! ! (54)= F k F k + F k − F k +1 F k + F k +1 ! = F k F k +1 F k +1 F k +2 ! . This result can be understood from observation that eigenvalues of matrix A are ϕ and ϕ ′ , and eigenvalues of A n are powers ϕ n , ϕ ′ n related with Fibonaccinumbers. As we have seen, eigenvalues of matrix A are ϕ , ϕ ′ , −
1. It implies thatfor A n , eigenvalues are ϕ n , ϕ ′ n , ( − n , and the matrix can be expressed byFibonacci divisor F (2) n conjugate to F , due to [4],( ϕ k ) n = ϕ k F ( k ) n + ( − k +1 F ( k ) n − , (55)( ϕ ′ k ) n = ϕ ′ k F ( k ) n + ( − k +1 F ( k ) n − , (56)where F ( k ) n = F nk /F k . Proposition 4.0.3
Arbitrary n th power of A matrix can be expressed interms of Fibonacci divisors F (2) n , A n = 15 (2 F (2) n − F (2) n − + 2( − n ) (2 F (2) n + 2 F (2) n − + 2( − n ) (3 F (2) n − F (2) n − − − n )( F (2) n + F (2) n − + ( − n ) (6 F (2) n − F (2) n − + ( − n ) (4 F (2) n − F (2) n − − ( − n )(3 F (2) n − F (2) n − − − n ) (8 F (2) n − F (2) n − − − n ) (7 F (2) n − F (2) n − + 2( − n ) roof 4.0.4 Let’s diagonalize the matrix A , φ = σ − A σ , where φ is the diagonal matrix and A = σ φ σ − . Taking the n th power of both sides gives, A n = ( σ φ σ − ) ( σ | {z } I φ σ − ) ... ( σ φ σ − ) ( σ | {z } I φ σ − ) Therefore, A n = σ φ n σ − . (57) By using the diagonalization principle, σ and σ − matrices can be obtainedas, σ = 12 − ϕ ′ − ϕ ϕ − ϕ ′ and, σ − = ϕ ′ +2)5( ϕ − ϕ ′ ) − ϕ ′ +2)5 ϕ ′ ( ϕ − ϕ ′ ) 2(2 ϕ ′ − ϕ ′ ( ϕ − ϕ ′ )35 35 − − ϕ +2)5( ϕ − ϕ ′ ) 4( ϕ +2)5 ϕ ( ϕ − ϕ ′ ) 2(1 − ϕ )5 ϕ ( ϕ − ϕ ′ ) = 25 √ ϕ ′ + 2 − − ϕ ) (2 + ϕ ) √
52 3 √ − √ − ( ϕ + 2) 2(1 − ϕ ′ ) − (2 + ϕ ′ ) Since eigenvalues of matrix A are ϕ , − , ϕ ′ , the diagonal matrix φ is, φ = ϕ ′ − ϕ ′ , (58) and an arbitrary n th power of this matrix is, φ n = ( ϕ ′ ) n − n
00 0 ( ϕ ′ ) n . (59)11 inally by using (57) , A n =15 (2 F (2) n − F (2) n − + 2( − n ) (2 F (2) n + 2 F (2) n − + 2( − n ) (3 F (2) n − F (2) n − − − n )( F (2) n + F (2) n − + ( − n ) (6 F (2) n − F (2) n − + ( − n ) (4 F (2) n − F (2) n − − ( − n )(3 F (2) n − F (2) n − − − n ) (8 F (2) n − F (2) n − − − n ) (7 F (2) n − F (2) n − + 2( − n ) is obtained. As we can expect, these results can be generalized to arbitrary matrix A n +1 . Since eigenvalues of A n +1 are powers ϕ n , ϕ ′ n , . . . , for A Nn +1 eigenvaluesare ϕ nN , ϕ ′ nN , . . . But these powers can be written in terms of Fibonaccidivisors as in (55), (56), and the matrix A Nn +1 itself can be represented byFibonacci divisors F ( n ) N .For powers of matrix A n +1 we have the following identities. Theorem 4.0.5
Invariants of A kn +1 matrix are found as, T r (cid:16) A kn +1 (cid:17) = F kn + k F k = F ( k ) n +1 , (60) det (cid:16) A kn +1 (cid:17) = ( − k n ( n +1)2 . (61) For k = 1 , it gives T r ( A n +1 ) = F n +1 ,det ( A n +1 ) = ( − n ( n +1)2 . Proof 4.0.6
Let’s diagonalize the general matrix A n +1 as, φ n +1 = σ − n +1 A n +1 σ n +1 where φ n +1 is diagonal and A n +1 = σ n +1 φ n +1 σ − n +1 . Taking the k th power of both sides gives, A kn +1 = ( σ n +1 φ n +1 σ − n +1 ) ( σ n +1 | {z } I φ n +1 σ − n +1 ) ... ( σ n +1 φ n +1 σ − n +1 ) ( σ n +1 | {z } I φ n +1 σ − n +1 )12 nd A kn +1 = σ n +1 φ kn +1 σ − n +1 . (62) By taking trace from both sides and using the cyclic permutation property oftrace,
T r ( A kn +1 ) = T r ( σ n +1 φ kn +1 σ − n +1 ) = T r ( σ − n +1 σ n +1 φ kn +1 ) = T r ( I φ kn +1 ) = T r ( φ kn +1 ) we get T r ( A kn +1 ) = T r ( φ kn +1 ) . The eigenvalues of matrix A n +1 in (51) , allows one to construct the diagonalmatrix φ n +1 and calculate T r ( A kn +1 ) = T r ϕ n . . . ϕ n − ϕ ′ . . . ϕ n − ϕ ′ . . . ... ... ... ... ... ... ... ... ... . . . ϕ ϕ ′ n − . . . ϕϕ ′ n −
00 0 0 . . . ϕ ′ n k . It gives
T r ( A kn +1 ) = T r ( ϕ n ) k . ϕ n − ϕ ′ ) k . ϕ n − ϕ ′ ) k . ... ... ... ... ... ... ... . ( ϕ ϕ ′ n − ) k . ϕϕ ′ n − ) k
00 0 0 . ϕ ′ n ) k and T r ( A kn +1 ) = ( ϕ n ) k + ( ϕ n − ϕ ′ ) k + . . . + ( ϕϕ ′ n − ) k + ( ϕ ′ n ) k , (63) or T r ( A kn +1 ) = ( ϕ k ) n + ( ϕ k ) n − ϕ ′ k + . . . + ϕ k ( ϕ ′ k ) n − + ( ϕ ′ k ) n . (64)13 he powers ( ϕ k ) n and ( ϕ ′ k ) n substituted from equations (55) and (56) give T r ( A kn +1 ) == (cid:16) ϕ k F ( k ) n + ( − k +1 F ( k ) n − (cid:17) + (cid:16) ϕ k F ( k ) n − + ( − k +1 F ( k ) n − (cid:17) ϕ ′ k + . . . + (cid:16) ϕ k F ( k )1 + ( − k +1 F ( k )0 (cid:17) ( ϕ ′ k ) n − + ( ϕ ′ k ) n = ϕ k (cid:16) F ( k ) n + F ( k ) n − ( ϕ ′ k ) + F ( k ) n − ( ϕ ′ k ) + . . . + F ( k )1 ( ϕ ′ k ) n − (cid:17) +( − k +1 (cid:16) F ( k ) n − + F ( k ) n − ( ϕ ′ k ) + F ( k ) n − ( ϕ ′ k ) + . . . + F ( k )0 ( ϕ ′ k ) n − (cid:17) +( ϕ ′ k ) n = ϕ k F kn F k + F ( n − k F k ( ϕ ′ k ) + F ( n − k F k ( ϕ ′ k ) + . . . + F k F k ( ϕ ′ k ) n − ! +( − k +1 F ( n − k F k + F ( n − k F k ( ϕ ′ k ) + F ( n − k F k ( ϕ ′ k ) + . . . + F F k ( ϕ ′ k ) n − ! +( ϕ ′ k ) n = F kn F k ϕ k + F ( n − k F k ( − k + F ( n − k F k ( − k ( ϕ ′ k ) + . . . + F k F k ( ϕ k )( ϕ ′ k ) n − + F ( n − k F k ( − k +1 + F ( n − k F k ( − k +1 ( ϕ ′ k ) + F ( n − k F k ( − k +1 ( ϕ ′ k ) + . . . + F F k ( − k +1 ϕ ′ n − + ( ϕ ′ k ) n = F kn F k ϕ k + F ( n − k F k ( − k + F ( n − k F k ( − k ( ϕ ′ k ) + . . . + F ( n − ( n − k F k ( − k ( ϕ ′ k ) n − + F ( n − k F k ( − k +1 + F ( n − k F k ( − k +1 ( ϕ ′ k )+ F ( n − k F k ( − k +1 ( ϕ ′ k ) + . . . + F k F k ( − k +1 ( ϕ ′ k ) n − + ( ϕ ′ k ) n = F kn F k ϕ k + F ( n − k F k (cid:16) ( − k + ( − k +1 (cid:17) + F ( n − k F k (cid:16) ( − k ϕ ′ k + ( − k +1 ϕ ′ k (cid:17) + F ( n − k F k (cid:16) ( − k ( ϕ ′ k ) + ( − k +1 ( ϕ ′ k ) (cid:17) + . . . + F k F k (cid:16) ( − k ( ϕ ′ k ) n − + ( − k +1 ( ϕ ′ k ) n − (cid:17) + ( ϕ ′ k ) n = F kn F k ϕ k + F ( n − k F k ( − k (1 + ( − F ( n − k F k ( − k ϕ ′ k (1 + ( − F ( n − k F k ( − k ( ϕ ′ k ) (1 + ( − . . . + F k F k ( − k ( ϕ ′ k ) n +2 (1 + ( − ϕ ′ k ) n = F kn F k ϕ k + ( ϕ ′ k ) n (56)= F kn F k ϕ k + ϕ ′ k F ( k ) n + ( − k +1 F ( k ) n − = F kn F k ϕ k + ϕ ′ k F kn F k + ( − k +1 F k ( n − F k = 1 F k (cid:16) F kn ϕ k + ϕ ′ k F kn + ( − k +1 F k ( n − (cid:17) = 1 F k ϕ − ϕ ′ h(cid:16) ϕ kn − ϕ ′ kn (cid:17) ϕ k + ϕ ′ k (cid:16) ϕ kn − ϕ ′ kn (cid:17) + ( − k +1 (cid:16) ϕ ( n − k − ϕ ′ ( n − k (cid:17)i = 1 F k ϕ − ϕ ′ h ϕ k ( n +1) − ϕ ′ kn ϕ k + ϕ ′ k ϕ kn − ϕ ′ k + kn + ( − k +1 ϕ ( n − k − ( − k +1 ϕ ′ ( n − k i = 1 F k ϕ − ϕ ′ (cid:20) ϕ k ( n +1) − ϕ ′ k ( n +1) − − ϕ ! kn ϕ k + − ϕ ! k ϕ kn + ( − k +1 ϕ ( n − k − ( − k +1 − ϕ ! ( n − k (cid:21) = 1 F k ϕ − ϕ ′ (cid:20) ϕ k ( n +1) − ϕ ′ k ( n +1) − ( − kn ϕ k (1 − n ) + ( − k ϕ k ( n − − ( − k ϕ k ( n − +( − k ( − k ( n − ϕ k (1 − n ) (cid:21) = 1 F k ϕ − ϕ ′ h ϕ k ( n +1) − ϕ ′ k ( n +1) − ( − kn ϕ k (1 − n ) + ( − k ( − kn ( − − k ϕ k (1 − n ) i = 1 F k ϕ − ϕ ′ h ϕ k ( n +1) − ϕ ′ k ( n +1) − ( − kn ϕ k (1 − n ) + ( − kn ϕ k (1 − n ) i = 1 F k ϕ − ϕ ′ h ϕ k ( n +1) − ϕ ′ k ( n +1) i = 1 F k ϕ k ( n +1) − ϕ ′ k ( n +1) ϕ − ϕ ′ = 1 F k F k ( n +1) = F k ( n +1) F k . o prove the relation for det (cid:16) A kn +1 (cid:17) , we take the determinant from both sidesin (62) , det (cid:16) A kn +1 (cid:17) = det (cid:16) σ n +1 φ kn +1 σ − n +1 (cid:17) . (65) By using property of determinants, det( AB ) = det( A ) det( B ) (66) we obtain, det (cid:16) A kn +1 (cid:17) = det ( σ n +1 ) det (cid:16) φ kn +1 (cid:17) det (cid:16) σ − n +1 (cid:17) ⇒ det (cid:16) A kn +1 (cid:17) = det ( σ n +1 ) det (cid:16) σ − n +1 (cid:17) det (cid:16) φ kn +1 (cid:17) ⇒ det (cid:16) A kn +1 (cid:17) = det (cid:16) σ n +1 σ − n +1 (cid:17) det (cid:16) φ kn +1 (cid:17) ⇒ det (cid:16) A kn +1 (cid:17) = det ( I ) det (cid:16) φ kn +1 (cid:17) ⇒ det (cid:16) A kn +1 (cid:17) = det (cid:16) φ kn +1 (cid:17) . Since the matrix φ kn +1 is known, the above equation becomes, det (cid:16) A kn +1 (cid:17) = ( ϕ n ) k ( ϕ n − ϕ ′ ) k ( ϕ n − ϕ ′ ) k . . . ( ϕ ϕ ′ n − ) k ( ϕϕ ′ n − ) k ( ϕ ′ n ) k = (cid:16) ϕ nk ϕ ( n − k ϕ ( n − k . . . ϕ k ϕ k (cid:17) (cid:16) ϕ ′ k ϕ ′ k . . . ϕ ′ ( n − k ϕ ′ ( n − k ϕ ′ nk (cid:17) = (cid:16) ϕ nk +( n − k +( n − k + ... +2 k + k (cid:17) (cid:16) ϕ ′ k +2 k + ... +( n − k +( n − k + nk (cid:17) = ϕ k [ n +( n − n − ... +2+1] ϕ ′ k [1+2+ ... +( n − n − n ] = ϕ k ( n ( n +1)2 ) ϕ ′ k ( n ( n +1)2 )= (cid:18) ϕ n ( n +1)2 (cid:19) k (cid:18) ϕ ′ n ( n +1)2 (cid:19) k = (cid:20) ( ϕϕ ′ ) n ( n +1)2 (cid:21) k ( ϕϕ ′ = − − k n ( n +1)2 . The above Theorem represnts Fibonacci divisors F ( k ) n +1 in terms of com-binatorial matrix A n +1 . Quantum calculus for such divisors was constractedrecently in [4]. As was shown, it is related with several problems from hy-drodynamics, quantum integrable systems and quantum information theory.This is why results of the present paper can be useful in the studies of thiscalculus and its applications. 16 Acknowledgements
One of the authors (O.K.P) would like to thanks Professor Johann Ciglerfor attracting our attention on equivalence of Golden binomials, introducedin [1] with Carlitz characteristic polynomials [3]. This work is supported byTUBITAK grant 116F206.
References [1] Pashaev O K and Nalci S 2012. Golden quantum oscillator and Binet-Fibonacci calculus
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Fibonacci Quarterly pp.81-89[4] Pashaev O K 2020. Quantum Calculus of Fibonacci Divisors and InfiniteHierarchy of Bosonic-Fermionic Golden Quantum Oscillators. arXiv:2010.12386