Generalized Fibonacci Polynomials and Some Identities

Generalized Fibonacci Polynomials and Some Identities

Omprakash Sikhwal1,

,1 Department of Mathematics, Mandsaur Institute of Technology,

Mandsaur (M.P.), India

Yashwant Vyas2

2 Department of Mathematics, Shri HarakChand Chourdia College,

Bhanpura (M.P.), India

Abstract:- The Fibonacci and Lucas polynomials are famous for possessing wonderful and amazing properties and identities. In this paper, Generalized Fibonacci polynomials are introduced and defined by

n

n

u x xu

n1 x u n2 x, n 2 with u 0x a and u 1 x 2a 1, where a is any integer. Further,

some basic identities are generated and derived by standard methods.

Keywords: Generalized Fibonacci polynomials, Generating function, Binets Formula

1. INTRODUCTION

   Fibonacci numbers are a popular topic for mathematical enrichment and popularization. They are famous for a host of interesting and surprising properties and show up in text books, magazine articles, and web sites. Various sequences of polynomials by the name of Fibonacci and Lucas polynomials occur in the literature over a century. The Fibonacci and Lucas polynomials are closely related and widely investigated. Fibonacci polynomials appear in different frameworks. These polynomials are of great importance in the study of many subjects such as algebra, geometry, combinatorics, approximation theory, statistics and number theory itself. Moreover these polynomials have been applied in every branch of mathematics.

   0 1

   0 1

   The Fibonacci polynomials satisfy the following recurrence formula:

   n 1 n n 1

   n 1 n n 1

   f x xf x f x , n 2

   with

   f x 0 , f x 1.

   (1.1)

   The Lucas polynomials [1] are defined by the recurrence formula

   n 1 n n 1

   n 1 n n 1

   0 1

   0 1

   l x xl x L x , n 2 with l x 2 , l x x

   (1.2)

   Generating function of Fibonacci polynomials is given by

   n

   n

   n0

   f x tn t 1 xt t2 1 .

   (1.3)

   Generating function of Lucas polynomials is given by

   ln

   n0

   x tn 2 xt 1 xt t2 1 .

   (1.4)

   Explicit sum formula for Fibonacci polynomials is given by

   n1

   2 n k 1

   fn x

   xn12k ,

   k 0 k

   (1.5)

   Explicit sum formula for Lucas polynomials is given by

   n

   2

   ln x

   k 0

   n n k

   n k

   xn2k ,

   k

   (1.6)

   k

   k

   n

   where a binomial coefficient and x is define as the greatest integer less than or equal to x .

   Fibonacci-Like polynomials [11] is defined by the recurrence relation:

   n n1 n2

   n n1 n2

   0 1

   0 1

   s x xs x s x , n 2. with s x 2 and s x 2x . (1.7)

   0

   0

   1

   1

   Generalized Fibonacci-Like polynomial [12] is defined by the recurrence relation:

   n n1 n2

   n n1 n2

   b x xb x b x, n 2.

   where b and s are integers.

   with

   b x 2b

   and b x s,

   (1.8)

   The Fibonacci and Lucas polynomials possess many fascinating properties which have been studied in [2] to [12]. In this paper, generalized Fibonacci-Like polynomials are introduced with some basic identities.

2. GENERALIZED FIBONACCI POLYNOMIALS

   n

   n

   1

   1

   Generalized Fibonacci polynomials u x are defined by the recurrence relation

   n n1 n2

   n n1 n2

   0

   0

   u x xu x u x , n 2. with where a is integer.

   u x a

   and u x 2a 1 , (2.1)

   The first few terms of generalized Fibonacci polynomials are as follows:

   0

   0

   u (x) a,

   1

   1

   u (x) 2a 1,

   2

   2

   u (x) (2a 1) x a,

   3

   3

   u (x) 2a 1 x2 ax 2a 1 ,

   4

   4

   u (x) 2a 1 x3 ax2 2 2a 1 x a,

   5

   5

   u (x) 2a 1 x4 ax3 32a 1 x2 2ax (2a 1),

   For x = 1 and a = 0, we obtain Fibonacci Sequence.

   and so on.

   The characteristic equation of recurrence relation (2.1) is 2 x 1 0 . Which has two real roots

   x x2 4

   and

   2

   x x2 4

   2

   Also,

   1,

   x,

   x2 4,

   2 2 x2 2.

   (2.2)

   Binets formula of generalized Fibonacci polynomials is given by

   n n

   n n x x2 4 x x2 4

   un (x) A B A 2 B 2

   (2.3)

   (2a 1) a

   a (2a 1)

   Here,

   A

   and B

   ,

   ,

   Also,

   AB

   (a2 3a 1)

   2

   A B u0 (x) a . (2.4)

   Generating function of generalized Fibonacci polynomials is given by

   n

   n

   un (x)t

   n0

   a (2a 1 ax)t

   1 xt t 2

   (2.5)

   Now we obtain hypergeometric representation of generating function.

   By generating function (2.5), we have

   n0

   u (x)tn a (2a 1 ax)t

   n 1 xt t 2

   a (2a 1 ax)t 1 (x t)t 1

   a (2a 1 ax)t (x t)ntn

   n0

   n

   n

   a (2a 1 ax)tt n

   n

   k

   k

   xnktk

   n0

   n

   n

   a (2a 1 ax)t

   k 0

   n! xnktnk

   n0 k 0 k !n k !

   a (2a 1 ax)t

   n k !

   x

   nt n2k

   n0 k 0

   k !n!

   xt n n k !

   a (2a 1 ax)t

   t2k

   n0

   n! k 0 k !

   a (2a 1 ax)t ext

   n k ! t 2 )k

   (

   (

   k 0

   u (x) n xt

   k !

   n k ! (t 2 )k

   n t

   n0 n!

   a (2a 1 ax)te

   n!

   n!

   k !

   k !

   k 0

   u (x) n xt n k 1 (t 2 )k

   n t

   n0 n!

   a (2a 1 ax)te

   k 0

   n 1 k !

   u (x) n xt (1) (t 2 )k

   n t a (2a 1 ax)te (n 1)k k

   n0 n!

   k 0

   (1)k k !

   Hence, un (x)

   n0

   n

   t xt

   t xt

   a (2a 1 ax)t e

   n!

   2 F1 n 1, 1; 1; t .

   (2.6)

   2

   2

3. SOME IDENTITIES OF GENERALIZED FIBONACCI POLYNOMIALS

   In this section, we present some recurrence relations and identities by generating function, and explicit sum formula.

   Theorem 3.1: Prove that

   un1(x) un1(x) xun (x), n 1.

   Proof: By generating function of generalized Fibonacci polynomials, we have

   (3.1)

   n0

   u (x)tn a (2a 1 ax)t 1 xt t 2 1

   n

   n

   Differentiating both sides with respect to t,

   we get

   n0

   nun

   (x)tn1 a (2a 1 ax)t x 2t 1 xt t 2 2 (2a 1 ax) 1 xt t 2 1

   1 xt t 2

   n0

   nun

   (x)tn1 a (2a 1 ax)t x 2t 1 xt t 2 1 (2a 1 ax)

   1 xt t 2 nu

   n0

   (x)tn1 x 2t u

   n n

   n n

   n0

   (x)tn (2a 1 ax)

   n n n n n

   n n n n n

   nu (x)tn1 nxu (x)tn nu (x)tn1 xu (x)tn 2u

   (x)tn1 (2a 1 ax)

   n0

   n0

   n0

   n0

   n0

   Now equating the coefficient of tn on both sides we get,

   (n 1)un1(x) nxun (x) (n 1)un1(x) xun (x) 2un1(x) (n 1)un1(x) (n 1)un1(x) (n 1)xun (x)

   un1(x) un1(x) xun (x)

   This is required result.

   Theorem 3.2: Prove that

   u' (x) xu' (x) u (x) u'

   (x), n 1

   (3.2)

   n1 n n n1

   Proof: By (3.1), we have

   un1(x) un1(x) xun (x), n 1.

   Differentiating both sides with respect to

   x, we get

   u

   u

   '

   n1

   (x) u'

   (x) xu' (x) u

   (x),

   n1

   n1

   n n

   n n

   u' (x) xu' (x) u (x) u'

   (x).

   n1 n n n1

   n

   n

   Theorem 3.3: Prove that

   n n n1

   n n n1

   nu (x) xu' (x) 2u'

   (x), n 1

   and

   '

   xu

   xu

   n1

   (x) (n 1)u

   n1

   (x) 2u' (x), n 1.

   Proof: By generating function of generalized Fibonacci polynomials, we have

   n0

   u (x)tn a (2a 1 ax)t 1 xt t 2 1

   n

   n

   Differentiating both sides with respect to t, we get

   nun (x)t

   n0

   n1 (2a 1 ax) 1 xt t 2

   1

   a (2a 1 ax)t x 2t 1 xt t 2

   2

   (3.3)

   Differentiating both sides with respect to

   x, we get

   n0

   n0

   u' (x)tn a (2a 1 ax)t 1 xt t 2 2 t at 1 xt t 2 1

   n

   n

   n

   n

   u' (x)tn1 a (2a 1 ax)t 1 xt t 2 2 a 1 xt t 2 1

   ' n1 2

   1

   1

   2 2

   un (x)t

   n0

   a 1 xt t

   a (2a 1 ax)t 1 xt t

   (3.4)

   Using (3.4) in (3.3), we get

   n1 2 1

   '

   n1 2

   1

   nun (x)t

   n0

   (2a 1 ax) 1 xt t

   x 2t un (x)t

   n0

   a(1 xt t ) .

   nu

   (x)tn1 (2a 1 ax) 1 xt t 2 1 x 2t u' (x)tn1 a x 2t (1 xt t 2 )1.

   n0

   n n

   n0

   Now equating the coefficient of tn1 on both sides, we get

   n n n1

   n n n1

   nu (x) xu' (x) 2u'

   (x).

   (3.5)

   Again equating the coefficient of tn on both sides, we get

   (n 1)u

   n1

   (x) xu'

   (x) 2u' (x),

   n1

   n1

   n

   n

   xu

   xu

   '

   n1

   (x) (n 1)u

   n1

   (x) 2u' (x).

   (3.6)

   n

   n

   n n1

   n n1

   n1

   n1

   Theorem 3.4: Prove that

   (n 1)u

   (x) u'

   (x) u'

   (x), n 1.

   Proof: By (3.1), we have

   un1(x) un1(x) xun (x), n 1.

   Differentiating both sides with respect to

   x, we get

   u

   u

   '

   n1

   (x) u'

   (x) xu' (x) u

   (x),

   n1

   n1

   n n

   n n

   xu' (x) u (x) u' (x) u'

   (x).

   (3.7)

   n n n1

   Using (3.5) in (3.7), we get

   n1

   n n1

   n n1

   nu (x) 2u'

   (x) u

   (x) u'

   (x) u'

   (x).

   n n1

   n n1

   n1

   n1

   nu (x) u (x) u'

   (x) 2u'

   (x) u'

   (x),

   n1

   n1

   n n n1

   n1

   n1

   n n1

   n n1

   (n 1)u

   (x) u'

   (x) u'

   (x).

   (3.8)

   Theorem 3.5: Prove that

   n n1

   n n1

   xu' (x) 2u'

   (x) (n 2)un

   (x), n 0.

   Proof: Using (3.5) in (3.8), we get

   n n1

   n n1

   (n 1)u

   (x) u'

   (x) 1 nu

   2

   (x) xu' (x) ,

   n n

   n n

   n n1

   n n1

   2(n 1)u

   (x) 2u'

   (x) nu

   (x) xu' (x) ,

   n n

   n n

   n n1

   n n1

   xu' (x) 2u'

   (x) nu

   n (x) (2n 2)un

   (x),

   n n1

   n n1

   xu' (x) 2u'

   (x) (n 2n 2)un

   (x),

   (3.9)

   n1

   n1

   Theorem 3.6: Prove that

   n n1

   n n1

   (n 1)xu' (x) nu'

   (x) (n 2)u'

   (x), n 1.

   Proof: Using (3.8) in (3.2), we get

   n1

   n1

   n n1

   n n1

   n1

   n1

   n1

   n1

   (n 1)u' (x) xu' (x) u' (x) u'

   (x) u'

   (x),

   n1

   n1

   (n 1)u'

   (x) (n 1)xu' (x) (n 1)u'

   (x) u'

   (x) u'

   (x),

   n n1

   n n1

   n1

   n1

   n1

   n1

   n1

   n1

   (n 1)u'

   (x) (n 1)u'

   (x) u'

   (x) u'

   (x) (n 1)xu' (x),

   n1

   n1

   n1

   n1

   n1

   n1

   n

   n

   nu

   nu

   '

   n1

   n1

   n1

   (x) (n 2)u'

   (x) (n 1)xu' (x),

   n1

   n1

   n

   n

   n n1

   n n1

   (n 1)xu' (x) nu'

   (x) (n 2)u'

   (x).

   (3.10)

   Theorem 3.7: (Explicit Sum Formula) The explicit sum formula for generalized Fibonacci polynomials is given by

   n

   n2k

   n2k

   2 n k

   un (x) a x .

   k 0 k

   (3.11)

   Proof: By generating function (2.5), we have

   n0

   u (x)tn a (2a 1 ax)t 1 xt t 2 1

   n

   n

   a (2a 1 ax)t 1 (x t)t 1

   a (2a 1 ax)t (x t)ntn

   n0

   a (2a 1 ax)t

   t n

   n

   k

   k

   xnktk

   n

   n

   n0

   k 0

   n

   n

   a (2a 1 ax)t

   n! xnktnk

   n0 k 0 k !n k !

   a (2a 1 ax)t

   n k !

   x

   nt n2k

   n0 k 0

   k !n!

   n

   2

   n k !

   a (2a 1 ax)t

   xn2ktn

   n

   n2k

   n2k

   2 n k

   un (x) a x .

   k 0 k

   n0 k 0 k !n 2k !

   Equating coefficients of tn on both sides, we get required explicit formula.

   Theorem 3.8: For positive integer n 0

   , prove that

   n n n 1

   4

   un( x ) ax

   2 F1 2 , 2 ;

   • n;

     x2 .

     (3.12)

     Proof. By explicit sum formula (3.11), it follows that

     n 2

     n

     n

     u ( x ) axn

     k 0

     n k !

     k ! n 2k !

     x2k

     n

     2 1k 1 n

     x2k

     axn n 2k

     k n

     k n

     k 0 n 12k 1 k !

     n 1k 22k n n 1

     2

     2

     2 x2 k

     k

     k

     axn k k

     k 0

     n 12k k !

     n

     n

     n n 1 4 k

     2 2 2 x2

     k

     k

     axn k k

     k 0

     n k !

     Hence, u ( x ) axn

     F n ,

     n 1 ;

   • n;

   4 .

   n 2 1 2 2 x2

   Theorem 3.9: For positive integer n 0

   , prove that

   t n

   c c c 1

   n 1

   n 2

   t 2

   cnun x

   a 1 xt

   3 F2

   , ,n 1; ; ;

   2 .

   (3.13)

   n0 n!

   2 2 2 2

   1 xt

   t n

   Proof. Multiplying both sides of the explicit sum formula by c

   n n!

   and summing between the limit

   n 0 to n , we obtain

   n0 k 0

   n0 k 0

   t n

   n

   2

   n k ! t n

   n

   n

   cnun

   n0

   x

   n! a k ! n 2k !

   c

   xn2k

   n!

   a

   n k ! c

   xntn2k

   n0 k 0 k ! n! n 2k !

   n2k

   xt n n k !

   2k

   2k

   a c 2k n

   n0

   n! k 0 k ! n 2k !

   c

   t 2k

   k

   k

   a 1 xt c2k k 0

   n k ! c t 2k ,

   k ! n 2k ! 2k

   t n

   c

   n k !

   t 2

   n!

   n!

   cnun x

   n0

   a 1 xt

   k ! n 2k ! c 2k 2

   k 0 1 xt

   k 0 1 xt

   a 1 xt c

   n k ! 22k c c 1

   t 2

   k

   k

   k !

   n 2k !

   2 2 2

   k 0

   k k 1 xt

   a 1 xt c

   n 1

   k

   k

   22k c c 1

   t 2

   k ! ,

   k

   k

   k

   k

   0 2k

   0 2k

   k k

   k k

   n 1

   2

   2 1 xt 2

   k

   k

   c c 1

   c

   2 2

   n 1 k

   t2

   a 1 xt

   k k k !

   k 0

   n 1 n 2

   1 xt 2

   2 2

   k k

   t n

   c c c 1

   n 1

   n 2

   t 2

   Hence, cnun x

   a 1 xt

   3 F2

   , , n 1; ; ; .

   2

   2

   n0 n!

   2 2 2 2

   1 xt

   th

   th

   Theorem 3.10 (Catalans Identity): Let un ( x) be the n term of generalized Fibonacci polynomials, then

   u2 (x) u (x)u

   (x)

   1nr

   (2a 1)u (x) au

   (x), n r 1

   (3.14)

   n nr nr

   a2 3a 1

   r r 1

   Proof: Using Binets formula (2.5), we have

   u2 (x) u (x)u

   (x) ( A n B n )2 ( A nr B nr )( A nr B nr )

   n nr nr

   AB n 2 r r r r

   AB 1nr r r 2

   2 2

   2 2

   (a 3a 1) 1nr r r

   2

   (a2 3a 1) 1nr

   r r 2

   2

   r r (2a 1)u (x) au (x) (2a 1)u (x) au (x)

   r r 1 r r 1

   Since

   2a 12 a(2a 1) a2

   1nr

   (a2 3a 1)

   u2 (x) u

   (x)u

   (x)

   (2a 1)u (x) au

   (x)2 , n r 1.

   n nr nr

   (a2 3a 1)

   r r 1

   th

   th

   Theorem 3.11( Cassinis Identity): Let un ( x) be the n term of generalized Fibonacci polynomials, then

   u2( x ) u ( x )u ( x ) ( 1)n1( a2 3a 1 ), n 1

   (3.15)

   n n1 n1

   Proof. If r = 1 in the Catalans Identity, then obtained required result.

   th

   th

   Theorem 3.12( dOcagnes Identity): Let un ( x) be the n term of generalized Fibonacci polynomials, then

   u ( x )u ( x ) u ( x )u ( x ) ( 1)n ( 2a 1)u ( x ) au ( x ) , m 1,n 0,m n.

   (3.16)

   m n1

   m1

   n mn mn1

   Proof: Using Binets formula (2.5), we have

   u (x)u

   (x) u

   (x)u (x) ( A n B m )(A n1 B n1) ( A m1 B m1)( A n B n )

   m n1

   m1

   n

   AB m n1 n1 m n m1 m1 n

   AB( )n mn mn mn mn

   AB(1)n mn mn

   (1)n

   (1)n

   2

   2

   (a 3a 1) mn mn

   2

   (a2 3a 1)(1)n

   mn mn

   mn mn

   (2a 1)u

   (x) au

   (x) (2a 1)u

   (x) au

   (x)

   Since,

   mn mn1 mn mn1 , we obtain

   2a 12 a(2a 1) a2

   (a2 3a 1)

   n

   n

   mn mn1

   mn mn1

   um( x )un1( x ) um1( x )un( x ) ( 1) ( 2a 1)u ( x ) au ( x ) , m 1,n 0,m n.

   th

   th

   Theorem 3.13 (Generalized Identity): Let un ( x) be the n term of generalized Fibonacci polynomials, then

   2 mr

   2 mr

   um (x)un (x) umr (x)unr (x) (a 3a 1) 1 (2a 1)ur (x) aur 1(x)(2a 1)unmr (x) aunmr 1(x), n m r 1

   (3.17)

   Proof: Using Binets formula (2.5), we have

   m m n n mr mr nr nr

   m m n n mr mr nr nr

   um (x)un (x) umr (x)unr (x) ( A B )(A B ) ( A B )(A B ),

   AB( r r

   m n n m

   ) r r

   AB1r ( r r )( m nr nr m )

   AB1r ( m m )( r r )( n pr n pr )

   AB1r ( m m )( r r )( n pr n pr )

   2

   2

   1 ( )( )( )

   1 ( )( )( )

   (a 3a 1) r m m r r n pr n pr

   ( )2

   Using subsequent results of Binets formula, we get

   r r

   (2a 1)u (x) au

   (x)

   nmr nmr

   (2a 1)u

   (x) au

   (x)

   Since,

   r r 1 , and

   (a2 3a 1)

   nmr nmr 1 . (a2 3a 1)

   2 mr

   2 mr

   r r 1 nmr nmr 1

   r r 1 nmr nmr 1

   um (x)un (x) umr (x)unr (x) (a 3a 1) 1 (2a 1)u (x) au (x)(2a 1)u (x) au (x), n m r 1

   The identity (3.13) provides Catalans identity, Cassinis and dOcagne and other identities.

4. CONCLUSION

In this paper, generalized Fibonacci polynomials is introduced and presented some basic results. Further some recurrence relations and identities are described with derivation by standard methods. The concept of generalized Fibonacci- Like polynomials can be extended in two and three variables with basic results and identities.

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