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Generalized Fibonacci Polynomials and Some Identities


Call for Papers Engineering Journal, May 2019

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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 nun0

    (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 1n 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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