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Generalized Fibonacci Polynomials Turkish Journal of Analysis and Number Theory, 2013, Vol. 1, No. 1, 43-47 Available online at http://pubs.sciepub.com/tjant/1/1/9 © Science and Education Publishing DOI:10.12691/tjant-1-1-9 Generalized Fibonacci Polynomials Yashwant K. Panwar1,*, B. Singh2, V.K. Gupta3 1Mandsaur Institute of Technology, Mandsaur, India 2School of Studies in Mathematics, Vikram University, Ujjain, India 3Govt. Madhav Science College, Ujjain, India *Corresponding author: [email protected] Abstract In this study, we present generalized Fibonacci polynomials. We have used their Binet’s formula and generating function to derive the identities. The proofs of the main theorems are based on special functions, simple algebra and give several interesting properties involving them. Keywords: generalized Fibonacci polynomials, Binet’s formula, generating function Cite This Article: Yashwant K. Panwar, B. Singh, and V.K. Gupta, “Generalized Fibonacci Polynomials.” Turkish Journal of Analysis and Number Theory 1, no. 1 (2013): 43-47. doi: 10.12691/tjant-1-1-9. Lucas sequences. That is fFnn(1) = and lLnn(1) = , 1. Introduction where Fn and Ln are the Fibonacci and Lucas numbers. Swamy [11] defined the Fibonacci Polynomials and Fibonacci polynomials are a great importance in obtained some more identities for these polynomials. mathematics. Large classes of polynomials can be defined Hogget and Lind [17] make a similar “symbolic by Fibonacci-like recurrence relation and yield Fibonacci substitution” of certain sequences into the Fibonacci numbers [15]. Such polynomials, called the Fibonacci polynomials, they extend these results to the substitution polynomials, were studied in 1883 by the Belgian of any recur rent sequence into any sequence of Mathematician Eugene Charles Catalan and the German polynomials obeying a recurrence relation with Mathematician E. Jacobsthal. polynomial coefficients. Since then many problems about the polynomials have been proposed in various issue of The polynomials fxn ( ) studied by Catalan are defined the Fibonacci Quarterly. Hoggatt, Philips and Leonard [16] by the recurrence relation have obtained some more identities involving Fibonacci Polynomials and Lucas polynomials. A. Lupas [3] present f++( x) = xf( x) + f( x) (1.1) n21 nn many interesting properties of Fibonacci and Lucas where fx( ) = 1 , fx( ) = x, and n ≥ 3 . Notice Polynomials. C. Berg [4] defined Fibonacci numbers and 1 2 orthogonal polynomials. S. Falcon and A. Plaza [13] that fFn (1) = , the nth Fibonacci number. defined the k-Fibonacci polynomials are the natural The Fibonacci polynomials studied by Jacobsthal were extension of the k-Fibonacci numbers and many of their defined by properties admit a straightforward proof and many relations for the derivatives of Fibonacci polynomials are Jnn( x) = J−−12( x) + xJ n( x) (1.2) proven. K. Kaygisiz and A. Sahin [10] present new generalizations of the Lucas numbers by matrix where Jx12( ) =1 = J( x) , and n ≥ 3 . representation, using Generalized Lucas Polynomials. G. Y. Lee and M. Asci [8], consider the Pascal matrix and The Pell polynomials pxare defined by n ( ) define a new generalization of Fibonacci polynomials called (p, q)-Fibonacci polynomials. They obtain = + pn( x) 2 xp nn−−12( x) p( x) (1.3) combinatorial identities and by using Riordan method they get a factorizations of Pascal matrix involving (p, q)- where px0 ( ) = 0 , px1 ( ) = 1, and n ≥ 2 . Fibonacci polynomials. Many authors have studied The Lucas polynomials ( ) nl x , originally studied in Fibonacci polynomials. In this paper, we present 1970 by Bicknell, are defined by generalization of Fibonacci and Lucas Polynomials by changing the initial terms but the recurrence relation is lnnn( x) = xl−−12( x) + l( x) (1.4) preserved. where lx0 ( ) = 2 , lx1 ( ) = x, and n ≥ 2 . It is well known that the Fibonacci polynomials and 2. Generalized Fibonacci Polynomials Lucas polynomials are closely related. Obviously, they have a deep relationship with the famous Fibonacci and The generalized Fibonacci polynomials defined by 44 Turkish Journal of Analysis and Number Theory = n++21 nn s if n 0 ℜ1 =x ℜ 11 +ℜ (3.4) = = ++ fn+1 ( x) sx if n 1 (2.1) ℜn21 =x ℜ nn +ℜ 2 22 xfnn( x) +≥ f−1 ( x) if n 2 Proof: Since ℜℜ12& are the roots of the If s=1, then we obtained classical Fibonacci polynomial characteristic equation (2.3), then sequence. 1 It is well known that the Fibonacci polynomials and ℜ=ℜ+11x 1 Lucas Polynomials are closely related. The 2 generalized Lucas polynomials defined by ℜ=ℜ+22x 2 20s if n = now, multiplying both sides of these equations by ℜℜnn& respectively, we obtain the desired result. ln ( x) = sx if n = 1 (2.2) 12 Proposition 4: For any integer n ≥ 1, xlnn−−12( x) +≥ l( x) if n 2 nn If s=1, then we obtained classical Lucas polynomial s(ℜ12 +ℜ) = fnn+− 1( xf) + 1( x) (3.5) sequence. In the 19th century, the French mathematician Binet Proof: By using Eq. (3.1) in the R.H.S. of Eq. (3.5) and devised two remarkable analytical formulas for the −1 taking in to account that ℜ=1 it is obtained Fibonacci and Lucas numbers. In our case, Binet’s ℜ2 formula allows us to express the generalized Fibonacci Polynomials in function of the roots ℜℜ& of the ℜnn++11 −ℜ ℜ nn −− 11 −ℜ 12 (RHS) = s 12+ s 12 following characteristic equation, associated to the ℜ12 −ℜ ℜ 12 −ℜ recurrence relation (2.1) & (2.2): s nnnn+−+−1111 =( ℜ1122 +ℜ −ℜ −ℜ ) 2 = + ℜ −ℜ x xt 1 (2.3) 12 s nn11 = ℜℜ+11 −ℜℜ+ 22 3. Properties of Generalized Fibonacci ℜ12 −ℜ ℜ 1 ℜ2 nn Polynomials =s( ℜ12 +ℜ ) Theorem 1: (Binet's formula). The nth generalized Proposition 5: For any integer n, Fibonacci Polynomials is given by 2 nn2 nn s (ℜ12 +ℜ ) ; if n is even ℜ −ℜ 22 2n (3.6) 12 x+4 fxn ( ) + 41 s( −=) fxn ( ) = s (3.1) ( ) 2 ℜ −ℜ s2 ℜnn −ℜ ; if n is odd 12 ( 12) where are the roots of the characteristic equation (3), Proof: From the Binet’s formula of generalized xx++2 4 xx−+2 4 Fibonacci Polynomials ℜ12 >ℜ and ℜ=1 and ℜ=2 . 2 2 2 s n fx2= ℜ 22nn −2 ℜ ℜ +ℜ Proof: we use the Principle of Mathematical Induction n ( ) 2 { 1( 12) 2} (PMI) on n. It is clear the result is true for n = 0 and n = 1 (ℜ12 −ℜ ) by hypothesis. Assume that it is true for i such that 0 ≤ i ≤ 2 r +1, then 22 2nn If n is even, ( x+44) fxn ( ) + s =( s ℜ+ℜ12 s) ii 2 ℜ12 −ℜ 22 2nn fxi ( ) = s If n is odd, ( x+44) fxn ( ) − s =( s ℜ12 −ℜ s) ℜ12 −ℜ nn It follows from definition of generalized Fibonacci Let us denote (ssℜ+ℜ12) by lxn ( ) . Polynomials and from equation (3.1), Then previous formula become: rr++22 n ℜ12 −ℜ x22+4 f( x) + 41 s 2( −=) lx 2( ) (3.7) fr++21( x) = xf tr( x) += f( x) s (3.2) ( ) nn ℜ12 −ℜ Thus, the formula is true for any positive integer n. Theorem 2: (Binet's formula). The nth generalized 4. Sums of Generalized Fibonacci Lucas Polynomials is given by Polynomials nn lxn ( ) = s( ℜ12 +ℜ ) (3.3) In this section, we study the sums of generalized Fibonacci Polynomials. This enables us to give in a Proposition 3: For any integer n ≥ 1, straightforward way several formulas for the sums of such Polynomials. Turkish Journal of Analysis and Number Theory 45 ≤≤− n q Lemma 6: For fixed integers p, q with 01qp , f33nq++( x) + f 3 nq + ( x) − fx q( ) −−( 1) f3−q( x) fx= (3.12) ∑ 3iq+ ( ) 3 the following equality holds i=0 sx+ 3 sx p n sfpn( +21) ( x) = lp ( x) fpn( ++) q( x) −−( 1) sfpn+ q ( x) f33n+−( xfxf) +− 3 nq( ) 3 ( x) (i) For q=0: fx= ∑ 3i ( ) 2 (3.8) i=0 sx+ s n Proof: From the Binet’s formula of generalized f++( x) + f( x) −+ s sx (ii) For q=1: fx( ) = 34nn 31 Fibonacci and Lucas Polynomials, ∑ 31i+ 2 i=0 sx+ s n lp ( xf) pn( ++1) q( x) fnn++( x) + f( x) −− s sx (iii) For q=2: fx= 35 32 ∑ 32i+ ( ) 2 pn( ++11) q pn( ++) q + ℜ −ℜ i=0 sx s =ss ℜnn +ℜ 12 ( 12) ℜ −ℜ (2) If m=2 then p=5 12 n q 2 f55nq++( x) + f 5 nq + ( x) − fx q( ) −−( 1) f5−q( x) s pn( ++22) q pppn++ q pn q pn( ++) q fx( ) = (3.13) = ℜ +−ℜ( 11) −−ℜ( ) −ℜ ∑ 5iq+ 53 1 1 22 i=0 sx++55 sx sx ℜ12 −ℜ 2 42 s pn( ++22) q pn( ++) q p ++ n + −++ = ℜ −ℜ +( −1) ℜpn q −ℜ pn q f55nn+ ( x) f 5( x) ( sx3 sx s) {1 2 ( 12)} (i) For q=0: fx( ) = ℜ12 −ℜ ∑ 5i 53 i=0 sx++55 sx sx p =sf( x) +−( 1) sf+ ( x) n 3 pn( +2) pn q f++( x) + f( x) ++− sx2 sx s (ii) For q=1: fx= 56nn 51 ∑ 51i+ ( ) 53 then, the equation becomes, i=0 sx++55 sx sx 2 n f57nn++( x) + f 52( x) −( sx ++ sx s) p (iii) For q=2: fx( ) = sf( x) = l( x) f( x) −−( 1) sf+ ( x) ∑ 52i+ 53 pn( +21) p pn( ++) q pn q i=0 sx++55 sx sx 2 n f58nn++( x) + f 53( x) −( sx −+ sx s) Proposition 7: For fixed integers p, q with (iv) For q=3: fx= ∑ 53i+ ( ) 53 01≤≤−qp , the following equality holds i=0 sx++55 sx sx 3 pp n f59nn++( x) + f 54( x) − sx ++2 sx s n − −− −− ( ) fxfxfxfxpn( ++1) q( ) q( ) ( 11) pnq+−( ) ( ) pq( ) (3.9) (v) For q=4: fx( ) = fx( ) = ∑ 54i+ 53 ∑ pi+ q p i=0 sx++55 sx sx i=0 lx( ) −−( 11) − p Corollary 7.2: Sum of even generalized Fibonacci Proof: Applying Binet’s formula of generalized polynomials Fibonacci Polynomials, If p=2m, then Eq.(3.9) is n q n f21mn( ++) q( x) + f22mn+− q( x) − fx q ( ) −−( 1) fm q ( x) fxpi+ q ( ) (3.14) ∑ ∑ fx2mi+ q ( ) = = − i 0 i=0 lx2m ( ) 2 n ℜpi++ q −ℜ pi q = s 12 ∑ ℜ −ℜ For example i=0 12 (1) If m=1 then p=2 s nn = ℜpi++ q −ℜ pi q ℜ −ℜ ∑∑12 n q 12ii=00= f22n++ q( x) + f 2 nq + ( x) − fx q( ) −−( 1) f2−q( x) fx+ ( ) = (3.15) pn++ q p q pn++ q p q ∑ 2iq 2 s ℜ −ℜ ℜ −ℜ = sx+−22 s = 1122− i 0 ℜ −ℜ ℜ−pp ℜ− 121211 n f+ ( x) −− f( x) sx s pq 22nn 2 = (−11) f( x) − f( x) + fx( ) +−( ) f( x) (i) For q=0: ∑ fx2i ( ) = p pnq+−pn( ++1) q q pq 2 (−−11) lxp ( ) + i=0 sx+−22 s − −− pq−− n fpn( ++1) q( x) fxq ( ) ( 11) fpn+ q ( x) ( ) fpq− ( x) f23nn++( x) −− f 21( x)
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