In this paper, we study h(x)-Lucas quaternion polynomials considering

several properties involving these polynomials and we present the exponential

generating functions and the Poisson generating functions of the h(x)-Lucas

quaternion polynomials. Also, by using Binet’s formula we give the Cassini’s

identity, Catalan’s identity and d’Ocagne’s identity of the h(x)-Lucas quaternion

polynomials

Kilic, Nayil

EKE Repository of Publications

The h(x)-Lucas quaternion polynomialsNayil KilicDepartment of MathematicsSinop University, Sinop, Turkeynayilkilic@gmail.comSubmitted May 16, 2017 — Accepted November 30, 2017AbstractIn this paper, we study h(x)-Lucas quaternion polynomials consideringseveral properties involving these polynomials and we present the exponentialgenerating functions and the Poisson generating functions of the h(x)-Lucasquaternion polynomials. Also, by using Binet’s formula we give the Cassini’sidentity, Catalan’s identity and d’Ocagne’s identity of the h(x)-Lucas quater-nion polynomials.Keywords: Lucas polynomials, recurrences, quaternion.MSC: 11B39, 11B37, 11R521. IntroductionThe Lucas sequence, {Ln}, is defined by the recurrence relation, for n > 1Ln+1 = Ln + Ln−1where L0 = 2, L1 = 1.In [13], Nalli and Haukkanen introduced the h(x)-Lucas polynomials.Definition 1.1 ([13]). Let h(x) be a polynomial with real coefficients. The h(x)-Lucas polynomials {Lh,n(x)}∞n=0 are defined by the recurrence relationLh,n+1(x) = h(x)Lh,n(x) + Lh,n−1(x), n ≥ 1, (1.1)with initial conditions Lh,0(x) = 2, Lh,1(x) = h(x).Annales Mathematicae et Informaticae47 (2017) pp. 119–128http://ami.uni-eszterhazy.hu119The quaternions are such numbers which extend the complex numbers. Theyare members of noncommutative algebra. A quaternion p is defined in the formp = a0 + a1i+ a2j + a3kwhere a0, a1, a2 and a3 are real numbers and i, j, k are standart orthonormal basisin R3 which satisfy the quaternion multiplication rules asi2 = j2 = k2 = −1, ij = −ji = k, jk = −kj = i ki = −ik = j.The conjugate of the quaternion p is denoted by p and p = a0 − a1i− a2j − a3k.We start by recalling some basic results concerning quaternion algebra H, itis well known that the algebra H = {a = a0e0 + a1e1 + a2e2 + a3e3|ai ∈ R, i ={0, 1, 2, 3}} of real quaternions define a four− dimensional vector space over Rhaving basis e0 ∼= 1, e1 ∼= i, e2 ∼= j and e3 ∼= k which satisfies the followingmultiplication rules.es2 = −1, s ∈ {1, 2, 3}, e1e2 = −e2e1 = e3, e2e3 = −e3e2 = e1, (1.2)e3e1 = −e1e3 = e2.In [8], Horodam defined the nth Lucas quaternions as follows.Definition 1.2 ([8]). The Lucas quaternion numbers that are given for the nthclassic Lucas Ln number are defined by the following recurrence relations:Tn = Ln + iLn+1 + jLn+2 + kLn+3where n = 0,∓1,∓2, . . . .The Lucas quaternions have been studied in several papers (see, for exam-ple [1, 2, 7, 10, 15]). Recently, in [2], Ari considered the h(x)-Lucas quaternionpolynomials, he derived the Binet formula and generating function of h(x)-Lucasquaternion polynomial sequence.In this paper, we study h(x)-Lucas quaternion polynomials considering severalproperties involving these polynomials and we present the exponential generatingfunctions and the Poisson generating functions of the h(x)-Lucas quaternion poly-nomials. Also, by using Binet’s formula we give the Cassini’s identity, the Catalan’sidentity and the d’Ocagne’s identity of the h(x)-Lucas quaternion polynomials.2. The h(x)-Lucas quaternion polynomials and somepropertiesLet ei (i = 0, 1, 2, 3) be a basis of H, which satisfy the multiplication rules (1.2). Leth(x) be a polynomial with real coefficients. In [2], Ari introduced the h(x)-Lucasquaternion polynomials as follows:120 N. KilicDefinition 2.1 ([2]). Let h(x) be a polynomial with real coefficients. The h(x)-Lucas quaternion polynomials {Th,n(x)}∞n=0 are defined by the recurrence relationTh,n(x) =3∑s=0Lh,n+s(x)es (2.1)where Lh,n(x) is the nth h(x)-Lucas polynomial.The conjugate of Th,n(x) is given byTh,n(x) = Lh,n(x)e0 − Lh,n+1(x)e1 − Lh,n+2(x)e2 − Lh,n+3(x)e3.For n = 0,Th,0(x) =3∑s=0Lh,s(x)es= Lh,0(x)e0 + Lh,1(x)e1 + Lh,2(x)e2 + Lh,3(x)e3= 2e0 + h(x)e1 + (h2(x) + 2)e2 + (h3(x) + 3h(x))e3.For n = 1,Th,1(x) =3∑s=0Lh,s+1(x)es= Lh,1(x)e0 + Lh,2(x)e1 + Lh,3(x)e2 + Lh,4(x)e3= h(x)e0 + (h2(x) + 2)e1 + (h3(x) + 3h(x))e2+ (h4(x) + 4h2(x) + 2)e3.From the recurrence relation (2.1), using the recurrence relation (1.1) and someproperties of summation formulas, we obtain thatTh,n+1(x) =3∑s=0Lh,s+1+n(x)es=3∑s=0(h(x)Lh,s+n(x) + Lh,s+n−1(x))es= h(x)3∑s=0Lh,s+n(x)es +3∑s=0Lh,s+n−1(x)es= h(x)Th,n(x) + Th,n−1(x)and soTh,n+1(x) = h(x)Th,n(x) + Th,n−1(x).The h(x)-Lucas quaternion polynomials 121In [13], authors studied some combinatorial properties of h(x)-Fibonacci and h(x)-Lucas polynomials and present properties of these polynomials. They obtained thefollowing Binet’s formula for Lh,n(x)Lh,n(x) = αn(x) + βn(x) (2.2)whereα(x) =h(x) +√h2(x) + 42, β(x) =h(x)−√h2(x) + 42(2.3)are roots of the characteristic equation y2 − h(x)y − 1 = 0 of the recurrence rela-tion (1.1).Ari in [2] calculated the Binet-style formula for Th,n(x),Th,n(x) = α∗(x)αn(x) + β∗(x)βn(x) (2.4)where α(x) and β(x) as in (2.3) and α∗(x) =3∑s=0αs(x)es, β∗(x) =3∑s=0βs(x)es.The following basic identities are needed for our purpose in proving.α(x) + β(x) = h(x), α(x)β(x) = −1, α(x)− β(x) =√h2(x) + 4 (2.5)andα(x)β(x)= −α2(x), β(x)α(x)= −β2(x).Also,1 + h(x)α(x) = α2(x), 1 + h(x)β(x) = β2(x), (2.6)and1 + α2(x) = α(x)√h2(x) + 4, 1 + β2(x) = −β(x)√h2(x) + 4. (2.7)The following Lemma, related with the h(x)-Lucas polynomials and it will beuseful in the proof of one property of the h(x)-Lucas quaternion polynomials in thenext Theorem.Lemma 2.2. For n ≥ 0,L2h,n(x) + L2h,n+1(x) = Lh,2n(x) + Lh,2n+2(x).Proof. Using (2.2) and (2.5), we getL2h,n(x) + L2h,n+1(x) = (αn(x) + βn(x))2 + (αn+1(x) + βn+1(x))2= α2n(x) + 2αn(x)βn(x) + β2n(x)+ α2n+2(x) + 2αn+1(x)βn+1(x) + β2n+2(x)= α2n(x) + β2n(x) + α2n+2(x) + β2n+2(x)= Lh,2n(x) + Lh,2n+2(x).So the proof is complete.122 N. KilicTheorem 2.3. For n ≥ 0, the following statements hold:(i) (Th,n(x))2+(Th,n+1(x))2 = (α2∗(x)α2n+1(x)−β2∗(x)β2n+1(x))(α(x)−β(x)).(ii)(Th,n(x))2+(Th,n+1(x))2(α(x)−β(x)) = (α2∗(x)α2n+1(x)− β2∗(x)β2n+1(x)).(iii) Th,n(x) + Th,n(x) = 2Lh,n(x)e0.(iv) (Th,n(x))2 = 2Lh,n(x)e0Th,n(x) − Th,n(x)Th,n(x) = Th,n(x)(2Lh,n(x)e0 −Th,n(x)).(v) Th,n(x)Th,n(x) = ((h(x))2 + 2)(Lh,2n+4(x) + Lh,2n+2(x)).(vi) Th,1(x)− α(x)Th,0(x) = −β∗(x)√h2(x) + 4.In particular Th,1(x)−α(x)Th,0(x)α(x)−β(x) = −β∗(x).(vii) Th,1(x)− β(x)Th,0(x) = α∗(x)√h2(x) + 4.In particular Th,1(x)−β(x)Th,0(x)α(x)−β(x) = α∗(x).Proof. (i) From (2.4), (2.5) and (2.7), we obtain(Th,n(x))2 + (Th,n+1(x))2= (α∗(x)αn(x) + β∗(x)βn(x))2 + (α∗(x)αn+1(x) + β∗(x)βn+1(x))2= α2∗(x)α2n(x)(1 + α2(x)) + β2∗(x)β2n(x)(1 + β2(x))= α2∗(x)α2n+1(x)√h2(x) + 4− β2∗(x)β2n+1(x)√h2(x) + 4= (α2∗(x)α2n+1(x)− β2∗(x)β2n+1(x))(α(x)− β(x)).(ii) The proof of (ii) follows immediately from (i).(iii) Using the definition of Th,n(x) and some computations, we haveTh,n(x) = Lh,n(x)e0 − Lh,n+1(x)e1 − Lh,n+2(x)e2 − Lh,n+3(x)e3= 2Lh,n(x)e0 −3∑s=0Lh,n+s(x)es= 2Lh,n(x)e0 − Th,n(x),and the result follows.(iv) By (iii), (iv) holds.(v) Using Definition 2.1, the definition of Th,n(x), Lemma 2.2 and (1.1) weobtainTh,n(x)Th,n(x) =3∑s=0Lh,n+s(x)esTh,n(x)=(Lh,n(x)e0 + Lh,n+1(x)e1 + Lh,n+2(x)e2 + Lh,n+3(x)e3)The h(x)-Lucas quaternion polynomials 123× (Lh,n(x)e0 − Lh,n+1(x)e1 − Lh,n+2(x)e2 − Lh,n+3(x)e3)= L2h,n(x) + L2h,n+1(x) + L2h,n+2(x) + L2h,n+3(x)= Lh,2n(x) + Lh,2n+2(x) + Lh,2n+4(x) + Lh,2n+6(x)= Lh,2n(x) + Lh,2n+2(x) + Lh,2n+4(x) + h(x)Lh,2n+5(x)+ Lh,2n+4(x)= 2Lh,2n+2(x) + h2(x)Lh,2n+2(x) + 2Lh,2n+4(x)+ h2(x)Lh,2n+4(x)=(2 + (h(x))2)(Lh,2n+2(x) + Lh,2n+4(x)).(vi) SinceLh,s+1(x)− β(x)Lh,s(x) = αs(x)(α(x)− β(x))andLh,s+1(x)− α(x)Lh,s(x) = βs(x)(α(x)− β(x)),using the definition of β∗(x), Definition2.1 and Eq.(2.5), we haveTh,1(x)− α(x)Th,0(x)= Lh,1(x)e0 + Lh,2(x)e1 + Lh,3(x)e2 + Lh,4(x)e3− α(x)(Lh,0(x)e0 + Lh,1(x)e1 + Lh,2(x)e2 + Lh,3(x)e3)=(Lh,1(x)− α(x)Lh,0(x))e0 +(Lh,2(x)− α(x)Lh,1(x))e1+(Lh,3(x)− α(x)Lh,2(x))e2 +(Lh,4(x)− α(x)Lh,3(x))e3= −β0(x)(α(x)− β(x))e0 − β1(x)(α(x)− β(x))e1− β2(x)(α(x)− β(x))e2 − β3(x)(α(x)− β(x))e3= −√h2(x) + 4(e0 + β1(x)e1 + β2(x)e2 + β3(x)e3)= −√h2(x) + 43∑s=0βs(x)es= −√h2(x) + 4β∗(x).which completes the first part of the proof of (vi). The proof of the remaining partcan be obtained from previous result.(vii) The proof is similar to part (vi) and thus, omitted.Theorem 2.4. For n ≥ 0,n∑k=0(nk)(h(x))kTh,k(x) = Th,2n(x).Proof. Using (2.4) and (2.6), we obtainn∑k=0(nk)(h(x))kTh,k(x) =n∑k=0(nk)(h(x))k[α∗(x)αk(x) + β∗(x)βk(x)]124 N. Kilic= α∗(x)n∑k=0(nk)(h(x))kαk(x)+ β∗(x)n∑k=0(nk)(h(x))kβk(x)= α∗(x)(1 + h(x)α(x))n + β∗(x)(1 + h(x)β(x))n= α∗(x)α2n(x) + β∗(x)β2n(x)= Th,2n(x).Theorem 2.5. The sum of the first m terms of the sequence {Th,m(x)}∞m=0 isgiven bym∑k=0Th,k(x) =Th,0(x)− Th,m(x)− Th,m+1(x)− α∗(x)β(x)− β∗(x)α(x)(1− α(x))(1− β(x)) .Proof. From (2.4), (2.5) and some calculations, we getm∑k=0Th,k(x) =m∑k=0(α∗(x)αk(x) + β∗(x)βk(x))= α∗(x)m∑k=0αk(x) + β∗(x)m∑k=0βk(x)= α∗(x)(1− αm+1(x)1− α(x))+ β∗(x)(1− βm+1(x)1− β(x))=α∗(x)− α∗(x)β(x)− α∗(x)αm+1(x) + α∗(x)αm(x)α(x)β(x)(1− β(x))(1− α(x))+β∗(x)− β∗(x)α(x)− β∗(x)βm+1(x) + β∗(x)α(x)β(x)βm(x)(1− β(x))(1− α(x))=Th,0(x)− Th,m(x)− Th,m+1(x)− α∗(x)β(x)− β∗(x)α(x)(1− α(x))(1− β(x)) .So the proof is complete.3. Exponential generating functions for the h(x)-Lucas quaternion polynomialsIn this section, we give the exponential generating functions for the sequence ofthe h(x)-Lucas quaternion polynomials. The exponential generating function of asequence {bk}∞k=0 is given byEG(bk, l) =∞∑k=0bklkk!.The h(x)-Lucas quaternion polynomials 125Theorem 3.1. The exponential generating function for the h(x)-Lucas quaternionpolynomials are∞∑k=0Th,k(x)k!lk = α∗(x)eα(x)l + β∗(x)eβ(x)l. (3.1)Proof. From the Binet-style formula for the h(x)-Lucas quaternion polynomials,we have∞∑k=0Th,k(x)k!lk =∞∑k=0(α∗(x)αk(x) + β∗(x)βk(x)) lkk!= α∗(x)∞∑k=0(α(x)l)kk!+ β∗(x)∞∑k=0(β(x)l)kk!= α∗(x)eα(x)l + β∗(x)eβ(x)l.4. Poisson generating functions for the h(x)-Lucasquaternion polynomialsIn this section, we present Poisson generating functions for the sequence of theh(x)-Lucas quaternion polynomials.Lemma 4.1. The Poisson generating functions for the h(x)-Lucas quaternion poly-nomials are∞∑k=0Th,k(x)k!lke−l =α∗(x)eα(x)l + β∗(x)eβ(x)lel. (4.1)Proof. Since PG(bn, x) = e−lEG(bn, x), we have the result by Theorem 3.1.5. Catalan’s, Cassini’s and d’Ocagne’s identity forthe h(x)-Lucas quaternion polynomialsIn this section, we compute Catalan’s identity, Cassini’s identity and d’Ocagne’sidentity for the h(x)-Lucas quaternion polynomials, we start with Catalan’s iden-tity.Theorem 5.1. For n ≥ m ≥ 1, Catalan identity for the h(x)-Lucas quaternionpolynomials isTh,n+m(x)Th,n−m(x)− T 2h,n(x) = (−1)n−m(αm(x)− βm(x))× 1(α∗(x)β∗(x)αm(x)− β∗(x)α∗(x)βm(x)).126 N. KilicProof. Using (2.4) and (2.5), we obtainTh,n+m(x)Th,n−m(x)− T 2h,n(x)=(α∗(x)αn+m(x) + β∗(x)βn+m(x))(α∗(x)αn−m(x) + β∗(x)βn−m(x))−(α∗(x)αn(x) + β∗(x)βn(x))2= α∗(x)β∗(x)αn+m(x)βn−m(x) + β∗(x)α∗(x)βn+m(x)αn−m(x)− α∗(x)β∗(x)αn(x)βn(x)− β∗(x)α∗(x)αn(x)βn(x)= α∗(x)β∗(x)(α(x)β(x))n(αm(x)βm(x)− 1)+ β∗(x)α∗(x)(α(x)β(x))n(βm(x)αm(x)− 1)= α∗(x)β∗(x)(−1)nαm(x)(αm(x)− βm(x)(α(x)β(x))m)+ β∗(x)α∗(x)(−1)nβm(x)(βm(x)− αm(x)(α(x)β(x))m)= (−1)n−m(αm(x)− βm(x))(α∗(x)β∗(x)αm(x)− β∗(x)α∗(x)βm(x)).So Theorem 5.1 is proved.Theorem 5.2. For any natural number n, Cassini identity for the h(x)-Lucasquaternion polynomials isTh,n+1(x)Th,n−1(x)− T 2h,n(x) = (−1)n−1(α(x)− β(x))×(α∗(x)β∗(x)α(x)− β∗(x)α∗(x)β(x)).Proof. Taking m = 1 in Catalan’s identity, the proof is completed.Theorem 5.3 (d’Ocagne’s identity). Suppose that n is a nonnegative integer num-ber and m any natural number. If m > n, thenTh,m(x)Th,n+1(x)− Th,m+1(x)Th,n(x)= (−1)n(α(x)− β(x))(β∗(x)α∗(x)βm−n(x)− α∗(x)β∗(x)αm−n(x)).Proof. From (2.4) and (2.5), we obtainTh,m(x)Th,n+1(x)− Th,m+1(x)Th,n(x)=(α∗(x)αm(x) + β∗(x)βm(x))(α∗(x)αn+1(x) + β∗(x)βn+1(x))−(α∗(x)αm+1(x) + β∗(x)βm+1(x))(α∗(x)αn(x) + β∗(x)βn(x))= α∗(x)β∗(x)αm(x)βn(x)(β(x)− α(x))+ β∗(x)α∗(x)βm(x)αn(x)The h(x)-Lucas quaternion polynomials 127×(α(x)− β(x))= α∗(x)β∗(x)αm−n(x)(α(x)β(x))n(β(x)− α(x))+ β∗(x)α∗(x)βm−n(x)×(α(x)β(x))n(α(x)− β(x))= (−1)n(α(x)− β(x))(β∗(x)α∗(x)βm−n(x)− α∗(x)β∗(x)αm−n(x)).So, the proof is complete.References[1] Akyigit, M., Kosal, H. H., Tosun, M., Fibonacci Generalized Quaternions,Advances in Applied Clifford Algebras, Vol. 24(3) (2014), 631–641.[2] Ari, K., On h(x)-Lucas quaternion polynomials, Ars Combinatoria, Vol. 121 (2015),291–303.[3] Catarino, P., A note on h(x)− Fibonacci quaternion polynomials, Chaos, SolitonsFractals, Vol. 77 (2015), 1–5.[4] Catarino, P., A note on certain matrices with h(x)− Fibonacci quaternion polyno-mials, Journal of Difference Equations and Applications, Vol. 22(2) (2016), 343–351.[5] Catarino, P., The h(x)− Fibonacci Quaternion Polynomials: Some CombinatorialProperties, Advances in Applied Clifford Algebras, Vol. 26 (2016), 71–79.[6] Falcón, S., On the generating matrices of the k− Fibonacci numbers, ProyeccionesJ. Math. Vol. 32(4) (2013), 347–357.[7] Halici, S., On Fibonacci quaternions, Advances in Applied Clifford Algebras,Vol. 22(2) (2012), 321–327.[8] Horodam, A. F., Complex Fibonacci numbers and Fibonacci quaternions, Am.Math. Mon. Vol. 70 (1963), 289–291.[9] Horodam, A. F., Recurrence relations, Ulam Q. Vol. 2(2) (1993), 23–33.[10] Iyer, M. R., Some results on Fibonacci quaternions, Fibonacci Q. Vol. 7(2) (1969),201–210.[11] Iyer, M. R., Note on Fibonacci quaternions, Fibonacci Q. 7(3) (1969), 225–229.[12] Kilic, E., Tasci, D., Haukkanen, P., On the generalized Lucas sequences byHessenberg matrices, Ars Combinatoria, Vol. 95 (2010), 383–395.[13] Nalli, A., Haukkanen, P., On generalized Fibonacci and Lucas polynomials,Chaos, Solitons Fractals, Vol. 42 (2009), 3179–3186.[14] Polatli, E., Kesim, S., A Note on Catalan’s identity for the k-Fibonacci quater-nions, Journal of Integer Sequences, Vol. 18 (2015), 1–4.[15] Tan, E., Yilmaz, S., Sahin, M., On a new generalization of Fibonacci Quaternions,Chaos, Solitons Fractals, Vol. 82 (2016), 1–4.128 N. Kilic

The h(x)-Lucas quaternion polynomials

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