The polynomial $X^{3}-X-1$ has a unique positive root known as plastic
number, which is denoted by $\rho$ and is approximately equal to $1.32471795$.
In this note we study the zeroes of the generalized polynomial
$X^{k}-\sum_{j=0}^{k-2}X^{j}$ for $k\geq 3$ and prove that its unique positive
root $\lambda_{k}$ tends to the golden ratio $\phi=\frac{1+\sqrt{5}}{2}$ as $k
\to \infty$. We also derive bounds on $\lambda_{k}$ in terms of Fibonacci
numbers.Comment: Publisher's pdf versio

Hare K.

Hoggatt V. E.

Pisot C.

Stewart I.

van der Laan D. H.

Wolfram D. A.

Zhu X.

Cogent Mathematics

arXiv

Vasileios Iliopoulos

Crossref

The plastic number and its generalized polynomial

Abstract: The polynomial X 3 − X − 1 has a unique positive root known as plastic number, which is denoted by and is approximately equal to 1.32471795. In this note, we study the zeroes of the generalised polynomial X k − ∑ k−2 j=0 X j , for k ≥ 3, and prove that its unique positive root k tends to the golden ratio = 1+ √ 5 2 as k → ∞. We also derive bounds on k in terms of Fibonacci numbers

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Iliopoulos, Vasileios

University of Essex Research Repository

Iliopoulos, Cogent Mathematics (2015), 2: 1023123http://dx.doi.org/10.1080/23311835.2015.1023123PURE MATHEMATICS | RESEARCH ARTICLEThe plastic number and its generalized polynomialVasileios  Iliopoulos1*Abstract: The polynomial X3 − X − 1 has a unique positive root known as plastic number, which is denoted by 휌 and is approximately equal to 1.32471795. In this note, we study the zeroes of the generalised polynomial Xk −∑k−2j=0 Xj, for k ≥ 3, and prove that its unique positive root 휆k tends to the golden ratio 휙 =1+√52 as k→ ∞. We also derive bounds on 휆k in terms of Fibonacci numbers.Subjects: Science;  Mathematics & Statistics; Advanced Mathematics; Number Theory; Pure MathematicsKeywords: Fibonacci; golden ratio; plastic numberAMS subject classifications:  11B39; 11B831. IntroductionThe recurrence Fn = Fn−1 + Fn−2, with initial values F0 = 0 and F1 = 1 yields the celebrated Fibonacci numbers. It is well known that for n ≥ 0where 휙 = 1+√52 is the positive root of the characteristic polynomial X2 − X − 1, known as golden ratio.One can readily generalise the recurrence and define the k ≥ 2 order Fibonacci sequence Fn = Fn−1 +⋯ + Fn−k, with initial conditions F0 = ⋯ = Fk−2 = 0 and Fk−1 = 1. The characteristic polynomial of this recurrence is Xk − Xk−1 −⋯ − X − 1. Its zeroes are much studied in literature: we refer to Martin (2004), Miles (1960), Miller (1971), Wolfram (1998), and Zhu and Grossman (2009), where it is proved that the unique positive root tends to 2, as k→∞. Series representations for this root are derived in Hare, Prodinger, and Shallit (2014) by Lagrange inversion theorem.Fn =휙n− (1 − 휙)n√5,*Corresponding author: Vasileios Iliopoulos, Department of Mathematical Sciences, University of Essex, Colchester, UKE-mail: viliop@essex.ac.ukReviewing editor:Yong Hong Wu, Curtin University of Technology, AustraliaAdditional information is available at the end of the articleABOUT THE AUTHORVasileios Iliopoulos obtained his PhD from the Department of Mathematical Sciences of University of Essex in 2014 and is currently working as a part-time lecturer there. His research interests are on analytic and enumerative combinatorics, operational research and elementary number theory.PUBLIC INTEREST STATEMENTIn this paper, the author considers a result of Siegel (1944), who proved that the positive root of the polynomial k(X) = Xk−Xk−1−1X−1, where k is an odd integer greater than or equal to 3, tends to the golden ratio 1+√52, as k→ ∞.The presented proof is elementary and simple. The author also obtains bounds on the positive zero of k(X) in terms of Fibonacci numbers.Received: 18 December 2014Accepted: 19 February 2015© 2015 The Author(s). This open access article is distributed under a Creative Commons Attribution (CC-BY) 4.0 license.Page 1 of 6Page 2 of 6Iliopoulos, Cogent Mathematics (2015), 2: 1023123http://dx.doi.org/10.1080/23311835.2015.1023123In this note, we turn our attention to the positive zero of the polynomial X3 − X − 1, known as plastic number, which will throughout be denoted by 휌 and is equal to 3�1 +3�1 +3√1… (Finch, 2003). The plastic number was introduced by van der Laan (1960). The recurrence relation is an = an−2 + an−3, with initial conditions a0 = a1 = a2 = 1 and defines the integer sequence, known as Padovan sequence Stewart (1996). Although the bibliography regarding the analysis of Fibonacci numbers is quite extensive, it seems not to be this case regarding the plastic number.In the next section, we examine a generalisation of the Padovan sequence and its associated characteristic polynomial and derive bounds on the unique positive root of the polynomial Xk − Xk−2 −⋯ − X − 1. The presented study utilises results from the theory of linear recurrences and elementary Calculus, where the nature of roots of this polynomial is investigated. Furthermore, the bounds on the largest root are developed using identities of Fibonacci numbers.2. The generalised sequenceConsider the recurrencefor k ≥ 3 and initial conditions a0 =⋯ = ak−1 = 1. For k = 3, we obtain as a special case the Padovan sequence. A lemma follows regarding the roots of its characteristic polynomial.Lemma 2.1 The polynomial k(X) = Xk− Xk−2 −⋯ − X − 1 has k simple roots. If k is odd, the poly-nomial has a unique real root 휆k ∈ (1,휙) and k − 1 complex roots. When k is even, the roots of the polynomial are 휆k ∈ (1,휙), −1 and k − 2 complex zeroes.Proof  It can be easily seen that neither 0 nor 1 are roots of k(X). Following Miles (1960) and Miller (1971), it is convenient to work with the polynomialDifferentiating Equation 1, we obtainEquation 2 is 0, at X = 0 or at the roots of the quadratic polynomial:Its discriminant can be easily computed to Δ = 5k2 − 4 > 0, for all k ≥ 3 and the two real roots of polynomial of (3) areWe identify the real roots by elementary means. Note that k(1) = 2 − k < 0 and k(휙) = 휙 and apply Descartes’ rule of signs to Equation 1, there is a unique positive root 휆k in (1,휙), and for k even, the unique negative root of the polynomial is −1. Also, the polynomial is monic and by Gauss’s lemma the root 휆k is irrational.Observe that 휆k ≠ 훽2(k), since 훽2(k) is negative for all k ≥ 3 and 휆k ≠ 훽1(k). For if they were equal, then by Rolle’s theorem there is at least one root 훼 of Equation 3 in (1, 훽1(k)), but 𝛽2(k) < 0 < 1 and considering the fundamental theorem of Algebra, which states that every polynomial with complex coefficients and degree k has k complex roots with multiplicities, we arrive in contradiction. This an =k∑l=2an−l(1)(X − 1)k(X) = Xk+1− Xk − Xk−1 + 1.(2)((X − 1)k(X))� = (k + 1)Xk − kXk−1 − (k − 1)Xk−2.(3)(k + 1)X2 − kX − (k − 1).(4)훽1,2(k) =k ±√5k2 − 42(k + 1).Downloaded by [University of Essex] at 07:56 26 March 2015 Page 3 of 6Iliopoulos, Cogent Mathematics (2015), 2: 1023123http://dx.doi.org/10.1080/23311835.2015.1023123shows that the polynomial (X − 1)k(X) has (k + 1) simple roots. We complete the proof noting that k(X) and (X − 1)k(X) are positive and increasing for X > 𝜆k and negative for 1 < X < 𝜆k. A direct consequence of Lemma 2.1 isCorollary 2.2 The polynomial k(X) is irreducible on the ring of integer numbers ℤ if and only if k is odd.Further, it is easy to prove that all complex zeroes of the polynomial are inside the unit circle. The next Lemma is from Miles (1960) and Miller (1971).Lemma 2.3 (Miles, 1960; Miller, 1971) For all the complex zeroes 휇 of the polynomial k(X), it holds that |𝜇| < 1.Proof Assume that there exists a complex 휇 (and hence 휇), with 1 < |𝜇| < 𝜆k. We have that (휇 − 1)k(휇) = 0 andApplying the triangle inequality to Equation 5, we deduce thatwhich contradicts Lemma 2.1. Assuming now that |𝜇| > 𝜆k, we havewhich is equivalent to k(|휇|) ≤ 0 and again we arrive in contradiction. Finally, by the same reason-ing it can be easily proved that there is no complex zero 휇, with either |휇| = 휆k or |휇| = 1. Lemma 2.3 implies that the solution of the generalised recurrence can be approximated bywith negligible error term. In Equation 6, C is a constant to be determined by the solution of a linear system of the initial conditions.We now consider, more carefully, Equation 4Observe that 훽1(k) =k+√5k2−42(k+1) is increasing and bounded sequence. Furthermore,Also, 훽2(k) =k−√5k2−42(k+1) is decreasing and bounded and(5)|휇k+1| = |휇k + 휇k−1 − 1|.(|𝜇| − 1)k(|𝜇|) < 0,|휇k| = ||||||k−2∑j=0휇j|||||| ≤k−2∑j=0|휇j|,(6)an ≈ C휆nk ,훽1,2(k) =k ±√5k2 − 42(k + 1).(7)limk→∞k +√5k2 − 42(k + 1)=12+�54= 휙.(8)limk→∞k −√5k2 − 42(k + 1)=12−�54= 1 − 휙.Downloaded by [University of Essex] at 07:56 26 March 2015 Page 4 of 6Iliopoulos, Cogent Mathematics (2015), 2: 1023123http://dx.doi.org/10.1080/23311835.2015.1023123From Equations 7 and 8, we deduce that two of the critical points of Equation 1, (recall that these are 0 with multiplicity (k − 2), 훽1(k) and 훽2(k)), converge to 휙 and 1 − 휙. A straightforward calculation can show that 훽1(k) are points of local minima of the function (X − 1)k(X) to the interval (1, 휆k), so 𝛽1(k) < 𝜆k < 𝜙 for all k ≥ 3 and by squeeze lemma we have that limk→∞ 휆k = 휙.We remark that 휌 is a Pisot–Vijayaraghavan number, a real algebraic integer having modulus greater to 1 where its conjugates lie inside the unit circle (Bertin, Decomps-Guilloux, Grandet-Hugot, Pathiaux-Delefosse, & Schreiber, 1992). These numbers are named after Pisot (1938) and Vijayaraghavan (1941), who independently studied them. Siegel (1944) considered several families of polynomials and showed that the plastic number is the smallest Pisot–Vijayaraghavan number. By Lemmas 2.1 and 2.3, the positive zeroes of the polynomial Xk −∑k−2j=0 Xj, where k is odd, are Pisot–Vijayaraghavan numbers. In case that k is even, the positive roots of the polynomial Xk −∑k−2j=0 Xj are Salem numbers (Salem, 1945). This family of numbers is closely related to the set of Pisot–Vijayaraghavan numbers. They are positive algebraic integers with modulus greater than 1, where its conjugates have modulus no greater than 1 and at least one root has modulus equal to 1.We have proved that for all k ≥ 3,Using the identity 5F2k = L2k − 4(−1)k (Hoggatt, 1969, Section 5), we have that for k = F2t+1where Ln is the nth Lucas number, defined by Ln = Ln−1 + Ln−2 for n ≥ 2, with initial conditions L0 = 2 and L1 = 1. Lucas numbers obey the following closed form expression for n ≥ 0, (Hoggatt, 1969)Now inequality 10 becomesActually, inequality 10 is valid when k +√5k2 − 42(k + 1) is quadratic irrational. A stronger result is the following Theorem.Theorem 2.4 For k ≥ 3, it holds thatProof For k = 3, we have thatwhere 휆3: = 휌. Since for k > 3, Fk+1Fk> 1 and Fk+1Fk+1> 1, by Lemma  2.1 it suffices to show that (Fk+1Fk+1− 1)k(Fk+1Fk+1)< 0 and (Fk+1Fk− 1)k(Fk+1Fk)> 0. Setting X = Fk+1Fk to Equation 1, we have to prove that(9)k +√5k2 − 42(k + 1)< 𝜆k < 𝜙.(10)𝜆F2t+1 >F2t+1 + L2t+12(F2t+1 + 1)Ln = 휙n+ (1 − 휙)n.𝜆F2t+1>F2t+1 + L2t+12(F2t+1 + 1)=F2(t+1)F2t+1 + 1.Fk+1Fk + 1< 𝜆k <Fk+1Fk.F4F3 + 1< 𝜌 <F4F3,Downloaded by [University of Essex] at 07:56 26 March 2015 Page 5 of 6Iliopoulos, Cogent Mathematics (2015), 2: 1023123http://dx.doi.org/10.1080/23311835.2015.1023123The previous inequality is the same asThe left-hand side of inequality 11 isThus, (11) is true for all k.In order to prove that Fk+1Fk+1< 𝜆k, we have to equivalently show thatWe then havewhich isUsing thatthe identitycan be easily proven by induction, and thus completes the proof. (Fk+1Fk)k−1((Fk+1Fk)2−Fk+1Fk− 1)> −1.(11)(Fk+1Fk)2−Fk+1Fk− 1 > −(FkFk+1)k−1.F2k+1 − FkFk+1 − F2kF2k=Fk+1(Fk+1 − Fk) − F2kF2k=Fk+1Fk−1 − F2kF2k=(−1)kF2k,(12)(Fk+1Fk + 1)2−Fk+1Fk + 1− 1 < −(Fk + 1Fk+1)k−1.(Fk+1Fk + 1)2−Fk+1Fk + 1− 1 =F2k+1 − FkFk+1 − Fk+1 − F2k − 2Fk − 1(Fk + 1)2=(−1)k − (Fk+1 + 2Fk + 1)(Fk + 1)2=(−1)k − Lk+1 − 1(Fk + 1)2,Lk+1Fk−1k+1 > (Fk + 1)k+1.Lk+1 = 휙k+1+ (1 − 휙)k+1 and Fk+1 =휙k+1− (1 − 휙)k+1√5F2(k+1) > (Fk + 1)2by Cassini’s identity (Hoggatt, 1969).Downloaded by [University of Essex] at 07:56 26 March 2015 Page 6 of 6Iliopoulos, Cogent Mathematics (2015), 2: 1023123http://dx.doi.org/10.1080/23311835.2015.1023123© 2015 The Author(s). This open access article is distributed under a Creative Commons Attribution (CC-BY) 4.0 license.You are free to: Share — copy and redistribute the material in any medium or format  Adapt — remix, transform, and build upon the material for any purpose, even commercially.The licensor cannot revoke these freedoms as long as you follow the license terms.Under the following terms:Attribution — You must give appropriate credit, provide a link to the license, and indicate if changes were made.  You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.  No additional restrictions  You may not apply legal terms or technological measures that legally restrict others from doing anything the license permits.Cogent Mathematics (ISSN: 2331-1835) is published by Cogent OA, part of Taylor & Francis Group. Publishing with Cogent OA ensures:• Immediate, universal access to your article on publication• High visibility and discoverability via the Cogent OA website as well as Taylor & Francis Online• Download and citation statistics for your article• Rapid online publication• Input from, and dialog with, expert editors and editorial boards• Retention of full copyright of your article• Guaranteed legacy preservation of your article• Discounts and waivers for authors in developing regionsSubmit your manuscript to a Cogent OA journal at www.CogentOA.comAcknowledgementsI thank the anonymous referees for their helpful suggestions.FundingThe authors received no direct funding for this research.Author detailsVasileios Iliopoulos1E-mail: viliop@essex.ac.uk1  Department of Mathematical Sciences, University of Essex, Colchester, UK.Citation informationCite this article as: The plastic number and its generalized polynomial, Vasileios Iliopoulos, Cogent Mathematics (2015), 2: 1023123.ReferencesBertin, M. J., Decomps-Guilloux, A., Grandet-Hugot, M., Pathiaux-Delefosse, M. & Schreiber, J. P. (1992). Pisot and Salem numbers. Basel: Birkhäuser. Finch, S. R. (2003). Mathematical constants. New York, NY: Cambridge University Press. Hare, K., Prodinger, H., & Shallit, J. (2014). Three series for the generalized golden mean. Fibonacci Quarterly, 52, 307–313.Hoggatt, V. E. Jr,  (1969). The Fibonacci and Lucas numbers. Boston, MA: Houghton Mifflin. Martin, P. A. (2004). The galois group of xn – xn–1 – … – x – 1. Journal of Pure and Applied Algebra, 190, 213–223.Miles, E. P., Jr (1960). Generalized Fibonacci numbers and associated matrices. American Mathematical Monthly, 67, 745–752.Miller, M. D. (1971). On generalized Fibonacci numbers. American Mathematical Monthly, 78, 1108–1109.Pisot, C. (1938). La répartition modulo 1 et les nombres algébriques. Annali di Pisa, 7, 205–248.Salem, R. (1945). Power series with integral coefficients. Duke Mathematical Journal, 12, 153–172.Siegel, C. L. (1944). Algebraic numbers whose conjugates lie in the unit circle. Duke Mathematical Journal, 11, 597–602.Stewart, I. (1996). Tales of a neglected number. Scientific American, 274, 102–103.van der Laan, D. H. (1960). Le Nombre Plastique: quinze leçons sur l’Ordonnance architectonique. Leiden: Brill.Vijayaraghavan, T. (1941). On the fractional parts of the powers of a number, II. Proceedings of the Cambridge Philosophical Society, 37, 349–357.Wolfram, D. A. (1998). Solving generalized Fibonacci recurrences. Fibonacci Quarterly, 36, 129–145.Zhu, X., & Grossman, G. (2009). Limits of zeros of polynomial sequences. Journal of Computational Analysis and Applications., 11, 140–158.Downloaded by [University of Essex] at 07:56 26 March 2015 

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