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U niversity of Nebraska at OmahaDi gitalCommons@UNOM athematics Faculty Publications P alindrome-Polynomials with Roots on the UnitC ircleJ ohn KonvalinaU niversity of Nebraska at Omaha, johnkon@unomaha.eduV alentin MatacheU niversity of Nebraska at Omaha, v matache@unomaha.eduF ollow this and addit ional works at:#

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.# (.$-*((*)-hThi s Ar ticle is brought to you for free and open access by the Department*! .# (.$-/'.2/'$.$*)-2)/.#*,$3 R ecommended CitationKon valina, John and Matache, Valentin, "Palindrome-Polynomials with R oots on the Unit Circle" (2004).M athematics FacultyP ublications. 44.

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Palindrome-Polynomials with Roots on the

Unit Circle

John Konvalina and Valentin Matache

Abstract

Given a polynomialf(x) of degreen, letfr(x) denote itsreciprocal, i.e., f r(x) =xnf(1=x). If a polynomial is equal to its reciprocal, we call it a palindromesince the coefficients are the same when read backwards or for- wards. In this mathematical note we show that palindromes whose coeffi- cients satisfy a certain magnitude-condition must have a root on the unit circle. More exactly our main result is the following. If a palindromef(x) of even degreenwith real coefficients²0;²1;:::;²nsatisfies the condition j²kj ¸ j²n=2jcos(¼=([n=2 n=2¡k] + 2)), for somek2 f0;1;:::n=2¡1g, thenf(x) has unimodular roots. In particular, palindromes with coefficients 0 and 1 always have a root on the unit circle.

R´esum´e

Soitf(x) un polynˆome de degr´en. Soitfr(x) =xnf(1=x). Le polynˆome f(x) s"appelle polynˆome r´eciproque sifr(x) =f(x). Dans cet article nous prouvons que les polynˆomes r´eciproques dont les coefficients poss`edent une certaine propri´et´e, ont des z´eros sur le cercle unit´e, c"est `a dire ont des zeros de valeur absolue 1. Notre principal r´esultat est le th´eor`eme suivant. Soit f(x) un polynˆome r´eciproque dont le degr´enest pair et dont les coefficients

0;²1;:::;²nsont des nombres r´eels tels quej²kj ¸ j²n=2jcos(¼=([n=2

n=2¡k]+2)), pour au moins une valeur dek2 f0;1;:::n=2¡1g. Tel pˆolynomef(x) pos`ede des z´eros de valeur absolue 1. Pour cosequence, chaque polynˆome r´eciproque avec des coefficients 0 et 1 a des z´eros sur le cercle unit´e.

2000Mathematics Subject Classification. Primary 12D10; Secondary 30C15.

1

1 INTRODUCTION

We arrived at an interesting geometric property of palindrome-polynomials, i.e. of polynomials whose coefficients are the same when read backwards or forwards, by investigating polynomials with coefficients 0 and 1 or (0;1)- polynomials. They are interesting because of their applications in various areas of pure and applied mathematics including algebra, number theory, combinatorics, and coding theory. Computer-assisted experiments lead us to conjecture that (0;1)-palindromes necessarily have at least one unimodular root, that is, have a root of absolute value 1 . In the sequel we will prove not only that this conjecture is true but that palindromes with real coefficients often have roots on the unit circle. This introductory section is dedicated to setting up notations and terminology. The second section contains the main results of the article. We begin by defining the notion of palindrome. Given a polynomialf(x) of degreen, letfr(x) denote itsreciprocal, i.e., f r(x) =xnf(1=x). If a polynomial is equal to its reciprocal, we call it apalindrome-polynomialor simply apalindrome. In the next section we establish rather easily the following.Letf(x)2R[X]be a palindrome having even degreenand null middle coefficient. Then the polynomialf(x)has unimodular roots,(Lemma 2). Note that the existence of unimodular roots is interesting only for palindromes of even degree, because if the degree is odd, obviously¡1 is a root. ByR[X] we denote the ring of polynomials of one variable with coefficients inR, the field of real numbers. If the middle coefficient is not null, a palindrome with real coefficients may lack unimodular roots, (consider e.g.f(x) =x2¡3x+ 1). It turns out that the relative size of the middle coefficient with respect to the other coefficients is important for the existence of unimodular roots, only the proof of this fact relies on deeper results on cosine polynomials going back to F´ejer, Riesz, and Szeg¨o. For each realxlet [x] denote the largest integer which is less than or equal tox. The main result proved in Section 2 is the following.

Letf(x) =Pn

k=0²kxk, with²j=²n¡j,8j= 0;1;:::;n=2be a palindrome having even degreen. If there existsk2 f0;1;:::;n=2¡1gsuch that j²kj ¸ j²n=2jcosà n=2 n=2¡k] + 2! thenf(x)has unimodular roots(Theorem 1). In particular, (0;1)-palindromes of any degree have roots on the unit circle, (Corollary 1). Another consequence of Theorem 1 is the following. Letf(x)be a palindrome with real coefficients, having even degreen. If

2j²nj ¸ j²n=2jthenf(x)has unimodular roots, (Corollary 2).

2

2 PALINDROMES AND THEIR ZEROS

Our approach is based on the following classical construction. Denoting u=x+ 1=xwe associate to each palindromef(x)2R[X] having even degreenthe polynomialg(u) uniquely determined by the following identity. f(x) =xn=2g(u) (1) Several comments are in order here. For eachk= 1;2;:::we denote¾k(x) = x k+1=xk. Fork= 0 we set¾0(x) = 1. Observe that¾1(x) =uand that the following identity holds. u k=¾k(x) +X

1·j·k=2¾

k¡2j(x)µk k= 2;3;:::(2)

Letf(x) =Pn

k=0²kxk, with²j=²n¡j,8j= 0;1;:::;n=2. Note that f(x) =xn=2h(x) whereh(x) =n=2X k=0² k¾n=2¡k(x):(3) Equations (2) and (3) prove thatg(u) exists and is uniquely determined. After these preparations we can prove the following lemma, which appears also in [3], in essentially the same form. We include it with proof, to make the paper self-contained.

Lemma 1

Letf(x)2R[X]andg(u)2R[U]be as above. The polynomial f(x)has a unimodular root if and only if the polynomialg(u)has a real root in the interval[¡2;2], respectively if and only if the following cosine polynomial has real zeros. '(x) =²n=2+n=2¡1X k=02²kcos((n=2¡k)x) (4) Proof.Ifz=eiµis a unimodular root off(x) thenu= 2cosµis a root ofg(u) and clearly¡2·u·2. Conversely ifg(u) has a root,u2[¡2;2], then there existsµ2Rsuch that 2cosµ=u. Letz=eiµ, thenzis a unimodular root off(x). The fact that the existence of unimodular roots for the palindrome under consideration is equivalent to the existence of zeros for the cosine polynomial in (4) is an immediate consequence of the considerations above, equality (3), and the fact that ifx=eiµthen¾k(x) = 2coskµifk¸1. Based on the previous remarks we can prove the existence of unimodular roots for arbitrary palindromes with real coe±cients, having even degree, and null middle-term. More exactly the following is true. 3

Lemma 2

Letf(x)2R[X]be a palindrome having even degreenand such that²n=2= 0. Then the polynomialf(x)has unimodular roots. Proof.By Lemma 1 it will suffice to show that a cosine polynomial of the form '(x) =n=2¡1X k=0² kcos((n=2¡k)x) with real coe±cients²k, necessarily has a zero in the interval [0;2¼]. The later is an immediate consequence of the fact that the integral of'(x) on [0;2¼] equals 0 and'(x) is a continuous function. As we observed in the introduction, if the middle coe±cient is not null, a palindrome with real coe±cients may fail to have unimodular roots. However, the relative size of the middle coe±cient with respect to the other coe±cients matters here. The crucial inequality needed in the proof of our next and main result is sometimes known as theEgerv´ary-Sz´asz Inequality. It states that a non-negative cosine polynomial of the form 1 2 +nX k=1a kcoskx an6= 0 has the property jakj ·cosµ¼ k2 f1;2;:::;ng (see[1],[4]).

Theorem 1

Letf(x)be a palindrome with real coefficients, having even de- green, such that for somek2 f0;1;:::;n=2¡1gthe following condition is satisfied j²kj ¸cosà n=2 n=2¡k] + 2! j²n=2j:(5)

Such a palindrome has unimodular roots.

Proof.The case²n=2= 0 is covered by Lemma 2. Let us consider the case n=26= 0. Sincef(x) and¡f(x) have the same roots we can assume without loss of generality that²n=2>0. Denote by'(x) the cosine polynomial in equality (4). Since the integral of'(x) on [0;2¼] is positive, it follows that '(x) attains positive values on that interval. Thus, the only way it can 4 have no zeros would be if it is a positive cosine polynomial. Assume by contradiction that this is the case, and observe that the following cosine polynomial,Á(x), is also positive.

Á(x) = 1=2 +n=2¡1X

k=0² k n=2cos((n=2¡k)x) Let±= minfÁ(x) :x2[0;2¼]g. Observe that 0< ± <1=2. Indeed, Á(x)¡1=2 is not the null function on [0;2¼] and has null integral on that interval, thusÁ(x)¡1=2, being a continuous function, must assume both positive and negative values. Therefore the following cosine polynomial is non-negative onR

1=2 +n=2¡1X

k=0² k

2²n=2(1=2¡±)cos((n=2¡k)x):

By the Egerv´ary-Sz´asz Inequality, in such a case one should have the following inequality satisfied for eachk2 f0;1;:::;n=2¡1g k n=2(1¡2±)¯

¯¯¯·cosÃ

n=2 n=2¡k] + 2! This leads to the fact that for eachk2 f0;1;:::;n=2¡1gone can write k n=2¯

¯¯¯·cosÃ

n=2quotesdbs_dbs33.pdfusesText_39

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