Zeros of Polynomial Functions

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Board of Directors The Central Board of Directors is at the top of the Reserve Bank s organisational structure Appointed by the Government under the provisions of the Reserve Zeros of Polynomial Functions Zeros of Polynomial Functions By: OpenStax College Precalculus A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions The bakery wants the volume of a small cake to be 351 cubic inches The cake is in the shape of a rectangular solid They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width What should the dimensions of the cake pan be? This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake In this section, we will discuss a variety of tools for writing polynomial functions and solving polynomial equations Evaluating a Polynomial Using the Remainder Theorem In the last section, we learned how to divide polynomials We can now use polynomial division to evaluate polynomials using the Remainder Theorem If the polynomial is divided by x – k, the remainder may be found quickly by evaluating the polynomial function at k, that is, f(k) Let’s walk through the proof of the theorem Recall that the Division Algorithm states that, given a polynomial dividend f(x) and a non-zero polynomial divisor d(x) where the degree of d(x) is less than or equal to the degree of f(x), there exist unique polynomials q(x) and r(x) such that f(x) = d(x)q(x) + r(x) If the divisor, d(x), is x − k, this takes the form f(x) = (x − k)q(x) + r Since the divisor x − k is linear, the remainder will be a constant, r And, if we evaluate this for x = k, we have 1/28 Zeros of Polynomial Functions f(k) = (k − k)q(k) + r = ⋅ q(k) + r =r In other words, f(k) is the remainder obtained by dividing f(x) by x − k A General Note The Remainder Theorem If a polynomial f(x) is divided by x − k, then the remainder is the value f(k) How To Given a Theorem function f, evaluate f(x) at x = k using polynomial the Remainder Use synthetic division to divide the polynomial by x − k The remainder is the value f(k) Using the Remainder Theorem to Evaluate a Polynomial Use the Remainder Theorem to evaluate f(x) = 6x4 − x3 − 15x2 + 2x − at x = To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by x − 2 −1 − 15 12 22 11 −7 14 32 16 25 The remainder is 25 Therefore, f(2) = 25 Analysis We can check our answer by evaluating f(2) f(x) = 6x4 − x3 − 15x2 + 2x − f(2) = 6(2)4 − (2)3 − 15(2)2 + 2(2) − = 25 2/28 Zeros of Polynomial Functions Try It Use the Remainder Theorem to evaluate f(x) = 2x5 − 3x4 − 9x3 + 8x2 + at x = − f( − 3) = − 412 Using the Factor Theorem to Solve a Polynomial Equation The Factor Theorem is another theorem that helps us analyze polynomial equations It tells us how the zeros of a polynomial are related to the factors Recall that the Division Algorithm tells us f(x) = (x − k)q(x) + r If k is a zero, then the remainder r f(x) = (x − k)q(x) is f(k) = and f(x) = (x − k)q(x) + or Notice, written in this form, x − k is a factor of f(x) We can conclude if k is a zero of f(x), then x − k is a factor off(x) Similarly, if x − k is a factor of f(x), then the remainder of the Division Algorithm f(x) = (x − k)q(x) + r is This tells us that k is a zero This pair of implications is the Factor Theorem As we will soon see, a polynomial of degree n in the complex number system will have n zeros We can use the Factor Theorem to completely factor a polynomial into the product of n factors Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial A General Note The Factor Theorem According to the Factor Theorem, k is a zero of f(x) if and only if (x − k) is a factor of f(x) How To Given a factor and a third-degree polynomial, use the Factor Theorem to factor the polynomial Use synthetic division to divide the polynomial by (x − k) Confirm that the remainder is Write the polynomial as the product of (x − k) and the quadratic quotient If possible, factor the quadratic 3/28 Zeros of Polynomial Functions Write the polynomial as the product of factors Using the Factor Theorem to Solve a Polynomial Equation Show that (x + 2) is a factor of x3 − 6x2 − x + 30 Find the remaining factors Use the factors to determine the zeros of the polynomial We can use synthetic division to show that (x + 2) is a factor of the polynomial −2 1 −6 −1 30 −2 16 − 30 − 15 The remainder is zero, so (x + 2) is a factor of the polynomial We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient: (x + 2)(x2 − 8x + 15) We can factor the quadratic factor to write the polynomial as (x + 2)(x − 3)(x − 5) By the Factor Theorem, the zeros of x3 − 6x2 − x + 30 are –2, 3, and Try It Use the Factor Theorem to find the zeros of f(x) = x3 + 4x2 − 4x − 16 given that(x − 2) is a factor of the polynomial The zeros are 2, –2, and –4 Using the Rational Zero Theorem to Find Rational Zeros Another use for the Remainder Theorem is to test whether a rational number is a zero for a ...LINEAR AND NON-LINEAR OPERATORS, AND THE DISTRIBUTION OF ZEROS OF ENTIRE FUNCTIONS A DISSERTATION SUBMITTED TO THE GRADUATE DIVISION OF THE UNIVERSITY OF HAWAI‘I AT M ¯ ANOA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY IN MATHEMATICS AUGUST 2013 By Rintaro Yoshida Dissertation Committee: George Csordas, Chairperson Thomas Craven Erik Guentner Marcelo Kobayashi Wayne Smith ACKNOWLEDGMENTS I would like to express my deepest gratitude to my advisor Dr. George Csordas for his patience, enthusiasm, and support throughout the years of my studies. The completion of this dissertation is mainly due to the willingness of Dr. Csordas in generously bestowing me countless hours of advice and insight. The members of the dissertation committee, who willingly accepted to take time out of their busy schedules to review, comment, and attend the defence regarding this work, deserve great recognition. I would like to thank Dr. Pavel Guerzhoy for his understanding when I decided to change my area of study. As iron sharpens iron, fellow advisees Dr. Andrzej Piotrowski, Dr. Matthew Chasse, Dr. Lukasz Grabarek, and Mr. Robert Bates have my greatest appreciation in allowing me to take part in honing my ability to understand our beloved theory of distribution of zeros of entire functions. I appreciate everyone in the Mathematics department, but in particular, Austin Anderson, Mike Andonian, John and Tabitha Brown, William DeMeo, Patricia Goldstein, Alex Gottlieb, Zach Kent, Sue Hasegawa, Mike Joyce, Shirley Kikiloi, Troy Ludwick, Alicia Maedo, John Marriot, Chi Mingjing, Paul Nguyen, Geoff Patterson, John Radar, Gretel Sia, Jacob Woolcutt, Diane Yap, and Robert Young for their congeniality. Financial support was received from the University of Hawai‘i Graduate Student Organization for travel expenses to Macau, China, and the American Institute of Mathematics generously provided funding for travel and accommodation expenses in hosting the workshop “Stability, hyperbolicity, and zero location of functions” in Palo Alto, California. My parents have been very patient during my time in graduate school, and I’m glad to have a sister who is a blessing. I am thankful for my closest friends in California, Kristina Aquino, Jeremy Barker, Rod and Aleta Bollins, Jason Chikami, Daniel and Amie Chikami, Jay Cho, Dominic Fiorello, Todd Gilliam, Jim and Betty Griset, Karl and Jane Gudino, Michael Hadj, Steve and Hoan Hensley, Ruslan Janumyan, Kevin Knight, Phuong Le, Lemee Nakamura, Barry and Irene McGeorge, Israel and Yoko Peralta, Shawn Sami, Roger Yang, Henry Yen, my church family in California at Bethany Bible Fellowship, my church ohana at Kapahulu Bible Church, and last but clearly not least, Yahweh. ii ABSTRACT An important chapter in the theory of distribution of zeros of entire functions pertains to the study of linear operators acting on entire functions. This dissertation presents new results involving not only linear, but also some non-linear operators. If {γ k } ∞ k=0 is a sequence of real numbers, and Q = {q k (x)} ∞ k=0 is a sequence of polynomials, where deg q k (x)= k, associate with the sequence {γ k } ∞ k=0 a linear operator T such that T [q k (x)]=γ k q k (x), k = 0, 1, 2, . . . . The sequence {γ k } ∞ k=0 is termed a Q-multiplier sequence if T is a hyperbolicity preserving operator. Some multiplier sequences are characterized Zeros of the Jones polynomial are dense in the complex plane Xian’a n Jin ∗ Fuji Zhang School of Mathematical Sciences, Xiamen University, Xiamen, Fujian 361005, P.R.China xajin@xmu.edu.cn fjzhang@xmu.edu.cn Fengming Dong Eng Guan Tay Mathematics and Mathematics Education, National Institute of Education Nanyang Technological University, Singapore 637616, Singapore fengming.dong@nie.edu.sg engguan.tay@nie.edu.sg Submitted: Apr 13, 2010; Accepted: Jun 28, 2010; Published: Jul 10, 2010 Mathematics S ubject Classification: 05C10,05C22,05C31,57M15,57M25,57M27,82B20 Abstract In this paper, we present a formula for computing the Tutte polynomial of the signed graph formed from a labeled graph by edge replacements in terms of the chain polynomial of the labeled graph. Then we define a family of ‘ring of tangles’ links and consider zeros of their Jones polynomials. By applying the formula obtained, Beraha-Kahane-Weiss’s theorem and Sokal’s lemma, we prove that zeros of Jones polynomials of (pretzel) links are dense in the whole complex plane. 1 Introduction Let L be an oriented link, and D be a diagram of L. Let V L (t) be the Jones polynomial [1] of L. The writhe w(D) of D is defined to be t he sum of signs of the crossings of L. Let [D] be the one-variable Kauffman bracket polynomial [2] of D in A (with the orientation of D coming from L ignored). Let f L (A) = (−A 3 ) −w(D) [D]. Then [2] V L (t) = f L (t −1/4 ). ∗ Corresponding author. the electronic journal of combinatorics 17 (2010), #R94 1 It is well known that there is a one-to-one correspondence between link diagrams and signed plane graphs via the medial construction [3]. Based on this correspondence, in [3] Kauffman converted the Kauffman bracket polynomial to the Tutte polynomial of signed graphs, which are not necessarily planar. Let G be a signed graph. We shall denote by Q[G] = Q[G](A, B, d) ∈ Z[A, B, d] the Tutte polynomial of G. To analyze zeros of Jones polynomials, it suffices for us to consider the Tutte polynomials of signed gr aphs. There have been some works on zeros of Jones polynomials [4 ]- [9]. For example, the distribution of zeros of Jones polynomials for prime knots with small crossing number has been depicted in [4] and [5]. See Fig. 1 for an example. Looking at these figures, one may be tempted to conclude that zeros of Jones polynomials do not exist in some regions, for example, a small circle region around z = 1 and a large area in the left half complex plane. But in this paper by considering zeros of pretzel links we shall show that, on the contrary, zeros of Jones po lynomial of knots are dense in the whole complex plane. We point out that Sokal proved that chromatic roots a r e dense in the whole complex plane [10], and Zhang and Chen proved eigenvalues of digraphs are dense in the whole complex plane [11]. Fig. 1: Zeros of 1288 prime alternating knots with crossing number 12 [4]. This paper contains two parts. The first part is on a formula of computing the Tutte polynomial of signed graphs formed by edge replacements via the chain polynomial [12]. This result generalizes our previous results in [13]. It is worth noting that there are two closely related results in [14] and [15]. In the second part, we consider the Tutte polynomial of ‘ring of tangles’ links which include the well known pretzel links. By applying Beraha- Kahane-Weiss’s Theorem and Sokal’s lemma, we prove that zeros of the Jones polynomial of pretzel knots are dense in the complex plane. the electronic journal of combinatorics 17 (2010), #R94 2 2 Tutte polynomials of signed graphs formed by edge replace- ments The chain polynomial was introduced by Read and Whitehead in [12] for studying the chromatic polynomials of homeomorphic graphs. It is defined on labeled graphs, i.e. graphs whose edges have been labeled with elements of a commutative ring with unity. Although in a labeled graph different edges can receive the same label, we usually denote the edges by the labels associated with them. Th Noname manuscript No. (will be inserted by the editor) Representation of non-negative Morse polynomial functions and applications in Polynomial Optimization Lê Công-Trình Received: date / Accepted: date Abstract In this paper we study the representation of Morse polynomial functions which are non-negative on a compact basic closed semi-algebraic set in Rn , and having only finitely many zeros in this set. Following C. BiviàAusina [2], we introduce two classes of non-degenerate polynomials for which the algebraic sets defined by them are compact. As a consequence, we study the representation of non-negative Morse polynomials on these kinds of nondegenerate algebraic sets. Moreover, we apply these results to study the polynomial Optimization problem for Morse polynomial functions. Mathematics Subject Classification (2000) 11E25 · 13J30 · 14H99 · 14P05 · 14P10 · 90C22 Keywords Sum of squares · Positivstellensatz · Polynomial Optimization · Local-global principle · Morse function · Non-degenerate polynomial map Contents 1 2 3 4 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Representation of non-negative Morse polynomial functions . . . . . . . . . . . Representation of Morse polynomial functions on non-degenerate algebraic sets Applications in polynomial Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 . 4 . 6 . 10 1 Introduction Let us denote by R[X] the ring of real polynomials in n variables x1 , · · · , xn , and by R[X]2 the set of all finitely many sums of squares (SOS) of polynoLê Công-Trình Department of Mathematics, Quy Nhon University 170 An Duong Vuong, Quy Nhon, Binh Dinh, Vietnam E-mail: lecongtrinh@qnu.edu.vn 2 Lê Công-Trình mials in R[X]. Let us fix a finite subset G = {g1 , · · · , gm } in R[X]. Let KG = {x = (x1 , · · · , xn ) ∈ Rn |g1 (x) ≥ 0, · · · , gm (x) ≥ 0} be the basic closed semi-algebraic set in Rn generated by G. Let m MG := { si gi |si ∈ R[X]2 } i=1 be the quadratic module in R[X] generated by G, and let σm sσ g1σ1 · · · gm |sσ ∈ TG := { R[X]2 } σ=(σ1 ,··· ,σm )∈{0,1}m denote the preordering in R[X] generated by G. It is clear that MG ⊆ TG , and if a polynomial belongs to TG (or MG ) then it is non-negative on KG . However the converse is not always true, that means there exists a polynomial which is non-negative on KG but it does not belong to TG (resp. MG ). The well-known examples (cf. [9]) are Motzkin’s polynomial, Robinson’s polynomial, etc. in the case G = ∅ (then KG = Rn and TG = MG = R[X]2 ). In 1991, Schm¨ udgen [16] showed, that if a polynomial is positive on a compact basic closed semi-algebraic set then it belongs to the corresponding preordering. After that, Putinar ([15], 1993) showed that if a polynomial is positive on a basic closed semi-algebraic set whose associated quadratic module is Archimedean, then it belongs to that quadratic module. If we allow the polynomial f having zeros in KG then the results above of Schm¨ udgen and Putinar are not true. Indeed, let us consider the set G = {(1−x2 )3 } in the ring R[x] of real polynomials in one variable. In this example, KG is the closed interval [−1, 1] ⊆ R which is compact. The polynomial f = 1 − x2 ∈ R[x] is non-negative on [−1, 1], and it has two zeros in [−1, 1]. It is not difficult to show that f∈ / TG = MG = {s0 + s1 (1 − x2 )3 |s0 , s1 ∈ R[x]2 }. Therefore a natural question is that under which conditions a polynomial which is non-negative on a basic closed semi-algebraic set belonging to the corresponding preordering (or quadratic module)? C. Scheiderer (2003 and 2005) has given the following local-global principles to answer this question. Theorem 1 ([17, Corollary 3.17]) Let G, KG and TG be as above, and let f ∈ R[X]. Assume that the following conditions hold true: (1) KG is compact; (2) f ≥ 0 on KG , and f has only finitely many zeros p1 , · · · , pr in KG ; (3) at each pi , f ∈ Tpi . Then f ∈ TG . Representation of Morse polynomial functions Recognizing circulant graphs of prime order in polynomial time ∗ Mikhail E. Muzychuk Netanya Academic College 42365 Netanya, Israel mikhail@netvision.net.il Gottfried Tinhofer Technical University of Munich 80290 M¨unchen, Germany gottin@mathematik.tu-muenchen.de Submitted: December 19, 1997; Accepted: April 1, 1998 Abstract A circulant graph G of order n is a Cayley graph over the cyclic group Z n . Equivalently, G is circulant iff its vertices can be ordered such that the cor- responding adjacency matrix becomes a circulant matrix. To each circulant graph we may associate a coherent configuration A and, in particular, a Schur ring S isomorphic to A. A can be associated without knowing G to be circu- lant. If n is prime, then by investigating the structure of A either we are able to find an appropriate ordering of the vertices proving that G is circulant or we are able to prove that a certain necessary condition for G being circulant is violated. The algorithm we propose in this paper is a recognition algorithm for cyclic association schemes. It runs in time polynomial in n. MR Subject Number: 05C25, 05C85, 05E30 Keywords: Circulant graph, cyclic association scheme, recognition algorithm ∗ The work reported in this paper has been partially supported by the German Israel Foundation for Scientific Research and Development under contract # I-0333-263.06/93 the electronic journal of combinatorics 3 (1996), #Rxx 2 1 Introduction The graphs considered in this paper are of the form (X, γ), where X is a finite set and γ is a binary relation on X which is not necessarily symmetric. Let G be a group and G =(X, γ) a graph with vertex set X = G and with adjacency relation γ defined with the aid of some subset C ⊂Gby γ = {(g, h):g,h ∈G∧gh −1 ∈ C}. Then G is called Cayley graph over the group G. Let Z n , n ∈ N, stand for a cyclic group of order n written additively. A circulant graph G over Z n is a Cayley graph over this group. In this particular case, the adjacency relation γ has the form γ = n−1  i=0 {i}×{i+γ(0)} where γ(0) is the set of successors of the vertex 0. Evidently, the set of successors γ(i) of an arbitrary vertex i satisfies γ(i)=i+γ(0). The set γ(0) is called the connection set of the circulant graph G. G is a simple undirected graph if 0 ∈ γ(0) and j ∈ γ(0) implies −j ∈ γ(0). There are different equivalent characterizations of circulant graphs. One of them is this: A graph G is a circulant graph iff its vertex set can be numbered in such a way that the resulting adjacency matrix A(G) is a circulant matrix. We call such a numbering a Cayley numbering. Still another characterization is: G is a circulant graph iff a cyclic permutation of its vertices exists which is an automorphism of G. Cayley graphs, and in particular, circulant graphs have been studied intensively in the literature. These graphs are easily seen to be vertex transitive. In the case of a prime vertex number n circulant graphs are known to be the only vertex transitive graphs. Because of their high symmetry, Cayley graphs are ideal models for commu- nication networks. Routing and weight balancing is easily done on such graphs. Assume that a graph G on the set V (G)={0, ,n−1} is given by its diagram or by its adjacency matrix, or by some other data structure commonly used in dealing with graphs. How can we decide whether G is a Cayley graph or not? In such a generality, this decision problem seems to be far from beeing tractable efficiently. A recognition algorithm for Cayley graphs would have to involve implicitly checking all finite groups of order n. In the special case of circulant graphs, or in any other case where the group G is given, we could recognize Cayley graphs by checking all different numberings of the vertex set and comparing ... Factors, and Graphs of Polynomial Functions Complex Factorization Theorem Find the Zeros of a Polynomial Function Find the Zeros of a Polynomial Function Find the Zeros of a Polynomial Function... the zeros of f(x) = 4x3 − 3x − p The Rational Zero Theorem tells us that if q is a zero of f(x), then p is a factor of –1 and q is a factor of 7/28 Zeros of Polynomial Functions p factor of constant... is a zero of f(x), then p is a factor of and q is a factor of p factor of constant term = q factor of leading coefficient = factor of factor of p The factors of are±1 and the factors of are±1