A Probabilistic Factorization Algorithm with Quadratic Forms of Negative Discriminant
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Solving Cubic Polynomials
Solving Cubic Polynomials 1.1 The general solution to the quadratic equation There are four steps to finding the zeroes of a quadratic polynomial. 1. First divide by the leading term, making the polynomial monic. a 2. Then, given x2 + a x + a , substitute x = y − 1 to obtain an equation without the linear term. 1 0 2 (This is the \depressed" equation.) 3. Solve then for y as a square root. (Remember to use both signs of the square root.) a 4. Once this is done, recover x using the fact that x = y − 1 . 2 For example, let's solve 2x2 + 7x − 15 = 0: First, we divide both sides by 2 to create an equation with leading term equal to one: 7 15 x2 + x − = 0: 2 2 a 7 Then replace x by x = y − 1 = y − to obtain: 2 4 169 y2 = 16 Solve for y: 13 13 y = or − 4 4 Then, solving back for x, we have 3 x = or − 5: 2 This method is equivalent to \completing the square" and is the steps taken in developing the much- memorized quadratic formula. For example, if the original equation is our \high school quadratic" ax2 + bx + c = 0 then the first step creates the equation b c x2 + x + = 0: a a b We then write x = y − and obtain, after simplifying, 2a b2 − 4ac y2 − = 0 4a2 so that p b2 − 4ac y = ± 2a and so p b b2 − 4ac x = − ± : 2a 2a 1 The solutions to this quadratic depend heavily on the value of b2 − 4ac. -
Algorithmic Factorization of Polynomials Over Number Fields
Rose-Hulman Institute of Technology Rose-Hulman Scholar Mathematical Sciences Technical Reports (MSTR) Mathematics 5-18-2017 Algorithmic Factorization of Polynomials over Number Fields Christian Schulz Rose-Hulman Institute of Technology Follow this and additional works at: https://scholar.rose-hulman.edu/math_mstr Part of the Number Theory Commons, and the Theory and Algorithms Commons Recommended Citation Schulz, Christian, "Algorithmic Factorization of Polynomials over Number Fields" (2017). Mathematical Sciences Technical Reports (MSTR). 163. https://scholar.rose-hulman.edu/math_mstr/163 This Dissertation is brought to you for free and open access by the Mathematics at Rose-Hulman Scholar. It has been accepted for inclusion in Mathematical Sciences Technical Reports (MSTR) by an authorized administrator of Rose-Hulman Scholar. For more information, please contact [email protected]. Algorithmic Factorization of Polynomials over Number Fields Christian Schulz May 18, 2017 Abstract The problem of exact polynomial factorization, in other words expressing a poly- nomial as a product of irreducible polynomials over some field, has applications in algebraic number theory. Although some algorithms for factorization over algebraic number fields are known, few are taught such general algorithms, as their use is mainly as part of the code of various computer algebra systems. This thesis provides a summary of one such algorithm, which the author has also fully implemented at https://github.com/Whirligig231/number-field-factorization, along with an analysis of the runtime of this algorithm. Let k be the product of the degrees of the adjoined elements used to form the algebraic number field in question, let s be the sum of the squares of these degrees, and let d be the degree of the polynomial to be factored; then the runtime of this algorithm is found to be O(d4sk2 + 2dd3). -
Elements of Chapter 9: Nonlinear Systems Examples
Elements of Chapter 9: Nonlinear Systems To solve x0 = Ax, we use the ansatz that x(t) = eλtv. We found that λ is an eigenvalue of A, and v an associated eigenvector. We can also summarize the geometric behavior of the solutions by looking at a plot- However, there is an easier way to classify the stability of the origin (as an equilibrium), To find the eigenvalues, we compute the characteristic equation: p Tr(A) ± ∆ λ2 − Tr(A)λ + det(A) = 0 λ = 2 which depends on the discriminant ∆: • ∆ > 0: Real λ1; λ2. • ∆ < 0: Complex λ = a + ib • ∆ = 0: One eigenvalue. The type of solution depends on ∆, and in particular, where ∆ = 0: ∆ = 0 ) 0 = (Tr(A))2 − 4det(A) This is a parabola in the (Tr(A); det(A)) coordinate system, inside the parabola is where ∆ < 0 (complex roots), and outside the parabola is where ∆ > 0. We can then locate the position of our particular trace and determinant using the Poincar´eDiagram and it will tell us what the stability will be. Examples Given the system where x0 = Ax for each matrix A below, classify the origin using the Poincar´eDiagram: 1 −4 1. 4 −7 SOLUTION: Compute the trace, determinant and discriminant: Tr(A) = −6 Det(A) = −7 + 16 = 9 ∆ = 36 − 4 · 9 = 0 Therefore, we have a \degenerate sink" at the origin. 1 2 2. −5 −1 SOLUTION: Compute the trace, determinant and discriminant: Tr(A) = 0 Det(A) = −1 + 10 = 9 ∆ = 02 − 4 · 9 = −36 The origin is a center. 1 3. Given the system x0 = Ax where the matrix A depends on α, describe how the equilibrium solution changes depending on α (use the Poincar´e Diagram): 2 −5 (a) α −2 SOLUTION: The trace is 0, so that puts us on the \det(A)" axis. -
Subclass Discriminant Nonnegative Matrix Factorization for Facial Image Analysis
Pattern Recognition 45 (2012) 4080–4091 Contents lists available at SciVerse ScienceDirect Pattern Recognition journal homepage: www.elsevier.com/locate/pr Subclass discriminant Nonnegative Matrix Factorization for facial image analysis Symeon Nikitidis b,a, Anastasios Tefas b, Nikos Nikolaidis b,a, Ioannis Pitas b,a,n a Informatics and Telematics Institute, Center for Research and Technology, Hellas, Greece b Department of Informatics, Aristotle University of Thessaloniki, Greece article info abstract Article history: Nonnegative Matrix Factorization (NMF) is among the most popular subspace methods, widely used in Received 4 October 2011 a variety of image processing problems. Recently, a discriminant NMF method that incorporates Linear Received in revised form Discriminant Analysis inspired criteria has been proposed, which achieves an efficient decomposition of 21 March 2012 the provided data to its discriminant parts, thus enhancing classification performance. However, this Accepted 26 April 2012 approach possesses certain limitations, since it assumes that the underlying data distribution is Available online 16 May 2012 unimodal, which is often unrealistic. To remedy this limitation, we regard that data inside each class Keywords: have a multimodal distribution, thus forming clusters and use criteria inspired by Clustering based Nonnegative Matrix Factorization Discriminant Analysis. The proposed method incorporates appropriate discriminant constraints in the Subclass discriminant analysis NMF decomposition cost function in order to address the problem of finding discriminant projections Multiplicative updates that enhance class separability in the reduced dimensional projection space, while taking into account Facial expression recognition Face recognition subclass information. The developed algorithm has been applied for both facial expression and face recognition on three popular databases. -
Polynomials and Quadratics
Higher hsn .uk.net Mathematics UNIT 2 OUTCOME 1 Polynomials and Quadratics Contents Polynomials and Quadratics 64 1 Quadratics 64 2 The Discriminant 66 3 Completing the Square 67 4 Sketching Parabolas 70 5 Determining the Equation of a Parabola 72 6 Solving Quadratic Inequalities 74 7 Intersections of Lines and Parabolas 76 8 Polynomials 77 9 Synthetic Division 78 10 Finding Unknown Coefficients 82 11 Finding Intersections of Curves 84 12 Determining the Equation of a Curve 86 13 Approximating Roots 88 HSN22100 This document was produced specially for the HSN.uk.net website, and we require that any copies or derivative works attribute the work to Higher Still Notes. For more details about the copyright on these notes, please see http://creativecommons.org/licenses/by-nc-sa/2.5/scotland/ Higher Mathematics Unit 2 – Polynomials and Quadratics OUTCOME 1 Polynomials and Quadratics 1 Quadratics A quadratic has the form ax2 + bx + c where a, b, and c are any real numbers, provided a ≠ 0 . You should already be familiar with the following. The graph of a quadratic is called a parabola . There are two possible shapes: concave up (if a > 0 ) concave down (if a < 0 ) This has a minimum This has a maximum turning point turning point To find the roots (i.e. solutions) of the quadratic equation ax2 + bx + c = 0, we can use: factorisation; completing the square (see Section 3); −b ± b2 − 4 ac the quadratic formula: x = (this is not given in the exam). 2a EXAMPLES 1. Find the roots of x2 −2 x − 3 = 0 . -
Quadratic Polynomials
Quadratic Polynomials If a>0thenthegraphofax2 is obtained by starting with the graph of x2, and then stretching or shrinking vertically by a. If a<0thenthegraphofax2 is obtained by starting with the graph of x2, then flipping it over the x-axis, and then stretching or shrinking vertically by the positive number a. When a>0wesaythatthegraphof− ax2 “opens up”. When a<0wesay that the graph of ax2 “opens down”. I Cit i-a x-ax~S ~12 *************‘s-aXiS —10.? 148 2 If a, c, d and a = 0, then the graph of a(x + c) 2 + d is obtained by If a, c, d R and a = 0, then the graph of a(x + c)2 + d is obtained by 2 R 6 2 shiftingIf a, c, the d ⇥ graphR and ofaax=⇤ 2 0,horizontally then the graph by c, and of a vertically(x + c) + byd dis. obtained (Remember by shiftingshifting the the⇥ graph graph of of axax⇤ 2 horizontallyhorizontally by by cc,, and and vertically vertically by by dd.. (Remember (Remember thatthatd>d>0meansmovingup,0meansmovingup,d<d<0meansmovingdown,0meansmovingdown,c>c>0meansmoving0meansmoving thatleft,andd>c<0meansmovingup,0meansmovingd<right0meansmovingdown,.) c>0meansmoving leftleft,and,andc<c<0meansmoving0meansmovingrightright.).) 2 If a =0,thegraphofafunctionf(x)=a(x + c) 2+ d is called a parabola. If a =0,thegraphofafunctionf(x)=a(x + c)2 + d is called a parabola. 6 2 TheIf a point=0,thegraphofafunction⇤ ( c, d) 2 is called thefvertex(x)=aof(x the+ c parabola.) + d is called a parabola. The point⇤ ( c, d) R2 is called the vertex of the parabola. -
The Determinant and the Discriminant
CHAPTER 2 The determinant and the discriminant In this chapter we discuss two indefinite quadratic forms: the determi- nant quadratic form det(a, b, c, d)=ad bc, − and the discriminant disc(a, b, c)=b2 4ac. − We will be interested in the integral representations of a given integer n by either of these, that is the set of solutions of the equations 4 ad bc = n, (a, b, c, d) Z − 2 and 2 3 b ac = n, (a, b, c) Z . − 2 For q either of these forms, we denote by Rq(n) the set of all such represen- tations. Consider the three basic questions of the previous chapter: (1) When is Rq(n) non-empty ? (2) If non-empty, how large Rq(n)is? (3) How is the set Rq(n) distributed as n varies ? In a suitable sense, a good portion of the answers to these question will be similar to the four and three square quadratic forms; but there will be major di↵erences coming from the fact that – det and disc are indefinite quadratic forms (have signature (2, 2) and (2, 1) over the reals), – det and disc admit isotropic vectors: there exist x Q4 (resp. Q3) such that det(x)=0(resp.disc(x) = 0). 2 1. Existence and number of representations by the determinant As the name suggest, determining Rdet(n) is equivalent to determining the integral 2 2 matrices of determinant n: ⇥ (n) ab Rdet(n) M (Z)= g = M2(Z), det(g)=n . ' 2 { cd 2 } ✓ ◆ n 0 Observe that the diagonal matrix a = has determinant n, and any 01 ✓ ◆ other matrix in the orbit SL2(Z).a is integral and has the same determinant. -
Finite Fields: Further Properties
Chapter 4 Finite fields: further properties 8 Roots of unity in finite fields In this section, we will generalize the concept of roots of unity (well-known for complex numbers) to the finite field setting, by considering the splitting field of the polynomial xn − 1. This has links with irreducible polynomials, and provides an effective way of obtaining primitive elements and hence representing finite fields. Definition 8.1 Let n ∈ N. The splitting field of xn − 1 over a field K is called the nth cyclotomic field over K and denoted by K(n). The roots of xn − 1 in K(n) are called the nth roots of unity over K and the set of all these roots is denoted by E(n). The following result, concerning the properties of E(n), holds for an arbitrary (not just a finite!) field K. Theorem 8.2 Let n ∈ N and K a field of characteristic p (where p may take the value 0 in this theorem). Then (i) If p ∤ n, then E(n) is a cyclic group of order n with respect to multiplication in K(n). (ii) If p | n, write n = mpe with positive integers m and e and p ∤ m. Then K(n) = K(m), E(n) = E(m) and the roots of xn − 1 are the m elements of E(m), each occurring with multiplicity pe. Proof. (i) The n = 1 case is trivial. For n ≥ 2, observe that xn − 1 and its derivative nxn−1 have no common roots; thus xn −1 cannot have multiple roots and hence E(n) has n elements. -
Prime Factorization
Prime Factorization Prime Number A number with only two factors: ____ and itself Circle the prime numbers listed below 25 30 2 5 1 9 14 61 Composite Number A number that has more than 2 factors List five examples of composite numbers What kind of number is 0? What kind of number is 1? Every human has a unique fingerprint. Similarly, every COMPOSITE number has a unique "factorprint" called __________________________ Prime Factorization the factorization of a composite number into ____________ factors You can use a _________________ to find the prime factorization of any composite number. Ask yourself, "what Factorization two whole numbers 24 could I multiply together to equal the given number?" If the number is prime, do not put 1 x the number. Once you have all prime numbers, you are finished. Write your answer in exponential form. 24 Expanded Form (written as a multiplication of prime numbers) _______________________ Exponential Form (written with exponents) ________________________ Prime Factorization Ask yourself, "what two 36 numbers could I multiply together to equal the given number?" If the number is prime, do not put 1 x the number. Once you have all prime numbers, you are finished. Write your answer in both expanded and exponential forms. Prime Factorization Ask yourself, "what two 68 numbers could I multiply together to equal the given number?" If the number is prime, do not put 1 x the number. Once you have all prime numbers, you are finished. Write your answer in both expanded and exponential forms. Prime Factorization Ask yourself, "what two 120 numbers could I multiply together to equal the given number?" If the number is prime, do not put 1 x the number. -
Nature of the Discriminant
Name: ___________________________ Date: ___________ Class Period: _____ Nature of the Discriminant Quadratic − b b 2 − 4ac x = b2 − 4ac Discriminant Formula 2a The discriminant predicts the “nature of the roots of a quadratic equation given that a, b, and c are rational numbers. It tells you the number of real roots/x-intercepts associated with a quadratic function. Value of the Example showing nature of roots of Graph indicating x-intercepts Discriminant b2 – 4ac ax2 + bx + c = 0 for y = ax2 + bx + c POSITIVE Not a perfect x2 – 2x – 7 = 0 2 b – 4ac > 0 square − (−2) (−2)2 − 4(1)(−7) x = 2(1) 2 32 2 4 2 x = = = 1 2 2 2 2 Discriminant: 32 There are two real roots. These roots are irrational. There are two x-intercepts. Perfect square x2 + 6x + 5 = 0 − 6 62 − 4(1)(5) x = 2(1) − 6 16 − 6 4 x = = = −1,−5 2 2 Discriminant: 16 There are two real roots. These roots are rational. There are two x-intercepts. ZERO b2 – 4ac = 0 x2 – 2x + 1 = 0 − (−2) (−2)2 − 4(1)(1) x = 2(1) 2 0 2 x = = = 1 2 2 Discriminant: 0 There is one real root (with a multiplicity of 2). This root is rational. There is one x-intercept. NEGATIVE b2 – 4ac < 0 x2 – 3x + 10 = 0 − (−3) (−3)2 − 4(1)(10) x = 2(1) 3 − 31 3 31 x = = i 2 2 2 Discriminant: -31 There are two complex/imaginary roots. There are no x-intercepts. Quadratic Formula and Discriminant Practice 1. -
Resultant and Discriminant of Polynomials
RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results are well-known 19th century mathematics, but I have not inves- tigated the history, and no references are given. 1. Resultant n m Definition 1.1. Let f(x) = anx + ··· + a0 and g(x) = bmx + ··· + b0 be two polynomials of degrees (at most) n and m, respectively, with coefficients in an arbitrary field F . Their resultant R(f; g) = Rn;m(f; g) is the element of F given by the determinant of the (m + n) × (m + n) Sylvester matrix Syl(f; g) = Syln;m(f; g) given by 0an an−1 an−2 ::: 0 0 0 1 B 0 an an−1 ::: 0 0 0 C B . C B . C B . C B C B 0 0 0 : : : a1 a0 0 C B C B 0 0 0 : : : a2 a1 a0C B C (1.1) Bbm bm−1 bm−2 ::: 0 0 0 C B C B 0 bm bm−1 ::: 0 0 0 C B . C B . C B C @ 0 0 0 : : : b1 b0 0 A 0 0 0 : : : b2 b1 b0 where the m first rows contain the coefficients an; an−1; : : : ; a0 of f shifted 0; 1; : : : ; m − 1 steps and padded with zeros, and the n last rows contain the coefficients bm; bm−1; : : : ; b0 of g shifted 0; 1; : : : ; n−1 steps and padded with zeros. In other words, the entry at (i; j) equals an+i−j if 1 ≤ i ≤ m and bi−j if m + 1 ≤ i ≤ m + n, with ai = 0 if i > n or i < 0 and bi = 0 if i > m or i < 0. -
Performance and Difficulties of Students in Formulating and Solving Quadratic Equations with One Unknown* Makbule Gozde Didisa Gaziosmanpasa University
ISSN 1303-0485 • eISSN 2148-7561 DOI 10.12738/estp.2015.4.2743 Received | December 1, 2014 Copyright © 2015 EDAM • http://www.estp.com.tr Accepted | April 17, 2015 Educational Sciences: Theory & Practice • 2015 August • 15(4) • 1137-1150 OnlineFirst | August 24, 2015 Performance and Difficulties of Students in Formulating and Solving Quadratic Equations with One Unknown* Makbule Gozde Didisa Gaziosmanpasa University Ayhan Kursat Erbasb Middle East Technical University Abstract This study attempts to investigate the performance of tenth-grade students in solving quadratic equations with one unknown, using symbolic equation and word-problem representations. The participants were 217 tenth-grade students, from three different public high schools. Data was collected through an open-ended questionnaire comprising eight symbolic equations and four word problems; furthermore, semi-structured interviews were conducted with sixteen of the students. In the data analysis, the percentage of the students’ correct, incorrect, blank, and incomplete responses was determined to obtain an overview of student performance in solving symbolic equations and word problems. In addition, the students’ written responses and interview data were qualitatively analyzed to determine the nature of the students’ difficulties in formulating and solving quadratic equations. The findings revealed that although students have difficulties in solving both symbolic quadratic equations and quadratic word problems, they performed better in the context of symbolic equations compared with word problems. Student difficulties in solving symbolic problems were mainly associated with arithmetic and algebraic manipulation errors. In the word problems, however, students had difficulties comprehending the context and were therefore unable to formulate the equation to be solved.