Low-Degree Polynomial Roots

Total Page:16

File Type:pdf, Size:1020Kb

Low-Degree Polynomial Roots Low-Degree Polynomial Roots David Eberly, Geometric Tools, Redmond WA 98052 https://www.geometrictools.com/ This work is licensed under the Creative Commons Attribution 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by/4.0/ or send a letter to Creative Commons, PO Box 1866, Mountain View, CA 94042, USA. Created: July 15, 1999 Last Modified: September 10, 2019 Contents 1 Introduction 3 2 Discriminants 3 3 Preprocessing the Polynomials5 4 Quadratic Polynomials 6 4.1 A Floating-Point Implementation..................................6 4.2 A Mixed-Type Implementation...................................7 5 Cubic Polynomials 8 5.1 Real Roots of Multiplicity Larger Than One............................8 5.2 One Simple Real Root........................................9 5.3 Three Simple Real Roots......................................9 5.4 A Mixed-Type Implementation................................... 10 6 Quartic Polynomials 12 6.1 Processing the Root Zero...................................... 14 6.2 The Biquadratic Case........................................ 14 6.3 Multiplicity Vector (3; 1; 0; 0).................................... 15 6.4 Multiplicity Vector (2; 2; 0; 0).................................... 15 6.5 Multiplicity Vector (2; 1; 1; 0).................................... 15 6.6 Multiplicity Vector (1; 1; 1; 1).................................... 16 1 6.7 A Mixed-Type Implementation................................... 17 2 1 Introduction Consider a polynomial of degree d of the form d X i p(y) = piy (1) i=0 where the pi are real numbers and where pd 6= 0. A root of the polynomial is a number r, real or non-real (complex-valued with nonzero imaginary part) such that p(r) = 0. The polynomial can be factored as p(y) = (y − r)mf(y), where m is a positive integer and f(r) 6= 0. The power m is the multiplicity of the root. If m = 1, the root is said to be a simple root. If m ≥ 2, the root is said to be a multiple root. The fundamental theorem of algebra states that p(y) has exactly d roots, counting the multiplicities. If the K PK−1 distinct roots are yk for 0 ≤ k < K and if mk is the multiplicity of root yk, then d = k=0 mk. We are assuming that the pi are real numbers, so if a root is non-real, say, r = u + {v, where v 6= 0, thenr ¯ = u − {v is also a root. The pair (r; r¯) is referred to as a complex conjugate pair of roots. The roots of polynomials of degrees 2, 3, or 4 can be written using closed-form algebraic expressions. The theory of polynomials shows us that there cannot be general closed-form algebraic expressions for the roots of p(y) for d ≥ 5. In practice, numerical root finders must be used to locate the roots. For degrees 2, 3, and 4, one may compute the roots using the closed-form expressions, but this approach is typically ill conditioned when using floating-point arithmetic. The main problem occurs in sign tests used to classify the roots, both in domain (real or non-real) and in multiplicity (determine m). This document provides a robust approach for computing the roots using the closed-form expressions. The assumption is that the polynomial coefficients are floating-point numbers. In an implementation, they are converted to exact rational representations so that the root classification is performed with exact rational arithmetic and the result is theoretically correct. Naturally, the coefficients themselves might have been computed in a manner that introduced floating-point rounding errors. The discussion here does not take the coefficient errors into account, because only you the programmer have|and can take advantage of|that knowledge. 2 Discriminants The first step in classification uses a concept called the discriminant of a polynomial p(y) and is defined as 2d−2 Y 2 d(d−1)=2 2d−2 Y ∆ = pd (ri − rj) = (−1) pd (ri − rj) (2) i<j i6=j where r0 through rd−1 are the roots of p(y) counting the multiplicities: a root r with multiplicity m occurs m times in the set fr0; : : : ; rd−1g. In the event a root exists with multiplicity 2 or larger, it is clear that ∆ = 0; thus, at its highest level the discriminant lets you know whether or not a polynomial has repeated roots. More information is actually known regarding the sign of ∆, • ∆ > 0: for some integer k with 0 ≤ k ≤ d=4, there are 2k distinct complex-conjugate pairs of roots and d − 4k distinct real roots. Each root has a multiplicity of 1. • ∆ < 0: for some integer k with 0 ≤ k ≤ (d − 2)=4, there are 2k + 1 distinct complex-conjugate pairs of roots and d − 4k − 2 distinct real roots. Each root has a multiplicity of 1. 3 • ∆ = 0: at least one root r has multiplicity m ≥ 2 and r may be either real or non-real. If r is non-real, then its conjugater ¯ also has multiplicity m. 2 2 The prototypical case is p(y) = p0 + p1y + p2y . The discriminant is ∆ = p1 − 4p0p2. If ∆ > 0, the square root is a positive real number and there are two distinct real roots, p p −p1 − ∆ −p1 + ∆ r0 = ; r1 = 2p2 2p2 The integer k in the sign tests is 0, so we have 2 distinct real roots and 0 distinct complex-conjugate pairs of roots. If ∆ < 0, the square root is a pure imaginary number and the roots are non-real, p p −p1 − { −∆ −p1 + { −∆ r0 = ; r1 = 2p2 2p2 The integer k in the sign tests is 0. If ∆ = 0, we have a repeated real root, −p1 r0 = r1 = 2p2 so its multiplicity is m = 2. Of course we usually do not know the roots of p(y), so Equation (2) is not immediately helpful. However, the discriminant is related to another concept called the resultant of two polynomials, which is the product of 0 Pd−1 i the differences between the roots of the polynomials. The first derivative of p(y) is p (y) = i=0 (i+1)pi+1y and the resultant of p(y) and p0(y) is denoted R(p; p0). It is the determinant of a (2d − 1) × (2d − 1) matrix called the Sylvester matrix, 2 3 pd pd−1 pd−2 : : : p1 p0 0 :::::: 0 6 0 pd pd−1 pd−2 : : : p1 p0 0 ::: 0 7 6 . 7 6 . 7 6 . 7 6 7 6 0 ::: 0 pd pd−1 pd−2 : : : p1 p0 7 R(p; p0) = det 6 7 (3) 6 dpd (d − 1)pd−1 (d − 2)pd−2 : : : p1 0 :::::: 0 7 6 7 6 0 dpd (d − 1)pd−1 (d − 2)pd−2 : : : p1 0 ::: 0 7 6 . 7 6 . 7 4 . 5 0 0 ::: 0 dpd (d − 1)pd−1 (d − 2)pd−2 : : : p1 The discriminant is then d(d−1)=2 0 ∆ = (−1) R(p; p )=pd (4) The discriminant for the degree-2 (quadratic) polynomial is 2 3 p2 p1 p0 6 7 ∆ = −1 det 6 2p p 0 7 p2 6 2 1 7 4 5 0 2p2 p1 (5) 2 = p1 − 4p0p2 4 The discriminant for the degree-3 (cubic) polynomial is 2 3 p3 p2 p1 p0 0 6 7 6 7 6 0 p3 p2 p1 p0 7 6 7 ∆ = −1 det 6 3p 2p p 0 0 7 p3 6 3 2 1 7 6 7 6 7 6 0 3p3 2p2 p1 0 7 (6) 4 5 0 0 3p3 2p2 p1 2 2 3 3 2 2 = −27p0p3 + 18p0p1p2p3 − 4p0p2 − 4p1p3 + p1p2 The discriminant for the degree-4 (quartic) polynomial is 2 3 p4 p3 p2 p1 p0 0 0 6 7 6 7 6 0 p4 p3 p2 p1 p0 0 7 6 7 6 7 6 0 0 p4 p3 p2 p1 p0 7 6 7 ∆ = 1 det 6 4p 3p 2p p 0 0 0 7 p4 6 4 3 2 1 7 6 7 6 7 6 0 4p4 3p3 2p2 p1 0 0 7 6 7 6 7 6 0 0 4p4 3p3 2p2 p1 0 7 (7) 4 5 0 0 0 4p4 3p3 2p2 p1 3 3 2 2 2 2 2 2 2 2 4 2 2 2 2 = 256p0p4 − 192p0p1p3p4 − 128p0p2p4 + 144p0p2p3p4 − 27p0p3 + 144p0p1p2p4 − 6p0p1p3p4 2 3 4 3 2 4 2 3 3 3 − 80p0p1p2p3p4 + 18p0p1p2p3 + 16p0p2p4 − 4p0p2p3 − 27p1p4 + 18p1p2p3p4 − 4p1p3 2 3 2 2 2 − 4p1p2p4 + p1p2p3 For larger d, the discriminant expressions become quite complicated. These occur in a simpler format for preprocessed polynomials. 3 Preprocessing the Polynomials Preprocessing steps may be applied that make the root classification and construction simpler. The steps are error free when performed with exact rational arithmetic. In the analysis of p(y), the polynomial is made monic by dividing through by its leading coefficient to obtain d−1 X i d q(y) = qiy + y (8) i=0 where qi = pi=pd for 0 ≤ i ≤ d−1. The polynomials p(y) and q(y) have the same set of roots and multiplicities. The monic polynomial is made depressed by eliminating the yd−1 term using a simple translation, y = 5 x − qd−1=d, to obtain d−2 X i d c(x) = cix + x (9) i=0 The coefficients ci are obtained from binomial expansions of powers of (x − qd−1=d). These will be listed for each low-degree polynomial discussed in later sections. It is possible that low-order coefficients of c(x) are zero, so the analysis must allow for this. The roots of c(x) of Equation (9) are computed and then inverse transformed to obtain roots to p(y). Specifically, ifx ^ is a root of c(x), the corresponding root of p(y) is y^ =x ^ − qd−1=d.
Recommended publications
  • On the Number of Cubic Orders of Bounded Discriminant Having
    On the number of cubic orders of bounded discriminant having automorphism group C3, and related problems Manjul Bhargava and Ariel Shnidman August 8, 2013 Abstract For a binary quadratic form Q, we consider the action of SOQ on a two-dimensional vector space. This representation yields perhaps the simplest nontrivial example of a prehomogeneous vector space that is not irreducible, and of a coregular space whose underlying group is not semisimple. We show that the nondegenerate integer orbits of this representation are in natural bijection with orders in cubic fields having a fixed “lattice shape”. Moreover, this correspondence is discriminant-preserving: the value of the invariant polynomial of an element in this representation agrees with the discriminant of the corresponding cubic order. We use this interpretation of the integral orbits to solve three classical-style count- ing problems related to cubic orders and fields. First, we give an asymptotic formula for the number of cubic orders having bounded discriminant and nontrivial automor- phism group. More generally, we give an asymptotic formula for the number of cubic orders that have bounded discriminant and any given lattice shape (i.e., reduced trace form, up to scaling). Via a sieve, we also count cubic fields of bounded discriminant whose rings of integers have a given lattice shape. We find, in particular, that among cubic orders (resp. fields) having lattice shape of given discriminant D, the shape is equidistributed in the class group ClD of binary quadratic forms of discriminant D. As a by-product, we also obtain an asymptotic formula for the number of cubic fields of bounded discriminant having any given quadratic resolvent field.
    [Show full text]
  • 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.
    [Show full text]
  • Robust Algorithms for Object Localization
    Robust Algorithms for Ob ject Lo calizati on y Dinesh Mano cha Aaron S. Wallack Department of Computer Science Computer Science Division University of North Carolina University of California Chap el Hill, NC 27599-3175 Berkeley, CA 94720 mano [email protected] wallack@rob otics.eecs.b erkeley.edu Abstract Ob ject lo calization using sensed data features and corresp onding mo del features is a fun- damental problem in machine vision. We reformulate ob ject lo calization as a least squares problem: the optimal p ose estimate minimizes the squared error discrepancy b etween the sensed and predicted data. The resulting problem is non-linear and previous attempts to estimate the optimal p ose using lo cal metho ds such as gradient descent su er from lo cal minima and, at times, return incorrect results. In this pap er, we describ e an exact, accurate and ecient algorithm based on resultants, linear algebra, and numerical analysis, for solv- ing the nonlinear least squares problem asso ciated with lo calizing two-dimensional ob jects given two-dimensional data. This work is aimed at tasks where the sensor features and the mo del features are of di erenttyp es and where either the sensor features or mo del features are p oints. It is applicable to lo calizing mo deled ob jects from image data, and estimates the p ose using all of the pixels in the detected edges. The algorithm's running time dep ends mainly on the typ e of non-p oint features, and it also dep ends to a small extent on the number of features.
    [Show full text]
  • Math 2250 HW #10 Solutions
    Math 2250 Written HW #10 Solutions 1. Find the absolute maximum and minimum values of the function g(x) = e−x2 subject to the constraint −2 ≤ x ≤ 1. Answer: First, we check for critical points of g, which is differentiable everywhere. By the Chain Rule, 2 2 g0(x) = e−x · (−2x) = −2xe−x : Since e−x2 > 0 for all x, g0(x) = 0 when 2x = 0, meaning when x = 0. Hence, x = 0 is the only critical point of g. Now we just evaluate g at the critical point and the endpoints: 2 g(−2) = e−(−2) = e−4 = 1=e4 2 g(0) = e−0 = e0 = 1 2 g(2) = e−1 = e−1 = 1=e: Since 1 > 1=e > 1=e4, we see that the absolute maximum of g(x) on this interval is at (0; 1) and the absolute minimum is at (−2; 1=e4). 2. Find all local maxima and minima of the curve y = x2 ln x. Answer: Notice, first of all, that ln x is only defined for x > 0, so the function x2 ln x is likewise only defined for x > 0. This function is differentiable on its entire domain, so we differentiate in search of critical points. Using the Product Rule, 1 y0 = 2x ln x + x2 · = 2x ln x + x = x(2 ln x + 1): x Since we're only allowed to consider x > 0, we see that the derivative is zero only when 2 ln x + 1 = 0, meaning ln x = −1=2. Therefore, we have a critical point at x = e−1=2.
    [Show full text]
  • The Fast Quartic Solver Peter Strobach AST-Consulting Inc., Bahnsteig 6, 94133 Röhrnbach, Germany Article Info a B S T R a C T
    View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector Journal of Computational and Applied Mathematics 234 (2010) 3007–3024 Contents lists available at ScienceDirect Journal of Computational and Applied Mathematics journal homepage: www.elsevier.com/locate/cam The fast quartic solver Peter Strobach AST-Consulting Inc., Bahnsteig 6, 94133 Röhrnbach, Germany article info a b s t r a c t Article history: A fast and highly accurate algorithm for solving quartic equations is introduced. This Received 27 February 2010 new algorithm is more than six times as fast and several times more accurate than the Received in revised form 14 April 2010 quasi-standard Companion matrix eigenvalue quartic solver. Moreover, the method is exceptionally robust in cases of extreme root spread. The new algorithm is based on MSC: a factorization of the quartic in two quadratics, which are solved in closed form. The 65E05 performance key at this point is a fixed-point iteration based fitting algorithm for backward Keywords: optimization of the underlying quartic-to-quadratic polynomial decomposition. Detailed Quartic function experimental results confirm our claims. Polynomial factorization ' 2010 Elsevier B.V. All rights reserved. Polynomial rooting Companion matrix Eigenvalues 1. Introduction Consider the practical rooting of quartic polynomials or functions of the type f .x/ D x4 C ax3 C bx2 C cx C d D x2 C αx C β x2 C γ x C δ D .x − x1/.x − x2/.x − x3/.x − x4/ : (1) A quartic solver is an algorithm that relates a set of real or complex conjugate or mixed real/complex conjugate roots fx1; x2; x3; x4g to a given set of real quartic coefficients fa; b; c; dg.
    [Show full text]
  • Supplement 1: Toolkit Functions
    Supplement 1: Toolkit Functions What is a Function? The natural world is full of relationships between quantities that change. When we see these relationships, it is natural for us to ask “If I know one quantity, can I then determine the other?” This establishes the idea of an input quantity, or independent variable, and a corresponding output quantity, or dependent variable. From this we get the notion of a functional relationship in which the output can be determined from the input. For some quantities, like height and age, there are certainly relationships between these quantities. Given a specific person and any age, it is easy enough to determine their height, but if we tried to reverse that relationship and determine age from a given height, that would be problematic, since most people maintain the same height for many years. Function Function: A rule for a relationship between an input, or independent, quantity and an output, or dependent, quantity in which each input value uniquely determines one output value. We say “the output is a function of the input.” Function Notation The notation output = f(input) defines a function named f. This would be read “output is f of input” Graphs as Functions Oftentimes a graph of a relationship can be used to define a function. By convention, graphs are typically created with the input quantity along the horizontal axis and the output quantity along the vertical. The most common graph has y on the vertical axis and x on the horizontal axis, and we say y is a function of x, or y = f(x) when the function is named f.
    [Show full text]
  • As1.4: Polynomial Long Division
    AS1.4: POLYNOMIAL LONG DIVISION One polynomial may be divided by another of lower degree by long division (similar to arithmetic long division). Example (x32+ 3 xx ++ 9) (x + 2) x + 2 x3 + 3x2 + x + 9 1. Write the question in long division form. 3 2. Begin with the x term. 2 x3 divided by x equals x2. x 2 Place x above the division bracket x+2 x32 + 3 xx ++ 9 as shown. subtract xx32+ 2 2 3. Multiply x + 2 by x . x2 xx2+=+22 x 32 x ( ) . Place xx32+ 2 below xx32+ 3 then subtract. x322+−+32 x x xx = 2 ( ) 4. “Bring down” the next term (x) as 2 x + x indicated by the arrow. +32 +++ Repeat this process with the result until xxx23x 9 there are no terms remaining. x32+↓2x 2 2 5. x divided by x equals x. x + x Multiply xx( +=+22) x2 x . xx2 +−1 ( xx22+−) ( x +2 x) =− x 32 x+23 x + xx ++ 9 6. Bring down the 9. xx32+2 ↓↓ 2 7. − x divided by x equals − 1. xx+↓ Multiply xx2 +↓2 −12( xx +) =−− 2 . −+x 9 (−+xx9) −−−( 2) = 11 −−x 2 The remainder is 11 +11 (x32+ 3 xx ++ 9) equals xx2 +−1, with remainder 11. (x + 2) AS 1.4 – Polynomial Long Division Page 1 of 4 June 2012 If the polynomial has an ‘xn ‘ term missing, add the term with a coefficient of zero. Example (2xx3 − 31 +÷) ( x − 1) Rewrite (2xx3 −+ 31) as (2xxx32+ 0 −+ 31) Divide using the method from the previous example. 2 22xx32÷= x 2xx+− 21 2 32 −= − 2xx( 12) x 2 x 32 x−12031 xxx + −+ 20xx32−=−−(22xx32) 2x2 2xx32− 2 ↓↓ 2 22xx÷= x 2xx2 −↓ 3 2xx( −= 12) x2 − 2 x 2 2 2 2xx−↓ 2 23xx−=−−22xx− x ( ) −+x 1 −÷xx =−1 −+x 1 −11( xx −) =−+ 1 0 −+xx1− ( + 10) = Remainder is 0 (2xx32− 31 +÷) ( x −= 1) ( 2 xx + 21 −) with remainder = 0 ∴(2xx32 − 3 += 1) ( x − 12)( xx + 2 − 1) See Exercise 1.
    [Show full text]
  • Positivity Conditions for Cubic, Quartic and Quintic Polynomials Arxiv:2008.10922V10 [Math.GM] 18 Sep 2020
    Positivity Conditions for Cubic, Quartic and Quintic Polynomials Liqun Qi,∗ Yisheng Song,y and Xinzhen Zhang,z August 17, 2021 Abstract We present a necessary and sufficient condition for a cubic polynomial to be positive for all positive reals. We identify the set where the cubic polynomial is nonnegative but not all positive for all positive reals, and explicitly give the points where the cubic polynomial attains zero. We then reformulate a necessary and sufficient condition for a quartic polynomial to be nonnegative for all posi- tive reals. From this, we derive a necessary and sufficient condition for a quartic polynomial to be nonnegative and positive for all reals. Our condition explic- itly exhibits the scope and role of some coefficients, and has strong geometrical meaning. In the interior of the nonnegativity region for all reals, there is an ap- pendix curve. The discriminant is zero at the appendix, and positive in the other part of the interior of the nonnegativity region. By using the Sturm sequences, we present a necessary and sufficient condition for a quintic polynomial to be positive and nonnegative for all positive reals. We show that for polynomials of a fixed even degree higher than or equal to four, if they have no real roots, then their discriminants take the same sign, which depends upon that degree only, except on an appendix set of dimension lower by two, where the discriminants attain zero. arXiv:2008.10922v10 [math.GM] 18 Sep 2020 Key words. Cubic polynomials, quartic polynomials, quintic polynomials, the Sturm theorem, discriminant, appendix. AMS subject classifications.
    [Show full text]
  • Quartic Equation of General Form
    EqWorld http://eqworld.ipmnet.ru Exact Solutions > Algebraic Equations and Systems of Algebraic Equations > Algebraic Equations > Quartic Equation of General Form 8. ax4 + bx3 + cx2 + dx + e = 0 (a ≠ 0). Quartic equation of general form. 1±. Reduction to an incomplete equation. The quartic equation in question is reduced to an incom- plete equation y4 + py2 + qy + r = 0.(1) with the change of variable b x = y − 4a 2±. Decartes–Euler solution. The roots of the incomplete equation (1) are given by 1 ¡p p p ¢ 1 ¡p p p ¢ y1 = z1 + z2 + z3 , y2 = z1 − z2 − z3 , 2 2 2 1 ¡ p p p ¢ 1 ¡ p p p ¢ ( ) y3 = 2 − z1 + z2 − z3 , y4 = 2 − z1 − z2 + z3 , where z1, z2, z3 are roots of the cubic equation z3 + 2pz2 + (p2 − 4r)z − q2 = 0,(3) which is called the cubic resolvent of equation (1). The signs of the roots in (2) are chosen so that p p p z1 z2 z3 = −q. The roots of the incomplete quartic equation (1) are determined by the roots of the cubic resolvent (3); see the table below. TABLE Relation between the roots of the incomplete quartic equation and the roots of its cubic resolvent Cubic resolvent (3) Quartic equation (1) All roots are real and positive* Four real roots All roots are real, Two pairs of complex conjugate roots one positive and two negative* One roots is positive Two real and two complex conjugate roots and two roots are complex conjugate 2 * By Vieta’s theorem, the product of the roots z1, z2, z3 is equal to q ≥ 0.
    [Show full text]
  • Fundamental Units and Regulators of an Infinite Family of Cyclic Quartic Function Fields
    J. Korean Math. Soc. 54 (2017), No. 2, pp. 417–426 https://doi.org/10.4134/JKMS.j160002 pISSN: 0304-9914 / eISSN: 2234-3008 FUNDAMENTAL UNITS AND REGULATORS OF AN INFINITE FAMILY OF CYCLIC QUARTIC FUNCTION FIELDS Jungyun Lee and Yoonjin Lee Abstract. We explicitly determine fundamental units and regulators of an infinite family of cyclic quartic function fields Lh of unit rank 3 with a parameter h in a polynomial ring Fq[t], where Fq is the finite field of order q with characteristic not equal to 2. This result resolves the second part of Lehmer’s project for the function field case. 1. Introduction Lecacheux [9, 10] and Darmon [3] obtain a family of cyclic quintic fields over Q, and Washington [22] obtains a family of cyclic quartic fields over Q by using coverings of modular curves. Lehmer’s project [13, 14] consists of two parts; one is finding families of cyclic extension fields, and the other is computing a system of fundamental units of the families. Washington [17, 22] computes a system of fundamental units and the regulators of cyclic quartic fields and cyclic quintic fields, which is the second part of Lehmer’s project. We are interested in working on the second part of Lehmer’s project for the families of function fields which are analogous to the type of the number field families produced by using modular curves given in [22]: that is, finding a system of fundamental units and regulators of families of cyclic extension fields over the rational function field Fq(t). In [11], we obtain the results for the quintic extension case.
    [Show full text]
  • (Trying To) Solve Higher Order Polynomial Equations. Featuring a Recall of Polynomial Long Division
    (Trying to) solve Higher order polynomial equations. Featuring a recall of polynomial long division. Some history: The quadratic formula (Dating back to antiquity) allows us to solve any quadratic equation. ax2 + bx + c = 0 What about any cubic equation? ax3 + bx2 + cx + d = 0? In the 1540's Cardano produced a \cubic formula." It is too complicated to actually write down here. See today's extra credit. What about any quartic equation? ax4 + bx3 + cx2 + dx + e = 0? A few decades after Cardano, Ferrari produced a \quartic formula" More complicated still than the cubic formula. In the 1820's Galois (age ∼19) proved that there is no general algebraic formula for the solutions to a degree 5 polynomial. In fact there is no purely algebraic way to solve x5 − x − 1 = 0: Galois died in a duel at age 21. This means that, as disheartening as it may feel, we will never get a formulaic solution to a general polynomial equation. The best we can get is tricks that work sometimes. Think of these tricks as analogous to the strategies you use to factor a degree 2 polynomial. Trick 1 If you find one solution, then you can find a factor and reduce to a simpler polynomial equation. Example. 2x2 − x2 − 1 = 0 has x = 1 as a solution. This means that x − 1 MUST divide 2x3 − x2 − 1. Use polynomial long division to write 2x3 − x2 − 1 as (x − 1) · (something). Now find the remaining two roots 1 2 For you: Find all of the solutions to x3 + x2 + x + 1 = 0 given that x = −1 is a solution to this equation.
    [Show full text]
  • Keshas Final Report
    Copyright by LaKesha Rochelle Whitfield 2009 Understanding Complex Numbers and Identifying Complex Roots Graphically by LaKesha Rochelle Whitfield, B.A. Report Presented to the Faculty of the Graduate School of The University of Texas at Austin in Partial Fulfillment of the Requirements for the Degree of Masters of Arts The University of Texas at Austin August 2009 Understanding Complex Numbers and Identifying Complex Roots Graphically Approved by Supervising Committee: Efraim P. Armendariz Mark L. Daniels Dedication I would like to dedicate this master’s report to two very important men in my life – my father and my son. My father, the late John Henry Whitfield, Sr., who showed me daily what it means to work hard to achieve something. And to my son, Byshup Kourtland Rhodes, who continuously encourages and believes in me. To you both, I love and thank you! Acknowledgements I would like to acknowledge my friends and family members. Thank you for always supporting, encouraging, and believing in me. You all are the wind beneath my wings. Also, thank you to the 2007 UTeach Mathematics Masters Cohort. You all are a uniquely dynamic group of individuals whom I am honored to have met and worked with. 2009 v Abstract Understanding Complex Numbers and Identifying Complex Roots Graphically LaKesha Rochelle Whitfield, M.A. The University of Texas at Austin, 2009 Supervisor: Efraim Pacillas Armendariz This master’s report seeks to increase knowledge of complex numbers and how to identify complex roots graphically. The reader will obtain a greater understanding of the history of complex numbers, the definition of a complex number and a few of the field properties of complex numbers.
    [Show full text]