DOCSLIB.ORG

On the Solution to Nonic Equations

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
On the Solution to Nonic Equations On the Solution to Nonic Equations By Dr. Raghavendra G. Kulkarni Abstract. This paper presents a novel decomposition tech- nique in which a given nonic equation is decomposed into quartic and quintic polynomials as factors, eventually lead- ing to its solution in radicals. The conditions to be satisfied by the roots and the coefficients of such a solvable nonic equation are derived. Introduction It is well known that the general polynomial equations of degree five and above cannot be solved in radicals, and have to be solved algebraically using symbolic coefficients, such as the use of Bring radicals for solving the general quintic equation, and the use of Kampe de Feriet functions to solve the general sextic equation [1, 2]. There is hardly any literature on the analytical solution to the polynomial equations beyond sixth degree. The solution to certain solvable octic equations is described in a recent paper [4]. This paper presents a method of determining all nine roots of a nonic equation (ninth-degree polynomial equation) in radicals. First, the given nonic equation is decomposed into two factors in a novel fashion. One of these factors is a fourth-degree polynomial, while the other one is a fifth-degree polynomial. When the fourth- degree polynomial factor is equated to zero, the four roots of the given nonic equation are extracted by solving the resultant quar- tic equation using the well-known Ferrari method or by a method given in a recent paper [3]. However, when the fifth-degree polyno- mial factor is equated to zero, the resultant quintic equation is also required to be solved in radicals, since the aim of this paper is to solve the nonic equation completely in radicals. For this purpose, suitable supplementary equations are employed so that the given [3] 4 Alabama Journal of Mathematics nonic equation can be successfully decomposed, and then the resul- tant quintic factor also can be further decomposed. The criteria to be satisfied by the roots and the coefficients of such a solvable nonic equation are derived. At the end, we solve a numerical example using the proposed method. Decomposition of a Nonic Equation Consider the following nonic equation in x : 9 8 7 6 5 4 3 2 x +a8x +a7x +a6x +a5x +a4x +a3x +a2x +a1x+a0 =0, (1) where a0,a1,a2,a3,a4,a5,a6,a7, and a8 are real coefficients. In the proposed method, we add a root “at the origin” to the above equation, which is equivalent to multiplying (1) by x. This action converts the nonic equation (1) to a decic equation as shown below: 10 9 8 7 6 5 4 3 2 x +a8x +a7x +a6x +a5x +a4x +a3x +a2x +a1x +a0x =0. (2) Our aim is to decompose the above decic equation into two quintic factors. For this purpose, consider another decic equation as shown below: 5 4 3 2 2 x + b4x + b3x + b2x + b1x + b0 ¡ 4 3 2 ¢2 (c4x + c3x + c2x + c1x + c0) =0, − (3) where b0,b1,b2,b3,b4, and c0,c1,c2,c3,c4 are the unknown coeffi- cients in the quintic and the quartic polynomial terms respectively, in (3). Notice that the above decic equation (3) can be easily de- composed into two quintic factors as shown below: 5 4 3 2 x + b4x + b3x + b2x + b1x + b0 £¡ 4 3 2 ¢ c4x + c3x + c2x + c1x + c0 − 5 ¡ 4 3 2 ¢¤ x + b4x + b3x + b2x + b1x + b0 ∗ £¡ 4 3 2 ¢ + c4x + c3x + c2x + c1x + c0 =0, ¡ (4) ¢¤ Spring/Fall 2008 5 where the symbol * is the multiplication symbol. Equation 4 is further simplified as: 5 4 3 2 x +(b4 c4)x +(b3 c3)x +(b2 c2)x − − − £ +(b1 c1)x + b0 c0] − − 5 4 3 2 x +(b4 + c4)x +(b3 + c3)x +(b2 + c2)x ∗ £ +(b1 + c1)x + b0 + c0]=0. (5) Thus if the decic equation (2) can be represented in the form of de- cic equation (3), then it can be easily decomposed into two quintic factors as given in (5). In order for (3) to represent decic equation (2), the coefficients of (2) and (3) must be equal. Note, however, that the coefficients of (3) are not given explicitly. Therefore, to facilitate comparison of coefficients of (2) and (3), the decic equa- tion (3) is expanded and rearranged in descending powers of x, as shown below: 10 9 2 2 8 7 x +2b4x + b +2b3 c x +2(b2 + b3b4 c3c4) x 4 − 4 − 2 ¡ 2 ¢ 6 + b +2b1 +2b2b4 c 2c2c4 x 3 − 3 − ¡ ¢ 5 +2 (b0 + b1b4 + b2b3 c1c4 c2c3) x − − 2 2 4 + b +2b0b4 +2b1b3 c 2c0c4 2c1c3 x 2 − 2 − − ¡ 3 ¢ +2 (b0b3 + b1b2 c0c3 c1c2) x − − 2 2 2 2 2 +(b +2b0b2 c 2c0c2)x +2(b0b1 c0c1)x + b c =0. 1 − 1 − − 0 − 0 (6) Equating the coefficients of (2) and (6), we obtain the following ten equations: 2b4 = a8 (7) 2 2 b +2b3 c = a7 (8) 4 − 4 6 Alabama Journal of Mathematics 2(b2 + b3b4 c3c4)=a6 (9) − 2 2 b +2b1 +2b2b4 c 2c2c4 = a5 (10) 3 − 3 − 2(b0 + b1b4 + b2b3 c1c4 c2c3)=a4 (11) − − 2 2 b +2b0b4 +2b1b3 c 2c0c4 2c1c3 = a3 (12) 2 − 2 − − 2(b0b3 + b1b2 c0c3 c1c2)=a2 (13) − − 2 2 b +2b0b2 c 2c0c2 = a1 (14) 1 − 1 − 2(b0b1 c0c1)=a0 (15) − b2 c2 =0. (16) 0 − 0 Equation 16 results into two expressions for c0 as: c0 = b0, and we choose: ± c0 = b0. (17) When (17) is used in the factored decic equation (5), we obtain the following expression containing the quartic polynomial and the quintic polynomial as factors: 4 3 2 x x +(b4 c4) x +(b3 c3) x − − £ +(b2 c2) x +(b1 c1)] − − 5 4 3 x +(b4 + c4) x +(b3 + c3) x ∗ £ 2 +(b2 + c2) x +(b1 + c1) x +2b0 =0. ¤ (18) Notice that the added root (“at the origin,” x =0)to the given nonic equation (1) is visible in the factored equation (18). To facilitate extraction of the nine roots of the given nonic (1), the quartic and the quintic factors in (18) are equated to zero leading to the following quartic and quintic equations respectively as shown below: Spring/Fall 2008 7 4 3 2 x +(b4 c4)x +(b3 c3)x − − +(b2 c2)x + b1 c1 =0 − − (19) 5 4 3 2 x +(b4 + c4)x +(b3 + c3)x +(b2 + c2)x +(b1 + c1)x +2b0 =0. (20) Once the unknown coefficients in (19) are determined, the quartic equation (19) can be solved in radicals by well-known meth- ods [2, 3]. However, the same cannot be said for the quintic equa- tion (20), since there is no solution in radicals for a general quintic equation. Thus to make quintic equation (20) solvable in radicals, we need to find some way to decompose it into factors. In the next section we introduce some supplementary equations to facilitate determination of unknowns as well as the decomposition of (20). Choice of Supplementary Equations Since general polynomial equations beyond fourth-degree can- not be solved in radicals, it is to be noted that, even though there are ten equations, (7) — (16), to determine the ten unknowns in the decic equation (3), some more equations (called supplementary equations) are required to determine all the ten unknowns and to decompose the quintic equation (20). The supplementary equa- tions introduced will decide the type of solvable nonic equation. Let the supplementary equations introduced here be as indicated below: c4 = b4 (21) − c1 = b1 (22) − Using these supplementary equations in the quartic and the quintic equations, (19) and (20), leads to the following equations: 4 3 2 x +2b4x +(b3 c3)x +(b2 c2)x +2b1 = 0 (23) − − 5 3 2 x +(b3 + c3)x +(b2 + c2)x +2b0 =0. (24) 8 Alabama Journal of Mathematics We decompose the quintic equation (24) as shown below: 3 2 2b0 x (x + b3 + c3)+(b2 + c2) x2 + =0. (25) b2+c2 ³ ´ The above equation can be decomposed if the following condition is satisfied. 2b0 b3 + c3 = (26) b2+c2 With the coefficients of the quintic equation (24) satisfying the condition (26), the equation (24) can be factored as shown below: 2 3 (x + b3 + c3)(x + b2 + c2)=0. (27) The roots of the factored quintic equation (27) can be easily de- termined in radicals. The next task is to determine the unknowns (b0,b1,b2,b3,b4,c0,c1,c2,c3,c4) using the equations, (7) — (16), and the supplementary equations (21), (22), and (26). Determination of Unknowns Using equations, (7), (8), and the supplementary equation (21), the following unknowns are determined as shown below: a8 b4 = 2 a8 c4 = − 2 a7 b3 = 2 . (28) The equations, (9) — (15), get converted to the following new equa- tions, (9A) — (15A) respectively, by substituting the values of b4,c4, and b3 from (28), as well as by eliminating c0 and c1 (through the use of (17) and (22)). a7 2b2 + a8 2 + c3 = a6 (9A) ¡ ¢ 2 a7 2 2b1 + a8(b2 + c2)+ c = a5 (10A) 4 − 3 2b0 + a7b2 2c2c3 = a4 (11A) − 2 2 b +2a8b0 + b1(a7 +2c3) c = a3 (12A) 2 − 2 Spring/Fall 2008 9 b0(a7 2c3)+2b1(b2 + c2)=a2 (13A) − 2b0(b2 c2)=a1 (14A) − 4b0b1 = a0 (15A) The supplementary equation (26) also gets modified as shown be- low: 4b0 a7 +2c3 = . (26A) b2+c2 Using (15A), b0 is eliminated from the above set of equations, (11A) — (14A) and (26A), to obtain a new set of equations, (11B) — (14B) and (26B), respectively, as shown below: a0 + a7b2 2c2c3 = a4 (11B) 2b1 − 2 2 a0a8 b2 c2 + + b1(a7 +2c3)=a3 (12B) − 2b1 a0 (a7 2c3)+2b1(b2 + c2)=a2 (13B) 4b1 − 2a1 b2 c2 = b1 (14B) − a0 a0 b2 + c2 = .
Load more

Recommended publications
  • Arxiv:1406.1948V2 [Math.GM] 18 Dec 2014 Hr = ∆ Where H Ouino 1 Is (1) of Solution the 1
    Solution of Polynomial Equations with Nested Radicals Nikos Bagis Stenimahou 5 Edessa Pellas 58200, Greece [email protected] Abstract In this article we present solutions of arbitrary polynomial equations in nested periodic radicals Keywords: Nested Radicals; Quintic; Sextic; Septic; Equations; Polynomi- als; Solutions; Higher functions; 1 A general type of equations Consider the following general type of equations ax2µ + bxµ + c = xν . (1) Then (1) can be written in the form b 2 ∆2 a xµ + = xν , (2) 2a − 4a where ∆ = √b2 4ac. Hence − 2 µ b ∆ 1 ν x = − + 2 + x (3) 2a r4a a Set now √d x = x1/d, d Q 0 , then ∈ + −{ } 2 µ b ∆ 1 x = − + + xν (4) s 2a 4a2 a arXiv:1406.1948v2 [math.GM] 18 Dec 2014 r hence µ/ν 2 ν b ∆ 1 ν x = − + 2 + x (5) s 2a r4a a Theorem 1. The solution of (1) is µ v 2 µ/ν 2 2 u b v ∆ 1 b ∆ 1 µ/ν b ∆ x = u− + u + v− + v + − + + ... (6) u 2a u4a2 a u 2a u4a2 a s 2a 4a2 u u u u r u u u t t t t 1 Proof. Use relation (4) and repeat it infinite times. Corollary 1. The equation ax6 + bx5 + cx4 + dx3 + ex2 + fx + g = 0 (7) is always solvable with nested radicals. Proof. We know (see [1]) that all sextic equations, of the most general form (7) are equivalent by means of a Tchirnhausen transform y = kx4 + lx3 + mx2 + nx + s (8) to the form 6 2 y + e1y + f1y + g1 = 0 (9) But the last equation is of the form (1) with µ = 1 and ν = 6 and have solution 2 1/6 2 2 f v ∆ 1 f ∆ 1 1/6 f ∆ y = 1 + u 1 v 1 + 1 1 + 1 + ..
    [Show full text]
  • ALGEBRAIC GEOMETRY (1) Compute the Points of Intersection
    ALGEBRAIC GEOMETRY HOMEWORK 5 (1) Compute the points of intersection of the circle x2 +y2 +x−y = 0 with the lemniscate (x2 +y2)2 = x2 −y2 using resultants (can you compute the affine points of intersections directly?), and compute the multiplicities. Explain which coordinate system you choose, how the equations look like, what the resultant is, and what its factors are. (2) Here you will learn how to solve cubics using resultants. Assume you want to solve the cubic equation x3 + ax2 + bx + c = 0. (a) Show that the substitution y = x + a/3 transforms the cubic into y3 + By + C = 0. (b) Now assume that we have to solve x3 + bx + c = 0. Using linear re- lations y = rx ∗ b will not make the linear term disappear (without bringing back the quadratic term). Thus we try to put y = x2 + cx + d and then eliminate x using resultants: type in p = polresultant(x^3+b*x+c,y-x^2-d*x-e,x) polcoeff(p,2,y) polcoeff(p,1,y) (you can find out what polcoeff is doing by typing ?polcoeff). Now we know that we can keep the quadratic term out of the equation by choosing e = 2b/3. With this value of e go back and recompute the resultant; this will now be a cubic in y without quadratic term. The coefficient of the linear term can be made to disappear by solving a quadratic equation (which is called the resolvent of the original cubic). (c) Thus you can solve the cubic by solving the resolvent, using the trans- formations above to turn the cubic into a pure cubic y3 + c, which in turn is easy to solve.
    [Show full text]
  • 04. Rukhin-Final5a-116-1.Qxd
    Volume 116, Number 1, January-February 2011 Journal of Research of the National Institute of Standards and Technology [J. Res. Natl. Inst. Stand. Technol. 116, 539-556 (2011)] Maximum Likelihood and Restricted Likelihood Solutions in Multiple-Method Studies Volume 116 Number 1 January-February 2011 Andrew L. Rukhin A formulation of the problem of considered to be known an upper bound combining data from several sources is on the between-method variance is National Institute of Standards discussed in terms of random effects obtained. The relationship between and Technology, models. The unknown measurement likelihood equations and moment-type Gaithersburg, MD 20899-8980 precision is assumed not to be the same equations is also discussed. for all methods. We investigate maximum likelihood solutions in this model. By representing the likelihood equations as Key words: DerSimonian-Laird simultaneous polynomial equations, estimator; Groebner basis; [email protected] the exact form of the Groebner basis for heteroscedasticity; interlaboratory studies; their stationary points is derived when iteration scheme; Mandel-Paule algorithm; there are two methods. A parametrization meta-analysis; parametrized solutions; of these solutions which allows their polynomial equations; random effects comparison is suggested. A numerical model. method for solving likelihood equations is outlined, and an alternative to the maximum likelihood method, the restricted Accepted: August 20, 2010 maximum likelihood, is studied. In the situation when methods variances are Available online: http://www.nist.gov/jres 1. Introduction: Meta Analysis and difficult when the unknown measurement precision Interlaboratory Studies varies among methods whose summary results may not seem to conform to the same measured property.
    [Show full text]
  • International Journal of Pure and Applied Mathematics ————————————————————————– Volume 18 No
    International Journal of Pure and Applied Mathematics ————————————————————————– Volume 18 No. 2 2005, 213-222 NEW WAY FOR A TWO-PARAMETER CANONICAL FORM OF SEXTIC EQUATIONS AND ITS SOLVABLE CASES Yoshihiro Mochimaru Department of International Development Engineering Graduate School of Science and Engineering Tokyo Institute of Technology 2-12-1, O-okayama, Meguru-ku, Tokyo 152-8550, JAPAN e-mail: [email protected] Abstract: A new way for a two-parameter canonical form of sextic equations is given, through a Tschirnhausian transform by multiple steps, supplemented with some solvable cases at most in terms of hypergeometric functions. AMS Subject Classification: 11D41 Key Words: sextic equation, Tschirnhausian transform 1. Introduction 6 6 6−n Actually the general sextic equation x + anx = 0 can be solved in nX=1 terms of Kamp´ede F´eriet functions, and it can be reduced to a three-parameter Received: October 21, 2004 c 2005, Academic Publications Ltd. 214 Y. Mochimaru 6 4 2 resolvent of the form y + ay + by + cy + c = 0 by Joubert [3]. Such a type of the sextic equation was treated by Felix Klein and a part was given in [4]. Other contributions to treatment on sextic equations were given e.g. by Cole [2] and Coble [1]. In this paper, a new way leading to a two-parameter canonical form of sextic equations is given. 2. Analysis 2.1. General Let the original given sextic equation be expressed as 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ ∗ x + a x + b x + c x + d x + e x + f = 0. (1) 3 Here the quantity 5a∗ − 18a∗b∗ + 27c∗ corresponding to the polynomial of the left hand side of equation (1) is invariant with respect to the parallel shift x → x+constant.
    [Show full text]
  • Analytic Solutions to Algebraic Equations
    Analytic Solutions to Algebraic Equations Mathematics Tomas Johansson LiTH{MAT{Ex{98{13 Supervisor: Bengt Ove Turesson Examiner: Bengt Ove Turesson LinkÄoping 1998-06-12 CONTENTS 1 Contents 1 Introduction 3 2 Quadratic, cubic, and quartic equations 5 2.1 History . 5 2.2 Solving third and fourth degree equations . 6 2.3 Tschirnhaus transformations . 9 3 Solvability of algebraic equations 13 3.1 History of the quintic . 13 3.2 Galois theory . 16 3.3 Solvable groups . 20 3.4 Solvability by radicals . 22 4 Elliptic functions 25 4.1 General theory . 25 4.2 The Weierstrass } function . 29 4.3 Solving the quintic . 32 4.4 Theta functions . 35 4.5 History of elliptic functions . 37 5 Hermite's and Gordan's solutions to the quintic 39 5.1 Hermite's solution to the quintic . 39 5.2 Gordan's solutions to the quintic . 40 6 Kiepert's algorithm for the quintic 43 6.1 An algorithm for solving the quintic . 43 6.2 Commentary about Kiepert's algorithm . 49 7 Conclusion 51 7.1 Conclusion . 51 A Groups 53 B Permutation groups 54 C Rings 54 D Polynomials 55 References 57 2 CONTENTS 3 1 Introduction In this report, we study how one solves general polynomial equations using only the coef- ¯cients of the polynomial. Almost everyone who has taken some course in algebra knows that Galois proved that it is impossible to solve algebraic equations of degree ¯ve or higher, using only radicals. What is not so generally known is that it is possible to solve such equa- tions in another way, namely using techniques from the theory of elliptic functions.
    [Show full text]
  • Closed-Form Solution of Polynomial Equations
    CLOSED-FORM SOLUTION OF POLYNOMIAL EQUATIONS ALEXANDER KHEYFITS Abstract. Complex analysis is used to find closed-form expres- sions of the roots of a univariate polynomial of any degree as inte- grals of elementary functions. 1. Introduction The roots of any univariate algebraic equation of degree four or less can be expressed explicitly through its coefficients as superpositions of four arithmetic operations and radicals, while for no n 5 there exists a general formula that represents all the n complex roots≥ of every n th degree polynomial through its coefficients by means of the arithmet− ic operations and radicals. The reader can consult, e.g., [4] or [7] for the history and the details of these results. The topic is of interest even today; see, e.g., a comprehensive current monograph of Teo Mora [9] and the very recent papers by Boyd [2] and Nash [10]. The Abel-Galois result did not, of course, stop scientists from search for other means to represent the roots of the quintics and more general equations. Already in 1858, Brioschi, Hermite, and Kronecker indepen- dently expressed the roots of a general quintic through elliptic modular functions; quarter century later Klein published his famous book [8]. Thus, in solving the quintics, mathematicians have followed the stan- arXiv:1804.00698v2 [math.CV] 29 Apr 2018 dard path of extending the existing toolbox. For instance, the equation x2 = 1 cannot be solved in real numbers, but it can be done if we extend− the available numbers to the imaginary numbers; to study the ordinary differential equation x2y′′ + xy′ +(x2 ν2)y = 0, which can- not be solved in elementary functions, the special− (Bessel) functions were introduced and included in the set of possible solutions.
    [Show full text]
  • A Combined Methodology for the Approximate Estimation of the Roots of the General Sextic Polynomial Equation
    A combined methodology for the approximate estimation of the roots of the general sextic polynomial equation Giorgos P. Kouropoulos ( [email protected] ) Technological Educational Institute of Athens https://orcid.org/0000-0003-2033-0034 Method Article Keywords: sextic polynomial, general sextic, hexic polynomial, radical solutions, estimation of roots, approximate estimation, Galois theory, regression analysis, statistics Posted Date: September 8th, 2021 DOI: https://doi.org/10.21203/rs.3.rs-882192/v2 License: This work is licensed under a Creative Commons Attribution 4.0 International License. Read Full License A combined methodology for the approximate estimation of the roots of the general sextic polynomial equation Giorgos P. Kouropoulos BSc in Energy Technology, Technological Educational Institute of Athens E-mail: [email protected] ORCID id: 0000-0003-2033-0034 Abstract In this article we present the methodology, according to which it is possible to derive approximate solutions for the roots of the general sextic polynomial equation as well as some other forms of sextic polynomial equations that normally cannot be solved by radicals; the approximate roots can be expressed in terms of polynomial coefficients. This methodology is a combination of two methods. The first part of the procedure pertains to the reduction of a general sextic equation H(x) to a depressed equation G(y), followed by the determination of solutions by radicals of G(y) which does not include a quintic term, provided that the fixed term of the equation depends on its other coefficients. The second method is a continuation of the first and pertains to the numerical correlation of the roots and the fixed term of a given sextic polynomial P(x) with the radicals and the fixed term of the sextic polynomial Q(x), where the two polynomials P(x) and Q(x) have the same coefficients except for the fixed term which might be different.
    [Show full text]
  • Solution of Polynomial Equations with Nested Radicals
    Solution of Polynomial Equations with Nested Radicals Nikos Bagis Stenimahou 5 Edessa Pellas 58200, Greece [email protected] Abstract In this article we present solutions of certain polynomial equations in periodic nested radicals. Keywords: Nested Radicals; Bring Radicals; Iterations; Quintic; Sextic; Equations; Polynomials; Solutions; Higher functions; 1 Introduction In (1758) Lambert considered the trinomial equation xm + q = x (1) and solve it giving x as a series of q. Euler, write the same equation into, the more consistent and symmetrical form xa xb = (a b)νxa+b, (2) − − using the transformation x x−b and setting m = ab, q = (a b)ν in (1). Euler gave his solution as → − 1 1 xn =1+ nν + n(n + a + b)µ2 + n(n + a +2b)(n +2a + b)+ 2 6 1 + n(n + a +3b)(n +2a +2b)(n +3a + b)ν4+ 24 arXiv:1406.1948v4 [math.GM] 17 Dec 2020 1 + n(n + a +4b)(n +2a +3b)(n +3a +2b)(n +4a + b)ν5 + ... (3) 120 The equation of Lambert and Euler can formulated in the next (see [4] pg.306- 307): Theorem 1. The equation aqxp + xq = 1 (4) 1 admits root x such that ∞ n Γ( n + pk /q)( qa)k xn = { } − , n =1, 2, 3,... (5) q Γ( n + pk /q k + 1)k! =0 Xk { } − where Γ(x) is Euler’s the Gamma function. d 1/d Moreover if someone defines √x := x , where d Q+ 0 , then the solution x of (4) can given in nested radicals: ∈ −{ } q q/p x = 1 aq 1 aq q/p 1 aq q/p√1 .
    [Show full text]
  • Calculus Glossary High School Level
    Calculus Glossary High School Level English - Spanish Updated November 2016 GLOSSARY ENGLISH LANGUAGE ARTS ENGLISH ‐ SPANISH THE STATE EDUCATION DEPARTMENT / THE UNIVERSITY OF THE STATE OF NEW YORK / ALBANY, NY 12234 P‐16 Office of Elementary, Middle, Secondary and Continuing Education and Office of Higher Education Office of Bilingual Education and Foreign Language Studies http://www.emsc.nysed.gov/biling/ THE UNIVERSITY OF THE STATE OF NEW YORK Regents of the University BETTY A. ROSA, Chancellor, B.A., M.S. in Ed., M.S. in Ed., M.Ed., Ed.D. ............ Bronx T. ANDREW BROWN, Vice Chancellor, B.A., J.D. ………………......................................... Syracuse NAN EILEEN MEAD, B.A. ………….................................................................................. Manhattan JOSEPHINE VICTORIA FINN, B.A., J.D. ……………………................................................... Albany BEVERLY L. OUDERKIRK, B.S., M.S. ............................................................................. Saratoga ELIZABETH. S. HAKANSON ……………........................................................................ Onondaga CATHERINE COLLINS, B.S., M.S., Ph.D. ....................................................................... West New York JAMES R. TALLON, R., B.A., M.A. ................................................................................. Binghamton ROGER TILLES, B.A., J.D.................................................................................................. Great Neck JUDITH JOHNSON, B.A., M.A., Ed.D. ...........................................................................
    [Show full text]
  • The Maximum Modulus of a Trigonometric Trinomial
    The maximum modulus of a trigonometric trinomial Stefan Neuwirth Abstract Let Λ be a set of three integers and let CΛ be the space of 2π-periodic functions with spectrum in Λ endowed with the maximum modulus norm. We isolate the maximum modulus points x of trigonometric trinomials T ∈ CΛ and prove that x is unique unless |T | has an axis of symmetry. This enables us to compute the exposed and the extreme points of the unit ball of CΛ, to describe how the maximum modulus of T varies with respect to the arguments of its Fourier coefficients and to compute the norm of unimodular relative Fourier multipliers on CΛ. We obtain in particular the Sidon constant of Λ. 1 Introduction Let λ1, λ2 and λ3 be three pairwise distinct integers. Let r1, r2 and r3 be three positive real numbers. Given three real numbers t1, t2 and t3, let us consider the trigonometric trinomial i(t1+λ1x) i(t2+λ2x) i(t3+λ3x) T (x)= r1 e + r2 e + r3 e (1) for x R. The λ’s are the frequencies of the trigonometric trinomial T , the r’s are the moduli or ∈ it1 it2 it3 intensities and the t’s the arguments or phases of its Fourier coefficients r1 e , r2 e and r3 e . −1 −e iπ/3 Figure 1: The unit circle, the hypotrochoid H with equation z =4e −i2x +e ix, the segment from 1 to the unique point on H at maximum distance and the segments from e iπ/3 to the two points− on H at maximum distance.
    [Show full text]
  • An Application of Moore's Cross-Ratio Group To
    AN APPLICATIONOF MOORE'SCROSS-RATIO GROUP TO THE SOLUTIONOF THE SEXTICEQUATION* BY ARTHUR B. COBLE f Introduction. By making use of the fact that all the double ratios of n things can be expressed rationally in terms of a properly chosen set of n — 3 double ratios, E. H. Moore\ has developed the "cross-ratio group," C„,, a Cremona group in Sn_3 of order n\ which is isomorphic with the permutation group of n things. H. E. Slaught§ has discussed the (75! in considerable detail. Both C6! and C6, have been listed by S. Kantor. || I have already pointed out^f that Cn, defines a form-problem which I shall denote by PCn,. The solution of PCn, carries with it the solution of the n-ic equation, and I have worked out in detail the application to the quintic equation. After the adjunction of the square root of the discriminant of the n-ic, a new form-problem, PCn]l2, can be enunciated whose underlying group is the even subgroup CnU2 of Cnl. It appears in the particular case, n= 5, that the solu- tion of PC5,I2 can be expressed rationally in terms of the solution of Klein's ^.-problem.** The transition from the one problem to the other is accomplished by a very direct process which is suggested by simple geometric considerations. Explicit formulée are always desirable, and these have been lacking hitherto in the solutions of the quintic equation in terms of the ^4-problem. By intro- ducing PC5!I2 as an intermediate stage I have obtained such formulae from the long known invariant theory of the binary quintic.
    [Show full text]
  • Arxiv:2009.11873V1 [Hep-Th]
    G4 Flux, Algebraic Cycles and Complex Structure Moduli Stabilization A. P. Brauna and R. Valandrob aDepartment of Mathematical Sciences, Durham University, Lower Mountjoy, Stockton Rd, Durham DH1 3LE, UK bDipartimento di Fisica, Università di Trieste, Strada Costiera 11, I-34151 Trieste, Italy and INFN, Sezione di Trieste, Via Valerio 2, I-34127 Trieste, Italy Abstract We construct G4 fluxes that stabilize all of the 426 complex structure moduli of the sextic Calabi-Yau fourfold at the Fermat point. Studying flux stabilization usually requires solving Picard-Fuchs equations, which becomes unfeasible for models with many moduli. Here, we instead start by considering a specific point in the complex structure moduli space, and look for a flux that fixes us there. We show how to construct such fluxes by using algebraic cycles and analyze flat directions. This is discussed in detail for the sextic Calabi-Yau fourfold at the Fermat point, and we observe that there appears to be tension between M2-tadpole cancellation and the requirement of stabilizing all moduli. Finally, we apply our results to show that even though symmetric fluxes allow to automatically solve most of the F-term equations, they typically lead to flat directions. arXiv:2009.11873v1 [hep-th] 24 Sep 2020 Contents 1 Introduction 2 2 Fluxes and Moduli Stabilization 4 3 Fermat Sextic Fourfold and Algebraic Cycles 7 3.1 Algebraic cycles at the Fermat point . .... 8 3.2 Somepropertiesofalgebraiccycles . ..... 10 3.3 Algebraic cycles and their Hodge loci . ..... 11 4 Residues of Rational Forms and Complex Structure Deformations 12 4.1 Middle cohomology from residues of rational forms .
    [Show full text]