4 Polynomial Functions
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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. -
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. -
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. -
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. -
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. -
Lesson 1.2 – Linear Functions Y M a Linear Function Is a Rule for Which Each Unit 1 Change in Input Produces a Constant Change in Output
Lesson 1.2 – Linear Functions y m A linear function is a rule for which each unit 1 change in input produces a constant change in output. m 1 The constant change is called the slope and is usually m 1 denoted by m. x 0 1 2 3 4 Slope formula: If (x1, y1)and (x2 , y2 ) are any two distinct points on a line, then the slope is rise y y y m 2 1 . (An average rate of change!) run x x x 2 1 Equations for lines: Slope-intercept form: y mx b m is the slope of the line; b is the y-intercept (i.e., initial value, y(0)). Point-slope form: y y0 m(x x0 ) m is the slope of the line; (x0, y0 ) is any point on the line. Domain: All real numbers. Graph: A line with no breaks, jumps, or holes. (A graph with no breaks, jumps, or holes is said to be continuous. We will formally define continuity later in the course.) A constant function is a linear function with slope m = 0. The graph of a constant function is a horizontal line, and its equation has the form y = b. A vertical line has equation x = a, but this is not a function since it fails the vertical line test. Notes: 1. A positive slope means the line is increasing, and a negative slope means it is decreasing. 2. If we walk from left to right along a line passing through distinct points P and Q, then we will experience a constant steepness equal to the slope of the line. -
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. -
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. -
9 Power and Polynomial Functions
Arkansas Tech University MATH 2243: Business Calculus Dr. Marcel B. Finan 9 Power and Polynomial Functions A function f(x) is a power function of x if there is a constant k such that f(x) = kxn If n > 0, then we say that f(x) is proportional to the nth power of x: If n < 0 then f(x) is said to be inversely proportional to the nth power of x. We call k the constant of proportionality. Example 9.1 (a) The strength, S, of a beam is proportional to the square of its thickness, h: Write a formula for S in terms of h: (b) The gravitational force, F; between two bodies is inversely proportional to the square of the distance d between them. Write a formula for F in terms of d: Solution. 2 k (a) S = kh ; where k > 0: (b) F = d2 ; k > 0: A power function f(x) = kxn , with n a positive integer, is called a mono- mial function. A polynomial function is a sum of several monomial func- tions. Typically, a polynomial function is a function of the form n n−1 f(x) = anx + an−1x + ··· + a1x + a0; an 6= 0 where an; an−1; ··· ; a1; a0 are all real numbers, called the coefficients of f(x): The number n is a non-negative integer. It is called the degree of the polynomial. A polynomial of degree zero is just a constant function. A polynomial of degree one is a linear function, of degree two a quadratic function, etc. The number an is called the leading coefficient and a0 is called the constant term. -
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. -
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. -
Polynomials.Pdf
POLYNOMIALS James T. Smith San Francisco State University For any real coefficients a0,a1,...,an with n $ 0 and an =' 0, the function p defined by setting n p(x) = a0 + a1 x + AAA + an x for all real x is called a real polynomial of degree n. Often, we write n = degree p and call an its leading coefficient. The constant function with value zero is regarded as a real polynomial of degree –1. For each n the set of all polynomials with degree # n is closed under addition, subtraction, and multiplication by scalars. The algebra of polynomials of degree # 0 —the constant functions—is the same as that of real num- bers, and you need not distinguish between those concepts. Polynomials of degrees 1 and 2 are called linear and quadratic. Polynomial multiplication Suppose f and g are nonzero polynomials of degrees m and n: m f (x) = a0 + a1 x + AAA + am x & am =' 0, n g(x) = b0 + b1 x + AAA + bn x & bn =' 0. Their product fg is a nonzero polynomial of degree m + n: m+n f (x) g(x) = a 0 b 0 + AAA + a m bn x & a m bn =' 0. From this you can deduce the cancellation law: if f, g, and h are polynomials, f =' 0, and fg = f h, then g = h. For then you have 0 = fg – f h = f ( g – h), hence g – h = 0. Note the analogy between this wording of the cancellation law and the corresponding fact about multiplication of numbers. One of the most useful results about integer multiplication is the unique factoriza- tion theorem: for every integer f > 1 there exist primes p1 ,..., pm and integers e1 em e1 ,...,em > 0 such that f = pp1 " m ; the set consisting of all pairs <pk , ek > is unique.