Need for Nice . . . Cubic Polynomials: . . . To Make It Simpler . . . What Is Known and . . . How to Generate “Nice” Analysis of the Problem Cubic Polynomials – with Using the Fact That . . . Using the General . . . Rational Coefficients, Towards a General . . . Resulting Algorithm . . . Rational Zeros and Rational Home Page Extrema: A Fast Algorithm Title Page ◭◭ ◮◮ Laxman Bokati 1 , Olga Kosheleva 2 , and Vladik Kreinovich 1 ◭ ◮ 1 Computational Science Program 2 Department of Teacher Education Page 1 of 14 University of Texas at El Paso El Paso, TX 79968, USA Go Back lbokati@miners.utep.edu, olgak@utep.edu, vladik@utep.edu Full Screen Close Quit
Need for Nice . . . Cubic Polynomials: . . . 1. Need for Nice Calculus-Related Examples To Make It Simpler . . . • After students learn the basics of calculus, they prac- What Is Known and . . . tice them graphing functions y = f ( x ). Analysis of the Problem Using the Fact That . . . • They find the roots (zeros), i.e., values where f ( x ) = 0. Using the General . . . • They find the extreme points, i.e., values where the Towards a General . . . derivative f ′ ( x ) is equal to 0. Resulting Algorithm . . . Home Page • They find out whether f ( x ) increases or decreases be- tween extreme points – by checking the sign of f ′ ( x ). Title Page • They use this information – plus the values of f ( x ) at ◭◭ ◮◮ several points x – to graph the function. ◭ ◮ • For this practice, students need examples for which Page 2 of 14 they can compute both the zeros and the extreme points. Go Back Full Screen Close Quit
Need for Nice . . . Cubic Polynomials: . . . 2. Cubic Polynomials: the Simplest Case When To Make It Simpler . . . Such an Analysis Makes Sense What Is Known and . . . • The simplest possible functions are polynomials. Analysis of the Problem Using the Fact That . . . • For linear functions, the derivative is constant, so there Using the General . . . are no extreme point. Towards a General . . . • For quadratic functions, there is an extreme point. Resulting Algorithm . . . Home Page • However, after studying quadratic equations, students already know how to graph the corresponding function. Title Page • So, for quadratic polynomials, there is no need to use ◭◭ ◮◮ calculus. ◭ ◮ • The simplest case when calculus tools are needed is the Page 3 of 14 case of cubic polynomials. Go Back Full Screen Close Quit
Need for Nice . . . Cubic Polynomials: . . . 3. To Make It Simpler For Students, It Is Desir- To Make It Simpler . . . able to Limit Ourselves to Rational Roots What Is Known and . . . • Students are much more comfortable with rational num- Analysis of the Problem bers than with irrational ones. Using the Fact That . . . Using the General . . . • Thus, it is desirable to have examples when all the Towards a General . . . coefficients, zeros, and extreme points of a are rational. Resulting Algorithm . . . • Good news is that when we know that the roots are Home Page rational, it is (relatively) easy to find these roots. Title Page • Indeed, for each rational root x = p/q of a polynomial ◭◭ ◮◮ a n · x n + a n − 1 · x n − 1 + . . . + a 0 with integer coefficients: ◭ ◮ – the numerator p is a factor of a 0 , and Page 4 of 14 – the denominator q is a factor of a n . Go Back • How can we find polynomials for which both zeros and Full Screen extreme points are rational? Close Quit
Need for Nice . . . Cubic Polynomials: . . . 4. What Is Known and What We Do To Make It Simpler . . . • An algorithm for generating such polynomials was re- What Is Known and . . . cently proposed. Analysis of the Problem Using the Fact That . . . • This algorithm, however, is not the most efficient one. Using the General . . . • For each tuple of the corresponding parameter values, Towards a General . . . it uses exhaustive trial-and-error search. Resulting Algorithm . . . • In this presentation, we produce an efficient algorithm Home Page for producing nice polynomials. Title Page • Namely, we propose simple computational formulas: ◭◭ ◮◮ – for each tuple of the corresponding parameters, these ◭ ◮ formulas produce a “nice” cubic polynomial; Page 5 of 14 – every “nice” cubic polynomial can be thus gener- Go Back ated. Full Screen • For each tuple, our algorithm requires the same con- stant number of elementary steps. Close Quit
Need for Nice . . . Cubic Polynomials: . . . 5. Analysis of the Problem To Make It Simpler . . . • A general cubic polynomial with rational coefficients What Is Known and . . . has the form a · X 3 + b · X 2 + c · X + d. Analysis of the Problem Using the Fact That . . . • Roots and extreme points of f ( x ) do not change if we Using the General . . . simply divide all its values by the same constant a . Towards a General . . . • Thus, it is sufficient to consider polynomials with only Resulting Algorithm . . . three parameters: X 3 + p · X 2 + q · X + r, where Home Page = b = c = d def def def Title Page p a, q a, r a. ◭◭ ◮◮ • We can further simplify the problem if we replace X = X + p ◭ ◮ 3 , then we get x 3 + α · x + β, where def with x Page 6 of 14 α = q − p 2 + 2 p 3 3 and β = r − p · q Go Back 27 . 3 Full Screen Close Quit
Need for Nice . . . Cubic Polynomials: . . . 6. Analysis of the Problem (cont-d) To Make It Simpler . . . • Let r 1 , r 2 , and r 3 denote rational roots of x 3 + α · x + β, What Is Known and . . . then, we have Analysis of the Problem Using the Fact That . . . x 3 + α · x + β = ( x − r 1 ) · ( x − r 2 ) · ( x − r 3 ) . Using the General . . . Towards a General . . . • So, r 1 + r 2 + r 3 = 0, α = r 1 · r 2 + r 2 · r 3 + r 1 · r 3 , and β = − r 1 · r 2 · r 3 . Resulting Algorithm . . . Home Page • Substituting r 3 = − ( r 1 + r 2 ) into these formulas, we Title Page get ◭◭ ◮◮ α = − ( r 2 1 + r 1 · r 2 + r 2 2 ) and β = r 1 · r 2 · ( r 1 + r 2 ) . ◭ ◮ Page 7 of 14 Go Back Full Screen Close Quit
Need for Nice . . . Cubic Polynomials: . . . 7. Using the Fact That the Extreme Points x 0 To Make It Simpler . . . Should Also Be Rational What Is Known and . . . • Differentiating and equating the derivative to 0, we get Analysis of the Problem Using the Fact That . . . 3 x 2 0 − ( r 2 1 + r 1 · r 2 + r 2 2 ) = 0 . Using the General . . . 0 − 3 y 2 − z 2 = 0, where • This is equivalent to 3 x 2 Towards a General . . . Resulting Algorithm . . . = r 1 + r 2 = r 1 − r 2 def def y and z . Home Page 2 2 Title Page • If we divide both sides of this equation by y 2 , we get = x 0 = z ◭◭ ◮◮ 0 − 3 − Z 2 = 0 , where X 0 def def 3 X 2 y and Z y . ◭ ◮ • One of the solution of above equation is easy to find: Page 8 of 14 namely, when X 0 = − 1, we get Z 2 = 0 and Z = 0. Go Back • This means that for every y , x 0 = − y , y and z = 0 Full Screen solve the above equation. Close Quit
Need for Nice . . . Cubic Polynomials: . . . 8. Using the Fact That the Extreme Points x 0 To Make It Simpler . . . Should Also Be Rational (cont-d) What Is Known and . . . • We can now reconstruct r 1 and r 2 from y and z as Analysis of the Problem r 1 = y + z and r 2 = y − z , Using the Fact That . . . • In our case, r 1 = r 2 = y , so α = − 3 y 2 and β = 2 y 3 . Using the General . . . Towards a General . . . • We can then: Resulting Algorithm . . . Home Page – shift by a rational number s , ( x → X = x + s ), and – multiply all the coefficients by an arbitrary rational Title Page number a . ◭◭ ◮◮ • Then, we get ◭ ◮ d = a · ( s 3 + 2 y 3 ) . c = a · (3 s 2 − 3 y 2 ) , b = 3 a · s, Page 9 of 14 Go Back Full Screen Close Quit
Need for Nice . . . Cubic Polynomials: . . . 9. Using the General Algorithm for Finding All To Make It Simpler . . . Rational Solutions to a Quadratic Equation What Is Known and . . . • We have already found a solution of the equation 3 X 2 0 − Analysis of the Problem 3 − Z 2 = 0 , corresponding to X 0 = − 1: then Z = 0. Using the Fact That . . . Using the General . . . • Every other solution ( X 0 , Z ) can be connected to this simple solution ( − 1 , 0) by a straight line. Towards a General . . . Resulting Algorithm . . . • A general equation of a straight line passing through Home Page the point ( − 1 , 0) is Z = t · ( X 0 + 1) . Title Page Z • When X 0 and Z are rational, t = X 0 + 1 is rational. ◭◭ ◮◮ • Substituting this expression for Z into the equation, ◭ ◮ 0 − 3 − t 2 · ( X 0 + 1) 2 = 0 . we get 3 X 2 Page 10 of 14 • Since X 0 � = − 1, we can divide both sides by X 0 + 1. Go Back then 3 · ( X 0 − 1) − t 2 · ( X 0 + 1) = 0 , hence Full Screen X 0 = 3 + t 2 6 t 3 − t 2 and Z = 3 − t 2 . Close Quit
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