The Value of Computational Thinking Across Grade Levels. Computational Thinking Module

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1 VC T A L VCTAL The Value of Computational Thinking Across Grade Levels Computational Thinking Module Polynomiography: Visual Displays of Solutions to Polynomial Equations STUDENT EDITION DIMACS

2 VC T A L VCTAL The Value of Computational Thinking Across Grade Levels Computational Thinking Module Polynomiography: Visual Displays of Solutions to Polynomial Equations STUDENT EDITION By Carl Anderberg Jonathan Choate Bahman Kalantari

3 Funded by the National Science Foundation, Award No. DRL This material was prepared with the support of the National Science Foundation. However, any opinions, findings, conclusions, and/or recommendations herein are those of the authors and do not necessarily reflect the views of the NSF. At the time of publishing, all included URLs were checked and active. We make every effort to make sure all links stay active, but we cannot make any guaranties that they will remain so. If you find a URL that is inactive, please inform us at info@comap.com. DIMACS Published by COMAP, Inc. in conjunction with DIMACS, Rutgers University. 016 COMAP, Inc. Printed in the U.S.A. COMAP, Inc. 175 Middlesex Turnpike, Suite 3B Bedford, MA ISBN: Front Cover Photograph Credit: ID Ilja Mašík Dreamstime.com

4 Polynomiography: Visual Displays of Solutions to Polynomial Equations SECTION 1 Iterations as a Form of Computational Thinking 1.1 Iterating Functions, Fixed Points, Seeds, Orbits, and Basins of Attraction With the advent of high powered digital computers, iteration has become a very useful mathematical technique for solving a variety of problems in particular, solving equations. In this module, we will learn how iteration can be used to approximate solutions to polynomial equations e.g., quadratic equations and consider their visualization through polynomiography, a software that visualizes polynomial equations. We can also use iteration to solve linear equations, even though we are all familiar with how to solve a linear equation of the form ax + b = 0. Let s assume we didn t know how to solve a linear equation and instead try to see if we can repeatedly compute values that approach the solution, x = b / a. If we want to check whether a particular value α is the solution, we simply plug it into the formula and check if aα + b = 0. Likewise, we can check if a value α is an approximate solution, i.e. if a α + b is nearly zero. Here is how iteration works: Take a function f (x) and an initial value for x 0. Then evaluate the function for that specific value. You can think of the function as a machine that takes an input x 0 and outputs f ( x 0 ). Next, you input the value f ( x 0 ) into this machine and wait for it to compute the new output. We can repeat this process over and over again; see Figure 1.1. x f f(x) Figure 1.1. Input becomes output, next output becomes input. Polynomiography 1

5 For example, suppose f ( x) = x 1 and we start with a seed, or initial, value of x =. You can now create a sequence whose first term is and the next term is f ( ) = 1 = 3. The third term is found by calculating f ( 3) = 3 1= 5. Continue to do this to create the sequence, 3, 5, 9, 17, 33. We now write this sequence as shown in Example 1.1. Example 1.1 Iterations of f (x) = x 1. Complete the following table: k x k Notice that we refer to the seed as the 0 th term and 3 as the 1 st term or first iterate. For a function f (x) and an initial number x 0, called seed, the sequence of numbers x 0, x 1 = f (x 0 ), x = f (x 1 ), x 3 = f (x ),...,is called the orbit of x 0. In each of the exercises below for the given function, compute the orbit, starting with the given seed under column 0. You may find it helpful to use a calculator or a spreadsheet. Exercise 1.1 f (x) = x + 5 k x 3 k Exercise 1. f (x) = 1 x + 4 k x 0 k Exercise 1.3 f (x) = 1 4 x 10 k x k 100 What is of interest to mathematicians is this: given a function and an initial value, or seed, what is the fate of the orbit? In Example 1.1, the sequence of values keeps getting Polynomiography

6 larger and larger and will grow without bound. The same is true for Exercise 1.1. Something different is happening in Exercise 1., where the orbit appears to be approaching 8. Something similar appears to be happening in Exercise 1.3, where the orbit appears to be approaching 40/3. When analyzing what happens when a function is iterated, it is useful to calculate what is known as the fixed point of the function. A number α is said to be a fixed point of f (x) if f (α ) = α 1 Referring back to Exercise 1., what is the fixed point of f ( x) = x + 4? Solving f ( α ) = α, you get 1 α = α α = 4 α = 8 A fixed point α is said to be an attracting fixed point if orbits that are generated for a seed near α converge to it (i.e., get closer and closer to α). The basin of attraction of a fixed point α, denoted by B(α), is the set of all seeds whose orbit gets attracted to α. Since the fixed point is 8, and 8 is the value that the orbit of 0 is approaching, we say that 8 is an attracting fixed point. In Exercise 1., f (x) = 1 x + 4, the basin of attraction for the fixed point 8 is the set of all real numbers. In other words, for any seed you use to iterate f (x) = 1 x + 4, f(x) will attract to the fixed point of 8. Going back to Example 1.1, find the fixed points of the function f ( x) = x 1. Solving f ( α ) = α, we get α = α 1 α =1 A fixed point α is said to be a repelling fixed point if orbits that are generated for a seed near α will not converge to it, no matter how close the seed may be taken (unless the seed is the fixed point itself). Polynomiography 3

7 For example, if the fixed point is 1 and the orbit of is not converging to 1, we say that 1 is a repelling fixed point. Exercise 1.4 Find the fixed point and the associated basin of attraction for the following functions. a. f (x) = 1 x + 10 x + 1 Let us consider analyzing the orbits created by iterating f ( x) =. x 1 α = α + 1 To find the fixed points, we solve α 1 for α. α (α 1) = α + 1 α α = α + 1 α α 1 = 0 (α 4)(α + 3) = 0 The fixed points are α = 4 and α = 3. Exercise 1.5 Given the function f (x) = x +1 x 1 a. Show that if you iterate f (x) = x +1 x 1, use your calculator or a spreadsheet to: b. Find the basins of attraction for B(4) and B( 3), α = 4 and α = 3 are fixed points. c. Things get strange for values of x close to 0.5. What happens to the orbits as you take values closer and closer to x = 0.5? 1. The Babylonian Iteration to Compute Square Roots The Babylonian way to approximate square roots was discovered thousands of years ago. It is an iterative method for approximating square roots. Here is an example of how to use that method. To find a good approximation for the square root of 10, use the following algorithm: Step 1. Make an initial guess, say 3. Step. Using 3 as an initial value, create the orbit of 3 using the function f (x) = 1 x + 10 x. Polynomiography 4

8 This function takes the average of x, the guess, and 10/x. If x < , then 10/x has to be greater than Therefore, the average will be between two values, one less than and one greater than and hence will be closer to the actual value of The first column in Table 1.1 shows the orbit of 3. Notice that the orbit very quickly converges to Also, included in Table 1.1 are the orbits for 1, 5, and 100. Notice that if the initial guess is negative, the orbit converges to the negative square root Table 1.1. The orbits of 3, 1, 5, and 100 The Babylonian Algorithm to Approximate Pick a seed and iterate the function f (x) = 1 x + a x This gives an approximation if exact value. a a is an irrational number; otherwise, it can give one an Exercise 1.6 Use the Babylonian Algorithm to find five figure approximations for the square roots of the following numbers. a. 7 b. 13 c.,435 d e. 4 Exercise 1.7 What happened in Exercise1.6e? Try the Babylonian Method for different negative numbers. What do you observe? Exercise 1.8 For each of the following functions, find the fixed points and then the basin of attraction for each fixed point. Polynomiography 5

9 a. f (x) = 1 x + 1 x b. f (x) = 3 x + 1 x c. f (x) = 3 4 x + 1 3x 3 d. f (x) = 4 5 x + 1 4x 4 Do you see a pattern in the preceding functions? Exercise 1.9 The equation in Exercise 1.8a finds 1. It is also finds the solutions to the polynomial equation x 1 = 0. What polynomial equations do the equations in Exercise 1.8b, c, and d solve? Exercise 1.10 Use the Babylonian Algorithm spreadsheet to find the square root of 40 to the nearest and answer the following: Find seeds that converge to the root after a. iterations b. 3 iterations c. 5 iterations Example 1. The plot in Figure 1. for the interval {0, 0} shows that seeds between 1 and 3 converge after 1 iteration and are colored blue between 3 and 6 converge after iterations and are colored red between 6 and 11 converge after 3 iterations and are colored green between 11 and 0 converge after 4 iterations and are colored orange Figure 1.. Basin of Attraction Color Plot Class Activity Different seeds within a given basin of attraction take different numbers of iterations to converge to the root. Use the Babylonian Algorithm spreadsheet to find Polynomiography 6

10 the positive square root of 5. Use your results to create a number line which covers the interval {5, 0} and then color the number line in the way shown in Example 1.. Use the table below and a centimeter ruler to help you create the colored number line. Number of Iterations Low High Polynomiography 7

11 SECTION Using Iteration to Solve Quadratic Equations.1 Roots of a Quadratic Polynomial Consider a quadratic polynomial p(x) = ax + bx + c, a 0. We say a number r is a zero or root of p ( x) = ax + bx + c if p ( r) = ar + br + c = 0. Note that testing to see if a number is a root is easy because we can plug it into the equation and see that it equals zero. The reverse process is not easy; that is, finding a root or, more precisely, approximating a root e.g., approximation of the square root of two. By solving a quadratic equation iteratively, we mean having a process that can compute a root as precisely as we wish. In order to approximate the roots of a quadratic polynomial through iterations, it is necessary to find fixed points.. Newton s Method for Solving Quadratic Equations In the 1700s, Sir Isaac Newton devised a way to find solutions to equations of the form f ( x) = 0. Newton s Method uses iteration to find solutions to equations such as quadratic equations and more general polynomial equations. If f(x) = ax + bx + c, the derivative of f (x) is f '( x) = ax + b. The derivative gives the instantaneous rate of change of the function. If you take a calculus course, you will learn how the derivative is defined and calculated. Newton s Method for solving equations of the form f (x) = 0 computes orbits of the function x = b a Thus, the Newton Function for f(x) is N(x): N ( x) = x ax + bx + c = ax + bx ax bx c = ax c ax + b ax + b ax + b How Newton s Method Works: f (x) is the slope of the tangent line at (x, f(x)). If the initial guess, or seed, equals x 0, then f (x 0 ) gives the slope of the tangent line at (x 0, f(x 0 )). The Newton function for f(x), N(x), is equal to Polynomiography 8

12 N(x) = x f (x) f '(x) Iterating N(x) yields the solutions to f(x) = 0. This is the procedure: 1. Pick a seed x 0 and calculate f(x 0 ).. The tangent line has equation y f(x 0 ) = f (x 0 )(x x 0 ). 3. The x-intercept of the tangent line, x 1, is equal to x 1 = x 0 f (x ) 0 f '(x 0 ) 4. Repeat to find x. 5. Use N(x n ) = x n + 1 until you get the desired accuracy. Figure.1. Newton s Method Diagram In order to solve x + 6x + 8 = 0.a = 1, b = 6, c = 8. In this case,, we can form the Newton Function using N(x) = x x + 6x + 8 x + 6 = x 8 x + 6 The first column in Table.1 shows the orbit of 3. Notice that the orbit very quickly converges to. Also included in Table.1 are the orbits for, 5, and 10. Notice that some orbits tend to one of the solutions and some to the other solution. Polynomiography 9

13 Table.1. Iterative solutions to x + 6x + 8 = 0 Remark. If we take the polynomial f ( x) = x, then Newton s Method gives a formula for approximation of N( x) = x + x = 1 x x + x = 1 x + x. and it gives: Surprisingly, this coincides with the Babylonian formula! Exercises 1. In the above example, what is the basin of attraction for the solution x =, x = 4?. Use Newton s Method and a spreadsheet to find the solutions and the corresponding basins of attraction for the following quadratic equations. You can use the Newton Quadratic Analysis spreadsheet to find the solutions and the corresponding basins of attraction. a. 3x + 10x + 6 = 0 b. x 6x + 1 = 0 c. x 5x + 6 = 0 d. x x + 6 = 0 Class Activity Use the Newton Quadratic Analysis spreadsheet to find the basin of attraction for the root x = 5 of the quadratic x x 15 = 0. Use your results to create a number line which covers the interval {5, 0} and then color the number line in the way shown in Example 1., showing the basin of attraction for the solution x = 5 of the quadratic equation. To do this, students will need to find intervals for which seeds in that interval converge after 1 iteration, after iterations, after 3 iterations, and after 4 iterations. They need to find the boundaries accurate to the nearest 100 th. Split the class into 4 groups and instruct each to find an interval. Polynomiography 10

14 .3 Solving Quadratic Equations and Complex Numbers Our goal here is to solve quadratic equations using iterative methods and then enjoy them through visualization with the polynomiography software. Before that, however, we need to develop some understanding of complex numbers and why we need to solve equations iteratively. First, we consider the solutions of a quadratic equation through the quadratic formula. In middle and high schools, we learn how to solve a linear equation of the form ax + b = 0, where a is nonzero. The solution is x = b a In an algebra course, we learn several different methods for solving quadratic equations ax + bx + c = 0 where a is nonzero. The numbers a, b, c are real numbers, called coefficients. Linear and quadratic equations are examples of polynomial equations. We can solve linear equations easily, and some quadratic equations can be solved by factoring. For example, you could solve x + 6x + 8 = 0 by noticing that the sum of the roots is 6 and their product is 8 so that x + 6x + 8 = ( x + )( x + 4) and then solving ( x + )( x + 4) = 0, which has solutions x = and x = 4. You can also solve quadratics by using the quadratic formula, which states that the solutions to the quadratic equation ax + bx + c = 0 are b + x = b 4ac a or b x = b 4ac a Solving x + 6x + 8 = 0 by the quadratic formula, you get and 6 + x = 6 x = = 6 = 6 + = = 4 6 = = 4 Polynomiography 11

15 The discriminant of a quadratic polynomial is Δ = b 4ac This can say a great deal about the solution of a quadratic equation. If Δ > 0, there are two real roots. If Δ = 0, there is only one real solution and it is said to be a repeated root or a double root. b x = a If Δ < 0, there are no real solutions. It turns out that there are two complex solutions. In Section 3, we will consider this case again after we have formally defined complex numbers. The quadratic formula gives exact answers in terms of a, b, c. If b 4ac is a perfect square, the roots are rational. If the discriminant is not a perfect square, the roots are irrational and you have to use technology to find approximate values for the square root of b 4ac. Actually, the calculator uses an iterative method, which had its roots in work done by the Babylonians..4 Complex Numbers We can think of complex numbers as a mechanism to turn a location (a, b) in the Euclidean plane into a number. This is done by identifying the point with the complex number a + ib where i = 1 so that i = 1. With this rule, we can add two complex numbers, subtract them, multiply them, and divide them. Examples follow. Example.1 Consider two complex numbers 1+ 3i and i. When multiplying or dividing, we have to keep in mind that i = 1. Four elementary operations can be defined. Examples: ( 1+ 3 i) + ( i) = (1 + ) + (3 1) i = 3 + i ( 1+ 3 i) ( i) = (1 ) + (3 + 1) i = 1+ 4i (1 + 3 i) ( i) = i + 6i 3i = + 5i + 3 = 5 + 5i Polynomiography 1

16 Next consider division of the two numbers. 1+ 3i i Exercises 1+ 3i + i = i + i 1+ 3i = ( )( + i) i ( )( + i) 1+ 7i = 5 = i 1. Let u = 3i, v = 3 + 4i and w = 5 i. Calculate the following: a. u + v b. uv c. u + 5w u d. w u e. v + v w. The conjugate of a + bi is equal to a bi. Calculate the following: a. (1 + i)(1 i) and (1 + i) + (1 i) b. (3 4i)(3 + 4i) and (3 4i) + (3 + 4i) c. i) What happens when you multiply a + bi and its conjugate a bi? ii) What happens when you add a + bi and its conjugate a bi? 3. Find two numbers a and b such that a. a + b = 6 and a 5 b = 5 b. a + b = and a 5 b = 6 4. Food for thought: Given any two real numbers n and m, can you always find real numbers a and b such that a + b = m and ab = n? Polynomiography 13

17 Plotting Complex Numbers Complex numbers, a + bi, can be plotted in an Argand diagram. The Argand diagram can also be referred to as the complex plane. In the complex plane, the traditional x-axis represents the real axis and the traditional y-axis represents the imaginary axis. See Figure. for an example. When you plot the complex number 4 + i in the complex plane, as shown in Figure., you go over 4 units on the Real axis and up units on the Imaginary axis. Figure.. Plot of 4 + i Historical Note: While Argand (1806) is generally credited with the discovery, the Argand diagram (also known as the Argand plane) was actually described by C. Wessel prior to Argand [1]. Historically, the geometric representation of a complex number as a point in the plane was important because it made the whole idea of a complex number more acceptable. In particular, this visualization helped imaginary and complex numbers become accepted in mainstream mathematics as a natural extension to negative numbers along the real line. Polynomiography 14

18 Exercises Graph the following complex numbers on the set of axes below i i 3. 1 i i Polynomiography 15

19 SECTION 3 Introduction to Polynomiography 3.1 What is Polynomiography? The importance of complex numbers is that everything that was said earlier about quadratics, roots, fixed points, fixed-point iterations, Newton s Method, and convergence extends to the domain of the complex numbers, and we suddenly get a much more beautiful and colorful world with many visual possibilities. The word polynomiography is a simple combination of polynomial and the suffix graphy. Dating back to the Babylonians and the Sumerians, solving a polynomial equation has resulted in numerous discoveries. The greatest mathematicians of each generation have investigated the problem for one mathematical reason or another. Attempts to solve polynomial equations inspired the discovery and study of the queen of computational algorithms, Newton s Method, which in turn is the foundation behind iterative methods and their visualization leading to fractal behavior, a term coined by Mandelbrot. In fact, according to historical experts, the very ideas of abstract thinking and the use of mathematical notation are largely a result of the study of polynomial equations []. Solving for the unknown is a problem faced by people every day for instance, in figuring out percentages, which essentially amounts to the solution of a linear equation. Computer technology and sophisticated mathematical algorithms have now resulted in a new appreciation for solving polynomial equations, going far beyond routine fractal images. We will show how polynomiography can produce spectacular and diverse images, called polynomiographs, that not only can be appreciated as art but also can induce striking appreciation of the connections between creativity in art and the intrinsic beauty of mathematics. Sample images can be found at A polynomial can be entered into the software either through its coefficients, which is normally the way we think of a polynomial, or through its roots. For instance, if we want a polynomial to have two roots, α and β, we consider p ( x) = ( x α) ( x β ) The polynomiography software allows us to select the roots with the click of the computer mouse on the screen by selecting locations as the roots of the polynomial equation. We can select these locations as arbitrary points or we can select them to form a desired pattern. In the latter case, the software subsequently turns the points into a polynomial equation, then an initial polynomiograph, a graphic showing the basins of attraction for the roots of the given polynomial, is rendered. In this sense, polynomiography turns the problem of root-finding upside down and into a problem in which the user can select the polynomial that will then undergo visualization through polynomiography. Polynomiography 16

20 Furthermore, the role of algorithms becomes more pronounced as it is one of the main sources of unveiling or creating beauty from such a simple-looking task as solving a polynomial equation. While a dictionary definition of a polynomial equation describes it as a linear combination of an internal power of a variable such as x, a modern view of solving a polynomial equation, as well as an informal description of polynomiography that brings it to the non-mathematician is the following from the book Polynomial Root- Finding and Polynomiography: Solving a polynomial equation could be considered as a game of hide-and-seek with a bunch of tiny dots on a painting canvas. We hide the dots behind a polynomial equation, we then seek them using a formula or an algorithm. Polynomiography is the algorithmic visualization of the process of searching for the dots, and painting the canvas along the way. The polynomiograph is produced through iteration functions which can serve as algorithms for approximation of polynomial roots, using a computer. Polynomiography is somewhat analogous to photography in which a photographer uses a camera and its lenses, together with his or her own creativity to produce an initial image, a photograph. A photographer may subsequently alter an initial photograph in order to create or enhance desired effects. Many parameters come into play to create the final image: the subject, the composition, the angle, the lighting, the settings, the lens, the filters, among others. In polynomiography, the polynomiographer uses the corresponding software and iteration functions which act as lenses, or paintbrushes, together with his or her creativity to produce an initial polynomiograph. The polynomiographer then may go through the same kind of decision making as the photographer: changing scale, isolating parts of the image, enlarging or reducing, adjusting values and color until the polynomiograph is resolved into a visually satisfying entity. Like a photographer, a polynomiographer can learn to create images that are aesthetically beautiful and individual, with or without the knowledge of mathematics or art. Like an artist and a painter, a polynomiographer can be creative in coloration and composition of images. Like a camera, or a paintbrush, polynomiography software can be made simple enough that even a child could learn to operate it. 3. How to Use the Polynomiography Software Step 1. Open the software and you get a screen like the one shown below. Polynomiography 17

21 Step. Enter into the polynomial window the polynomial you want to evaluate. Polynomiography uses z as a variable as well as x. So to solve x 3x 10 = 0, enter x 3x 10 into the polynomial window. Step 3. Press the green play key and you get a plot of the solutions of x 3x 10 = 0 in the complex plane. Step 4. Go to VIEW/ axis / standard and a set of axes will be added to the plot. Polynomiography 18

22 Step 5. Go to VIEW / roots and the two real solutions will be highlighted. Step 6. To change the colors used, go to COLORING. In the plot below COLORING /color pallet, rainbow has been selected. Polynomiography 19

23 If COLORING / Color Palette II- / Blue-Red, you will get 3.3 How Polynomiography Plots Basins of Attraction When we used Newton s Method and Excel to calculate the square root of 16, we calculated it with an accuracy of 4 decimal places using the function as shown in Table 3.1. N f (x) = x x 16 = x + 16 x x Notice that the number of iterations it took would be different for different initial guesses. Table 3.1. Different results using Newton s Method to find the square root of 16 Table 3.1 shows the following: It takes 4 iterations to get an approximation accurate to 4 places with an initial guess of 10. It takes 3 iterations to get an approximation accurate to 5 places with an initial guess of 6. It takes 4 iterations to get an approximation accurate to 5 places with an initial guess of. Polynomiography 0

24 It takes 3 iterations to get an approximation accurate to 5 places with an initial guess of 6. You can argue that the basin of attraction for using Newton s Method is the set of all nonzero real numbers. An interesting question is What elements in the basin of attraction will converge after a given number of iterations? For example, guesses of 8 and 6 both converge after 4 iterations. Polynomiography will help answer this question. Figure 3.1(a) shows the polynomiograph for x 16. Any guess in the pink region will produce an orbit that converges after 6 iterations. Figure 3.1(b) shows the polynomiograph produced for x Any guess in the blue region will produce an orbit that converges after 4 iterations. Figure 3.1 (a) x 16 Figure 3.1 (b) x + 16 Figure 3.. The Cartesian graph shows that the graph of f(x)= x + 16 does not have x-intercepts so f(x) must have imaginary roots. The polynomiograph shows that the imaginary roots are 4i and 4i. Polynomiography 1

25 Figure 3.3. The Cartesian graph shows that the graph of f(x) = x 16 has x-intercepts at ( 4, 0) and (4, 0). The polynomiograph shows that the real roots are 4 and 4. Exercises Use polynomiography to answer the following questions. 1. Find the solutions and the corresponding basins of attraction for the following quadratic equations. a. x 3x 15 = 0 b. x 8x +16 = 0 c. x 3x + 15 = 0. A given quadratic equation can have two real solutions, one repeated real solution, or two complex solutions. Make up examples of each type, plot their solutions using polynomiography, and write a summary of how the plots differ. Polynomiography

26 3. Match each Cartesian plot with the corresponding polynomiograph. Cartesian Plot 1 Polynomiograph 1 Cartesian Plot Polynomiograph Cartesian Plot 3 Polynomiograph 3 Cartesian Plot 4 Polynomiograph 4 Polynomiography 3

27 3. Use polynomiography to find the roots of x 4 4x 3 14x + 85x 116 = 0. Polynomiography 4

28 SECTION 4 Cayley s Discovery, the Fundamental Theorem of Algebra, and Creating Your Own Polynomials This section focuses on seeing how the images you get when solving a cubic and higher degree polynomials are different from those you get when solving a cubic. Using the Newton Cubic Analysis spreadsheet to analyze the solutions to z 3 z = 0, students can discover the fractal nature of the basin of attractions for the three roots. To learn more about the solutions to z 3 z = 0, consult the UMAP module Newton s Fractals included with the resources for this module. After exposure to the fractal nature of the basins of attraction for cubics, the module moves on to the version of the Fundamental Theorem of Algebra for polynomials with real coefficients, which states that any n th -degree polynomial with real coefficients has n roots with imaginary roots occurring in conjugate pairs. 4.1 Basins of Attraction for Solutions of Cubics As the previous examples have shown, if a quadratic has real solutions a and b, then when Newton s Method is used to solve the corresponding quadratic x (a + b)x + ab = 0, the basin of attraction for x = a is the set {x x < (a + b)/} and the basin of attraction for x = b is the set {x x > (a + b)/). In the mid 1800s, Arthur Cayley discovered that the basins of attraction for the solutions of cubics were far more complicated than those for the solutions of quadratics. Here s what he found. Consider solving x 3 x = 0, which has solutions x = 1, 0, and 1. Note that when you take calculus you will learn how to compute derivatives, but for now: The derivative f (x) of f(x) = ax 3 + bx + cx + d is f (x) = 3ax + bx + c. The Newton function for the general cubic is N(x) = x ax3 + bx + cx + d 3ax + bx + c In this case, f (x) = 3x 1, so the Newton Function for f(x) = x 3 x is N(x) = x x3 x 3x 1 Something interesting happens when you examine initial guesses on the interval ( 1.5, 1.5). The Newton Cubic Analysis spreadsheet allows you to do this. You can enter an interval (min, max) and the spreadsheet will divide the interval into 10 initial values x = min, min + d, min + d, min + 3d,... min + 9d, b where d = (max min)/10 and then calculate 10 iterations to see which root the guess converges to. Letting min = 1.5 and Polynomiography 5

29 max = 1.5, one gets the result shown in Table 4.1. It looks like if an initial guess between 1.5 and.6 is used, N(x) converges to 1; between.3 and.3, N(x) converges to 0; and between.6 and 1.5, N(x) converges to 1. Table 4.1. Iterates of N(x) for seeds in the interval ( 1.5, 1.5) Cayley discovered that if you examined the iterates of N(x) for seeds in the interval (.447,.65), see Table 4., you get a surprising result. In that interval, there are seeds that will converge to each of the three roots. Table 4.. Iterates of N(x) for seeds in the interval (.447,.65) In fact, if you examined an even smaller interval, such as (.4565,.4435), you again find seeds that converge to all three roots, as shown in Table 4.3. Table 4.3. Iterates of N(x) for seeds in the interval (.4565,.4435) Caley also discovered that there were basins of attraction inside basins of attraction inside... It took years and the work of Fatou and Julia in the early 1900s to understand quantitatively what was going on, But it wasn t until 1980 that people first saw the pictures of what was going on. Polynomiography creates the pictures that Cayley, Fatou, and Julia never saw. When one uses Newton s Method in the complex plane, the images Polynomiography 6

30 are often fractal since there will be basins of attraction inside basins of attraction inside basins of attraction... One can see the fractal nature of the basins of attraction for solutions of cubics using polynomiography. Figure 4.1 shows the image polynomiography creates if one enters the polynomial z 3 z. Figure 4.1. Polynomiograph of z 3 z. Exercise 4.1 Graph x 3 x in polynomiography. Zoom in on one of the blue bulbs and then zoom in on the red bulb at the tip of the blue bulb. Next, zoom in on the blue bulb on the tip of the red bulb. Describe what you are seeing. 4. The Fundamental Theorem of Algebra We saw from the above discussion that the quadratic formula is not always helpful in giving us a decimal approximation to the solution of a quadratic equation. However, the existence of the quadratic formula inspired scholars to solve cubic equations. A cubic equation is p( x) = ax 3 + bx + cx + d, a 0 The fundamental theorem of algebra tells us that any quadratic or cubic polynomial will always have a root, possibly a complex root, no matter what we select as coefficients. We have already seen this in the case of quadratic and cubic polynomials with real coefficients. But the fundamental theorem says a polynomial will always have a root no matter what the degree of the polynomial is. Polynomiography 7

31 There is a formula for the solutions of cubic equations as well. However, the formula is complicated and not easy to keep in one s head! There is also a closed formula in the case of quartic equations, equations of the form p( x) = ax 4 + bx 3 + cx + dx + e, a 0 However, in the case of quintic equations, degree 5 or higher, one needs to use other methods, including iterative ones. This says that for some polynomials, the only way to approximate their solutions is to use iterative methods. From the above, we also see that if we increase the degree of the polynomial from 4 to 5 and higher it becomes cumbersome to use the letter of the alphabet; there is a more convenient way to denote a polynomial equation. Formally, a polynomial of degree n > 0 is an expression of the form p( x) = a n x n + a n 1 x n a 1 x + a 0, a n 0 Solving a polynomial equation is to find one or more roots i.e., solutions to p(x) = 0 When creating polynomiographs of polynomials in this module, we will only consider the cases in which the coefficients are real. Of course, a real number is always a complex number because the complex numbers comprise a superset of the reals. However, a complex number that has an imaginary part is not a real number. So for quadratics we have considered those with real coefficients. We saw that there are three types: a double root, two different real roots, or a pair of conjugate roots. For a cubic polynomial with real coefficients, we can have several cases such as: 1. A single real triple root. A double real root and another real root 3. A real root and a pair of complex conjugates 4. Three distinct real roots. Polynomiography 8

32 Figure 4. (a) A Single Real Triple Root Figure 4. (b) A Double Real Root and Another Real Root Figure 4. (c) A Real Root and a Pair of Complex Roots Figure 4. (d) Three Distinct Real Roots For a quartic (degree 4) polynomial with real coefficients, we can have even more cases, including the following: 1. A single real quadruple root. Four distinct real roots. 3. A triple real root and a distinct second root 4. Two real double roots 5. A double real root and a pair of conjugate roots 6. Two distinct pairs of conjugate pairs Polynomiography 9

33 Figure 4.3. Polynomiographys for these cases, a c (top row) and d f (bottom row). 4.3 Creating Designer Polynomials If a quadratic has roots x = a and x = b, then the polynomial with these roots can be created by expanding (x a)(x b). Therefore, the polynomial p(x) = x (a + b)x + ab has roots a and b. For example, p(x) = x 5x + 6 has roots x = 3 and x =. If a quadratic polynomial has no real roots, then it has complex roots of the form x = a + bi and x = a bi; then the polynomial with these roots can be created by expanding (x (a + bi))(x (a bi)). Therefore, the polynomial p(x) = x [(a + bi) + (a bi)] x + (a + bi)(a bi) = x ax + (a + b ) has roots a + bi and a bi. For example, p(x) = x 8x + 5 has roots x = 4 3i and x = 4 + 3i. Since the Fundamental Theorem of Algebra states that any n th degree polynomial with real coefficients has n roots and that the complex roots come in conjugate pairs, the preceding can be used to build polynomials with the desired number of real and complex roots. For example, to create a polynomial with four complex roots and two distinct real roots, do the following: 1. Select the two real roots, say x = 3 and x = 4.. Select the two conjugate pairs x = +/ 3i and x = 3 +/ 5i. 3. Create the quadratic polynomials that have the conjugate pairs as roots (x ( + 3i))(x ( 3i)) = x 4x + 13 (x ( 3 + i))(x ( 3 3i)) = x + 6x + 34 Polynomiography 30

34 4. The desired polynomial is p(x) = (x + 3)(x 4)(x 4x + 13)(x + 6x + 34) In order to create a polynomial in standard form given its factored form, you can use a computer algebra system such as WolframAlpha ( which is free and very easy to use to expand a factored polynomial. Using a computer algebra system, one gets p(x) in standard polynomial form. Here is what you get if you used WolframAlpha to expand (x + 3)(x 4)(x 4x + 13)(x + 6x + 34) Figure 4.4 is a polynomiograph of x 6 + x 5 + 9x 4 105x 3 + 4x + 54x 5,304, showing the 6 roots. Figure 4.4. Plot of x 6 + x 5 + 9x 4 105x 3 + 4x + 54x 5,304. Note that given the preceding, one can argue that any n th -degree polynomial can be factored into a product of linear and quadratic factors. Polynomiography 31

35 Exercises 1. At the end of Section 4., the number of different possible roots for 4 th -degree polynomials was enumerated; they were: a. A single quadruple real root b. Four distinct real roots. c. A triple real root and a distinct second root d. Two real double roots e. A double real root and a pair of conjugate roots f. Two distinct pairs of conjugate pairs Using the above, create examples of each of the 6 possible cases.. Make a list of the possible different types of roots for a 5 th -degree polynomial which have exactly one pair of complex roots. Construct a polynomial for each and then create its polynomiograph. 3. Create a polynomial with degree greater than 8 and less than 18 which has at least two pairs of conjugate imaginary roots. Graph it using polynomiography. 4. Split your class into groups of 4. Have each group member write down the day of the month he or she was born, creating a list of 4 numbers. Use these numbers to create a 3 rd - degree polynomial in standard form. Plot the polynomial solutions using polynomiography. Have the groups compare their results. Next, combine the groups of 4 into groups of 8 and create 7 th -degree polynomials. Plot the polynomial solutions using polynomiography. Have the groups compare their results. Polynomiography 3

36 Assessment Questions 1. Create a spreadsheet that uses Newton s Method to find the roots of y = 6x! + 11x +. Use the Newton Cubic Analysis spreadsheet to find the roots of y = x 3 1x 453x + 3, Here s a polynomiograph of. y = z 1 + z 4z Using the zoom features of the software, find the polynomiograph below. 4. Given the roots 8, 3, ±9i, and 6 ± 11i, find the polynomial for the given roots in factored and then in standard form. 5. Use the polynomiography software to create a piece of art. The degree of the polynomial that you use must be greater than 7 and must have at least two pairs of imaginary roots. Explain all the mathematics in your art. Polynomiography 33

37 References [1] [] V. Katz, A History of Mathematics, 3rd ed. New York: Pearson, 009. Polynomiography 34