Exercises for Chaos Under Control

Chapter 8: Newton's Method for Digging Up Roots

6. Refer to the Figure.

(a) Sketch the first two points, x₁ and x₂, generated by applying Newton's method to the guess x₀.

(b) Locate two nearby starting points, A and B, such that Newton's method from A leads to the left root (denoted L), and Newton's method from B leads to the right root (denoted R). Sketch the first few steps of applying Newton's method to both of these guesses.

7.[A] Let f(x) be the function defined by f(x) = x².

(a) Show that Newton's method for finding the roots of f(x) is equivalent to the iterative scheme x_n+1 = x_n/2. Answer

(b) Show x = 0 is the root of f(x).

(d) Repeat (c) taking as starting points x₀ = -4, ±2, ±1, and ±1/3. What do you conclude happens in general? Answer

(e) Note that iterating Newton's method for x² is different from iterating x² itself. What is the basin of attraction of x = 0 when iterating x²? Answer

8.[A] Let f(x) be the function defined by f(x) = x² - 1.

(a) Show that Newton's method for finding the roots of f(x) is equivalent to the iterative scheme x_n+1 = (x_n² + 1)/(2x_n). Answer

(b) Show x = 1 and x = -1 are fixed points for this Newton map.

(c) Using a calculator, plot the values of x_n+1 for x_n = ±1/8, ±1/4, ±1/2, ±3/4, ±1, ±5/4, ±3/2, ±2, ±3. Sketch the graph of x_n+1 as a function of x_n. Answer

(d) From the graph in (c), deduce x_n = 1 and x_n = -1 are stable fixed points of the Newton map. Answer

(e) Using graphical iteration, find the basins of attraction of these roots. Answer

(f) Is your answer to (e) consistent with Cayley's result for z² - 1?

9.[A] Let f(x) = x² + 1.

(a) Show this f(x) has no roots that are real numbers.

(b) Show that Newton's method for finding the roots of f(x) is equivalent to the iterative scheme x_n+1 = (x_n² - 1)/(2x_n).

(c) Using a calculator, plot the values of x_n+1 for x_n = ±1/8, ±1/4, ±1/2, ±3/4, ±1, ±5/4, ±3/2, ±2, ±3. Sketch the graph of x_n+1 as a function of x_n. Answer

(d) From the graph of (c), deduce that the Newton map has no fixed points. Answer

(e) Using a calculator, determine the first ten iterates of Newton's method starting with x₀ = 0.333. Answer

(f) Repeat (e) starting with x₀ = 0.334. Answer

(g) What do the results of (e) and (f) suggest to you about the dynamics of this map? Try other starting values, being careful to avoid anything leading to x = 0. Answer

10.[A] This is a generalization of the previous three exercises. Consider the family of functions f_c(x) = x² + c, for c and x real numbers.

(a) Show f_c(x) has two real roots when c < 0. What are these roots? Answer

(b) Using a calculator, sketch the graph of the Newton map for c = -1/2. Answer

(d) Repeat (b) and (c) for c = -3/2. Answer

(e) What do you think happens for any c < 0? Answer

(f) Repeat (b) and (c) for c = 1/2. Answer

(g) What do you think happens for any c > 0? Answer

11.[A] Suppose f(x^*) = 0 and f'(x^*) not equal to 0. Show x^* is a fixed point for the associated Newton map. Answer

12.[A] Show that (z+d)³ = z³ + 3z²d + 3zd² + d³.

13.[A] Show that Equation (4) can be rewritten as

z_n+1 = (2*z_n)/3 + 1/(3*z_n²).

14.[A] The complex number 1/(A + B*i) is not in standard form: it has the i downstairs. To put it in standard form multiply both numerator and denominator by A - B*i. Show that

1/(A + B*i) = (A - B*i)/(A² + B²).

(Thus, 1/(A+B*i) has the real part A/(A² + B²) and the imaginary part -B/(A² + B²).)

15.[A] With the help of exercise 14, show that Newton's method for finding the roots of z³ - 1 is equivalent to a set of transformations

x_n+1 = (2*x_n/3) + (x_n² + y_n²)/D

and

y_n+1 = (2*y_n/3) - 2*x_n*y_n/D

where D = 3*((x_n² - y_n²)² + 4*x_n²*y_n²) = 3*(x_n² + y_n²)².

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