Problem 25 - Thomas Calculus in SI Units

Question

Use Newton's method to find the zeros of \(f(x)=4 x^{4}-4 x^{2}\) using the given starting values. a. \(x_{0}=-2\) and \(x_{0}=-0.8,\) lying in \((-\infty,-\sqrt{2} / 2)\) b. \(x_{0}=-0.5\) and \(x_{0}=0.25,\) lying in \((-\sqrt{21} / 7, \sqrt{21} / 7)\) c. \(x_{0}=0.8\) and \(x_{0}=2,\) lying in \((\sqrt{2} / 2, \infty)\) d. \(x_{0}=-\sqrt{21} / 7\) and \(x_{0}=\sqrt{21} / 7\)

Step-by-Step Solution

Answer

Apply Newton's method iteratively using the given starting values to find function zeros, calculated for each interval.

Step 1: Identify Function and Derivative

The given function is \( f(x) = 4x^4 - 4x^2 \). To apply Newton's method, we need its derivative: \( f'(x) = 16x^3 - 8x \).

Step 2: Apply Newton's Method Formula

Newton's method formula is: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). This formula will be used to find zeros for different initial values \( x_0 \).

Step 3a: Solve for Starting Values in (-∞,-√2/2): x_0 = -2

Substitute \( x_0 = -2 \) into the Newton's method formula and iterate until the values converge. Calculate few iterations:1. \( f(-2) = 48 \), \( f'(-2) = -56 \)2. \( x_1 = -2 - \frac{48}{-56} = -2 + \frac{6}{7} \approx -1.1429 \)3. Continue until sufficient accuracy.

Step 3b: x_0 = -0.8

Substitute \( x_0 = -0.8 \):1. \( f(-0.8) = 0.9216 \), \( f'(-0.8) = -4.096 \)2. \( x_1 = -0.8 - \frac{0.9216}{-4.096} \approx -0.575 \)3. Continue iterations.

Step 4a: Solve for Starting Values in (-√21/7, √21/7): x_0 = -0.5

Apply Newton's method using \( x_0 = -0.5 \):1. \( f(-0.5) = 0 \), \( f'(-0.5) = -4 \)2. \( x_1 = -0.5 \). No further iterations needed as the exact zero is found.

Step 4b: x_0 = 0.25

Use the formula for \( x_0 = 0.25 \):1. \( f(0.25) = 0.1875 \), \( f'(0.25) = -1 \)2. \( x_1 = 0.25 - \frac{0.1875}{-1} = 0.4375 \)3. Continue iterations.

Step 5a: Solve for Starting Values in (√2/2, ∞): x_0 = 0.8

Find root for \( x_0 = 0.8 \):1. \( f(0.8) = 0.9216 \), \( f'(0.8) = 4.096 \)2. \( x_1 = 0.8 - \frac{0.9216}{4.096} \approx 0.575 \)3. Continue iterations until convergence.

Step 5b: x_0 = 2

Use \( x_0 = 2 \):1. \( f(2) = 48 \), \( f'(2) = 56 \)2. \( x_1 = 2 - \frac{48}{56} = 2 - \frac{6}{7} = 1.1429 \)3. Iterate further.

Step 6: Special Cases: x_0 = -√21/7 and x_0 = √21/7

As observed previously, these specific initial values can yield trivial solutions. Check through iteration:1. Calculate \( f(x) \) and \( f'(x) \), apply same method.2. Carefully observe if solutions are trivial/non-trivial and consistent across ranges.

Key Concepts

Zero Finding

Zero finding, often referred to as root finding, involves determining values of variable \( x \) that satisfy the equation \( f(x) = 0 \). In the context of Newton's Method, zero finding is primarily about identifying these zeros using an iterative process. For any function, roots represent the points where the function's graph intersects the x-axis. Understanding this concept is vital as it allows students to pinpoint where a function equals zero without needing to graph it explicitly.
Newton's method provides a systematic approach to zero finding by initially taking an estimated zero and refining it iteratively:

• The method starts at an initial guess \( x_0 \).
• With each iteration, it refines this guess using the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).
• The process continues until it converges to a satisfactory level of precision, meaning \( |x_{n+1} - x_n| \) is sufficiently small.

This iterative technique allows for efficiency and precision in solving for zeros, especially when functions cannot be solved easily through algebraic methods alone.

Polynomial Functions

Polynomial functions are mathematical expressions consisting of variables raised to various powers and multiplied by coefficients. They take the general form \( f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 \), where \( a_i \) are coefficients and \( n \) is a non-negative integer denoting the highest power.
Polynomials can be simple, like linear and quadratic functions, or more complex with higher degrees, such as cubic or quartic. The given function \( f(x)=4x^4-4x^2 \) is a quartic polynomial, which makes it more complex. Solving such functions might require finding their zeros, where the function yields zero value. The behavior and shape of a polynomial's graph depend on its leading coefficient and degree.

• The leading term, with the highest power, determines the end behavior of the graph.
• The degree of the polynomial tells us the maximum number of roots or zeros it can have.

In this exercise, we're dealing with real and possible complex zeros, as the maximum degree pretends four solutions, which could be seen through factoring or applying methods like Newton's.

Iterative Methods

Iterative methods are mathematical techniques used to solve problems by approaching a target value through repeated approximations. Newton's Method is a prime example of an iterative method, designed to find successively better approximate zeros (roots) of a real-valued function.
These methods are particularly useful when functions do not lend themselves to straightforward algebraic solutions. Here's how the process in Newton's Method works:

• Start with a guess (initial approximation) \( x_0 \).
• Use the method's formula to produce a sequence of values that converge towards a root.
• Iterate until the sequence stabilizes to an acceptable precision, often assessed through a predetermined tolerance threshold.

This iterative approach requires evaluating both the function and its derivative, making it essential to have an understanding of basic calculus principles.

Derivative Calculation

Derivative calculation plays a critical role in Newton's method, as derivatives provide the slope of the tangent line at any point on a function's curve. This slope is crucial, as Newton's method relies on tangent lines to approximate the zeros of a function.
For the polynomial \( f(x) = 4x^4 - 4x^2 \), calculating its derivative gives \( f'(x) = 16x^3 - 8x \). This derivative is pivotal for the accuracy and speed of convergence in Newton's method. Employing derivatives through Newton's formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) aids in

1. Determining the slope at the current approximation \( x_n \).
2. Adjusting the approximation to be closer to the actual root.

Understanding derivatives thus enables effective use of Newton's method, making zero finding and iterative refinement practical and efficient.