Problem 10 - Thomas Calculus in SI Units

Question

Apply Newton's method to \(f(x)=x^{1 / 3}\) with \(x_{0}=1\) and calculate \(x_{1}, x_{2}, x_{3},\) and \(x_{4}\). Find a formula for \(\left|x_{n}\right| .\) What happens to \(\left|x_{n}\right|\) as \(n \rightarrow \infty ?\) Draw a picture that shows what is going on.

Step-by-Step Solution

Answer

The sequence \(|x_n|\) oscillates between 2 and 4, and does not converge.

Step 1: Understand Newton's Method

Newton's Method is an iterative technique for finding approximate solutions to the equation \(f(x) = 0\). The general formula for Newton's Method is \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). We will use this formula to calculate subsequent values of \(x_n\).

Step 2: Find the Derivative

Calculate the derivative of the function \(f(x) = x^{1/3}\). The derivative is \(f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}\).

Step 3: Calculate x_1

Using the initial value \(x_0 = 1\), apply Newton's Method: \[x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{1^{1/3}}{\frac{1}{3}(1)^{2/3}} = 1 - 3 = -2\]

Step 4: Calculate x_2

Now use \(x_1 = -2\) to find \(x_2\): \[x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = -2 - \frac{(-2)^{1/3}}{\frac{1}{3}(-2)^{2/3}}\]Calculate \((-2)^{1/3} = -\sqrt[3]{2}\) and \((-2)^{2/3} = \sqrt[3]{4}\): \[x_2 = -2 - \frac{-\sqrt[3]{2}}{\frac{1}{3}\sqrt[3]{4}} = -2 - (-6) = 4\]

Step 5: Calculate x_3

Using \(x_2 = 4\): \[x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 4 - \frac{4^{1/3}}{\frac{1}{3}4^{2/3}}\]Where \(4^{1/3} = \sqrt[3]{4}\) and \(4^{2/3} = \sqrt[3]{16}\): \[x_3 = 4 - \frac{\sqrt[3]{4}}{\frac{1}{3}\sqrt[3]{16}} = 4 - 6 = -2\]

Step 6: Calculate x_4

Using \(x_3 = -2\) once again: \[x_4 = x_3 - \frac{f(x_3)}{f'(x_3)} = -2 - \frac{(-2)^{1/3}}{\frac{1}{3}(-2)^{2/3}} = -2 - (-6) = 4.\]

Step 7: Check the Pattern

Observe the calculated sequence: \(x_0 = 1\), \(x_1 = -2\), \(x_2 = 4\), \(x_3 = -2\), \(x_4 = 4\). Notice the oscillation between -2 and 4 after the initial point.

Step 8: Analyze \(|x_n|\) as \(n \rightarrow \infty\)

The sequence of values \(x_n\) alternates between -2 and 4, indicating that \(|x_n|\) will also alternate between 2 and 4 for all \(n > 1\). Thus, \(|x_n|\) does not converge as \(n \rightarrow \infty\).

Step 9: Visualization

A graphical representation helps to understand the oscillatory nature. Plot \(f(x) = x^{1/3}\) and the tangent lines at calculated \(x_n\). Notice that the tangent at -2 leads to 4 and vice versa. This visual cue explains the oscillation.

Key Concepts

Iterative Technique

Newton's Method is a prime example of an iterative technique used to find approximate solutions to equations of the form \(f(x) = 0\). Essentially, this method involves initializing with a guess—known as \(x_0\)—and then improving this guess through successive iterations. These iterations are calculated using the formula:

• \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\)

By performing this calculation repeatedly, the goal is to hone in on the solution. Each iteration aims to bring the guess closer to the actual root of the function.
Newton's Method is particularly powerful because it can rapidly converge to a root when initiated close to the actual solution. However, its performance highly depends on the initial guess and the nature of the function.

Derivative Calculation

A crucial element of Newton's Method is the derivative calculation, which is an essential step for determining the behavior of the function near the current guess. For the function \(f(x) = x^{1/3}\), the derivative is calculated as follows:

• \(f'(x) = \frac{1}{3}x^{-2/3}\)

Understanding this derivative tells us how the function increases or decreases near any given point—critical for adjusting our guesses in Newton's method.
Without the precise calculation of this derivative, the iterative process stands faulty. It directly impacts the adjustments made to each successive approximation \(x_n\). In functions where derivatives get complex, precision becomes even more important to ensure convergence.

Convergence Analysis

Convergence analysis is the key to understanding how effective Newton's Method will be for a specific function.
With the exercise involving \(f(x) = x^{1/3}\), we see that instead of converging to a single fixed point, the sequence oscillates between -2 and 4. The pattern of alternation does not settle as \(n\) approaches infinity:

• \(x_0 = 1\)
• \(x_1 = -2\)
• \(x_2 = 4\)
• \(x_3 = -2\)
• \(x_4 = 4\)

This lack of a stable limit means Newton's Method doesn't provide a meaningful solution in this case.
Such behavior highlights a limitation: if the derivative reaches zero or the initial guess isn't close enough to the root, convergence might fail or lead to erratic results, as seen here. Beyond the choice of initial guess, the nature of the function, like multi-peaked functions, affects convergence significantly.

Graphical Representation

Graphical representation aids immensely in grasping why iteration methods behave the way they do. For this problem, plotting \(f(x) = x^{1/3}\) with its tangent lines provides clarity.

• Tangents at a guess point show how next steps are calculated.
• Oscillation visible between -2 and 4 for tangible iterations.

Visualizing the tangent lines connecting one point to the next offers insights that calculation alone might obscure.
When you see the plot, it's easier to understand why the sequence alternates—the slope of the tangents sends the sequence back to a previous point rather than moving towards a single point. Graphical tools are thus invaluable, not just for solution verification, but also for diagnosing why a series might not converge.