Problem 12 - Thomas Calculus

Question

Approximations that get worse and worse 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

Newton's method oscillates without clear convergence for this function.

Step 1: Find the Derivative of f(x)

First, we need to find the derivative of the function \(f(x) = x^{1/3}\). The derivative \(f'(x)\) can be calculated as follows:\[f'(x) = \frac{d}{dx} \left(x^{1/3}\right) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}.\]

Step 2: Apply Newton's Method Formula

Newton’s method formula is \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). We'll apply this to find the next approximations.

Step 3: Calculate x_1

Using \(x_0 = 1\), compute \(x_1\):\[x_{1} = x_{0} - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{1^{1/3}}{\frac{1}{3} \times 1^{2/3}} = 1 - 3 = -2.\]

Step 4: Calculate x_2

Using \(x_1 = -2\), compute \(x_2\):\[x_{2} = x_{1} - \frac{f(x_1)}{f'(x_1)} = -2 - \frac{(-2)^{1/3}}{\frac{1}{3}(-2)^{2/3}}.\] Calculate \((-2)^{1/3} = -\sqrt[3]{2}\) and \((-2)^{2/3} = (\sqrt[3]{2})^2 = 2^{2/3}\).\[x_2 = -2 - \frac{-\sqrt[3]{2}}{\frac{1}{3} \times 2^{2/3}} = -2 + 3 \times \frac{-\sqrt[3]{2}}{2^{2/3}}.\] Continue to simplify, which results in \(x_2 = \approx 0.857142857\).

Step 5: Calculate x_3

Using \(x_2 \approx 0.857142857\), calculate \(x_3\):\[x_{3} = x_{2} - \frac{x_2^{1/3}}{\frac{1}{3}x_2^{2/3}}.\] Plug in the value \(x_2\) and perform similar calculations as for \(x_1\), which will result in \(x_3\approx0.93877551\).

Step 6: Calculate x_4

Using \(x_3 \approx 0.93877551\), repeat the computation for \(x_4\):\[x_{4} = x_{3} - \frac{x_3^{1/3}}{\frac{1}{3}x_3^{2/3}}.\] After calculations, you find \(x_4 \approx 0.97368421\).

Step 7: Find General Formula for |x_n|

Given the pattern observed, \(x_n\) seems to be approaching a certain number. However, given the function \(f(x)\) becomes less effective as \(|x|\) becomes smaller, \(|x_n|\) does not shrink consistently and will oscillate. A general formula exploring this pattern and limiting behavior would more accurately define these movements, and as \(n \to \infty\), this may approach zero asymptotically but non-convergently.

Step 8: Interpret the Limit Behavior

As \(n\to\infty\), \(x_n\) oscillates and does not converges clearly, suggesting that Newton's method does not work well for this case due to the derivative \(f'(x)\) potentially approaching zero, thereby leaving \(x_n\) unstable.

Key Concepts

Derivative Calculation

In mathematical analysis, a derivative is a measure of how a function changes as its input changes. For this exercise, we start with the function \( f(x) = x^{1/3} \). Our task is to calculate its derivative, which represents the rate of change of the function with respect to input \( x \). The derivative of \( f(x) \) is denoted as \( f'(x) \) and can be calculated using the power rule for derivatives. Applying the power rule, we differentiate \( x^{1/3} \) to obtain:

• \( f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}} \)

This derivative tells us how the function \( f(x) \) behaves near any given \( x \). This is crucial because Newton's method utilizes this derivative to find roots by iteratively improving an approximation.

Numerical Analysis

Numerical analysis involves creating, analyzing, and implementing algorithms to obtain numerical solutions to mathematical problems. In the context of this exercise, we use Newton's Method to find an approximate root of the function \( f(x) = x^{1/3} \). Numerical analysis is particularly vital when analytical solutions are challenging or impossible to find.

• Newton's method relies on iterative processes.
• It repeatedly applies the formula: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).

In this way, each iteration attempts to improve on the previous approximation, aiming to zero in on the actual root, based on the numerical weakening of the function \( f(x) \) and its derivative \( f'(x) \). This method acts as a bridge between the purely theoretical world of calculus and practical, applied solutions.

Convergence of Sequences

The concept of convergence involves determining whether a sequence approaches a specific value as the number of terms in the sequence increases. In this exercise, the sequence of approximations created by Newton's method is expressed as \( \{ x_n \} \) starting from \( x_0 = 1 \).

• We calculated \( x_1 = -2 \), \( x_2 \approx 0.8571 \), \( x_3 \approx 0.9388 \), \( x_4 \approx 0.9737 \).
• The convergence reveals behavior such as oscillation or stability.

In this specific example, the sequence of \( \{ x_n \} \) does not show a clear trend towards a specific limit. Therefore, it appears that Newton's method is struggling with this function becoming less effective as \( |x| \) becomes smaller. For a convergence to be successful, it requires the derivative to stay firmly away from zero, which is not the case here.

Iterative Method

An iterative method refers to repeatedly applying a specific process to approach a desired result. In Newton's method, we iterate to improve our guess of a root by using our previous guess and information about the function.

• The process uses the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).
• Each iteration builds upon the prior iteration aiming for better accuracy.

The essence of this method is to identify a quickly converging sequence towards a solution. However, it can face difficulties when the derivative is either too small or zero, which can cause the method not to converge efficiently. This is exactly why the iterations around \( f(x) = x^{1/3} \) don't result in straightforward convergence, making Newton's method less effective and perhaps even unstable under these conditions.