Problem 25 - Calculus of a Single Variable

Question

Use Newton’s Method to show that the equation \(x_{n+1}=x_{n}\left(2-a x_{n}\right)\) can be used to approximate 1\(/ a\) when \(x_{1}\) is an initial guess of the reciprocal of \(a .\) Note that this method of approximating reciprocals uses only the operations of multiplication and subtraction. (Hint: Consider \(f(x)=\frac{1}{x}-a.\))

Step-by-Step Solution

Answer

Yes, the equation \(x_{n+1}=x_{n}\left(2-a x_{n}\right)\) can be derived from Newton’s Method and used to approximate 1/a, and it uses only the operations of multiplication and subtraction. This application of Newton's method is special due to its lean operations, avoiding the usually computationally expensive division operation.

Step 1: Formulate the Function

First, the function is identified. This is \(f(x)=\frac{1}{x}-a.\)

Step 2: Determine the Derivative of the Function

Newton's Method needs the derivative of the function. The derivative of \(f(x)\) is \(f'(x)=-\frac{1}{x^{2}}\).

Step 3: Newton’s Method: Iteration Formula

The general iteration formula of Newton's Method is \(x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\). If we plug \(f(x)\) and \(f'(x)\) into Newton’s formula, we will get \(x_{n+1}=x_{n}-\frac{\frac{1}{x_n}-a}{-\frac{1}{x_{n}^{2}}} = x_{n}\left(2-a x_{n}\right)\). This is the equation given in the problem, proving that this formula can be used.

Step 4: Operations Being Used

Looking at the final formula closely, it is noted that only multiplication and subtraction are used, no division is involved in it. This satisfies the condition in the problem.

Key Concepts

Iteration Formula: Key to Newton's Method

Newton's Method is a powerful technique for finding successively better approximations to the roots (or zeroes) of a real-valued function. A critical component of Newton's Method is the use of the iteration formula, which can be expressed generally as:

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

The iteration formula represents the core iterative step that refines our initial guess to get closer to the actual root. Using this method involves iterating: repeatedly applying the formula using the result from the previous step.
This process depends on two crucial elements: having a defined function, \( f(x) \), and its derivative, \( f'(x) \). For the exercise at hand, starting with the function \( f(x) = \frac{1}{x} - a \), Newton’s Method allows us to derive an iteration formula:

• \( x_{n+1} = x_{n} ( 2 - a x_{n}) \)

Here, you see the iteration doesn't require complex operations beyond basic multiplication and subtraction, which are computationally simple and efficient. This simplicity is particularly useful for approximating reciprocals in a practical setting without performing division.

Derivative Calculation: Foundation of the Iteration Formula

The calculation of derivatives is a fundamental aspect of Newton’s Method, as it is used to derive the iteration formula. In the exercise, we work with the function \( f(x) = \frac{1}{x} - a \). To apply Newton's Method, it is necessary to calculate the derivative of this function:

• The derivative, \( f'(x) = -\frac{1}{x^2} \), represents the slope of \( f(x) \) at any point \( x \).

This derivative explains how changes in \( x \) affect the function value \( f(x) \). By substituting both \( f(x) \) and \( f'(x) \) into the Newton's iteration formula, you obtain:

• \( x_{n+1} = x_{n} - \frac{\frac{1}{x_n} - a}{-\frac{1}{x_{n}^{2}}} \)
• Which simplifies to \( x_{n+1} = x_{n} ( 2 - a x_{n}) \).

Understanding the role of the derivative in this process is crucial. It provides the necessary adjustment to our guess, ensuring rapid convergence towards the accurate reciprocal value. Proper derivative calculation underpins the effectiveness of Newton’s Method.

Function Approximation: Achieving Precise Results

Function approximation with Newton's Method involves taking an initial guess (\( x_1 \)) and refining it to get closer to the target value – in this case, the reciprocal of \( a \). This is especially useful when exact calculations are impractical, and a good approximation is sufficient.
The task in this exercise is to approximate \( \frac{1}{a} \) by iterating the initially guessed value using the derived formula:

• \( x_{n+1} = x_{n} ( 2 - a x_{n}) \)

This iteration continues until \( x_{n} \) changes very little between steps, indicating convergence.
Newton's Method is particularly effective because its convergence is usually quadratic, meaning the approximation becomes very accurate in just a few iterations. By limiting operations to multiplication and subtraction, the method becomes computationally efficient, which is significant in digital computations where minimizing complex operations is beneficial.
Approximating functions using such methods not only consolidates the understanding of calculus concepts but also equips you with practical tools for solving real-world problems where precision and efficiency are paramount.