Problem 23 - Essential Calculus Early Transcendentals

Question

(a) Apply Newton's method to the equation \(x^{2}-a=0\) to derive the following square-root algorithm (used by the ancient Babylonians to compute \(\sqrt{a}\) ): $$x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right)$$ (b) Use part (a) to compute \(\sqrt{1000}\) correct to six decimal places.

Step-by-Step Solution

Answer

\(\sqrt{1000} \approx 31.622777\).

Step 1: Understand Newton's Method

Newton's method is an iterative process used to find successively better approximations to the roots of a real-valued function. Given an equation of the form \(f(x) = 0\), Newton's method uses the iterative formula:\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]where \(f'(x)\) is the derivative of \(f(x)\).

Step 2: Define the Function and Its Derivative

For our problem, we want to find a root of the equation \(x^2 - a = 0\), which corresponds to finding \(\sqrt{a}\). Define the function as \(f(x) = x^2 - a\). The derivative of this function with respect to \(x\) is \(f'(x) = 2x\).

Step 3: Apply Newton's Method Formula

Substitute \(f(x) = x^2 - a\) and \(f'(x) = 2x\) into Newton's method formula:\[x_{n+1} = x_n - \frac{x_n^2 - a}{2x_n}\]Simplify:\[x_{n+1} = x_n - \frac{x_n}{2} + \frac{a}{2x_n}\]\[x_{n+1} = \frac{2x_n - x_n + \frac{a}{x_n}}{2}\]\[x_{n+1} = \frac{x_n + \frac{a}{x_n}}{2}\]This is the square-root algorithm derived: \[x_{n+1} = \frac{1}{2}\left( x_n + \frac{a}{x_n} \right)\]

Step 4: Compute \(\sqrt{1000}\) with Initial Guess

To approximate \(\sqrt{1000}\) using this formula, start with an initial guess, say \(x_0 = 31\) because \(31^2 = 961\) is close to 1000. The iteration formula becomes:\[x_{n+1} = \frac{1}{2}\left( x_n + \frac{1000}{x_n} \right)\]

Step 5: Iterative Calculations

Using the initial guess \(x_0 = 31\), perform the iterations:1. \(x_1 = \frac{1}{2}(31 + \frac{1000}{31}) \approx 31.48387096\)2. \(x_2 = \frac{1}{2}(31.48387096 + \frac{1000}{31.48387096}) \approx 31.62202862\)3. \(x_3 = \frac{1}{2}(31.62202862 + \frac{1000}{31.62202862}) \approx 31.6227766\)4. \(x_4 = \frac{1}{2}(31.6227766 + \frac{1000}{31.6227766}) \approx 31.6227766\)The changes are now very small, indicating convergence to six decimal places.

Key Concepts

Square-Root Algorithm

The square-root algorithm is a method used to calculate the square root of a number. This algorithm has ancient origins, believed to be used by the Babylonians. The goal is to find the square root of a number \( a \) by solving the equation \( x^2 = a \) or equivalently \( x^2 - a = 0 \). One elegant way to do this is by applying Newton's Method.
The algorithm works as follows:

• Start with an initial guess \( x_0 \) for the square root of \( a \).
• Use the formula \( x_{n+1} = \frac{1}{2} \left(x_n + \frac{a}{x_n}\right) \) to get closer to the true square root.
• Repeat the process iteratively until satisfied with the precision.

This algorithm is specifically useful because it converges quickly to the accurate square root, meaning only a few iterations are typically needed. Each iteration refines your estimate, getting you closer to the exact square root.

Iterative Process

An iterative process involves repeating a sequence of operations to get closer to a desired result. Newton's Method, also known as the Newton-Raphson method, is a perfect example of this type of process.
Here's how it works in the context of finding roots:

• Define the function \( f(x) \). In our case, it's \( f(x) = x^2 - a \).
• Calculate the derivative \( f'(x) \). For our function, that's \( f'(x) = 2x \).
• Apply Newton's iteration formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).

The power of the iterative process lies in its ability to systematically refine estimates. Each iteration uses the most recent approximation to generate a more accurate guess.
Over time, this leads to convergence, meaning the estimates stabilize around the true root. This method is effective because it uses calculus to inform each subsequent guess, making it a powerful tool for solving equations.

Approximation of Roots

Approximation refers to finding a value that is close to the exact mathematical solution. In the context of finding roots, this concept often involves iterative methods like Newton's Method, where the root of an equation can be estimated to increasing degrees of precision.
To approximate the root of the equation \( x^2 - a = 0 \):

• Begin with an initial guess. For instance, guessing \( x_0 = 31 \) to find \( \sqrt{1000} \).
• Apply an iterative formula. With each iteration, the approximation moves closer to the actual root.
• Continue the iterations until the changes are minimal, indicating that further iterations would not significantly alter the result.

Approximation is crucial in contexts where exact solutions may be complex or impossible to derive algebraically. By utilizing numerical methods, we can estimate square roots, thus providing practical solutions that are sufficiently accurate for many purposes, such as engineering or scientific calculations.