Problem 25
Question
The mechanic's rule for approximating square roots states that \(\sqrt{a} \approx x_{n+1},\) where $$ x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right), \quad n=1,2,3, \ldots $$ and \(x_{1}\) is any positive approximation to \(\sqrt{a} .\) (a) Apply Newton's Method to $$ f(x)=x^{2}-a $$ to derive the mechanic's rule. (b) Use the mechanic's rule to approximate \(\sqrt{10} .\)
Step-by-Step Solution
Verified Answer
The mechanic's rule is derived using Newton's Method and approximates \( \sqrt{10} \) as \( 3.1623 \).
1Step 1: Define the Function
Given the function for which we want to find the square root, we define it as \( f(x) = x^2 - a \). This function will help us in both deriving the mechanic's rule and applying Newton's Method.
2Step 2: Apply Newton's Method Formula
For Newton's Method, we start with the iteration formula: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). Evaluate \( f(x) \) and its derivative \( f'(x) \): \( f(x) = x^2 - a \) and \( f'(x) = 2x \).
3Step 3: Substitute into Newton's Method
Substitute \( f(x) = x^2 - a \) and \( f'(x) = 2x \) into the Newton's Method formula:\[x_{n+1} = x_n - \frac{x_n^2 - a}{2x_n} = x_n - \frac{x_n}{2} + \frac{a}{2x_n} = \frac{2x_n - x_n + \frac{a}{x_n}}{2} = \frac{x_n + \frac{a}{x_n}}{2}.\]This matches the mechanic's rule.
4Step 4: Approximate \( \sqrt{10} \) Using Mechanic's Rule
Choose an initial approximation. Let's take \( x_1 = 3 \), as \( 3^2 = 9 \) is close to 10.Calculate subsequent values using:\[ x_{n+1} = \frac{1}{2} \left( x_n + \frac{10}{x_n} \right) \]- \( x_2 = \frac{1}{2} (3 + \frac{10}{3}) \approx 3.1667 \)- \( x_3 = \frac{1}{2} (3.1667 + \frac{10}{3.1667}) \approx 3.1623 \) Continue until desired accuracy.
Key Concepts
Newton's MethodIterative MethodsMechanic's Rule
Newton's Method
Newton's Method, also known as Newton-Raphson, is a powerful technique for finding successively better approximations to the roots (or zeros) of a real-valued function. It's like a toolset for solving equations where traditional algebra might be tricky. The method starts with a guess, then refines that guess by utilizing the function and its derivative.
The iteration formula for Newton's Method is:
Newton's Method speeds up the convergence to the actual root, making it an efficient choice in numerical methods, especially for well-behaved functions where derivatives are easy to calculate.
The iteration formula for Newton's Method is:
- \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
Newton's Method speeds up the convergence to the actual root, making it an efficient choice in numerical methods, especially for well-behaved functions where derivatives are easy to calculate.
Iterative Methods
Iterative methods, like the mechanic's rule for approximating square roots, are processes where a sequence of improving approximations of a value is generated. The key goal is to reach a solution by repetition, honing in gradually with each step until the desired proximity is achieved.
The idea is to use each approximation to get to the next, more accurate approximation. This is typically done by following a simpler formula derived from the original problem, such as in the mechanics rule:
The idea is to use each approximation to get to the next, more accurate approximation. This is typically done by following a simpler formula derived from the original problem, such as in the mechanics rule:
- \[ x_{n+1} = \frac{1}{2}\left(x_n + \frac{a}{x_n}\right) \]
Mechanic's Rule
The mechanic's rule, often used for approximating square roots, is a simplified version of Newton's Method. It specifically addresses the root-finding iteration for equations of the form \( x^2 = a \). The formula is beautifully simple, yet effective:
For example, to approximate \( \sqrt{10} \), we might start with \( x_1 = 3 \). Applying the mechanic's rule iteratively gives us refinements like \( x_2 \approx 3.1667 \) and \( x_3 \approx 3.1623 \). Continue this cycle, and you'll rapidly zero in on the accurate square root. The beauty of this method lies in its simplicity—using basic arithmetic to achieve impressive precision.
- \[ x_{n+1} = \frac{1}{2}\left(x_n + \frac{a}{x_n}\right) \]
For example, to approximate \( \sqrt{10} \), we might start with \( x_1 = 3 \). Applying the mechanic's rule iteratively gives us refinements like \( x_2 \approx 3.1667 \) and \( x_3 \approx 3.1623 \). Continue this cycle, and you'll rapidly zero in on the accurate square root. The beauty of this method lies in its simplicity—using basic arithmetic to achieve impressive precision.
Other exercises in this chapter
Problem 24
Find: (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open inte
View solution Problem 25
Let \(g(x)=x e^{x}-e^{x} .\) Find \(f(x)\) so that \(f^{\prime}(x)=g^{\prime}(x)\) and \(f(1)=2\)
View solution Problem 25
Find the absolute maximum and minimum values of \(f \), if any, on the given interval, and state where those values occur. \(f(x)=2 x^{3}-6 x+2 ;(-\infty,+\inft
View solution Problem 25
Use the given derivative to find all critical points of \(f\) and at each critical point determine whether a relative maximum, relative minimum, or neither occu
View solution