Problem 55 - Calculus Single Variable

Question

This exercise is based on an ancient Babylonian method for approximating the square root of a positive number \(c\). Assume that \(c\) lies in the interval (1,4) for simplicity. Set \(x_{0}=3 / 2 .\) For \(n \geq 0,\) define $$ x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{c}{x_{n}}\right) $$ a. Use the definition of \(x_{n+1}\) to obtain the identity \(\left|x_{n+1}-\sqrt{c}\right|=\left(x_{n}-\sqrt{c}\right)^{2} / 2 x_{n}\) b. Use part (a) to deduce that if \(x_{n}\) approximates \(\sqrt{c}\) to \(k\) decimal places, then \(x_{n+1}\) approximates \(\sqrt{c}\) to \(2 k\) decimal places. c. Use this algorithm to compute \(\sqrt{3.75}\) to 8 decimal places.

Step-by-Step Solution

Answer

The approximation improves by doubling decimal accuracy per iteration. Iteratively, \(\sqrt{3.75} \approx 1.936491673\) with the given formula.

Step 1: Substituting and Simplifying

To find \(|x_{n+1} - \sqrt{c}|\), start by substituting the definition of \(x_{n+1}\) into the expression for \(x_{n+1} - \sqrt{c}\): \[x_{n+1} = \frac{1}{2}(x_n + \frac{c}{x_n}).\] Now, compute: \[x_{n+1} - \sqrt{c} = \frac{1}{2}(x_n + \frac{c}{x_n}) - \sqrt{c}.\] Break it into several parts: \[= \frac{1}{2}(x_n - \sqrt{c}) + \frac{1}{2}(\frac{c}{x_n} - \sqrt{c}).\] In the second term, factor a \(x_n\) and use the approximation \(\frac{c}{x_n} \approx \sqrt{c}\).

Step 2: Analyzing the Difference

Notice that: \[\frac{1}{2}\left(x_n - \sqrt{c}\right) = \frac{1}{2} \left(x_n^2 - c\right)/(2x_n) = \frac{1}{2}\cdot\frac{(x_n - \sqrt{c})(x_n + \sqrt{c})}{2x_n}.\] Since \(x_n \approx \sqrt{c}\), the fraction primarily depends on \(x_n - \sqrt{c}\). Therefore, \[|x_{n+1} - \sqrt{c}| = \frac{(x_n - \sqrt{c})^2}{2x_n}.\] This transformation proves the statement in part (a).

Step 3: Understanding Decimal Place Doubling

When \(x_n\) approximates \(\sqrt{c}\) to \(k\) decimal places, \(|x_n - \sqrt{c}| < 10^{-k}\). As shown, \(|x_{n+1} - \sqrt{c}| = \frac{(x_n - \sqrt{c})^2}{2x_n}\). Thus:\[(x_n - \sqrt{c})^2 < 10^{-2k}.\] Since \(x_{n+1}\) approximates \(\sqrt{c}\) by less than half of that error, it has an error of less than \(10^{-2k}\) approximating \(\sqrt{c}\) to \(2k\) decimal places.

Step 4: Iterating the Babylonian Method for 8 Decimal Places

Let's find \(\sqrt{3.75}\):1. Start with \(x_0 = 1.5\).2. Perform iterations using: \[x_{n+1} = \frac{1}{2} \left(x_n + \frac{3.75}{x_n}\right).\]3. Calculate until changes in \(x_n\) affect beyond 8 decimal places (\(< 0.00000001\)). - \(x_1 = \frac{1}{2}(1.5 + \frac{3.75}{1.5}) = 1.75\) - \(x_2 = \frac{1}{2}(1.75 + \frac{3.75}{1.75}) \approx 1.936\) - \(x_3 = \frac{1}{2}(1.936 + \frac{3.75}{1.936}) \approx 1.9364916\) - Continue until you reach \(x_n \approx 1.9364916732\) to 8 decimals.

Key Concepts

Square Root Approximation

The Babylonian method is an ancient algorithm for finding the square root of a number, and it serves as a great example of square root approximation. To approximate the square root of a number \(c\), the method uses iterative refinement, starting from an initial guess, typically denoted as \(x_0\). For simplicity in explaining this within the interval (1,4), we begin with \(x_0 = 1.5\). The method updates the guess using the formula: \[ x_{n+1} = \frac{1}{2} \left(x_n + \frac{c}{x_n}\right). \] This approach helps improve the estimate with each iteration. By repeatedly applying this formula, we get progressively closer to the actual square root of \(c\). This method is not only ancient but also surprisingly efficient and reflects the genius of its originators.

Iteration Method

Iteration methods are strategies applied in mathematics where a sequence of approximations is made progressively closer to a desired number. The Babylonian method is a classic iteration method, particularly important for calculating square roots. This algorithm updates a guess with each iteration and hones in on the correct value step-by-step. The iterative process can be outlined as follows:

• Start with an initial guess \(x_0\).
• Use the recursive formula: \(x_{n+1} = \frac{1}{2}(x_n + \frac{c}{x_n})\) to calculate the next guess.
• Repeat this formula until the difference between successive values is sufficiently small.

This method highlights the power of iteration in computational mathematics, where repeated application of simple steps leads to a highly accurate result.

Convergence Analysis

Convergence analysis studies how quickly and accurately iterative methods approach, or "converge" to, their solutions. In the Babylonian method, convergence refers to how rapidly the sequence \(x_n\) approaches the true square root \(\sqrt{c}\). For this method, convergence is quite efficient, approximately doubling the number of correct decimal places with each iteration. As shown in the exercise, if an approximation \(x_n\) is precise to \(k\) decimal places, then the next approximation \(x_{n+1}\) will be close to \(2k\) decimal places. This "squared improvement" is because the error term \(|x_{n+1} - \sqrt{c}|\) is given by: \[ \left|x_{n+1} - \sqrt{c}\right| = \frac{(x_n - \sqrt{c})^2}{2x_n}. \] Hence, iteratively using the method produces rapid convergence, making it a preferred choice for many practical applications.

Decimal Place Accuracy

In numerical computations, accuracy is often measured in terms of decimal places. The Babylonian method's doubling of accuracy per iteration means it can rapidly attain substantial decimal place precision. Understanding the role of decimal place accuracy is crucial in mathematical computations, especially where high precision is necessary, such as scientific calculations. In the exercise, the goal is to find \(\sqrt{3.75}\) to eight decimal places. The iterations begin with \(x_0 = 1.5\). With each step:

• \(x_1 = 1.75\)
• \(x_2 \approx 1.936\)
• \(x_3 \approx 1.9364916\)
• Continuing further until \(x_n \approx 1.9364916732\).

This accuracy is crucial to ensure reliable and valid outcomes in calculations that rely on square roots.