Question

Approximating square roots Let \(a>0\) be given and suppose we want to approximate \(\sqrt{a}\) using Newton's method. a. Explain why the square root problem is equivalent to finding the positive root of \(f(x)=x^{2}-a\) b. Show that Newton's method applied to this function takes the form (sometimes called the Babylonian method) $$x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right), \text { for } n=0,1,2, \ldots$$ c. How would you choose initial approximations to approximate \(\sqrt{13}\) and \(\sqrt{73} ?\) d. Approximate \(\sqrt{13}\) and \(\sqrt{73}\) with at least 10 significant digits.

Short answer

Answer: Finding the square root of a number \(a\) is equivalent to finding the positive root of the function \(f(x) = x^2 - a\) because both problems involve finding a number \(x\) such that \(x^2 = a\). The positive root of the function \(f(x) = x^2 - a\) is the value of \(x\) that makes \(f(x)\) equal to zero, i.e., \(x^2 - a = 0\). Adding \(a\) to both sides, we get \(x^2 = a\), which is the same as finding the square root of \(a\).

Step-by-step solution

(a) Explain the square root problem's equivalence to finding the positive root of \(f(x)=x^{2}-a\)

To find the square root of a number \(a\), we need to find a number \(x\) such that \(x^2 = a\). Now, let's consider the function \(f(x) = x^2 - a\). We can see that the positive root of this function is the value of \(x\) that makes \(f(x)\) equal to zero; in other words, \(x^2 - a = 0\). Adding \(a\) to both sides, we have \(x^2 = a\). So, we can see that finding the square root of \(a\) is equivalent to finding the positive root of \(f(x) = x^2 - a.\)

(b) Show that Newton's method applied to this function takes the Babylonian method's form

Newton's method is an iterative method for finding the roots of a function. The general formula for Newton's method is given by: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$$ In our case, \(f(x) = x^2 - a\). Let's calculate its derivative \(f'(x)\): $$f'(x) = 2x.$$ Now, let's apply Newton's method to the function \(f(x)\): $$x_{n+1} = x_n - \frac{x_n^2 - a}{2x_n}.$$ By simplifying the expression, we arrive at the Babylonian method's equation: $$x_{n+1} = \frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right).$$

(c) Choose initial approximations for approximating \(\sqrt{13}\) and \(\sqrt{73}\)

For finding good initial approximations, we usually take a number that is close to the square root of the given number. In this case, we can use the following approximations for \(\sqrt{13}\) and \(\sqrt{73}\): For \(\sqrt{13}\), we can see that \(3^2 = 9 < 13 < 16 = 4^2\), so a good initial approximation would be \(x_0 = 3.5\). For \(\sqrt{73}\), we can see that \(8^2 = 64 < 73 < 81 = 9^2\), so a good initial approximation would be \(x_0 = 8.5\).

(d) Approximate \(\sqrt{13}\) and \(\sqrt{73}\) with at least 10 significant digits

Using the Babylonian method and the initial approximations found in the previous step, we will iterate the method until obtaining the desired accuracy: For \(\sqrt{13}\): Let's start with the initial approximation \(x_0 = 3.5\). Now, we iterate the Babylonian method: 1. \(x_1 = \frac{1}{2}(3.5 + \frac{13}{3.5}) = 3.6071428571\) 2. \(x_2 = \frac{1}{2}(3.6071428571 + \frac{13}{3.6071428571}) = 3.6055514730\) 3. \(x_3 = \frac{1}{2}(3.6055514730 + \frac{13}{3.6055514730}) = 3.6055512755\) Since there is no observable difference after the 10th significant digit, we can conclude that \(\sqrt{13} \approx 3.6055512755\). For \(\sqrt{73}\): Let's start with the initial approximation \(x_0 = 8.5\). Now, we iterate the Babylonian method: 1. \(x_1 = \frac{1}{2}(8.5 + \frac{73}{8.5}) = 8.5441176471\) 2. \(x_2 = \frac{1}{2}(8.5441176471 + \frac{73}{8.5441176471}) = 8.5440037391\) 3. \(x_3 = \frac{1}{2}(8.5440037391 + \frac{73}{8.5440037391}) = 8.5440037453\) Since there is no observable difference after the 10th significant digit, we can conclude that \(\sqrt{73} \approx 8.5440037453\).

Key concepts

Approximating Square Roots

When it comes to understanding numbers, the concept of a square root is fundamental. But how do you find the square root of a non-perfect square without a calculator? This is when approximating square roots becomes a very useful skill.

Approximating a square root essentially means finding a value that, when squared, comes close to the number we started with. For the task of discovering what number multiplied by itself gives us a certain value (like \( \sqrt{a} \)), we utilize iterative methods, which gradually improve guesses until they reach a satisfactory level of accuracy.

The process starts with an initial 'guess' and systematically refines this guess by following a particular mathematical approach. Newton's method, a classic iterative method for root finding, applies beautifully here, especially considering that the square root of a number is actually the positive root of the function \( f(x) = x^2 - a \). By leveraging this method, we can hone in on the square root of any positive number, regardless of how complicated it may seem at first glance.

When we employ Newton’s method for this task, we follow specific steps that bring us closer to our value with each iteration. This underscores the power of iterative methods in mathematics - by following simple steps repeatedly, we can solve otherwise complex problems with great precision.

Iterative Methods for Root Finding

Iterative methods are at the heart of computational mathematics and allow us to tackle problems that, at first glance, might seem unsolvable. These methods include a broad family of algorithms designed to find numerical solutions to various types of equations, including polynomials, transcendental, and nonlinear systems.

Newton’s method, also known as the Newton-Raphson method, is perhaps one of the most powerful and widely-used algorithms for finding roots of real-valued functions. Beginning with a tentative guess, Newton's method refines this guess by applying a specific formula repeatedly.

The general iterative formula for Newton's method is given by \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) where \( f(x) \) is the function we're finding the root of, and \( f'(x) \) is its derivative. This process continues until the result stabilizes, within a specified degree of accuracy. In use for square roots, the method converges rapidly, which means very accurate results can be obtained in just a few steps, as demonstrated in the exercise above where \( \sqrt{13} \) and \( \sqrt{73} \) were approximated with high precision.

The Babylonian Method

One of the earliest known algorithms for approximating square roots hails from ancient Mesopotamia: the Babylonian method, also known as Heron's method. This technique is impressively efficient and has stood the test of time, being a special case of Newton’s method for finding square roots.

The Babylonian method of finding a square root works by making a guess at the square root, computing an average, and then repeating. The iterative formula reflecting this ancient wisdom is \( x_{n+1} = \frac{1}{2}\left(x_{n} + \frac{a}{x_{n}}\right) \), where \( a \) is the number we want to find the square root of, and \( x_n \) is our current guess.

This method aligns perfectly with what Newton’s method gives us for the function \( f(x) = x^2 - a \), which is another way of reinforcing how common mathematical truths can be discovered independently across different cultures and eras. It lays out a simple yet effective strategy for root approximation that can be executed with pen and paper, or programmed into a computer for finding roots with incredible accuracy, as seen with \( \sqrt{13} \) and \( \sqrt{73} \) in the given exercise.