Problem 32 - Single Variable Calculus

Question

Approximating the \(k\) th Root of a Positive Number a. Apply Newton's method to the solution of the equation \(f(x)=x^{k}-A=0\) to show that an approximation of \(\sqrt[k]{A}\) can be found by using the iteration $$ x_{n+1}=\frac{1}{k}\left[(k-1) x_{n}+\frac{A}{x_{n}^{k-1}}\right] $$ b. Use this iteration to find \(\sqrt[10]{50}\) accurate to four decimal places.

Step-by-Step Solution

Answer

We derived the Newton's method iteration formula for approximating the k-th root of A as \(x_{n+1} = \frac{1}{k}\left[(k-1) x_{n}+\frac{A}{x_{n}^{k-1}}\right]\). Using this formula, we found the 10th root of 50, accurate to four decimal places, is approximately 1.9608.

Step 1: Understanding Newton's method

Newton's method is used to find the root of a function f(x) iteratively. It starts from an initial guess x₀ and refines the guess by following iterations: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) Here, we want to find the k-th root of A such that f(x) = \(x^k -A\). Our task is to obtain the Newton's method iteration formula for this specific function.

Step 2: Calculate the derivative of f(x)

First, we need to find the derivative of \(f(x) = x^k - A\). It can be easily obtained using basic differential calculus rules: \[ f'(x) = \frac{d}{dx} (x^k - A) = kx^{k-1} \]

Step 3: Substitute f(x) and f'(x) into Newton's method

Now, we can substitute \(f(x)\) and \(f'(x)\) into the Newton's method formula: \(x_{n+1} = x_n - \frac{x_n^k - A}{kx_n^{k-1}}\) We can simplify the expression above as: \(x_{n+1} = \frac{1}{k}\left[(k-1) x_{n}+\frac{A}{x_{n}^{k-1}}\right]\) Now we have derived the Newton's method iteration formula for approximating the k-th root of A.

Step 4: Use the iteration formula to find the 10th root of 50

Let's now use the derived iteration formula to find the 10th root of 50 (\(k=10\), \(A=50\)), accurate to four decimal places. We will use an initial guess of \(x_0 = 1\). Repeat the iteration until the difference between consecutive guesses is less than 0.0001: 1. \(x_1 = \frac{1}{10}\left[(10-1) 1+\frac{50}{1^{10-1}}\right] = 5.4\) 2. \(x_2 = \frac{1}{10}\left[(10-1) 5.4+\frac{50}{5.4^{9}}\right] \approx 2.2667\) 3. \(x_3 = \frac{1}{10}\left[(10-1) 2.2667+\frac{50}{2.2667^{9}}\right] \approx 1.9362\) 4. \(x_4 = \frac{1}{10}\left[(10-1) 1.9362+\frac{50}{1.9362^{9}}\right] \approx 1.9606\) 5. \(x_5 = \frac{1}{10}\left[(10-1) 1.9606+\frac{50}{1.9606^{9}}\right] \approx 1.9608\) The difference between \(x_4\) and \(x_5\) is less than 0.0001, so we can stop the iteration here. Thus, the 10th root of 50, accurate to four decimal places, is approximately 1.9608.

Key Concepts

k-th Root Approximation

Determining the k-th root of a number is an important task in mathematics, where you want to find a number that, when raised to the power of k, equals the given number. For instance, finding the 10th root of 50 means looking for a number which, when raised to the power of 10, equals 50. This process is a pivotal aspect of various mathematical fields, including scientific calculations and engineering.

The challenge is often how to approximate this root, especially when the exact root is not an integer or a simple fraction. Newton's method provides an elegant solution to find these approximations by refining an initial guess iteratively until a satisfactory level of accuracy is achieved. By setting our function as \( f(x) = x^k - A \) where \( A \) is the given number, we can apply Newton's method to derive a powerful formula for root approximation. This approach transforms the complex problem of taking roots into a series of simpler arithmetic operations.

Iteration Formula

In mathematical terms, an iteration formula is a recursive method used to generate successive approximations to a solution. Specifically for Newton's method, the iteration formula helps refine the guess of a root by considering both the function and its derivative.

For the k-th root approximation using Newton's method, the specific iteration formula derived is:

• \( x_{n+1} = \frac{1}{k}\left[(k-1) x_{n} + \frac{A}{x_{n}^{k-1}}\right] \)

This formula kicks off with an initial guess and keeps refining that guess. The recursive aspect means each new term \( x_{n+1} \) is reliant on the previous term \( x_n \). As you repeat the process, you edge closer to the actual root. The beauty of this method lies in its efficiency and accuracy, allowing for precise approximations by repeated application of simple mathematical operations.

Differential Calculus

Differential calculus plays a crucial role in deriving the iteration formula for Newton's method. In calculus, the derivative tells us the rate at which a function is changing at any given point. It's essentially the mathematical tool for understanding how a function behaves as it changes.

To derive the iteration formula for the k-th root approximation, we first consider the function \( f(x) = x^k - A \). Applying the rules of differential calculus, the derivative is calculated as:

• \( f'(x) = kx^{k-1} \)

This derivative \( f'(x) \) represents how the function \( f(x) \) changes as \( x \) changes, and it is key to constructing the Newton's method iteration. By incorporating \( f'(x) \) into the iteration process, we create a pathway to fine-tune our approximations more effectively. Understanding this relationship is vital in both theoretical and real-world applications of calculus.

Numerical Analysis

Numerical analysis is the branch of mathematics that focuses on finding approximate solutions to complex mathematical problems using computational algorithms. It is indispensable in situations where exact solutions are hard or impossible to obtain.

The use of Newton's method for the k-th root approximations is a classic example of numerical analysis in action. The method provides a way to tackle problems by iteratively approaching the true value rather than aiming for direct computation. This approach is not only practical but often necessary when dealing with real numbers and complex functions. By selecting a reasonable initial guess and iteratively updating it, numerical methods bring solutions within grasp.

The elegance of numerical analysis lies in allowing us to solve problems that are otherwise impractical or impossible to handle analytically, transforming mathematical problems into computationally solvable tasks. This approach is widely used in engineering, physics, and computer science for modeling and simulations.