Question

The roots of \(f(x)\) are known or are easily found. Use 5 iterations of Newton's Method with the given initial approximation to approximate the root. Compare it to the known value of the root. $$ f(x)=\sin x, x_{0}=1 $$

Short answer

The approximated root is \( x \approx 0 \) after 5 iterations, which matches the known root.

Step-by-step solution

Understand the Problem

We need to approximate the root of the equation \( f(x) = \sin x \) using Newton's Method, starting with an initial approximation \( x_0 = 1 \). We will perform 5 iterations and compare the result with the known root, which is \( x = 0 \).

Recall Newton's Method Formula

Newton's Method is an iterative formula given by \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). We need to identify \( f(x) \) and its derivative \( f'(x) \) for our function.

Calculate Derivative of \( f(x) \)

The function given is \( f(x) = \sin x \). The derivative of this function is \( f'(x) = \cos x \).

Apply Newton's Method Iterations

Starting with \( x_0 = 1 \):- **Iteration 1:** \[ x_1 = x_0 - \frac{\sin x_0}{\cos x_0} = 1 - \frac{\sin(1)}{\cos(1)} \approx 1 - 1.557 \approx -0.557 \]- **Iteration 2:** \[ x_2 = x_1 - \frac{\sin x_1}{\cos x_1} \approx -0.557 - \frac{\sin(-0.557)}{\cos(-0.557)} \approx -0.557 + 0.612 \approx 0.055 \]- **Iteration 3:** \[ x_3 = x_2 - \frac{\sin x_2}{\cos x_2} \approx 0.055 - \frac{\sin(0.055)}{\cos(0.055)} \approx 0.055 - 0.055 \approx 0.000 \]- **Iteration 4:** \[ x_4 = x_3 - \frac{\sin x_3}{\cos x_3} \approx 0.000 \] (No significant change)- **Iteration 5:** \[ x_5 = x_4 - \frac{\sin x_4}{\cos x_4} \approx 0.000 \] (Still no significant change)

Compare with Known Root

The known root of \( f(x) = \sin x \) is \( x = 0 \). After 5 iterations of Newton's Method, our approximation \( x_5 \approx 0 \), which matches the known root very closely.

Key concepts

Calculus

Calculus is a branch of mathematics that studies changes between values that are related by a function. It serves as the foundation for understanding complex concepts like limits, derivatives, and integrals. In the context of Newton's Method, calculus is crucial because the method relies on the use of derivatives. A derivative represents the rate of change of a function and helps find the slope of the tangent line to the function's graph at a given point.

To utilize Newton's Method effectively, we first need to find the derivative of the function for which we are trying to find the root. This derivative plays a critical role in the iterative formula of Newton's Method, guiding the adjustment of approximations closer to the actual root with each iteration.

In our exercise, we start with the function \( f(x) = \sin x \). Its derivative, \( f'(x) = \cos x \), tells us how the sine function is changing at any point along its curve. Understanding these changes allows us to apply Newton's Method and approach the root more precisely.

Numerical Analysis

Numerical analysis is a field of mathematics that focuses on algorithms to find approximate solutions to a wide variety of mathematical problems. When it comes to finding roots of equations, numerical analysis provides us with techniques like Newton's Method to obtain approximate solutions for complex equations that may not have a simple algebraic solution.

In practical scenarios, especially with functions that are difficult or impossible to solve analytically, numerical methods are vital. They give us the tools to calculate approximate solutions with a high degree of accuracy. These types of methods, like Newton's Method, are versatile and can be implemented in various software used for scientific and engineering computations.

Our given problem involves using Newton's Method to approximate the root of the equation \( f(x) = \sin x \). By applying this iterative method, we can comfortably find a precise estimate of the root, even when the exact analytical solution is not directly accessible.

Root Approximation

Root approximation is all about finding an approximate value for the roots of a function where it equals zero. In mathematics, a root is a solution to the equation \( f(x) = 0 \). Approximating these values accurately is essential because exact solutions are not always obtainable, especially for complex functions.

Newton's Method is a widely used technique for root approximation. It is particularly known for its efficiency and rapid convergence to an actual root. The method starts with an initial guess and repeatedly improves the approximation by calculating the slope of the tangent to the curve until the changes become insignificant.

In our exercise, we used an initial approximation \( x_0 = 1 \) and applied Newton's Method through five iterations. This process allowed our estimate to converge to the known root \( x = 0 \) of \( f(x) = \sin x \), showcasing the effectiveness of the technique in approximating the root accurately.

Iterative Methods

Iterative methods are mathematical procedures that generate a sequence of improving approximate solutions for a problem. These methods are used when a direct solution is difficult or impossible to obtain. Instead of finding an exact answer in one step, iterative methods refine approximate answers over time, usually through a repetitive process.

Newton's Method is a prime example of an iterative method. Each iteration uses the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) to compute a new approximation, \( x_{n+1} \), from the current one, \( x_n \).

• Start with an initial guess.
• Use the formula to find a better approximation.
• Repeat until the approximation is sufficiently accurate.

In the provided exercise, you can see the power of Newton's Method as it takes us from an initial approximation \( x_0 = 1 \) to a refined result \( x_5 \approx 0 \) through a series of five iterations. The method effectively reduced the error in each step, aligning closer and closer with the exact root of the function.