Problem 4 - Thomas Calculus

Question

Use Newton's method to estimate the two zeros of the function \(f(x)=2 x-x^{2}+1 .\) Start with \(x_{0}=0\) for the left-hand zero and with \(x_{0}=2\) for the zero on the right. Then, in each case, find \(x_{2}\) .

Step-by-Step Solution

Answer

The approximations for \(x_2\) are -0.5833 and 2.5833.

Step 1: Define Newton's Method Formula

Newton's Method is an iterative process for approximating the roots (or zeros) of a real-valued function. The formula for the iteration is given by \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). Here, \( f(x) = 2x - x^2 + 1 \).

Step 2: Compute the Derivative

To apply Newton's Method, we need the derivative of \( f(x) \). Compute this derivative: \( f'(x) = 2 - 2x \).

Step 3: First Zero Approximation (Starting Point \(x_0 = 0\))

With the starting point \(x_0 = 0\), compute the next iteration using Newton's formula: - Evaluate \( f(x_0) = f(0) = 2 \times 0 - 0^2 + 1 = 1 \).- Evaluate \( f'(x_0) = f'(0) = 2 \times 1 = 2 \).- Compute \( x_1 = 0 - \frac{1}{2} = -0.5 \).

Step 4: Continue to Find \(x_2\) for the First Zero

Using \(x_1 = -0.5\), perform another iteration:- Evaluate \( f(x_1) = f(-0.5) = 2 \times (-0.5) - (-0.5)^2 + 1 = 0.25 \).- Evaluate \( f'(x_1) = f'(-0.5) = 2 + 1 = 3 \).- Compute \( x_2 = -0.5 - \frac{0.25}{3} \approx -0.5833 \).

Step 5: Second Zero Approximation (Starting Point \(x_0 = 2\))

With the starting point \(x_0 = 2\), compute the next iteration using Newton's formula:- Evaluate \( f(x_0) = f(2) = 2 \times 2 - 2^2 + 1 = 1 \).- Evaluate \( f'(x_0) = f'(2) = 2 - 4 = -2 \).- Compute \( x_1 = 2 - \frac{1}{-2} = 2.5 \).

Step 6: Continue to Find \(x_2\) for the Second Zero

Using \(x_1 = 2.5\), perform another iteration:- Evaluate \( f(x_1) = f(2.5) = 2 \times 2.5 - (2.5)^2 + 1 = 0.25 \).- Evaluate \( f'(x_1) = f'(2.5) = 2 - 5 = -3 \).- Compute \( x_2 = 2.5 - \frac{0.25}{-3} \approx 2.5833 \).

Key Concepts

Function Zeros

In mathematics, finding the zeros of a function is a fundamental concept. A zero of a function, also known as a root, is a value of the input variable (like \( x \)) for which the function evaluates to zero. For the function \( f(x) = 2x - x^2 + 1 \), a zero would be any \( x \) such that \( f(x) = 0 \). Finding the zeros helps in understanding the behavior of functions and is crucial for various applications in calculus and algebra. When dealing with polynomial functions like the one given here, zeros are the points where the graph of the function intersects the x-axis. Identifying these points gives insights into the function's structure, its increasing or decreasing nature, and the regions of interest in its graph.

Iterative Process

Newton's Method is an example of an iterative process, a technique used in numerical analysis to approach the desired solution through repeated approximations. The process begins with an initial guess, often denoted as \( x_0 \), which should be as close as possible to the actual root for better efficiency. In each iteration, the guess is revised using the formula:\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]This formula helps in converging towards the actual root if the function behaves well in the vicinity of the initial guess. Iterative processes rely on repeating this step until the result meets the desired level of accuracy, measured by how small the difference between consecutive approximations becomes.

Derivative Computation

The derivative is a critical component in Newton's Method as it provides the slope of the tangent line at a given point on the function. For the function \( f(x) = 2x - x^2 + 1 \), we compute its derivative as \( f'(x) = 2 - 2x \). Without the derivative, it would not be possible to calculate the tangent line, which is necessary for projecting the next guess for the root approximation. The derivative essentially guides the corrections on each iterative step, ensuring the process heads in the direction of a root. Accurate computation of the derivative is therefore essential in applying Newton's Method correctly.

Root Approximation

Root approximation in Newton's Method involves using the iterative formula to get continually closer to the function's zeros. Starting with a chosen \( x_0 \), the root approximation involves calculating subsequent points like \( x_1 \), \( x_2 \), and so on, until the change between successive approximations is negligible, signifying convergence. For example, starting at \( x_0 = 0 \) for our function, the next approximation \( x_1 \) is calculated as \( -0.5 \), and then \( x_2 \) as approximately \( -0.5833 \). This sequence gets us closer to the actual zero of the function each time. Precision is key, and maintaining accuracy with each step ensures the approximation converges efficiently to the true root.