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Magazine Chapter 4: Problem 5
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. $$ f(x)=x^{2}+x-2, x_{0}=0 $$
Short Answer
Expert verified
Newton's Method approximates the root to be 1 after 5 iterations.
Step by step solution
01
Identify the function and its derivative
The given function is \( f(x) = x^2 + x - 2 \). To apply Newton's Method, we first find the derivative of this function. The derivative is calculated as \( f'(x) = 2x + 1 \).
02
Apply Newton's Method formula
Newton's Method formula is \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). We will use this formula iteratively for 5 iterations starting from the initial guess \( x_0 = 0 \).
03
Perform the first iteration
Substitute \( x_0 = 0 \) into the formula. \[ x_1 = 0 - \frac{f(0)}{f'(0)} = 0 - \frac{0^2 + 0 - 2}{2 \times 0 + 1} = 0 + 2 = 2 \]. Thus, \( x_1 = 2 \).
04
Perform the second iteration
Now use \( x_1 = 2 \) in Newton's formula. \[ x_2 = 2 - \frac{f(2)}{f'(2)} = 2 - \frac{2^2 + 2 - 2}{2 \times 2 + 1} = 2 - \frac{4 + 2 - 2}{5} = 2 - \frac{4}{5} = 1.2 \]. Thus, \( x_2 = 1.2 \).
05
Perform the third iteration
Use the value \( x_2 = 1.2 \). \[ x_3 = 1.2 - \frac{f(1.2)}{f'(1.2)} = 1.2 - \frac{1.2^2 + 1.2 - 2}{2 \times 1.2 + 1} = 1.2 - \frac{1.44 + 1.2 - 2}{3.4} = 1.2 - \frac{0.64}{3.4} \approx 1.0118 \]. Thus, \( x_3 \approx 1.0118 \).
06
Perform the fourth iteration
Use \( x_3 \approx 1.0118 \). \[ x_4 = 1.0118 - \frac{f(1.0118)}{f'(1.0118)} \]. Calculate \( f(1.0118) \approx 0.0358 \) and \( f'(1.0118) \approx 3.0236 \). Therefore, \[ x_4 \approx 1.0118 - \frac{0.0358}{3.0236} \approx 1.0002 \]. Thus, \( x_4 \approx 1.0002 \).
07
Perform the fifth iteration
Use \( x_4 \approx 1.0002 \). \[ x_5 = 1.0002 - \frac{f(1.0002)}{f'(1.0002)} \]. Calculate \( f(1.0002) \approx 0.0004 \) and \( f'(1.0002) \approx 3.0004 \). Therefore, \[ x_5 \approx 1.0002 - \frac{0.0004}{3.0004} \approx 1.0000 \]. Thus, \( x_5 \approx 1.0000 \).
08
Compare with the known root
The quadratic equation \( x^2 + x - 2 = 0 \) has roots \( x = 1 \) and \( x = -2 \). After 5 iterations, Newton's Method approximates one of the roots as \( x_5 \approx 1.0000 \), which closely aligns with the actual root \( x = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
iteration process
The iteration process is a core part of Newton's Method and many numerical methods. It involves repeating a series of steps to get closer to an answer. In the context of finding roots using Newton's Method, we start with an initial guess. This guess is denoted as \( x_0 \). Using a formula that incorporates this guess and the function's derivative, we compute a new approximation. This new point is \( x_1 \).
- We then take \( x_1 \) and go through the process again, which gives us \( x_2 \).
- This cycle repeats for a specified number of iterations, in this case, five times, continuing to use the result from the previous step as the starting point for the next one.
- The idea behind iteration is that with each step, the approximation becomes more accurate.
derivatives
Derivatives play a crucial role in Newton's Method. They describe how a function changes as its input changes, giving us a way to measure and utilize the slope of a function at any given point. In simpler terms, derivatives tell us how steep a graph is at different points, which is essential for the tangent line approach used in Newton's Method.
- The formula we use to find the derivative is based on the definition of a derivative: \( f'(x) \).
- For our function \( f(x) = x^2 + x - 2 \), the derivative \( f'(x) = 2x + 1 \) gives us the slope at any point \( x \).
- We then use this slope in the iteration process to help adjust our guesses toward the root.
root approximation
Root approximation is the process of finding a value that is close to the actual root of a function. In terms of mathematics, a root is the solution to the equation \( f(x) = 0 \). Sometimes, these roots can't be found by simple algebraic manipulation, and this is where methods like Newton's Method shine.
- We start with an educated guess, refine it through iteration, and get progressively closer to the actual root.
- Each iteration produces a better approximation of the root.
- The more iterations we perform, the closer we get to the actual root value.
quadratic equations
Quadratic equations are a key focus of Newton's Method exercises, especially because they often have solutions that can easily showcase the method's effectiveness. A quadratic equation is usually written in the form \( ax^2 + bx + c = 0 \).
- The example function from the problem \( f(x) = x^2 + x - 2 \) follows this structure.
- Generally, quadratic equations have two roots or solutions.
- These roots can be found using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
