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\(5-8\) Use Newton's method with the specified initial approximation \(x_{1}\) to find \(x_{3},\) the third approximation to the root of the given equation. (Give your answer to four decimal places.) \(x^{5}-x-1=0, \quad x_{1}=1\)

Short Answer

Expert verified
The third approximation \( x_3 \) is approximately 1.1785.

Step by step solution

01

Understanding Newton's Method

Newton's method is an iterative approach to find successively better approximations to the roots of a real-valued function. The 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 and \( f'(x) \) is its derivative.
02

Differentiating the Function

Given the function \( f(x) = x^5 - x - 1 \), we need to find its derivative. By applying basic differentiation rules, we get: \[ f'(x) = 5x^4 - 1 \]. This will be used in our Newton's formula for iteration.
03

Applying Newton's Method for x_2

Start with the initial approximation \( x_1 = 1 \). Substitute into Newton's formula to find \( x_2 \): \[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 1 - \frac{1^5 - 1 - 1}{5 \cdot 1^4 - 1} = 1 - \frac{-1}{4} = 1.25 \].
04

Applying Newton's Method for x_3

Now use \( x_2 = 1.25 \) to find \( x_3 \). Substitute \( x_2 \) into Newton's formula: \[ x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.25 - \frac{1.25^5 - 1.25 - 1}{5 \cdot 1.25^4 - 1} \]. Calculate the values: \[ f(1.25) \approx 0.80176, \quad f'(1.25) \approx 11.207, \] \[ x_3 = 1.25 - \frac{0.80176}{11.207} \approx 1.1785 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Iterative Methods
Iterative methods are techniques used to approximate solutions to complex mathematical problems through successive iterations. In simpler words, you start with an initial guess and make it better with each step until it is close enough to the actual solution. There are various iterative methods, but Newton's method is widely known for its efficiency in finding roots of real-valued functions.

This method involves repeatedly applying a formula that relies on both the function and its derivative. Every iteration results in a value closer to the actual root. This process continues until the difference between two successive values is negligibly small. This approach is particularly useful for complex equations where finding an exact solution might be challenging or impossible otherwise.

To sum up, iterative methods are excellent for solving problems iteratively by employing an initial guess and refining it through repetitive application of mathematical processes. This method is perfect in scenarios where problem-solving requires trial and adjustment over brute force calculations.
Root Approximation
Root approximation aims to find the zeroes, or roots, of a function. In mathematical terms, if you have a function like \( f(x) = 0 \), solving it means finding the value of \( x \) that makes the equation true. Newton's method excels in this domain by helping approximate these roots precisely and efficiently.

For example, when the equation is as complex as \( x^5 - x - 1 = 0 \), trying to solve it directly can be tiresome or overly complicated. Newton's method simplifies this by starting with an approximate value, known as the initial approximation, and refining it to get as close as possible to the actual root within desired decimal places. This principle is crucial in numerous scientific and engineering fields, where precision is vital.

Root approximation through iterative methods, like Newton's, allows us to deal with real-world issues effectively. It brings practicality to the solving of equations that arise in various industrial, scientific, and technological innovations.
Differentiation
Differentiation is a fundamental concept in calculus, dealing with the rate at which things change. When applied to functions, differentiation helps us understand how the output of the function changes as the inputs change, i.e., it helps find the slope of the tangent to the curve at any point.

In Newton's method, differentiation plays a pivotal role because it involves the derivative of the function you're dealing with. To use the method, you must differentiate the function to find \( f'(x) \), which acts as a guide for adjusting your current approximation towards the function's root.

For instance, with the function \( f(x) = x^5 - x - 1 \), differentiating gives \( f'(x) = 5x^4 - 1 \). This derivative is crucial because it influences how Newton's formula adjusts your guess. Differentiation not only enhances precision but also ensures stability in Newton's iterative method, making the approximation faster and more accurate. Ultimately, differentiation provides the essential mathematical toolkit needed for effective root approximation using iterative methods.

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