/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Estimate the value of the radica... [FREE SOLUTION] | 91Ó°ÊÓ

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Estimate the value of the radical accurate to four decimal places by using three iterations of Newton's method to solve the equation \(f(x)=0\) with initial estimate \(x_{0}\). \sqrt{6} ; \quad f(x)=x^{2}-6 ; x_{0}=2.5

Short Answer

Expert verified
The estimate of \(\sqrt{6}\) using three iterations of Newton's method with initial estimate \(x_0 = 2.5\) and function \(f(x) = x^2 - 6\) is approximately \(2.2358\).

Step by step solution

01

Function and its Derivative

Given function: \(f(x) = x^2 - 6\) Derivative of the function: \(f'(x) = 2x\)
02

First iteration of Newton's Method

Newton's method equation: \(x_{1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})}\) With the initial estimate \(x_0 = 2.5\), we have \(x_{1} = 2.5 - \frac{(2.5)^2 - 6}{2(2.5)} \) Calculate the value of \(x_1\): \(x_{1} = 2.5 - \frac{1.25}{5}\) \(x_{1} = 2.5 - 0.25\) \(x_{1} = 2.25\)
03

Second iteration of Newton's Method

Using the updated value for x: \(x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})}\) Plug in the value of \(x_1 = 2.25\): \(x_{2} = 2.25 - \frac{(2.25)^2 - 6}{2(2.25)} \) Calculate the value of \(x_2\): \(x_{2} = 2.25 - \frac{0.0625}{4.5}\) \(x_{2} = 2.25 - 0.01389\) \(x_{2} = 2.23611\)
04

Third iteration of Newton's Method

Using the updated value for x: \(x_{3} = x_{2} - \frac{f(x_{2})}{f'(x_{2})}\) Plug in the value of \(x_2 = 2.23611\): \(x_{3} = 2.23611 - \frac{(2.23611)^2 - 6}{2(2.23611)} \) Calculate the value of \(x_3\): \(x_{3} = 2.23611 - \frac{0.00150}{4.47222}\) \(x_{3} = 2.23611 - 0.000336\) \(x_{3} = 2.235774\)
05

Round the final estimate to four decimal places

The final estimate of the square root of 6 after three iterations of Newton's method is: \(x_3 = 2.2358\) (rounded to four decimal places) So, the estimate of \(\sqrt{6}\) using Newton's method is approximately \(2.2358\).

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Key Concepts

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

Numerical Methods
Numerical methods are algorithms used for calculating approximate solutions to mathematical problems that may not have exact answers or are difficult to compute with analytical methods. For example, finding the square roots of numbers, solving differential equations, and evaluating integrals often rely on numerical techniques because they involve continuous processes that are not always easy to express with a simple formula.

These techniques are essential in fields like engineering, physics, and finance where precise calculations are necessary for design, simulations, and predictions. Newton's Method, as illustrated in our exercise, is a prime example of a numerical method applied to find the root of a function - that is, the solution to an equation of the form \( f(x) = 0 \).
Root Approximation
Root approximation is the process of finding an estimate for the root, or solution, of an equation. It's very common in calculus to encounter equations that don't have simple algebraic solutions. Numerical methods, including Newton's Method, provide a way to progressively approximate the root with increasing accuracy.

The goal is to start with an initial guess and refine this guess iteratively until a sufficiently accurate value is found. In our exercise, we're approximating the square root of 6 to four decimal places, which means we're looking for a number that, when squared, comes very close to 6. The process involves using an iterative algorithm that progressively hones in on the true value of the root.
Calculus
Calculus is the mathematical study of change and motion. It's divided into two main branches - differential calculus and integral calculus. Differential calculus deals with the concept of a derivative, which measures how a function changes as its input changes.

In the context of our problem, we use calculus to both define the function whose root we are approximating (\( f(x) = x^2 - 6 \)) and to find its derivative (\( f'(x) = 2x \)). The derivative is a crucial part of Newton's Method because it provides the slope of the tangent line to the function at any given point, which we need for the formula used in each iteration.
Derivative
The derivative of a function at a point is the rate at which the value of the function is changing at that point. It's the fundamental tool of differential calculus. When we interpret the derivative geometrically, it represents the slope of the tangent line to the function's graph at a specific point.

In Newton's Method, we use the function's derivative to help find where the function crosses the x-axis, which is the root of the function. The formula for Newton's Method uses the derivative to adjust successive guesses as seen in the given solution where the derivative \( f'(x) = 2x \) helps us to revise our estimate for \( \(x^2 - 6 = 0 \) \) closer to the actual square root of 6.
Iterative Algorithms
Iterative algorithms are step-by-step procedures that repeat a series of computations to get closer to a desired result. Each repetition, known as an iteration, builds upon the previous one to improve the accuracy of the solution. Iterative methods are common in algorithms that require approximation, optimization, or when direct methods would be too computationally expensive.

Newton's Method is based on an iterative algorithm; it begins with an initial guess and then refines this guess repeatedly. The key feature of such an algorithm is the feedback loop where the result of one iteration informs the next, incorporating the function and its derivative. This concept is demonstrated in the given exercise where we saw how each new approximation of \( \(x_{n+1} \) \) gets you closer to the actual square root value with every iteration.

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Most popular questions from this chapter

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