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Chapter 4: Problem 6

Use Newton's method to find the negative fourth root of 2 by solving the equation \(x^{4}-2=0 .\) Start with \(x_{0}=-1\) and find \(x_{2}\) .

Short Answer

Expert verified
The approximate negative fourth root of 2 using Newton's method is \( x_2 \approx -0.794 \).

Step by step solution

01

Understand Newton's Method Formula

Newton's method is an iterative formula used to approximate roots of a function. It 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. We will use this formula to find the approximate fourth root of 2.
02

Define the Function and Its Derivative

In this case, the function is \( f(x) = x^4 - 2 \). The derivative of this function is \( f'(x) = 4x^3 \). We'll use these in Newton's formula.
03

Calculate First Iteration (x1)

Begin with the initial estimate \( x_0 = -1 \). Substitute it into Newton's formula:\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = -1 - \frac{(-1)^4 - 2}{4(-1)^3} = -1 - \frac{1 - 2}{-4} = -1 + \frac{1}{4} = -\frac{3}{4} \]Thus, the first iteration gives us \( x_1 = -\frac{3}{4} \).
04

Calculate Second Iteration (x2)

Use \( x_1 = -\frac{3}{4} \) for the next iteration. Substitute into the formula:\[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = -\frac{3}{4} - \frac{\left(-\frac{3}{4}\right)^4 - 2}{4\left(-\frac{3}{4}\right)^3} \]Calculate \( f(x_1) = \left(-\frac{3}{4}\right)^4 - 2 = \frac{81}{256} - 2 = -\frac{191}{256} \).Calculate \( f'(x_1) = 4\left(-\frac{3}{4}\right)^3 = -\frac{27}{16} \).Then,\[ x_2 = -\frac{3}{4} - \frac{-\frac{191}{256}}{-\frac{27}{16}} = -\frac{3}{4} + \frac{191}{256} \times \frac{16}{27} = -\frac{3}{4} + \frac{191}{432} \approx -0.793981481 \]Thus, the second iteration gives us \( x_2 \approx -0.793981481 \).

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Key Concepts

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

Iterative Formula
Newton's method relies on an iterative formula often used to find or approximate roots or zeros of a real-valued function. The idea is straightforward yet powerful:
  • The method starts with an initial guess, a point where you believe the root might be close.
  • It uses the function's behaviour at that point to adjust the guess and get closer to the actual root.
This is done by repeatedly applying Newton's iterative formula:\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]The formula helps refine the guess with each iteration. Here:
  • \( f(x) \) is the original function you are examining.
  • \( f'(x) \) is the derivative of that function.
The result, \( x_{n+1} \), is usually a better approximation of the root compared to \( x_n \). Keep iterating until the result is accurate enough!
This step-by-step improvement highlights why iteration is vital in converging on the correct solution.
Derivative Calculation
For Newton's method to work effectively, you need the derivative of the function, because it explains how the function grows or shrinks.
  • In the given exercise, the function is \( f(x) = x^4 - 2 \). It describes the relation of \( x \) to the equation whose root we are trying to find.
  • The derivative, \( f'(x) = 4x^3 \), indicates the function's slope or rate of change at any point \( x \).
Having the derivative enables the iterative formula to adjust the initial guess efficiently. Here's how it fits in: when you plug \( x_n \) into the derivative \( f'(x_n) \), you get the slope which helps determine how much to adjust \( x_n \).
This adjustment is key, allowing each new estimate to land closer to the root based on the change rates. Calculating the derivative correctly is fundamental for iteration.
Approximating Roots
Using Newton's method is all about approximating roots accurately by refining estimates over several iterations. The process is practical, especially when dealing with equations like our example, \( x^4 - 2 = 0 \).
  • Start with an initial approximation. We began with \( x_0 = -1 \).
  • Use the iterative formula and calculate the next approximations. For instance, \( x_1 \) becomes closer to the root, with \( x_1 = -\frac{3}{4} \).
  • Continue iterating for better precision, as shown with \( x_2 \approx -0.793981481 \), demonstrating that each round can dramatically refine the result.
These steps illustrate how Newton's method is capable of homing in on root values with high accuracy. It's particularly useful when exact algebraic solutions are difficult to determine.
Each iteration gets you a step closer to the truth, making this method invaluable for mathematicians and scientists.

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Most popular questions from this chapter

Distance between two ships At noon, ship \(A\) was 12 nautical miles due north of ship \(B .\) Ship \(A\) was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship \(B\) was sailing east at 8 knots and continued to do so all day. \begin{equation} \begin{array}{l}{\text { a. Start counting time with } t=0 \text { at noon and express the dis- }} \\ \quad {\text { tance } s \text { between the ships as a function of } t \text { . }} \\ {\text { b. How rapidly was the distance between the ships changing at }} \\ \quad {\text { noon? One hour later? }} \\\ {\text { c. The visibility that day was } 5 \text { nautical miles. Did the ships }} \\ \quad {\text { ever sight each other? }} \\ {\text { d. Graph } s \text { and } d s / d t \text { together as functions of } t \text { for }-1 \leq t \leq 3} \\ \quad {\text { using different colors if possible. Compare the graphs and }} \\ \quad {\text { reconcile what you see with your answers in parts (b) }} \\ \quad {\text { and (c). }} \\ {\text { e. The graph of } d s / d t \text { looks as if it might have a horizontal }} \\\ \quad {\text { asymptote in the first quadrant. This in turn suggests that }} \\\ \quad {d s / d t \text { approaches a limiting value as } t \rightarrow \infty . \text { What is this }} \\ \quad {\text { value? What is its relation to the ships' individual speeds? }} \end{array} \end{equation}

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