/*! 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 2 Use Newton's method to estimate ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use Newton's method to estimate the one real solution of \(x^{3}+3 x+1=0 .\) Start with \(x_{0}=0\) and then find \(x_{2}\) .

Short Answer

Expert verified
\( x_2 = -\frac{29}{90} \) after following two iterations with Newton's method.

Step by step solution

01

Define the Functions

Newton's method requires you to define the function and its derivative. Here, the function is given as \[ f(x) = x^3 + 3x + 1 \] and its derivative is \[ f'(x) = 3x^2 + 3 \] .
02

Initial Guess

The initial guess is provided as \( x_0 = 0 \). This is the first approximation of the root.
03

Apply Newton's Method Formula for First Iteration

Apply Newton's method: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] For \( x_1 \): Substitute \( x_0 = 0 \) into the function and its derivative: \[ f(0) = 0^3 + 3(0) + 1 = 1 \] \[ f'(0) = 3(0)^2 + 3 = 3 \]. Compute \( x_1 \): \[ x_1 = 0 - \frac{1}{3} = -\frac{1}{3} \] .
04

Apply Newton's Method Formula for Second Iteration

Use the result from the first iteration: \( x_1 = -\frac{1}{3} \). First, we calculate the function and derivative: \[ f\left(-\frac{1}{3}\right) = \left( -\frac{1}{3} \right)^3 + 3\left(-\frac{1}{3}\right) + 1 = -\frac{1}{27} - 1 + 1 = -\frac{1}{27} \] \[ f'\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right)^2 + 3 = \frac{1}{3} + 3 = \frac{10}{3} \]. Finally, compute \( x_2 \): \[ x_2 = -\frac{1}{3} - \frac{-\frac{1}{27}}{\frac{10}{3}} = -\frac{1}{3} + \frac{1}{27} \cdot \frac{3}{10} \] \[ x_2 = -\frac{1}{3} + \frac{1}{90} = -\frac{30}{90} + \frac{1}{90} = -\frac{29}{90} \].
05

Simplify and Confirm Second Iteration Result

Simplifying \(-\frac{29}{90}\) gives the second approximation. Ensure that each arithmetic operation in the formula was correctly applied, leading to the result: \( x_2 = -\frac{29}{90} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Numerical Analysis
Numerical analysis is an essential field in mathematics that focuses on finding approximate solutions to mathematical problems. These problems might be complex, and finding an exact solution can be challenging or even impossible. By using numerical methods, we can arrive at solutions that are close enough for practical purposes. These numerical solutions allow us to understand and predict how systems behave.
In our example, Newton's method is a technique used within numerical analysis for finding the roots of a real-valued function. It specifically targets problems that arise in various fields, from engineering to physics. Since exact solutions can be challenging to obtain, numerical analysis offers a bridge to practical results.
  • Efficient for approximating complex equations.
  • Provides insight into system behavior without exact solutions.
  • Bridges theoretical concepts with practical applications.
Understanding numerical analysis means recognizing the usefulness of approximations in real-world scenarios. Newton's method exemplifies how numerical analysis can apply this to root finding, making it a cornerstone of computational mathematics.
Root Finding Algorithms
Root finding algorithms are techniques used to find values of a variable that satisfy an equation, typically where the function equals zero. These algorithms are integral in computational problem-solving and are widely used in both pure and applied mathematics.
Among these algorithms, Newton’s method stands out due to its efficiency and speed in converging to a solution. It is part of a broader family of methods but specifically works well under the condition that the function is well-behaved, locally.
  • Effective for continuous functions.
  • Rapid convergence makes it popular in iterative methods.
  • Requires the function to be differentiable.
In the exercise provided, Newton's method is employed to find the root of the equation: \[ f(x) = x^3 + 3x + 1 = 0 \] By iteratively adjusting the approximation based on the value of the function and its derivative, Newton's method refines the solution towards the true root. Understanding its use helps solve complex equations efficiently.
Iterative Methods
Iterative methods are approaches to find solutions by repeatedly refining approximations. They are central to computational mathematics and play a crucial role in scenarios where direct solutions are not feasible, or the system is too large.
In essence, iterative methods continually update an estimated solution until it meets a specific criterion for accuracy or reaches the desired level of precision.
  • Works incrementally to improve results.
  • Ideal for solving large, complex systems.
  • Simple iterations can lead to profound accuracy.
With Newton's method, each iteration involves recalculating the function and its derivative at the current approximation. In the exercise: - Start with an initial guess, - Apply the formula repeatedly. Each cycle brings us closer to the solution. The steps are straightforward, leveraging past approximations to hone in on the root. By understanding iterative methods, we gain practical tools that turn abstract mathematical problems into solvable puzzles. Newton's method, in particular, highlights how powerful these iterative techniques can be, especially in root finding scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The derivative \(d t / d x\) in Example 4 a. Show that $$f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}$$ is an increasing function of \(x .\) b. Show that $$g(x)=\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}$$ is a decreasing function of \(x .\) c. Show that $$\frac{d t}{d x}=\frac{x}{c_{1} \sqrt{a^{2}+x^{2}}}-\frac{d-x}{c_{2} \sqrt{b^{2}+(d-x)^{2}}}$$ is an increasing function of \(x .\)

Use a CAS to solve the initial value problems in Exercises \(107-110\) . Plot the solution curves. $$y^{\prime}=\frac{1}{x}+x, \quad y(1)=-1$$

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x-\sin x, \quad 0 \leq x \leq 2 \pi$$

Suppose that \(0 < f^{\prime}(x) < 1 / 2\) for all \(x\) -values. Show that \(f(-1) < \) \(f(1) < 2+f(-1)\)

Tin pest When metallic tin is kept below \(13.2^{\circ} \mathrm{C},\) it slowly becomes brittle and crumbles to a gray powder. Tin objects eventually crumble to this gray powder spontaneously if kept in a cold climate for years. The Europeans who saw tin organ pipes in their churches crumble away years ago called the change tin pest because it seemed to be contagious, and indeed it was, for the gray powder is a catalyst for its own formation. A catalyst for a chemical reaction is a substance that controls the rate of reaction without undergoing any permanent change in itself. An autocatalytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at first if the amount of catalyst is small and slowly again at the end, when most of the original substance is used up. But in between, when both the substance and its catalyst product are abundant, the reaction proceeds at a faster pace. In some cases, it is reasonable to assume that the rate \(v=d x / d t\) of the reaction is proportional both to the amount of the original substance present and to the amount of product. That is, \(v\) may be considered to be a function of \(x\) alone, and $$v=k x(a-x)=k a x-k x^{2},$$ where $$\begin{aligned} x &=\text { the amount of product } \\ a &=\text { the amount of substance at the beginning } \\ k &=\text { a positive constant. } \end{aligned}$$ At what value of \(x\) does the rate \(v\) have a maximum? What is the maximum value of \(v ?\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.