/*! 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
\( x_2 \approx -\frac{9}{28} \)

Step by step solution

01

Write the Function and Derivative

The function we want to find the root for is given by \( f(x) = x^3 + 3x + 1 \). To use Newton's method, we also need the derivative, which is \( f'(x) = 3x^2 + 3 \).
02

Apply Newton's Method Formula

Newton's method uses the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) to find successively better approximations of the root. We'll use this formula to find \( x_1 \) from \( x_0 = 0 \).
03

Calculate x_1

Compute the values of the function and its derivative at \( x_0 = 0 \): \( f(0) = 0^3 + 3(0) + 1 = 1 \) and \( f'(0) = 3(0)^2 + 3 = 3 \). Substituting into Newton's formula, we get \( x_1 = 0 - \frac{1}{3} = -\frac{1}{3} \).
04

Calculate x_2

Using the value \( x_1 = -\frac{1}{3} \), compute the function and derivative values: \( f\left(-\frac{1}{3}\right) = \left(-\frac{1}{3}\right)^3 + 3\left(-\frac{1}{3}\right) + 1 = -\frac{1}{27} - 1 + 1 = -\frac{1}{27} \) and \( f'\left(-\frac{1}{3}\right) = 3\left(\frac{1}{9}\right) + 3 = 3\frac{1}{9} + 3 = 3\frac{10}{9} =\frac{28}{9} \). Using Newton's formula again, \( x_2 = -\frac{1}{3} - \frac{-\frac{1}{27}}{\frac{28}{9}} = -\frac{1}{3} + \frac{1}{27} \times \frac{9}{28} = -\frac{1}{3} + \frac{1}{84} = -\frac{28}{84} + \frac{1}{84} = -\frac{27}{84} = -\frac{9}{28}. \)

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.

Root-finding algorithms
Root-finding algorithms are mathematical methods used to determine where a function equals zero, symbolically represented as finding the 'root' of an equation. These algorithms are crucial for solving equations that may not be easily simplified or solved by algebraic manipulation. One such algorithm is Newton's method, a powerful tool for estimating roots of differentiable functions.

Newton's method starts with an initial guess (known as an approximation) and iteratively refines this guess to get closer to the actual root. It does so by leveraging the function's derivative to calculate the tangent line at a given point. The place where this tangent line crosses the x-axis provides a new, better approximation of the root.

One of the advantages of Newton’s method is its speed of convergence; it can quickly hone in on a root compared to other methods like bisection or secant methods. However, it requires a good initial guess and the function's derivative, limiting its use in certain scenarios.
Derivatives in calculus
Derivatives are a fundamental concept in calculus that represent the rate of change of a function with respect to one of its variables. In essence, a derivative gives us the slope of the function at any given point. This is especially useful when working with Newton’s method, which relies on derivatives to find function zeroes.

For a given function, say \( f(x) = x^3 + 3x + 1 \), the derivative, \( f'(x) = 3x^2 + 3 \), provides the slope of the tangent to the curve at any point \( x \). This tangent line helps us estimate where the function crosses the x-axis, moving closer to the root with each iteration.

Calculating the derivative correctly is essential for the success of Newton’s method. It not only defines the tangent, but it also influences how quickly and accurately the method converges to a root. Thus, understanding and computing derivatives is a skill every student needs to master for effective application of calculus techniques.
Approximate solutions
Approximate solutions are estimates of solutions that aren’t exact but close enough to give meaningful insight or practical utility. In mathematical contexts, particularly numerical methods like Newton’s method, approximations allow us to work with complex equations that might be challenging or even impossible to solve analytically.

The process of approximating begins with an initial guess, which can be refined over iterations to approach the real solution. For example, starting with \( x_0 = 0 \), Newton's method yields subsequent approximations. First, calculating \( x_1 = -\frac{1}{3} \), and then further refining to \( x_2 = -\frac{9}{28} \). Each step integrates the function and derivative values to move closer to the actual root of the equation \( x^3 + 3x + 1 = 0 \).

The accuracy of approximations depends on the number of iterations performed and the characteristics of the function itself. Often, three to five iterations suffice to achieve a desirable level of accuracy in practical applications.
Polynomial equations
Polynomial equations are algebraic expressions containing variables raised to whole number exponents. A common form is \( ax^n + bx^{n-1} + ... + k = 0 \). Solving these equations often involves finding roots where the polynomial equals zero. This is fundamental in fields like physics, engineering, and economics.

The given example, \( x^3 + 3x + 1 = 0 \), is a cubic polynomial due to the variable raised to the power of three. Polynomial equations like these can have multiple roots, real or complex, and determining them exactly isn't always feasible using simple algebraic techniques. This is where methods like Newton's method become extremely useful.

These equations play a vital role in mathematical modeling of real-world scenarios. Understanding and solving polynomial equations is crucial for applying mathematical theory to practice. Whether it's optimizing processes or predicting outcomes, the ability to tackle these equations provides clarity and solutions to complex problems.

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

Constructing cones \(A\) right triangle whose hypotenuse is \(\sqrt{3} \mathrm{m}\) long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.

Solve the initial value problems in Exercises \(71-90\) . $$\frac{d^{2} y}{d x^{2}}=2-6 x ; \quad y^{\prime}(0)=4, \quad y(0)=1$$

Right, or wrong? Say which for each formula and give a brief reason for each answer. \begin{equation}\begin{array}{l}{\text { a. }{\int x \sin x d x=\frac{x^{2}}{2} \sin x+C}} \\ {\text { b. } \int x \sin x d x=-x \cos x+C} \\\ {\text { c. } \int x \sin x d x=-x \cos x+\sin x+C}\end{array}\end{equation}

Tin pest When metallic tin is kept below \(13.2^{\circ} \mathrm{C},\) it slowly becomes britle 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 present 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 \begin{equation}v=k x(a-x)=k a x-k x^{2}\end{equation} where \begin{equation} \begin{aligned} x &=\text { the amount of product } \\ a &=\text { the amount of substance at the beginning } \\ k &=\text { a positive constant. } \end{aligned} \end{equation} At what value of \(x\) does the rate \(v\) have a maximum? What is the maximum value of \(v\) ?

Solve the initial value problems in Exercises \(71-90\) . $$\frac{d s}{d t}=\cos t+\sin t, \quad s(\pi)=1$$

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.