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Chapter 4: Problem 3

Use Newton's method to estimate the two zeros of the function \(f(x)=x^{4}+x-3 .\) Start with \(x_{0}=-1\) for the left-hand zero and with \(x_{0}=1\) for the zero on the right. Then, in each case, find \(x_{2}\)

Short Answer

Expert verified
For the left-hand zero, \( x_2 \) is 3. For the right-hand zero, \( x_2 \) is approximately 1.121.

Step by step solution

01

Understanding Newton's Method

Newton's Method is used to find successively better approximations to the roots (or zeroes) of a real-valued function. It uses the formula: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) to iteratively find closer approximations.
02

Calculate the derivative of the function

Given the function \( f(x) = x^4 + x - 3 \), we need to find its derivative to use in Newton's formula. The derivative is \( f'(x) = 4x^3 + 1 \).
03

First Iteration for Left-hand Zero

Starting with \( x_0 = -1 \), calculate \( f(-1) \ = (-1)^4 + (-1) - 3 = 1 - 1 - 3 = -3 \ \) and \( f'(-1) = 4(-1)^3 + 1 = -4 + 1 = -3 \). Use Newton's formula: \( x_1 = -1 - \frac{-3}{-3} = 0 \).
04

Second Iteration for Left-hand Zero

Use \( x_1 = 0 \) to find \( x_2 \). Calculate \( f(0) = 0^4 + 0 - 3 = -3 \) and \( f'(0) = 4(0)^3 + 1 = 1 \). Use Newton's formula: \( x_2 = 0 - \frac{-3}{1} = 3 \).
05

First Iteration for Right-hand Zero

Starting with \( x_0 = 1 \), calculate \( f(1) = 1^4 + 1 - 3 = -1 \) and \( f'(1) = 4(1)^3 + 1 = 5 \). Use Newton's formula: \( x_1 = 1 - \frac{-1}{5} = 1.2 \).
06

Second Iteration for Right-hand Zero

Use \( x_1 = 1.2 \) to find \( x_2 \). Calculate \( f(1.2) = (1.2)^4 + 1.2 - 3 \approx 0.6896 \) and \( f'(1.2) = 4(1.2)^3 + 1 \approx 8.728 \). Use Newton's formula: \( x_2 = 1.2 - \frac{0.6896}{8.728} \approx 1.121 \).

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Key Concepts

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

Numerical Analysis
Numerical analysis is a branch of mathematics that deals with algorithms for solving mathematical problems numerically. This means instead of finding a solution through algebraic manipulation, numerical methods use approximations and iterations to find solutions as close as possible to the real answer.
A common application of numerical analysis is finding roots of functions, which means solving for x values that make the function equal zero.
  • Newton's Method is one such algorithm; it iteratively finds successively better approximations of roots.
  • This method is particularly useful when dealing with complex functions, which might be difficult to solve using direct algebraic methods.
Each step involves calculations using current estimates and improving on them based on the derivative of the function. By applying Newton鈥檚 Method multiple times, we converge closer to the actual root with each iteration.
Root-Finding Algorithms
Root-finding algorithms are techniques used for determining the zeros of a function, where the function crosses the x-axis. These points are important because they demonstrate where the function reaches a value of zero, often reflecting equilibrium in real-world applications.
Newton's Method is one of the more efficient algorithms for root-finding due to its rapid convergence under certain conditions.
  • The method starts with an initial guess, which is crucial for its success. A good initial guess leads to quicker convergence.
  • The formula for Newton's Method is: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
Here, \( x_n \) is the current approximation, and \( x_{n+1} \) is the next approximation. By repeatedly applying this formula, we use the tangent line at the current point to find a better guess, honing in on the true zero of the function.
Calculus
Calculus is the mathematical study of change and is integral to Newton's Method. It provides the toolkit for understanding how a function behaves, notably through derivatives that indicate how the function's rate of change or slope varies at any given point.
To apply Newton's Method, it is essential to compute the derivative of the function first.
  • In the exercise, the function \( f(x) = x^4 + x - 3 \) has a derivative \( f'(x) = 4x^3 + 1 \).
  • This derivative is crucial because it is used in the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).
By knowing the rate at which the function changes, we can adjust our guesses more accurately towards the root. Thus, calculus enables us to use derivatives to guide our iterative process, making each step more precise.

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Most popular questions from this chapter

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 \(v\) ?

Find the values of constants \(a, b,\) and \(c\) so that the graph of \(y=a x^{3}+b x^{2}+c x\) has a local maximum at \(x=3,\) local minimum at \(x=-1,\) and inflection point at \((1,11) .\)

Find the curve \(y=f(x)\) in the \(x y\) -plane that passes through the point \((9,4)\) and whose slope at each point is 3\(\sqrt{x} .\)

Find a positive number for which the sum of its reciprocal and four times its square is the smallest possible.

\begin{equation} \begin{array}{l}{\text { a. The function } y=\tan x+3 \cot x \text { has an absolute minimum }} \\ {\text { value on the interval } 0 < x < \pi / 2 . \text { Find it. }} \\ {\text { b. Graph the function and compare what you see with your }} \\ {\text { answer in part (a). }}\end{array} \end{equation}
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