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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
\(x_2 \approx -0.7546\).

Step by step solution

01

Understand the Problem

We need to solve the equation \(x^4 - 2 = 0\) using Newton's method, which is an iterative method for finding successively better approximations to the roots of a real-valued function. We start with an initial approximation, \(x_0 = -1\), and compute up to \(x_2\).
02

Define the Function and Its Derivative

Let \(f(x) = x^4 - 2\). The derivative of this function is \(f'(x) = 4x^3\). We will use Newton's formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) to compute the next approximation.
03

Calculate First Iteration (x_1)

Starting with \(x_0 = -1\), apply Newton's formula:\[x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = -1 - \frac{(-1)^4 - 2}{4(-1)^3} = -1 - \frac{1 - 2}{-4} = -1 - \frac{-1}{-4} = -1 + \frac{1}{4} = -\frac{3}{4}\] Thus, \(x_1 = -\frac{3}{4}\).
04

Calculate Second Iteration (x_2)

Now use \(x_1 = -\frac{3}{4}\) to find \(x_2\):\[x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = -\frac{3}{4} - \frac{\left(-\frac{3}{4}\right)^4 - 2}{4\left(-\frac{3}{4}\right)^3}\]Calculate separately:- \(\left(-\frac{3}{4}\right)^4 = \left(\frac{81}{256}\right)\)- \(\left(-\frac{3}{4}\right)^3 = -\frac{27}{64}\)Plug these into the equation:\[x_2 = -\frac{3}{4} - \frac{\frac{81}{256} - 2}{4 \times -\frac{27}{64}} = -\frac{3}{4} - \frac{\frac{81}{256} - \frac{512}{256}}{-\frac{108}{64}}\]Simplify the fraction:\[x_2 = -\frac{3}{4} - \frac{-\frac{431}{256}}{-\frac{108}{64}} = -\frac{3}{4} + \frac{431 \times 64}{256 \times 108}\]Calculate the fraction part:\[= -\frac{3}{4} + \frac{27584}{27648} \approx -0.75 + 0.9977 \approx -0.7523\]Thus, \(x_2 \approx -0.7546\).

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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 like magical tools in mathematics. They help us find solutions to problems that might be really hard, or even impossible, to solve with just algebra. Sometimes, equations are too complex, and we need a way to tackle them step by step to get an answer. This is where methods like Newton's method come into play.

Newton's method is an iterative numerical technique. It means we start with a guess and keep refining that guess until we get close to the actual answer. This method is particularly useful for finding roots of an equation, and it's based on calculus principles.

Here’s how it works:
  • We start with an initial guess. In the problem, this is called \(x_0\).
  • Next, we apply a formula that uses our guess, calculates where we are wrong, and adjusts the guess accordingly.
  • This process is repeated, improving our accuracy with each step.
Newton's method can be accurate very quickly, but we need a good starting point and the function has to have certain properties like a continuous derivative.
Roots of Equations
Roots of equations are the values of \(x\) that make an equation true. For example, in the equation \(x^4 - 2 = 0\), we want to find \(x\) such that when we plug it back into the equation, it equals zero. Simply put, the root of the equation is the solution.

Finding the roots is crucial in many fields such as engineering, physics, and algebra, because many real-world problems can be formulated as equations that need solving. There are different types of roots:
  • Real roots: These are solutions that are real numbers.
  • Imaginary roots: These involve complex numbers where real solutions can't be found.
In our example, we are looking specifically for a negative real root of the equation \(x^4 = 2\). By solving for this root using Newton's method, we iteratively refine our guesses until we find \(x_2\) as approximately \(-0.7546\). This means \(x \approx -0.7546\) is a value that makes the equation true.
Calculus
Calculus is like the backbone of many numerical methods, including Newton's method. In calculus, we study things that change in a dynamic way, and it involves concepts such as derivatives and integrals.

For Newton's method, the derivative of a function plays a key role. The derivative tells us how the function is changing at any point, and this information helps in refining our guesses effectively. In our example, we have the function \(f(x) = x^4 - 2\) and its derivative \(f'(x) = 4x^3\).
  • Being able to find the derivative provides us with the slope of the tangent to the curve at a given point. This slope is used in adjusting the guesses to get closer to the root.
  • The method of calculating the next approximation using \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) involves these calculus principles.
By understanding the role calculus plays in Newton's method, students can appreciate why this powerful branch of mathematics is essential for solving problems that change and move, helping us get solutions to complex equations.

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Most popular questions from this chapter

Use I'Hópital's rule to find the limits. $$\lim _{x \rightarrow \infty} \frac{\ln (x+1)}{\log _{2} x}$$

Sketch a smooth connected curve \(y=f(x)\) with \(\begin{aligned}f(-2) &=8 \\\f(0) &=4 \\\f(2) &=0 \\\f^{\prime}(x) &>0 \quad \text { for } \quad|x|>2\end{aligned}\) \(\begin{aligned}&f^{\prime}(2)=f^{\prime}(-2)=0\\\&f^{\prime}(x)<0 \text { for }|x|<2\\\&f^{\prime \prime}(x)<0 \text { for } x<0\\\&f^{\prime \prime}(x)>0 \quad \text { for } \quad x>0\end{aligned}\)

a. Find the local extrema of each function on the given interval, and say where they occur. b. Graph the function and its derivative together. Comment on the behavior of \(f\) in relation to the signs and values of \(f^{\prime}\) $$f(x)=\sqrt{3} \cos x+\sin x, \quad 0 \leq x \leq 2 \pi$$

Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(x-\left(\frac{1}{2}\right)^{x}\) b. \(x^{2}+2^{x}\) c. \(\pi^{x}-x^{-1}\)

a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2} \mathrm{cm} / \mathrm{sec}, \quad \frac{r_{0}}{2} \leq r \leq r_{0},$$ where \(r_{0}\) is the rest radius of the trachea in centimeters and \(c\) is a positive constant whose value depends in part on the length of the trachea. Show that \(v\) is greatest when \(r=(2 / 3) r_{0} ;\) that is, when the trachea is about \(33 \%\) contracted. The remarkable fact is that X-ray photographs confirm that the trachea contracts about this much during a cough. b. Take \(r_{0}\) to be 0.5 and \(c\) to be 1 and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see with the claim that \(v\) is at a maximum when \(r=(2 / 3) r_{0}\).

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