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Chapter 10: Problem 10

Use Newton's method to approximate the root of each equation, beginning with the given \(x_{0}\) and continuing until two successive approximations agree to three decimal places. Carry out the calculation "by hand" with the aid of a calculator, rounding to three decimal places. $$ \begin{array}{l} e^{x}+4 x-4=0 \\ x_{0}=0 \end{array} $$

Short Answer

Expert verified
The approximate root is 0.562.

Step by step solution

01

Understand Newton's Method

Newton's Method is an iterative numerical approach used to approximate the roots of a real-valued function. It uses the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] to obtain a sequence of approximations \( x_0, x_1, x_2, \ldots \) that converge to a root of \( f(x) = 0 \).
02

Define the Function and Its Derivative

Given the function \( f(x) = e^x + 4x - 4 \), we need to find its derivative to apply Newton's method. The derivative is: \[ f'(x) = e^x + 4 \]
03

Apply Initial Guess

Start with the initial guess \( x_0 = 0 \). Evaluate both \( f(x_0) \) and \( f'(x_0) \): \[ f(0) = e^0 + 4 \times 0 - 4 = 1 - 4 = -3 \] \[ f'(0) = e^0 + 4 = 1 + 4 = 5 \] Then, Calculate the next approximation using Newton鈥檚 formula: \[ x_1 = 0 - \frac{-3}{5} = 0.6 \]
04

Calculate Second Approximation

Use the result \( x_1 = 0.6 \) to find the next approximation. First, compute \( f(x_1) \) and \( f'(x_1) \): \[ f(0.6) = e^{0.6} + 4 \times 0.6 - 4 \approx 1.822 + 2.4 - 4 = 0.222 \] \[ f'(0.6) = e^{0.6} + 4 \approx 1.822 + 4 = 5.822 \] Now find \( x_2 \) using Newton鈥檚 formula: \[ x_2 = 0.6 - \frac{0.222}{5.822} \approx 0.6 - 0.038 = 0.562 \] Round \( x_2 \) to three decimal places: \( x_2 = 0.562 \).
05

Calculate Third Approximation

Use the result \( x_2 = 0.562 \) to compute the next approximation. Calculate \( f(x_2) \) and \( f'(x_2) \): \[ f(0.562) = e^{0.562} + 4 \times 0.562 - 4 \approx 1.754 + 2.248 - 4 = 0.002 \] \[ f'(0.562) = e^{0.562} + 4 \approx 1.754 + 4 = 5.754 \] Now find \( x_3 \) using Newton鈥檚 formula: \[ x_3 = 0.562 - \frac{0.002}{5.754} \approx 0.562 - 0.0003 = 0.562 \] Round this to three decimal places: \( x_3 = 0.562 \).
06

Check Convergence

The latest approximations \( x_2 = 0.562 \) and \( x_3 = 0.562 \) agree to three decimal places. Therefore, the approximate root is \( x = 0.562 \).

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Key Concepts

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

Numerical Approximation
Numerical approximation techniques are essential in mathematics and engineering, as they allow us to find approximate solutions to problems where exact solutions are difficult or impossible to obtain. These approaches are fascinating because they provide a practical method for solving real-world problems. Newton's Method is a particularly effective numerical approximation technique for finding roots of functions.
  • This method relies on the idea that we can refine our guesses for the root of a function using the function's derivative.
  • With it, we make an initial guess, and iteratively apply a formula to get closer to the actual root.
  • Each iteration yields a new approximation that theoretically gets closer to the true solution, even if we don't know what it is outright.
What makes numerical approximations exciting is that they bridge the gap between the theoretical world of calculus and practical applications like engineering or physics. You often encounter functions that are too complex to solve analytically, and here numerical methods shine, offering solutions that are good enough for practical purposes.
Iterative Method
Iterative methods are procedures that generate sequences of approximations to a solution. In many situations, this is more efficient than trying to find an exact solution directly. Newton's Method is a classic example of an iterative method used widely in numerical analysis.
  • The process starts from an initial approximation, known as the initial guess.
  • Each subsequent approximation is derived from the previous one, using a specific formula that depends on the function and its derivative.
  • The method proceeds step by step, evaluating and adjusting the approximation until it converges to a satisfactory level of accuracy.
Iterative methods are powerful because they can handle functions without neat, closed-form solutions, allowing us to reach practical answers. In Newton's Method, we use the tangent line at each point to predict where the function might reach zero, adjusting our approximations until they stabilize and provide a consistent result.
Calculus
Calculus is the mathematical study of continuous change and has two principal branches, differentiation and integration. Newton's Method brilliantly applies the concept of differentiation to find the roots of equations, showcasing calculus's versatility.
  • Differentiation, a key concept in calculus, allows us to find the rate of change of a function, which is fundamental to Newton's Method.
  • By calculating the derivative, we can determine the slope of the tangent line to the function at given points.
  • This derivative is crucial as it helps adjust our approximations by showing how the function behaves around our current guesses.
Understanding calculus, particularly differentiation, is essential to grasping why Newton's Method is so effective. This approach embodies the practical application of calculus concepts, relying on the accurate computation of derivatives to successively hone in on the function's roots. By mastering these foundational calculus concepts, students can appreciate the depth of tools available for tackling complex mathematical challenges.

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Most popular questions from this chapter

a. Find the Taylor polynomial at \(x=0\) that approximates \(f(x)=\cos x\) on the interval \(-1 \leq x \leq 1\) with error less than 0.0002 b. Graph \(f(x)=\cos x\) and the Taylor polynomial on \([-2 \pi, 2 \pi]\) by \([-2,2] .\) Use TRACE to estimate the maximum difference between the two curves for \(-1 \leq x \leq 1\).

Let \(a\) be a given number and suppose that the function \(f\) is \(n+1\) times differentiable between \(a\) and \(x\). Since \(a\) and \(x\) are now fixed, we will use \(t\) for a variable taking values between them. $$ \begin{aligned} &\text { Show that }\\\ &\begin{array}{l} \int_{a}^{x} f^{(k+1)}(t) \cdot(x-t)^{k} d t \\ =\frac{1}{k+1}(x-a)^{k+1} f^{(k+1)}(a) \\ \quad+\frac{1}{k+1} \int_{a}^{x} f^{(k+2)}(t) \cdot(x-t)^{k+1} d t \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { for } 1

What is the internal rate of return if you borrow \(\$ 100\) and pay back \(\$ 110\) a year later?

(Requires sequence and series operations) a. Use your calculator to find \(e^{2}\) rounded to six decimal places. b. The Taylor series for \(e^{x}\) evaluated at \(x=2\) gives \(e^{2}=\sum_{n=0}^{\infty} \frac{2^{n}}{n !}\). Set your calculator to find the sum of this series up to any number of terms. How many terms are required for the sum (rounded to six decimal places) to agree with your answer to part (a)?

True or False: If the partial sums \(S_{n}\) of an infinite series all satisfy \(S_{n} \leq 10,\) then the sum \(S\) also satisfies \(\quad S \leq 10\).
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