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Chapter 4: Problem 7

Write the formula for Newton's method and use the given initial approximation to compute the approximations \(x_{1}\) and \(x_{2}\). $$f(x)=e^{-x}-x ; x_{0}=\ln 2$$

Short Answer

Expert verified
Question: Using Newton's method, find the first two approximations, \(x_{1}\) and \(x_{2}\), for the given function \(f(x) = e^{-x} - x\) with the initial approximation \(x_{0} = \ln 2\). Answer: The first two approximations using Newton's method in the given exercise are: \(x_1 = \ln 2 + \frac{\frac{1}{2} - \ln 2}{\frac{3}{2}}\) and \(x_2 = x_1 - \frac{e^{-x_1} - x_1}{-e^{-x_1} - 1}\).

Step by step solution

01

Write the formula for Newton's method

Newton's method is an iterative process to find the roots of a continuous and differentiable function. The formula for Newton's method is given by: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ where \(x_{n+1}\) is the \((n+1)^{th}\) approximation, \(x_n\) is the \(n^{th}\) approximation, \(f(x_n)\) is the value of the function at \(x_n\), and \(f'(x_n)\) is the value of the derivative at \(x_n\).
02

Write down the function and find its derivative

We are given the function: $$f(x) = e^{-x} - x$$ To find the derivative, we'll differentiate the function with respect to \(x\): $$f'(x) = -e^{-x} - 1$$
03

Find the first approximation (\(x_1\)) using the given initial approximation

We are given the initial approximation: $$x_0 = \ln 2$$ Now, we'll use the formula for Newton's method to find the first approximation: $$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$$ Evaluate the function and its derivative at the given initial approximation: $$f(\ln 2) = e^{-(\ln 2)} - \ln 2 = \frac{1}{2} - \ln 2$$ $$f'(\ln 2) = -e^{-(\ln 2)} - 1 = -\frac{1}{2} - 1$$ Now, we'll substitute the values into the formula for the first approximation: $$x_1 = \ln 2 - \frac{\frac{1}{2} - \ln 2}{-\frac{1}{2}-1} = \ln 2 + \frac{\frac{1}{2} - \ln 2}{\frac{3}{2}}$$
04

Find the second approximation (\(x_2\)) using the first approximation

We'll now use the first approximation, \(x_1\), to find the second approximation, \(x_2\): $$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}$$ Evaluate the function and its derivative at \(x_1\): $$f(x_1) = e^{-x_1} - x_1$$ $$f'(x_1) = -e^{-x_1} - 1$$ Now, we'll substitute the values into the formula for the second approximation: $$x_2 = x_1 - \frac{e^{-x_1} - x_1}{-e^{-x_1} - 1}$$ Thus, the first two approximations using Newton's method in the given exercise are: $$x_1 = \ln 2 + \frac{\frac{1}{2} - \ln 2}{\frac{3}{2}}$$ $$x_2 = x_1 - \frac{e^{-x_1} - x_1}{-e^{-x_1} - 1}$$

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