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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: Use Newton's method to find the subsequent approximations \(x_{1}\) and \(x_{2}\) for the function \(f(x) = e^{-x} - x\), given the initial approximation \(x_{0}=\ln{2}\). Answer: Using Newton's method, we found the subsequent approximations to be \(x_1 \approx 0.8332\) and \(x_2 \approx 0.7280\).

Step by step solution

01

Write the formula for Newton's method

Newton's method formula is: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
02

Find the derivative of the function

Firstly, we have to find the derivative of the function \(f(x) = e^{-x} - x\). This will give us the value of \(f'(x)\). $$f'(x) = \frac{d}{dx}(e^{-x} - x) = -e^{-x} - 1$$
03

Substitute the initial approximation \(x_0\)

Now, we will substitute the given initial approximation \(x_0 = \ln{2}\) into the Newton's method formula and calculate \(x_1\). $$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = \ln{2} - \frac{e^{-\ln{2}} - \ln{2}}{-e^{-\ln{2}} - 1}$$
04

Simplify and calculate the value of \(x_1\)

Using properties of logarithms and simplifying the expression, we get: $$x_1 = \ln{2} - \frac{e^{-\ln{2}} - \ln{2}}{-e^{-\ln{2}} - 1} = \ln{2} - \frac{\frac{1}{2} - \ln{2}}{-\frac{1}{2} - 1}$$ Now, we calculate the value of \(x_1\): $$x_1 = \ln{2} - \frac{\frac{1}{2} - \ln{2}}{-\frac{1}{2} - 1} \approx 0.8332$$
05

Calculate the value of \(x_2\)

Similarly, substitute the calculated value of \(x_1\) into the Newton's method formula to find the value of \(x_2\): $$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.8332 - \frac{e^{-0.8332} - 0.8332}{-e^{-0.8332} - 1}$$ Now, we calculate the value of \(x_2\): $$x_2 = 0.8332 - \frac{e^{-0.8332} - 0.8332}{-e^{-0.8332} - 1} \approx 0.7280$$ So, we have found the values of the approximations \(x_1 \approx 0.8332\) and \(x_2 \approx 0.7280\).

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