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Chapter 4: Problem 8

Write the formula for Newton's method and use the given initial approximation to compute the approximations \(x_{1}\) and \(x_{2}\). $$f(x)=x^{3}-2 ; x_{0}=2$$

Short Answer

Expert verified
Question: Find the first two approximations of the solution of the equation \(f(x) = x^3 - 2\) using Newton's method with the initial approximation \(x_{0} = 2\). Answer: Using Newton's method, the first two approximations for the solution are \(x_1 = 1.5\) and \(x_2 \approx 1.5185\).

Step by step solution

01

Write the formula for Newton's method

Newton's method formula is given by: $$x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}$$ where \(x_{n+1}\) is the next approximation, \(x_{n}\) is the current approximation, \(f(x_n)\) is the function evaluated at the current approximation, and \(f'(x_{n})\) is the derivative of the function evaluated at the current approximation.
02

Calculate the derivative of the given function

We are given the function: $$f(x) = x^{3} - 2$$ Now we need to find the derivative of this function with respect to \(x\). The derivative of \(f(x)\) is: $$f'(x) = 3x^{2}$$
03

Use the initial approximation to find the next approximation

The initial approximation \(x_{0}\) is given as \(2\). We will use the formula for Newton's method to find \(x_{1}\) as follows: $$x_{1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})} = 2 - \frac{f(2)}{f'(2)}$$ First, we need to calculate the value of \(f(x_0)\) and \(f'(x_0)\): $$f(2) = 2^{3} - 2 = 6$$ $$f'(2) = 3 \times 2^{2} = 12$$ Now, we can plug these values back into the formula to find \(x_{1}\): $$x_{1} = 2 - \frac{6}{12} = 2 - 0.5 = 1.5$$
04

Use the first approximation to find the second approximation

Now that we have \(x_{1}\), we can use Newton's method formula again to find \(x_{2}\): $$x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})} = 1.5 - \frac{f(1.5)}{f'(1.5)}$$ Calculate the value of \(f(x_1)\) and \(f'(x_1)\): $$f(1.5) = 1.5^{3} - 2 = -0.125$$ $$f'(1.5) = 3 \times (1.5)^{2} = 6.75$$ Now, we can plug these values back into the formula to find \(x_{2}\): $$x_{2} = 1.5 - \frac{-0.125}{6.75} \approx 1.5185$$ So, the first two approximations using Newton's method for the given function and initial approximation are: $$x_{1} = 1.5$$ $$x_{2} \approx 1.5185$$

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