Newton’s Method

Problem 176: Use Newton’s Method to approximate the positive root of the equation

(1) $\begin{equation*} f(x) = x^2 - 2 .\end{equation*}$

Solution: We know that the actual positive root of $f(x) = \sqrt{2} \approx 1.14421$ to 5-decimal places. We will use

(2) $\begin{equation*} x_{n+1} & = x_n - \frac{f(x_n)}{f'(x_n)} \\ & = x_n - \frac{x_n^2 - 2}{2x_n} \\ & = \frac{x_n}{2} + \frac{1}{x_n},\end{equation*}$

which is the Newton’s Method applied to our function $f(x)$ .

We start with the guess $x_0=1$ . Then

(3) $\begin{align*} x_0 & = 1 \\ x_1 & = \frac{x_0}{2} + \frac{1}{x_0} = \frac{1}{2} + 1 = 1.5 \\ x_2 & = \frac{x_1}{2} + \frac{1}{x_1} = \frac{1.5}{2} + \frac{1}{1.5} = 1.41667 \\ x_3 & = \frac{x_2}{2} + \frac{1}{x_2} = \frac{1.41667}{2} + \frac{1}{1.41667} = 1.41422 \\ x_4 & = \frac{x_3}{2} + \frac{1}{x_3} = \frac{1.41422}{2} + \frac{1}{1.41422} = 1.41421 \\ x_5 & = \frac{x_4}{2} + \frac{1}{x_4} = \frac{1.41421}{2} + \frac{1}{1.41421} = 1.41421 .\end{align*}$

Note that we get an exact approximation to 5-decimal places after the 4th iterations.

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