Newton's method of approximation

**zalar81957** · August 1st 2012, 11:30 AM

Approximate

by applying Newton's Method to the equation

Round your answer to 4 decimal places.

**x3bnm** · August 1st 2012, 01:35 PM

Procedure for Newton’s Method

1. Guess a first approximation to a solution of the equation $f(x) = 0$ . A graph
of $y = f(x)$ may help.
2. Use the first approximation to get a second, the second to get a third, and so
on, using the formula

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \text{ if } f'(x_n) \neq 0 .....................(1)$

Example:
So to find a positive root of this equation $f(x) = x^2 -2$

we have to guess a value of $x_n$ . It doesn't have to be exact.

Let $x = x_n$ so $f(x) = x^2 - 2$ becomes $f(x_n) = (x_n)^2 - 2$ if we plugin this into (1) and after simplification we get:

$\begin{aligned}x_{n+1} = & x_n - \frac{(x_n)^2 - 2}{2 * x_n} \\ x_{n+1} = & \frac{x_n}{2} + \frac{1}{x_n} \text{ ..............................(2)}\end{aligned}$

The graph of $f(x) = x^2 - 2$ is:

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Figure 1

Let's guess a approximate value where the graph of $f(x) = y = x^2 - 2 = 0$ so from the graph(Figure 1) say $x_0 = 1$ where $n = 0$ and plug in into equation (2) and solve for $x_1$ and then plug-in $x_1$ and find $x_2$ and this goes on until you get a good approximation up to a particular decimal places:

$\begin{array}{|c|c|c|}\hline \text{ value of x} & \text{Error} & \text{Number of Correct Digits} \\ \hline x_0 = 1 & -0.41421 & 1 \\ \hline x_1 = 1.5 & 0.08579 & 1 \\ \hline x_2 = 1.41667 & 0.00246 & 3 \\ \hline x_3 = 1.41422 & 0.00001 & 5 \\ \hline \end{array}$

And that's it. You approximated the positive root of $f(x) = x^2 - 2$ to five decimal places.

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