Newton's method of approximation

Re: Newton's method of approximation

__Procedure for Newton’s Method__

1. Guess a first approximation to a solution of the equation $\displaystyle f(x) = 0$. A graph

of $\displaystyle y = f(x)$ may help.

2. Use the first approximation to get a second, the second to get a third, and so

on, using the formula

$\displaystyle 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 $\displaystyle f(x) = x^2 -2$

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

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

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

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

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Figure 1

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

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

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