use-newton-raphson-method-to-approximate-the-positive-root-x-2-1-0-correct-to-4-decimal-places-perform-3-iteration-only-setting-with-x-2-

Question Number 40366 by mondodotto@gmail.com last updated on 20/Jul/18

use newton raphson method

to approximate the positive root

x^{2} - 1 = 0 correct to 4 decimal places

perform 3 iteration only

setting with x = 2

Commented by math khazana by abdo last updated on 20/Jul/18

l e t p (x) = x^{2} - 1 a n d x_{0} = 2 \Rightarrow

x_{n + 1} = x_{n} - \frac{p (x_{n})}{p^{^{'}} (x_{n})} \Rightarrow x_{1} = x_{0} - \frac{p (x_{0})}{p^{^{'}} (x_{0})}

= 2 - \frac{3}{4} = \frac{5}{4}

x_{2} = x_{1} - \frac{p (x_{1})}{p^{^{'}} (x_{1})} = \frac{5}{4} - \frac{{(\frac{5}{4})}^{2} - 1}{2 . \frac{5}{4}} = \frac{5}{4} - \frac{9}{16 . \frac{5}{2}}

= \frac{5}{4} - \frac{9}{40} = \frac{41}{40}

x_{3} = x_{2} - \frac{p (x_{2})}{p^{^{'}} (x_{2})} = \frac{41}{40} - \frac{{(\frac{41}{40})}^{2} - 1}{2 . \frac{41}{40}}

= \frac{41}{40} - \frac{41^{2} - 40^{2}}{40^{2} . \frac{41}{20}} = \frac{41}{40} - \frac{81}{40^{2}} . \frac{20}{41} = \frac{41}{40} - \frac{1620}{41 . 40^{2}}

= \frac{41}{40} - \frac{1620}{41.1600} = \frac{41}{40} - \frac{162}{41.160} = \dots .