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

$$\mathbf{use}\mathbf{newton}\mathbf{raphson}\mathbf{method}$$$$\mathbf{to}\mathbf{approximate}\mathbf{the}\mathbf{positive}\mathbf{root}$$$${\mathit{x}}^2-1=0\mathbf{correct}\mathbf{to}4\mathbf{decimal}\mathbf{places}$$$$\mathbf{perform}3\mathbf{iteration}\mathbf{only}$$$$\mathbf{setting}\mathbf{with}\mathit{x}=2$$

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

$$letp\left(x\right)=x^2-1and{x}_0=2\Rightarrow$$$$x_{n+1}=x_n-\frac{p\left(x_n\right)}{p^{{}^{\prime }}\left(x_n\right)}\Rightarrow x_1=x_0-\frac{p\left(x_0\right)}{p^{{}^{\prime }}\left(x_0\right)}$$$$=2-\frac{3}{4}=\frac{5}{4}$$$$x_2=x_1-\frac{p\left(x_1\right)}{p^{{}^{\prime }}\left(x_1\right)}=\frac{5}{4}-\frac{{\left(\frac{5}{4}\right)}^2-1}{2.\frac{5}{4}}=\frac{5}{4}-\frac{9}{16.\frac{5}{2}}$$$$=\frac{5}{4}-\frac{9}{40}=\frac{41}{40}$$$$x_3=x_2-\frac{p\left(x_2\right)}{p^{{}^{\prime }}\left(x_2\right)}=\frac{41}{40}-\frac{{\left(\frac{41}{40}\right)}^2-1}{2.\frac{41}{40}}$$$$=\frac{41}{40}-\frac{{41}^2-{40}^2}{{40}^2.\frac{41}{20}}=\frac{41}{40}-\frac{81}{{40}^2}.\frac{20}{41}=\frac{41}{40}-\frac{1620}{41.{40}^2}$$$$=\frac{41}{40}-\frac{1620}{41.1600}=\frac{41}{40}-\frac{162}{41.160}=....$$

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