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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

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

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

let p(x)=x^2 −1 and x_0 =2 ⇒  x_(n+1) =x_n  −((p(x_n ))/(p^′ (x_n ))) ⇒x_1 =x_0 −((p(x_0 ))/(p^′ (x_0 )))  =2−(3/4) =(5/4)  x_2 =x_1 −((p(x_1 ))/(p^′ (x_1 ))) =(5/4) −((((5/4))^2 −1)/(2.(5/4)))=(5/4) −(9/(16.(5/2)))  =(5/4) −(9/(40)) = ((41)/(40))  x_3 =x_2  −((p(x_2 ))/(p^′ (x_2 ))) = ((41)/(40)) −(((((41)/(40)))^2 −1)/(2.((41)/(40))))  =((41)/(40)) −((41^2 −40^2 )/(40^2  .((41)/(20)))) =((41)/(40)) −((81)/(40^2 )) .((20)/(41)) =((41)/(40)) −((1620)/(41.40^2 ))  =((41)/(40)) −((1620)/(41.1600)) =((41)/(40)) −((162)/(41.160)) =....

$${let}\:{p}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{1}\:{and}\:{x}_{\mathrm{0}} =\mathrm{2}\:\Rightarrow \\ $$$${x}_{{n}+\mathrm{1}} ={x}_{{n}} \:−\frac{{p}\left({x}_{{n}} \right)}{{p}^{'} \left({x}_{{n}} \right)}\:\Rightarrow{x}_{\mathrm{1}} ={x}_{\mathrm{0}} −\frac{{p}\left({x}_{\mathrm{0}} \right)}{{p}^{'} \left({x}_{\mathrm{0}} \right)} \\ $$$$=\mathrm{2}−\frac{\mathrm{3}}{\mathrm{4}}\:=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${x}_{\mathrm{2}} ={x}_{\mathrm{1}} −\frac{{p}\left({x}_{\mathrm{1}} \right)}{{p}^{'} \left({x}_{\mathrm{1}} \right)}\:=\frac{\mathrm{5}}{\mathrm{4}}\:−\frac{\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}.\frac{\mathrm{5}}{\mathrm{4}}}=\frac{\mathrm{5}}{\mathrm{4}}\:−\frac{\mathrm{9}}{\mathrm{16}.\frac{\mathrm{5}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{4}}\:−\frac{\mathrm{9}}{\mathrm{40}}\:=\:\frac{\mathrm{41}}{\mathrm{40}} \\ $$$${x}_{\mathrm{3}} ={x}_{\mathrm{2}} \:−\frac{{p}\left({x}_{\mathrm{2}} \right)}{{p}^{'} \left({x}_{\mathrm{2}} \right)}\:=\:\frac{\mathrm{41}}{\mathrm{40}}\:−\frac{\left(\frac{\mathrm{41}}{\mathrm{40}}\right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}.\frac{\mathrm{41}}{\mathrm{40}}} \\ $$$$=\frac{\mathrm{41}}{\mathrm{40}}\:−\frac{\mathrm{41}^{\mathrm{2}} −\mathrm{40}^{\mathrm{2}} }{\mathrm{40}^{\mathrm{2}} \:.\frac{\mathrm{41}}{\mathrm{20}}}\:=\frac{\mathrm{41}}{\mathrm{40}}\:−\frac{\mathrm{81}}{\mathrm{40}^{\mathrm{2}} }\:.\frac{\mathrm{20}}{\mathrm{41}}\:=\frac{\mathrm{41}}{\mathrm{40}}\:−\frac{\mathrm{1620}}{\mathrm{41}.\mathrm{40}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{41}}{\mathrm{40}}\:−\frac{\mathrm{1620}}{\mathrm{41}.\mathrm{1600}}\:=\frac{\mathrm{41}}{\mathrm{40}}\:−\frac{\mathrm{162}}{\mathrm{41}.\mathrm{160}}\:=.... \\ $$

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