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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-

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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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