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Question Number 12612 by frank ntulah last updated on 26/Apr/17

Use Newtown Raphson method to find aproximate value of X=(√(((7/(x+1))))) ,starting with x_0 =2. perform 4 iteration and all iteration should be presented in 4 decimal places

$$\mathrm{Use}\:\mathrm{Newtown}\:\mathrm{Raphson}\:\mathrm{method}\:\mathrm{to}\:\mathrm{find}\:\mathrm{aproximate} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\boldsymbol{{X}}=\sqrt{\left(\frac{\mathrm{7}}{\boldsymbol{{x}}+\mathrm{1}}\right)}\:,\mathrm{starting}\:\mathrm{with}\:\boldsymbol{{x}}_{\mathrm{0}} =\mathrm{2}. \\ $$$$\boldsymbol{{perform}}\:\mathrm{4}\:\boldsymbol{{iteration}}\:\boldsymbol{{and}}\:\boldsymbol{{all}}\:\boldsymbol{{iteration}}\: \\ $$$$\boldsymbol{{should}}\:\boldsymbol{{be}}\:\boldsymbol{{presented}}\:\boldsymbol{{in}}\:\mathrm{4}\:\boldsymbol{{decimal}}\:\boldsymbol{{places}} \\ $$

Answered by mrW1 last updated on 26/Apr/17

f(x)=(√(7/(x+1)))−x f′(x)=(1/(2(√(7/(x+1)))))×((−7)/((x+1)^2 ))−1 ((f(x))/(f′(x)))=−(((√(7/(x+1)))−x)/((1/(2(√(7/(x+1)))))×(7/((x+1)^2 ))+1))=−(((7/(x+1))−x(√(7/(x+1))))/((7/(2(x+1)^2 ))+(√(7/(x+1))))) =−((14(x+1)−2x(√(7(x+1)^3 )))/(7+2(√(7(x+1)^3 ))))=g(x) x_(n+1) =x_n −((f(x_n ))/(f′(x_n )))=x_n −g(x_n ) x_0 =2 x_1 =x_0 −g(x_0 )=2−g(2)=1.6234 x_2 =x_1 −g(x_1 )=1.6234−g(1.6234)=1.6310 x_3 =x_2 −g(x_2 )=1.6310−g(1.6310)=1.6311 x_4 =x_2 −g(x_3 )=1.6311−g(1.6311)=1.6311 ⇒x≈1.6311

$${f}\left({x}\right)=\sqrt{\frac{\mathrm{7}}{{x}+\mathrm{1}}}−{x} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{\mathrm{7}}{{x}+\mathrm{1}}}}×\frac{−\mathrm{7}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{1} \\ $$$$\frac{{f}\left({x}\right)}{{f}'\left({x}\right)}=−\frac{\sqrt{\frac{\mathrm{7}}{{x}+\mathrm{1}}}−{x}}{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{\mathrm{7}}{{x}+\mathrm{1}}}}×\frac{\mathrm{7}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{1}}=−\frac{\frac{\mathrm{7}}{{x}+\mathrm{1}}−{x}\sqrt{\frac{\mathrm{7}}{{x}+\mathrm{1}}}}{\frac{\mathrm{7}}{\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+\sqrt{\frac{\mathrm{7}}{{x}+\mathrm{1}}}} \\ $$$$=−\frac{\mathrm{14}\left({x}+\mathrm{1}\right)−\mathrm{2}{x}\sqrt{\mathrm{7}\left({x}+\mathrm{1}\right)^{\mathrm{3}} }}{\mathrm{7}+\mathrm{2}\sqrt{\mathrm{7}\left({x}+\mathrm{1}\right)^{\mathrm{3}} }}={g}\left({x}\right) \\ $$$${x}_{{n}+\mathrm{1}} ={x}_{{n}} −\frac{{f}\left({x}_{{n}} \right)}{{f}'\left({x}_{{n}} \right)}={x}_{{n}} −{g}\left({x}_{{n}} \right) \\ $$$$ \\ $$$${x}_{\mathrm{0}} =\mathrm{2} \\ $$$${x}_{\mathrm{1}} ={x}_{\mathrm{0}} −{g}\left({x}_{\mathrm{0}} \right)=\mathrm{2}−{g}\left(\mathrm{2}\right)=\mathrm{1}.\mathrm{6234} \\ $$$${x}_{\mathrm{2}} ={x}_{\mathrm{1}} −{g}\left({x}_{\mathrm{1}} \right)=\mathrm{1}.\mathrm{6234}−{g}\left(\mathrm{1}.\mathrm{6234}\right)=\mathrm{1}.\mathrm{6310} \\ $$$${x}_{\mathrm{3}} ={x}_{\mathrm{2}} −{g}\left({x}_{\mathrm{2}} \right)=\mathrm{1}.\mathrm{6310}−{g}\left(\mathrm{1}.\mathrm{6310}\right)=\mathrm{1}.\mathrm{6311} \\ $$$${x}_{\mathrm{4}} ={x}_{\mathrm{2}} −{g}\left({x}_{\mathrm{3}} \right)=\mathrm{1}.\mathrm{6311}−{g}\left(\mathrm{1}.\mathrm{6311}\right)=\mathrm{1}.\mathrm{6311} \\ $$$$\Rightarrow{x}\approx\mathrm{1}.\mathrm{6311} \\ $$

Commented by frank ntulah last updated on 27/Apr/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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