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Question Number 37818 by prof Abdo imad last updated on 17/Jun/18

let 0<u_0 <1  and u_(n+1) =(√((1+u_n )/2))  study the convergence of u_n

$${let}\:\mathrm{0}<{u}_{\mathrm{0}} <\mathrm{1}\:\:{and}\:{u}_{{n}+\mathrm{1}} =\sqrt{\frac{\mathrm{1}+{u}_{{n}} }{\mathrm{2}}} \\ $$ $${study}\:{the}\:{convergence}\:{of}\:{u}_{{n}} \: \\ $$ $$ \\ $$

Commented bymath khazana by abdo last updated on 19/Jun/18

let f(x) =(√((1+x)/2))  with x≥0   f(x)=x ⇒((1+x)/2) =x^2  ⇒1+x =2x^2  ⇒  2x^2 −x−1 =0   Δ =1+8 =9  x_1 =((1+3)/4) =1  and x_2 =((1−3)/4) =−(1/2)  we have  u_(n+1) =f(u_n ) and f is continue ⇒lim_(n→+∞) u_n =1  let find another form of u_n  we put  u_0 =cosα  ⇒ u_1 =(√((1+cos(α))/2))  =cos((α/2))   let prove that u_n = cos((α/2^n ))  relation true for n=0  let suppose u_n =cos((α/2^n )) ⇒  u_(n+1) =(√((1+u_n )/2)) =(√((1+cos((α/2^n )))/2))  =cos((α/2^(n+1) )) . its now clear that  lim_(n→+∞)  u_n =1 .

$${let}\:{f}\left({x}\right)\:=\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{2}}}\:\:{with}\:{x}\geqslant\mathrm{0}\: \\ $$ $${f}\left({x}\right)={x}\:\Rightarrow\frac{\mathrm{1}+{x}}{\mathrm{2}}\:={x}^{\mathrm{2}} \:\Rightarrow\mathrm{1}+{x}\:=\mathrm{2}{x}^{\mathrm{2}} \:\Rightarrow \\ $$ $$\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{1}\:=\mathrm{0}\:\:\:\Delta\:=\mathrm{1}+\mathrm{8}\:=\mathrm{9} \\ $$ $${x}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{3}}{\mathrm{4}}\:=\mathrm{1}\:\:{and}\:{x}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{3}}{\mathrm{4}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\:{we}\:{have} \\ $$ $${u}_{{n}+\mathrm{1}} ={f}\left({u}_{{n}} \right)\:{and}\:{f}\:{is}\:{continue}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {u}_{{n}} =\mathrm{1} \\ $$ $${let}\:{find}\:{another}\:{form}\:{of}\:{u}_{{n}} \:{we}\:{put} \\ $$ $${u}_{\mathrm{0}} ={cos}\alpha\:\:\Rightarrow\:{u}_{\mathrm{1}} =\sqrt{\frac{\mathrm{1}+{cos}\left(\alpha\right)}{\mathrm{2}}}\:\:={cos}\left(\frac{\alpha}{\mathrm{2}}\right) \\ $$ $$\:{let}\:{prove}\:{that}\:{u}_{{n}} =\:{cos}\left(\frac{\alpha}{\mathrm{2}^{{n}} }\right) \\ $$ $${relation}\:{true}\:{for}\:{n}=\mathrm{0} \\ $$ $${let}\:{suppose}\:{u}_{{n}} ={cos}\left(\frac{\alpha}{\mathrm{2}^{{n}} }\right)\:\Rightarrow \\ $$ $${u}_{{n}+\mathrm{1}} =\sqrt{\frac{\mathrm{1}+{u}_{{n}} }{\mathrm{2}}}\:=\sqrt{\frac{\mathrm{1}+{cos}\left(\frac{\alpha}{\mathrm{2}^{{n}} }\right)}{\mathrm{2}}} \\ $$ $$={cos}\left(\frac{\alpha}{\mathrm{2}^{{n}+\mathrm{1}} }\right)\:.\:{its}\:{now}\:{clear}\:{that} \\ $$ $${lim}_{{n}\rightarrow+\infty} \:{u}_{{n}} =\mathrm{1}\:. \\ $$

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