Question Number 55641 by gunawan last updated on 01/Mar/19
$$\mathrm{Studies}\:\mathrm{of}\:\mathrm{convergences} \\ $$$$\mathrm{the}\:\mathrm{numbers}\:\mathrm{real}\:\mathrm{sequence}\:\left\{{x}_{{n}} \right\}, \\ $$$$\mathrm{with}\:{x}_{\mathrm{1}} =\mathrm{1}\:\mathrm{and}\:{x}_{{n}+\mathrm{1}} =\frac{{x}_{{n}} ^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}_{{n}} },\:{n}\geqslant\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 01/Mar/19
![we have x_(n+1) =f(x_n ) with f(x)=((x^2 +2)/(2x)) =(x/2) +(1/x) with x>0 we havef^′ (x)=(1/2)−(1/x^2 ) =((x^2 −2)/(2x^2 )) and f^′ (x)=0 ⇔x =+^− (√2) x 0 (√2) +∞ f^′ (x) − + f(x) +∞ deccf((√2)) incre +∞ f((√2)) =(1/( (√2))) +(1/( (√2))) =(√2) f is continue on ]0,+∞[ let find the fix point f(x)=x ⇔(x/2) +(1/x) =x ⇔(1/x) =(x/2) ⇒2=x^2 ⇒x =(√2) because x >0 (x_n ) is convergent and lim_(n→+∞) x_n =(√2).](https://www.tinkutara.com/question/Q55676.png)
$${we}\:{have}\:{x}_{{n}+\mathrm{1}} ={f}\left({x}_{{n}} \right)\:{with}\:{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} \:+\mathrm{2}}{\mathrm{2}{x}}\:=\frac{{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{{x}}\:\:{with}\:{x}>\mathrm{0} \\ $$$${we}\:{havef}^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\frac{{x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}} }\:\:\:\:{and}\:{f}^{'} \left({x}\right)=\mathrm{0}\:\:\Leftrightarrow{x}\:=\overset{−} {+}\sqrt{\mathrm{2}} \\ $$$${x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${f}^{'} \left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$${f}\left({x}\right)\:\:\:\:\:\:\:+\infty\:{deccf}\left(\sqrt{\mathrm{2}}\right)\:\:{incre}\:\:+\infty \\ $$$${f}\left(\sqrt{\mathrm{2}}\right)\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:=\sqrt{\mathrm{2}} \\ $$$$\left.{f}\:{is}\:{continue}\:{on}\:\right]\mathrm{0},+\infty\left[\:\:\:\:\:{let}\:{find}\:{the}\:{fix}\:{point}\right. \\ $$$${f}\left({x}\right)={x}\:\Leftrightarrow\frac{{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{{x}}\:={x}\:\Leftrightarrow\frac{\mathrm{1}}{{x}}\:=\frac{{x}}{\mathrm{2}}\:\Rightarrow\mathrm{2}={x}^{\mathrm{2}} \:\Rightarrow{x}\:=\sqrt{\mathrm{2}}\:\:\:\:{because}\:{x}\:>\mathrm{0} \\ $$$$\left({x}_{{n}} \right)\:{is}\:{convergent}\:\:{and}\:{lim}_{{n}\rightarrow+\infty} {x}_{{n}} =\sqrt{\mathrm{2}}. \\ $$