Question Number 209342 by essaad last updated on 07/Jul/24
Answered by Berbere last updated on 07/Jul/24
![U_(n+1) =f(U_n ) x→^f (x/2)+(x^2 /4);f increase f([0,1])=[0,(3/4)]⊂[0,1] ⇒U_n ∈[0,1] U_(n+1) −U_n =−(U_n /2)+(U_n ^2 /4)=((U_n (U_n −2))/4)<0 U_n decrease Bounded Cv to l∈[0,1] since f is continus in[0,1]⇒l=f(l)⇔l=(l/2)+(l^2 /4) ⇒(l/2)=(l^2 /4)⇒l(2−l)=0⇒l=0](Q209346.png)
$${U}_{{n}+\mathrm{1}} ={f}\left({U}_{{n}} \right) \\ $$$${x}\overset{{f}} {\rightarrow}\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}};{f}\:{increase} \\ $$$${f}\left(\left[\mathrm{0},\mathrm{1}\right]\right)=\left[\mathrm{0},\frac{\mathrm{3}}{\mathrm{4}}\right]\subset\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\Rightarrow{U}_{{n}} \in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${U}_{{n}+\mathrm{1}} −{U}_{{n}} =−\frac{{U}_{{n}} }{\mathrm{2}}+\frac{{U}_{{n}} ^{\mathrm{2}} }{\mathrm{4}}=\frac{{U}_{{n}} \left({U}_{{n}} −\mathrm{2}\right)}{\mathrm{4}}<\mathrm{0} \\ $$$${U}_{{n}} \:{decrease}\:{Bounded}\:{Cv}\:{to}\:{l}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${since}\:{f}\:{is}\:{continus}\:{in}\left[\mathrm{0},\mathrm{1}\right]\Rightarrow{l}={f}\left({l}\right)\Leftrightarrow{l}=\frac{{l}}{\mathrm{2}}+\frac{{l}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\frac{{l}}{\mathrm{2}}=\frac{{l}^{\mathrm{2}} }{\mathrm{4}}\Rightarrow{l}\left(\mathrm{2}−{l}\right)=\mathrm{0}\Rightarrow{l}=\mathrm{0} \\ $$