Arithmetic Questions

Question Number 209342 by essaad last updated on 07/Jul/24

Answered by Berbere last updated on 07/Jul/24

$${U}_{{n}+\mathrm{1}} ={f}\left({U}_{{n}} \right) \\$$$${x}\overset{{f}} {\rightarrow}\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}};{f}\:{increase} \\$$$${f}\left(\left[\mathrm{0},\mathrm{1}\right]\right)=\left[\mathrm{0},\frac{\mathrm{3}}{\mathrm{4}}\right]\subset\left[\mathrm{0},\mathrm{1}\right] \\$$$$\Rightarrow{U}_{{n}} \in\left[\mathrm{0},\mathrm{1}\right] \\$$$${U}_{{n}+\mathrm{1}} −{U}_{{n}} =−\frac{{U}_{{n}} }{\mathrm{2}}+\frac{{U}_{{n}} ^{\mathrm{2}} }{\mathrm{4}}=\frac{{U}_{{n}} \left({U}_{{n}} −\mathrm{2}\right)}{\mathrm{4}}<\mathrm{0} \\$$$${U}_{{n}} \:{decrease}\:{Bounded}\:{Cv}\:{to}\:{l}\in\left[\mathrm{0},\mathrm{1}\right] \\$$$${since}\:{f}\:{is}\:{continus}\:{in}\left[\mathrm{0},\mathrm{1}\right]\Rightarrow{l}={f}\left({l}\right)\Leftrightarrow{l}=\frac{{l}}{\mathrm{2}}+\frac{{l}^{\mathrm{2}} }{\mathrm{4}} \\$$$$\Rightarrow\frac{{l}}{\mathrm{2}}=\frac{{l}^{\mathrm{2}} }{\mathrm{4}}\Rightarrow{l}\left(\mathrm{2}−{l}\right)=\mathrm{0}\Rightarrow{l}=\mathrm{0} \\$$