Question Number 209232 by alcohol last updated on 04/Jul/24
![u_0 = a, u_(n+1) = (√(u_n v_n )) v_0 = b ∈ ]0,1[ , v_(n+1) = (1/(2(u_n +v_n ))) • show that a≤u_n ≤u_(n+1) ≤v_n ≤v_(n+1) ≤b • show that v_n − u_n ≤ ((a+b)/2^n )](https://www.tinkutara.com/question/Q209232.png)
$${u}_{\mathrm{0}} \:=\:{a},\:{u}_{{n}+\mathrm{1}} \:=\:\sqrt{{u}_{{n}} {v}_{{n}} } \\ $$$$\left.{v}_{\mathrm{0}} \:=\:{b}\:\in\:\right]\mathrm{0},\mathrm{1}\left[\:,\:{v}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}\left({u}_{{n}} +{v}_{{n}} \right)}\right. \\ $$$$\bullet\:{show}\:{that}\:{a}\leqslant{u}_{{n}} \leqslant{u}_{{n}+\mathrm{1}} \leqslant{v}_{{n}} \leqslant{v}_{{n}+\mathrm{1}} \leqslant{b} \\ $$$$\bullet\:{show}\:{that}\:{v}_{{n}} \:−\:{u}_{{n}} \:\leqslant\:\frac{{a}+{b}}{\mathrm{2}^{{n}} } \\ $$
Answered by Berbere last updated on 04/Jul/24
$${u}_{\mathrm{0}} ={a}\geqslant\mathrm{0}\:\:\:{if}\:{a}<\mathrm{0}\:{U}_{\mathrm{1}} {Will}\:{not}\:{defined} \\ $$$${v}_{\mathrm{0}} ={b}>{a} \\ $$$${we}\:{will}\:{show}\:\:{u}_{{n}} >\mathrm{0}\:{and}\:{v}_{{n}} >\mathrm{0}\:\:\:….{P}\left({n}\right) \\ $$$${u}_{\mathrm{0}} ={a}>\mathrm{0}\:\&\:{v}_{\mathrm{0}} ={b}>\mathrm{0}\:\:\: \\ $$$${p}\left({n}+\mathrm{1}\right)..\left\{{u}_{{n}+\mathrm{1}} \sqrt{{u}_{{n}} {v}_{{n}} }>\mathrm{0};{v}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}\left({u}_{{n}} +{v}_{{n}} \right)}>\mathrm{0}\right\} \\ $$$${a}=\frac{\mathrm{8}}{\mathrm{10}};{b}=\frac{\mathrm{9}}{\mathrm{10}} \\ $$$${u}_{\mathrm{1}} \overset{?} {\leqslant}{v}_{\mathrm{1}} \Rightarrow\sqrt{\frac{\mathrm{8}.\mathrm{9}}{\mathrm{100}}}\leqslant\frac{\mathrm{1}}{\mathrm{2}\left(\frac{\mathrm{17}}{\mathrm{10}}\right)}=\frac{\mathrm{5}}{\mathrm{17}} \\ $$$$\Rightarrow\mathrm{8}<\sqrt{\mathrm{72}}\leqslant\frac{\mathrm{50}}{\mathrm{17}}<\mathrm{3}\:{false}!\:{tchek}\:{Quation}\:{sir}! \\ $$$$ \\ $$$$ \\ $$