Question Number 190464 by alcohol last updated on 03/Apr/23
$$\begin{cases}{{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −\mathrm{3}\:}\\{{v}_{{n}+\mathrm{1}} \:=\:\mathrm{4}{v}_{{n}} }\end{cases}\::\:{u}_{\mathrm{0}} \:=\:{v}_{\mathrm{0}} \:=\:\mathrm{1} \\ $$$${w}_{{n}} \:=\:\frac{\mathrm{1}−{u}_{{n}} }{{v}_{{n}} } \\ $$$$−\:{show}\:{that}\:{w}_{{n}} \:{is}\:{bounded} \\ $$$$−\:{find}\:{a},{b}\in\mathbb{R}\:{such}\:{that}\:{a}\:\leqslant\:{w}_{{n}} \:\leqslant\:{b} \\ $$
Answered by mr W last updated on 03/Apr/23
$${u}_{\mathrm{0}} =\mathrm{1} \\ $$$${u}_{\mathrm{1}} =\mathrm{1}−\mathrm{3} \\ $$$${u}_{\mathrm{2}} =\mathrm{1}−\mathrm{3}×\mathrm{2} \\ $$$$… \\ $$$${u}_{{n}} =\mathrm{1}−\mathrm{3}×{n}=\mathrm{1}−\mathrm{3}{n} \\ $$$$ \\ $$$${v}_{\mathrm{0}} =\mathrm{1} \\ $$$${v}_{\mathrm{1}} =\mathrm{1}×\mathrm{4} \\ $$$${v}_{\mathrm{2}} =\mathrm{1}×\mathrm{4}^{\mathrm{2}} \\ $$$$… \\ $$$${v}_{{n}} =\mathrm{1}×\mathrm{4}^{{n}} =\mathrm{4}^{{n}} \\ $$$${w}_{{n}} =\frac{\mathrm{1}−\left(\mathrm{1}−\mathrm{3}{n}\right)}{\mathrm{4}^{{n}} }=\frac{\mathrm{3}{n}}{\mathrm{4}^{{n}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{w}_{{n}} =\mathrm{0} \\ $$$$\mathrm{0}\leqslant{w}_{{n}} \leqslant\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${for}\:{a}\leqslant{w}_{{n}} \leqslant{b}:\: \\ $$$${a}=\mathrm{0},\:{b}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by alcohol last updated on 03/Apr/23
$${alright}\:{thank}\:{you} \\ $$$${what}\:{about}\:{the}\:{bounds}? \\ $$$$ \\ $$
Commented by mr W last updated on 03/Apr/23
$${i}\:{misread}.\:{w}_{{n}} =\frac{\mathrm{3}{n}}{\mathrm{4}^{{n}} }. \\ $$
Commented by alcohol last updated on 03/Apr/23
$${thank}\:{you} \\ $$$${w}_{{n}} \:=\:\frac{\mathrm{3}{n}}{\mathrm{4}^{{n}} }\:{right}\:? \\ $$$${please}\:{how}\:{did}\:{you}\:{get}\:\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} ? \\ $$
Commented by mr W last updated on 03/Apr/23
$${w}_{\mathrm{0}} =\mathrm{0}={min} \\ $$$${w}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{4}}={max} \\ $$$$\mathrm{0}\leqslant{w}_{{n}} \leqslant\frac{\mathrm{3}}{\mathrm{4}} \\ $$