Question Number 36920 by maxmathsup by imad last updated on 07/Jun/18
$${let}\:\alpha\:{from}\:{R}\:{and}\:\:{u}_{{n}} \:−\mathrm{2}{cos}\left(\alpha\right){u}_{{n}−\mathrm{1}} \:+{u}_{{n}−\mathrm{2}} =\mathrm{0}\:\:\:{withn}\geqslant\mathrm{2} \\ $$$${find}\:{u}_{{n}} \:{and}\:{study}\:{its}\:{convrgence}. \\ $$
Commented by math khazana by abdo last updated on 10/Jun/18
$$\left({e}\right)\Rightarrow\:{u}_{{n}+\mathrm{2}} \:−\mathrm{2}\:{cos}\alpha\:{u}_{{n}+\mathrm{1}} \:+{u}_{{n}} =\mathrm{0} \\ $$$${the}\:{caracteristic}\:{equation}\:{is} \\ $$$${x}^{\mathrm{2}} \:−\mathrm{2}{cos}\alpha\:{x}\:+\mathrm{1}\:=\mathrm{0} \\ $$$$\Delta^{'} \:={cos}^{\mathrm{2}} \alpha\:−\mathrm{1}\:=−{sin}^{\mathrm{2}} \alpha\:=\left({isin}\alpha\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${x}_{\mathrm{1}} ={cos}\alpha\:+{isin}\alpha\:={e}^{{i}\alpha} \\ $$$${x}_{\mathrm{2}} ={cos}\alpha\:−{isin}\alpha\:={e}^{−{i}\alpha} \\ $$$${u}_{{n}} =\:{a}\:{e}^{{in}\alpha} \:\:+{be}^{−{in}\alpha} \\ $$$${u}_{\mathrm{0}} ={a}+{b} \\ $$$${u}_{\mathrm{1}} ={a}\:{e}^{{i}\alpha} \:+{b}\:{e}^{−{i}\alpha} \:={a}\:{e}^{{i}\alpha} \:+\left({u}_{{o}} −{a}\right){e}^{−{i}\alpha} \\ $$$$=\mathrm{2}{ia}\:{sin}\left(\alpha\right)\:+{u}_{\mathrm{0}} \:{e}^{−{i}\alpha} \Rightarrow\mathrm{2}{ia}\:{sin}\alpha\:={u}_{\mathrm{1}} −{u}_{\mathrm{0}} {e}^{−{i}\alpha} \\ $$$${if}\:{sin}\alpha\neq{o}\:{we}\:{get}\:\:{a}\:=\frac{{u}_{\mathrm{1}} \:−{u}_{{o}} {e}^{−{i}\alpha} }{\mathrm{2}{isin}\alpha} \\ $$$${b}\:={u}_{\mathrm{0}} −{a}\:={u}_{\mathrm{0}} \:\:−\frac{{u}_{\mathrm{1}} \:−{u}_{\mathrm{0}} \:{e}^{−{i}\alpha} }{\mathrm{2}{isin}\alpha}\:\:…. \\ $$