Question Number 67023 by mathmax by abdo last updated on 21/Aug/19
$${find}\:{the}\:{sequence}\:{U}_{{n}} \:{wich}\:{verify}\:\:{U}_{{n}} +{U}_{{n}+\mathrm{1}} ={sin}\left({n}\right)\:\:\forall{n}\:{from}\:{n} \\ $$
Commented by mathmax by abdo last updated on 23/Aug/19
$${we}\:{have}\:{u}_{{n}} \:+{u}_{{n}+\mathrm{1}} ={sin}\left({n}\right)\:\Rightarrow\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({u}_{{k}} \:+{u}_{{k}+\mathrm{1}} \right)\left(−\mathrm{1}\right)^{{k}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {sin}\left({k}\right) \\ $$$$\Rightarrow{u}_{\mathrm{0}} +{u}_{\mathrm{1}} −{u}_{\mathrm{1}} −{u}_{\mathrm{2}} +{u}_{\mathrm{2}} \:+{u}_{\mathrm{3}} +….+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({u}_{{n}−\mathrm{1}} \:+{u}_{{n}} \right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:{sin}\left({k}\right)\:\Rightarrow{u}_{\mathrm{0}} +\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {u}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {sin}\left({k}\right)\:\Rightarrow \\ $$$${u}_{\mathrm{0}} −\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {sin}\left({k}\right)\:=\left(−\mathrm{1}\right)^{{n}} \:{u}_{{n}} \:\Rightarrow{u}_{{n}} =\left(−\mathrm{1}\right)^{{n}} {u}_{\mathrm{0}} \:−\left(−\mathrm{1}\right)^{{n}} \sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {sin}\left({k}\right) \\ $$$$\left.{l}\left.{et}\:{find}\:\:{w}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {sin}\right){k}\right) \\ $$$${w}_{{n}} ={Im}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {e}^{{ik}} \right)\:={Im}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \left(−{e}^{{i}} \right)^{{k}} \right) \\ $$$${and}\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−{e}^{{i}} \right)^{{k}} \:=\frac{\mathrm{1}−\left(−{e}^{{i}} \right)^{{n}+\mathrm{1}} }{\mathrm{1}+{e}^{{i}} }\:=\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {e}^{{i}\left({n}+\mathrm{1}\right)} }{\mathrm{1}+{e}^{{i}} } \\ $$$$=\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {e}^{{i}\left({n}+\mathrm{1}\right)} }{\mathrm{1}+{cos}\left(\mathrm{1}\right)+{isin}\left(\mathrm{1}\right)}\:=\frac{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)−{isin}\left(\mathrm{1}\right)\right)\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {e}^{{i}\left({n}+\mathrm{1}\right)} \right)}{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \mathrm{1}} \\ $$$$=\frac{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)−{isin}\left(\mathrm{1}\right)\right)\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {cos}\left({n}+\mathrm{1}\right)+{i}\left(−\mathrm{1}\right)^{{n}} {sin}\left({n}+\mathrm{1}\right)\right)}{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \mathrm{1}} \\ $$$$=\frac{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {cos}\left({n}+\mathrm{1}\right)\right)+{i}\left(−\mathrm{1}\right)^{{n}} {sin}\left({n}+\mathrm{1}\right)\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)−{isin}\left(\mathrm{1}\right)\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {cos}\left({n}+\mathrm{1}\right)\right)+….}{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \left(\mathrm{1}\right)} \\ $$$$\Rightarrow{w}_{{n}} =\frac{\left.\left(\left.−\mathrm{1}\right)^{{n}} {sin}\left({n}+\mathrm{1}\right)\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)−{sin}\left(\mathrm{1}\right)\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {cos}\right){n}+\mathrm{1}\right)\right)}{\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{1}\right)\:} \\ $$$$\Rightarrow{u}_{{n}} =\left(−\mathrm{1}\right)^{{n}} {u}_{\mathrm{0}} −\left(−\mathrm{1}\right)^{{n}} {w}_{{n}} \\ $$$$ \\ $$
Answered by mind is power last updated on 23/Aug/19
$${let}\:{v}_{{n}+\mathrm{1}} +{v}_{{n}} ={cos}\left({n}\right) \\ $$$${and}\:{Zn}={v}_{{n}} +{iu}_{{n}} \\ $$$${z}_{{n}+\mathrm{1}} +{z}_{{n}} ={cos}\left({n}\right)+{isin}\left({n}\right)={e}^{{in}} \\ $$$$==>\frac{{z}_{{n}+\mathrm{1}} }{{e}^{{i}\left({n}\right)} }+\frac{{z}_{{n}} }{{e}^{{in}} }=\mathrm{1}={e}^{{i}} \frac{{z}_{{n}+\mathrm{1}} }{{e}^{{i}\left({n}+\mathrm{1}\right)} }+\frac{{Z}_{{n}} }{{e}^{{in}} }=\mathrm{1} \\ $$$${let}\:{W}_{{n}} =\frac{{z}_{{n}} }{{e}^{{in}} }===>{e}^{{i}} {W}_{{n}+\mathrm{1}} +{Wn}=\mathrm{1} \\ $$$${eix}+\mathrm{1}=\mathrm{0}==>{x}=−\frac{\mathrm{1}}{{e}^{{ix}} }=−{e}^{−{ix}} \\ $$$${W}_{{n}} ={c}\left(−\mathrm{1}\right)^{{n}} {e}^{−{inx}} +{t}_{{n}} \\ $$$${withe}\:{tn}\:{particular}\:{solution}\:{of}\:{e}^{{i}} {t}_{{n}+\mathrm{1}} +{t}_{{n}} =\mathrm{1} \\ $$$${let}\:{t}_{{n}} ={an}+{b}==>{e}^{{i}} \left({an}+{a}+{b}\right)+{an}+{b}=\mathrm{1}\:\:\forall{n}\in{IN} \\ $$$$==>{an}\left({e}^{{i}} +\mathrm{1}\right)=\mathrm{0}\:\:{and}\:{b}\left({e}^{{i}} +\mathrm{1}\right)=\mathrm{1}==>{a}=\mathrm{0}{and}\:{b}=\frac{\mathrm{1}}{\mathrm{1}+{e}^{{i}} } \\ $$$$==>{t}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}+{e}^{{i}} } \\ $$$${W}_{{n}} =\left(−\mathrm{1}\right)^{{n}} {ce}^{−{in}} +\frac{\:\mathrm{1}}{\mathrm{1}+{e}^{{i}} }.{c}\in{IC}. \\ $$$$==>{Z}_{{n}} ={e}^{{in}} {W}_{{n}} ={c}\left(−\mathrm{1}\right)^{{n}} +\frac{{e}^{{in}} }{\mathrm{1}+{e}^{{i}} }={c}\left(−\mathrm{1}\right)^{{n}} +\frac{{e}^{{in}} +{e}^{{i}\left({n}−\mathrm{1}\right)} }{\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{1}\right)} \\ $$$${u}_{{n}} ={IM}\left({Z}_{{n}} \right)={k}\left(−\mathrm{1}\right)^{{n}} +\frac{\mathrm{1}}{\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{1}\right)}\left[{sin}\left({n}\right)+{sin}\left({n}−\mathrm{1}\right)\right] \\ $$$${sin}\left({a}\right)+{sin}\left({b}\right)=\mathrm{2}{cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right){sin}\left(\frac{{a}+{b}}{\mathrm{2}}\right).{and}\:\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{1}\right)=\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${we}\:{find}\:{u}_{{n}} ={k}\left(−\mathrm{1}\right)^{{n}} +\frac{\mathrm{2}{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){sin}\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)}={k}\left(−\mathrm{1}\right)^{{n}} +\frac{{sin}\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 23/Aug/19
$${thank}\:{you}\:{sir}. \\ $$