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Question Number 67023 by mathmax by abdo last updated on 21/Aug/19
find the sequence U_n wich verify U_n +U_(n+1) =sin(n) ∀n from n
$${find}\:{the}\:{sequence}\:{U}_{{n}} \:{wich}\:{verify}\:\:{U}_{{n}} +{U}_{{n}+\mathrm{1}} ={sin}\left({n}\right)\:\:\forall{n}\:{from}\:{n} \\ $$
Commented by mathmax by abdo last updated on 23/Aug/19
we have u_n +u_(n+1) =sin(n) ⇒Σ_(k=0) ^(n−1) (u_k +u_(k+1) )(−1)^k =Σ_(k=0) ^n (−1)^k sin(k) ⇒u_0 +u_1 −u_1 −u_2 +u_2 +u_3 +....+(−1)^(n−1) (u_(n−1) +u_n ) =Σ_(k=0) ^n (−1)^k sin(k) ⇒u_0 +(−1)^(n−1) u_n =Σ_(k=0) ^n (−1)^k sin(k) ⇒ u_0 −Σ_(k=0) ^n (−1)^k sin(k) =(−1)^n u_n ⇒u_n =(−1)^n u_0 −(−1)^n Σ_(k=0) ^n (−1)^k sin(k) let find w_n =Σ_(k=0) ^n (−1)^k sin)k) w_n =Im(Σ_(k=0) ^n (−1)^k e^(ik) ) =Im(Σ_(k=0) ^n (−e^i )^k ) and Σ_(k=0) ^n (−e^i )^k =((1−(−e^i )^(n+1) )/(1+e^i )) =((1+(−1)^n e^(i(n+1)) )/(1+e^i )) =((1+(−1)^n e^(i(n+1)) )/(1+cos(1)+isin(1))) =(((1+cos(1)−isin(1))(1+(−1)^n e^(i(n+1)) ))/((1+cos(1))^2 +sin^2 1)) =(((1+cos(1)−isin(1))(1+(−1)^n cos(n+1)+i(−1)^n sin(n+1)))/((1+cos(1))^2 +sin^2 1)) =(((1+cos(1))(1+(−1)^n cos(n+1))+i(−1)^n sin(n+1)(1+cos(1))−isin(1)(1+(−1)^n cos(n+1))+....)/((1+cos(1))^2 +sin^2 (1))) ⇒w_n =(((−1)^n sin(n+1)(1+cos(1))−sin(1)(1+(−1)^n cos)n+1)))/(2+2cos(1) )) ⇒u_n =(−1)^n u_0 −(−1)^n w_n
$${we}\:{have}\:{u}_{{n}} \:+{u}_{{n}+\mathrm{1}} ={sin}\left({n}\right)\:\Rightarrow\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({u}_{{k}} \:+{u}_{{k}+\mathrm{1}} \right)\left(−\mathrm{1}\right)^{{k}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {sin}\left({k}\right) \\ $$$$\Rightarrow{u}_{\mathrm{0}} +{u}_{\mathrm{1}} −{u}_{\mathrm{1}} −{u}_{\mathrm{2}} +{u}_{\mathrm{2}} \:+{u}_{\mathrm{3}} +….+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({u}_{{n}−\mathrm{1}} \:+{u}_{{n}} \right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:{sin}\left({k}\right)\:\Rightarrow{u}_{\mathrm{0}} +\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {u}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {sin}\left({k}\right)\:\Rightarrow \\ $$$${u}_{\mathrm{0}} −\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {sin}\left({k}\right)\:=\left(−\mathrm{1}\right)^{{n}} \:{u}_{{n}} \:\Rightarrow{u}_{{n}} =\left(−\mathrm{1}\right)^{{n}} {u}_{\mathrm{0}} \:−\left(−\mathrm{1}\right)^{{n}} \sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {sin}\left({k}\right) \\ $$$$\left.{l}\left.{et}\:{find}\:\:{w}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {sin}\right){k}\right) \\ $$$${w}_{{n}} ={Im}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {e}^{{ik}} \right)\:={Im}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \left(−{e}^{{i}} \right)^{{k}} \right) \\ $$$${and}\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−{e}^{{i}} \right)^{{k}} \:=\frac{\mathrm{1}−\left(−{e}^{{i}} \right)^{{n}+\mathrm{1}} }{\mathrm{1}+{e}^{{i}} }\:=\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {e}^{{i}\left({n}+\mathrm{1}\right)} }{\mathrm{1}+{e}^{{i}} } \\ $$$$=\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {e}^{{i}\left({n}+\mathrm{1}\right)} }{\mathrm{1}+{cos}\left(\mathrm{1}\right)+{isin}\left(\mathrm{1}\right)}\:=\frac{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)−{isin}\left(\mathrm{1}\right)\right)\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {e}^{{i}\left({n}+\mathrm{1}\right)} \right)}{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \mathrm{1}} \\ $$$$=\frac{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)−{isin}\left(\mathrm{1}\right)\right)\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {cos}\left({n}+\mathrm{1}\right)+{i}\left(−\mathrm{1}\right)^{{n}} {sin}\left({n}+\mathrm{1}\right)\right)}{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \mathrm{1}} \\ $$$$=\frac{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {cos}\left({n}+\mathrm{1}\right)\right)+{i}\left(−\mathrm{1}\right)^{{n}} {sin}\left({n}+\mathrm{1}\right)\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)−{isin}\left(\mathrm{1}\right)\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {cos}\left({n}+\mathrm{1}\right)\right)+….}{\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \left(\mathrm{1}\right)} \\ $$$$\Rightarrow{w}_{{n}} =\frac{\left.\left(\left.−\mathrm{1}\right)^{{n}} {sin}\left({n}+\mathrm{1}\right)\left(\mathrm{1}+{cos}\left(\mathrm{1}\right)\right)−{sin}\left(\mathrm{1}\right)\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {cos}\right){n}+\mathrm{1}\right)\right)}{\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{1}\right)\:} \\ $$$$\Rightarrow{u}_{{n}} =\left(−\mathrm{1}\right)^{{n}} {u}_{\mathrm{0}} −\left(−\mathrm{1}\right)^{{n}} {w}_{{n}} \\ $$$$ \\ $$
Answered by mind is power last updated on 23/Aug/19
let v_(n+1) +v_n =cos(n) and Zn=v_n +iu_n z_(n+1) +z_n =cos(n)+isin(n)=e^(in) ==>(z_(n+1) /e^(i(n)) )+(z_n /e^(in) )=1=e^i (z_(n+1) /e^(i(n+1)) )+(Z_n /e^(in) )=1 let W_n =(z_n /e^(in) )===>e^i W_(n+1) +Wn=1 eix+1=0==>x=−(1/e^(ix) )=−e^(−ix) W_n =c(−1)^n e^(−inx) +t_n withe tn particular solution of e^i t_(n+1) +t_n =1 let t_n =an+b==>e^i (an+a+b)+an+b=1 ∀n∈IN ==>an(e^i +1)=0 and b(e^i +1)=1==>a=0and b=(1/(1+e^i )) ==>t_n =(1/(1+e^i )) W_n =(−1)^n ce^(−in) +(( 1)/(1+e^i )).c∈IC. ==>Z_n =e^(in) W_n =c(−1)^n +(e^(in) /(1+e^i ))=c(−1)^n +((e^(in) +e^(i(n−1)) )/(2+2cos(1))) u_n =IM(Z_n )=k(−1)^n +(1/(2+2cos(1)))[sin(n)+sin(n−1)] sin(a)+sin(b)=2cos(((a−b)/2))sin(((a+b)/2)).and 2+2cos(1)=2+2cos(2.(1/2))=4cos^2 ((1/2)) we find u_n =k(−1)^n +((2cos((1/2))sin(n−(1/2)))/(4cos^2 ((1/2))))=k(−1)^n +((sin(n−(1/2)))/(2cos((1/2))))
$${let}\:{v}_{{n}+\mathrm{1}} +{v}_{{n}} ={cos}\left({n}\right) \\ $$$${and}\:{Zn}={v}_{{n}} +{iu}_{{n}} \\ $$$${z}_{{n}+\mathrm{1}} +{z}_{{n}} ={cos}\left({n}\right)+{isin}\left({n}\right)={e}^{{in}} \\ $$$$==>\frac{{z}_{{n}+\mathrm{1}} }{{e}^{{i}\left({n}\right)} }+\frac{{z}_{{n}} }{{e}^{{in}} }=\mathrm{1}={e}^{{i}} \frac{{z}_{{n}+\mathrm{1}} }{{e}^{{i}\left({n}+\mathrm{1}\right)} }+\frac{{Z}_{{n}} }{{e}^{{in}} }=\mathrm{1} \\ $$$${let}\:{W}_{{n}} =\frac{{z}_{{n}} }{{e}^{{in}} }===>{e}^{{i}} {W}_{{n}+\mathrm{1}} +{Wn}=\mathrm{1} \\ $$$${eix}+\mathrm{1}=\mathrm{0}==>{x}=−\frac{\mathrm{1}}{{e}^{{ix}} }=−{e}^{−{ix}} \\ $$$${W}_{{n}} ={c}\left(−\mathrm{1}\right)^{{n}} {e}^{−{inx}} +{t}_{{n}} \\ $$$${withe}\:{tn}\:{particular}\:{solution}\:{of}\:{e}^{{i}} {t}_{{n}+\mathrm{1}} +{t}_{{n}} =\mathrm{1} \\ $$$${let}\:{t}_{{n}} ={an}+{b}==>{e}^{{i}} \left({an}+{a}+{b}\right)+{an}+{b}=\mathrm{1}\:\:\forall{n}\in{IN} \\ $$$$==>{an}\left({e}^{{i}} +\mathrm{1}\right)=\mathrm{0}\:\:{and}\:{b}\left({e}^{{i}} +\mathrm{1}\right)=\mathrm{1}==>{a}=\mathrm{0}{and}\:{b}=\frac{\mathrm{1}}{\mathrm{1}+{e}^{{i}} } \\ $$$$==>{t}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}+{e}^{{i}} } \\ $$$${W}_{{n}} =\left(−\mathrm{1}\right)^{{n}} {ce}^{−{in}} +\frac{\:\mathrm{1}}{\mathrm{1}+{e}^{{i}} }.{c}\in{IC}. \\ $$$$==>{Z}_{{n}} ={e}^{{in}} {W}_{{n}} ={c}\left(−\mathrm{1}\right)^{{n}} +\frac{{e}^{{in}} }{\mathrm{1}+{e}^{{i}} }={c}\left(−\mathrm{1}\right)^{{n}} +\frac{{e}^{{in}} +{e}^{{i}\left({n}−\mathrm{1}\right)} }{\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{1}\right)} \\ $$$${u}_{{n}} ={IM}\left({Z}_{{n}} \right)={k}\left(−\mathrm{1}\right)^{{n}} +\frac{\mathrm{1}}{\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{1}\right)}\left[{sin}\left({n}\right)+{sin}\left({n}−\mathrm{1}\right)\right] \\ $$$${sin}\left({a}\right)+{sin}\left({b}\right)=\mathrm{2}{cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right){sin}\left(\frac{{a}+{b}}{\mathrm{2}}\right).{and}\:\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{1}\right)=\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${we}\:{find}\:{u}_{{n}} ={k}\left(−\mathrm{1}\right)^{{n}} +\frac{\mathrm{2}{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){sin}\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)}={k}\left(−\mathrm{1}\right)^{{n}} +\frac{{sin}\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 23/Aug/19
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$

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