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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
findthesequenceUnwichverifyUn+Un+1=sin(n)nfromn
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
wehaveun+un+1=sin(n)k=0n1(uk+uk+1)(1)k=k=0n(1)ksin(k)u0+u1u1u2+u2+u3+.+(1)n1(un1+un)=k=0n(1)ksin(k)u0+(1)n1un=k=0n(1)ksin(k)u0k=0n(1)ksin(k)=(1)nunun=(1)nu0(1)nk=0n(1)ksin(k)letfindwn=k=0n(1)ksin)k)wn=Im(k=0n(1)keik)=Im(k=0n(ei)k)andk=0n(ei)k=1(ei)n+11+ei=1+(1)nei(n+1)1+ei=1+(1)nei(n+1)1+cos(1)+isin(1)=(1+cos(1)isin(1))(1+(1)nei(n+1))(1+cos(1))2+sin21=(1+cos(1)isin(1))(1+(1)ncos(n+1)+i(1)nsin(n+1))(1+cos(1))2+sin21=(1+cos(1))(1+(1)ncos(n+1))+i(1)nsin(n+1)(1+cos(1))isin(1)(1+(1)ncos(n+1))+.(1+cos(1))2+sin2(1)wn=(1)nsin(n+1)(1+cos(1))sin(1)(1+(1)ncos)n+1))2+2cos(1)un=(1)nu0(1)nwn
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))))
letvn+1+vn=cos(n)andZn=vn+iunzn+1+zn=cos(n)+isin(n)=ein==>zn+1ei(n)+znein=1=eizn+1ei(n+1)+Znein=1letWn=znein===>eiWn+1+Wn=1eix+1=0==>x=1eix=eixWn=c(1)neinx+tnwithetnparticularsolutionofeitn+1+tn=1lettn=an+b==>ei(an+a+b)+an+b=1nIN==>an(ei+1)=0andb(ei+1)=1==>a=0andb=11+ei==>tn=11+eiWn=(1)ncein+11+ei.cIC.==>Zn=einWn=c(1)n+ein1+ei=c(1)n+ein+ei(n1)2+2cos(1)un=IM(Zn)=k(1)n+12+2cos(1)[sin(n)+sin(n1)]sin(a)+sin(b)=2cos(ab2)sin(a+b2).and2+2cos(1)=2+2cos(2.12)=4cos2(12)wefindun=k(1)n+2cos(12)sin(n12)4cos2(12)=k(1)n+sin(n12)2cos(12)
Commented by mathmax by abdo last updated on 23/Aug/19
thank you sir.
thankyousir.

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