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Question-7063




Question Number 7063 by Tawakalitu. last updated on 08/Aug/16
Commented by Tawakalitu. last updated on 08/Aug/16
Find the product of the question above.  Please.
Findtheproductofthequestionabove.Please.
Commented by Yozzii last updated on 08/Aug/16
Restating question:  Π_(k=0) ^n cos(2^k θ)=?
Restatingquestion:nk=0cos(2kθ)=?
Commented by Yozzii last updated on 08/Aug/16
u=cosθ cos2θ cos4θ cos8θ ×...× cos2^(n−1) θ cos2^n θ  Let u(n)=Π_(k=0) ^n cos(2^k θ)  Observing the result of u(n) for n=0,1,2,3  u(0)=cosθ=(1/2^0 )cosθ  u(1)=cosθcos2θ=(1/2)(cos3θ+cosθ)  u(2)=cosθcos2θcos4θ=0.5(0.5[cos7θ+cosθ+cos5θ+cos3θ])=(1/2^2 )[cosθ+cos3θ+cos5θ+cos7θ]  u(3)=cosθcos2θcos4θcos8θ=(1/2^2 )[cos7θcos8θ+cos5θcos8θ+cos3θcos8θ+cosθcos8θ]  u(3)=(1/2^3 )(cos15θ+cosθ+cos13θ+cos3θ+cos11θ+cos5θ+cos9θ+cos7θ)  u(3)=(1/2^3 )(cosθ+cos3θ+cos5θ+cos7θ+cos9θ+cos11θ+cos13θ+cos15θ)  I conject that u(n)=(1/2^n )Σ_(i=1) ^2^n  cos[(2i−1)θ]  u(n)sinθ=(1/2^n )Σ_(i=1) ^2^n  sinθcos(2i−1)θ    cos(2i−1)θsinθ=0.5{sin(2i)θ−sin(2i−2)θ}  Let f(i)=sin2iθ   1≤i≤2^n   i∈N    ∴ u(n)sinθ=(1/2^(n+1) )Σ_(i=1) ^2^n  {f(i)−f(i−1)}     u(n)sinθ=(1/2^(n+1) )({f(1)−f(0)}+{f(2)−f(1)}+{f(3)−f(2)}+...+{f(2^n −2)−f(2^n −3)}+{f(2^n −1)−f(2^n −2)}+{f(2^n )−f(2^n −1)})  u(n)sinθ=(1/2^(n+1) )[f(2^n )−f(0)]  u(n)=((sin2×2^n θ−sin(0×θ))/(2^(n+1) sinθ))  u(n)=((sin2^(n+1) θ)/(2^(n+1) sinθ))  ⇒Π_(k=0) ^n cos(2^k θ)=((sin2^(n+1) θ)/(2^(n+1) sinθ))    (θ≠mπ)  n=0,1,2,3,...  By induction, this formular is correct.
u=cosθcos2θcos4θcos8θ××cos2n1θcos2nθLetu(n)=nk=0cos(2kθ)Observingtheresultofu(n)forn=0,1,2,3u(0)=cosθ=120cosθu(1)=cosθcos2θ=12(cos3θ+cosθ)u(2)=cosθcos2θcos4θ=0.5(0.5[cos7θ+cosθ+cos5θ+cos3θ])=122[cosθ+cos3θ+cos5θ+cos7θ]u(3)=cosθcos2θcos4θcos8θ=122[cos7θcos8θ+cos5θcos8θ+cos3θcos8θ+cosθcos8θ]u(3)=123(cos15θ+cosθ+cos13θ+cos3θ+cos11θ+cos5θ+cos9θ+cos7θ)u(3)=123(cosθ+cos3θ+cos5θ+cos7θ+cos9θ+cos11θ+cos13θ+cos15θ)Iconjectthatu(n)=12n2ni=1cos[(2i1)θ]u(n)sinθ=12n2ni=1sinθcos(2i1)θcos(2i1)θsinθ=0.5{sin(2i)θsin(2i2)θ}Letf(i)=sin2iθ1i2niNu(n)sinθ=12n+12ni=1{f(i)f(i1)}u(n)sinθ=12n+1({f(1)f(0)}+{f(2)f(1)}+{f(3)f(2)}++{f(2n2)f(2n3)}+{f(2n1)f(2n2)}+{f(2n)f(2n1)})u(n)sinθ=12n+1[f(2n)f(0)]u(n)=sin2×2nθsin(0×θ)2n+1sinθu(n)=sin2n+1θ2n+1sinθnk=0cos(2kθ)=sin2n+1θ2n+1sinθ(θmπ)n=0,1,2,3,Byinduction,thisformulariscorrect.
Commented by Tawakalitu. last updated on 09/Aug/16
Thanks so much. i really appreciate your effort.
Thankssomuch.ireallyappreciateyoureffort.
Answered by Yozzii last updated on 08/Aug/16
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