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Question Number 6311 by FilupSmith last updated on 23/Jun/16
According to WolframAlpha: Π_(k=0) ^n (1−x^((−1)^k ) )=(1−x)^(⌊(n/2)⌋+1) (((x−1)/x))^(⌊((n−1)/2)⌋+1) Can anyone work out how?
AccordingtoWolframAlpha:∏nk=0(1−x(−1)k)=(1−x)⌊n2⌋+1(x−1x)⌊n−12⌋+1Cananyoneworkouthow?
Commented by FilupSmith last updated on 24/Jun/16
Amazing!
Amazing!
Commented by Yozzii last updated on 23/Jun/16
Let Π_(k=0) ^n (1−x^((−1)^k ) )=φ, n∈Z^≥ (x≠0) −−−−−−−−−−−−−−−−−−−−−−−− (I) For n being even, let n=2r, r∈Z^≥ . r=(n/2). As k varies from k=0 to k=2r, r terms in the product will have k=2t−1(t∈N)⇒(1−x^((−1)^k ) )=1−(1/x)=((x−1)/x), and the remaining r+1 terms will have k=2t⇒ (1−x^((−1)^k ) )=1−x. So, we may write φ={Π_(k∈E,0≤k≤n) (1−x^((−1)^k ) )}{Π_(k∈O,0≤k≤n) (1−x^((−1)^k ) )} φ=(1−x)^(r+1) ×(((x−1)/x))^r or φ=(1−x)^((n/2)+1) (((x−1)/x))^(n/2) −−−−−−−−−−−−−−−−−−−−−−−−−− (II) If n is odd then let n=2r+1, r∈Z^≥ . There will be r+1 terms with odd k and r+1 terms with even k. So we can write, similarly to above, φ=(1−x)^(r+1) (((x−1)/x))^(r+1) or φ=(1−x)^(((n−1)/2)+1) (((x−1)/x))^(((n−1)/2)+1) −−−−−−−−−−−−−−−−−−−−−−−−− Now, if n is even then truly ⌊(n/2)⌋=(n/2) ⇒⌊(n/2)⌋+1=(n/2)+1. So, this is an identity for part (I) for the term (1−x)^((n/2)+1) . Also, n−1=2r+1 for some r∈Z^≥ . ⇒((n−1)/2)=r+(1/2)⇒⌊((n−1)/2)⌋=r⇒⌊((n−1)/2)⌋+1=r+1=((n−2)/2)+1=(n/2). Hence, we can use ⌊((n−1)/2)⌋+1 instead of (n/2) for the term (((x−1)/x))^(n/2) . −−−−−−−−−−−−−−−−−−−−−−−−− If n is odd then for n=2r+1 ⇒(n/2)=r+(1/2)⇒⌊(n/2)⌋=r⇒⌊(n/2)⌋+1=r+1=((n−1)/2)+1. Therefore, this is an identity for the term (1−x)^(((n−1)/2)+1) in part (II). For n=2r+1⇒n−1=2r⇒((n−1)/2)=r ⇒⌊((n−1)/2)⌋=r⇒⌊((n−1)/2)⌋+1=r+1=((n−1)/2)+1. Thus, for the term in (((x−1)/x))^(((n−1)/2)+1) we can write (((x−1)/x))^(⌊((n−1)/2)⌋+1) instead. −−−−−−−−−−−−−−−−−−−−−−−−− Notice in the both cases, (I) and (II), we can write (1−x)^(⌊(n/2)⌋+1) and (((x−1)/x))^(⌊((n−1)/2)⌋+1) in the result of φ for any n∈Z^≥ . ∴ Generally, for x≠0 and n∈Z^≥ Π_(k=0) ^n (1−x^((−1)^k ) )=(1−x)^(⌊(n/2)⌋+1) (((x−1)/x))^(⌊((n−1)/2)⌋+1) . (Shown)
Let∏nk=0(1−x(−1)k)=ϕ,n∈Z⩾(x≠0)−−−−−−−−−−−−−−−−−−−−−−−−(I)Fornbeingeven,letn=2r,r∈Z⩾.r=n2.Askvariesfromk=0tok=2r,rtermsintheproductwillhavek=2t−1(t∈N)⇒(1−x(−1)k)=1−1x=x−1x,andtheremainingr+1termswillhavek=2t⇒(1−x(−1)k)=1−x.So,wemaywriteϕ={∏k∈E,0⩽k⩽n(1−x(−1)k)}{∏k∈O,0⩽k⩽n(1−x(−1)k)}ϕ=(1−x)r+1×(x−1x)rorϕ=(1−x)n2+1(x−1x)n2−−−−−−−−−−−−−−−−−−−−−−−−−−(II)Ifnisoddthenletn=2r+1,r∈Z⩾.Therewillber+1termswithoddkandr+1termswithevenk.Sowecanwrite,similarlytoabove,ϕ=(1−x)r+1(x−1x)r+1orϕ=(1−x)n−12+1(x−1x)n−12+1−−−−−−−−−−−−−−−−−−−−−−−−−Now,ifniseventhentruly⌊n2⌋=n2⇒⌊n2⌋+1=n2+1.So,thisisanidentityforpart(I)fortheterm(1−x)n2+1.Also,n−1=2r+1forsomer∈Z⩾.⇒n−12=r+12⇒⌊n−12⌋=r⇒⌊n−12⌋+1=r+1=n−22+1=n2.Hence,wecanuse⌊n−12⌋+1insteadofn2fortheterm(x−1x)n/2.−−−−−−−−−−−−−−−−−−−−−−−−−Ifnisoddthenforn=2r+1⇒n2=r+12⇒⌊n2⌋=r⇒⌊n2⌋+1=r+1=n−12+1.Therefore,thisisanidentityfortheterm(1−x)n−12+1inpart(II).Forn=2r+1⇒n−1=2r⇒n−12=r⇒⌊n−12⌋=r⇒⌊n−12⌋+1=r+1=n−12+1.Thus,forthetermin(x−1x)n−12+1wecanwrite(x−1x)⌊n−12⌋+1instead.−−−−−−−−−−−−−−−−−−−−−−−−−Noticeinthebothcases,(I)and(II),wecanwrite(1−x)⌊n2⌋+1and(x−1x)⌊n−12⌋+1intheresultofϕforanyn∈Z⩾.∴Generally,forx≠0andn∈Z⩾∏nk=0(1−x(−1)k)=(1−x)⌊n2⌋+1(x−1x)⌊n−12⌋+1.(Shown)
Commented by FilupSmith last updated on 23/Jun/16
S=(1−x)(1−(1/x))(1−x)... S=(1−x)(((x−1)/x))(1−x)... n=0 ⇒1−x n=1 ⇒(((1−x)(x−1))/x) n=2 ⇒(((1−x)^2 (x−1))/x) n=3 ⇒(((1−x)^2 (1−x)^2 )/x^2 ) etc
S=(1−x)(1−1x)(1−x)…S=(1−x)(x−1x)(1−x)…n=0⇒1−xn=1⇒(1−x)(x−1)xn=2⇒(1−x)2(x−1)xn=3⇒(1−x)2(1−x)2x2etc

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