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According-to-WolframAlpha-k-0-n-1-x-1-k-1-x-n-2-1-x-1-x-n-1-2-1-Can-anyone-work-out-how-

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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?
$$\mathrm{According}\:\mathrm{to}\:\mathrm{WolframAlpha}: \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\left(\mathrm{1}−{x}^{\left(−\mathrm{1}\right)^{{k}} } \right)=\left(\mathrm{1}−{x}\right)^{\lfloor\frac{{n}}{\mathrm{2}}\rfloor+\mathrm{1}} \left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\lfloor\frac{{n}−\mathrm{1}}{\mathrm{2}}\rfloor+\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{anyone}\:\mathrm{work}\:\mathrm{out}\:\mathrm{how}? \\ $$
Commented by FilupSmith last updated on 24/Jun/16
Amazing!
$$\mathrm{Amazing}! \\ $$
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)
$${Let}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\left(\mathrm{1}−{x}^{\left(−\mathrm{1}\right)^{{k}} } \right)=\phi,\:{n}\in\mathbb{Z}^{\geqslant} \:\left({x}\neq\mathrm{0}\right) \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\left(\mathrm{I}\right)\:{For}\:{n}\:{being}\:{even},\:{let}\:{n}=\mathrm{2}{r},\:{r}\in\mathbb{Z}^{\geqslant} . \\ $$$${r}=\frac{{n}}{\mathrm{2}}.\:{As}\:{k}\:{varies}\:{from}\:{k}=\mathrm{0}\:{to}\:{k}=\mathrm{2}{r}, \\ $$$${r}\:{terms}\:{in}\:{the}\:{product}\:{will} \\ $$$${have}\:{k}=\mathrm{2}{t}−\mathrm{1}\left({t}\in\mathbb{N}\right)\Rightarrow\left(\mathrm{1}−{x}^{\left(−\mathrm{1}\right)^{{k}} } \right)=\mathrm{1}−\frac{\mathrm{1}}{{x}}=\frac{{x}−\mathrm{1}}{{x}}, \\ $$$${and}\:{the}\:{remaining}\:{r}+\mathrm{1}\:{terms}\:{will}\:{have} \\ $$$${k}=\mathrm{2}{t}\Rightarrow\:\left(\mathrm{1}−{x}^{\left(−\mathrm{1}\right)^{{k}} } \right)=\mathrm{1}−{x}. \\ $$$${So},\:{we}\:{may}\:{write}\: \\ $$$$\phi=\left\{\underset{{k}\in\mathbb{E},\mathrm{0}\leqslant{k}\leqslant{n}} {\prod}\left(\mathrm{1}−{x}^{\left(−\mathrm{1}\right)^{{k}} } \right)\right\}\left\{\underset{{k}\in\mathbb{O},\mathrm{0}\leqslant{k}\leqslant{n}} {\prod}\left(\mathrm{1}−{x}^{\left(−\mathrm{1}\right)^{{k}} } \right)\right\} \\ $$$$\phi=\left(\mathrm{1}−{x}\right)^{{r}+\mathrm{1}} ×\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{{r}} \:{or} \\ $$$$\phi=\left(\mathrm{1}−{x}\right)^{\frac{{n}}{\mathrm{2}}+\mathrm{1}} \left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\frac{{n}}{\mathrm{2}}} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\left(\mathrm{II}\right)\:{If}\:{n}\:{is}\:{odd}\:{then}\:{let}\:{n}=\mathrm{2}{r}+\mathrm{1},\:{r}\in\mathbb{Z}^{\geqslant} . \\ $$$${There}\:{will}\:{be}\:{r}+\mathrm{1}\:{terms}\:{with}\:{odd}\:{k} \\ $$$${and}\:{r}+\mathrm{1}\:{terms}\:{with}\:{even}\:{k}. \\ $$$${So}\:{we}\:{can}\:{write},\:{similarly}\:{to}\:{above}, \\ $$$$\phi=\left(\mathrm{1}−{x}\right)^{{r}+\mathrm{1}} \left(\frac{{x}−\mathrm{1}}{{x}}\right)^{{r}+\mathrm{1}} \\ $$$${or}\:\phi=\left(\mathrm{1}−{x}\right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}+\mathrm{1}} \left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}+\mathrm{1}} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${Now},\:{if}\:{n}\:{is}\:{even}\:{then}\:{truly}\:\lfloor\frac{{n}}{\mathrm{2}}\rfloor=\frac{{n}}{\mathrm{2}} \\ $$$$\Rightarrow\lfloor\frac{{n}}{\mathrm{2}}\rfloor+\mathrm{1}=\frac{{n}}{\mathrm{2}}+\mathrm{1}.\:{So},\:{this}\:{is}\:{an}\:{identity}\:{for}\:{part}\:\left(\mathrm{I}\right)\: \\ $$$${for}\:{the}\:{term}\:\left(\mathrm{1}−{x}\right)^{\frac{{n}}{\mathrm{2}}+\mathrm{1}} .\: \\ $$$${Also},\:{n}−\mathrm{1}=\mathrm{2}{r}+\mathrm{1}\:{for}\:{some}\:{r}\in\mathbb{Z}^{\geqslant} . \\ $$$$\Rightarrow\frac{{n}−\mathrm{1}}{\mathrm{2}}={r}+\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\lfloor\frac{{n}−\mathrm{1}}{\mathrm{2}}\rfloor={r}\Rightarrow\lfloor\frac{{n}−\mathrm{1}}{\mathrm{2}}\rfloor+\mathrm{1}={r}+\mathrm{1}=\frac{{n}−\mathrm{2}}{\mathrm{2}}+\mathrm{1}=\frac{{n}}{\mathrm{2}}. \\ $$$${Hence},\:{we}\:{can}\:{use}\:\lfloor\frac{{n}−\mathrm{1}}{\mathrm{2}}\rfloor+\mathrm{1}\:{instead}\:{of}\:\frac{{n}}{\mathrm{2}} \\ $$$${for}\:{the}\:{term}\:\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{{n}/\mathrm{2}} . \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${If}\:{n}\:{is}\:{odd}\:{then}\:{for}\:{n}=\mathrm{2}{r}+\mathrm{1}\: \\ $$$$\Rightarrow\frac{{n}}{\mathrm{2}}={r}+\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\lfloor\frac{{n}}{\mathrm{2}}\rfloor={r}\Rightarrow\lfloor\frac{{n}}{\mathrm{2}}\rfloor+\mathrm{1}={r}+\mathrm{1}=\frac{{n}−\mathrm{1}}{\mathrm{2}}+\mathrm{1}. \\ $$$${Therefore},\:{this}\:{is}\:{an}\:{identity}\:{for} \\ $$$${the}\:{term}\:\left(\mathrm{1}−{x}\right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}+\mathrm{1}} \:{in}\:{part}\:\left(\mathrm{II}\right). \\ $$$${For}\:{n}=\mathrm{2}{r}+\mathrm{1}\Rightarrow{n}−\mathrm{1}=\mathrm{2}{r}\Rightarrow\frac{{n}−\mathrm{1}}{\mathrm{2}}={r} \\ $$$$\Rightarrow\lfloor\frac{{n}−\mathrm{1}}{\mathrm{2}}\rfloor={r}\Rightarrow\lfloor\frac{{n}−\mathrm{1}}{\mathrm{2}}\rfloor+\mathrm{1}={r}+\mathrm{1}=\frac{{n}−\mathrm{1}}{\mathrm{2}}+\mathrm{1}. \\ $$$${Thus},\:{for}\:{the}\:{term}\:{in}\:\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}+\mathrm{1}} \\ $$$${we}\:{can}\:{write}\:\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\lfloor\frac{{n}−\mathrm{1}}{\mathrm{2}}\rfloor+\mathrm{1}} \:{instead}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${Notice}\:{in}\:{the}\:{both}\:{cases},\:\left(\mathrm{I}\right)\:{and}\:\left(\mathrm{II}\right),\: \\ $$$${we}\:{can}\:{write}\:\left(\mathrm{1}−{x}\right)^{\lfloor\frac{{n}}{\mathrm{2}}\rfloor+\mathrm{1}} \:{and}\:\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\lfloor\frac{{n}−\mathrm{1}}{\mathrm{2}}\rfloor+\mathrm{1}} \\ $$$${in}\:{the}\:{result}\:{of}\:\phi\:{for}\:{any}\:{n}\in\mathbb{Z}^{\geqslant} . \\ $$$$\therefore\:{Generally},\:{for}\:{x}\neq\mathrm{0}\:{and}\:{n}\in\mathbb{Z}^{\geqslant} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\left(\mathrm{1}−{x}^{\left(−\mathrm{1}\right)^{{k}} } \right)=\left(\mathrm{1}−{x}\right)^{\lfloor\frac{{n}}{\mathrm{2}}\rfloor+\mathrm{1}} \left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\lfloor\frac{{n}−\mathrm{1}}{\mathrm{2}}\rfloor+\mathrm{1}} .\:\left({Shown}\right) \\ $$
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
$${S}=\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\left(\mathrm{1}−{x}\right)… \\ $$$${S}=\left(\mathrm{1}−{x}\right)\left(\frac{{x}−\mathrm{1}}{{x}}\right)\left(\mathrm{1}−{x}\right)… \\ $$$${n}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−{x} \\ $$$${n}=\mathrm{1} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}−{x}\right)\left({x}−\mathrm{1}\right)}{{x}} \\ $$$${n}=\mathrm{2} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left({x}−\mathrm{1}\right)}{{x}} \\ $$$${n}=\mathrm{3} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$${etc} \\ $$

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