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Question Number 155183 by mathdanisur last updated on 26/Sep/21
Answered by aleks041103 last updated on 26/Sep/21
1−tan^2 (x/2^k )=((cos^2 (x/2^k )−sin^2 (x/2^k ))/(cos^2 (x/2^k )))=((cos(x/2^(k−1) ))/(cos^2 (x/2^k ))) Π_(k=1) ^n cos(x/2^(k−1) )=cosx cos(x/2) ... cos(x/2^(n−1) )= =(1/(sin(x/2^(n−1) )))sin(x/2^(n−1) )cos(x/2^(n−1) )cos(x/2^(n−2) )...cos(x/2)cosx= =(1/(sin(x/2^(n−1) ))) (1/2)sin(x/2^(n−2) )cos(x/2^(n−2) )...cos(x/2)cosx= =(1/(2^2 sin(x/2^(n−1) )))sin(x/2^(n−3) )Π_(k=0) ^(n−3) cos(x/2^k )=...= =(1/(2^m sin(x/2^(n−1) )))sin(x/2^(n−1−m) )Π_(k=0) ^(n−1−m) cos(x/2^k )= =(1/(2^(n−1) sin(x/2^(n−1) )))sinxcosx=((sin(2x))/(2^n sin((x/2^(n−1) )))) Π_(k=1) ^n cos(x/2^k )=Π_(k=1) ^n cos((x/2)/2^(k−1) )= =((sin(2(x/2)))/(2^n sin(((x/2)/2^(n−1) ))))=((sinx)/(2^n sin(x/2^n ))) ⇒Π_(k=1) ^n ((cos(x/2^(k−1) ))/(cos^2 (x/2^k )))=(((Π_(k=1) ^n cos(x/2^(k−1) )))/((Π_(k=1) ^n cos(x/2^k ))^2 ))= =(((sin(2x))/(2^n sin((x/2^(n−1) ))))/((((sinx)/(2^n sin(x/2^n ))))^2 ))=((sin(2x)2^(2n) sin^2 (x/2^n ))/(2^n sin((2x)/2^n )sin^2 x))= =((2sinx cosx 2^n sin^2 (x/2^n ))/(2sin(x/2^n )cos(x/2^n )sin^2 x))=((2^n cosx sin(x/2^n ))/(cos(x/2^n )sinx))= =2^n ((tan((x/2^n )))/(tan(x))) ⇒Ω=lim_(x→0) (x/(tan(x)))lim_(n→∞) ((tan(x/2^n ))/(x/2^n ))=1
$$\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }=\frac{{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }−{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }}{{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }}=\frac{{cos}\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }}{{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }={cosx}\:{cos}\frac{{x}}{\mathrm{2}}\:…\:{cos}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }= \\ $$$$=\frac{\mathrm{1}}{{sin}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }}{sin}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }{cos}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }{cos}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{2}} }…{cos}\frac{{x}}{\mathrm{2}}{cosx}= \\ $$$$=\frac{\mathrm{1}}{{sin}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }}\:\frac{\mathrm{1}}{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{2}} }{cos}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{2}} }…{cos}\frac{{x}}{\mathrm{2}}{cosx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} {sin}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }}{sin}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{3}} }\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{3}} {\prod}}{cos}\frac{{x}}{\mathrm{2}^{{k}} }=…= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{m}} {sin}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }}{sin}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}−{m}} }\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}−{m}} {\prod}}{cos}\frac{{x}}{\mathrm{2}^{{k}} }= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} {sin}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }}{sinxcosx}=\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}^{{n}} {sin}\left(\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\frac{{x}}{\mathrm{2}^{{k}} }=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\frac{{x}/\mathrm{2}}{\mathrm{2}^{{k}−\mathrm{1}} }= \\ $$$$=\frac{{sin}\left(\mathrm{2}\left({x}/\mathrm{2}\right)\right)}{\mathrm{2}^{{n}} {sin}\left(\frac{{x}/\mathrm{2}}{\mathrm{2}^{{n}−\mathrm{1}} }\right)}=\frac{{sinx}}{\mathrm{2}^{{n}} {sin}\frac{{x}}{\mathrm{2}^{{n}} }} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{cos}\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }}{{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }}=\frac{\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right)}{\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\frac{{x}}{\mathrm{2}^{{k}} }\right)^{\mathrm{2}} }= \\ $$$$=\frac{\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}^{{n}} {sin}\left(\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }\right)}}{\left(\frac{{sinx}}{\mathrm{2}^{{n}} {sin}\frac{{x}}{\mathrm{2}^{{n}} }}\right)^{\mathrm{2}} }=\frac{{sin}\left(\mathrm{2}{x}\right)\mathrm{2}^{\mathrm{2}{n}} {sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{n}} }}{\mathrm{2}^{{n}} {sin}\frac{\mathrm{2}{x}}{\mathrm{2}^{{n}} }{sin}^{\mathrm{2}} {x}}= \\ $$$$=\frac{\mathrm{2}{sinx}\:{cosx}\:\mathrm{2}^{{n}} \:{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{n}} }}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}^{{n}} }{cos}\frac{{x}}{\mathrm{2}^{{n}} }{sin}^{\mathrm{2}} {x}}=\frac{\mathrm{2}^{{n}} {cosx}\:{sin}\frac{{x}}{\mathrm{2}^{{n}} }}{{cos}\frac{{x}}{\mathrm{2}^{{n}} }{sinx}}= \\ $$$$=\mathrm{2}^{{n}} \frac{{tan}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}{{tan}\left({x}\right)} \\ $$$$\Rightarrow\Omega=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}}{{tan}\left({x}\right)}\underset{{n}\rightarrow\infty} {{lim}}\frac{{tan}\left({x}/\mathrm{2}^{{n}} \right)}{{x}/\mathrm{2}^{{n}} }=\mathrm{1} \\ $$
Commented by aleks041103 last updated on 26/Sep/21
There is an even easier method tan(2x)=((2tan(x))/(1−tan^2 (x)))⇒1−tan^2 x=2((tan(x))/(tan(2x))) ⇒Π_(k=1) ^n (1−tan^2 (x/2^k ))=2^n Π_(k=1) ^n ((tan((x/2^k )))/(tan((x/2^(k−1) )))) telescopic sum: ⇒Π_(k=1) ^n (1−tan^2 (x/2^k ))=((2^n tan(2^(−n) x))/(tan(x))) ... ⇒Ω=1
$${There}\:{is}\:{an}\:{even}\:{easier}\:{method} \\ $$$${tan}\left(\mathrm{2}{x}\right)=\frac{\mathrm{2}{tan}\left({x}\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \left({x}\right)}\Rightarrow\mathrm{1}−{tan}^{\mathrm{2}} {x}=\mathrm{2}\frac{{tan}\left({x}\right)}{{tan}\left(\mathrm{2}{x}\right)} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }\right)=\mathrm{2}^{{n}} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{tan}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)}{{tan}\left(\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right)} \\ $$$${telescopic}\:{sum}: \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }\right)=\frac{\mathrm{2}^{{n}} {tan}\left(\mathrm{2}^{−{n}} {x}\right)}{{tan}\left({x}\right)} \\ $$$$… \\ $$$$\Rightarrow\Omega=\mathrm{1} \\ $$
Commented by mathdanisur last updated on 26/Sep/21
Very nice Ser, thank you
$$\mathrm{Very}\:\mathrm{nice}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$

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