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Question Number 119479 by 675480065 last updated on 24/Oct/20
Π_(k=1) ^∞ cos((x/2^k )) = ?
$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right)\:=\:? \\ $$
Answered by Olaf last updated on 25/Oct/20
sin2θ = 2sinθcosθ cosθ = (1/2).((sin2θ)/(sinθ)), θ ≠ kπ ⇒ cos((x/2^k )) = (1/2).((sin((x/2^(k−1) )))/(sin((x/2^k )))) And Π_(k=1) ^n cos((x/2^k )) = (1/2^n )Π_(k=1) ^n ((sin((x/2^(k−1) )))/(sin((x/2^k )))) Π_(k=1) ^n cos((x/2^k )) = (1/2^n ).((sinx)/(sin((x/2^n )))) (telescopic product) Π_(k=1) ^∞ cos((x/2^k )) = sinx.lim_(n→∞) (1/(2^n sin((x/2^n )))) 2^n sin((x/2^n )) ∼_∞ 2^n ×(x/2^n ) = x Π_(k=1) ^∞ cos((x/2^k )) = ((sinx)/x)
$$\mathrm{sin2}\theta\:=\:\mathrm{2sin}\theta\mathrm{cos}\theta \\ $$$$\mathrm{cos}\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{sin2}\theta}{\mathrm{sin}\theta},\:\theta\:\neq\:{k}\pi \\ $$$$\Rightarrow\:\mathrm{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right)}{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)} \\ $$$$\mathrm{And}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right)}{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }.\frac{\mathrm{sin}{x}}{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)} \\ $$$$\left(\mathrm{telescopic}\:\mathrm{product}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:=\:\mathrm{sin}{x}.\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} \mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)} \\ $$$$\mathrm{2}^{{n}} \mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)\:\underset{\infty} {\sim}\mathrm{2}^{{n}} ×\frac{{x}}{\mathrm{2}^{{n}} }\:=\:{x} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:=\:\frac{\mathrm{sin}{x}}{{x}} \\ $$
Commented by 675480065 last updated on 25/Oct/20
Thanks sir. can i apply complex numbers to it?
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{can}\:\mathrm{i}\:\mathrm{apply}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{to}\:\mathrm{it}? \\ $$
Answered by Bird last updated on 25/Oct/20
let A_n =Π_(k=1) ^n cos((x/2^k )) snd B_n =Π_(k=1) ^n sin((x/2^k )) we hsve A_n .B_n =Π_(k=1) ^n cos((x/2^k ))sin((x/2^k )) =(1/2^n )Π_(k=1) ^n sin((x/2^(k−1) )) =(1/2^n )Π_(k=0) ^(n−1) sin((x/2^k )) =(1/2^n )((sinx)/(sin((x/2^n ))))×Π_(k=1) ^n sin((x/2^k )) =((sinx)/(2^n sin((x/2^n ))))×B_n we hsve B_n ≠0 ⇒ A_n =((sinx)/(2^n sin((x/2^n )))) but sin((x/2^n ))∼(x/2^n ) ⇒2^n sin((x/2^n ))∼x(n→∞) ⇒lim_(n→∞) A_n =((sinx)/x) ⇒ Π_(k=1) ^∞ cos((x/2^k ))=((sinx)/x)
$${let}\:{A}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:{snd} \\ $$$${B}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:{we}\:{hsve} \\ $$$${A}_{{n}} .{B}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} {cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right){sin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\prod_{{k}=\mathrm{1}} ^{{n}} {sin}\left(\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} {sin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\frac{{sinx}}{{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}×\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right) \\ $$$$=\frac{{sinx}}{\mathrm{2}^{{n}} {sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}×{B}_{{n}} \:\:{we}\:{hsve}\:{B}_{{n}} \neq\mathrm{0}\:\Rightarrow \\ $$$${A}_{{n}} =\frac{{sinx}}{\mathrm{2}^{{n}} \:{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}\:\:{but} \\ $$$${sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)\sim\frac{{x}}{\mathrm{2}^{{n}} }\:\Rightarrow\mathrm{2}^{{n}} {sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)\sim{x}\left({n}\rightarrow\infty\right) \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} {A}_{{n}} =\frac{{sinx}}{{x}}\:\Rightarrow \\ $$$$\prod_{{k}=\mathrm{1}} ^{\infty} {cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)=\frac{{sinx}}{{x}} \\ $$
Commented by 675480065 last updated on 25/Oct/20
Thanks Sir. Plz i want to learn this topic. how can i get notes on it sir. pls help me if u have Thanks
$$\mathrm{Thanks}\:\mathrm{Sir}. \\ $$$$\mathrm{Plz}\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{learn}\:\mathrm{this}\:\mathrm{topic}. \\ $$$$\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{get}\:\mathrm{notes}\:\mathrm{on}\:\mathrm{it}\:\mathrm{sir}. \\ $$$$\mathrm{pls}\:\mathrm{help}\:\mathrm{me}\:\mathrm{if}\:\mathrm{u}\:\mathrm{have}\: \\ $$$$\mathrm{Thanks} \\ $$
Answered by Bird last updated on 25/Oct/20
A_n =Π_(k=1) ^n cos((x/2^k )) ⇒ A_n =Π_(k=1) ^n ((e^(i(x/2^k )) +e^(−((ix)/2^k )) )/2) =(1/2^n )Π_(k=1) ^n e^((ix)/2^k ) Π_(k=1) ^n (1+e^(−((2ix)/2^k )) ) =(1/2^n )e^(ixΣ_(k=1) ^n (1/2^k )) Π_(k=1) ^n (1+cos(((2x)/2^k ))−isin(((2x)/2^k ))) =(1/2^n ) e^(ixΣ_(k=0) ^(n−1) (1/(2^(k+1) ))) ×Π_(k=1) ^n (1+cos((x/2^(k−1) ))−isin((x/2^(k−1) ))) =(1/2^n ) e^(((ix)/2)×(1/(1−(1/2)))) ×Π_(k=0) ^(n−1) (1+cos((x/2^k ))−isin((x/2^k ))) =(1/2^n ) e^(ix) ×Π_(k=0) ^(n−1) {2cos^2 ((x/2^(k+1) ))−2isin((x/2^(k+1) ))cos((x/2^(k+1) ))} ....be continued..i think this method give the answer ...
$${A}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:\Rightarrow \\ $$$${A}_{{n}} \:=\prod_{{k}=\mathrm{1}} ^{{n}} \frac{{e}^{{i}\frac{{x}}{\mathrm{2}^{{k}} }} +{e}^{−\frac{{ix}}{\mathrm{2}^{{k}} }} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\prod_{{k}=\mathrm{1}} ^{{n}} {e}^{\frac{{ix}}{\mathrm{2}^{{k}} }} \:\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+{e}^{−\frac{\mathrm{2}{ix}}{\mathrm{2}^{{k}} }} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{e}^{{ix}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}^{{k}} }} \:\:\:\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+{cos}\left(\frac{\mathrm{2}{x}}{\mathrm{2}^{{k}} }\right)−{isin}\left(\frac{\mathrm{2}{x}}{\mathrm{2}^{{k}} }\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:{e}^{{ix}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} \:}} \:×\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+{cos}\left(\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right)−{isin}\left(\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:{e}^{\frac{{ix}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}} ×\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{1}+{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)−{isin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:{e}^{{ix}} ×\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left\{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}^{{k}+\mathrm{1}} }\right)−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}^{{k}+\mathrm{1}} }\right){cos}\left(\frac{{x}}{\mathrm{2}^{{k}+\mathrm{1}} }\right)\right\} \\ $$$$….{be}\:{continued}..{i}\:{think}\:{this} \\ $$$${method}\:{give}\:{the}\:{answer}\:… \\ $$

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