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Question Number 48066 by rahul 19 last updated on 18/Nov/18
(√(1/2)).(√((1/2)+(1/2)(√(1/2)))).(√((1/2)+(1/2)(√((1/2)+(1/2)(√(1/2))))))......∞=?
$$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}.\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}.\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}……\infty=? \\ $$
Commented by MJS last updated on 18/Nov/18
=(2/π) approximated it but I can′t prove it
$$=\frac{\mathrm{2}}{\pi} \\ $$$$\mathrm{approximated}\:\mathrm{it}\:\mathrm{but}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{prove}\:\mathrm{it} \\ $$
Commented by maxmathsup by imad last updated on 19/Nov/18
we have (√(1/2))=cos((π/4)) ,(√((1/2)+(1/2)(√(1/2))))=(√((1+cos((π/4)))/2))=cos((π/8)) ... S_n =cos((π/4))cos((π/8)).cos((π/(16)))...cos((π/2^n ))=Π_(k=2) ^n cos((π/2^k )) W_n =Π_(k=2) ^n sin((π/2^k )) ⇒S_n .W_n =Π_(k=2) ^n cos((π/2^k ))sin((π/2^k )) =(1/2^(n−1) )Π_(k=2) ^n sin((π/2^(k−1) ))=(1/2^(n−1) )Π_(k=2) ^n Π_(k=2) ^n sin((π/2^(k−1) )) =(1/2^(n−1) )Π_(k=2) ^n Π_(k=1) ^(n−1) sin((π/2^k )) =(1/2^(n−1) )Π_(k=2) ^n ((Π_(k=2) ^n sin((π/2^k )))/(sin((π/(2n))))) =(1/(sin((π/2^n )))) (1/2^(n−1) ) .W_n ⇒ S_n =(1/(2^(n−1) sin((π/2^n )))) =(2/(2^n sin((π/2^n )))) ∼ (2/(2^n .(π/2^n ))) (n→+∞) ⇒lim_(n→+∞) S_n =(2/π) .
$${we}\:{have}\:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}={cos}\left(\frac{\pi}{\mathrm{4}}\right)\:,\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}=\sqrt{\frac{\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{2}}}={cos}\left(\frac{\pi}{\mathrm{8}}\right)\:… \\ $$$${S}_{{n}} ={cos}\left(\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\pi}{\mathrm{8}}\right).{cos}\left(\frac{\pi}{\mathrm{16}}\right)…{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)=\prod_{{k}=\mathrm{2}} ^{{n}} \:{cos}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right) \\ $$$${W}_{{n}} =\prod_{{k}=\mathrm{2}} ^{{n}} \:{sin}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)\:\Rightarrow{S}_{{n}} .{W}_{{n}} =\prod_{{k}=\mathrm{2}} ^{{n}} {cos}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right){sin}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\prod_{{k}=\mathrm{2}} ^{{n}} \:{sin}\left(\frac{\pi}{\mathrm{2}^{{k}−\mathrm{1}} }\right)=\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\prod_{{k}=\mathrm{2}} ^{{n}} \:\:\prod_{{k}=\mathrm{2}} ^{{n}} \:{sin}\left(\frac{\pi}{\mathrm{2}^{{k}−\mathrm{1}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\prod_{{k}=\mathrm{2}} ^{{n}} \:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)\:\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\prod_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\prod_{{k}=\mathrm{2}} ^{{n}} \:{sin}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:=\frac{\mathrm{1}}{{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}\:\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\:.{W}_{{n}} \:\Rightarrow \\ $$$$ \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} \:{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}\:=\frac{\mathrm{2}}{\mathrm{2}^{{n}} \:{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}\:\sim\:\frac{\mathrm{2}}{\mathrm{2}^{{n}} .\frac{\pi}{\mathrm{2}^{{n}} }}\:\left({n}\rightarrow+\infty\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:{S}_{{n}} =\frac{\mathrm{2}}{\pi}\:. \\ $$
Commented by rahul 19 last updated on 20/Nov/18
thank you prof Abdo����
Commented by Abdo msup. last updated on 20/Nov/18
you are welcome sir.
$${you}\:{are}\:{welcome}\:{sir}. \\ $$
Answered by arcana last updated on 19/Nov/18
cos(θ)=(√(1/2)) (√(1/2))∙(√((1/2)+(1/2)(√(1/2))))=cos(θ)(√((1+cos(θ))/2))=cos(θ)cos(θ/2) cos(θ)cos(θ/2)(√((1+cos(θ/2))/2))=cos(θ)cos(θ/2)cos(θ/4) asi podemos escribirlo como cos(θ)∙cos(θ/2)∙cos(θ/4)∙...=Π_(i=0) ^∞ cos((θ/2^i )) si multiplicamos y dividimos por sin((θ/2^i )) ((cos(θ)∙cos(θ/2)∙cos(θ/4)∙...∙(1/2)2cos(θ/2^i )∙sin(θ/2^i ))/(sin(θ/2^i ))) usando sin(θ/2^(i−1) )=2sin(θ/2^i )cos(θ/2^i ) ((cos(θ)∙cos(θ/2)∙cos(θ/4)∙...∙(1/(2 ))∙(1/2)2cos(θ/2^(i−1) )∙sin(θ/2^(i−1) ))/(sin(θ/2^i ))) ((cos(θ)∙cos(θ/2)∙cos(θ/4)∙...∙(1/2^2 )∙(1/2)2cos(θ/2^(i−2) )∙sin(θ/2^(i−2) ))/(sin(θ/2^i ))) ⋮ ((sin(2θ))/(2^(i+1) sin(θ/2^i )))=((sin(2θ))/(2^(i+1) ((sin(θ/2^i ))/(θ/2^i ))∙(θ/2^i ))) luego lim_(i→∞) ((sin(2θ))/(2^(i+1) ((sin(θ/2^i ))/(θ/2^i ))∙(θ/2^i )))=((sin(2θ))/(2^(i+1) ∙(θ/2^i )lim_(i→∞) ((sin(θ/2^i ))/((θ/2^i ))))) =((sin(2θ))/(2θ))=(1/(2θ)) ya que θ=(π/4) ⇒Π_(i=0) ^∞ cos((θ/2^i ))=(1/(π/2))=(2/π)
$${cos}\left(\theta\right)=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\centerdot\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}={cos}\left(\theta\right)\sqrt{\frac{\mathrm{1}+{cos}\left(\theta\right)}{\mathrm{2}}}={cos}\left(\theta\right){cos}\left(\theta/\mathrm{2}\right) \\ $$$${cos}\left(\theta\right){cos}\left(\theta/\mathrm{2}\right)\sqrt{\frac{\mathrm{1}+{cos}\left(\theta/\mathrm{2}\right)}{\mathrm{2}}}={cos}\left(\theta\right){cos}\left(\theta/\mathrm{2}\right){cos}\left(\theta/\mathrm{4}\right) \\ $$$$\mathrm{asi}\:\mathrm{podemos}\:\mathrm{escribirlo}\:\mathrm{como} \\ $$$${cos}\left(\theta\right)\centerdot{cos}\left(\theta/\mathrm{2}\right)\centerdot{cos}\left(\theta/\mathrm{4}\right)\centerdot…=\underset{{i}=\mathrm{0}} {\overset{\infty} {\prod}}{cos}\left(\frac{\theta}{\mathrm{2}^{{i}} }\right) \\ $$$$\mathrm{si}\:\mathrm{multiplicamos}\:\mathrm{y}\:\mathrm{dividimos}\:\mathrm{por}\:{sin}\left(\frac{\theta}{\mathrm{2}^{{i}} }\right) \\ $$$$\frac{{cos}\left(\theta\right)\centerdot{cos}\left(\theta/\mathrm{2}\right)\centerdot{cos}\left(\theta/\mathrm{4}\right)\centerdot…\centerdot\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2}{cos}\left(\theta/\mathrm{2}^{{i}} \right)\centerdot{sin}\left(\theta/\mathrm{2}^{{i}} \right)}{{sin}\left(\theta/\mathrm{2}^{{i}} \right)} \\ $$$$\mathrm{usando}\:{sin}\left(\theta/\mathrm{2}^{{i}−\mathrm{1}} \right)=\mathrm{2}{sin}\left(\theta/\mathrm{2}^{{i}} \right){cos}\left(\theta/\mathrm{2}^{{i}} \right) \\ $$$$\frac{{cos}\left(\theta\right)\centerdot{cos}\left(\theta/\mathrm{2}\right)\centerdot{cos}\left(\theta/\mathrm{4}\right)\centerdot…\centerdot\frac{\mathrm{1}}{\mathrm{2}\:}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2}{cos}\left(\theta/\mathrm{2}^{{i}−\mathrm{1}} \right)\centerdot{sin}\left(\theta/\mathrm{2}^{{i}−\mathrm{1}} \right)}{{sin}\left(\theta/\mathrm{2}^{{i}} \right)} \\ $$$$\frac{{cos}\left(\theta\right)\centerdot{cos}\left(\theta/\mathrm{2}\right)\centerdot{cos}\left(\theta/\mathrm{4}\right)\centerdot…\centerdot\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2}{cos}\left(\theta/\mathrm{2}^{{i}−\mathrm{2}} \right)\centerdot{sin}\left(\theta/\mathrm{2}^{{i}−\mathrm{2}} \right)}{{sin}\left(\theta/\mathrm{2}^{{i}} \right)} \\ $$$$\vdots \\ $$$$\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}^{{i}+\mathrm{1}} {sin}\left(\theta/\mathrm{2}^{{i}} \right)}=\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}^{{i}+\mathrm{1}} \frac{{sin}\left(\theta/\mathrm{2}^{{i}} \right)}{\theta/\mathrm{2}^{{i}} }\centerdot\left(\theta/\mathrm{2}^{{i}} \right)} \\ $$$${luego} \\ $$$$ \\ $$$$\underset{{i}\rightarrow\infty} {\mathrm{lim}}\:\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}^{{i}+\mathrm{1}} \frac{{sin}\left(\theta/\mathrm{2}^{{i}} \right)}{\theta/\mathrm{2}^{{i}} }\centerdot\left(\theta/\mathrm{2}^{{i}} \right)}=\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}^{{i}+\mathrm{1}} \centerdot\left(\theta/\mathrm{2}^{{i}} \right)\underset{{i}\rightarrow\infty} {\mathrm{lim}}\frac{{sin}\left(\theta/\mathrm{2}^{{i}} \right)}{\left(\theta/\mathrm{2}^{{i}} \right)}} \\ $$$$=\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}\theta}=\frac{\mathrm{1}}{\mathrm{2}\theta}\:\:\:\mathrm{ya}\:\mathrm{que}\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\underset{{i}=\mathrm{0}} {\overset{\infty} {\prod}}{cos}\left(\frac{\theta}{\mathrm{2}^{{i}} }\right)=\frac{\mathrm{1}}{\pi/\mathrm{2}}=\frac{\mathrm{2}}{\pi} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by rahul 19 last updated on 19/Nov/18
thank you so much Arcana ,����
Commented by ajfour last updated on 19/Nov/18
superb!
$${superb}! \\ $$

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