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Question-9617

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Question Number 9617 by tawakalitu last updated on 20/Dec/16
Commented by sou1618 last updated on 21/Dec/16
m=(√(1/2))×(√((1/2)+(1/2)(√(1/2))))×(√((1/2)+(1/2)(√((1/2)+(1/2)(√(1/2))))))×... put a_n : a_1 =(√(1/2)),a_(n+1) =(√((1/2)+(1/2)a_n )) m=a_1 ×a_2 ×a_3 ×.... put M_n :M_n =Π_(k=1) ^n a_k (m=lim_(n→∞) M_n ) a_(n+1) ^2 =((1+a_n )/2) ⇉ cos^2 θ=((1+cos2θ)/2) a_1 =(√(1/2))=cos(1/4)π a_2 =(√((1+cos(π/4))/2))=cos(1/8)π a_3 =(√((1+cos(π/8))/2))=cos(1/(16))π ... M_n =cos(π/2^2 )×cos(π/2^3 )×cos(π/2^4 )×....×cos(π/2^(n+1) ) M_n ×sin(π/2^(n+1) )=cos(π/2^2 )×cos(π/2^3 )×....×cos(π/2^n )×(cos(π/2^(n+1) )×sin(π/2^(n+1) )) sinθcosθ=(1/2)sin2θ M_n ×sin(π/2^(n+1) )=cos(π/2^2 )×cos(π/2^3 )×....×(cos(π/2^n )×(1/2)sin(π/2^n )) M_n ×sin(π/2^(n+1) )=cos(π/2^2 )×cos(π/2^3 )×....×((1/2))^2 sin(π/2^(n−1) ) ... M_n ×sin(π/2^(n+1) )=((1/2))^n sin(π/2)=2^(−n) M_n =(2^(−n) /( sin(π×2^(−(n+1)) ) )) m=lim_(n→∞) ((π×2^(−(n+1)) ×(π^(−1) ×2))/( sin(π×2^(−(n+1)) ) )) n→∞⇔π×2^(−(n+1)) →0 m=lim_(x→0) (x/(sinx))×(2/π) m=(2/π) . I found this “Vieta′s Formula for Pi” on the web.
$${m}=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}×\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}×\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}×… \\ $$$$\mathrm{put}\:{a}_{{n}} :\:{a}_{\mathrm{1}} =\sqrt{\frac{\mathrm{1}}{\mathrm{2}}},{a}_{{n}+\mathrm{1}} =\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{a}_{{n}} } \\ $$$${m}={a}_{\mathrm{1}} ×{a}_{\mathrm{2}} ×{a}_{\mathrm{3}} ×…. \\ $$$$\mathrm{put}\:{M}_{{n}} :{M}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{a}_{{k}} \:\:\:\:\:\:\left({m}=\underset{{n}\rightarrow\infty} {{lim}M}_{{n}} \right) \\ $$$${a}_{{n}+\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{1}+{a}_{{n}} }{\mathrm{2}}\:\rightrightarrows\:{cos}^{\mathrm{2}} \theta=\frac{\mathrm{1}+{cos}\mathrm{2}\theta}{\mathrm{2}} \\ $$$${a}_{\mathrm{1}} =\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}={cos}\frac{\mathrm{1}}{\mathrm{4}}\pi \\ $$$${a}_{\mathrm{2}} =\sqrt{\frac{\mathrm{1}+{cos}\left(\pi/\mathrm{4}\right)}{\mathrm{2}}}={cos}\frac{\mathrm{1}}{\mathrm{8}}\pi \\ $$$${a}_{\mathrm{3}} =\sqrt{\frac{\mathrm{1}+{cos}\left(\pi/\mathrm{8}\right)}{\mathrm{2}}}={cos}\frac{\mathrm{1}}{\mathrm{16}}\pi \\ $$$$… \\ $$$${M}_{{n}} ={cos}\frac{\pi}{\mathrm{2}^{\mathrm{2}} }×{cos}\frac{\pi}{\mathrm{2}^{\mathrm{3}} }×{cos}\frac{\pi}{\mathrm{2}^{\mathrm{4}} }×….×{cos}\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$${M}_{{n}} ×{sin}\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }={cos}\frac{\pi}{\mathrm{2}^{\mathrm{2}} }×{cos}\frac{\pi}{\mathrm{2}^{\mathrm{3}} }×….×{cos}\frac{\pi}{\mathrm{2}^{{n}} }×\left({cos}\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }×{sin}\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\right) \\ $$$${sin}\theta{cos}\theta=\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}\theta \\ $$$${M}_{{n}} ×{sin}\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }={cos}\frac{\pi}{\mathrm{2}^{\mathrm{2}} }×{cos}\frac{\pi}{\mathrm{2}^{\mathrm{3}} }×….×\left({cos}\frac{\pi}{\mathrm{2}^{{n}} }×\frac{\mathrm{1}}{\mathrm{2}}{sin}\frac{\pi}{\mathrm{2}^{{n}} }\right) \\ $$$${M}_{{n}} ×{sin}\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }={cos}\frac{\pi}{\mathrm{2}^{\mathrm{2}} }×{cos}\frac{\pi}{\mathrm{2}^{\mathrm{3}} }×….×\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} {sin}\frac{\pi}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$… \\ $$$$ \\ $$$${M}_{{n}} ×{sin}\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} {sin}\frac{\pi}{\mathrm{2}}=\mathrm{2}^{−{n}} \\ $$$${M}_{{n}} =\frac{\mathrm{2}^{−{n}} }{\:{sin}\left(\pi×\mathrm{2}^{−\left({n}+\mathrm{1}\right)} \right)\:} \\ $$$${m}=\underset{{n}\rightarrow\infty} {{lim}}\frac{\pi×\mathrm{2}^{−\left({n}+\mathrm{1}\right)} ×\left(\pi^{−\mathrm{1}} ×\mathrm{2}\right)}{\:{sin}\left(\pi×\mathrm{2}^{−\left({n}+\mathrm{1}\right)} \right)\:} \\ $$$${n}\rightarrow\infty\Leftrightarrow\pi×\mathrm{2}^{−\left({n}+\mathrm{1}\right)} \rightarrow\mathrm{0} \\ $$$${m}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}}{{sinx}}×\frac{\mathrm{2}}{\pi} \\ $$$${m}=\frac{\mathrm{2}}{\pi}\:. \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{this}\:“{Vieta}'{s}\:{Formula}\:{for}\:{Pi}''\:\mathrm{on}\:\mathrm{the}\:\mathrm{web}. \\ $$
Commented by tawakalitu last updated on 21/Dec/16
i really appreciate your effort sir. God bless you.
$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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