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Question-190324




Question Number 190324 by Shrinava last updated on 31/Mar/23
Commented by senestro last updated on 01/Apr/23
how can we solve it?
$${how}\:{can}\:{we}\:{solve}\:{it}? \\ $$
Answered by witcher3 last updated on 01/Apr/23
S=Π_(k=1) ^(4n+2) (1/2)(e^(4i((πk)/(4n+3))) +1).e^(−((2π)/(4n+3))(((4n+3)/2)(4n+1)))   =(1/2^(4n+2) )Π_(t∈S_k ) (1+t^2 ),s_k ={e^((2iπk)/(4n+3)) ,k∈{0,4n+2}}  Z^(4n+3) =1⇔z∈S_k   Z^(8n+6) −1=(1/(16^n .4 ))Π_(t∈S_k −{0}) (Z^2 −t^2 )⇔((Z^(8n+6) −1)/(z^2 −1))=P(z)  Π_(k=1) ^(4n+2) (e^((4ikπ)/(4n+3)) +1)=P(i)=1  S=(1/(16^n .4))  ⇒Ω=(1/4)
$$\mathrm{S}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{4n}+\mathrm{2}} {\prod}}\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{4i}\frac{\pi\mathrm{k}}{\mathrm{4n}+\mathrm{3}}} +\mathrm{1}\right).\mathrm{e}^{−\frac{\mathrm{2}\pi}{\mathrm{4n}+\mathrm{3}}\left(\frac{\mathrm{4n}+\mathrm{3}}{\mathrm{2}}\left(\mathrm{4n}+\mathrm{1}\right)\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4n}+\mathrm{2}} }\underset{\mathrm{t}\in\mathrm{S}_{\mathrm{k}} } {\prod}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right),\mathrm{s}_{\mathrm{k}} =\left\{\mathrm{e}^{\frac{\mathrm{2i}\pi\mathrm{k}}{\mathrm{4n}+\mathrm{3}}} ,\mathrm{k}\in\left\{\mathrm{0},\mathrm{4n}+\mathrm{2}\right\}\right\} \\ $$$$\mathrm{Z}^{\mathrm{4n}+\mathrm{3}} =\mathrm{1}\Leftrightarrow\mathrm{z}\in\mathrm{S}_{\mathrm{k}} \\ $$$$\mathrm{Z}^{\mathrm{8n}+\mathrm{6}} −\mathrm{1}=\frac{\mathrm{1}}{\mathrm{16}^{\mathrm{n}} .\mathrm{4}\:}\underset{\mathrm{t}\in\mathrm{S}_{\mathrm{k}} −\left\{\mathrm{0}\right\}} {\prod}\left(\mathrm{Z}^{\mathrm{2}} −\mathrm{t}^{\mathrm{2}} \right)\Leftrightarrow\frac{\mathrm{Z}^{\mathrm{8n}+\mathrm{6}} −\mathrm{1}}{\mathrm{z}^{\mathrm{2}} −\mathrm{1}}=\mathrm{P}\left(\mathrm{z}\right) \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{4n}+\mathrm{2}} {\prod}}\left(\mathrm{e}^{\frac{\mathrm{4ik}\pi}{\mathrm{4n}+\mathrm{3}}} +\mathrm{1}\right)=\mathrm{P}\left(\mathrm{i}\right)=\mathrm{1} \\ $$$$\mathrm{S}=\frac{\mathrm{1}}{\mathrm{16}^{\mathrm{n}} .\mathrm{4}} \\ $$$$\Rightarrow\Omega=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by Shrinava last updated on 03/Apr/23
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Commented by witcher3 last updated on 05/Apr/23
withe Pleasur God bless You
$$\mathrm{withe}\:\mathrm{Pleasur}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{You} \\ $$

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