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Question Number 76514 by mathmax by abdo last updated on 28/Dec/19
calculate Π_(k=0) ^(n−1) (α^(k ) +α^−^k ) with α root of x^2 +2x+5 =0
$${calculate}\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\alpha^{{k}\:} \:+\overset{−^{{k}} } {\alpha}\right)\:{with}\:\alpha\:{root}\:{of}\:{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\:=\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 28/Dec/19
let A_n =Π_(k=0) ^(n−1) (α^k +α^−^k ) let solve x^2 +2x+5=0 Δ^′ =1−5 =−4 =(2i)^2 /⇒ α =−1+2i and α^− =−1−2i α =(√5)e^(−i arctan(2)) and α^− =(√5)e^(i arctan(2)) ⇒ α^k +α^−^k =((√5))^k e^(−ik arctan(2)) +((√5))^k e^(ik arctan(2)) =((√5))^k (2cos(k arctan2) ⇒ A_n =Π_(k=0) ^(n−1) 2((√5))^k cos(k arctan(2)) =2^n Π_(k=0) ^(n−1) ((√5))^k Π_(k=0) ^(n−1) cos(k arctan(2)) =2^n ((√5))^(Σ_(k=0) ^(n−1) k) Π_(k=0) ^(n−1) cos(k arctan(2)) =2^n ((√5))^(((n−1)n)/2) Π_(k=0) ^(n−1) cos(k arctan(2))
$${let}\:{A}_{{n}} =\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\alpha^{{k}} \:+\overset{−^{{k}} } {\alpha}\right)\:\:{let}\:{solve}\:{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{5}=\mathrm{0} \\ $$$$\Delta^{'} =\mathrm{1}−\mathrm{5}\:=−\mathrm{4}\:=\left(\mathrm{2}{i}\right)^{\mathrm{2}} /\Rightarrow\:\alpha\:=−\mathrm{1}+\mathrm{2}{i}\:\:{and}\:\:\overset{−} {\alpha}=−\mathrm{1}−\mathrm{2}{i} \\ $$$$\alpha\:=\sqrt{\mathrm{5}}{e}^{−{i}\:{arctan}\left(\mathrm{2}\right)} \:\:{and}\:\overset{−} {\alpha}=\sqrt{\mathrm{5}}{e}^{{i}\:{arctan}\left(\mathrm{2}\right)} \:\Rightarrow \\ $$$$\alpha^{{k}} \:+\overset{−^{{k}} } {\alpha}\:=\left(\sqrt{\mathrm{5}}\right)^{{k}} \:{e}^{−{ik}\:{arctan}\left(\mathrm{2}\right)} \:+\left(\sqrt{\mathrm{5}}\right)^{{k}} \:{e}^{{ik}\:{arctan}\left(\mathrm{2}\right)} \\ $$$$=\left(\sqrt{\mathrm{5}}\right)^{{k}} \:\left(\mathrm{2}{cos}\left({k}\:{arctan}\mathrm{2}\right)\:\Rightarrow\:{A}_{{n}} =\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \mathrm{2}\left(\sqrt{\mathrm{5}}\right)^{{k}} \:{cos}\left({k}\:{arctan}\left(\mathrm{2}\right)\right)\right. \\ $$$$=\mathrm{2}^{{n}} \:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\sqrt{\mathrm{5}}\right)^{{k}} \:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{cos}\left({k}\:{arctan}\left(\mathrm{2}\right)\right) \\ $$$$=\mathrm{2}^{{n}} \:\left(\sqrt{\mathrm{5}}\right)^{\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} {k}} \:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{cos}\left({k}\:{arctan}\left(\mathrm{2}\right)\right) \\ $$$$=\mathrm{2}^{{n}} \:\left(\sqrt{\mathrm{5}}\right)^{\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}} \:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{cos}\left({k}\:{arctan}\left(\mathrm{2}\right)\right) \\ $$
Answered by john santu last updated on 28/Dec/19
(x+1)^2 +4=0→α=2i−1 α+α^(−1) =2i−1+(1/(2i−1))=2i−1+((−2i−1)/((2i−1)(−2i−1))) =((10i−5−2i−1)/5)=((8i−6)/5) α^2 +α^(−2) =(((8i−6)/5))^2 −2
$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}=\mathrm{0}\rightarrow\alpha=\mathrm{2}{i}−\mathrm{1} \\ $$$$\alpha+\alpha^{−\mathrm{1}} =\mathrm{2}{i}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{i}−\mathrm{1}}=\mathrm{2}{i}−\mathrm{1}+\frac{−\mathrm{2}{i}−\mathrm{1}}{\left(\mathrm{2}{i}−\mathrm{1}\right)\left(−\mathrm{2}{i}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{10}{i}−\mathrm{5}−\mathrm{2}{i}−\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{8}{i}−\mathrm{6}}{\mathrm{5}} \\ $$$$\alpha^{\mathrm{2}} +\alpha^{−\mathrm{2}} =\left(\frac{\mathrm{8}{i}−\mathrm{6}}{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{2} \\ $$

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