Question Number 190360 by mnjuly1970 last updated on 01/Apr/23
$$ \\ $$$${if}\:\:\:{x}+\:\frac{\mathrm{1}}{{x}}\:=\:\varphi\:\left(\:\:{Golden}\:{ratio}\right) \\ $$$$\:\:\:\:\:\Rightarrow\:\:\:{x}^{\:\mathrm{2000}} +\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2000}} }=? \\ $$$$ \\ $$$$ \\ $$
Answered by Frix last updated on 01/Apr/23
$${x}=\mathrm{e}^{\pm\mathrm{i}\frac{\pi}{\mathrm{5}}} \\ $$$${x}^{\mathrm{2000}} =\mathrm{e}^{\pm\mathrm{400}\pi\mathrm{i}} =\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$
Commented by mnjuly1970 last updated on 02/Apr/23
$${please}\:{solution} \\ $$
Answered by BaliramKumar last updated on 02/Apr/23
$${x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\frac{\sqrt{\mathrm{5}}\:+\:\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:……\left({i}\right) \\ $$$${x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\frac{\sqrt{\mathrm{5}}\:−\:\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:……..\left({ii}\right) \\ $$$${x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}\:−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:……….\left({iii}\right) \\ $$$$\left({ii}\right)×\left({iii}\right) \\ $$$$\left({x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\left({x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\:=\:−\:\left(\frac{\sqrt{\mathrm{5}}\:−\:\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\:+\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\frac{\sqrt{\mathrm{5}}\:−\:\mathrm{3}}{\mathrm{2}} \\ $$$${x}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\:\:=\:\frac{\sqrt{\mathrm{5}}\:−\:\mathrm{3}}{\mathrm{2}}\:−\left({x}\:+\:\frac{\mathrm{1}}{{x}}\right) \\ $$$${x}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\:\:=\:\frac{\sqrt{\mathrm{5}}\:−\:\mathrm{3}}{\mathrm{2}}\:−\left(\frac{\sqrt{\mathrm{5}}\:+\:\mathrm{1}}{\mathrm{2}}\right)\:=\:−\:\mathrm{2} \\ $$$$\therefore\:{x}^{\mathrm{5}} \:=\:−\:\mathrm{1} \\ $$$$ \\ $$$${x}^{\mathrm{2000}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2000}} }\:=\:\left({x}^{\mathrm{5}} \right)^{\mathrm{400}} \:+\:\frac{\mathrm{1}}{\left({x}^{\mathrm{5}} \right)^{\mathrm{400}} } \\ $$$${x}^{\mathrm{2000}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2000}} }\:=\:\left(−\mathrm{1}\right)^{\mathrm{400}} \:+\:\frac{\mathrm{1}}{\left(−\mathrm{1}\right)^{\mathrm{400}} }\:=\:\mathrm{2} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 02/Apr/23
$${thx}\:{sir} \\ $$
Answered by BaliramKumar last updated on 02/Apr/23
![x + (1/x) = ϕ x + (1/x) = (((√5) + 1)/2) [ϕ = (((√5) + 1)/2)] x + (1/x) = 2cos((π/5)) [cos((π/5)) = (((√5) + 1)/4)] square both side x^2 + (1/x^2 ) + 2 = 4cos^2 ((π/5)) x^2 + (1/x^2 ) = 4cos^2 ((π/5)) − 2 = 2[2cos^2 ((π/5))− 1] x^2 + (1/x^2 ) = 2cos2((π/5)) x^n + (1/x^n ) = 2cos(((nπ)/5)) x^(2000) + (1/x^(2000) ) = 2cos(((2000π)/5)) = 2cos(400π) x^(2000) + (1/x^(2000) ) = 2×1 = 2 [cos2nπ = 1] x^(2023) + (1/x^(2023) ) = 2cos(((2023π)/5)) ⇒ 2cos(404π + ((3π)/5)) ⇒ 2cos(((3π)/5)) ⇒ 2cos(π − ((2π)/5)) ⇒ −2cos(((2π)/5)) ⇒ −2cos72° = −2sin18° = −2((((√5) − 1)/4)) ⇒ − ((((√5) − 1)/2))](https://www.tinkutara.com/question/Q190409.png)
$${x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\varphi \\ $$$${x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\frac{\sqrt{\mathrm{5}}\:+\:\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\varphi\:=\:\frac{\sqrt{\mathrm{5}}\:+\:\mathrm{1}}{\mathrm{2}}\right] \\ $$$${x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{5}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{cos}\left(\frac{\pi}{\mathrm{5}}\right)\:=\:\frac{\sqrt{\mathrm{5}}\:+\:\mathrm{1}}{\mathrm{4}}\right] \\ $$$${square}\:{both}\:{side} \\ $$$${x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\:\mathrm{2}\:=\:\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right) \\ $$$${x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\:=\:\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right)\:−\:\mathrm{2}\:=\:\mathrm{2}\left[\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right)−\:\mathrm{1}\right] \\ $$$${x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\:=\:\mathrm{2}{cos}\mathrm{2}\left(\frac{\pi}{\mathrm{5}}\right) \\ $$$${x}^{{n}} \:+\:\frac{\mathrm{1}}{{x}^{{n}} }\:=\:\mathrm{2}{cos}\left(\frac{{n}\pi}{\mathrm{5}}\right) \\ $$$${x}^{\mathrm{2000}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2000}} }\:=\:\mathrm{2}{cos}\left(\frac{\mathrm{2000}\pi}{\mathrm{5}}\right)\:=\:\mathrm{2}{cos}\left(\mathrm{400}\pi\right) \\ $$$${x}^{\mathrm{2000}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2000}} }\:=\:\:\mathrm{2}×\mathrm{1}\:=\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{cos}\mathrm{2}{n}\pi\:=\:\mathrm{1}\right] \\ $$$$ \\ $$$$ \\ $$$${x}^{\mathrm{2023}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2023}} }\:=\:\mathrm{2}{cos}\left(\frac{\mathrm{2023}\pi}{\mathrm{5}}\right)\: \\ $$$$\Rightarrow\:\:\mathrm{2}{cos}\left(\mathrm{404}\pi\:+\:\frac{\mathrm{3}\pi}{\mathrm{5}}\right)\:\Rightarrow\:\mathrm{2}{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{5}}\right) \\ $$$$\Rightarrow\:\mathrm{2}{cos}\left(\pi\:−\:\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:\Rightarrow\:−\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right) \\ $$$$\Rightarrow\:−\mathrm{2}{cos}\mathrm{72}°\:=\:−\mathrm{2}{sin}\mathrm{18}°\:=\:−\mathrm{2}\left(\frac{\sqrt{\mathrm{5}}\:−\:\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\:−\:\left(\frac{\sqrt{\mathrm{5}}\:−\:\mathrm{1}}{\mathrm{2}}\right) \\ $$
Commented by mnjuly1970 last updated on 02/Apr/23
$${thx}\:{very}\:{nice} \\ $$