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Question Number 77591 by naka3546 last updated on 08/Jan/20
x + (1/x) = 1 x^(100) + (1/x^(100) ) = ? x^(2020) + (1/x^(2020) ) = ?
$${x}\:+\:\frac{\mathrm{1}}{{x}}\:\:=\:\:\mathrm{1} \\ $$$${x}^{\mathrm{100}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{100}} }\:\:=\:\:? \\ $$$${x}^{\mathrm{2020}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2020}} }\:\:=\:\:? \\ $$
Commented by jagoll last updated on 08/Jan/20
(1/x^2 )+x^2 =−1 (1/x^4 )+x^4 =−1 (1/x^8 )+x^8 = −1 (1/x^(2n) )+x^(2n) =−1 so (1/x^(100) )+x^(100) =−1
$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+{x}^{\mathrm{4}} =−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{8}} }+{x}^{\mathrm{8}} \:=\:−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}{n}} }+{x}^{\mathrm{2}{n}} =−\mathrm{1}\:{so}\:\frac{\mathrm{1}}{{x}^{\mathrm{100}} }+{x}^{\mathrm{100}} =−\mathrm{1} \\ $$
Commented by abdomathmax last updated on 08/Jan/20
x+(1/x)=1 ⇒x^2 +1=x ⇒x^2 −x+1=0 Δ=1−4=−3 ⇒x_1 =((1+i(√3))/2) =e^((iπ)/3) x_2 =((1−i(√3))/2) =e^(−((iπ)/3)) x=x_1 ⇒x^(100) +x^(−100) = e^((i100π)/3) +e^(−((i100π)/3)) =2cos(((100π)/3)) =2cos(((99π+π)/3))=2cos(33π +(π/3)) = 2cos(π+(π/3))=−2×(1/2) =−1 x=x_2 ⇒x^(100) +x^(−100) = e^(−((i100π)/3)) +e^((i100π)/3) =−1 for x^(2020) +(1/x^(2020) ) x=x_1 ⇒x^(2020) +x^(−2020) =e^(i((2020π)/3)) +e^(−((i2020π)/3)) =2cos(((2020π)/3))=2cos(((2019π+π)/3)) =2cos(((2019π)/3)+(π/3))=...
$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{1}\:\Rightarrow{x}^{\mathrm{2}} +\mathrm{1}={x}\:\Rightarrow{x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\mathrm{1}−\mathrm{4}=−\mathrm{3}\:\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{\frac{{i}\pi}{\mathrm{3}}} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{−\frac{{i}\pi}{\mathrm{3}}} \: \\ $$$${x}={x}_{\mathrm{1}} \:\Rightarrow{x}^{\mathrm{100}} \:+{x}^{−\mathrm{100}} =\:{e}^{\frac{{i}\mathrm{100}\pi}{\mathrm{3}}} \:+{e}^{−\frac{{i}\mathrm{100}\pi}{\mathrm{3}}} \\ $$$$=\mathrm{2}{cos}\left(\frac{\mathrm{100}\pi}{\mathrm{3}}\right)\:=\mathrm{2}{cos}\left(\frac{\mathrm{99}\pi+\pi}{\mathrm{3}}\right)=\mathrm{2}{cos}\left(\mathrm{33}\pi\:+\frac{\pi}{\mathrm{3}}\right) \\ $$$$=\:\mathrm{2}{cos}\left(\pi+\frac{\pi}{\mathrm{3}}\right)=−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\:=−\mathrm{1} \\ $$$${x}={x}_{\mathrm{2}} \:\Rightarrow{x}^{\mathrm{100}} \:+{x}^{−\mathrm{100}} =\:{e}^{−\frac{{i}\mathrm{100}\pi}{\mathrm{3}}} +{e}^{\frac{{i}\mathrm{100}\pi}{\mathrm{3}}} =−\mathrm{1} \\ $$$${for}\:{x}^{\mathrm{2020}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2020}} } \\ $$$${x}={x}_{\mathrm{1}} \:\Rightarrow{x}^{\mathrm{2020}} \:+{x}^{−\mathrm{2020}} ={e}^{{i}\frac{\mathrm{2020}\pi}{\mathrm{3}}} \:+{e}^{−\frac{{i}\mathrm{2020}\pi}{\mathrm{3}}} \\ $$$$=\mathrm{2}{cos}\left(\frac{\mathrm{2020}\pi}{\mathrm{3}}\right)=\mathrm{2}{cos}\left(\frac{\mathrm{2019}\pi+\pi}{\mathrm{3}}\right) \\ $$$$=\mathrm{2}{cos}\left(\frac{\mathrm{2019}\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{3}}\right)=… \\ $$
Commented by naka3546 last updated on 08/Jan/20
n= 6 ⇒ the answer is not −1
$${n}=\:\mathrm{6}\:\:\Rightarrow\:\:{the}\:\:{answer}\:\:{is}\:\:{not}\:\:−\mathrm{1} \\ $$
Commented by jagoll last updated on 08/Jan/20
how you get 6 sir?
$${how}\:{you}\:{get}\:\mathrm{6}\:{sir}? \\ $$
Commented by naka3546 last updated on 08/Jan/20
prove it if the formula is right for all n .
$${prove}\:\:{it}\:\:{if}\:\:{the}\:\:{formula}\:\:{is}\:\:{right}\:\:{for}\:\:{all}\:\:{n}\:. \\ $$
Answered by MJS last updated on 08/Jan/20
x=(1/2)±((√3)/2)i ⇒ (1/x)=(1/2)∓((√3)/2)i z=x^m +(1/x^m ) x^(6n) =1 ⇒ z=2 x^(6n+1) =x ⇒ z=1 x^(6n+2) =−(1/x)=−x^� ⇒ z=−1 x^(6n+3) =−1 ⇒ z=−2 x^(6n+4) =−x ⇒ z=−1 x^(6n+5) =(1/x)=x^� ⇒ z=1 100=16×6+4 ⇒ z=−1 2020=336×6+4 ⇒ z=−1
$${x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\:\Rightarrow\:\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$${z}={x}^{{m}} +\frac{\mathrm{1}}{{x}^{{m}} } \\ $$$${x}^{\mathrm{6}{n}} =\mathrm{1}\:\Rightarrow\:{z}=\mathrm{2} \\ $$$${x}^{\mathrm{6}{n}+\mathrm{1}} ={x}\:\Rightarrow\:{z}=\mathrm{1} \\ $$$${x}^{\mathrm{6}{n}+\mathrm{2}} =−\frac{\mathrm{1}}{{x}}=−\bar {{x}}\:\Rightarrow\:{z}=−\mathrm{1} \\ $$$${x}^{\mathrm{6}{n}+\mathrm{3}} =−\mathrm{1}\:\Rightarrow\:{z}=−\mathrm{2} \\ $$$${x}^{\mathrm{6}{n}+\mathrm{4}} =−{x}\:\Rightarrow\:{z}=−\mathrm{1} \\ $$$${x}^{\mathrm{6}{n}+\mathrm{5}} =\frac{\mathrm{1}}{{x}}=\bar {{x}}\:\Rightarrow\:{z}=\mathrm{1} \\ $$$$\mathrm{100}=\mathrm{16}×\mathrm{6}+\mathrm{4}\:\Rightarrow\:{z}=−\mathrm{1} \\ $$$$\mathrm{2020}=\mathrm{336}×\mathrm{6}+\mathrm{4}\:\Rightarrow\:{z}=−\mathrm{1} \\ $$

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