Question-213796

Question Number 213796 by efronzo1 last updated on 17/Nov/24

Answered by A5T last updated on 17/Nov/24

x_0 =k⇒x_1 =((1+k)/(1−k))⇒x_2 =((−1)/k)⇒x_3 =((k−1)/(1+k))⇒x_4 =((2k)/2)=k ⇒x_(4n) =k=2022

$${x}_{\mathrm{0}} ={k}\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}\Rightarrow{x}_{\mathrm{2}} =\frac{−\mathrm{1}}{{k}}\Rightarrow{x}_{\mathrm{3}} =\frac{{k}−\mathrm{1}}{\mathrm{1}+{k}}\Rightarrow{x}_{\mathrm{4}} =\frac{\mathrm{2}{k}}{\mathrm{2}}={k} \\ $$$$\Rightarrow{x}_{\mathrm{4}{n}} ={k}=\mathrm{2022} \\ $$

Answered by Frix last updated on 17/Nov/24

n≥0: x_(4n) =x_0 x_(4n+1) =((1+x_0 )/(1−x_0 )) x_(4n+2) =−(1/x_0 ) x_(4n+3) =((x_0 −1)/(x_0 +1)) It′s easy...

$${n}\geqslant\mathrm{0}: \\ $$$${x}_{\mathrm{4}{n}} ={x}_{\mathrm{0}} \\ $$$${x}_{\mathrm{4}{n}+\mathrm{1}} =\frac{\mathrm{1}+{x}_{\mathrm{0}} }{\mathrm{1}−{x}_{\mathrm{0}} } \\ $$$${x}_{\mathrm{4}{n}+\mathrm{2}} =−\frac{\mathrm{1}}{{x}_{\mathrm{0}} } \\ $$$${x}_{\mathrm{4}{n}+\mathrm{3}} =\frac{{x}_{\mathrm{0}} −\mathrm{1}}{{x}_{\mathrm{0}} +\mathrm{1}} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{easy}… \\ $$

Answered by golsendro last updated on 17/Nov/24

x_n = ((1+x_(n−1) )/(1−x_(n−1) )) ; n−1 ≥ 0 x_1 = ((2023)/(−2021)) , x_2 =((1−((2023)/(2021)))/(1+((2023)/(2021)))) = ((−1)/(2022)) x_3 = ((1−(1/(2022)))/(1+(1/(2022)))) = ((2021)/(2023)) , x_4 = ((1+((2021)/(2023)))/(1−((2021)/(2023))))=2022 ⇒ 4 ∣2024 ; x_(2024) = x_0 =x_4 =x_8 ...=2022

$$\:\:\:\mathrm{x}_{\mathrm{n}} \:=\:\frac{\mathrm{1}+\mathrm{x}_{\mathrm{n}−\mathrm{1}} }{\mathrm{1}−\mathrm{x}_{\mathrm{n}−\mathrm{1}} }\:;\:\mathrm{n}−\mathrm{1}\:\geqslant\:\mathrm{0} \\ $$$$\:\:\mathrm{x}_{\mathrm{1}} \:=\:\frac{\mathrm{2023}}{−\mathrm{2021}}\:,\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{1}−\frac{\mathrm{2023}}{\mathrm{2021}}}{\mathrm{1}+\frac{\mathrm{2023}}{\mathrm{2021}}}\:=\:\frac{−\mathrm{1}}{\mathrm{2022}} \\ $$$$\:\mathrm{x}_{\mathrm{3}} \:=\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2022}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2022}}}\:=\:\frac{\mathrm{2021}}{\mathrm{2023}}\:,\:\mathrm{x}_{\mathrm{4}} =\:\frac{\mathrm{1}+\frac{\mathrm{2021}}{\mathrm{2023}}}{\mathrm{1}−\frac{\mathrm{2021}}{\mathrm{2023}}}=\mathrm{2022} \\ $$$$\:\Rightarrow\:\mathrm{4}\:\mid\mathrm{2024}\:;\:\mathrm{x}_{\mathrm{2024}} \:=\:\mathrm{x}_{\mathrm{0}} =\mathrm{x}_{\mathrm{4}} =\mathrm{x}_{\mathrm{8}} …=\mathrm{2022} \\ $$

Answered by mr W last updated on 17/Nov/24

let x_n =tan θ_n tan θ_(n+1) =x_(n+1) =((1+tan θ_n )/(1−tan θ_n ))=tan (θ_n +(π/4)) ⇒θ_(n+1) =θ_n +(π/4) ⇒θ_n =θ_(n−1) +(π/4)=θ_(n−2) +((2π)/4)=...=θ_0 +((nπ)/4) ⇒x_n =tan (θ_0 +((nπ)/4))=((tan θ_0 +tan ((nπ)/4))/(1−tan θ_n tan ((nπ)/4))) ⇒x_n =((x_0 +tan ((nπ)/4))/(1−x_0 tan ((nπ)/4)))=((2022+tan ((nπ)/4))/(1−2022 tan ((nπ)/4))) x_(2024) =((2022+tan ((2024π)/4))/(1−2022 tan ((2024π)/4))) =((2022+0)/(1−2022×0))=2022

$${let}\:{x}_{{n}} =\mathrm{tan}\:\theta_{{n}} \\ $$$$\mathrm{tan}\:\theta_{{n}+\mathrm{1}} ={x}_{{n}+\mathrm{1}} =\frac{\mathrm{1}+\mathrm{tan}\:\theta_{{n}} }{\mathrm{1}−\mathrm{tan}\:\theta_{{n}} }=\mathrm{tan}\:\left(\theta_{{n}} +\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Rightarrow\theta_{{n}+\mathrm{1}} =\theta_{{n}} +\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\theta_{{n}} =\theta_{{n}−\mathrm{1}} +\frac{\pi}{\mathrm{4}}=\theta_{{n}−\mathrm{2}} +\frac{\mathrm{2}\pi}{\mathrm{4}}=…=\theta_{\mathrm{0}} +\frac{{n}\pi}{\mathrm{4}} \\ $$$$\Rightarrow{x}_{{n}} =\mathrm{tan}\:\left(\theta_{\mathrm{0}} +\frac{{n}\pi}{\mathrm{4}}\right)=\frac{\mathrm{tan}\:\theta_{\mathrm{0}} +\mathrm{tan}\:\frac{{n}\pi}{\mathrm{4}}}{\mathrm{1}−\mathrm{tan}\:\theta_{{n}} \mathrm{tan}\:\frac{{n}\pi}{\mathrm{4}}} \\ $$$$\Rightarrow{x}_{{n}} =\frac{{x}_{\mathrm{0}} +\mathrm{tan}\:\frac{{n}\pi}{\mathrm{4}}}{\mathrm{1}−{x}_{\mathrm{0}} \mathrm{tan}\:\frac{{n}\pi}{\mathrm{4}}}=\frac{\mathrm{2022}+\mathrm{tan}\:\frac{{n}\pi}{\mathrm{4}}}{\mathrm{1}−\mathrm{2022}\:\mathrm{tan}\:\frac{{n}\pi}{\mathrm{4}}} \\ $$$${x}_{\mathrm{2024}} =\frac{\mathrm{2022}+\mathrm{tan}\:\frac{\mathrm{2024}\pi}{\mathrm{4}}}{\mathrm{1}−\mathrm{2022}\:\mathrm{tan}\:\frac{\mathrm{2024}\pi}{\mathrm{4}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2022}+\mathrm{0}}{\mathrm{1}−\mathrm{2022}×\mathrm{0}}=\mathrm{2022} \\ $$

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