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

Answered by A5T last updated on 17/Nov/24

$$x_0=k\Rightarrow x_1=\frac{1+k}{1-k}\Rightarrow x_2=\frac{-1}{k}\Rightarrow x_3=\frac{k-1}{1+k}\Rightarrow x_4=\frac{2k}{2}=k$$$$\Rightarrow x_{4n}=k=2022$$

Answered by Frix last updated on 17/Nov/24

$$n⩾0:$$$$x_{4n}=x_0$$$$x_{4n+1}=\frac{1+x_0}{1-x_0}$$$$x_{4n+2}=-\frac{1}{x_0}$$$$x_{4n+3}=\frac{x_0-1}{x_0+1}$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{easy}...$$

Answered by golsendro last updated on 17/Nov/24

$${\mathrm{x}}_{\mathrm{n}}=\frac{1+{\mathrm{x}}_{\mathrm{n}-1}}{1-{\mathrm{x}}_{\mathrm{n}-1}};\mathrm{n}-1⩾0$$$${\mathrm{x}}_1=\frac{2023}{-2021},{\mathrm{x}}_2=\frac{1-\frac{2023}{2021}}{1+\frac{2023}{2021}}=\frac{-1}{2022}$$$${\mathrm{x}}_3=\frac{1-\frac{1}{2022}}{1+\frac{1}{2022}}=\frac{2021}{2023},{\mathrm{x}}_4=\frac{1+\frac{2021}{2023}}{1-\frac{2021}{2023}}=2022$$$$\Rightarrow 4\mid 2024;{\mathrm{x}}_{2024}={\mathrm{x}}_0={\mathrm{x}}_4={\mathrm{x}}_8...=2022$$

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

$$let{x}_n=\tan {\theta }_n$$$$\tan {\theta }_{n+1}=x_{n+1}=\frac{1+\tan {\theta }_n}{1-\tan {\theta }_n}=\tan ({\theta }_n+\frac{\pi }{4})$$$$\Rightarrow {\theta }_{n+1}={\theta }_n+\frac{\pi }{4}$$$$\Rightarrow {\theta }_n={\theta }_{n-1}+\frac{\pi }{4}={\theta }_{n-2}+\frac{2\pi }{4}=...={\theta }_0+\frac{n\pi }{4}$$$$\Rightarrow x_n=\tan ({\theta }_0+\frac{n\pi }{4})=\frac{\tan {\theta }_0+\tan \frac{n\pi }{4}}{1-\tan {\theta }_n\tan \frac{n\pi }{4}}$$$$\Rightarrow x_n=\frac{x_0+\tan \frac{n\pi }{4}}{1-x_0\tan \frac{n\pi }{4}}=\frac{2022+\tan \frac{n\pi }{4}}{1-2022\tan \frac{n\pi }{4}}$$$$x_{2024}=\frac{2022+\tan \frac{2024\pi }{4}}{1-2022\tan \frac{2024\pi }{4}}$$$$=\frac{2022+0}{1-2022\times 0}=2022$$

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