Given $a_0 > 0$ and $c > 0$, the sequence $(a_n)$ is defined by \[a_{n+1}=\frac{a_n+c}{1-ca_n}\quad\text{for }n=1,2,\dots\] Is it possible that $a_0, a_1, \dots , a_{1989}$ are all positive but $a_{1990}$ is negative?
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fractals
19.04.2013 20:57
Let $a = \arctan c$ and $\theta_n = \arctan a_n$. Then $\theta_{n + 1} = \theta_n + a$, an arithmetic sequence. From here it is easy to see that the answer is yes.
isaacmeng
03.04.2022 10:23
Baltic Way 1990.3. Given $a_0\in\mathbb R_+$ and $c\in\mathbb R_+$, the sequence $(a_n)$ is defined by \[a_{n+1}=\frac{a_n+c}{1-ca_n}\]for all $n\in\mathbb Z_{\ge 0}$. Determine if it is possible that $a_0, \dots, a_{1989}$ are all positive but $a_{1990}$ is negative? Solution. If we take $a_0=\tan\theta_0=\tan\frac{\pi}{3981}$ and $c=\tan\gamma=\tan\frac{\pi}{3980}$, it is easy to see that \[0<\theta_0<\theta_0+\gamma<\dots<\theta_{1989}+\gamma<\frac{\pi}2<\theta_{1990}+\gamma<\pi.\]Hence \[\tan(\theta_0), \tan(\theta_0+\gamma), \dots, \tan(\theta_0+1989\gamma)>0, \tan (\theta_0+1990\gamma)<0.\]It is also easy to show by induction that \[a_n=\tan(\theta_0+n\gamma)\]for all $n\in\mathbb Z_+$, since we had the recursion \[a_{n+1}=\frac{\tan(\theta_0+n\gamma)+\tan\gamma}{1-\tan(\theta_0+n\gamma)\tan(\gamma)}=\tan(\theta_0+n\gamma+\gamma),\]this finishes the proof.