Suppose $b_1, b_2, b_3, \dots \in \Bbb{R}$ satisfy the Riccati-type recurrence $$b_{k+1}=\frac{1+kb_k}{k-b_k},\quad k\ge 1.$$
Is it true that such a sequence reaches infinitely many positive as well as negative values? It appears to be so.
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asked May 4, 2018 at 16:22
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1 Answer
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Yes, choose $a_1$ such that $\frac{1}{b_1}=\tan(a_1)$, and let $a_{k+1}=a_k-\arctan\frac{1}{k}$, then we have $\frac{1}{b_k}=\tan(a_k)$, since $\sum_k \arctan\frac{1}{k}=\infty$, the conjecture follows.
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$\begingroup$ Note that if $b_1=0$ we can take $a_1=\frac{\pi}{2}$ $\endgroup$– BonbonCommented May 4, 2018 at 16:45
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$\begingroup$ Not sure if this proves the claim. $\endgroup$ Commented May 4, 2018 at 17:23
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$\begingroup$ @T.Amdeberhan where makes it "not sure"? You mean $b_i$ may be zero ? If you mean this, its my fault. I should write it as $b_i=cot(a_i)=tan(\frac{\pi}{4}-a_i)$. $\endgroup$– BonbonCommented May 4, 2018 at 19:19
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4$\begingroup$ It's very nice. If you need more details, the sequence $a_k$ is decreasing to $-\infty$ with jumps going to 0, so its tangent achieves infinitely many change signs. The trick is quite natural: your homographies can be viewed in the projective line (given by rotations), and with a coordinate changes this can be viewed as standard rotations, which after taking these parameters (on the universal covering), yield translations. $\endgroup$– YCorCommented May 4, 2018 at 19:45
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