Timeline for Riccati-type recurrence: infinitely many sign changes?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 5, 2018 at 13:47 | vote | accept | T. Amdeberhan | ||
| May 4, 2018 at 20:48 | comment | added | YCor | The argument gives you in addition an estimate in the number of consecutive steps with a given sign. | |
| May 4, 2018 at 20:38 | comment | added | Bonbon | @T.Amdeberhan As is explained by YCor in his comments before, $arctan\frac{1}{k}\rightarrow 0$ as $k\rightarrow \infty$, while the sign changes once you go $\frac{\pi}{2}$, $a_k$ can't just skip a $\frac{\pi}{2}$ and keep the sign of $b_k$, | |
| May 4, 2018 at 20:30 | comment | added | T. Amdeberhan | Thanks, but it is possible that the $a_k\rightarrow-\infty$ and yet $b_k>0$ all the time. This needs justification, I presume. | |
| May 4, 2018 at 19:52 | comment | added | Bonbon | Yes, a great comment! | |
| May 4, 2018 at 19:45 | comment | added | YCor | 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. | |
| May 4, 2018 at 19:19 | comment | added | Bonbon | @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 $b_i=cot(a_i)=tan(\frac{\pi}{4}-a_i)$. | |
| May 4, 2018 at 19:11 | history | edited | YCor | CC BY-SA 4.0 |
latexified tan
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| May 4, 2018 at 17:23 | comment | added | T. Amdeberhan | Not sure if this proves the claim. | |
| May 4, 2018 at 16:45 | comment | added | Bonbon | Note that if $b_1=0$ we can take $a_1=\frac{\pi}{2}$ | |
| May 4, 2018 at 16:35 | history | answered | Bonbon | CC BY-SA 4.0 |