Timeline for Spirals in Apollonian circle-packings
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 10, 2014 at 15:01 | answer | added | Hao Chen | timeline score: 5 | |
| Nov 10, 2014 at 14:42 | history | edited | James Propp | CC BY-SA 3.0 |
Added forward pointer to follow-up question
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| Nov 5, 2014 at 2:25 | comment | added | Noam D. Elkies | Thanks. Meanwhile I did verify this, and will post some more details soon. For now you can zoom in on the picture at math.harvard.edu/~elkies/mo186137.pdf to see a few iterations of the limiting configuration. BTW there's the additional small subtlety that conformal maps don't preserve circle centers, but again the discrepancy disappears in the limit. | |
| Nov 5, 2014 at 2:13 | comment | added | James Propp | I like Noam's argument. Can anyone complete the proof by showing that the angle in question is irrational? | |
| Nov 5, 2014 at 2:11 | comment | added | James Propp | I should point out a subtlety that some readers may have missed: conformal maps are only locally angle-preserving, so the angular distribution of the unit vectors in the general case cannot be obtained by applying a rotation or other simple map to the unit vectors in the special case that Noam proposes. However, since the points $P_n$ and $Q_n$ all approach $P_{\infty}$, and since the inversive map preserves angles in the vicinity of $P_{\infty}$, we can ignore this distortion for purposes of the asymptotic angular distribution of the vectors. | |
| Nov 3, 2014 at 23:44 | comment | added | Noam D. Elkies | [cont'd] a positive root of $r^2 + r^{-2} = 2(r+r^{-1}+1)$, which sauf erreur is $$ r_0 = \frac12\Bigl(1 + \sqrt{5} - \sqrt{2+\sqrt{20}}\Bigr) = 0.346014339\ldots $$ or equivalently the root $r_0^{-1}$ obtained by changing $-\sqrt\cdots$ to $+\sqrt\cdots$. Presumably this spiral yields irrational angles, and thus exact equidistribution. | |
| Nov 3, 2014 at 23:43 | comment | added | Noam D. Elkies | All such configurations are inversively equivalent, and any inversion is angle-preserving and a local homothecy. So it's enough to consider the case that $C_1,C_2,C_3$ have radii chosen so that the radii of $C_1,C_2,C_3,C_4$ are in geometric progression, because then by induction the same is true for all the $C_n$, whence $P_n$ and $Q_n$ lie on actual spirals. If I did this right the common ratio of this progression is [cont'd] | |
| Nov 3, 2014 at 22:54 | history | asked | James Propp | CC BY-SA 3.0 |