Timeline for distance formula in Farey graph?
Current License: CC BY-SA 2.5
12 events
when toggle format | what | by | license | comment | |
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Dec 13, 2013 at 0:56 | comment | added | Alexey Ustinov | @Ricardo Andrade: Sorry, now I realize this rule. | |
Dec 12, 2013 at 17:42 | comment | added | Ricardo Andrade | Dear @Alexey Ustinov, please do not edit more than three old posts each day, as they get bumped to the front page: see this meta thread. You have edited over ten old posts in the past 24 hours. Also, this question could use several tags which are more relevant than the the tag 'continued-fractions'. | |
S Dec 12, 2013 at 9:35 | history | suggested | Alexey Ustinov |
The Tag "continued fractions" was added
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Dec 12, 2013 at 4:08 | review | Suggested edits | |||
S Dec 12, 2013 at 9:35 | |||||
Jul 27, 2010 at 17:20 | vote | accept | Kevin M Pilgrim | ||
Jul 27, 2010 at 17:20 | vote | accept | Kevin M Pilgrim | ||
Jul 27, 2010 at 17:20 | |||||
Jul 25, 2010 at 2:04 | answer | added | Ian Agol | timeline score: 10 | |
Jul 23, 2010 at 19:35 | answer | added | Sam Nead | timeline score: 2 | |
Jul 23, 2010 at 17:56 | answer | added | Allen Hatcher | timeline score: 8 | |
Jul 22, 2010 at 20:45 | comment | added | S. Carnahan♦ | Let $a$ and $b$ be integers such that $ap+bq=1$. Then the distance is the length of the continued fraction expansion of $(ar+bs)/(ps-qr)$, where $\infty$ has length zero. Dually, it is the number of Ford circles that intersect the vertical line with real part $(ar+bs)/(ps-qr)$. | |
Jul 22, 2010 at 20:34 | comment | added | Robin Chapman | When $p/q=1/0$ then this distance is essentially the number of steps in the Euclidean algorithm for $r$ and $s$ (and the general case reduces to this via $SL_2(\mathbb{Z})$-equivariance). Now I don't know if there is a "formula" for the number of steps in the Euclidean algorithm starting at $r$ and $s$ ( save perhaps for the phrase "the number of steps in the Euclidean algorithm starting at $r$ and $s$:-)). | |
Jul 22, 2010 at 20:26 | history | asked | Kevin M Pilgrim | CC BY-SA 2.5 |