Timeline for Is it possible for the repeated doubling of a non torsion point of an elliptic curve stays bounded in the affine plane?
Current License: CC BY-SA 2.5
9 events
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| Mar 28, 2010 at 11:44 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
typo
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| Mar 28, 2010 at 11:31 | comment | added | Robin Chapman | In that case should the ``elliptic logarithm'' of a rational point have this property then the answer to the original question would be `yes'. I don't think this can be the case but proving it suddenly looks like hard work :-( | |
| Mar 28, 2010 at 11:31 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
edited to say answer was wrong
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| Mar 28, 2010 at 11:22 | comment | added | Douglas Zare | However, $2^n \xi$ doesn't need to meet every neighborhood of the origin if you only assume $\xi$ is irrational. That $\xi$ is irrational just means the "digits" of the binary expansion aren't preperiodic, not that $\xi$ is normal base 2. | |
| Mar 28, 2010 at 10:59 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
typo
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| Mar 28, 2010 at 10:57 | comment | added | Kevin Buzzard | @Robin: you're right---I misread the question. Thanks. | |
| Mar 28, 2010 at 10:32 | comment | added | Robin Chapman | Kevin, repeated doubling of a point doesn't produce the subgroup generated by that point. What is needed is a theorem to the effect that if $\xi$ is irrational then the set of $2^n\xi$ in $\mathbb{R}/\mathbb{Z}$ meets every neighbourhood of the origin in $\mathbb{R}/\mathbb{Z}$. I'm sure this is true but can't see an immediate proof. | |
| Mar 28, 2010 at 10:30 | comment | added | defgh | I mean the absolute. Thanks. You've answered my question. | |
| Mar 28, 2010 at 10:15 | history | answered | Kevin Buzzard | CC BY-SA 2.5 |