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May 15, 2016 at 12:59 vote accept WhatsUp
May 13, 2016 at 0:24 answer added Joe Silverman timeline score: 5
May 12, 2016 at 23:05 comment added WhatsUp @ACL You are right. I just figured it out myself. Do you have idea for the denominator?
May 12, 2016 at 23:01 comment added ACL There is a uniform bound for $h(P)-\hat h(P)$; apply it to the point nP and use that $\hat h(nP)=n^2\hat h(P)$.
May 12, 2016 at 22:55 comment added WhatsUp @ACL Could you tell me where in the book of Silverman is this result stated? The canonical height, by definition, is the limit of $h(2^n P)/4^n$, which is different from what you write. Besides, his Corollary 6.4 of Chapter VIII has a big O constant that depends on $m$.
May 12, 2016 at 22:43 comment added ACL The answer to the second question is YES, and is certainly explained in books on elliptic curves such as Joe Silverman's : the canonical height of $P$, $\hat h(P)$, is the limit of $h(x(nP))/n^2$, when $n\to\infty$.
May 12, 2016 at 22:21 history asked WhatsUp CC BY-SA 3.0