bio | website | maths.bris.ac.uk/~mahah |
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visits | member for | 5 years, 2 months |
seen | Dec 12 at 20:15 | |
stats | profile views | 3,959 |
Dec 18 |
awarded | Nice Question |
Oct 15 |
awarded | Yearling |
Oct 13 |
awarded | Custodian |
Oct 13 |
reviewed | Approve books on very large scale linear optimization |
Sep 24 |
revised |
Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
added 14 characters in body |
Sep 24 |
answered | Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large |
Sep 23 |
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Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
HJRW - this (= going over a ball in the Cayley graph) is actually how I am producing drafts of posters right now. I still think it would be nice to know a priori how far one needs to go. |
Sep 23 |
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Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
Well, one can take care of the region $\Re(y)>1$ in the obvious way, and then apply a Monte Carlo method like this one to the rest. Still, this isn't really that satisfying. |
Sep 22 |
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Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
This is a good idea, but I'd still like to have an explicit bound. $2 A(S)/\epsilon$ (where $A(S)$ is presumably the Euclidean area of $S$) can be pretty large. |
Sep 22 |
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Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
In the case of a box $\lbrack 0,N\rbrack \times \lbrack 0,N\rbrack$ (a good example, actually), the hyperbolic area of $S$ is $\infty$, yet the question still makes sense. So, you really have to know what $S$ is. (And as I said, you may assume it is a box.) |
Sep 22 |
asked | Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large |
Sep 15 |
awarded | Popular Question |
Jul 23 |
awarded | Popular Question |
Jul 11 |
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Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
The point is that I should have been a bit more precise about the wording - I think most have understood the "right meaning" by now, but it is worth repeating. Sean Eberhard's reply below contains a clean formulation. |
Jul 8 |
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Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
... but if anybody can get $< (1-\epsilon) p$ for quadruple exponentiation, Kate and I will be very happy. |
Jul 8 |
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Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
Yes, we know that - that gives that there are at most $(1-\epsilon) p$ solutions. The answer below gives $<\epsilon p$ (without attempting to follow closely the group-theoretical ideas that give $(1-\epsilon) p$). Notice that the supposedly non-group-theoretical proof for two iterations (and possibly that for three iterations) is soft enough that it can be adapted even if there are $o(p)$ breaking points. |
Jul 8 |
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Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
...and actually, you mean either "if one disproves a slightly weaker statement than the above", or, what is the same, "if one proves a slightly stronger statement than the negative of the statement above". I'm trying to keep "positive" and "negative" answers straight... |
Jul 8 |
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Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
Kate, do you have to give everything away? :) |
Jul 8 |
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Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
On your second comment: which answer is positive and which answer is negative? I take you mean to say that, if $2^{2^{2^{2^x}}} = x \mod p$ had $> (1-\epsilon) p$ fixed points, then the discrete logarithm would be easy to compute? (Agreed.) |
Jul 8 |
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Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
Actually, Andreas, your example (the group $H$) strengthens the case for an algebraic approach. Its soficity corresponds to $2 (x+1) = 2x + 2$, which is, um, true. Or do you have a different soficity-algebra dictionary in mind? Please explain. |