bio | website | math.harvard.edu/~elkies |
---|---|---|
location | Harvard University, Cambridge, MA 02138 | |
age | 48 | |
visits | member for | 3 years, 11 months |
seen | 7 hours ago | |
stats | profile views | 23,390 |
[Not sure yet what to use this space for...]
Apr 12 |
comment |
Models for the moduli space $\overline{M}_{1,n}$
I thought it was only rank $9$ (with further tricks to find rank $10$ and $11$ even though $M_{1,11}$ and $M_{1,12}$ are not rational). So it's $T_1$ through $T_{18}$, not $T_{20}$; and moreover the first $8$ parameters can be removed thanks to the ${\rm PGL}_3$ automorphisms of the projective plane, leaving a rational parametrization of the moduli space $\overline{M}_{1,10}$ by $T_9,\ldots,T_{18}$. |
Apr 12 |
comment |
Models for the moduli space $\overline{M}_{1,n}$
I think it's rational only for $n \leq 10$ and is birationally parametrized by configurations of $n-1$ points in the plane and the cubic(s) through them. (The last point is obtained using the divisor $O(1)$ on the cubic.) |
Apr 4 |
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Diophantine equations and the numbers $4,7,8$
If $x,y,z$ are positive then $x^n + y^n + z^n \geq 3 (xyz)^{n/3}$ (AM-GM inequality), so once $n>3$ it's a finite search. |
Apr 3 |
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Finding extrema of a cubic equation
1) The problem has been concocted to have a rational solution. Finding rational roots of a cubic (e.g. by factoring) is fair game. Or 2) divide by $2$ to get $216/(10-x)^3 = 125/x^3$ and extract cube roots. |
Mar 28 |
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Are there any Algebraic Geometry Theorems that were proved using Combinatorics?
(A generic smooth hypersurface, I suppose you mean: there are certainly hypersurfaces of each degree and dimension with nontrivial automorphism groups.) |
Mar 22 |
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About distinct eigenvalues of a graph
Wait, a disconnected graph does not satisfy this condition for any polynomial $p$, and can easily have all eigenvalues distinct (e.g. the graph with 3 vertices and 1 edge). [Presumably $A$ is the adjacency matrix, though the problem statement never actually defines $A$.] |
Mar 22 |
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Enumeration of $0-1$ matrices with determinant $1$
Why is that a reasonable guess? There are $2^{n^2}$ zero-one matrices, each with $|{\rm disc}| < n^{n/2}$, so I'd expect $2^{n^2 - O(n\log n)}$. |
Mar 20 |
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Can you find squares in this class?
@Michael Stoll For $p=997$, your ratpoints program doesn't take long to find the minimal solution $(l,m) = (130792, 148329)$. |
Mar 18 |
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Dynamics of electrons on a sphere
The answer to Q1 must be Yes, because the differential equation has a unique solution so it must retain the initial symmetry. For Q2 the particles must approach a stable local minimum, but in general they don't have to find a global minimum, and the probability of success might depend also on the strength of the damping. (Is a regular $n$-gon on the equator locally stable once $n>3$?) |
Mar 15 |
revised |
Structure of sign changes under the heat flow
Explain why $u$ solves the heat equation. |
Mar 14 |
awarded | Enlightened |
Mar 14 |
awarded | Nice Answer |
Mar 14 |
awarded | Necromancer |
Mar 14 |
awarded | Revival |
Mar 14 |
answered | Degree 17 number fields ramified only at 2 |
Mar 12 |
revised |
Structure of sign changes under the heat flow
Change notation from $N_u(\ldots)$ to $V_u(\ldots)$ to conform with OP's notation |
Mar 12 |
answered | Structure of sign changes under the heat flow |
Mar 12 |
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The Maximal $\ell_2$ norm of a signed sum of vectors
There are $2^{n-1}$ possibilities, so for $n=3$ it's just the maximum of four candidates, which is easy to compute. For example, let $G$ be the Gram matrix with $(i,j)$ entry $G_{ij} = v_i \cdot v_j$. Then the maximum norm is at most $\|G\|^{1/2}$ where $\|G\| := \sum_{i,j=1}^3 |G_{ij}|$. Equality holds unless all $G_{i,j}$ entries are nonzero and an odd number of the entries above the diagonal are negative, in which case the maximum is the square root of $\|G\| - 2 \min_{i,j} |G_{ij}|$. |
Mar 11 |
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$x^4+y^4$ powerful for relatively prime $x,y$
Yes I checked, but not that way: you don't want to wait for gp to count to $10^{16}$, let alone factor every number of at most $16$ digits! Much better to try all coprime $(x,y)$ of opposite parity with $x<y$ and $x^4 + y^4 < 10^{16}$; that's only 40 million or so factorizations, which take a few hours to try (and as expected find nothing). Still it's hopeless to reach $3 \cdot 10^{36}$ this way... Now that it's a couple of months since I posted this question, I should post a partial answer evaluating different strategies, the best of which might make the computation barely feasible. |
Mar 11 |
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$2^n$-1 consisting only of small factors
According to that Wikipedia article, the result on $2^n-1$ is actually an earlier theorem of Bang which Zsigmondy generalized. |