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visits member for 4 years, 6 months
seen Nov 21 at 1:58

Graduate student at UC Berkeley.


Sep
24
awarded  Autobiographer
Dec
10
awarded  Scholar
Dec
10
comment Representation-theoretic operations on modular forms
I'm aware AB isn't a eigenform: that's why this question is interesting. Is it an integer combination of eigenforms? It sounds like very little is known in this direction.
Dec
9
awarded  Yearling
Dec
9
awarded  Student
Dec
9
asked Representation-theoretic operations on modular forms
Dec
1
answered Are there any fast algorithms for factoring integers that don't work by searching for smooth numbers?
Nov
28
awarded  Critic
Nov
26
awarded  Citizen Patrol
Nov
26
answered Is there adequate test statistics for the outlier in the set of data
Oct
14
comment Rational points or a Weierstrass model for degree 8 elliptic curve
It might help if you explain where this equation comes from. This doesn't look like an elliptic curve but some sort of parametrization where fibers are curves. Maybe take some fibers and see how they work? Also, assuming Birch and Swinnerton-Dyer you can do calculations over finite fields to help create bounds on rank.
Jun
22
answered Does Godel's incompleteness theorem admit a converse?
Jan
29
answered Concentration bounds for sums of random variables of permutations
Dec
12
comment how to find/define eigenvectors as a continuous function of matrix?
So consider the map $det(M-\lambdaI)$. This is a continuous map from $\mathbb{C}\times D$ to $\mathbb{C}$, with nice properties of the derivative. Apply the Inverse Function Theorem.
Dec
12
awarded  Supporter
Dec
12
answered how to find/define eigenvectors as a continuous function of matrix?
Dec
9
comment Sieve of Erathostenes: removing consecutive items
That is true. I should think harder about what $L(n)$ means, and hopefully come up with something useful.
Dec
9
revised Sieve of Erathostenes: removing consecutive items
added 2 characters in body; edited body; added 26 characters in body
Dec
9
comment Sieve of Erathostenes: removing consecutive items
Well, if that is the case then $(a-b)p>q$ because there is a prime between $2p$ and $p$. And so one number in the range $ap$ to $bp$ must be divisible by $q$.
Dec
8
answered Sieve of Erathostenes: removing consecutive items