bio | website | |
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location | ||
age | ||
visits | member for | 3 years, 11 months |
seen | 2 days ago | |
stats | profile views | 433 |
Graduate student at UC Berkeley.
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 |
Dec 2 |
awarded | Editor |