Impact
~11k
people reached
- 0 posts edited
- 0 helpful flags
- 23 votes cast
Sep
8 |
awarded | Taxonomist |
Jul
10 |
awarded | Nice Question |
Feb
27 |
revised |
A generalization of the Powers-Stormer inequality
edited tags |
Feb
27 |
asked | A generalization of the Powers-Stormer inequality |
Sep
24 |
awarded | Autobiographer |
Jul
2 |
awarded | Curious |
May
22 |
comment |
A measure of closure under sumset?
Thanks, quid, for your answer, and the links to Hamidoune's work. |
May
22 |
accepted | A measure of closure under sumset? |
May
22 |
revised |
A measure of closure under sumset?
added 138 characters in body |
May
21 |
asked | A measure of closure under sumset? |
Apr
22 |
awarded | Yearling |
Oct
9 |
accepted | Multivariate Hensel's Lemma, but with only one polynomial |
Oct
8 |
comment |
Multivariate Hensel's Lemma, but with only one polynomial
I have a quick clarification question: When you say "Suppose $Q(X_1,\ldots,X_n)$ is in $I^t$", you mean suppose there is are $\beta_1,\ldots,\beta_n\in R$ such that $Q(\beta_1,\ldots,\beta_n)\in I^t$? And then every time you write $\frac{dQ}{dX_1}$, you mean it is evaluated at the $\beta_1,\ldots,\beta_n$? If that's the case, then I interpret the mod $I^{t+1}$ solution as $\beta_1 + a_1,\ldots,\beta_n+a_n$. Thanks! |
Oct
3 |
revised |
Multivariate Hensel's Lemma, but with only one polynomial
edited title |
Oct
3 |
revised |
Multivariate Hensel's Lemma, but with only one polynomial
deleted 4 characters in body |
Oct
3 |
accepted | What is the relationship between singular value decomposition and solving linear systems? |
Oct
3 |
accepted | Inequalities and bounds for relating p-norms (Reference request) |
Oct
3 |
asked | Multivariate Hensel's Lemma, but with only one polynomial |
Jul
30 |
accepted | Can formal power series become polynomial often, when composed with polynomials? |
Jul
30 |
comment |
Can formal power series become polynomial often, when composed with polynomials?
Thanks David! This will take me a little while to parse, so i might add follow up questions in the comments. But for now I accept this! |