7h
answered Can we add an uncountable number of positive elements, and can this sum be finite?
9h
comment How can a society deal with berserkers?
Dwarf fortress has a straightforward solution: if he commits crimes while berserk, you drag the dwarf to jail, chain him up, hit him with a hammer a few times, then release him (or bury him, depending on how heavy the hammer was).
9h
answered Should I free long-lived memory that would normally be freed at the very end of the program?
11h
comment Writing integers in ring of integers of number fields
Also, you can't always have $a_i \in \mathbb{N}$: you can have $\mathbb{Z}[b]$ be a proper subset of $\mathcal{O}_K$. I think there are number fields where $\mathbb{Z}[b]$ is a proper subring for every choice of $b$... although the actual theorem I'm remembering is that the span of $\{ 1, b, \ldots, b^{[K:\mathbb{Q}] - } \}$ is a proper subgroup of $\mathcal{O}_K$ for all $b$.
11h
comment Writing integers in ring of integers of number fields
I would surely call it a restricted form of $b$-adic notation.
1d
answered Proving a Set is a Vector Space
1d
answered Learning the exponents in a sum of two modular roots of unity
1d
comment Learning the exponents in a sum of two modular roots of unity
You can find $a+b$ (if you can compute discrete logarithms), because $f(x)^2 - f(x^2) = 2 x^{a+b}$.
1d
comment Learning the exponents in a sum of two modular roots of unity
@Hao: As an extreme counterexample, suppose $q$ is prime and $n = q-1$. Then $1 + x^{b-a}$ is never going to have order coprime to $n$. Still, there may be something one can do with $(1+x^{b-a})^n$ and suitable $q$.
1d
comment How do mathematicians find the underlying idea?
Even without the underlying idea, this particular instance seems straightforward: just write down the triangle inequality in full generality, and then figure out where the thing you want to know about has to fit in, then figure out where the things you already know have to go, and check if you get a result you're looking for.
1d
comment Learning the exponents in a sum of two modular roots of unity
Ah, fair enough. :(
1d
comment Learning the exponents in a sum of two modular roots of unity
Ah, if $n$ is odd, it easy to find $a+b$, by computing $$ f(1+n) \equiv 2 + (a+b)n \pmod{n^2} $$
1d
revised Learning the exponents in a sum of two modular roots of unity
added 218 characters in body
1d
answered Learning the exponents in a sum of two modular roots of unity
1d
comment Learning the exponents in a sum of two modular roots of unity
Easy with a sparse polynomial representation, but is it actually easy with a black box representation of elements of the field?
1d
comment Learning the exponents in a sum of two modular roots of unity
If $n \mid p-1$, then $\mathbb{Z}[\zeta_n] \subseteq \mathbb{Z}_p$, and the restriction to evaluating in $\mathbb{Z}/p^n$ allows us to get an arbitrary amount of $p$-adic precision. This approach should actually work if the goal was to find $f(\zeta_n)$; but the point I wanted to make is that an approach for $\mathbf{Z}[\zeta_n]$ is probably easier to find (and surely much easier to analyze), and once you have it, can then work out how to do a practical computation with finite quotients.
1d
comment Why should quaternions exist?
$x^2 = x = 1$ shouldn't have any solutions either then?
1d
comment Learning the exponents in a sum of two modular roots of unity
I think you're allowing enough information to evaluate $f$ at any $n$-th root of unity in any ring at all. Methods for finding $a,b$ are probably easier to first work out for $\mathbf{Z}[\zeta_m]$ with $m \mid n$ before worrying about finite fields.
1d
comment Recovering a basis from an isomorphism with the dual space.
$\phi^{-1}$ is an isomorphism $V^* \to V$. Also $\phi^*$ is an isomorphism $V^* \to V^{**}$. So this really isn't even a different problem!
Aug
27
answered What allows us divide/multiply dx in calculus?
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