@AlecTeal Please do not flag issues in the chat if they are not important. Every time you flag something it goes to EVERY moderator across the entire network and that's rude. Flags should be used for serious issues. Do it again and you might get suspended.
These are the current numbers ("current" because they can obviously be changed, but there aren't any plans to do so; they have been like this for a while):
The threshold is six, meaning the net flag count of the message has to reach six. Net flag count means the number of flags minus the number ...
I imagine the difficulty is comparable to finding a generator for $\Bbb F_q^\times$ itself, which is at least as difficult as finding a generator for $(\Bbb Z/p\Bbb Z)^\times$ ($p$ prime), which is difficult enough that there are only algorithms
I think there are formulas one can derive for various (congruence) classes of $p$ as rational functions of $p$
Does anyone know a workaround on how to delete or edit a comment which did not parse correctly and the edit/delete button is being covered up by links to other questions on the far right?
It's definitely not dead, there are plenty of people posting daily
next to nobody is online right now tho
the room is about general discussions regarding maths, so linear algebra questions are welcome, as long as they can be reasonably answered in the chat, for questions which need a long answer the main site would probably be better
Would someone like to help me writing down a formal proof dealing different cases of this claim math.stackexchange.com/q/1482309/275935 ? I wish to learn writing down formal proofs and not just trivial hints.
Could someone help me writing down a formal proof dealing with different cases of this claim math.stackexchange.com/q/1482309/275935 ? On my current stage of learning I wish to and need to learn writing down formal proofs and not just trivial hints.
i think the upvote/downvote function in MSE is already misused sometimes...presenting informal/trivial hints get upvotes...presenting newbie formal proofs get downvotes without any comment
Well clearly, don't experiment with drugs if you're pressed for time, individual reactions can be wildly varied
And your doctor is the only person who can answer that question anyways
I just made a joke about how my experience with college was, everyone was partying all their time away and then abusing adderall to make up for it. I think you are taking it too seriously in search of a magic solution.
Let me be clear, Adderall is exactly equivalent to "speed", a harsh street drug. Unless you have severe ADHD, I think it's not any kind of solution for trying to learn in the long term. I think you're already on the right track just being disciplined enough to study.
The only "magic bullet" I can think of for you here is sleep.
ha thanks for your idea :) in fact i also asked taking Provigil (modafinil) to a doctor and he also refused. Adderall and Provigil are controlled substances in Taiwn.
@JoeStavitsky I'm a horrible teacher but if nobody else answers: It's by definition in that situation, remember that the substitution only applies when your expression can be mapped to $\sqrt{a^2-u^2}$ and then look closely at the variable a
@JoeStavitsky I would like to answer but I will almost surely confuse you because I don't understand what's being asked precisely. I will point out that the 3 comes from $a^2$ being 9 just in case that helps though, sorry if it's painfully obvious already though too.
It is going alright, probably just going to be in here for a little while, procrastinating for a bit before starting dif geom hw. But just got a computer in my office, so now it is more convenient to come to chat, so I should be coming more. Still have not done much algebraic topology, been busier expected, and bad time management skills :) But I did bring student geometry topology seminar back to life here. @BalarkaSen How about you?
The thing I am suggesting is the following : if $X$ is a collection of a bunch of data points on $\Bbb R^2$, denote $X_\delta$ to be the $\delta$-thickening of $X$. In persistent homology, you vary $\delta$ and see which betti numbers persist more.
so for example say after you compute the homology at $\delta_1$, you find a $1$-cycle. Eventually once you keep growing $\delta$, you'll get to some $\delta_2$ where the $1$-cycle closes up
Could someone give some healthy comments on this newbie answer math.stackexchange.com/a/1473377/275935 ? My answer only got downvotes without any comments or editings for now.
@BalarkaSen Cool, know anything about bounded cohomology? Seems like it has some interesting applications. I think my goal for studying algebraic topology on my own will be to get to a point where I feel like I understand some of the bounded cohomology stuff.
Has anyone ever considered the cardinality of the set of all subsequences of a fixed sequence? The problem is if we consider subsequences with the same terms are the same element of the set, than investigating the cardinality will be very non-trivial.
The main thing that's important is the Frobenius digraph, which is this tool for talking about how sylow subgroups of solvable groups interact with one another
1. one way to characterize solvable groups is that a group is solvable if and only if it has all its Hall subgroups. (Hall subgroups are the same thing as Sylow subgroups but for multiple primes, e.g. if a group's order was $2\times 3^2\times 5$, it's $\{2,3\}$-Hall subgroup would be a subgroup of order $2\times 3^2$.)
So one thing that's neat about that is that any induced subgraph of the prime graph of a solvable group is the prime graph of a corresponding Hall subgroup
Well, the second cool thing is a theorem that somebody proved about solvable groups: "a solvable group with a disconnected prime graph is either a Frobenius group or a 2-Frobenius group"
So, Frobenius groups are basically semidirect products $K\rtimes C$ where $C$ acts fixed point freely on $K$. (i.e. if $k\in K, c\in C$, then $c^{-1}kc=k$ iff $k=1$)
2-Frobenius, if you aren't familiar, are groups with a subgroup that's Frobenius, and if you quotient the whole group by its Frobenius kernel, you get another Frobenius group.
so kind of like one Frobenius group on top of the other
In any case, the main thing is that we have this very specific description of groups with disconnected prime graphs
if there is an edge $\{p,q\}$ that is missing from the prime graph of a solvable group, then, necessarily, there is a $\{p,q\}$-Hall subgroup of $G$ that is either Frobenius or $2$-Frobenius
we're interested in edges that aren't in the prime graph, so we look at the complement of the prime graph. Every edge corresponds to a hall subgroup that's either Frobenius or 2-frobenius
the interesting thing about Frobenius groups is that the kernel and complement have to be coprime. Thus for each edge, one vertex corresponds to the kernel, the other to the complement
so, the basic idea is, every edge in the complement corresponds to a Frobenius group, and that means one vertex on the edge corresponds to a Frobenius kernel. It would behoove us to keep track of the information of which vertex that is.
So, we orient the edge, and pointing it at the kernel
(a similar rule is made for the $2$-Frobenius case, we point the edge towards the kernel of the quotient Frobenius group)
that orientation of the complement is called the Frobenius digraph
So, the paper is basically about using the Frobenius digraph to prove a bunch of open stuff about solvable groups, the first of which is an if-and-only-if characterization theorem of the prime graphs of solvable groups
How can I make $(z^2+a^2)(z^2+b^2)(z^2+c^2)$ into complete square form?
Consider $z$ to be complex number, a, b, and c to be constants.
What have I tried?
I have expanded the into polynomial but I don't have any clue about making it into the complete square form below
$$x+ax = \left(x + \frac {a...
Well I initially saw it when I was flipping through "Groups acting on the circle" by Ghys and it gives an invariant (bounded Euler class) to study groups acting on the circle. I don't really know much about. It seems to capture a lot of ideas of similiar to quasi-isometry (and a lot of other "quasi-__"), where you are more concerned with "approximate" invarients
I'm interested : subgroups of $\text{homeo}(S^1)$ and $\text{diffeo}(S^1)$ are hard.
@PaulPlummer By the way, I pinged you a while ago (but, naturally, you didn't get the ping): do you know anything about the recent development on the study of the structure of the Cremona group? I have been told that some decent results have been discovered using geometric group theory.
The cremona group $Cr_n(k)$ is defined to be the group of birational automorphisms of $\Bbb A_n^k$. Equivalently, it's the group $Aut(k(X_1, X_2, \cdots, X_n)/k)$
:) I guess I am in a similar boat, have some more stuff to learn. Maybe someday we can work on a project involving some of this stuff. Looks like an interesting opportunity for geometric group theorists and algebraic geometers(maybe even number theorists) to work together @BalarkaSen
