@DannyCheuk I would like to tell you something, too.
That really upset me.
Please, refrain from making major edits on posts that are not "yours", that is, changing notation, formulas, or things like that. If you'd like to make major changes to a post, ask the poster first!
Well I have already ask the support team to delete my account
Isn't that irreversible?
I spent too much time online and on non-academic stuff, last school year I literally accomplished nothing, and now I'm forced to transfer to a new school and start things all over again
I'm just trying to literally ... eliminate ... everything that's spending my time
@HerpDerp Since I noticed you were trying to give away your reputation via bounties, I figured you wanted to handle that first. If not, I will delete your account.
@skullpatrol the nomenclature "$x$ has a domain" does not really say anything in itself. Nothing belongs to $x$. It really means that whatever function $f$, sometimes called $f(x)$, has a domain and that domain restricts that $x\in\mathbb{R}$. It is very contextual.
Guys, In a group, when checking the associativity of the operation $a(bc)=(ab)c$, do $a,b,c$ have to be all distinct or they can be similar. Because there are some groups like $\{0\}$ or $\{1,-1\}$ that have less than 3 elements
Okay, the definition says any a,b,c so a,b,c can be the same
Just in case: is there any way of quickly finding the set of integer $n$ for which $\frac{\prod_i (a_i+b_in)}{c_i+d_in} \in \mathbb{Z}$, where $a_i,b_i,c_i,d_i \in \mathbb{Z}$?
@PeterTamaroff I'm having a hard time going from ${\bf B}\star {\bf 1}={\bf B}+[n=1]$ to $\sum\limits_{k = 0}^{n}\binom {n+1}k {{B_k}} = 0$. How should I be looking at it?
@PeterTamaroff Yeah, it's equal to $\sum_{k=0}^n \frac{(-1)^k}{k+1} \left( {\bf H}^n \star \mu \right)_k$, and I don't see how to deal with the ${\bf H}^n$ easily.
@AntonioVargas Ah, OK. But I don't see how that is true. I am getting $${B_n} = \sum\limits_{k = 0}^n {\frac{{{{\left( { - 1} \right)}^k}}}{{k + 1}}} {\left( {{\mu _j} \star {j^n}} \right)_k}$$
@skullpatrol the nomenclature "$x$ has a domain" does not really say anything in itself. Nothing belongs to $x$. It really means that whatever function $f$, sometimes called $f(x)$, has a domain and that domain restricts that $x\in\mathbb{R}$. It is very contextual.
@robjohn I convinced the Homotopy Theory chat room owner to embed your mathjax link into the room description so it doesn't have to be continuously re-pinned.
A room for anyone interested in homotopy theory, or any nearby fields (e.g. category theory, algebraic geometry). To activate chatjax in this room go to math.ucla.edu/~robjohn/math/mathjax.html
@skullpatrol I would prefer they link through the meta post. That way I can move that page if need be. Besides, stackexchange.com links are just better.
