(Un-)Countable union of open sets - Mathematics Stack Exchange
Jun 4, 2012 · A remark: regardless of whether it is true that an infinite union or intersection of open sets is open, when you have a property that holds for every finite collection of sets (in this case, the union …
modular arithmetic - Prove that that $U (n)$ is an abelian group ...
Prove that that $U(n)$, which is the set of all numbers relatively prime to $n$ that are greater than or equal to one or less than or equal to $n-1$ is an Abelian ...
Mnemonic for Integration by Parts formula? - Mathematics Stack …
Nov 11, 2018 · The Integration by Parts formula may be stated as: $$\\int uv' = uv - \\int u'v.$$ I wonder if anyone has a clever mnemonic for the above formula. What I often do is to derive it from the …
For what $n$ is $U_n$ cyclic? - Mathematics Stack Exchange
When can we say a multiplicative group of integers modulo $n$, i.e., $U_n$ is cyclic? $$U_n=\\{a \\in\\mathbb Z_n \\mid \\gcd(a,n)=1 \\}$$ I searched the internet but ...
Newest Questions - Mathematics Stack Exchange
3 days ago · Mathematics Stack Exchange is a platform for asking and answering questions on mathematics at all levels.
Homotopy groups U(N) and SU(N): $\\pi_m(U(N))=\\pi_m(SU(N))$
Oct 3, 2017 · Yes, that's right, and yes, $\pi_1$ should be $\mathbb {Z}$ for all $N$ in the table.
optimization - Minimizing KL-divergence against un-normalized ...
Jun 10, 2024 · Minimizing KL-divergence against un-normalized probability distribution Ask Question Asked 1 year, 5 months ago Modified 1 year, 5 months ago
If a series converges, then the sequence of terms converges to $0$.
@NeilsonsMilk, ah, it did not even occur to me that this involves a step. See, where I learned mathematics, it is not unusual to first define when a sequence converges to zero (and we have a …
Intuitive proof that $U(n)$ isn't isomorphic to $SU(n) \\times S^1$
Jan 5, 2016 · The "larger" was because there are multiple obvious copies of $U (n)$ in $SU (n) \times S^1$. I haven't been able to get anywhere with that intuition though, so it ...
$\\sum a_n$ converges $\\implies\\ \\sum a_n^2$ converges?
May 14, 2015 · $\sum a_n$ convergent implies that $a_n\rightarrow 0$, then you always have $a_n\in [0,1]$ for $n$ large.