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12h
comment How to prove $\sinh^{-1}x=\ln\left(x+\sqrt{x^2+1}\right)$
Fix some real number $x\in\Bbb R$, you are looking for a formula for $y\in\Bbb R$ such that $\sinh(y)=x$. If you write $Y=e^y$, then you are trying to solve the equation $$\frac{Y-\frac1Y}2=x$$ i.e. $$Y^2-2Yx-1=0$$ and you can apply the quadratic formula to get the (right) $Y$, thus $y$.
2d
comment Are closed simple curves with this property necessarily circles?
Many curves have this property, for instance ellpises, or squares, or regular polygons drawn from the $(2n)$-th roots of unity...
May
19
comment Which other “exotic” permutation-related things exist?
I apologize, this is very basic, and I don't doubt you are aware of it, but it relates to $S_4$. There are 3 ways to cut a 4 element set in two equal halves, hence there exists a (clearly) surjective morphism $S_4\to S_3$.
May
15
comment compact operators surjection
@r.pomegranate finite rank continuous linear maps are always compact.
May
15
comment compact operators surjection
@r.pomegranate $A(B_X)$ is open by the open mapping theorem, and contains zero, so there is some positive real number $c$ such that $cB_X$ is contained in $A(B_X)$.
May
15
comment compact convergence for a series in complex space
Compact convergence means that the series $\sum f_n\big|_K$ converges uniformly for every compact set $K\subset D$. But the series already converges normally on $D$, hence uniformly on $D$ and thus uniformly on every compact subset of $D$.
May
15
comment compact convergence for a series in complex space
The series already converges normally on $D$.
May
13
comment Prove that $ A = - A^{\top} $ and $ \text{rank}(A) \leq 1 $ imply $ A = \mathbf{0} $.
Antisymmetric real matrices have even rank.
May
13
comment $ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}} $ appears to disagree with $\int_0^1 \frac{dx}{4-x^2}$
I didn't take offence, but most people on here (not just you) put effort into their answers, are equally deserving of receiving upvotes yet don't ask for them. It's bad form in my opinion.
May
13
comment $ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}} $ appears to disagree with $\int_0^1 \frac{dx}{4-x^2}$
What do you mean by "please do the honors"?
May
12
comment What is the homeomorphism type of the surface given by the polygonal presentation $aaa$?
It's not a surface: any neighborhood of the arrowhead looks like three sheets of paper attached along their sides, in other words, the space isn't locally homeomorphic to $\Bbb R^2$.
May
11
comment unitary transformation of $\mathbb{C^2} \otimes \mathbb{C^2}$ that preserves the decomposability
A what? ${}{}$
May
10
comment unitary transformation of $\mathbb{C^2} \otimes \mathbb{C^2}$ that preserves the decomposability
There's permutation of factors $x\otimes y\mapsto y\otimes x$, and composites of this and the previous. These might be the only such transformations.
May
10
comment unitary transformation of $\mathbb{C^2} \otimes \mathbb{C^2}$ that preserves the decomposability
Any isometry of the form $u\otimes v$ with $u,v\in U(2)$ preserves decomposability, and if $u_t,v_t$ are one parameter subgroups of $U(2)$, then $u_t\otimes v_t$ is a one parameter subgroup that preserves decomposability.
May
7
comment $P$ and $Q$ are unitarily equivalent iff dimensions of ranges and kernels are the same
It's false.${}$
May
4
comment $\mathbb{T}^2 \not \cong \mathbb{R}P^2$ without homologies or fundamental group
What techniques do you consider to be elementary?
May
4
comment $\mathbb{T}^2 \not \cong \mathbb{R}P^2$ without homologies or fundamental group
Would you accept this? If you remove a point from any torus, you are left with a space homotopy equivalent to a wedge of two circles, whereas if you remove any point from the projective plane, you are left with a space homotopy equivalent to a circle. These two spaces are not homotopy equivalent, and thus the torus and the projective plane cannot be homeomorphic.
May
4
comment $\mathbb{T}^2 \not \cong \mathbb{R}P^2$ without homologies or fundamental group
What does $\simeq$ mean to you? Homeomorphism? Homotopy equivalence?
May
1
comment Show that $|e^z -1| \leq e^{|z|}-1$ for any z
It's easy if you use the power series of the exponential.
May
1
comment Lattice orders and number of elements in a set
@CKKOY Not only is there no reason to assume $x\neq y$ in the definition of their infinimum or supremum, but the property $x\vee x=x=x\wedge x$ is usually included (under the name idempotence) when one defines a lattice as a set with two binary operations $\vee$ and $\wedge$ satisfying some axioms.