I... don't know, but the way he asks questions seemed to just want an answer and no thing else, which is quite different from other users who does similar things...
The list of questions looks so random and pairwise unrelated to each other. Even JEE and Ultradark's questions follows a certain theme. I think my reaction is definitely too overeactive in response to that weird feeling
@user76284 I guess the only fair thing I can say is I never seen such a question pattern where each question does not seemed to be remotely related to the previous one, and possibly that weird feeling lead me to misinterpret as repetative and trivia like (regardless of the truth that none of the questions are actually trivial)
If you're looking for themes, the most recent ones are related to regularization of divergent series. The others are related to representations of ordinals and generalizing their arithmetic to subtraction and division (Grothendieck group of a commutative monoid and field of fractions, respectively).
The game theory stuff is a third theme, though somewhat related to the ordinals (through nimber arithmetic).
They finally added a built-in ComplexPlot function to Mathematica. Nice.
Had no background on that level of complex analysis, but if those nontrivial poles are only dense, then in theory certain transcendentals may be able to allow some contours to slip past
still, that they have something to do with the nontrivial zeros seemed to suggests it is going to be riemannian hypothesis difficulty in getting past them
Another way to look at it is to think about the Riemann sphere if you're familiar with that. $z$ always rotates counterclockwise with respect to the center of the shaded region.
My short reasoning is this: On the Riemann sphere, $z$ is rotating counterclockwise with respect to the shaded region, and clockwise with respect to the unshaded region. Therefore, its image $w$ will also be rotating clockwise with respect to the unshaded region.
Another way to think about it is to draw an arrow perpendicular to the direction of motion, say to the right, which for $z$ would point into the unshaded region. Such an arrow will continue to point into the unshaded region for $w$.
You can see that moving counterclockwise (say, around the red equator) with respect to 0 (i.e. the interior region) is the same as moving clockwise with respect to infinity (i.e. the exterior region).
According to jstor.org/stable/44237555 "Borel has shown that there exists a kind of analytic continuation which differs from the usual kind (the theory of the usual kind is due to Weierstrass), and Borel made use of examples such as the ones studied here in showing that in some cases it is possible to continue beyond what Weierstrass called natural boundaries".
Where can I read more about this type of continuation? They cite Borel's 1917 Leçons sur les fonctions monogènes d'une variable complexe. Is there an English translation somewhere?
@BalarkaSen and I came across this amazing thoerem
continuous_if :
∀ {α : Type u_1} {β : Type u_2} [_inst_1 : topological_space α] [_inst_2 : topological_space β]
{p : α → Prop} {f g : α → β} {h : Π (a : α), decidable (p a)},
(∀ (a : α), a ∈ frontier {a : α | p a} → f a = g a) →
continuous f → continuous g → continuous (λ (a : α), ite (p a) (f a) (g a))
I think thinking of the crux of $\Pi_{\leq 1} X$ being a category as a fact about concordance of configurations of points on $[0, 1]$ has some merit to it - for example $A_\infty$-spaces are something you get by forgetting about homotopy-coherence
Remember, an important part mental health is therapeutic behavioural exercises, for example offline trolling of humours institutions like scientology, there is one in almost every city, and they are there to make life a pleasure to live, remember it's ok to hurt their feelings and play tricks on them, they are differently abled, and much like a goldfish resumes orbit seconds after being frightened by something, so will they too meg, so will they too.
the idea is that there exists a monoid $H$ such that the monoid algebra $k[H]$ is a domain, but $H$ has some strange properties such that $H$ can't be embedded into a group
now if we had an embedding $k[H] \hookrightarrow D$ into a division ring, then that would induce an embedding $H \hookrightarrow D^\times$ into a group
but I don't want to spell out the details of the contruction of $H$ and of the proof that $k[H]$ is a domain here
unfortunately the cohorts that come here aren't interested in proofs so the lecturers quite often just glide over proofs and go straight to applications and "exam questions"
i did also plan on taking the undergrad funktionentheorie course
I have a function p in $L^{\infty}(0,+\infty)$ that is measurable and bounded all most everywhere, I want to obtain that there exist a,b >0 such that $a\leq p(x)\leq b$ what can be a and b?
@ÍgjøgnumMeg still working on what ironic means but do recommend trolling the church of scientology in a literal sense it's one of my favourite drinking games
Rant: Say I have a convex polyhedron, and I barycentrically triangulate it. Let $\Delta$ be the "fundamental triangle", reflections along whose edges generate the full thing. The three vertices each will have different "link structures", aka one there will have $p$ triangles meeting at, another there will have $q$ triangles meeting at and the remaining one will have $r$ triangles meeting at. If you "bloat this up" to a geodesic map on $S^2$, you'll have a tessellation with $(p, q, r)$ triangles.
The symmetry group of my guy will be the triangle group $\langle a, b, c | a^2, b^2, c^2, (ab)^p, (bc)^q, (ac)^r\rangle$, right?
Of course, there is a severe restriction on $p, q, r$ coming from Gauss-Bonnet. Namely, $1/p + 1/q + 1/r > 1$.
I think $[p, q, r]$ is what they call a Schlafli symbol
A cube has Schlafli symbol $[2, 3, 4]$, so the full octahedral symmetry group is $\langle a, b, c | a^2, b^2, c^2, (ab)^2, (bc)^3, (ac)^4 \rangle$
This is isomorphic to $S_4 \times \Bbb Z_2$ (because the rotation group is $S_4$ and you're taking the preimage under the double cover $O(3) \to SO(3)$ which is isomorphic to the trivial cover $SO(3) \times \Bbb Z_2$ group theoretically, as principal $\Bbb Z_2$-bundles)
Er, I meant $2p, 2q, 2r$ is the link structures, the number of triangles meeting at the various vertices. The spherical angles at those vertices is $2\pi/2p, 2\pi/2q, 2\pi/2r$ i.e., $\pi/p, \pi/q, \pi/r$ thereof
It seems by convention you remove the $\pi/2$ angles when considering the Schlafli symbol and write it the order of barycenter of 2-simplex, then barycenter of 1-symplex, then barycenter of 0-simplex. So the correct notation for Schlafli symbol of the cube would be $[4, 3]$
Ah, here's another way to think about this. The reflections can be realized in $\Bbb R^3$ by a collection of three 2-planes which pass through those edges, instead of passing to the spherical map. The angles between the (unit vector orthogonal to the) planes are $\pi - \pi/p, \pi - \pi/q, \pi - \pi/r$. Assign three vertices corresponding to these three planes, and join them by a labelled edge with label $m$ if they make an angle of $\pi - \pi/m$ with $m > 2$.
Usually you don't label the edge if it corresponds to $m = 3$ because it's quite common. This is the Coxeter-Dynkin diagram of the polyhedron
This generalizes to regular polytopes, since their symmetry groups are generated by reflections as well (this is a theorem of Coxeter, but it's not hard to believe)
So if we have a regular $n$-tope, it's fundamental chamber is an $n$-simplex and we have a corresponding hyperplane arrangement of $n$ planes in $\Bbb R^n$. If $v_1, \cdots, v_n$ are the unit normal vectors, then the matrix $M = ((v_i \cdot v_j))_{1 \leq i, j \leq n}$ encoding the angles between these hyperplanes is a positive definite symmetric matrix. This is the necessary and sufficient condition for the Dynkin diagram to be realizable by a polytope
I get how they can bust people for insider trading but how to they prevent people from insider pre-trading
?
for example project Jedi, all of the quality flock control programs that could be developed from the research of mk ultra, seems like an unfair advantage
My favorite new fact for tonight is: The Jewish Kabbalah acknowledges the existence of infinity as being the only entity to have truly existed for eternity, which sets it apart from all other monotheistic belief systems in that if no entity existed when God was created, it stands to reason that we should be in an infinite regress of God production, thereby nullifying God as the supreme entity, but in the Kabbalah, they acknowledge the existence of "Ein Sof" which is unusual
because the Old Testament from the Christian Bible was from Jewish scripture, and it clearly states at the beginning of Genesis that there was nothing prior to the point of God's creation
@skillpatrol - of course. If I say, in multiple places, "I believe X", you can then say "It's a fact that Joe believes X." The underlying belief doesn't need to be proven true or false.
Let $K$ be some compact Hausdorff space. Recall that if $C(K)$ is finite dimensional, then $K$ is finite. The standard way of proving this is by contrapositive and an application of Urysohn's lemma. I was wondering, is there a way of proving this based on the idea of showing that $K$ is a discrete? I.e., first show that $C(K)$ being finite dimensional implies $K$ is a discrete space?
Definitely since I'm trying to make sense of the definition! So I have a smooth bundle $E\to M$ and a connection on it is a linear map $\nabla\colon\Gamma^\infty(E)\to\Gamma(E\otimes T^\ast M)$ satisfying $\nabla(\psi f)=(\nabla\psi)f+\psi\otimes\mathrm{d}f$. I don't have an intuitive idea of $E\otimes T^\ast M$, but this is fine
@AlessandroCodenotti I don't know if this is helpful or not but while studying somethings about connections of vector bundles, I thought of $E$ as $TM$ and then like tried to understand it using this basic example.
