Ah, I feel your pain. I've been fighting a cold too! It feels like I have waterfalls pouring out of my eyes and nose, @teadawg1337. We're all gonna make it. ^_^
@Chris'ssis There's an urban legend among Japanese students about the guy name Kunihiko Chikaya. His calculus problems specially integrals are very awesome & fantastic. I've tried to make contact with him many times, but he always ignores my messages. You might be interested in knowing his integral problems.
@Alyosha I'm not very good at thinking categorically, to be honest. If you've seen MacLane's category theory for the working mathematician, there's a nice discussion in it about what it means for functors to be adjoint.
Personal Background
I am by no means any expert at integration, but I've have done a fair share of it. Integration is like a hoby to me. Instead of solving puzzles, or riddles is solve integrals. As an example I am in the process of writing some personal notes on integration. Hopefully at least ...
@Anastasiya-Romanova秀 As regards all those very good at integrals, series and limits I'd ask them if they see in this more an art than a simple math job. I consider what I do is an art first. What's your opinion?
@KonradVoelkel This theorem is given: Let the algebraic curve $f(x_0, x_1, x_2) \in K[x_0, x_1, x_2]$. The inflection points are the non-singular points of the curve that are the intersection points with the hessian.
We have the curve $x^3+y^3+z^3=0$ and the hessian is equal to $216 \cdot x \cdot y \cdot z$.
So we have to solve the system $216 \cdot x \cdot y \cdot z=0$ and $x^3+y^3+z^3=0$.
From $216 \cdot x \cdot y \cdot z=0$ we have that $x=0$ or $y=0$ or $z=0$.
Then we will have the equation $a^3=-b^3$, where $a,b \in \{x,y,z\}$, right? But how can we find now the inflection points?
@evinda They work on the assumption that the arithmetic operations (arr[i]/exp)%n and the copying of array elements take $O(1)$ time. As long as you're within machine types, that is the case, but then $\log n$ is also bounded - oh, wait, $n$ itself is also bounded - so talking about algorithmic complexity doesn't strictly make sense. If you assume that the arithmetic operations take bounded time, you have an $O(n)$ algorithm. If you take the time the arithmetic operations actually take ...
... into account, then it's $O(n\log n)$ or worse - but to be fair, the comparisons in a comparison sort then also cannot be assumed $O(1)$, so it's still a gain in algorithmic complexity.
Hi guys, probability question here. I was trying to calculate, $f_{X,Y}(x,y)$ for two dependent continuous random variables. So I thought about calculating $d/dx d/dy P(X\leq x, Y \leq y)$, my solution manual however calculates : $d/dx d/dy P(X\leq x, Y \geq y)$ and then claims that this is equal to $-f_{X,Y}(x,y)$. I'm not sure if this is general true, or if I have to see this from the given situation.
I understand why they are doing it, because that expression is much simpler in the given situation, but I don't see why it is true, that $d/dx d/dy P(X<x, Y>y) = - f_{X,Y} (x,y)$
@anon given an open cover O of some connected space X, Cech nerve of O is the simplical complex constructed as follows : place the nodes(0-simplicias) corresponding to the sets in O, an edge (1-simplicia) between two nodes if the sets intersect and an n-simplex if O_1 \cap O_2 \cap ... \cap O_{n+1} is nonempty.
@anon X is the inverse limit of a bunch of S^1. take an open cover O_0 of the lowermost S^1. construct O_n, an open cover of the nth copy of S^n by taking preimage of sets in O_{n-1} by taking preimage of them under the covering map.
declined
I'd like to nominate Jonas Meyer
According to the Citizenship score query mentioned in Behaviour's answer he is one of the few users having considerably high citizenship score ($37$) as well as good activity on meta.
It's great, yesterday I gave a talk on quantum groups using the language of category theory & hall algebras, it was the hardest thing I've ever done to prepare for, I stayed awake working for 30 hours non-stop, nearly died but it went great in the end!
@robjohn I got comment like this in my official nomination by Ellie Kesselman: "Anastasiya, this is tacky behavior, asking for badge earning help on Math SE in your Quora profile! ..."
Thank you so much for the Quora users who have already helped me to earn a gold badge: Publicity for my post: The Monster Integral. Right now, I'm trying to earn other gold badges for my posts: The Wicked Integral, Integral Contest & Surely You're Joking, Mr. Feynman! and Famous Question for my post: The Monster Integral (>‿◠)✌ You actually link to your Quora profile from your Math SE profile too!
@MikeMiller so Drinfeld and others in the 80's derived quantum groups from things like the Yang-Baxter equation, and were led to formulate a bunch of axioms based on this. It's basically taking the algebraic group RxR, turning it into the algebra R[X,Y] of functions in two indeterminates, noticing XY = YX is a defining relation, so we add non-commutativity by instead assuming XY = qYX and call the result Rq[X,Y] the quantum plane q-deformation
If you formulate that stuff in terms of algebraic axioms on generators it becomes crazy, but Ringel noticed some of the axioms are very similar to the axioms you get using Hall algebras, in fact the positive part of a quantum group (a subset of the quantum group generators) it's nearly the same as the axioms in a Hall algebra, so Ringel modified the axioms by defining a twisted Hall algebra (THA) & then proved there is an isomorphism between the positive part of a quantum group & a THA
Yeah same Drinfeld, notice that Teichmuller theory can basically be derived from Polyakov's path integral derivation of the Bosonic string (I'm just learning it, fucking crazy), you literally just derive a Teichmuller space as a side-consequence, and you can relate this to quantum groups which is probably how Drinfeld did his thing
The Grothedieck-Teichmuller, AFAIK it, is an extensions of the work done by Grothedieck in Equisse d'un Programme to realize Gal(\bar Q/Q) geometrically/topologically
What do you guys think of my question? http://math.stackexchange.com/questions/1063863/solving-special-function-equations-using-lie-symmetries Imagine taking the Laplace equation, knowing your coordinate system is ellipsoidal, knowing instantly that this will give you 3 separation equations (you can write down based on knowing it's ellipsoidal), then knowing those 3 ode's are basically just an un-factored Lie algebra representation of some conformal symmetry, allowing you to write down the solution using Lie groups instantly, I mean wow! That whole Lebedev book on special functions is just …
The essence of Grothendieck's work was that the absolute galois group acts on a bunch of graphs transitively. in particular, Beyli proved that any algebraic curve over \bar Q can be realized as a branched cover over P^1 - {0, 1, \infty}. so grothendieck defined graphs corresponding to algebraic curves on \bar Q by looking at the preimage of the interval [0, 1] under the covering map that sends the riemann surface onto P^1.
@evinda Whether it's wrong depends on what you count for the complexity. Note however, that the algorithmic complexity is not so important for real-life situations, for normal data sizes, $\log n$ is so small that the smaller constants of a good $O(n\log n)$ algorithm more than compensate the $\log$ factor over a complicated $O(n)$ algorithm. Whether the given algorithm can beat a good merge sort or intro sort on normal input sizes is doubtful.
It's amazingly interesting! I've been looking for over a year just to know how to formulate that question! I just hoped I'd find something like this and I finally found it! Special functions with baby pictures to remember how to derive them :D
what grothdieck proved is that Gal(\bar Q/Q) acts on the set of all these graphs, which is amazing actually since no geometric picture of this was known before
@Anastasiya-Romanova秀 This is part of the reason that I thought your previous query was a bad idea. Some people don't like advertising on other sites to get rep, badges, etc.
@bolbteppa as far as i have heard, drinfeld did some very strong work based on this stuff which explicitly gives an injection of Gal(\bar Q/Q) onto some braid group
Mentioning Riemann surfaces, I still can't answer this simple question http://math.stackexchange.com/questions/776127/branch-cuts-of-fz-gz-sqrtpz-sqrtqz :( Nobody in my college has any idea how to do it, nobody! We were talking about it yesterday!
@evinda You can either describe an algorithm and prove that it runs in $O(n)$, or you can assume that no such algorithm exists and derive a contradiction. The first way is much easier generally, if there exists such an algorithm.
One place to start is to take the reflection group and notice how it is related to the Roots of a lie algebra, i.e. the Weyl group. Further notice that the scattering of two particles creates a "><" shape, >< , which partitions the plane of scattering into separate areas, so you can apply the reflection group to it. Further notice scattering of particles must satisfy the Yang-Baxter equation, which is a Braid equation. In other words, coxeter groups, lie algebras, Weyl groups, braids,
scattering of particles, all related, I believe they are joined by ADE diagrams which is a DEEP subject
Screw it, I have a bounty up for my branches question! http://math.stackexchange.com/questions/776127/branch-cuts-of-fz-gz-sqrtpz-sqrtqz Make 50 easy points with a deep description guys, be my guest ;)
Wow
So maps from the Galois group to the General linear group resemble monodromy representations? Interesting
@Anastasiya-Romanova秀 Not exactly. He is not advertising on another site. He is just putting something in his profile for that site. It is not the best in taste, but better received than advertising elsewhere.
Monodromy comes up heavily in conformal field theory, and it is related to the geometric langlands only recently by Frenkel and others, I would love to see if that's a point of linking them
@bolbteppa one should believe if you replace the natural notion of monodromy by some kind of "etale monodromy", there would be a good way to attack Langlands
@bolbteppa haha! nice.
actually i happen to know a bit about the connection of monodromies with CFT
i met some physicist a few weeks ago who told me the stuff Vafa did on spheres, asking about what happens in torus. i figured that this connects to monodromies ;)
@robjohn But what does Publication mean then if I don't share it to public? I believe almost all the users who earn Publicity badge putting the link in his/ her own site.
i.e., if you mark 3 points on a sphere, and if rotating the sphere leaving one of the marked point fixed, some "phase change" happens on the sphere. so vafa explained the identity $\tau_{123} = \tau_1 \tau_2 \tau_3$ where $\tau_i$ is a rotation of the sphere with $i$ fixed.
he asked me what happens when you do this in, say, a torus.
@Anastasiya-Romanova秀 The Publicity badge is for putting links to questions/answers elsewhere to get more people interested in the site, not to publicize an election.
apparently, this looks a bit like monodromies, e.g., let a point loop around the marked points. you can actually think of the marked points as branch points of some branched cover over P^1.
@evinda I have no idea what the numbers should mean. I would have thought "discovery time" would mean something like "number of steps until the vertex is first visited" or so, but apparently it means something else. What? Also, what is a "finish time", and how are the kinds of edges defined?
@robjohn No answer yet to my question, do you think I better still wait and then maybe post it to MO, or perhaps I should ask another question so that I have one for each problem?
@DanielFischer The discovery time d[v] is the number of nodes discovered or finished before first seeing v. The finishing time f[v] is the number of nodes discovered or finished before finishing the expansion of v (return from DFS-Visit). We can define four edge types in terms of the depth-first forest Gπ produced by a depth-first search on a graph G. 1. Tree edges are edges in the depth-first forest Gπ. Edge (u, v) is a tree edge if v was first discovered by exploring edge (u, v). A tree edge always describes a relation
Shall I apply again the Depth-first search? Have I calculated the finish and discovery time wrong?
@evinda That's too complicated for me. But I can tell that something is wrong, since the vertex $b$ can never be reached in a DFS starting from a different vertex. It has no incoming edges.
The function is not given explicitly, it's notes for proving the existence of a solution for pde given a vector valued function $a$. So all that is known really is that it is from $\mathbb{R}$ to $\mathbb{R}^{n}$.
@JorgeFernández I will ask the question differently.
@JorgeFernández If you have an operator $\langle A(u), v \rangle := \int_{\Omega}a(x,u,\nabla u)\cdot \nabla v dx$ then does it follow that $\langle A(u),v \rangle = \sum_{i=1}^{n}\int_{\Omega}a_{i}(x,y,\nabla u)\cdot \frac{\partial v}{\partiall x_{i}}dx$?
@PedroTamaroff $g(z) = f(e^z)$ is entire with $\lvert g(z)\rvert \leqslant \lvert z\rvert$. Hence $g(z) = c\cdot z$ for a $c$ with $\lvert c\rvert \leqslant 1$. Now, $c\neq 0$ would mean that $\log$ is entire.
Monodromy is what happens when a path goes around a point, i.e. the monodromy in an ode comes about by integrating around the path of a singular point of the ode, but if you rotate the whole space I have no idea if that's monodromy?
@bolbteppa: It's a measurement of how something changes when you go around a closed path, coming back to your starting point. It might be a function, it might be a vector field, ...
If you take the general solution of a 2nd order ode, the linear combination of the two solutions, you integrate around a singular point and the 'new' solutions change into linear combinations of your original ones, so you get a monodromy matrix relating the solutions
I don't think you would have invented any of these concepts unless you had the ode's and their linear combinations as was derived in the 19'th century to motivate it, I know the way you guys think about it is ultimately right/correct though
i'll fight with you to convince you the other way @bolbteppa. the ideas of monodromies came from looping around the branch points in the riemann surface of algebraic curves
Historically speaking, I think you're right, @bolbteppa. Parallel transport in differential geometry is, of course, the solution of a differential equation as well.
@TedShifrin Hello!!! Having the equation $a^3=-b^3$, where $a,b \in \{x,y,z\}$ how can we find now the inflection points in $\mathbb{P}^2(\mathbb{C})$? Could you give me a hint?
@evinda: You already said that one of $x$, $y$, $z$ had to be $0$. So how many points $[0,?,?]$ are there in $\Bbb P^2(\Bbb C)$ that satisfy the equation?
Because there are usually only 3 singularities for an ode (the studied ones) they define them on a Riemann surface so that everything can be handled at once, and the natural thing to do is to look at monodromies on a Riemann surface, hence monodromy is intimately tied up with Riemann surfaces through ode's :)
I just understood why Dirac wanted to formulate quantum mechanics in projective space instead of real space, man it's so obvious! You get those phase factors and projective geometry just stops the phases from distinguishing wave functions!
If only I had a feel for projective geometry now :(
interesting. a guy from the homotopy chatroom gives an excellent idea about thinking about the solenoid as a universal cover of some space with Cech \pi_1 being Z_p instead of a space with Cech \pi_1 Z_p.
declined
I'd like to nominate Jonas Meyer
According to the Citizenship score query mentioned in Behaviour's answer he is one of the few users having considerably high citizenship score ($37$) as well as good activity on meta.
