@evinda The $y+x^2-1 = y +\langle x-1\rangle$ is wrong, on the left, you have a polynomial, on the right a coset of an ideal. To show that $\langle a,b\rangle = \langle c,d\rangle$, you show that $a,b \in \langle c,d\rangle$ and $c,d \in \langle a,b\rangle$. Here we have $a = c$, which makes half of the thing trivial. Then you write $y = (y+x^2-1) - (x-1)(x+1)$ to show $y\in \langle x-1,y+x^2-1\rangle$, and the remaining bit should be a straightforward modification.
@evinda Yes, that is correct. Now you need to see that $\mathbb{C}[X,Y]/\langle X-1,Y\rangle \cong \mathbb{C}[Y]/\langle Y\rangle$ (which, by the way is $\cong \mathbb{C}$). If it were $\mathbb{C}[X,Y]/\langle X,Y\rangle$, would you see the isomorphism?
Choose a random polynomial $P\in\mathbb{Z}[x]$ of degree $n$ and coefficients $\leq n$ and $\geq-n$.
Let $r_1,\ldots,r_n$ be the roots of $P$ and consider $$G=\operatorname{Gal}(\mathbb{Q}(r_1,\ldots,r_n)/\mathbb{Q})$$
What is the probability, as $n\to\infty,$ that $G$ is solvable? (I assume 0...
@Semiclassical Take it on my word as a non-expert who has for some reason decided he knows what the answer should be that the unsolvable ones are dense
@Studentmath I find the stuff that comes later a lot more interesting; the basics are just the language one needs... but I'm not supposed to say this out loud
@evinda No, you can just go ahead and define a map and check that it is a homomorphism. The universal property just makes it easy, you know that you only need to define the values at $X$ and $Y$, and everything else comes for free.
@evinda Well, whether it suffices to see that $Y$ is in the image depends on what you (officially) know. But it's pretty easy to show that the map is surjective directly, we have a canonical injection $\mathbb{C}[Y] \hookrightarrow \mathbb{C}[X,Y]$.
It is well-known that, as polynomials of degree exceeding 4, there exist quintics whose roots cannot be solved for by radicals (Abel-Ruffini theorem). So we can divide the set of rational quintics into those which are solvable by radials and those which not. Obviously, both subsets are infinite ...
@evinda We can just regard a polynomial in $Y$ as a polynomial in $X$ and $Y$ (that happens to not include any positive powers of $X$). To see that it is sufficient that $Y$ is in the image, note that the image is a subring of $\mathbb{C}[Y]$ containing $\mathbb{C}$ (since $\phi$ is the identity on $\mathbb{C}$) and $Y$. But by definition, $\mathbb{C}[Y]$ is the smallest ring containing $\mathbb{C}$ and $Y$.
@robjohn The difficulty I had yesterday with a question I could not see on the front page was due to a fault of mine: one of the tags, (calculus), was by mistake in my list of ignored tags.
@MikeMiller: i'd rather avoid placing it in the context of computing a specific limit, though. i should find a way to properly replace the reference to cardinality with density instead
as an extension, i'm also curious what happens if you consider a 'random' polynomial of arbitrary degree. but the notion of density in that case is even worse
To show that no square is isometric to equilateral triangle, is it enough to just go: A square has 2 pairs of dots at maximal distance (the non-adacent vertices), and an equilateral triangle has 3 (each pair of vertices)?
Last bothering for the night: The interoir is defined as the set of points, so that there is an open ball in the space centered at that point. Why $IntQ=IntN=\emptyset$? If I take x=1, there is an open ball with radius 5 in $\Bbb N$ which holds {1} itself for example
So be careful. The interior is defined for subsets of a topological/metric space. In the case of $\Bbb Q$ and $\Bbb N$, you're probably considering them as subsets of
Hi guys, I have a basic probability question. Let A a vector with x^2 entries each of one has a positive integer value. Let C a vector when I get arbitrary elements from A with the constraint that they can be at most x. (e.g., x-2 from entry 1 and 2 from entry 3 or 1 from entry 1 and stop) How many possible C are there? I would say (x+1)^x^2. Am I correct?
All I need to ask is, how do we tell which is the $P$ node and which is the $Q$ node for optimal control?? Do we always take $C^+$ curves from the $P$ node?
@Ted I only had time to do the second one, but it was fun indeed! Gave thought to the other two just to make sure I still recall how to do it, but didn't do the technical stuff sadly. When will you teach it?
That stuff is this week's homework, @Studentmath. I'm doing $aX+bY$ for $X$ and $Y$ standard normal rv's tomorrow. ... Then stuff on conditional expectation and variance next week. (I found a problem on the web that shows up on actuarial exams, so that should convince them they should know how to do it. :D)
It's actually interesting ... and important for computing curvature from representation theory. (Yikes, no, I am not an algebraist.) Nor have I thought about it in a number of years. I need to make sure I scan those lecture notes before I burn my office.
I don't know, @Studentmath. I think my department head thinks I'm doing too much hard stuff, but, as usual, I feel I owe the top half of the class a real class.
Because if the vector bundle has a $G$-action, then it must be an associated bundle for some representation :P
I think there's a limit to how much one can motivate students to care/work, plus people care and work less as more is on the internet and as degrees become a necessity more than something you want
I don't even see why it should be true. You mean that if I have a vector bundle $E$ on a manifold $M$ with a $G$-action on $E$, I should be able to obtain $E$ by taking the associated bundle of a principal bundle and a representation of $G$. But the action of $G$ on the latter thing fixes fibers, and there's no reason it should on the former...
Given an $n$-plane bundle one can take the associated $k$-plane Grassmannianization. I was amused to note that a smooth $k$-distribution is the same as a smooth map to this Grassmannian...
It seems people frequently like to talk about smooth distributions/forms/whatever in terms of how they act on other smooth things (e.g., a smooth $n$-form is one that spits out a smooth function whenever you give it smooth vector fields) instead of just smooth maps, when they always seem to be interpretable as the latter...
That's fair. But it seems rare to even bring up the latter. Petersen even likes to define a smooth metric as one that operates smoothly on smooth things and never talk about sections or anything.
I suppose I don't need to convince you that Petersen's presentation isn't the best, though.
ACtually, @Mike, if you look at my notes, you'll see — with my algebro-geometric influence — I emphasize vector bundles in general way more than most courses.
This reminds me of Tchebotarev density theorem type results — e.g., "almost all" integer polynomials are reducible mod every $p$ (shockingly). I would think we'd get an adequate answer for it on MSE, without MO.
Using definitions I would go: Obviously $Diam(A)\le Diam(ClA)$ as $A\subset ClA$. On the other hand, assume $Diam(ClA)>Diam(A)$, then the two points are in the boundary. Yet we know that if $x\in \delta A$ then $d(x,A)=0$, so I can take $y'in A$ so that $d(x,y)\to 0$ and.. then again I am back to using epsilon in an informal way.
I'm ok with what you're thinking, @Studentmath. If you had $\text{diam}(\text{Cl }(A))-\text{diam}(A)>\epsilon$, there would have to be points in $\text{Cl }(A)$ at least $\epsilon/2$ away from $A$.
well, @UserX, I never set out to have enemies. But I got personally attacked for correcting someone's error-prone mathematics, so apparently I have an enemy.
@UserX "Should this site be used as a homework help site that answers every question that gets asked, or should question-askers show some modicum of effort and good faith in asking questions?"
Rather frequently I see someone posting their calculus homework verbatim with no thoughts on it whatsoever and within seconds of me seeing the question, a full answer shows up.
I remember that when I was a kid, CA had both Lincoln's and Washington's birthdays as state holidays. Maybe they were both federal holidays in those days.
:18570124: with regards to putting my question on MSE v. MO, I could definitely foresee putting an extension of that question on MO, depending on what i learn from the MSE responses
I can't have a strategy of not eating as that won't feed my hunger, so that's out. If I choose frozen food it's less work but less taste. If I chose that I would prefer to change my strategy. The other two are equivalent. I guess the Nash equilibrium is on eggs and toast.
Huh. if $l_\infty$ is the infinite bounded series of real numbers with the supremem metric, and $A$ is the subset of such series, so that they turn to $0$'s after some $n$ elements. Shouldn't $IntA=\emptyset$? Any positive radius $r$ I take from any $x_i$ will include some infinie bounded series that don't turn to $0$'s after some $n$ elements, but instead could be all $r$'s..
Choose a random polynomial $P\in\mathbb{Z}[x]$ of degree $n$ and coefficients $\leq n$ and $\geq-n$.
Let $r_1,\ldots,r_n$ be the roots of $P$ and consider $$G=\operatorname{Gal}(\mathbb{Q}(r_1,\ldots,r_n)/\mathbb{Q})$$
What is the probability, as $n\to\infty,$ that $G$ is solvable? (I assume 0...
