But yeah the reason I'm not fond of it is that it only does projective varieties, not affine at all, so that makes it much up kinda weirdly with how we're doing things
The conversions from the various coordinate systems to cartesian coordinates in 3-spaceare given as follows (note that the author is indicated where necessary): bipolarcylindrical: (Spiegel) a sinh(v) x = ---------------- cosh(v) - cos(u) a sin(u) y = ---------------- cosh(v) - cos(u) z = w
why would it ever be necessary? and seriously why name it after the first person to print the functions in a paywalled journal it's so much more logical to name things according to associative meaning
look if he keep asking you to compute something contact the admins and they will arrange for a new word or category theory terminology to be invented and copy and pasted to him, if he continues after that just ban the guy from chat you shouldn't have to put up with that
If we define closure of a set as the smallest closed set containing that set, then how do we see closure actuallu exists and it is the intersection of all closed sets containing our original set?
$f(x+\mathbb Z)=e^{2\pi x i}$ is an isomorphism. $\mathbb Z$ is the integers, a subgroup of $\mathbb R$, and $x \in \mathbb R$ en.wikipedia.org/wiki/Group_isomorphism
@user330477 the intersection of all closed sets containing ur original set is 1. closed (obviously) 2. contained in all closed sets containing ur set (also obvious) so it's the smallest
@EricSilva How do show that closure actually exists if it is defined to be the smallest closed containing the given set? This is to be done before the other question.
@TedShifrin I don't actually know what a closed subset really is. (In the sense that I may read the definition but the intuition won't come up, so I'll have no grip to work with compactness and stuff)
Is it necessarily true that if you have some surjective morphism $f:V\to W$ where $V$ is a variety and $W$ is some irreducible variety, that some irreducible component of $V$ must map onto $W$?
I would know it's true if I knew the image of a variety was a variety, which isn't true in general but I think might hold under assumption of surjectivity
I'm assuming your varieties are not irreducible. As stated, if you take the disjoint union of two affine lines covering $\mathbb{P}^1$ then you get a counterexample.
If you want this to be connected, then you can just imagine drawing some affine line joining the two lines, mapping to a point on $\mathbb{P}^1$
@AkivaWeinberger Probably no one cares, but I just stumbled across Ramsey's Theorem. I think it is conceptually related to that puzzle I was posing. I'll have to dig into this later and get an intuitive grasp on Ramsey Theory and then dig into my puzzle again.
(Oddly, my puzzle is the reverse of Ramsey's Theorem in a way, because the order I describe definitely appears in small instances, and I can't prove whether it also appears in large instances.)
Yeah I actually noticed how awkward this is when teaching algebraic geometry recently - in fact when I say that $\mathbb{A}^1 - \{0\}$ is a variety is isomorphic to ... you should be a little bit cautious here too - $\mathbb{A}^1 -\{0\}$ apriori is just a set - so it doesn't really make sense to say "isomorphic"
For the purpose of your question you can replace $\mathbb{A}^1 - \{0\}$ by $V(xy-1)$ in $ \mathbb{A}^2$ and do the same argument
but perhaps to make things a little bit clear -- you might learn later the definition of varieties in general (or maybe just quasi-projective varieties - which are open subsets of projective varieties), and morphisms between varieties (quasi-proj varieties) --- then with this you can genuinely say that $\mathbb{A}^1 -\{ 0\}$ is isomorphic (as varieties) to the affine variety $V(xy-1) \subset \mathbb{A}^2$
Oh and - $V(xy-1) \rightarrow \mathbb{A}^1$ is given by the projection map $(x,\frac{1}{x}) \mapsto x$
Then, actually, you can just take the union with the line $\{ (0,y) \} \subset \mathbb{A}^2$ and project to $\mathbb{A}^1$ -- maybe this is simpler
Alright so, just to be sure of what you're saying, we send an element of $V(xy-1)$ to $x$, this is a polynomial map since it restricts a projection. Then you take $V((x-1)(y-1) - 1)$ to $x$ as well, and yeah rip
The idea was that you defined the dimension of an irreducible variety to be the length of the longest chain of irreducible subvarieties, and wanted to show that if you have a surjective morphism, then the dimension of the domain is larger than that of the image
And we were like, okay well, take a chain $W \supsetneq W_{n-1} \supsetneq \ldots \supsetneq W_0 = *$, then $f^{-1}(W_{n-1})$ is a variety... not an irreducible one though... maybe one of its irreducible components surjects?
Let $(X,\mathscr T)$ be a topological space and let $a,b \in X$. A simple chain connecting $a$ and $b$ is a finite set $U_1,U_2,...,U_n$ of open sets such that $a\in U_1\setminus U_2$, $b\in U_n\setminus U_{n-1}$, and each $i,j=1,2,...,n, U_i\cap U_j \neq \emptyset$ iff $|i-j|\leq 1.$
Let $\m...
How do I find the area in between these two curves $f(x) = sin(2x)$ and $g(x)= sin(-2x)$ between $ -\pi \leq x \leq \pi$
$\int_{a}^{b} |f(x)-g(x)| dx$
Is the best way to approach this problem kind of graphically. Graphically thinking I know that there will be 4 equivalent regions and thus If I find the intersection and integrate w/ those bounds then multiply by 4 I shoulda arrive upon the answer.
@CaptainAmerica16 super speaking of super, because government superannuation funds are legal does this also mean that armed robbery is now legal provided nobody notices it's happening? I mean chloroform or ether seems the go and I realise that drugging someone is wrong and of course, id never need to for any sexual assaulty reasons
Hey guys, if X_1, X_2 are independent random variables, their joint density is the product of the densities. Then for any random variable X=(X_1,X_2), the distribution of X would simply depend on the limits of integration of the joint density right? But for any X, the joint density that we integrate over would be the same. Is that correct?
I have another inane question about Math.SE... There's quite a few questions which go like "my friend gave me this problem". Do people really do that? Give their friends math problems to solve?
Or is this a cultural thing? Maybe "my friend asked me to solve this" means something else entirely? I'm lost
@YuriyS My suspicion is that most of those are really homework and the person is trying to hide it.
But on the other hand, I have on several occasions ended up scribbling on napkins and similar over dinner due to people posing various mathematical riddles for the others.
oh btw if you evaluate for $k=\varphi(n^2+n-1)+1$, the first value greater than the maximum element of the finite set the relation specifies, watch how enormous the values get so quickly as n is increased. That's why Euler's theorem is so important to large number factorization
Please check this: any abelian group $G$ is solvable.
I am using this definition of solvable group, as given in Dummit and Foote's text: A group $G$ is solvable if there is a chain of subgroups $1=G_0\trianglelefteq G_1\trianglelefteq \ldots\trianglelefteq G_s=G$ such that $G_{i+1}/G_i$ is abelian for $i=0,1,2,\ldots,s-1$.
So, since if $G$ abelian, then $1\trianglelefteq G$ and $G/1$ abelian hence solvable. Am I right?
Hi, Does anyone know what it does mean if there is a propability with multiple variables, sperated by commas, before the condition ? I have searched for conditional densitiy probability but I cannot find anything...
$P(x_k, m | Z_{0:k}, U_{0:k}, x_0)$
Or for which different key word could I continue my search ?
@Anush are you referring to a total quantity of binary digits necessary? I recommend looking at Kummer's theorem for $p=2$, also the summation of the digits of each element of the subset will be related to the 2-adic valuation of both $n$ and $k$ I would predict. But the selection of "any" $k$ elements will involve the binomial coefficient no doubt, but I wont guarantee Kummer's theorem will apply directly
$C = A+D$, $A$ being square matrix and $D$ a diagonal matrix. Is there any easy way to compute $C^{-1}$ from $A^{-1}$ and $D$
Edit 2: (important edit)
Iam interested in this question, because my matrix $A$ is huge and so is $C$. So computing inverse of $C$ is not practical, but luckily the matri...
@Ted also, even if I understand compactness, I’ll probably need much more topology and complex analysis to get why a compact (and closed-bounded, which is Heine-Borel that I’ll still have to learn) set must have a minimum/maximum on its image.
Can I have a hint on this question? "Assume that $f$, an orientation-preserving homeomorphism of the circle, has a unique fixed point. Find the $\omega(x)$ and $\alpha(x)$ limit sets for $x\in \mathbb{S}^1$."
I figure that if $p$ is the fixed point, then $\omega(p)=\{p\}$. But I am not sure what I can say about the non-fixed points.
A product of nonzero integers whose absolute values are $<p$ will have the property that all its prime factors are $≤p−1$.
I know that Composite Number has Prime Factor not Greater Than its Square Root. But how this solve the above problem.
@LucasHenrique That is not a precise statement. Anything can be covered by a single open set: the whole space. It is about reducing arbitrary open covers to finite open covers.
And Heine-Borel is precisely the statement that that is equivalent to "closed and bounded" in $\Bbb R$, or even $\Bbb R^n$.
For a fixed real $R$ and a complex polynomial $f$, we choose the region $S = \{z \in \Bbb C: |z| \leq R\}$. Why does $\{|z|: x \in S \land z = f(x)\}$ have a minimum?
@MikeMiller yep. I’m sorry, I have no idea of how topology works.
Suppose F is an abelian group. Suppose $F$ injects into $F'$ and we know that there is a well-defined homomorphism between $F''$ and $F' $ from $F'$ to $F''$. Also, we know that $F$ $\cong$ $F''$; then how to show that it is left split.
Actually I want to know only that why they say via split exact we have direct sum decomposition right because of map W(Z) to W(Q) to W(F_(p)) is exact then can we say if f is map from W(Z) to W(Q) and
and t is map from W(Q) to W(R) ; also note that f is injective then can we say tq is the identity on W(Q) @TobiasKildetoft
@FizzleDizzle Off the top of my head I don't see a quick trick to do it. I first tried w/o knowing $p$, ($u_4 = 2p^3 + ...$)
that looked crazy, so then I just tried $p=-1$ and that did not get $u_4 = -1$
then I tried $p=-2$ and that worked, and I had the other three values on the way, added them and got...
And thought, you're right, it felt very laborious, it took about 3 minutes, about 20 seconds worried that computing by hand would be a killer but in the end it turned out not terrible. Also maybe 10 seconds worried about the negative numbers which are scary because they make things flip around so wildly and easy to make mistakes.
I didn't see a pattern that would lead to an easy formula for partial sums. But if you're doing these all the time I wouldn't be surprised if there were one.
Let $(X,\mathscr T)$ be a topological space and let $a,b \in X$. A simple chain connecting $a$ and $b$ is a finite set $U_1,U_2,...,U_n$ of open sets such that $a\in U_1\setminus U_2$, $b\in U_n\setminus U_{n-1}$, and each $i,j=1,2,...,n, U_i\cap U_j \neq \emptyset$ iff $|i-j|\leq 1.$
Let $\m...
A product of nonzero integers whose absolute values are $<p$ will have the property that all its prime factors are $≤p−1$.
I know that Composite Number has Prime Factor not Greater Than its Square Root. But how this solve the above problem.
