@PhilipHoskins Typically the proof doesn't really use the surjectivity (supposing we are talking about the version for complete metric TVSs), all you need is that for some $n$ the set $\overline{T(nV)}$ has nonempty interior for some $n$, for any neighbourhood $V$ of $0$ in the domain. By homogeneity, $\overline{T(V)}$ has nonempty interior.
@PhilipHoskins Yes, Functional Analysis. I don't think he has any version of it in RCA.
Ah that makes sense. I almost got Harry Boas' copy of that book when he had a big book give-away, but my friend snatched it right before I noticed it was right in front of me :(
But Lax's book seems pretty comprehensive, so I'll look it up in the morning.
Yes, it's a different parametrization. The latter is not what I call a regular curve. Indeed, because $y=|x|$ is not a differentiable $1$-manifold, there's no way to parametrize it by a function whose derivative is everywhere nonzero.
Think about driving along that path. You have no choice but to slow down, stop, turn (while stationary), then restart. You cannot come in to that corner with nonzero speed, i.e., with a nonzero velocity vector, or else you'll continue in that direction, rather than turning $90^\circ$.
With regard to your question 4 hours ago, do whatever exercises interest you. In some sense, you really could start chapters earlier ... but you don't have those.
Because the implicit function theorem is the key to understanding them.
In econ and max/min we want to think of sets implicitly. But when we want to integrate on them, we need to parametrize them. So we need to understand that one can go back and forth between those two viewpoints, at least in theory.
I didnt see a big deal made of the implicit function theorem in John Lee's book, although I remember him saying it was an extremely important result in mathematics.
Notation and Defination:
$G$ is a Strongly Regular Graph( not a complete , cycle graph) with
Every two adjacent vertices have $\lambda$ common neighbours.
Every two non-adjacent vertices have $\mu$ common neighbours.
$r$ regular and total $n$ vertices .
A graph of this kind is someti...
@DanielFischer: As a comment on a question on MSE, someone said that Milnor won a fields medal in 62 in part for giving the first example of a pair of smooth manifolds that are homeomorphic but not diffeomorphic. Is this what "On manifolds homeomorphic to the $7$-sphere" is about?
@DanielFischer: It's a straightforward computation that shows the group property of flows on an open set $U$, namely if we have $\varphi: U \times (-S,S) \to V \subset M$ and $\varphi: V \times (-T,T) \to M$, then $\varphi^u$ is defined for $|u| < S+T$ and $$\varphi^{t+s} = \varphi^t \circ \varphi^s$$ on $U$. Now I am supposed to show a corollary: Assuming $U$ is open and $\varphi^t$ exists on $U$, then $\varphi^t(U)$ is open and $\varphi^t$ is a diffeomorphism onto its image.
@DanielFischer: I suspect the idea is to use $\varphi^{-t} \circ \varphi^{t} = \operatorname{id}$, but how can we know that $\varphi^{-t}|_{\varphi^t(U)}$ is well-defined? Does that somehow follow from the group property too or is there a different idea behind it?
@Huy Can you clarify the notation? At the moment, you're using $\varphi$ for two entirely different things. [Well, not really entirely different, but ...] I guess the setting is that $\varphi$ is defined on an open neighbourhood $W$ of $M\times \{0\}$ in $M\times \mathbb{R}$ and the premises are 1. $U\times (-S,S) \subset W$, 2. $\varphi(U\times (-S,S)) \subset V$, 3. $V\times (-T,T) \subset W$?
@DanielFischer: I think so, yes. I think maybe I could say that $\varphi^t \circ \varphi^s$ is defined on $U$ for $|s| < S, \, |t| < T$, and the group property says that then $\varphi^u$ is defined on $U$ for $|u| < S+T$ and $\varphi^{t+s} = \varphi^t \circ \varphi^s$.
That's a bit shorter, less explicit but it should be the same as you said if I'm not mistaken.
@Huy The key is that if $(x,r) \in W$, then $(\varphi^r(x),-r) \in W$. Now take small neighbourhoods to see that $\varphi^t$ is - on $U$ of course - a local diffeomorphism (with local inverses $\varphi^{-t}$). Glue together.
there are lots of instances in mathematics where you break up a "large mathematical structure" into smaller "indivisible" structures. i would not bother much about this similarity.
yes, you're right, but i like to think about prime factorization in the ambient Z with Z given a ring structure. i don't think much number theory can be done by considering FTA as a special case of JH.
@Balarka So it does imply some kind of prime factorization but it cannot give me interesting results .. Or does it? Remember I am not someone who knows a lot about this topic of the subject
@Rememberme: You had to find the link to the post again AFTER I told you to not just read/cite wiki out of context, which implies that you were for sure not reading that post just then.
the reformulation would give Kostka numbers as the sizes (number of right cells) of those intersections (the lemma just says that such an intersection is not empty)
@robjohn is this kind of series obvious in any way? $$2 \coth \left(2^{\pi/4+1}\right) \text{csch}\left(2^{\pi/4+1}\right)+2^2 \coth \left(2^{\pi/4+2}\right) \text{csch}\left(2^{\pi/4+2}\right)+2^3 \coth \left(2^{\pi/4+3}\right) \text{csch}\left(2^{\pi/4+3}\right)+\cdots$$
From the little I picked up from here and there : every group appears as fundamental group of n-folds for n > 3. so classifying n-manifolds for n > 3 like we do for n = 2 or n = 3 (considerably harder) is about almost impossible. however, there is a thing called the h-cobordism theorem, which works for smooth n-manifolds for n > 4, which makes it easy to deal with higher dimensional manifolds.
As 4-folds lies in the intersection of the two, they are very hard (impossible?) to classify.
Addendum: The first case.
For $x\le 1$ as mentioned ALL of the expressions in the $|\cdot|$ are negative, so the expression takes the form:
$$-(x+1)+x-3(x-1)+2(x-2)$$
$=-x-2$ after tidying up. This is valid as long as $-\infty<x\le -1$ - which the diagram confirms.
It is easy to see that $x+2...
I'm going to write some "A-level logic" here, but I think it will help.
Take: $$f(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots$$
I noticed that:
$f(0)=a_0$
$f'(0)=a_1$
$f''(0)=2a_2$
$f'''(0)=3*2a_3$
$f''''(0)=4*3*2a_4$
$\cdots$
$f^{(n)}(0)=n!a_n$
It follow from this that $a_n=\frac{f^{(n)}(0)}{n!}$ - yo...
Were just taken down..... oh god, who have I pissed off now? Is Timbuc unbanned?
(note : h-cobordism theorem for 4-folds is true in the topological category, but afaik, is false in the smooth category. that's why things in the 4th dimension are so exotic)
fair enough. if someone downvotes in a period of a few minutes/seconds, only then the bot will be able to recognize that as a serial downvote, if i recall correctly.
$\mathbb{R}_{\ge a}$ is VERY standard for $[a,+\infty)\subset\mathbb{R}$ and $\mathbb{R}_{> a}$ for $(a,+\infty)\subset\mathbb{R}$
This is a very obvious "There is no other sensible interpretation" convetion, if I give you $\mathbb{R}_{\le -5}$ you know immediately I mean $(-\infty,-5]$
I have ...
That question.
$R_{>0}$ is OBVIOUS, there is no sane way of interpreting that which isn't $(0,+\infty)$
@skillpatrol: if you ain't got nothing better to do with your life than trying to show a 7th grader how much smarter you are than him, that's your problem not mine
They've replaced the Ghost thing in Destiny, but NOT the multiplayer announcer. Who sounds so depressed and bored, from his voice alone he ought to be on a suicide watch.
@skillpatrol I think that stuff is automatically sent to some users. I post some problems on main once in a while out of curiosity more than to find solutions.
@Huy That was his first paper on the subject. He did plenty further work later, culminating (post-medal) in Kervaire-Milnor, groups of homotopy spheres.
@BalarkaSen It's impossible to classify $n$-manifolds for $n>3$ for the reason you mentioned. One usually restricts to highly connected $n$-manifolds, or puts some restrictions on the fundamental group. The h-cobordism theorem's failure makes it hard to classify different smooth structures on the same 4-manifold; in high dimensions this is classified by surgery theory, in 4 it's... not.
There is a classification of simply connected topological 4-manifolds up to homeomorphism. There is not of s.c. smooth 4-manifolds up to diffeomorphism. (Or even up to homeomorphism, yet - the 11/8 conjecture is the last thing in the way.)
otoh there was a full classification of s.c. 5-manifolds up to diffeomorphism in the 60s.
@MikeMiller: Let $a: (x,y) \mapsto (x+1, -y)$ on $\mathbb{R}^2$. Why is $\mathbb{R}^2 / \langle a \rangle = [0,1) \times [0, \infty)$? Shouldn't it be $[0,2) \times [0,\infty)$? Or what would the representant for $P = (\frac{3}{2}, \frac{1}{2})$ be?
Yes. It's not $[0,1)$ bexause it's not. You have to do better than just picking a point for each wquivalence class; this is a topological space, and it has a topology.
Hi!!! Could I ask you something? We consider the points A, B, C with A =(1, 1, 0), B = (1,0,1) , and C = (0, 1, 0). Find a non-zero vector U=<U_1,U_2,U_3> that is perpendicular to the vectors AC and BC. I have tried the followig:
AC is the vector (-1,0,0) and the vector BC is (-1,1,-1). So that U is perpendicular to these 2 vectors, it has to hold $U \cdot AC=0 \Rightarrow -U_1=0$ and $U \cdot BC=0 \Rightarrow B=C$. So the vector U is $(0,1,1)$.
I went looking for a definition of "quantum algebra". According to Wikipedia, Quantum algebra "is one of the top-level mathematics categories used by the arXiv". And that's it. Okay...
@Hippalectryon I think I said that once here: some time ago I went to a company where I worked for some years, but at the beginning I met some that told me I won't resists 3 days on that position. A few months later I was teaching the whole department teh stuff they had to do. The same story will happen with math in a way or other, and one has to be very stong, extremely strong especially when you see around you so many people ready
to discourage you, to be like them, without no high dream, aim, to be just another average person.
I often read here people saying that I compare myself to Ramanujan. And what? I'm going to be as great as him in terms of integrals, series and limits.
The thing that makes no sense to me is to have a little dream when you can choose to be great, not the fact that you compare yourself with anyone.
I actually see in Ramanujan a reference point in terms of performance. I'm not afraid to have such reference points, but it makes me life entirely noble.
Notation and Defination:
$G$ is a Strongly Regular Graph( not a complete , cycle graph) with
Every two adjacent vertices have $\lambda$ common neighbours.
Every two non-adjacent vertices have $\mu$ common neighbours.
$r$ regular and total $n$ vertices .
A graph of this kind is someti...
