« first day (3157 days earlier)      last day (54 days later) » 

12:14 AM
@MatheinBoulomenos nice
 
12:25 AM
quick question... IF we have eight people showing up for free concert tickets and we want to figure out how many ways can exactly 3 of them get tickets, isn't that just 8 choose 3 or $\binom{8}{3}$?
 
yes
 
ok so if we wanted to find out how many subsets of size k are there from a set of size n isn't it just $\binom{n}{k}$ or am I missing something here since we need different subsets of size k ... maybe it's $\binom{n}{k_{1}}\binom{n-k_{1}}{k_{2}}$ and so forth?
 
12:48 AM
it is just $\binom n k$
 
ah got it. I've been typing a lot and staying up late these past couple of days X_X
so my thinking is like weeeeee
 
oh oh ... it's Demonark
 
 
3 hours later…
whooaaa
 
 
2 hours later…
5:57 AM
 
do you know tensor product?
 
In above argument, we can conclude that $K_1K_2$ is spanned by $\alpha_i\beta_j$ over F, because, closed set under addition and multiplication of $\sum\alpha_i\beta_j$ is a field, right?
@LeakyNun no!
 
there's no need to shout that
 
ok :)
 
@Silent yes
 
6:02 AM
they span K_1 K_2 because they include the generators $\alpha_i$, $\beta_j$
so the ring they generate contains $F[\alpha_i, \beta_j]$
which is $K_1 K_2$
 
@Silent Yeah
Well, $\sum a_{ij}\alpha_i\beta_j$
 
sorry
 
It's a field, it contains $\alpha$ and $\beta$, and anything that contains $\alpha$ and $\beta$ contains $\sum a_{ij}\alpha_i\beta_j$
Therefore it's the field generated by $\alpha$ and $\beta$
 
Are we implicitly assuming that $K_1,K_2$ are contained in some larger field $L$? Because otherwise we need to check closure under inversion as well, perhaps
 
We can use the same trick as the last time we talked about this
 
6:06 AM
leaky, does tensor product help in field theory?
@AkivaWeinberger oh yes
 
If $A=\sum a\alpha\beta$ then $x\mapsto Ax$ is an $F$-linear map
and we have a generating set with $mn$ things
which means there's a basis with at most $mn$ things
so it's a finite-dimensional space
so that's a surjective map
so $1=Ax$ has a solution
so $1/A$ is in our set
 
thank you!!
 
On the other hand, I don't think that this works if $K_1$ and $K_2$ are infinite-dimensional. Like, I don't think you can write $\dfrac1{\pi+e}$ as a finite sum $\sum a_{ij}\alpha_i\beta_j$ where $a_{ij}\in\Bbb Q$, $~\alpha_i\in\Bbb Q(\pi)$, and $\beta_j\in\Bbb Q(e)$
(assuming $\pi$ and $e$ are algebraically independent, which they probably are)
I don't have a proof of this
 
6:37 AM
@AkivaWeinberger how do we know that it is injective? i
I think that we can say that $1/A$ exists in larger field $L$ and hence it is injective. Our motive was to show that $1/A$ is indeed in $K_1K_2$.
Oh! 1/A exists in $K_1K_2$ already, since it is field, and our motive is to show 1/A lies in set $\{\sum a_{ij}\alpha_i\beta_j}$, right?
* I meant $\{\sum a_{ij}\alpha_i\beta_j\}$ above
 
7:00 AM
Oh I think you're right that they need to be in a larger field actually
 
yeah. i just saw it here as well but do not understand why do we need larger field!
 
Imagine if $K_1=\Bbb Q(x)/\langle x^2-2\rangle$ and $K_2=\Bbb Q(y)/\langle y^2-2\rangle$
These are both isomorphic to $\Bbb Q(\sqrt2)$
 
@Silent because there can be more than one way to extend two fields
 
If we were embedding these into $\Bbb C$, for example, we'd have to choose $x=\sqrt2$ or $x=-\sqrt2$
and similarly for $y$
which means either $x=y$ or $x=-y$
 
if $K_1 = K_2 = \Bbb Q(\sqrt[3]2) \subset \Bbb C$ then $[K_1 K_2 : \Bbb Q] = 3$
but if $L_1 = \Bbb Q(\omega \sqrt[3]2)$ and $L_2 = \Bbb Q(\sqrt[3]2)$ then $[L_1 L_2 : \Bbb Q] = 6$
despite the fact that $K_1/\Bbb Q \cong L_1 / \Bbb Q$ and $K_2/\Bbb Q = L_2/\Bbb Q$
 
7:04 AM
If we don't know what we're embedding them into, and we just have $K_1K_2$ as $\Bbb Q(x,y)/\langle x^2-2,y^2-2\rangle$,
 
(with tensor product you can show that these two ways are the only ways to embed two copies of $\Bbb Q(\sqrt[3]2)/\Bbb Q$ into a larger field)
 
then $x+y$ has no inverse.
Because $(x+y)(x-y)=x^2-y^2=2-2=0$.
 
@LeakyNun OMG!
 
Oh no!
 
In this abstract version of $K_1K_2$, both $x+y$ and $x-y$ are nonzero, so that means $x+y$ is a zero divisor
whereas if they were embedded in a larger field, either $x+y$ or $x-y$ would have to be zero (we don't know which if we don't know what the embeddings are)
 
7:06 AM
@Silent if you have two finite separable extensions of a field, say $K_1/F$ and $K_2/F$
then by primitive element theorem, $K_1 = F(\alpha)$ and $K_2 = F(\beta)$
i.e. $K_1 = F[X]/(p(X))$ and $K_2 = F[X]/(q(X))$
then we can construct the tensor product $K_1 \otimes_F K_2$ explicitly as $F[X,Y]/(p(X),q(Y))$
so Akiva's "abstract $K_1 K_2$" is actually $K_1 \otimes_\Bbb Q K_2$
 
ok, i have saved it to look at it in future!
 
now you can prove that $K_1 \otimes_F K_2$ has dimension $mn$ over $F$ and decomposes as a product of fields
those fields are all the ways that $K_1/F$ and $K_2/F$ can be embedded into a larger extension
(they may repeat)
 
@AkivaWeinberger Why is $$x\to Ax$ injective? Was my argument correct?
 
@Silent yes
 
It's injective because it's in a larger field
 
7:16 AM
corollary: in a finite extension every sub-ring-extension is a field
 
and multiplication by something nonzero is an injective map from a field to itself
 
oh wow:) feeling happy!
 
because if $Ax=Ay$ then $A(x-y)=0$ and multiplication by $1/A$ (which might not be in $\{\sum a\alpha\beta\}$ but is in $L$) gives $x-y=0$ and thus $x=y$
Unrelated thought
In the quaternions, $ij=-ji$ is a direct consequence of $(ij)^2=-1$
That is, once we have $i^2=j^2=k^2=-1$ and $ij=k$, the rest of the multiplication table is forced
 
7:39 AM
@AkivaWeinberger why pointwise multipliciation in amplitude/frequency correspond to convolution in ??/time ?
 
 
4 hours later…
11:25 AM
In Hatcher's book on Alg. Top., he calls $e_{\alpha}^n$ a cell. I tried searching through his book for an explanation of this, but I couldn't find anything. Would someone mind explaining what $e_{\alpha}^n$ is; i.e., what is the definition of $e_{\alpha}^n$?
Also, what is an attaching map?
 
12:10 PM
0
Q: About lattice of finitely generated projective module

maths studentLet $A$ be euclidian ring and $K$ be its field of a fraction.Let $(V,B)$ be a nonzero IPS (inner product space) over $K$. A finitely generated A-submodule $L ⊆ V $is said to be an $A$ -lattice in $V$ if $L$ contains a $K$-basis of $V$ . As we have already observed, $L$ must be $A$-free since it ...

 
12:21 PM
@LeakyNun Is it? I didn't know that
@user193319 On page 5 where he introduces cell complexes, he explains that $e_\alpha^n$ is an open $n$-disk
 
@AkivaWeinberger i.e. $\mathcal F\{f\} \times \mathcal F\{g\} = \mathcal F\{f \ast g\}$
 
You're taking the disjoint union of a bunch of (closed) disks, and then quotienting the boundaries of those disks together under an equivalence relation defined in terms of the "attaching maps"
The (open) interiors of those closed disks are your cells
Do you know what it means to quotient by an equivalence relation?
@LeakyNun Remind me why that is? I think I knew that at one point but forgot it
 
well that's what I'm asking you :P
 
By $\cal F$ you're not talking about the series, right? You have the continuous transform
 
I think both work
Fourier is a very general phenomenon
 
12:28 PM
Hm, what's $\sin(2x)*\sin(3x)$?
 
well the theorem says 0
 
That'd be $\int_{-\pi}^\pi\sin(2x)\sin(3(-x))dx=0$, right?
 
$\int \sin(2y) \sin(3(x-y)) ~\mathrm dy$
 
Oh right
What's the range of integration? Still one period ($-\pi$ to $\pi$)?
 
sure
they're elements of $C(S^1)$
you're integrating over $S^1$
over the normalized Haar measure, depending on who you ask
@AkivaWeinberger I can tell you more about this general Fourier if you're interested :P
 
12:32 PM
So what's $\sum(a_{1n}\sin nx+b_{1n}\cos nx)*\sum(a_{2n}\sin nx+b_{2n}\cos nx)$?
 
just use the formula lol
 
Yeah but I'm hoping it comes out to something nice
 
$\sum (a_{1n} a_{2n} \sin(nx) + b_{1n} b_{2n} \cos(nx))$
 
Is it really?
 
well the theorem says so
I'm sure you can distribute and verify
everything is (bi)linear so check on basis
 
12:34 PM
So that would imply that $\sin x*\cos x=0$
but that doesn't look like it's right
 
why not?
oh it isn't right
interesting
maybe it would be right if we used the "right" (i.e. complex) Fourier transform
 
I think $\sin x*\sin x=-\cos x$, $~\sin x*\cos x=\sin x$, and $\cos x*\cos x=\cos x$
 
$\sin x \ast \cos x = \pi \sin x$
 
yeah let's switch to the correct basis
$\exp(inx)$
 
12:37 PM
OK right so what's $e^{ix}*e^{-ix}$
And what's $e^{ix}*e^{ix}$
Given $\sin x*\sin x=-\pi\cos x$, $~\sin x*\cos x=\pi\sin x$, and $\cos x*\cos x=\pi\cos x$
$(\cos x+i\sin x)*(\cos x-i\sin x)=0$
$(\cos x+i\sin x)*(\cos x+i\sin x)=2\pi(\cos x+i\sin x)$
 
$\exp(imx) \ast \exp(inx) = \displaystyle \int_0^\tau \exp(i((m-n)y + nx)) ~ \mathrm dy = \exp(inx) \int_0^\tau \exp(i(m-n)y) ~ \mathrm dy = \tau \delta_{mn} \exp(inx)$
qed
that's why we use the correct basis
also the correct thing to integrate over is $\mathrm dy/\tau$
$\exp(imx) \ast \exp(inx) = \displaystyle \int_0^\tau \exp(i((m-n)y + nx)) ~ (\mathrm dy/\tau) = \exp(inx) \int_0^\tau \exp(i(m-n)y) ~ (\mathrm dy/\tau) = \delta_{mn} \exp(inx)$
this is much more beautiful
 
Arright so given $\sum c_{1n}e^{inx}$ and $\sum c_{2n}e^{inx}$, we can find $\sum c_{1n}c_{2n}e^{inx}$ by convolving them
 
right
so now the problem is... why?
@AkivaWeinberger so are you interested in the general phenomenon?
 
Well we just proved it didn't we
 
yeah but that's just a proof
is there a 3b1b-style explanation of all these
 
12:48 PM
We need an intuitive description of convolution don't we
of two functions $\Bbb R\to\Bbb C$
 
well convolution is just multiplication... wait
 
Two periodic functions to be clear
 
I was referring to how you multiply polynomials together
or more elementarily, how you multiply numbers together
wait... if multiplication corresponds to convolution, then convolution must also correspond to multiplication?
 
What, like how if $p(x)=\sum a_nx^n$ and $q(x)=\sum b_nx^n$ then $\sum a_nb_nx^n$ can be found by doing $p(e^{it})*q(e^{it})$ and then substituting in $t=\frac1i\ln x$?
 
oh that's interesting
 
12:51 PM
Hm and then $\langle p,q\rangle$ is $p(e^{it})*q(e^{it})$ evaluated at zero
where I have the definition of inner product on polynomials that makes the $x^n$ orthonormal to each other
 
we should ask 3b1b to do a video on this :P too many coincidences
 
Hm that doesn't actually work if they're Laurent polynomials
 
but if they're square-summable like we require elements of $L^2$ to be...
 
Hm question
 
what is $\delta_{am} \ast \delta_{bn}$
exactly $\delta_{(a+b)n}$
and it corresponds to $\exp(iax) \exp(ibx) = \exp(i(a+b)x)$
 
12:55 PM
Can I simplify $\int_{-\pi}^\pi p(e^{it})*q(e^{it})dt$
 
well that would be $(\int p(e^{it}) dt) (\int q(e^{it}) dt)$
 
Oh wait
Oh sorry never mind about the Laurent polynomials
I was thinking $x=0$, not $t=0$
$t=0$ makes $x=1$ and we don't have any problems
@LeakyNun This'll end up being $a_0b_0$
 
what is $x$?
@AkivaWeinberger yes
 
$p(x)=\sum a_nx^n$
Whatever ignore this
Notice by the way that $p(0)$ doesn't necessarily equal $a_0$
for example if $p(x)=x+2+\frac1x$ or something
 
of course it does
 
12:58 PM
$p(0)$ doesn't exist
 
oh
 
so you'll have to do $\int_{-\pi}^\pi p(e^{it})dt$ to get at it
Or, equivalently by a change of variables,
$\oint p(x)/x~dx$ around the origin
which we also know from the residue theorem
 
ok then $\int_0^\tau p(e^{it}) \ast q(e^{it}) (\mathrm dt/\tau)$ must be $\sum a_n b_{-n}$ right
 
No it's $a_0b_0$ we just did this
 
well I was wrong
 
1:01 PM
Oh
Did you mean $\cdot$
and not convolution there
I see you've embraced tauism by the way
 
$\tau$
@AkivaWeinberger
 
I don't watch black mirror though
 
Well now you've seen something of it
 
1:50 PM
In the last line of above proof, how can I be sure that there is no other way of splitting that polynomial, which can be split in a subfield of K?
 
$K[x]$ is a unique factorization domain so there's only one way to split $f$ into irreducible factors
 
wow
thanks
 
(as is $E[x]$)
 
2:18 PM
@Semiclassical I don't know about you, but Poincare-Bendixson theorem is actually not that bad
 
2:28 PM
Is a map a distorted version of the sphere
or is the sphere a distorted version of the map
 
2:48 PM
Is there a closed form for the number of $n\times n$ matrices with integer coefficients from $\{-m,\cdots,m\}$ whose determinant is $\pm a\in\mathbb{Z}$?
 
No idea but that sounds like a very interesting question
 
Hello?
 
♫ Hello ♫
 
any one who can tell me why $C_c(U)$ is a separable space? where $U$ is a open set in $R^n$
 
What's $C_c$ mean?
 
2:56 PM
@LarryEppes bump functions
 
I know that C[a,b] is separable space.
 
@AkivaWeinberger compact support
 
yes
 
@LarryEppes or just draw a triangle _/\_
continuous functions are plenty
 
Why would that give you a countable dense subspace @LeakyNun
 
2:58 PM
a triangle? but could this countable dense subset in C_c(U)?
 
oh... sorry I misread the question
 
What's the difference between C_c and C
 
C_c is all the function that have a compact support
 
@LarryEppes because in general any subspace of a separable metric space is separable
 
yes, the separable points is in the larger space that's right
 
3:00 PM
15
Q: Prove that a subset of a separable set is itself separable

PatrickThe problem statement, all variables and given/known data: Show that if $X$ is a subset of $M$ and $(M,d)$ is separable, then $(X,d)$ is separable. [This may be a little bit trickier than it looks - $E$ may be a countable dense subset of $M$ with $X\cap E = \varnothing$.] Definitions Per our boo...

 
This is true for metric spaces but not topological spaces in general I think
 
but C(U) is a metric space
 
but if the dense subset still in the C_c(U)?
 
@LarryEppes just look at my link
 
yes, I will read a minute, txh
 
3:03 PM
You could probably find a countable set of piecewise linear stuff
Piecewise linear stuff whose graphs are like polyhedra with vertices at rational coordinates
Something like that
 
@AkivaWeinberger This is very false for arbitrary topological spaces, for every space $X$ you can put a topology on $X\cup\{p\}$ such that $\{p\}$ is dense, if you start with a nonseparable $X$ then you have a counterexample
 
Oh lol
What if we want it to be Hausdorff
 
Sorgenfrey plane
 
Oh neat
Is there a name for a separable space all of whose subspaces are separable?
"Completely separable" or something like that?
 
hereditarily separable
In set theoretic topology there was (there is?) a lot of interest in S-spaces, which are regular and hereditarily separable but not Lindelöf spaces and L-spaces, which are regular and hereditarily Lindelöf but not separable spaces
What is the weakest $P$ such that $P$+separable$\implies$hereditarily separable? $P=$metric space works but I wonder if it can be weakened
 
3:13 PM
What's Lindelöf again?
 
Every (open) cover has countable subcover.
 
Every open cover has a countable subcover, it's a weaker version of compactness
 
Oh interesting
 
that's way too many adjectives
 
Is $\Bbb R^n$ Lindelöf? I'm guessing yes but I don't see why
 
3:14 PM
yes because it's second countable
 
@LeakyNun "Oh interesting" is one adjective
 
Isn't there also a concept of $\aleph_n-$Lindelöf too?
 
Oh right yeah there's another countable
 
I seem to recall something about that.
 
@Rithaniel One would assume, it seems easy enough to define
 
3:16 PM
Indeed, but is it "useful" enough to warrant studying?
 
@Rithaniel Probably by logicians or some such species
and set theorists
 
(there's a lot of very cool point set topology in that blog)
 
True enough
What do logicians actually work with, generally?
 
logic, duh
 
3:19 PM
I understand the post! thanks so much!
 
@AlessandroCodenotti I'm not gonna read that whole thing, but I didn't expect the existence of S-spaces to be independent of ZFC
 
I forgot too much about topology
 
@LarryEppes nice
 
Yeah, but $\aleph_n-$Lindelöf being useful to Logicians seems to imply something is going on there which I'm not aware of.
 
@AkivaWeinberger Yeah apparently people expected S-spaces and L-spaces to be very symmetric but now it doesn't seem to be the case
I think that "least $\kappa$ such that every open cover of $X$ has a subcover of cardinality $\kappa$" should even have a name. There are a lot of similar cardinal functions with names
 
3:21 PM
@Rithaniel I wouldn't know enough to comment
 
Ah, gotcha
 
Ah, here it is! It is called the Lindelöf degree of $X$, denote with $L(X)$
 
So model theory sounds like the sort of thing that logicians would study
but it also sounds like the sort of thing that set theorists would study
so I'm not entirely sure where the distinction is
Maybe it's more of a sliding scale
And mathematical computer science can blur into logic as well
 
@Rithaniel Found one special case of that on OEIS without a formula: oeis.org/A057981
 
or at least that's what I know from reading Turing, I don't know anything that's more current than then 30s
 
3:24 PM
I'm actually getting ready to apply to grad school for the spring of 2020. I'm trying to collect as much data on different fields of math as I can as I think about specializations.
@Semiclassical: Ooooo
 
Thinking something that behaves like a halting problem:
Let $S$ be a sequence of n numbers. The assigned task is to predict the n+1 th number
 
The extension of $S$ behaves in a way such that given any fixed polynomial $P$ that fit the n numbers, the n+1 th number is always different from P(n+1)
 
Here's an idea
 
unimodular means determinant has magnitude 1
 
3:26 PM
Take your favorite machine learning model for sequence prediction
"Long short-term memory recurrent neural networks" or what have you
Let $a_n$ be a sequence of bits, defined like this
$a_0=1$
 
Model theory is about studying models of theories (no really), a lot of set theory is about studying models of $ZF(C)$, a particular first-order theory, but there's also a lot more being done in set theory
 
$a_n$, $n>0$, is: train your model on the sequence $(a_0,a_1,\dots,a_{n-1})$ and let it make a prediction for $a_n$
$a_n$ is defined to be the opposite of its prediction
 
@Rithaniel doesn't look like OEIS has anything for the case where the determinant has magnitude 2
 
Ah yeah, 40 is the value I got for 2x2 matrices with elements in $\{-1,0,1\}$ and deteriminant -1 or 1.
 
So basically $a$ is defined so the machine learning algorithm is always wrong
 
3:28 PM
I should have checked OEIS first, myself.
 
the case where the elements are all {0,1} may be more studied
 
What I want to know is, would $a$ look pseudorandom? Or would it have some sort of pattern that the network would fail to pick up on
I guess it would depend on the exact model we use
 
Yeah, the 0,1 case is similar to matrices over $\mathbb{Z}/2\mathbb{Z}$
 
Well every neural network T has some kind of deterministic rules in it, so $a$ can be a formula that can never be captured by said neural network
 
$a$ is uniquely determined once we decide on a network
(unless the network has some sort of random initialization to it, in which case we have to specify the network as well as the random "seed")
 
3:30 PM
Well, I think we can have a neural network that given 1,2,3,4,5 will never produce the outcome 6, where a can be 1,2,3,4,5,6
 
I was thinking of a sequence of bits
so 0s and 1s
so we can say, "Whatever you predict the next bit is, it's the other one"
 
Heh, the two conversations converge on binary :P
2
 
By the way, I once looked up how LSTM neural networks work
Didn't understand it at all
You give it something analogous to "short-term memory" and then… I dunno, wire the pieces up in a way that looks like a rocket engine diagram
 
neutral networks are kinda magic in most of the case, because there are simply too many coupled systems that does stuff on the input and parameters
 
(Though the convergence is more like a pair of skew lines: their projections intersect but the lines themselves don’t )
 
3:33 PM
By "rocket engine diagram" I mean something this
Basically: complicated
 
Ok, it is MUCH worse than that
 
Heh, that's funny. "I found this: Image not found"
 
Dunno why the link wouldn't work
 
ok, that's indeed look like an engineering diagram or sort
Also regarding this:
8 mins ago, by Akiva Weinberger
What I want to know is, would $a$ look pseudorandom? Or would it have some sort of pattern that the network would fail to pick up on
I think it will be pseudorandom, cause say the network predicts (given bits),1,0,0,1,0,1,... then a has to be (given bits),0,1,1,0,1,0,.., so there is a perfect anti correlation to the prediction dependent on the seed and settings of the network (because that controls the prediction)
 
Sometimes I am wondering: If our computers are analogue and does not suffer from the problems of analogue computers in history, will circuit diagrams for e.g. addition have to be that complicated
 
Didn't Fermat make an analogue calculator
or someone similar
Well actually it was "digital" in that it worked with digits
but it had a gear with ten things on it to represent the digits 0 to 1
(Teeth?)
I remember learning about another version of that that was actually sold commercially around WWII
By the way
What the engine actually looks like
(The nozzle isn't pointing down, which I can only assume means they will not go to space today)
 
Probably they are test firing it or something
 
@Semiclassical Accurate
@Secret Yeah I think that's what it was
Though subtraction is basicaly just addition with an extra step
'cause $(2^n-1)-b$ is easy to find (flip all the bits of $b$)
Call that $\bar b$
Then you can just do $a+\bar b$ to find $a-b$
well actually $a+\bar b+1$
and throw away the "carry" bit (the $2^n$s place)
 
 
3 hours later…
6:55 PM
I’ve got two statements in probability theory. I’m pretty well-convinced that the first, at least, is false. But I don’t know about the second.
Let $X_k$, for $k=1$ to $n$, be a set of random variables.
First statement: if $P(X_k=x)=P(X_k=-x)$ for all k, then $$P(X_1=x_1,\cdots,X_n=x_n)=P(X_1=-x_1,\ldots,X_n=-x_n).$$
Second statement: if $P(X_j=x,X_k=y)=P(X_j=-x,X_k=-y)$ for all $j,k$ then $$P(X_1=x_1,\cdots,X_n=x_n)=P(X_1=-x_1,\ldots,X_n=-x_n).$$
The premise of the second implies the premise of the first (take $j=k$) so the second statement would follow from the first.
But right now I don’t think the first is true, and so don’t have much opinion at all on the second
(I think the first statement is true if the only possible outcomes are $\pm 1$ but that’s not generic.)
 
Problem: For $n \in \Bbb{N}$, let $X_n(\Bbb{Z})$ be the simplicial complex whose vertex set is $\Bbb{Z}$ and such that vertices $v_0,...,v_k$ span a $k$-simplex if and only if $|v_i-v_j| \le n$ for all $i,j$. Using induction, show that $X_n(\Bbb{Z})$ is contractible by showing that it deformation retracts onto $X_{n-1}(\Bbb{Z})$.
I've been thinking about this problem for at least a week and have made no progress. I could use some help.
$X_1(\Bbb{Z})$ is contractible because it is a connected graph.
 
7:13 PM
I typed all of the above on mobile. Didn’t expect it to be as long as a question on the main site, lol, but maybe I should move it to there
 
 
2 hours later…
9:55 PM
@AkivaWeinberger wow that's interesting
 
10:08 PM
0
Q: Vector field over $\Bbb Q^2$

UltradarkConsider the vector field restricted to the rationals, $\vec F_\Bbb {Q^2}=(x,y).$ This is a vector field $\vec F:\Bbb Q^2\to \Bbb Q^2.$ Is this vector field weak with respect to the same vector field over the reals: $\vec F_\Bbb {R^2}=(x,y),$ $\vec F: \Bbb R^2\to \Bbb R^2?$ Edit: A weak vector f...

1 more re-open vote needed
 
10:18 PM
It's really unclear what you are asking
what is a weak vector field w.r.t another vector field?
 
A weak vector field has less information than a smooth vector field
 
@Ultradark Is your vector field continuous?
 
the one in $\Bbb R^2$ is
How does that video relate?
 
I mean, the point seems to be that you’re talking about vector fields on (for instance) integer pairs rather than the entirety of R^2
Or on the rationals, for another
 
Yes I'm talking about a vector field on the rationals
 
10:47 PM
-1
Q: Vector field over $\Bbb Q^2$

UltradarkConsider the vector field restricted to the rationals, $\vec F_\Bbb {Q^2}=(x,y).$ This is a vector field $\vec F:\Bbb Q^2\to \Bbb Q^2.$ Is this vector field weak with respect to the same vector field over the reals: $\vec F_\Bbb {R^2}=(x,y),$ $\vec F: \Bbb R^2\to \Bbb R^2?$ Edit: A weak vector f...

 
11:13 PM
@AkivaWeinberger IIRC GRUs and MGUs are simpler than LSTMs and show comparable performance.
 

« first day (3157 days earlier)      last day (54 days later) »