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

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 Let 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 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 1:35 PM @LeakyNun 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 The 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\implieshereditarily 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 also this: oeis.org/… 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 Googling LSTM gives me this 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^ns 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 $X_{n-1}(\Bbb{Z})$.
$X_1(\Bbb{Z})$ 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

Consider 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

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

Consider 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.