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13:27
@Semiclassical That's not "everywhere"
EDIT: Sorry, misread "somewhere" for "everywhere"
As long as the domain doesn't contain loops around the origin you should be ON
*OK
14:15
i got no cancer in my head, so i am happy
I suck at proof language:
Def. 1. All elements in a Cayley table

Given a magma $(M,\circ)$, Let $S=\{s_1,s_2,\dots,s_n\}$ be the ordered set obtained by imposing an order $\mathcal{O}$ onto $M$. Then the cayley table $T$ is given by $T=\{\ x\lvert \forall (a,b) \in S \times S, a\circ b=x\}$. The tranpose is defined by $T^\textrm{T}=\{\ x\lvert \forall (b,a) \in S \times S, a\circ b=x\}$.


Def 2. Rows and columns and collections

A row $R_i \subset T$ is given by $R_i= \{\ x\lvert \exists i \in S,\forall (a,i) \in S \times S, a\circ i=x\}$. Similarly, a column $C_i=\{ x\lvert \exists i \in S,\forall (i,b) \in S \ti
@Null Was that a concern? What happened?
Actually, by just looking at the above stuff, is it clear to anyone what I am trying to formulate here?
Sometimes people said I have the correct idea, but it seems I can never get it down in the correct language for those within the field to understand
14:31
@AkivaWeinberger i have strange physical experiences. One doctor said now, that it is very unlikely that it is organic. Kinda feel happy, but sad too, because now my stubborn GF might have right and it is indeed psychological. On the other hand I can't take the doctor serious, because he trusted a urin-test for THC. It was positive, but I definitly didn't smoke for years. Well, got to wait for the more saying MRT (had only a CT), I really don't know what I wish haha.
Who wants to loose an argument with a psychology student >:|
but at least cancer is impossible
i guess that is something
On a sidenote: my brain looks very uninteresting, got the pictures on a CD^^
15:06
Sorry to hear that, I hope you find out what's causing it and I'm glad to hear that it's not cancer
Can anyone recommend some books on set theory, mostly for the purpose of starting to read undergrad maths. Thanks.
@JonathanRichardLombardy I think Halmos has a classic book called "naive set theory". I also recommend not spending too much time studying straight up set theory unless you're interested in it for its own sake ;)
If you Google "good set theory book" you get a few things from Amazon as well as three Math SE posts also asking for references
(I'd offer a real answer if I knew)
I agree with Danu concerning Halmos, which is a nice and soft introductiob, I wouldn't read a set theory book to get a feel for undergrad maths though
15:22
@JonathanRichardLombardy You don't need set theory to do that. Don't get any book. Just start doing math.
Let $\frac{p_n}{q_n}=[1;2,2,2\dots 2]$ then $\lim_{n\to\infty}\left(q_n^2\left|\sqrt{2}-\frac{p_n}{q_n}\right|\right)=\frac{‌​\sqrt{2}}{4}$
I have a proof for that but it's horrible
I'm trying to generalize that to any quadratic irrationals
@DHMO hi!
@Ramanujan sorry I'm busy today
@DHMO OK,i will discuss tomorrow
thanks
15:29
this gives a fine tuning on the order of rational approximations, in the sense that $\left|\frac{1+\sqrt{5}}{2}-\frac{p}{q}\right|$ can only get near $\frac{1}{\sqrt{5}q^2}$ but $\left|\sqrt{2}-\frac{p}{q}\right|$ can get as close as $\frac{1}{\frac{\sqrt{2}}{4}q^2}$
@Sophie What exactly are you trying to prove ?
Suppose we have two vector subspaces of dimension $3$.
Which is the dimension of the intersection? Is there a formula to calculate it?
Depends which subspaces you have @MaryStar
We have $$U_1=\begin{bmatrix} \begin{pmatrix} 3\\ 2\\ 2\\ 1 \end{pmatrix}, \begin{pmatrix} 3\\ 3\\ 2\\ 1 \end{pmatrix}, \begin{pmatrix} 2\\ 1\\ 2\\ 1 \end{pmatrix} \end{bmatrix}, \ \
U_2=\begin{bmatrix} \begin{pmatrix} 1\\ 0\\ 4\\ 0 \end{pmatrix}, \begin{pmatrix} 2\\ 3\\ 2\\ 3 \end{pmatrix}, \begin{pmatrix} 1\\ 2\\ 0\\ 2 \end{pmatrix} \end{bmatrix}$$

I have written them in the form
$$U_1=\{(x,y,2z,z) \mid x, y, z\in \mathbb{R}\}=\{x(1,0,0,0)+y(0,1,0,0)+z(0,0,2,1)\mid x, y, z\in \mathbb{R}\}$$
Try and compute the dimension of the sum
15:44
$U_1+U_2$ consists of all the elements $u_1+u_2$ where $u_1\in U_1$ and $u_2\in U_2$, right?

So, the sum is the following, or not?

$$(1,0,0,0)+(1,0,0,0)=(2,0,0,0), \ \ (1,0,0,0)+(0,1,0,1)=(1,1,0,1), \ \ (1,0,0,0)+(0,0,1,0)=(1,0,1,0), \\
(0,1,0,0)+(1,0,0,0)=(1,1,0,0), \ \ (0,1,0,0)+(0,1,0,1)=(0,2,0,1), \ \ (0,1,0,0)+(0,0,1,0)=(0,1,1,0), \\
(0,0,2,1)+(1,0,0,0)=(1,0,2,1), \ \ (0,0,2,1)+(0,1,0,1)=(0,1,2,2), \ \ (0,0,2,1)+(0,0,1,0)=(0,0,3,1)$$

@Astyx
15:54
@Astyx I want $$\liminf_{p,q\in\mathbb{Z},q\to\infty}\left(q^{\mu(\varrho)}\left|\varrho-\frac‌​{p}{q}\right|\right)$$
I can calculate that for any quadratic irrational using a very inefficient method
 
1 hour later…
17:04
I would like to (seriously) study Algebraic Geometry, and I am looking for a study partner. I imagine, for example, that we meet on Skype (at least) once a week and discuss the exercises and maybe lectures contained in the 2002 lecture notes by Andreas Gathmann. Where, besides here, is a good place to post this?
17:16
@MaryStar Your first statement is correct, I don't follow the second one
I calculated all the possible sums of the vectors of $U_1$ and $U_2$. @Astyx
You mean the basis ?
I think you forgot the zero vectors then @MaryStar
what are other forms of addition, in contrast to pointwise addition?
@Null What is that supposed to mean ?
there is pointwise addition of for example functions
but what would a notpointwise addition be??
17:24
By pointwise addition, you mean $\forall x, (f+g)(x) = f(x) + g(x)$ ?
yes
You can definea binary operator on functions by, let's say $f\oplus g = 2f + 2g$ and call it addition (even though this might seem veyr silly)
And then you wouldn't have $\forall x, (f\oplus g)(x) = f(x) + g(x)$
mmh, so what's the point in "pointwise"?
(let's say you're not interrested in the actual values the function takes, but want to establish a structure in which you can best study functions)
why is it called that way
17:27
What do you mean ?
Oh because the addition of functions "acts" like the addition of "points" over the codomain of the function
Does that make sense to you @Null ?
@Astyx yes
Good
i might add, my problem was not that I didn't understood pointwise , but I just didn't see an example of not pointwise :)
@Astyx I don't know if this is the basis... I just added every vector of $U_1$ with every vector of $U_2$.
No you haven't @MaryStar
17:34
Why not? @Astyx
$U_1$ has infinetly many vectors
Ahh... So, is the sum defined as follows: $a_1(1,0,0,0)+a_2 (0,1,0,0)+a_3 (0,0,2,1)+b_1(1,0,0,0)+b_2(0,1,0,1)+b_3(0,0,1,0)$ @Astyx
Yes, where the $a$'s and $b$'s are real @MaryStar
define $\sin(x)=\cos(x)\tan(x)$, $\cos(x)=\frac{\sin(x)}{\tan(x)}$, $\tan(x)=\frac{\sin(x)}{\cos(x)}$. Those are circular definitions for circular functions
17:44
Let k be an algebraically closed field, X c k^n an algebraic variety and O_P(X) the local ring of regular functions at the point P and m the maximal ideal of O_P(X). Is O_P(X) / m isomorphic to k?
@Sophie …Was that a pun
17:59
@Astyx And how do we find a basis for that?
The same way you do for any other subspace
Do we write that as a matrix and then we apply the Gauss algorithm? @Astyx
Well you have a family of vecctor that generates your subspace
You can remove a few to get a basis
(in a non-arbitrary way, evidently)
Hey ted here?
Not yet
18:09
hi chat
Hi @Semi
How are you ?
fine, fine. about to eat lunch
I found that a basis for the intersection, it is 2. So $\text{dim}(U_1+U_2)=\text{dim}(U_1)+\text{dim}(U_2)-\text{dim}(U_1\cap U_2)=3+3-2=4$.

Is each subspace od dimension 4 equal to $\mathbb{R}^4$ ? @Astyx
In $\Bbb R^4$, yes
More generally, any subspace whiches dimension is the same as the dimension of the vector space it lives in is that space
If I want my solution to a pde on an unbounded domain to be steady-state, does that mean I have to ensure that my solution is bounded?
18:15
Steady-state = time-independent.
Yes. I'm working on a Laplace equation on an unbounded domain. the variable t isn't even part of the problem.
I know it means time independent. But, do I need to make sure that my solution is bounded?
I don't -think- being steady-state requires that.
Too bad. I have no idea how to apply the BCs in that case.
18:18
As an example, the electric potential of an infinite line charge along the $z$-axis is of the form $\log(x^2+y^2)$.
That's certainly time-independent, but it's unbounded at infinity.
I' trying to solve $u_{xx}(x,y) + u_{yy}(x,y) = 0$ where B.C. is $u(x,0) = \begin{cases} 1 & \text{if}\, -1<x<0 \\ 1-x & \text{if} 0<x<1 \\ 0 & \text{otherwise}\end{cases}$.
Ah. So Dirichlet BC.
Need to use Fourier transform
Yes
There's definitely a bounded solution to that.
What I don't know is if it's strictly speaking the only one.
So, when applying hte B.C.'s, I can assume my solution must be bounded in order to get rid of the $B \exp(|\alpha|y)$ part of the solution to the transformed problem: $U(\alpha, y) = A\exp(-|\alpha|y)+B\exp(|\alpha|y)$?
18:22
I suspect it's not, since I could set this up in such a way that there's an electric field at $y=\infty$ i.e. $u_y\to $const. as $y\to \infty$.
Yeah.
So, how do I apply the B.C.'s to that form of the solution?
You mean, if it had that boundary condition?
yes
The ones I listed above
to the soution I mentioned
$U(\alpha, y) = A \exp (-|\alpha|y) + B \exp(|\alpha|y)$.
Hang on. Do you mean with the assumption that it's bounded at infinity, or that $u_y\to$ const. at infinity?
The second one is what I introduced as an example that's different than the usual 'bounded' assumption.
I mean the assumption that it's bounded at infinity. I.e., I can make $B=0$ in order to get rid of the possibility that it will become unbounded at infinity
18:25
Ah. Then yeah, $B=0$.
Otherwise, I have only one BC to take care of two constants.
And then the rest is standard.
Ok. From the solution in the back, I thought so, but I wasn't entirely sure.
Thanks.
The remaining question is whether they clearly indicated that boundary condition.
Sounds like they didn't, which is annoying.
They didn't, which is why I chose to ask someone who knows stuff ;)
18:28
Hi @Alessandro
Hi @Balarka!
By way of comparison: I distinctly remember that one of the problems we had on our grad-level electrodynamics final amounts to solving Laplace's equation on $r>1$ with boundary conditions imposed at both $r=R$ and $r=\infty$.
And with those boundary conditions depending on the azimuthal angle $\theta$.
that the heck is an azimuthal?
Angle from the z-axis
What is the answer 6 / 2 x (1+2)? is it 9 or 1?
18:30
i.e. the extra angle going from cylindrical to spherical coodinates.
Hi balarka
Hi, anybody willing to help?
balarka hereeee?
@Astyx I see... Thank you!! I want to find also a vector subspace $C$ of $\mathbb{R}^4$ such that $U_1\oplus C=\mathbb{R}^4$. This means that the sum of $U_1$ and $C$ must the $\mathbb{R}^4$ and the intersection $U_1\cap C$ must be empty, or not? How can we find such a $C$ ? Do we have to extend $U_1$ to a basis of $\mathbb{R}^4$ ?
@Victor It depends on how you group things. It can be disambiguated as either (6/2)*(1+2) or as 6 / (2*(1+2)).
18:31
@Adeek Yep
Just reading it, I'd probably interpret it as the former. But it could be either, and I'd probably insist that whoever wrote it change it to be clear what's meant.
Whether or not there's a 'correct' interpretation, it's written in a confusing way.
@Mary take a basis on $U$ and extend it to a basis of $\mathbb{R}^4$, what can you say about the span of those vector in the basis of $\mathbb{R}^4$ but not in the basis of $U$?
@Mary take a basis of $U$ and extend it to a basis of $\mathbb{R}^4$, what can you say about the span of those vector in the basis of $\mathbb{R}^4$ but not in the basis of $U$?
I aced geometry classs @BalarkaSen
Thanks a lot for your help
good to hear
To extend $U_1$ to a basis of $\mathbb{R}^4$ we add the vector $(0,0,0,1)$. So is $C=\{(0,0,0,1)\}$ ?
So, we get that $U_1\oplus C=\{(1,0,0,0), (0,1,0,0), (0,0,2,1), (0,0,0,1)\}$ ? Is this $\mathbb{R}^4$ ? @Alessandro
18:38
That's a basis of $\mathbb{R}^4$, but yeah, looks good to me
Ah ok... Thank you!!
I have also an other question... I want to calculate the limit $$\lim_{x\rightarrow \infty}x^{100}\left [\frac{1}{x}\right ]$$

Does it hold that $0<\frac{1}{x}<1$ ? If yes, it holds that $\left [\frac{1}{x}\right ]=0$ or not? Then $x^{100}\left [\frac{1}{x}\right ]=0$, and therefore the limit is $0$.

But this holds only when $x\rightarrow +\infty$, or not? What happens if $x\rightarrow -\infty$ ?
18:54
@TedShifrin Yes indeed!
Well, there's two aspects to that question. One is whether the limit is well-defined, and the other is what its value is.
Hello all. Can someone help me with fourier analysis? How do i find the fourier series representation of (-1)^n * impulse(n-2m)
If one takes for granted the former, then it really is as simple as plugging in $x=0$; the function is well-defined there, after all.
over a summation of m from -inf to +inf
On the other hand, I think there's quite a bit of subtlety in whether said limit exists in the first place. @JorgeFernándezHidalgo
18:56
@Semiclassical not only must it be proven that the limit exists, it must also be proven that the function is continuous at $0$ to solve it like that. ( or prove it is continuous in all the domain)
I mean, the answerer just went like "Duh, it's defined at $0$, plug it in!"
I mostly don't like the question in the first place, though. There's no effort shown in recognizing why the problem is complicated.
Yeah.
and everyone just went like "oh, genius"
lol
I can't generate the energy to be annoyed by it, though.
18:59
yeah, in retrospect maybe the OP just wanted us to analyse his solution.
Possibly. In which case, SA's point that said manipulations are quite silly is understandable.
But eh. I'm the kind of person who likes interesting computations rather than arguments about continuity :)
well yeah, I guess.
I'll ask a friend if he can calculate it using iteresting computations
About the only thing 'interesting' it reminds me of is the Jacobi-Anger expansion of $e^{i z\cos\theta}$
but that's a stretch, and it depends very much on your definition of 'interesting' :P
I have heard of anger management, but not of anger expansions.
I wouldn't recommend it in any case
@Balarka I just read that supposedly any Riemannian manifold has an open, dense set $O$ over which $TX$ is trivial. Is this easy to see? Do you know about it?
19:14
I am not sure. Without using the Riemannian structure at all, can't you just take the top-dimensional cell of the manifold (assuming it's compact here), modulo a cell decomposition?
@BalarkaSen All the top-dimensional cells, you mean? Hmm, sounds plausible...
Yeah, union of all of them, sorry.
Those will be disjoint disks, and it looks dense enough to me
I'll take it! :D Thanks
To walk five minutes to a see a talk, or not to.
What's the talk on?
19:17
Well, here's the abstract (it's part of a workshop on campus, so I wanted to go to a few of them)
Eigenvalue
nope :D
It's a two minute walk, so that's not really an impediment (besides the fact that it's really cold out)
lol
That's a feature not a bug for me :P
Current temperature in Minneapolis: 4 degrees Fahrenheit.
at 8am tomorrow its forecast to be -10 :)
It's pretty cold in here too (and is supposed to get colder) but ofc nothing compared to that
19:28
On the other hand, my sister's new husband is coming over this weekend. So I get to 'welcome' him to Minnesota winter :)
how cold is that in sensible units?
15 degrees C.
-16 degrees C, you mean
19:29
oh you were talking to SemiC.
going to -23 C by 8am tomorrow
ah
So you're as far above freezing as we are below it :)
it's also decently windy right now
that's cold indeed, we only go down to -2 or -3°C during the night at the moment here
19:36
Hey!
19:54
If anyone would like to go over a simple number theory problem, I'm trying to evaluate this Euler product: $$\left(\prod_{p\,\in\,\text{primes}}{\frac{(p-1)^{n-1}}{p^n}}\right)^n$$
It represents the probability that every element of a set of $n$ random integers contains at least one prime factor unique to the set
@Kri sup
Could you not rewrite this as $(1-1/p)^n \cdot 1/(p-1)$ per term
I believe so, let me check. I'm fairly sure I have rewritten it similarly to that while trying to get it into a zeta function
And then use that $(1-1/p)$ is the inverse of the term in the euler product you get from summing all natural numbers
@MickLH Yeah, like that
hmm ok I will put more effort into that approach, I must need food because I'm not thinking efficiently lol, thanks
19:58
Doesn't that Euler product diverge ?
I'm not sure though, I'm watching telly at the same time
@MickLH doesn't $\frac{(p-1)^{n-1}}{p^n}$ go to zero?
then the whole product must too
@BalarkaSen How are your exams going?
they're over
How did your exams go?
20:00
alrightish
what exams did you have?
and hi
@Sophie @Astyx I very well may have made a mistake earlier on in the process, I'm not working from a rigorous derivation. Let me consume calories and then come back and go over the logic behind the product more carefully
bunch of stuff
That's a very concise answer
Things, would have been even shorter
20:02
insha'Allah the crusade of answers should be called the jihad of answers
Not sure that term is very appropriate
maybe "the anschluss" would be a little excessive, but jihad? That's okay
crusade has two meanings
one is a religious military campaign the other a vigorous campaign
relating the two definitions would mean religious military is vigorous
Reading up about how to crusades went
I cannot draw that conclusion
That is, only if "campaign" is invertible
Let's just pretend I never said that
Today I read that Sophus Lie was hicking naked in the rain and got captured as they thought his phd dissertation was spy stuff
20:13
yep
@BalarkaSen have you seen hilbert-poincare series before?
you know when you take a $\Bbb Z$-graded algebra and take the generating function of the dimensions
Apparently my topology professor wants to introduce an undergrad algebraic topology course next year, I sure hope he manages to do that
good afternoon
20:15
sure
@Alessandro Very nice.
Hi @meow
you are familar with the path algebra of a quiver?
Hi @zach (force of habit)
hi @Alessandro
20:16
but you know what a quiver is?
basically a quiver is a category but you forget about it
So only directed graph
So he did the right thing
He forgot about quivers
@AliCaglayan I wonder when people started explaining it that way (I mean, when did categories become more common than directed graphs?)
@TobiasKildetoft only because I know @BalarkaSen knows what a category is
20:17
Also, quivers need not be the underlying set of arrows of a category
ok forget I mentioned categories I realised I dug myself a hole
A quiver is a directed graph
so why not just say "directed graph"?
In mathematics, a quiver is a directed graph where loops and multiple arrows between two vertices are allowed, i.e. a multidigraph. They are commonly used in representation theory: a representation V of a quiver assigns a vector space V(x) to each vertex x of the quiver and a linear map V(a) to each arrow a. In category theory, a quiver can be understood to be an underlying structure of a category, but without identity morphisms and composition. That is, there is a forgetful functor from Cat to Quiv. Its left adjoint is a free functor which, from a quiver, makes the corresponding free category...
but the point is you can come up with two notions of representations of a quiver
i'm not gonna read it tho. what's the big thing you wanted to tell me?
Assigning a vector space to each vertex
and linear maps to edges
But you can also define something called a path algebra
20:20
gotcha
@AliCaglayan That excerpt is false in claiming that the quiver underlies a category (though it does make precise what it means just after that)
and the representation of a path algebra is the same as the representation of a quiver
i see. why do i care?
because the hilbert series of path algebras is interesting?
but if you don't care then don't worry about it
why so?
i am really just asking why i'd want to care :)
20:23
how can I prove that eigenvectors with distinct eigenvalues are orthogonal?
I guess representations of quivers are interesting
I read through the solution here web.mit.edu/18.06/www/Fall07/pset9-soln.pdf but didn't quite understand it.
then this approach of finding a graded representation is interesting
That's not true @AbhishekBhatia do you have some assumption on your matrix? Symmetric or something?
I'll believe you that it's interesting for some people.
20:26
we say a path algebra generated by identity paths at a vertex $P_Q$ and some edge is
1. $p_i^2=p_i, \ p_ip_j=0$ for $i\ne j$
2. $p_i$ act as identity on the right
3. $p_i$ act as identity on the left
@AbhishekBhatia Right, there is an extra assumption there
but the only reason I find it interesting it was an exercise in a book I am reading
I am as much familiar with quivers as I am with Grothendieck topoi to appreciate all this. If you had told me something more fundamental about them which I could connect to what I know and think of as interesting, then I would certainly have been more interested. Just clarifying what I meant by "why do I care?".
but I will see if I can motivate it otherwise it is just a singleton curiosity
20:29
hello everyone
so I am facing a problem in designing an algorithm
@BalarkaSen you don't have to be an expert in quivers to understand the problem, when I first read about it quivers were just introduced
the thing is I have to find out the probability with which elements of a certain set are of type 1 or type 2 (they can only be of type 1 or type 2)
@BalarkaSen let me see if I can motivate this better
and the algorithm can generate random elements and test their types
I more-or-less understand what you said so far, just that I find it abstract and not very motivated.
20:32
would it be sufficient to sample this on 30 elements and say that the probability with which an element is type 1 is the proportion of type 1 elements in those 30 elements?
@BalarkaSen Well, path algebras generalize easily by quotienting certain types of ideals (called admissible). And these quotients give all finite dimensional algebras up to Morita equivalence (as long as the field is algebraically closed)
@BalarkaSen I think a general notion is that quivers arise naturally in algebraic geometry when talking about certain structures. Studying representations of quiver is therefore important if one wants to say anything about the structure it represents. Graded representations have a nice gist to them so applying that to quiver representations says nice things about certain structures.
@TobiasKildetoft Ah.
@AliCaglayan That's not very specific though, but ok.
@BalarkaSen I can't possibly be, I would have to know what I am talking about. I was only telling you these things because I found them interesting in their own abstract right. But now trying to come up with a motivational example is out of my scope.
Sure, I am not asking you to.
20:42
The only reason I know about quivers is because a book on representation theory said so
@TobiasKildetoft Wait a tick, I misread. Oh, so you mean, any finite dimensional algebra appears as these fellows upto Morita equivalence?
@BalarkaSen Right, over an algebraically closed field (possibly with some other small detail that I always forget)
ah, ah, got it. Now that's interesting.
And an ideal being admissible just means that it corresponds to putting relations on the quiver, which are not allowed to kill any edges and which must contain all paths of sufficiently large length.
(that last is to ensure the algebra becomes finite dimensional)
Hm, ok.
@AkivaWeinberger what was the context?
he's making a bad pun
. math.stackexchange.com/questions/2059001/… My approach would have been: 252-198=54. Then look if 54 has an inverse in $\mathbb{Z}_{198}$. What does he make there?
It's not clear (to me) what the asker wanted to do, but the answers agree with you @null

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