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12:00 AM
@amanuel2 If $a=r\sin\varphi$ and $b=r\cos\varphi$ then $b/a=\frac{r\sin\varphi}{r\cos\varphi}=\tan\varphi$
 
@AkivaWeinberger can u edit ur format please? I cant read it well thanks
 
@amanuel2 $\LaTeX$ in chat: tinyurl.com/cfqcvpc
but here it is without Latex
@amanuel2 If a=r·sin(φ) and b=r·cos(φ) then b/a=[r·cos(φ)]/[r·sin(φ)]=tan(φ)
(Whoops the link wasn't clickable: tinyurl.com/cfqcvpc)
 
12:56 AM
Ughhh
My name is Anita Hint on this haha
Fix $\kappa > \omega_{1}$. The club filter on $\kappa$ is not an ultrafilter . So do I prove that $\kappa$ is not an ultrafilter using the definitions of filter ? :/
The sets are huge if it's a filter though. Maybe the club filter on $\kappa$ is ideal ... Meaning small
Ya seems legit 😜🤪
 
 
2 hours later…
2:56 AM
Whelp
Israeli Elections 3: For Real This Time (Hopefully) will probably be in March
Also Bibi's indicted
(or, re: elections, maybe not… it depends on if someone can form a coalition in the next three weeks)
 
@AkivaWeinberger I am sooooo stuck. ;/ .
Maybe I should post in main ??? It's like I got an idea but then it just fizzles out
 
I don't know what a club filter
is
I vaguely remember it being an acronym for something
Closed least upper bound? Or something like tjat
Don't know what that means though
 
3:16 AM
Question: If we have that $a,b,c\in M$ where $M$ is an infinite magma, and $ab=ac$ implies that $b=c$ (similarly, $ba=ca$ implies that $b=c$), can we say that action by $a$ permutes the elements of $M$?
Wait, just answered my own question
If $M=\mathbb{Z}$ then $a=2$ is a counter example.
 
3:30 AM
More generally, in infinite magmas, any map of the form $a \cdot$ whose image biject with a subset of itself is not a permutation
You cannot do that in finite magmas as injections are surjective and hence bijections/permutations
 
Ah, yeah, I hadn't even connected the words 'action' and 'function' in my mind.
 
Club filter is the intersections are also club which is closed and unbounded but someone suggested looking at a URL which I saw already so I deleted my question and would just attempt and list as a reference being used
eats a celery stick
 
 
2 hours later…
6:08 AM
I am a sceptic. My mother gets nauseous if she looks at any kind of algebra or you begin to explain a concept to her. collectively for the most part others consider the interest to be a mental illness or syndrome. I think again Yale have done the classic bogus stats trick of hand picking their sample population from the "nice" suburbs
 
6:48 AM
It's good to know that there is mathematics tag on Skeptics Stack Exchange.
 
 
1 hour later…
8:09 AM
@usukidoll a filter $F$ on $\kappa$ is an ultrafilter iff for all $A\subseteq\kappa$ either $A\in F$ or $\kappa\setminus A\in F$. Can you find an $A$ such that neither $A$ nor its complement is a club?
 
 
1 hour later…
9:37 AM
hello
 
 
3 hours later…
1:00 PM
Hi @Ryan
 
Is applying the Euler-Maclaurin formula to the geometric series meaningless or impossible?
 
Hi @Balarka
 
 
1 hour later…
2:34 PM
Hi @Alessandro
 
How are you?
 
3:07 PM
I have a question: Why do we need the (previous map) Image to be contained in the (latter map) Kernel in the chain complex?
I've studied some algebraically topology and I am aware it has some geometric aspects.
But if we focus on the algebraic aspects, the only advantages I know is that one can do the exact triple, connect the right exact sequence of tensor products by Torsion functor to obtain a long exact one...
What will happen if the image is not contained in the kernel? Will we lose track of some informations in the chain?
 
how would you define homology?
 
The quotient defining homology of the complex doesn't make sense anymore to begin with which seems a big issue
 
It is a two stage construction. The first stage contains geometric information: Define ffee groups of simplices and map each one to a lower dimension one by boundary map, and the boundary map, the alternating sum of faces, contains the geometric information.
If you mean the homology for a chain complex: that's just purely algebraic.
I know we need the image to be contained in kernel to define homology.
But do we have to limit ourselves on just homology of the chain?
 
3:31 PM
we defined chain complexes because we cared about homology, by and large
 
Given a ring without identity, what needs to be true for it to be possible to "add" an identity element and have $R\bigcup\{1\}$ still be a ring
(Related question: can you construct a ring where every element except the identity is nilpotent?)
 
For your second question: the trivial ring?
 
there are notions more general than chain complexes (eg see matrix factorizations by eg eisenbud) but i think ive only seen it show up in commutative algebra / mirror symmetry stuff (i.e. its not as prominent as chain complexes/homological. algebra which shows up in almost all math having something to do with algebra)
 
Okay, non-trivial rings, then (I always fail to eliminate that as a possibility)
(I also forget to check it as a possibility)
 
3:48 PM
@loch Glad to know someone has studied it because I care about the pure algebraically aspects of it. Thank you;) I will look up for some references.
 
is this limit $\lim_{x\to 1} x^{\frac{1}{1-x}}$ exists ?
 
dsm
ok, very simple question, but I'm having a sizeable brain fart. why is $$\frac{\partial}{\partial \vec{r}}\frac{1}{2}(\vec{\Omega}\times\vec{r})^2=(\vec{\Omega}\times\vec{r})\times\vec{\Omega}?$$ not seeing how that last part turns into a cross with $\Omega$
 
Is the following answer okay? It's somewhat Socratic as I'm having trouble articulating what I mean. Here it is:
0
A: Prove that exist $x\in \left\{ 1,...,14 \right\}$ such that $\sigma(x)=x$, where $\sigma\in S_{14}$ and $|\sigma|=28$?

ShaunYou already have the divisors of $28$. What are the partitions of $14$ into those divisors, potentially including $1$, $4$, and $14$, such that the disjoint cycles of elements of $S_{14}$ form elements of order $28$ with cyclic decompositions composed of those divisors? You'll find that you'll a...

 
@Rithaniel it can be done for rings without identity, not sure about rings with identity
 
@WilliamSun note that matrix factorizations do come from geometry in some sense and aren't the same as "chain complexes where you don't need image inside kernel", which is better known perhaps as a "graded group with a self-map $d$"
@Rithaniel R must be trivial for this to hold
 
3:55 PM
Please help :)
 
Pick an element x of R. If x is non-trivial then 1+x is not x, so is in R. We also have (1+x)-x is in R, because both elements are in R. So 1 was in R all along.
 
Yes I am studying graded groups right now. Thank you for commenting.
@Rithaniel If you want the ring with identity to contain just nilpotent elements and $1$ then the ring has a unique prime ideal. If you also want the nilradical, which is just that prime ideal to be large enough so that it is $A-{1}$ we have to eliminate the fields, and all $Z/nZ$'s.
And it won't be an integral domain of course.
Oh wait someone just gave an answer above, ignore me.
 
4:34 PM
Yeah, I figured it out separately. If $1$ is a unit and $a$ is nilpotent, then $1+a$ is a unit, and therefore not nilpotent. So, since the only non-nilpotent is the identity, we see that $1+a=1$ and thus $a=0$
 
 
1 hour later…
5:46 PM
@Rithaniel I still maintain this is way too much work man
If the inclusion R -> R cup 1 is a ring homomorphism it is a group homomorphism on the underlying additive groups
That's not possible for cardinality reasons; cosets of a subgroup partition a group into pieces, but the "cosets" here must be R and 1
Those are only the same cardinality if R = 0
 
Evening
 
6:05 PM
Hi @ÍgjøgnumMeg
 
6:16 PM
What would you recommend for "more than introductory" linear algebra books?
 
Can someone ELI5 what "sample path properties" means?
 
@FuzzyPixelz either "linear algebra done right" by axler or "linear algebra done wrong" by treil
 
@MikeMiller What's there to see after that? Something more advanced?
 
@MikeMiller Nice, actually. So if you wanted to alter a ring to add a identity, you would have to at least double the number of elements in the ring?
 
@Rithaniel Yes, and actually usually more.
@FuzzyPixelz what exactly do you want to do with it
it's not really clear to me what it would mean to look for more advanced linear algebra than that
 
6:30 PM
I want to read more on Eigen-things, Matrix canonical forms.. and I haven't been able to find a reference to discuss precisely these topics in more detail, and include many problems
 
??? Every linear algebra text will talk about eigenvectors as well as canonical forms (including Jordan form). Some talk about different kinds of canonical forms, like polar decomposition, that others don't. But these are rather random topics and usually they will be introduced t you as you need them.
It's completely unclear to me what you're really looking for so I doubt I can help
 
Yeah, you ended up targeting the underlying topic when I was still focused on the the related question. Though, it's an interesting question: Given a ring $R$, what is the smallest ring with identity $R_1$ such that $R\subseteq R_1$
 
6:45 PM
(Unrelated: I think it's a pet peeve of mine when people refer to "rings without identity" as rngs)
 
@FuzzyPixelz: The best way to learn about canonical forms is to learn a bit more abstract algebra. Artin's Algebra book does the linear algebra using it. However, check out Friedberg/Insel/Spence for a self-contained, more advanced linear algebra treatment.
@Rithaniel: I don' t know who invented it, but it's cute and clear enough.
 
Indeed, I won't fault you for that, but now I gotta look up a different word when I wanna know about general rings.
 
Personally, I don't think I've ever cared about a rng.
 
Can you tile a space with copies of $\Bbb R^2?$
 
I hate cute names.
 
6:51 PM
Only ugly names allowed
 
@MikeM: But it's informative.
@Rithaniel: OK, you can have my name.
 
Also, yeah, rngs don't come up very often. I would believe that
 
I mean, some people want to be able to work with $2\Bbb Z$ as a ring.
 
Hey, Ted is an awesome name.
 
@TedShifrin Next semester I'll start calc 3 with a little linear algebra. What is a linear map, what is a matrix, solving Ax = b when you can and multiplying matrices.
No more than a week. Just enough to state the chain rule and the definition of a derivative.
 
7:00 PM
I'll be curious to find out how it goes.
 
7:18 PM
May someone check this proof, please?
0
Q: Basis of a topology, and new construction of topology.

topologicalmagicianLet $X$ be a set and suppose $\mathbb{B}$ is a collection of subsets of $X$. $\mathbb{B}$ is a basis for some topology on $X$ $\textbf{if and only if}$ (1)$\bigcup_{B\in \mathbb{B}}B = X$ (Covering property) (2) If $B_1, B_2 \in \mathbb{B}$ and $x\in B_1 \cap B_2$ then $\exists$ $B_3 \in \ma...

 
7:41 PM
@AlessandroCodenotti So-so
What about you
 
hi all
 
Hi a @Balarka, @anakhro
 
What's up, T-dog?
 
7:56 PM
Almost lunchtime. What's up with you?
 
Was just discussing with a friend about uncountable sums.
 
@TedShifrin I think it should be fine. Certainly I'm not losing anything by this change --- I'm cutting some days on conic sections and quadric surfaces, which are supposed to make things more geometric but people just seem to walk away trying to memorize names.
 
And how uncountable sums of positive numbers never converge.
 
I don't even know how to uncountably sum.
(And I split my infinitive on porpoise.)
 
sup of finite sums is what we were discussing.
 
7:59 PM
@BalarkaSen hi
 
If $I$ is uncountably infinite, then $\sum_{i\in I}a_i = \sup\{\sum_{i\in J}a_i \mid J\subset I\text{ is finite}\}$.
 
Are all the numbers nonnegative?
Does this even work for conditionally convergent countable if you reorder? No.
 
The theorem is that does not converge if $a_i > 0$ for all $i\in I$.
 
OK, so if all the terms are nonnegative, at least it makes sense to me in the countable case.
 
Yes, exactly.
In a measure theoretic way, it's like saying the uncountable union of disjoint positive measure sets has measure $\infty$.
 
8:03 PM
We only talk about countable additivity, of course.
 
@TedShifrin do you ever concern yourself with the philosophy of mathematics?
 
Not really.
 
Why not?
(not that you ought to)
I am just curious. :P
 
I took a few philosophy courses in college but ultimately it wasn't for me.
 
How about the philosophy of mathematics education?
 
8:09 PM
@skullpetrol I think that philosophy of mathematics deals a lot with mathematical pedagogy.
@TedShifrin have you heard of the debate between mathematical platonism/realism vs. formalism?
 
As skull suggests, a lot of the math ed scholarship becomes somewhat philosophical. So, I've been exposed to that when I've been on math ed Ph.D. committees. But I haven't read any of the literature.
 
Hi @Ted
 
I have been reading about how to revive philosophy of mathematics.
And it has been pretty cool.
 
Is it super-clear why every Riemannian metric on a manifold can be locally written as a positive linear combination of rank 1 symmetric tensors?
When I mean linear combination I mean constant coefficients, do not depend on the underlying point
Otherwise I could do parametrized LDL decomposition
This feels like some linear algebra. I have a smooth map $A : \Bbb R^m \to \text{Sym}^+_m(\Bbb R)$ and I want to find some $P \in M_{m \times n}(\Bbb R)$ such that near a neighborhood of a point $A = PDP^T$ where $D = \text{Diag}(f_1(x), \cdots, f_n(x))$ for some positive smooth functions $f_i$ on that neighborhood
I would expect $n$ to be much larger than $m$ in general
 
@BalarkaSen do you just want to diagonalize
 
8:24 PM
\o
 
That's not possible if I am demanding my basechange matrix to not depend on the underlying point
This is weirder than diagonalization, $P$ is not square
 
Can someone please verify the correctness of this answer I've given? Anyway, in general, I will appreciate if you provide some feedback
1
A: Concrete example of latent variables and observables plugged into the Bayes' rule for VAE

nbroLet's assume the probability distributions are Gaussian (or normal) distributions. In other words, in the Bayes' rule \begin{align} p(z|x)=\frac{p(x|z)p(z)}{p(x)} \tag{1}\label{1} \end{align} The posterior $p(z|x)$, the likelihood $p(x|z)$, the prior $p(z)$ and the evidence (or marginal) $p(x)$...

 
I want to write the metric as a sum $\sum f_i v_iv_i^T$ where $v_i$ are fixed vectors in $\Bbb R^m$
 
Oh I’m not reading the tex
I’m looking at the original question
 
Well then the answer is no, I don't want to diagonalize :p
 
8:27 PM
In particular, I am not completely sure 1. which concrete values for x and z can we choose to calculate the posterior, 2. if the posterior density value is ensured to be a density value
 
@BalarkaSen I'm reading some topology stuff by Dranishnikov and some closely related topology stuff by Gromov and the latter is the more readable one so...
 
Oof
 
@BalarkaSen are you a platonist/realist or a formalist?
 
Platonist, I think
But I do respect formalism
 
@BalarkaSen It's quite interesting though
 
8:43 PM
@BalarkaSen have you heard the one quote:
"the typical working mathematician is a platonist on weekdays and a formalist on Sundays. That is, when he is doing mathematics he is convinced that he is dealing with an objective reality whose properties he is attempting to determine. But then, when challenged to give a philosophical account of this reality, he finds it easiest to pretend that he does not believe in it after all"
 
Made some ordinal hyperoperators:$$\alpha\uparrow^\mu\beta=\begin{cases}\alpha\cdot\beta,&\mu=0\\1,&\mu>0\land\beta=0\\\displaystyle\sup_{\mu'<\mu\\\beta'<\beta}\max\{\alpha\uparrow^{\mu'}\alpha\uparrow^\mu\beta',\alpha\uparrow^\mu\beta'\uparrow^{\mu'}\alpha\},&\text{else}\end{cases}$$
They have lots of nice properties, namely a nice base-changing formula:$$\alpha\uparrow^\mu(\beta+\gamma)=(\alpha\uparrow^\mu\beta)\uparrow^\mu\gamma$$for limit ordinals $\beta,\gamma$ and $\mu\ge2$.
It also nicely extends things such as$$\varepsilon_1=\sup\{\varepsilon_0,\varepsilon_0^{\varepsilon_0},\varepsilon_0^{\varepsilon_0^{\varepsilon_0}},\dots\}$$to$$\zeta_1=\sup\{\zeta_0,{}^{\zeta_0}\zeta_0,{}^{{}^{\zeta_0}\zeta_0}\zeta_0,\dots\}$$where ${}^ab=b\uparrow^2a$.
Likely no-one cares lol but just in case someone finds it interesting :3
 
@anakhro see also: how physicists use QM
 
who knows what the typical working mathematician thinks
 
@Semiclassical can you elaborate?
 
I mean that there's a bit of a gap between how people -say- they use QM, and what kinds of intuitions they fall back on
 
8:57 PM
Oh, is there?
 
do you mean formalized mathematical QM?
 
I thought that QM was fairly well-understood but it was just the string theory and attempts to unify QM and CM that fell on less than honourable ideas.
 
9:29 PM
i mean, it depends on who you talk with
but I'm altogether dubious about how philosophically consistent people are
 
Can you give an example of this phenomenon?
 
9:51 PM
@anakhro would this^ qualify as an example?
 
10:38 PM
@skullpetrol I don't think statistical mechanics had anything to do with their deaths.
Also, this doesn't sound like what semiclassical is describing.
 
11:41 PM
You know how they say that axioms of algebraic structures are like taking quotients of some freeer structure?
0
Q: Can you get from an $R$-semimodule (minus 1 axiom) to an $R$-module by quotienting?

Shine On You Crazy Diamondhttps://en.wikipedia.org/wiki/Semimodule So every $R$-semimodule $N$ over a ring $R$ is a module. The axioms are: $$ \begin{matrix} (1) & r(m + n) = rm + rn \\ (2) & (r+s)m = r m + s m \\ (3) & (rs)m = r(sm) \\ (4) & 1m = m \\ (5) & 0m = r0 = 0 \\ \end{matrix} $$ for all $r, s \in R, \ \ m,n ...

 
@ShineOnYouCrazyDiamond hi
 
@Ultradark
Good research question for you
:D
I can't figure it out
 
0
Q: Prove that this unique three manifold exists, and is precisely $\Bbb R^3$

UltradarkSay you take a disk model of $\Bbb R^2$ and deform it into a square model of $\Bbb R^2$ and then glue six copies together such that they form the six faces of a cube. If each face of the cube is projected towards the opposite face, can a unique three manifold in the interior of the cube be built ...

What do you think?
 
I think we each have a matching close vote
lol
 
yeah I just got a close vote
 

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